JEROME J. CONNOR, Sc.D., Massachusetts Institute of Technology, is Professor of Civil Engineering at Massachusetts Institute of Technology. He has been active in teaching and research in structural analysis and mechanics
at the U.S. Army Materials and Mechanics Research Agency and for some years at M.I.T. His primary interest is in computer based analysis methods, and his current research is concerned with the dynamic analysis of prestressed concrete reactor vessels and the development of finite element models for fluid flow problems. Dr. Connor is one of the original developers of ICES-STRUDL, and has published extensively in the structural field.
ANALYSIS OF STRUCTURAL MEMBER
SYSTEMS JEROME J. CONNOR Massachusetts Institute of Technology
THE RONALD
PRESS COMPANY • NEW YORK
Copyright ©
1976 by
Ttrn RONALD PRESS COMPANY
All Rights Reserved
No part of this book may be reproduced in any form without permission in writing from the publisher.
Library of Congress Catalog Card Number: 74—22535 PRINTED IN ThE UNITCD STATES OF AMERICA
Preface
With the development over the past decade of computer-based analysis methods, the teaching of structural analysis subjects has been revolutionized. The traditional division between structural analysis and structural mechanics became no longer necessary, and instead of teaching a preponderance of solution details it is now possible to focus on the underlying theory. What has been done here is to integrate analysis and mechanics in a systematic presentation which includes the mechanics of a member, the matrix formulation of the equations for a system of members, and solution techniques. The three fundamental steps in formulating a problem in solid mechanics—. enforcing equilibrium, relating deformations and displacements, and relating forces and deformations—form the basis of the development, and the central theme is to establish the equations for each step and then discuss how the complete set of equations is solved. In this way, a reader obtains a more unified view of a problem, sees more clearly where the various simplifying assumptions are introduced, and is better prepared to extend the theory. The chapters of Part I contain the relevant topics for an essential background in linear algebra, differential and matrix transformations. Collecting this material in the first part of the book is convenient for the continuity of the mathematics presentation as well as for the continuity in the following development. Part II treats the analysis of an ideal truss. The governing equations for
small strain but arbitrary displacement are established and then cast into matrix form. Next, we deduce the principles of virtual displacements and virtual forces by manipulating the governing equations, introduce a criterion for evaluating the stability of an equilibrium position, and interpret the governing equations as stationary requirements for certain variational principles. These concepts are essential for an appreciation of the solution schemes described in the following two chapters. Part III is concerned with the behavior of an isolated member. For completeness, first are presented the governing equations for a deformable elastic solid allowing for arbitrary displacements, the continuous form of the principles of virtual displacements and virtual forces, and the stability criterion. Unrestrained torsion-flexure of a prismatic member is examined in detail and then an approximate engineering theory is developed. We move on to restrained torsion-flexure of a prismatic member, discussing various approaches for including warping restraint and illustrating its influence for thin-walled iii
PREFACE
and closed sections. The concluding chapters treat the behavior of planar and arbitrary curved members. How one assembles and solves the governing equations for a member sysopen
tern is discussed in Part IV. First, the direct stiffness method is outlined; then a general formulation of the governing equations is described. Geometrically nonlinear behavior is considered in the last chapter, which discusses member force-displacement relations, including torsional-flexural coupling, solution schemes, and linearized stability analysis. The objective has been a text suitable for the teaching of modern structural member system analysis, and what is offered is an outgrowth of lecture notes
developed in recent years at the Massachusetts Institute of Technology. To the many students who have provided the occasion of that development, I am deeply appreciative. Particular thanks go to Mrs. Jane Malinofsky for her patience in typing the manuscript, and to Professor Charles Miller for his encouragement. JEROME J. CONNOR
Cambridge, Mass. January, 1976
Contents
I—MATHEMATICAL PRELIMINARiES 1
Introduction to Matrix Algebra 1—i 1—2 1—3 1—4 1—5 1—6 1—7
1—8
Definition of a Matrix Equality, Addition, and Subtraction of Matrices Matrix Multiplication Transpose of a Matrix Special Square Matrices Operations on Partitioned Matrices Definition and Properties of a Determinant Cofactor Expansion Formula
Cramer's Rule 1—10 Adjoint and Inverse Matrices 1—11 Elementary Operations on a Matrix 1—12 Rank of a Matrix 1—13 Solvability of Linear Algebraic Equations 1—9
2
2—2 2—3 2—4
2—5
5 8 10
12 16 19 21
22 24 27 30
Introduction Second-Order Characteristic-Value Problem Similarity and Orthogonal Transformations The nth-Order Symmetrical Characteristic-Value Problem Quadratic Forms
46 46 48 52 55 57
Relative Extrema for a Function 3—1
3—2
3—3
4
5
Characteristic-Value Problems and Quadratic Forms 2—1
3
3
Relative Extrema for a Function of One Variable Relative Extrema for a Function of n Independent Variables Lagrange Multipliers
66 66 71
75
Differential Geometry of a Member Element 4—1
4—2
Parametric Representation of a Space Curve Arc Length V
81 81
82
CONTENTS 4—3 4—4 4—5 4—6
4—7 4—8
5
Unit Tangent Vector Principal Normal and Binormal Vectors Curvature, Torsion, and the Frenet Equations Summary of the Geometrical Relations for a Space
85 86 88
Curve
91
Local Reference Frame for a Member Element Curvilinear Coordinates for a Member Element
92 94
Matrix Transformations for a Member Element 5—1
5—2 5—3
Rotation Transformation Three-Dimensional Force Transformations Three-Dimensional Displacement Transformations
100 100 103 109
Il—ANALYSIS OF AN IDEAL TRUSS 6
Governing Equations for an Ideal Truss 6—1
General
6—2
Elongation—Joint Displacement Relation for a Bar General Elongation—Joint Displacement Relation Force-Elongation Relation for a Bar General Bar Force—Joint Displacement Relation Joint Force-Equilibrium Equations Introduction of Displacement Restraints; Governing Equations Arbitrary Restraint Direction Initial Instability
6—3
6—4 6—5
6—6 6—7
6—8 6—9
7
132 134 137
Variational Principles for an Ideal Truss 7—1
General
7—2
Principle of Virtual Displacements Principle of Virtual Forces Strain Energy; Principle of Stationary Potential
7—3 7—4
7—5
7—6
8
115 116 120 125 130 130
152 152 153 159
Energy
162
Complementary Energy; Principle of Stationary Complementary Energy Stability Criteria
165 169
Displacement Method—Ideal Truss 8—1
General
8—2
Operation on the Partitioned Equations The Direct Stiffness Method
8—3
178 178 178
180
CONTENTS 8—4
8—5
9
Incremental Formulation; Classical Stability Criterion Linearized Stability Analysis
191
200
Force Method—Ideal Truss 9—1
General
210
9—2
Governing Equations—Algebraic Approach Governing Equations—Variational Approach Comparison of the Force and Mesh Methods
211
9—3
9—4
216 217
Ill—ANALYSIS OF A MEMBER ELEMENT 10
Governing Equations for a Deformable Solid 10—1
General
10—2
Summation Convention; Cartesian Tensors Analysis of Deformation; Cartesian Strains Analysis of Stress Elastic Stress-Strain Relations Principle of Virtual Displacements; Principle of Stationary Potential Energy; Classical Stability Criteria Principle of Virtual Forces; Principle of Stationary Complementary Energy
10—3 10—4
10—5 10—6
10—7
11
229 230 232 240 248
253
257
St. Venant Theory of Torsion-Flexure of Prismatic Members 11—1
11—2 11—3 11—4
11—5 11—6 11—7
12
229
Introduction and Notation The Pure-Torsion Problem Approximate Solution of the Torsion Problem for Thin-Walled Open Cross Sections Approximate Solution of the Torsion Problem for Thin-Walled Closed Cross Sections Torsion-Flexure with Unrestrained Warping Exact Flexural Shear Stress Distribution for a Rectangular Cross Section Engineering Theory of Flexural Shear Stress Distribution in Thin-Walled Cross Sections
271 271
273 281
286 293 303
306
Engineering Theory of Prismatic Members 12—1
12—2
Introduction Force-Equilibrium Equations
330 330 331
CONTENTS 12—3
12—4 12—5 12—6
13
13—2 13—3
13—4 13—5 13—6
13—7 13—8 13—9
371
371 Introduction Displacement Expansions; Equilibrium Equations 372 Force-Displacement Relations—Displacement Model 375 Solution for Restrained Torsion—Displacement Model 379 Force-Displacement Relations—Mixed Formulation 383 Solution for Restrained Torsion—Mixed Formulation 389 Application to Thin-Walled Open Cross. Sections -395 405 Application to Thin-Walled Closed Cross Sections Governing Equations—Geometrically Nonlinear Restrained Torsion 414
Planar Deformation of a Planar Member 14—1
14—2 14—3
14—4
14—5
14—6
14—7 14—8
15
333 339 340 349
Restrained Torsion-Flexure of a Prismatic Member 13—1
14
Force-Displacement Relations; Principle of Virtual Forces Summary of the Governing Equations Displacement Method of Solution—Prismatic Member Force Method of Solution
Introduction; Geometrical Relations Force-Equilibrium Equations Force-Displacement Relations; Principle of Virtual Forces Force-Displacement Relations—Displacement Expansion Approach; Principle of Virtual Displacements Cartesian Formulation Displacement Method of Solution—Circular Member Force Method of Solution Numerical Integration Procedures
425 425 427
429
435 445 449 458 473
Engineering Theory of an Arbitrary Member 15—1
15—2 15—3
15—4 15—5 15—6 15—7
15—8
Introduction; Geometrical Relations Force-Equilibrium Equations Force-Displacement Relations—Negligible Warping Restraint; Principle of Virtual Forces Displacement Method—Circular Planar Member Force Method—Examples Restrained Warping Formulation Member Force-Displacement Relations—Complete End Restraint Generation of Member Matrices
485 485 488 490 .493 499 507 511
517
CONTENTS
Member Matrices—Prismatic Member 15—10 Member Matrices—Thin Planar Circular Member 15—11 Flexibility Matrix—Circular Helix 15—12 Member Force-Displacement Relations—Partial End Restraint 15—9
520 524 531
535
tV—ANALYSIS OF A MEMBER SYSTEM 16
Direct Stiffness Method—Linear System 16—1
16—2 16—3 16—4
17
Introduction Member Force-Displacement Relations System Equilibrium Equations Introduction of Joint Displacement Restraints
545 546 547 548
General Formulation—Linear System
17—4
Introduction Member Equations System Force-Displacement Relations System Equilibrium Equations
17—5
Introduction of Joint Displacement Restraints;
17—1
17—2 17—3
Governing Equations Network Formulation 17—7 Displacement Method 17—8 Force Method 17—9 Variational Principles 17—10 Introduction of Member Deformation Constraints 17—6
18
545
554 554 555 557 559 560 562 565 567 570 573
Analysis of Geometrically Nonlinear Systems 18—1
18—2 18—3 18—4
Index
Introduction Member Equations—Planar Deformation Member Equations—Arbitrary Deformation Solution Techniques; Stability Analysis
585 585 585 591
597
605
Part I MATHEMATICAL PRELIMINARIES
1
Introduction to Matrix Algebra 1—1.
DEFINITION OF A MATRIX
An ordered set of quantities may be a one-dimensional array, such as
a two-dimensional array, such as a11, a12, . . ,a1, a21, a22, . , .
.
.
ami,
a two-dimensional array, the first subscript defines the row location of an element and the second subscript its column location. A two-dimensional array having ,n rows and n columns is called a matrix of order m by n if certain arithmetic operations (addition, subtraction, multiplication) associated with it are defined. The array is usually enclosed in square brackets and written as* a11
a12
a21
a22
a,,,1
Note
am2
-
-
a1,,
-
a2,,
=
=
a
a,,,,,
that the first term in the order pertains to the number of rows and the
second term to the nuiñber of columns. For convenience, we refer to the order of a matrix as simply m x n rather than of order m by n. *
In print, a matrix is represented by a boldfaced letter. 3
INTRODUCTION TO MATRIX ALGEBRA
4
CHAP. 1
A matrix having only one row is called a row matrix. Similarly, a matrix having only one column is called a column matrix or column vector.* Braces
instead ofbrackets are commonly used to denote a column matrix and the column subscript is eliminated. Also, the elements are arranged horizontally instead of vertically, to save space. The various column-matrix notations are: C11
C1
C21
C2
{c1, c2,.
.
.
{c1}
,
=c
If the number of rows and the number of columns are equal, the matrix is said to be square. (Special types of square matrices are discussed in a later section.) Finally, if all the elements are zero, the matrix is called a null matrix, and is represented by 0 (boldface, as in the previous case). Example 3
1—1
x 4 Matrix
2—1
4 3
—7
2
4
1
—3
2
—8 1
1 x 3 Row Matrix [3
2]
4
3 x 1 Column Matrix f3] or
2
2
4Jor{3,4,2}
Square Matrix 5
[2
7
[0 [o
0
2 x 2 Null Matrix o
* This is the mathematical definition of a vector. In mechanics, a vector is defined as a quantity having both magnitude and direction. We will denote a mechanics vector quantity, such as force or moment, by means of an italic letter topped by an arrow, e.g., F. A knowledge of Vector algebra is assumed in this text. For a review, see Ref. 2 (at end of chapter, preceding Problems).
MATRIX MULTIPLICATION
SEC. 1—3. 1—2.
EQUALITY, ADDITION, AND SUBTRACTION OF MATRICES
Two matrices, a and b, are equal if they are of the same order and if corresponding elements are equal:
a=
when
b
If a is of order m x n, the matrix equation
a=b corresponds to mn equations:
= =
=
1,
2,. .
1,
2,.. .
.
,m ,
Addition and subtraction operations are defined only for matrices of the same
order. The sum of two m x n matrices, a and b, is defined to be the m x n matrix + +
=
+
—
=
— bLJ]
Similarly,
For example, if
[1
2
ii —d
then
[1
[0 b=[3
[1
—1
i 0
1
—1
1
and
—1
3
2
—1
—1
It is obvious from the example that addition is commutative and associative:
1—3.
a+b=b+a
(1—6)
a+(b+c)=(a+b)+c
(1—7)
MATRIX MULTIPLICATION
The product of a scalar k and a matrix a is defined to be the matrix in which each element of a is multiplied by k. For example, if
k=5
and
then
ka=[[—10 10
+35 5
INTRODUCTION TO MATRIX ALGEBRA
6
CHAP. 1
Scalar multiplication is commutative. That is,
ka = ak = {ka11]
To establish the definition of a matrix multiplied by a column matrix, we consider a system of m linear algebraic equations in n unknowns, x1, .x2
+ a12x2 +
+ +
a21x1 + a22x2 + l2miXi
+
am2x2
C1
=
C2
+
+
This set can be written as alkxk
C1
i=
1, 2, .
.
. ,rn
where k is a dummy index. Using column matrix notation, (1—9) takes the form
i= Now, we write (1—9) as a matrix product:
=
i= 1,2,..,,rn
{c1}
(1—11)
1,2
Since (1—10) and (1—Il) must be equivalent, it follows that the definition equation for a matrix multiplied by a column matrix is ax =
ulkxk}
j = 1, 2,. .
.
,m
This product is defined only when the column order of a is equal to the row
order of x. The result is a column matrix, the row order of which is equal to that of a. In general, if a is of order r x s, and x of order s x 1, the product ax is of orderr x 1. Example
1—2 1
a=
11
8
2 x={3}
-4j 1(1)(2) + (—1)(3) 4
+ (3)(3)
9
MATRIX MULTIPLICATION
SEC. 1—3.
We consider next the product of two matrices. This product is associated with a linear transformation of variables. Suppose that the n original variables x1, x2,. . . ,x,, in (1—9) are expressed as a linear combination of s new variables Y1,Y2, . . . ,ys:
k=
Xk =
1,
2,. .
.
i
1,
,n
(1—13)
1=
Substituting for Xk in (1—10),
=
2,. .
.
,m
Interchanging the order of summation, and letting
i = 1,2 j—
k=i
in
(1—14)
the transformed equations take the form
=
1,2,.. .,
i
Noting (1—12), we can write (1—15) as
py =
C
where p is in x .s and y is S x 1. Now, we also express the transformation of variables, in matrix form,
which defines
x = by
where b is n x s. Substituting for x in (1—11),
aby=c and requiring (1—16) and (1—18) to be equivalent, results in the following definition equation for the product, ab:
= ab
=
[bkJ] = [pt,]
k
1,2,.
. .
,n
This product is defined only when the column order of a is equal to the row order of b. In general, if a is of order r x n, and b of order n x q, the product ab is of order r x q. The element at the ith row and jth column of the product is obtained by multiplying corresponding elements in the ith row of the first matrix and the jth column of the second matrix.
INTRODUCTION TO MATRIX ALGEBRA
8
CHAP. 1
Example 1—3
(1)(1) + (0)(O)
(IXI) + (O)(1)
(1)(O) +
1)
ab = (—l)(1) + (1)(O) (—1)(l) + (1)(l) (—1)(O) + (O)(1) + (2)(O)
(O)(1) + (2)(l)
(0)(0) + (2)(—1)
[+1
+1
0
—l
ab=J_1
0
—1
+4
[
0
(1)(— 1) + (01(3)
(—1)(—1) + (1)(3) (0)(—1)
+ (2)(3)
+2 —2 +6
If the product ab is defined, a and b are said to be confbrmable in the order stated. One should note that a and b will be conformable in either order only when a is in x n and b is n x in. In the previous example, a and b are conformable but b and a are not since the product ha is not defined. When the relevant products are defined, multiplication of matrices is associative,
a(bc) =
(ab)c
(1—20)
and distributive, a(b + c) = ab + ac (b + c)a = ha + Ca
but, in general, not commutative, ab
ba
(1—22)
Therefore, in multiplying b by a, one should distinguish preinultiplication, ab, from postrnultiplication ha. For example, if a and b are square matrices of order 2, the products are [a11 [a21
a121[bij a22j[b21
b121
[b11
b121[aji b22j[a21
aizl
[b21
When ab
1—4.
=
ha,
—
b22j
a22]
—
[aitbji
+ a12b21
a11b12 + a12b22
[a21b11 + a22b21
a21b12 + a22b22
[bjjaj1 + b12a21
b11a12 + b12a22
[b21a11 + b22a21
b21a12 + b22a22
the matrices are said to commute or to be permutable.
TRANSPOSE OF A MATRIX
is defined as the matrix obtained from a by The transpose of a = interchanging rows and columns. We shall indicate the transpose of a by
TRANSPOSE OF A MATRIX
SEC. 1—4
aT
=
9
{a79]:
a
a11
a12
a1,
021
a22
a2, (1—23)
=
=
amj
a,,,
am2
a21 012
= [a79] =
022
am2
a,,, The element, a79, at the ith row and jth column of aT, where now i varies from 1 to n and j from 1 to m, is given by (1—24)
a79 =
where
is the element at the jth row and ith column of a. For example,
[3
2 T
r3
a =[2
1
7
5
1
4
Since the transpose of a column matrix is a row matrix, an alternate notation for a row matrix is
a,] =
[a1, a2
(1—25)
We consider next the transpose matrix associated with the product of two matrices. Let
p==ab (a) where a is m x n and b is n x s. The product, p, is m x s and the element, Pu,
=
m Ilukbkf
(b)
—1 .1 —
The transpose of p will be of order s x m and the typical element is (c)
p79 =
where now I =
1,
s and j = 1, 2,. .
2
. ,m.
Using (1—24) and (b), we can
write (c) as p79 =
k1
It follows from (d) that
=
1,
j =—
k1 =
(ab)T
=
2,.
S
(d)
bTaT
Equation (1—26) states that the transpose of a product is the product of the
INTRODUCTION TO MATRIX ALGEBRA
10
CHAP. 1
transposed matrices in reversed order. This rule is also applicable to multiple
products. For example, the transpose of abc is (abc)T = cT(ab)T
Example
cTbTaT
(1—27)
1—4
ab =
(ab)T = [4
13
13
6]
6
Alternatively, aT
= [2
—1]
= (ab)T = bTaT = [2
1—5.
= [4
—1]
13
6]
SPECIAL SQUARE MATRICES
If the numbers of rows and of columns are equal, the matrix is said to be square and of order n, where n is the number of rows. The elements (i = 1, 2,. .. , n) lie on the principal diagonal. If all the elements except the principal-diagonal elements are zero, the matrix is called a diagonal matrix. We will use d for diagonal matrices. If the elements of a diagonal matrix are all unity, the diagonal
matrix is referred to as a unit matrix. A unit matrix is usually indicated by where n is the order of the matrix. Example
1—5
Square
Matrix, Order 2 [1
7
[3
2
Diagonal Matrix, Order 3
[2 [o
0
0
5
0
0
3
Unit Matrix, Order 2
12[
LO
0 I
SPECIAL SQUARE MATRICES
SEC. 1—5.
We
introduce the Kronecker delta notation:
oij=0
(1—28)
i—j
+1
With this notation, the unit matrix can be written as
i,j = 1, 2
=
(1—29)
n
Also, the diagonal matrix, d, takes the form
d=
(1—30)
are the principal elements. If the principal diagonal elements . , are all equal to k, the matrix reduces to where d1, d2,. .
=
=
(1—31)
and is called a scalar matrix.
Let a be of order rn x n. One can easily show that multiplication of a by a conformable unit matrix does not change a: a
(1—32)
Ima = a
A unit matrix is commutative with any square matrix of the same order. Similarly, two diagonal niatrices of order n are commutative and the product is a diagonal matrix of order a. Premultiplication of a by a conformable diagonal matrix d multiplies the ith row of a by and postmultiplication multiplies the jth column by Example
1—6
[2 [o
01[2
01[3
01
[6
0
—i][o 5j[O 5j[O _ij[o —5 [2 01[3 'l_[ 6 2 —
ij [2 7] — [—2 —7
[3 11[2 [2
[6
01
7j[0 _1j[4
A square matrix a for which = property that a = If
—' —7
=
is called symmetrical and has the j) and the principal diagonal elements all equal zero, the matrix is said to be skew-symmetrical. In this case, aT = — a. Any square matrix can be reduced to the sum of a symmetrical matrix and a skew-symmetrical matrix: (i
a=b+c = =
+ —
(1-33)
CHAP. 1
INTRODUCTION TO MATRIX ALGEBRA
12
The product of two symmetrical matrices is symmetrical only when the matrices
are commutative.* Finally, one can easily show that products of the type (aTa)
(aTba)
(aaT)
where a is an arbitrary matrix and b a symmetrical matrix, result in symmetrical matrices.
A square matrix having zero elements to the left (right) of the principal diagonal is called an upper (lower) triangular matrix. Examples are: Upper Triangular Matrix
352 071 004 Lower Triangular Matrix
300 570 214 Triangular matrices are encountered in many of the computational procedures developed for linear systems. Some important properties of triangular matrices are: The transpose of an upper triangular matrix is a lower triangular matrix and vice versa. The product of two triangular matrices of like structure is a triangular matrix of the same structure.
1.
2.
[a11
0 1[b11 I
[a21
1-6.
0
1
I=
b22j
[aijbij
0
[a21b11 + a22b21
a22b22
OPERATIONS ON PARTITIONED MATRICES
Operations on a matrix of high order can be simplified by considering the matrix to be divided into smaller matrices, called .subina.trices or cells. The partitioning is usually indicated by dashed lines. A matrix can be partitioned in a number of ways. For example,
a
a11
012
0131
a21
a22
023
031
a32
a33J
a11
a12
013
032
a33
= a1 031
=
a11
a12
a13
a31
a32
a33
Note that the partition lines are always straight and extend across the entire matrix. To reduce the amount of writing, the submatrices are represented by *
See Prob. 1—7.
SEC. 1—6.
OPERATtONS ON PARTITIONED MATRICES
single symbol. We will use upper case letters to denote the submatrices whenever possible and omit the partition lines. a
Example 1-1 We represent
[au
a12
a13
a=Ia,i
a22
a23
as
[A11 A121 [A21 A22J
a
or
a = [A11
A12
[A21
A22
where Ia11
A11
= [a21
a121
A12 =
I
A21
=
[a31
Ia13 I
La23
A22 = [a33]
a32]
If two matrices of the same order are identically partitioned, the rules of matrix addition are applicable to the submatrices. Let [A11 [A23
[B11
A121 I
A22J
8121 B22j
[823
(134)
where BLJ and A13 are of the same order. The sum is a
+b =
[A11 + 8fl
+ B121 A22 + B22j A12
LA2I + B21
(1-35)
The rules of matrix multiplication are applicable to partitioned matrices provided that the partitioned matrices are conformable for multiplication. In general, two partitioned matrices are conformable for multiplication if the partitioning of the rows of the second matrix is identical to the partitioning of the columns of the first matrix. This restriction allows us to treat the various submatrices as single elements provided that we preserve the order of multiplication. Let a and b be two partitioned matrices: a b
[A131t
= 1, 2,.. I = 1,2
= [B1d
,
M M
(1—36)
k= 1,2,...,S
We can write the product as
C = ab = [CIk]
,,...,
M 1
C when
—
ik
i
.
i1,
—
the row partitions of b are consistent with the column partitions of a.
INTRODUCTION TO MATRIX ALGEBRA
14
As an
CHAP. 1
illustration, we consider the product
ab =
au
a12
a13
h1
1221
a22
a23
h2
1233
a32
033
b3
Suppose we partition a with a vertical partition between the second and third columns,
a=
1211
1212
a13
a21
a22
a23
a31
a32
a33
= [A11A12]
For the rules of matrix multiplication to be applicable to the submatrices of a, we must partition b with a horizontal partition between the second and third rows. Taking
the product has the form
= [A,1A12]
= A11B11 + A12B21
The conformability of two partitioned matrices does not depend on the horizontal partitioning of the first matrix or the vertical partitioning of the second matrix. To show this, we consider the product ab
a12
a13
£121
1222
a23
1231
a32
1233
b11
b12 1322
b31
b32
Suppose we partition a with a horizontal partition between the second and third rows: a
a11
1212
C1j3
1221
a22
1223
1231
a32
a33
r A11 =
Since the column order of A11 and A21 is equal to the row order of b, no partitioning of b is required. The product is ab =
[A111
[A11b
LA2ijb = [A21b
As an alternative, we partition b with a vertical partition. b12
b=
b21
b22
= [811B12]
b31
In this case, since the row order of B11 and B12 is the same as the column
OPERATIONS ON PARTITIONED MATRICES
SEC. 1—6.
order of a, no partitioning of a is necessary and the product has the form
ab =
a[B11B12]
= [aBj1
aBi2]
To transpose a partitioned matrix, one first interchanges the off-diagonal submatrices and then transposes each submatrix. If A1,,
A11
A12
A21
A22
Arnt
Am2
AT1 AT
AT1 AT
.
.
.
AT
AT
AT
.
.
.
AT
a= then
A particular type of matrix encountered frequently is the quasi-diagonal matrix. This is a partitioned matrix whose diagonal submatrices are square of various orders, and whose off-diagonal submatrices are null matrices. An example is 0
a= 0
0 a22
0
a32
a33
a11
which can be written in partitioned form as a
= [Ai A2]
where
a23]
A2 = [a22
A1 = [a11]
a33
a32
and 0 denotes a null matrix. The product of two quasi-diagonal matrices of like structure (corresponding diagonal submatrices are of the same order) is a quasi-diagonal matrix of the same structure. A1
0
...
0
B1
0
...
0
A1B1
0
...
0
0
A
0
are of the same order. A and We use the term quasi to distinguish between partitioned and unpartitioned matrices having the same form. For example, we call (1—40)
a lower quasi-triangular matrix.
INTRODUCTION TO MATRIX ALGEBRA
16
1—7.
Cl-lAP. 1
DEFINITION AND PROPERTIES OF A DETERMINANT
The concept of a determinant was originally developed in connection with the solution of square systems of linear algebraic equations. To illustrate how this concept evolved, we consider the simple case of two equations:
a11x1 + a21X1
a12x2
= + a22x2 =
C2
Solving (a) for x3 and x2, we obtain (a11a22 — a12a21)x1
c2a12
=
(a11a22 — a12a21)x2
The scalar quantity, a1 1a22 —
a21 a2
—c1a21
+ c2a11
defined as the determinant of the second-
order square array (i,j 1, 2). The determinant of an array (or matrix) is usually indicated by enclosing the array (or matrix) with vertical lines: a11
a12
a21
a22
= al =
a31a22
—
a12a21
We use the terms array and matrix interchangeably, since they are synonymous. Also, we refer to the determinant of an eth-order array as an nth-order determinant. It shou'd be noted that determinants are associated only with square arrays, that is, with square matrices. The determinant of a third-order array is defined as +a11a22a33
a11
a12
a13
a21
a22
a23 = —a12a21a33
a31
a32
a33
+ a12a23a31
(1—42)
+a13a21a32 — a13a22a31
This number is the coefficient of x1, x2, and x3, obtained when the third-order system ax c is solved successively for x1, x2. and x3. Comparing (l—41) and (1—42), we see that both expansions involve products which have the following properties: 1.
2.
Each product contains only one clement from any row or column and no element occurs twice in the same product. The products differ only in the column subscripts. The sign of a product depends on the order of the column subscripts, e.g., +a11a22a33 and —a11a23a32,
These properties are associated with the arrangement of the column subscripts
and can be conveniently is
described using
the concept of a permutation, which
discussed below.
A set of distinct integers is considered to be in natural order if each integer is followed only by larger integers. A rearrangement of the natural order is called a permutation of the set. For example, (1, 3, 5) is in natural order and
DEFINITION AND PROPERTIES OF A DETERMINANT
SEC. 1—7.
(1,5,3) is a permutation of(1, 3,5). If an integer is followed by a smaller integer, the pair is said to form an inversion. The number of inversions for a set is defined
as the sum of the inversions for each integer. As an illustration, we consider the set (3, 1, 4, 2). Working from left to right, the integer inversions are: Integer
Inversions
Total
3
(3, 1)(3, 2)
2
None (4,2) None
0
1
4 2
1
0 3
This set has three inversions. A permutation is classified as even (odd) if the total number of inversions for the set is an even (odd) integer. According to this convention, (1, 2, 3) and (3, 1, 2) are even permutations and (1, 3, 2) is an odd permutation. Instead of cbunting the inversions, we can determine the number of integer interchanges required to rearrange the set in its natural order since an even (odd) number of interchanges corresponds to an even (odd) number of inversions. For example, (3,2, 1) has three inversions and requires one interchange. Working with interchanges rather than inversions is practical only when the set is small. Referring back to (1—41) and (1—42), we see that each product is a permutation
of the set of column subscripts and the sign is negative the permutation is odd. The number of products is equal to the number of possible permutations of the column subscripts that can be formed. One can easily show that there are possible permutations for a set of n distinct integers. We let , n) and define , ce,,) be a permutation of the set (1, 2,. . .
.
.
.
as
•
•
+
I
when
—
1
when
.
is
. ,
an even permutation (1—43)
.
..
,
a,,)
is an odd permutation
Using (1—43), the definition equation for an ,ith-order determinant can be written as a11
a12
a1,,
a21
a22
a2,,
=
(1—44)
1
where
the summation is taken over all possible permutations of (1, 2,
Factorial n =
= n(n
—
1)(n
—
2)
.
• (2)(1).
.
.
, n).
INTRODUCTION TO MATRIX ALGEBRA
18
CHAP. 1
Example 1—8 The permutations for n =
3
are
cxi—1
x23
a33 a32
=2
1
=3
a1=1 z1=2
a3=1
a32 a3=1
e123=+1 e132=—1
e231=+1 e312=+1 e321—-—1
Using (1—44), we obtain a11a22a33 — a11a23a32
a11
a12
a13
a21
a22
a23 = —a12a21a33 + a12a23a31
a32
a33
+a13a21a32 — a13a22a31
This result coincides with (1—42).
The following properties of determinants can be established* from (1—44): 1.
2. 3.
4. 5.
6.
7.
If all elements of any row (or column) are zero, the determinant is zero. The value of the determinant is unchanged if the rows and columns are interchanged; that is, aT! = a!. If two successive rows (or two successive columns) are interchanged, the sign of the determinant is changed. If all elements of one row (or one column) are multiplied by a number k, the determinant is multiplied by k. If corresponding elements of two rows (or two columns) are equal or in a constant ratio, then the determinant is zero. If each element in one row (or one column) is expressed as the sum of two terms, then the determinant is equal to the sum of two determinants, in each of which one of the two terms is deleted in each element of that row (or column).
If to the elements of any row (column) are added k times the corresponding elements of any other row (column), the determinant is unchanged.
We demonstrate these properties for the case of a second-order matrix. Let a
=
[a31
[a21
a22
The determinant is a! = a11a22 — a12a21
Properties 1 and 2 are obvious. It follows from property 2 that laTl * See
Probs. 1—17, 1—18, 1—19.
a!. We
COFACTOR EXPANSION FORMULA
SEC. 1—8.
illustrate the third by interchanging the rows of a: [a21
a' =
a22
a12 = a21a12 — a11a22 = —Ia! a'!
Property 4 is also obvious from (b). To demonstrate the fifth, we take
a21 =
a22 = ka12
ka11
Then a! = Next,
a11(kaj2)
a12(ka1j)
=
0
let a12 = b12 + c12
+ c11
a11
According to property 6,
hi + ci
al
where ibi
b11
b12
a21
a22
= ci
a21
a22
result can be obtained by substituting for O.ii and a12 in (b). Finally, to illustrate property 7, we take This
b12 = a12 + ka22 b21 = a21 b22 = a7, Then, ibi
1-8.
=
(a11 + ka21)a22 — (a12 + ka22)a21
=
a!
COFACTOR EXPANSION FORMULA
in the square matrix, a, If the row and column containing an element, are deleted, the determinant of the remaining square array is called the minor of and is denoted by The cofactor of is related to denoted by the minor of by (1—45) = (— As an illustration, we take
a= The values of
and
328 1
7
4
531
associated with a23 and a22 are
M23. =
= —1
A23 = (— 1)5M23 = +
M22 =
= —37
A22 = (—1)4M22 =
1
—37
INTRODUCTION TO MATRIX ALGEBRA
20
CHAP. 1
Cofactors occur naturally when (.1 —44) is expanded9 in terms of the elements
of a row or column. This leads to the following expansion formula, called Laplace's expansion by cofactors or simply Laplace's expansion: a1kAIk
=
(1 —46)
akJAkJ
Equation (1—46) states that the determinant is equal to the sum of the products of the elements of any single row or column by their cofactors. Since the determinant is zero if two rows or columns are identical, if follows that
=
k1
0
(147) 0
k
s
I
The above identities are used to establish Cramer's rule in the following section.
Example
1—9
We apply (1—46) to a third-order array and expand with respect to the first row:
(1) a11
a12
a13
(121
a23
a23
a31
a32
a33 2
=
a22
023
a33
023
+
a11(a22a33 — a23a32)
+
a31
a52(—a21a33
+ 0j3(—
a22 1) 035
(133
+ a23a31) +
a53(a21a32
a32
— 022035)
To illustrate (1 —47), we take the cofactors for the first row and the elements of the second row:
=
a21(a22a33
—
a23a32)
+ a22(—a21a33 + a23a31) +
a23(a21a32
—
a22a31)
0
(2) Suppose the array is triangular in form, for example, lower triangular. Expanding with respect to the first row, we have
a21 031
0
0
a22
0
a32
= a11
(122
0
032
033
=
(a51)(a22a33)
=
a11a22a33
033
Generalizing this result, we find that the determinant of a triangular matrix is equal to the product of the diagonal elements. This result is quite useful.
* See
Probs. 1—20, 1—21.
f See Ref. 4, sect. 3 15, for a discussion of the general Laplace expansion method. The expansion in terms of cofactors for a iow Or a COlUmn is a special case of the general method.
CRAMER'S RULE
SEC. 1—9.
The evaluation of a determinant, using the definition equation (1—44) or the cofactor expansion formula (1—46) is quite tedious, particularly when the array
is large. A number of alternate and more efficient numerical procedures for evaluating determinants have been developed. These procedures are described in References 9—13.
Suppose a square matrix, say c, is expressed as the product of two square matrices,
c='ab and we want cJ. It can be shown* that the determinant of the product of two square matrices is equal to the product of the determinants:
ci =
(1—48)
a! hi
Whether we use (1—48) or first multiply a and b and then determine lab! depends
on the form and order of a and b. If they are diagonal or triangular, (1—48) is quite efficient. t
Example
1—10
[1
r2
31
5] a! =
hi
3
4
=
Ic!
=
—20
Alternatively,
[[11
c
and
29J [1
a=[0
—20
r2 0 b__[1
31
5] bi = 8
a! = 5
cj =
Ic! = +40
Determining c first, we obtain
rs
121
= [5 20]
1—9.
and
ci = +40
CRAMER'S RULE
We consider next a set of n equations in n unknowns:
= *
j
= 1, 2, .
.
.
, ii
See Ref. 4, section 3—16.
t See Prob. 1 —25 for an important theoretical application of Eq. 1—48.
(a)
INTRODUCTION TO MATRIX ALGEBRA
22
CHAP. 1
Multiplying both sides of (a) by Air, where r is an arbitrary integer from 1 to n,
and summing with respect to j,
we
obtain (after interchanging the order of
summation)
=
Xk
k1
j=1
k and equals al when r =
Now, the inner sum vanishes when r
k.
This
follows from (1—47). Then, (b) reduces to lalxr
=
The expansion on the right side of (c) differs from the expansion al
=
ajrAj.
only in that the rth column of a is replaced by c. Equation (c) leads to Cramer's rule, which can be stated as follows:
A set of n linear algebraic equations in n unknowns, ax = c, has a n) is unique solution when 0. The expression for Xr (r = 1, 2 the ratio of two determinants; the denominator is al and the numerator
is the determinant of the matrix obtained from a by replacing the rth column by c.
If jaf = 0, a is said to be singular. Whether a solution exists in this ease will depend on c. All we can conclude from Cramer's rule is that the solution, if it exists, will not be unique. Singular matrices and the question of solvability are discussed in Sec. 1 —13. 1—10.
We
ADJOINT AND INVERSE MATRICES
have shown in the previous section that the solution to a system of n
equations in n unknowns,
i,j
1, 2,..., n
can be expressed as 1
(note
that we have taken r =
I
1, 2,. . ., ii
in Eq. c of Sec. 1—9). Using matrix notation,
(b) takes the form [Au]T{cj}
Equation (e) leads naturally to the definition of adjoint and inverse matrices.
ADJOINT AND INVERSE MATRICES
SEC. 1—10.
23
We define the adjoint and inverse matrices for the square matrix a of order n as
adjoint a = Adj a =
a1
inverse a =
(1—49)
Adj a
(1—50)
Note that the inverse matrix is defined only for a nonsingular square matrix. Example
1—11
We determine the adjoint and inverse matrices for
123 412
a= 2
3
1
The matrix of cofactors is 5
0
—10
—1
—10
+7
—7
+5
—1
Also, al = —25. Then —i
5
Adja
—10
+ 1/25 + 2/5
+7/25
0
+2/5
—7/25
+ 1/25
—1/5
=
—-- Adj
a
—7
—10 +5 +7 —1
0
a=
—
1/5
Using the inverse-matrix notation, we can write the solution of (a) as
x= Substituting for x in (a) and c in (d), we see that a1 has the property that
a1a = aa'
= Equation (1—51) is frequently taken as the definition of the inverse matrix instead of (1—50). Applying (1—48) to (i—Si), we obtain
It follows that (1—Si) is valid only when 0. Multiplication by the inverse matrix is analogous to division in ordinary algebra. If a is symmetrical,, then a is also symmetrical. To show this, we take the transpose of (1—5 1), and use the fact that a =. aT: 1
(a_la)T =
INTRODUCTION TO MATRIX ALGEBRA
24
Premultiplication by a'
1
CHAP. 1
results in —
a"'
and therefore a1 is also symmetrical. One can also show* that, for any nonsingular square matrix, the inverse and transpose operations can be interchanged:
bT,_t =
(1—52)
We consider next the inverse matrix associated with the product of two square matrices. Let
c=
ab
where a and b are both of order n x n and nonsingular. Premultiplication and then b1 results in by
a'c = b
(b'a'')c = It follows from the definition of the inverse matrix that
(ab)1 =
(1—53)
In general, the inverse of a multiple matrix product is equal to the product of the inverse matrices in reverse order. For example,
= The determination of the inverse matrix using the definition equation (1 —50)
is too laborious when the order is large. A number of inversion procedures based on (1—51) have been developed. These methods are described in Ref. 9—13. 1—11.
ELEMENTARY OPERATIONS ON A MATRIX
The elementary operations on a matrix are:' 1.
2. 3.
The interchange of two rows or of two columns. The multiplication of the elements of a row or a column by a number other than zero. The addition, to the elements of a row or column, of k times the corresponding element of another row or column.
These operations can be effected by premultiplying (for row operation) or postmultiplying (for column operation) the matrix by an appropriate matrix, called an elementary operation matrix. We consider a matrix a of order x n. Suppose that we want to interchange rowsj and k. Then, we premultiply a by an rn x in matrix obtained by modifying the mth-order unit matrix, I,,,, in the following way: 1. 2. *
Interchange Interchange
See Prob. 1—28.
and 5k• and
SEC. 1—11.
ELEMENTARY OPERATIONS ON A MATRIX
25
For example, if a is 3 x 4, premultiplication by
001 010 100 interchanges rows 1 and 3 and postmultiplication by
1000 0001 0010 0100 interchanges columns 2 and 4. This simple example shows that to interchange rows, we first interchange the rows of the conformable unit matrix and premultiply. Similarly, to interchange columns, we interchange columns of the conformable unit matrix and postmultiply. The elementary operation matrices for operations (2) and (3) are also obtained by operating on the corresponding conformable unit matrix. The matrix which multiplies row j by is an mth order diagonal matrix having d1 = 1 for i j and = Similarly, postmultiplication by an nth order diagonal matrix having = 1 for i j and = will multiply thejth column by Suppose that we want to add times row jto row k. Then, we insert in the kth row and jth column of and premultiply. To add z times column jto column k, we put in the jth row and kth column of and postmu-ltiply.
We let e denote an elementary operation matrix. Then, ea represents the result of applying a set of elementary operations to the rows of a. Similarly, ac represents the result of applying a set of elementary operations to thc columns of a. In general, we obtain e by applying the same operations to the conformable unit matrix. Since we start with a unit matrix and since the elementary operations, at most, change the value of the determinant by a nonzero scalar factor,* it follows that e will always be nonsingular.
Example We
1—12
illustrate these operations on a third matrix:
a=
1
1/2
1/5
3
7
2
1
5
—2 We
first: 2.
Add (—3) times the first row to the second row. Add (2) times the first row to the third row.
See
properties of determinants (page 18).
1.
*
INTRODUCTION TO MATRIX ALGEBRA
26
These
CHAP. 1
operations are carried out by premultiplying by
100 0 201
—3
1
and the result is 1
1/2
1/5
0
11/2
7/5
0
2
27/5
Continuing, we multiply the second row by 2/11: 1
0
0
1
1/2
1/5
0
2,'Il
0
0
11/2
7/5
0
0
1
0
2
27/5
=
1
1/2
1/5
0
1
14/55
0
2
27/5
Next, we add (—2) times the second row to the third row: 0
1
0
1/2
1
1
1/5
1
0
1
1
14/55
0
269/55
Finally, we multiply the third row by 55/269. The complete set of operations is
100 010 0
0
55/269
1
0
0
1
0
0110
100
0
11/21/5
012/110 —310 372
—2
=
0
0
1
2
1
1
1/2
1/5
0
1
14/55 =b
00
0
—2
1
1
5
1
This example illustrates the reduction of a square matrix to a matrix using elementary operations on rows, and is the basis for the Gauss elimination solution scheme (Refs. 9, 11, 13). We write the result as
ea =
b
where e is the product of the four operation matrices listed above:
—6/11
e
+ 1870/2959
0
0
2/11
0
—220/2959
55/269
We obtain e by applying successive operations, starting with a unit matrix. This is more convenient than listing and then multiplying the operation matrices for the various steps. The form of e after each step is listed below: Initial
100 0 0 001 1
Step 2
Step 1
100 0 201
—3
1
1
—6/11 2
0
0
2/11
0
0
1
RANK OF A MATRIX
SEC. 1—12. Step 3
0
0
2/11
0
—4/11
0
1
—6/11
27
Step 4
[1
0
0
2/11
0 55/269
—6/11 }
+34/11
L+187o/2959
—220/2959
Two matrices are said to be equivalent if one can be derived from the other by any finite number of elementary operations. Referring to Example 1 —12, the matrices 1
1/2
1/5
3
7
2
—21
5
and
1
1/2
0
1
00
1/5 14/55 1
are equivalent. Tn general, a and b are equivalent if b = paq
(1—54)
where p and q are nonsinqular. This follows from the fact that the elementary operation matrices are nonsingular.
1-12.
RANK OF A MATRIX
The rank, r, of a matrix is defined as the order of the largest square array,
formed by deleting certain rows and columns, which has a nonvanishing deter-
minant. The concept of rank is quite important since, as we shall see in the next section, the solvability of a set of linear algebraic equations is dependent on the rank of certain matrices associated with the set.
Let a be of order in x n. Suppose the rank of a is r. Then a has r rows which are linearly independent, that is, which contain a nonvanishing determinant of order r, and the remaining rn — r rows are linear combinations of these r rows. Also, it has n — r columns which are linear combinations of r linearly independent columns. To establish this result, we suppose the determinant associated with the first r rows and columns does not vanish. If a is of rank r, one can always rearrange the rows and columns such that this condition is satisfied. We consider the (r + 1)th-order determinant associated with the first r rows and columns, row p, and column q where r < p in, r < q n. a11
a12
a21
a22
an
ar2
aIr
01q
azq (1—55)
arr
0rq apq
We multiply the elements in rowj by (j 1, 2,. . . ,r) and subtract the result from the last row. This operation will not change the magnitude of Ar+t (see Sec. 1—7). In particular, we determine the constants such that the first r elements
CHAP. 1
INTRODUCTION TO MATRIX ALGEBRA
28
in the last row vanish: a11
021
012
022
a1,
a2r
-
=
a,2
0p2
(1—56)
apr
Equation (1—56) has a unique solution since the coefficient matrix is nonsingular. Then (1 —55) reduces to a11
012
a21
022
a1,
Ar+i
(1—57)
0
a,2
0rr
0
0
where = apq —
(1—58)
Orq]
Applying Laplace's expansion formula to (1—57), we see that A,÷1 vanishes when a is of rank r, A,÷1 vanishes for all combinations of p and q. It
follows that apq
= [aiq. 02p
:
.,
m
(1—59)
Apr
Combining (1—56) and (1—59), we have a11
a21
012
022
0r1
0p1
4p1
r
r+2
m (1—60)
Equation (1—60) states that the last m — r rows of a are linear combinations
of the first r rows. One can also show* that the last n — r columns of a are linear combinations of the first r columns.
Example
1—13
Consider the 3 x 4 matrix
1234
a=21 32 5
? See Prob. 1—39.
7
12
14
RANK OF A MATRIX
SEC. 1—12.
29
We see that a is at least of rank 2 since the determinant associated with the first two rows
and columns is finite, Then, the first two rows are linearly independent. We consider the determinant of the third-order array consisting of columns 1, 2, and q: 1
2
ajq
2
1
a2q
5
7
a3q
Solving the system,
+ 223 = 5 22 = 7 we obtain
If a is of rank 2, A3 must vanish. This requires a3q =
2 101q
+
=
22a2q
3ajq
+
a3q
q = 3,4 Since a33 and (134 satisfy this requirement, we conclude that a is of rank 2. The rows are related by (third row) = + 3 (first row) + (second row)
One can show* that the elementary operations do not change the rank of a matrix. This fact can be used to dctcrmine the rank of a matrix. Suppose b defined by (1—61) is obtained by applying elementary operations to a. We know
that band a have the same rank. It follows that a is of rank p. A matrix having the form of b is called an echelon matrix. When a is large, it is more efficient to reduce it to an echelon matrix rather than try to find the largest nonvanishing determinant: (pxpt b12
...
I
a11
0
(121
b2p
II,
B12 (1—61)
0
0
Example
1—14
[i a=)2 [5 First, we eliminate
Prob. 1—40.
3
4
1
3
2
7
12
12
and a31, using the first row: 1
* See
2
2
3
4
0
—3 —3 —6
0
—3 —3
—8
INTRODUCTION TO MATRIX ALGEBRA
30
CHAP. 1
Next, we eliminate aW, using the second row: —1
2
4
3
—3 —3 —6
0
0
0
0
—2
At this point, we see that r = 3. To obtain b, we multiply the second row by — 1/3, the third row by — 1/2, and interchange the third and fourth columns:
1243 0010
b= 0
1
2
1
Suppose a is expressed as the product of two rectangular matrices: a
=
(rnxn) (nxs) b c
(1—62)
One can show* that the rank of a cannot be greater than the minimum value of r associated with b and C: ruin [r(b), r(c)]
r(a)
(1—63)
As an illustration, consider the product a
[—1/2 — [—1/2
+1/2 +1/2
01
1]
Since each matrix is of rank 2, the rank of a will be we obtain
[0
0
I
Evaluating the product,
0 1
It follows that a is of rank 1. 1—13.
SOLVABILITY OF LINEAR ALGEBRAiC EQUATIONS
We consider first a system of two equations in three unknowns:
[:: ::
(1-64)
Suppose a is of rank 2 and a11
a21 * See
Prob. 1—44.
a22
0
(1—65)
SEC. 1—13.
SOLVABILITY OF LINEAR ALGEBRAIC EQUATIONS
If a is of rank 2, we can always renumber the rows and columns such that (1—65) is satisfied. We partition a and x, [a11 [a21
a
X1
a12
a131
a22
a23j
[A1
A2]
(1—66)
çx1
1x2 and write (1—64) as A1X1 + A2X2 = c. Next, we transfer the term involving
X2 to the right-hand side: A1X1 =
c — A2X2
(1—67)
0, it follows from Cramer's rule that (1—67) has a unique solution Since jA1j for X1. Finally, we can write the solution as
= Aj'(c
—
A2X2)
(1—68)
Since X2 is arbitrary, the system does not have a unique solution for a given c. The order of X2 is generally called the defect of the system. The defect for this system is 1.
If a is of rank 1, the second row is a scalar multiple, say A, of the first row. Multiplying the second equation in (1—64) by 1/A, we have a12x2 + a13x3 = a11x1 + a12x2 + a13x3 =
a11x1 +
C1
c2/A
(1—69)
2cr, the equations are inconsistent and no solution exists. Then, when a is of rank 1, (1—64) has a solution only if the rows of c are related in the same manner as the rows of a. If this condition is satisfied, the two equations in (1—69) are identical and one can be disregarded. Assuming that 0, the solution is Tf c2
x1 =
(1/a11)(c1
—
a12x2
—
a13x3)
(1—70)
The defect of this system is 2. The procedure followed for the simple case of 2 equations in 3 unknowns is also applicable to the general case of in equations in n unknowns: a11
a12
a1,,
x1
C1
(221
a is of rank in, there exists an mth order array which has a nonvanishing
determinant. We rearrange the columns such that the first in columns are
INTRODUCTION TO MATRIX ALGEBRA
32
CHAP. 1
linearly independent. Partitioning a and x, a12
a11
ai,m+1 a2m÷1
az,,,
am2
{x1
A2
A1 (mxm)
] (1—72)
•.. x4 =
Xm+i
Xm
= [
a,,,,
am,m÷1
amni
X2
az,,
{ X1 (m
1)
X2 }
((n—rn)x 1)
we write (1—71) as
c — A2X2
A3X1
(1—73)
0, (1—73) can be solved for X1 in terms of c and X2. The defect of the set is n — m, that is, the solution involves n — m arbitrary constants represented by X2. Suppose a is of rank r where r < m. Then, a has r rows which contain an rth-order array having a nonvanishing determinant. The remaining m — r rows are linear combinations of these r rows. For (1—71) to be consistent, that is, have a solution, the relations between the rows of c must be the same as those for a. The defect for this case is n — r. Since IA1I
Example
1—15
As an illustration, consider the third-order system
a11x1 + a12x2 + a13x3 = a21x1 + a22x2 + a73x3 = a31x1 + a32x3 + Suppose
C1
(a)
= C3
that r = 2 and the rows of a are related by (third row) =
(first row) +
(second row)
(b)
For (a) to be consistent, the elements of c must satisfy the requirement, A1C1
+ 22C2
To show this, we multiply the first equation in (a) by these equations the third equation. Using (b), we obtain
0=
C3 —
— I12C2
(c)
the second by —A2, and add to
(d)
Unless the right-hand side vanishes, the equations are contradictory or inconsistent and no solution exists. When e 0, (c) is identically satisfied and we see that (a) has a nontrivial 0) only when r < 3. The general case is handled in the same manner.* solution (x
See
Prob. 1—45.
REFERENCES
33
In general, (1 —71) can be solved when r < ,n if the relations between the rows of a and c are identical. We define the augmented matrix, ci, for (1—71) as a11
a12
C1
=
a2,,
a,flfl
afl,2
[a
cJ
(1—74)
Cm
When the rows of a and c are related in the same way, the rank of tz is equal to
the rank of a. It follows that (1—71) has a solution only if the rank of the augmented matrix is equal to the rank of the coefficient matrix:
=
r(a)
(1—75)
Note that (1—75) is always satisfied when r(a) = m for arbitrary c. We can determine r(cz) and i(a) simultaneously using elementary operations on provided that we do not interchange the elements in the last column. The
reduction can be represented as (1-76)
> r(a) and has a nonvanishing element, is of rank r(a). If no solution exists. r(a), (1—71) contains r independent equations involving n unWhen
where
knowns. The remaining m — r equations are linear combinations of these r equations and can be disregarded. Thus, the problem reduces to first finding r@) and then solving a set of r independent equations in n unknowns. The complete problem can be efficiently handled by using the Gauss elimmation procedure (Refs. 9, 11, 13).
REFERENCES 1.
2. 3.
4. 5.
6. 7, 8.
R. A., W. J. DUNCAN and A. R. cOLLAR: Elementary Matrices, cambridge University Press, London, 1963. THOMAS, G. B., JR.: Calculus and Analytical Geometry. Addison-Wesley Publishing Co., Reading, Mass., 1953. BODBWIG, E.: Matrix calculus, Interscience Publishers, New York, 1956. HOUN, F. E.: Elementary Matrix Algebra, Macmillan Co., New York, 1958. HADLEY, G.: Linear Algebra, Addison-Wesley Publishing Co., Reading, Mass., 1961. HOUSEHOLDER, A. S.: The Theory of Matrices in Numerical Analysis, Blaisdell, Waltham, Mass., 1964. NOBLE, B.: Applied Linear Algebra, Prentice-Hall, New York, 1969. HIL DEBRAND, F. B.: Methods of Applied Mathematics, Prentice-Hall, New York, 1952.
9.
Faddeeva, V. N.: Computational Methods of Linear Algebra, Dover Publications, New York, 1959.
INTRODUCTION TO MATRIX ALGEBRA
34
CHAP. 1
RALSTON, A. and H. S. WILF: Mathematical Methods for Digital Computers, Vol. 1, New York, 1960. 11. RALSTON, A. and H. S. WILF: Mathematical Methods for Digital Computers, Vol. 2, Wiley, New York, 1967. 10.
Wiley,
12.
BEREZIN, I. S. and N. P. ZHIDKOV: Computing Methods, Vols. I and 2, Addison-Wesley
Publishing Co., Reading, Mass., 1965. 13. 14. 15.
FORSYTHE, G. E., and C. B. MALER: Computer Solution of Linear Algebraic Systems, Prentice-Hall, New York, 1967. VARGA, R. S.: Matrix Iterative Analysis, Prentice-Hall, New York, 1962. CONTE, S. D.: Elementary Numerical Analysis, McGraw-Hill, New York, 1965.
PROBLEMS 1—1.
Carry out the indicated operations:
(a) 4
1
0
2
5
.3
321 +713 510 056 (b) [2
1
6j[ 3
1
(c)
—1
2
[1 21 [2 3[3 4j+2[[—i0 ii 3j4[l
[i
(d)
—2152
[—3
4J 55
(e)
[—i
l][4
1
2
—3j [2
3
[
1—2.
[4
11
[—i
I
[2
3j
2
—3
[
Expand the following products: [a1, a2,
.
.
.
,
{b1, b2,
a2, .
.
.
,
[b1, b2
.
(b)
{ai, (c)
[c1
01
[a11
a12
[o
c2j
[a21
a22
.
.
,
b,j
3
PROBLEMS
35
(d)
1—3.
[Cii
a121
0
[021
022j [0
C2
Show that the product of
Sl= a1+ a2 + 03 = S2
+ b2 + b3
=
bk
=
can be written as 3
3
S1S2 I
1
Generalize this result for the sum of n elements. 1—4. Suppose the elements of a and b are functions of y. Let cia —
[dv j
dy
db
—
[dblk]
dy
—
[dy j
Using (1—19), show that if c
= ab
then dc dy 1—5.
db dy
da dy
=a—+—b
Consider the triple product, abc. When is this product defined? Let p
= abc
What is the order of p? Determine Determine an expression for aT. case where c 1—6. Evaluate the following products:
for the
(a) F
[—2
41 [1
21 [5
1
[4
1
ij [2 sj
(b)
1—7.
where a is a square matrix. (a + b)(a + b) Show that the product of two symmetrical matrices is symmetrical
only when they are commutative. 1—8. Show that the following products are symmetrical: (a)
aTa (b)
aTba
where b is symmetrical
bTaTcab
where c is symmetrical
(c)
INTRODUCTION TO MATRtX ALGEBRA
36
CHAP. 1
1-9. Evaluate the following matrix product, using the indicated submatrices: 1
3
3
1
51
4
1—10. Let c = ab. Show that the horizontal partitions of c correspond to those of a and the vertical partitions of c correspond to those of b. Hint: See
Eq. (1—37).
1—11. A matrix is said to be symmetrically partitioned if the locations of the row and column partitions coincide. For example, a11
012
0131
a21
a22 a32
a23 a33
a31
is symmetrically partitioned and
is
a11
a12
a13
a31
a32
033
unsymmetrically partitioned. 1 symmetrical partitions.
Suppose we partition a square matrix with
N—
i,j =
a (a)
ab
If a and b are symmetrically partitioned, show that CJk, AJk, the same order. Illustrate for the case of one partition, e.g., [A11
[A2,
—
1—13.
have
Consider the product of two square nth order matrices.
a (b)
.
Ars =
c= (a)
2,. ., N
Deduce that the diagonal submatrices are square and Ars,
the same order. (b) If a = aT, deduce that 1—12.
1,
are of
A12 A22
Suppose we symmetrically partition c. What restrictions are placed on the partitions of a and b? Does it follow that we must also partition a and b symmetrically? Hint: See Prob. 1—10. Consider the triple product, C=
a symmetric rth-order square matrix and a is of order r x ii. Suppose we symmetrically partition c. The order of the partitioned matrices are indicated in parentheses. (pxp)
(pxq)
(nxn) —
[C11
C12
—
[c21 (q>
c22 (qxg)
PROBLEMS (a)
Show that the following partitioning of a is consistent with that of c. (r x n)
(r x q)
A2]
(j, k = 1,2) in terms of A1, A2, and b. Let d = [Di] be a quasi-diagonal matrix. Show that
Express
1—14.
da = bd = when
(r X p1
={A1
a
(b)
37
the matrices are
1—iS.
conformably
[DJAJk] [BJkDk]
partitioned.
Determine the number of inversions and interchanges for the
following sets. (a)
(4,3,1,2)
(b)
(3, 4, 2, 1)
1—16. 1—17.
How many permutations Consider the terms
does (1,2, 3, 4, 5) have?
1a1122a1h3
The first subscripts in (a) are in natural order. We obtain (b) by rearranging (a)
such that the second subscripts are in natural order. For example, rearranging e231 a12 a23 a31
= (2, 3, 1)
e312a31a12a23
fi = (3, 1,2)
we obtain
Show that if is an even permutation, (fl1, permutation. Using this result, show that
is
also an even
= p
and, in general,
al = 1—18.
Consider the terms ,
Suppose that
=
=
=
Then, (b) takes the form
Show that (c)
= —(a)
Generalize this result and establish that the sign of a determinant is reversed
when two rows are interchanged. 1—19. Consider the third-order determinant
a! =
INTRODUCTION TO MATRIX ALGEBRA
38
Suppose
CHAP. 1
the second row is a multiple of the first row: =
= 0. (Hint: = Show that properties 5 and 7 of Sec. 1—7.
Generalize this result and establish
—
1—20. Suppose all the integers of a set are in natural order except for one integer, say n, which is located at position p. We can put the set in natural order by successively interchanging adjacent integers. For example,
3 l2—* 132—* 123 231 —*21 3—* 123 Show that In — p1 adjacent interchanges (called transpositions) are required. It follows that the sign of the resulting set is changed by 1) In —
1—21.
We can write the expansion for the third-order determinant as
\
\j
(
i=1
)
/
k
t,J, .
.
— —
Using the result of the previous problem,
= and (a) reduces to 3
3
—
= Following this approach, establish Laplace's cofactor expansion formula for
an nth-order determinant. 1—22. Use Laplace's expansion formula to show that 1
0
0
o
o
b11
b12
...i
b22
bni 1—23.
0
0
9
k12
(pxp) a
a11 b2P
0n1
-
012
(un
a22
(12,7
0n2
Consider the quasi-diagonal matrix, (pxp)
d_[D1 [o(qxp)
0
D2
(qxq)
By expressing d as 0
[o
Iqj[0
D2
(pxn)
bFa
(nXn)
= al
PROBLEMS
that dl = D11 D21. Verify this result for
show
1100 2
d—
0
3
0
0021 0053
Generalize for [A,
d
Let
1—24.
(pxp)
(pxq)
[Gii
g
0
[G2,
—
G22 (qxq)
Show that = jG1
G221
Generalize for a quasi-triangular matrix whose diagonal submatrices are square, of various orders. 1—25. Suppose we express a as the product of a lower triangular matrix, g, and an upper triangular matrix, b. a11
a12
a21
a22
=
az,,
g11
0
0
g21
g22
0
1n2
Yflfl
'1n2
b11
b12 b22
?
0
0
bflfl
introduce symmetrical partitions after row (and column) p and write the product as We
(pxp)
(pxp)
(pxq)
[A1,
[G1,
A121
A22J — [G2,
LA21
(qxp)
(qxp)
(qxq)
(pxq)
(pxp)
0
1[Bi,
G22J[0 (qxq)
(qxp)
(pxq) B12
B22
(qxq)
Note that the diagonal submatrices of g and b are triangular in form. (a) Show that = A.11 G11B12 = A12 G21B11 = A21
+ G22B22 = A22
G21B12 (b)
Show that A111 =
1G111
BijI
and al = 1G111 G22j 1B111 1B221 (c)
Suppose we require that
By taking p = 1, 2,.
.
.
,n
— 1,
deduce that this requirement leads to the
INTRODUCTION TO MATRIX ALGEBRA
40
CHAP. 1
following n conditions on the elements of a: a11
a12
2
aj1
aj2
The determinant of the array contained in the firstj rows and columns is called
the jth-order discriminant. 1—26. Does the following set of equations have a unique solution?
1—27.
1
2
1
3
3
x1
3
5
X2
7
ii
x3
1—28.
=
3
5
Determine the adjoint matrix for
a= Does a
2
123 1
3
5
3
7
11
[1
3
[3
2
exist?
T = Find the inverse of
Show that b1'
bT,
(a)
(b)
[2 4
[i
S
(c)
[1
[3
31[2
4
2J[l
s
(d)
[2 O1[2 [o
3j[l
s
Let a12
a
= 31
[A11
A12
= [A2,
A22
33
32
and B32
—
[8,, 82, where the order of BJk is the same as AJ1. Starting with the condition
aa'
13
PROBLEMS
41
determine the four matrix equations relating BJk and Aft (j, k
result to find the inverse of
1—3!.
1, 2). Use this
124 212 121
Find the inverse of A lq
L0
Note that A is (p x q). 1-32. Find the inverse of 0 D2 1—33.
Use the results of Probs. 1—31 and 1—32 to find the inverse of b
—
[B11
B12
[o
B22
where B11, B22 are square and nonsingular. Hint: write has b 1—34.
[B11
01[1
0
ij[o
[o
j[o B22
1
Consider the 3 x 4 matrix
1121 1211
a=
2
3
1
2
Determine the elementary row operation matrix which results in a21 = = Oand a11 022 033 = +1. 1—35. Let (pxp)
[A11
a
[A2,
a31
=
(p'
A12 A22
(qxq)
0. Show that the following elementary operations 0 and on the partitioned rows of a reduce a to a triangular matrix. Determine
where
°1 [ L0
01 [A11 Iqj [A21
01
cj [A21 lqj [0 A
—
1—36.
AIA
Aizi
AW
—
A22J —
[o
Iq
\1
Suppose we want to rearrange the columns of a in the following way: 1
2
3
a= 2
1
3
3
4
5
col2—+coll col3—*col2
INTRODUCTION TO MATRIX ALGEBRA
42
(a)
CHAP. 1
Show that postmultiplication byIl(which is called a permutation matrix) results in the desired column rearrangement: o
0
11
1
0
01
o
i oj
H=
Note that we just rearrange the corresponding columns of 13. (b)
rearranges the rows of a in the
Show that pre;nultiplication by same way.
(c)
(d) (e)
Show that 11TH 13. Generalize for the case where a is n x n. Show that
=
a!
2.. Show that a is of rank 1 when 1—37. Let a be of order 2 x n, where n the second row is a multiple of the first row. Also, show that when r = 1, the nth columns are multiples of the first column. second, third 1—38. Determine the rank of (a) 1
3
7
5
2
4
—4 —10
3
(b) 2
3
—1
4
6
—2
—2 —3
1
1
2 —1
1—39. Let a be of order in x ii and rank r. Show that a has n — r columns which are linear combinations of r linear independent columns. Verify for
1234
a= 2
5
1
3
2
7
12
14
1—40. Using properties 3, 4, and 7 of determinants (see Sec. 1—7), deduce that the elementary operations do not change the rank of a matrix. For con-
venience, consider the first r rows and columns to be linearly independent. 1—41. Find the rank of a by reducing it to an echelon matrix.
a—
1122 2132 7797 4
2
1
PROBLEMS 1—42.
43
Show that c is at most of rank 1. a1
a2 C :
When will r(c) 1—43.
0? Consider the product,
r b11b12
a12
a11
b is of rank 1 and b11
Then, we can write
0.
cblk) (j)2kj
Show that the second, third,. , nth columns of c are multiples of the first column and therefore r(c) 1. When will r(c) = 0? Suppose r(a) = 1 and a11 0. In this case, we can write .
(b)
j = 2,3,...,m
= Show that the second, third
row and therefore r(c) 1 —44. Consider the product a11
a12
a21
a22
a,,1
a,,,2
mth rows of c are multiples of the first When will r(c) = 0?
1.
a1.,
'
b11
b12
b21
h22
h2
Let
= B1
-
{b11b21
-
-
-
.. ..
-
-
Using (b), we can write (a) as A1
A2 Am
Suppose r(a) = r4, r(b) rb. For convenience, we assume the first r0 rows of a and the first r1 columns of b are linearly independent. Then,
j
A3 = p=
1
= ra + 1, ra + 2, -. , m .
INTRODUCTION TO MATRIX ALGEBRA
44
CHAP. 1
rb
Bk
=
(a)
.
.
,
1
+ 2,. .
Show that rows ra + 1,
the first rows. (b) Show that columns Tb + 1, of the first columns. (c)
+ 1, Tb + 2, .
k—
XkqBq q
of c are linear combinations of
, in
.
+ 2,.. . , n of c are linear combinations
From (a) and (b), what can you conclude about an upper bound on r(c)?
(d)
To determine the actual rank of e, we must find the rank of A1
A1B1
A1B2
A2
A2B1
A2B2
BB [1 2 •..
A1Br1,
•.
rb]
A2Brb
AraB2
Tb. What canyou conclude about
Suppose ra (e)
(f)
A1 is orthogonal
toB1,B2 Br? Utilize these results to find the rank of
Suppose ra =
=
rb
—1/2
1/2
0
1
0
—1/2
1/2
1
1
0
1
—1
1
1
1
1
2
Show that r(e) =
s.
s.
1
Verify for
U 1—45.
Consider the m x n system a12 021
022
Gm i
2
X1
C1
0
C2
X,,
0R111
Let
=
.
.
j
.
1,
2,. .
.
, fl2
Using (b), we write (a) as in = 1, 2 (c) Now, suppose a is of rank r and the first r rows are linearly independent. Then,
A3x =
A1
(a)
j
k = r + 1, r + 2
=
m
Show that the system is consistent only if
k=
Ck
i•
+ 1, r + 2,.. .
, in
p=1
Note that this requirement is independent of whether in < n or in > n.
PROBLEMS
(b)
45
If m < n and r = m, the equations are consistent for an arbibrary c. Is this also true when rn > n and r = n? Illustrate for
[i [i
—1
11
2
4]
fc1
X2
and
1—46.
1
1
C1
—1
2
c2
1
4
c3
Consider the following system of equations:
x1 + X2 + 2x3 +
2x4
4
2x1+x2+3x3+2x4=6 3x1-l-4x2+2x3+x4=9 7x1 +7x2+9.x3+7x4=23 Determine whether the above system is consistent using elementary operations on the augmented matrix. (b) Find the solution in terms of x4. (a)
2
Characteristic-Value Problems and Quadratic Forms 2—i.
INTRODUCTION
Consider the second-order homogeneous system,
(ajj
2)x1 + at2xz
0
a25x1 + (a22 — A)x2 = 0
where A is a scalar. Using matrix notation, we can write (2—i) as ax
(2—2)
Ax
or (a — 212)x
0
(2—3)
The values of 2. for which nontrivial solutions of (2—i) exist are called the characteristic values of a. Also, the problem of finding the characteristic values and corresponding nontrivial solutions of (2—i) is referred to as a second-order characteristic-value problem,* problem occurs naturally in the free-vibration The
analysis of a linear system. We illustrate for the system shown in Fig. 2—1. The equations of motion for the case of no applied forces (the free-vibration case) are d2y2
m1
*
d2y
+ k2(y2 —
+ k1y1 —
k2(y2
—
0 y1)
=
0
Also called "eigenvalue" problem in some texts. The term "cigenvalue" is a hybrid of the
German term Elgenwerte and English "value." 46
47
NTRODUCT$ON
SEC. 2—1.
Assuming a solution of the form —
A
y2zize
iwt
A
and substituting in (a) lead to the following set of algebraic equations relating the frequency, w, and the amplitudes, A1, A2:
(k1 + k2)A1 — k2A2 = in1w2A1 —k2A1 + k2A2 = m2w2A2
We can transform (c) to a form similar to that of(2—1) by defining new amplitude measures,* 2—co2
A2 = and the final equations are k2 k1+k2_A — —== 2= —
rn1
k2
A1 +
'fl2
A2 = 2A2
characteristic values and corresponding nontrivial solutions of (e) are related to the natural frequencies and normal mode amplitudes by (d). Note that the coefficient matrix in (e) is symmetrical. This fact is quite significant, The
as we shall see in the following sections.
]
I
Fig. 2—1. A system with two degrees of freedom.
Although the application to dynamics is quite important, our primary reason for considering the characteristic-value problem is that results obtained for the characteristic value problem provide the basis for the treatment of quadratic * See Prob. 2—i.
PROBLEMS
48
CHAP. 2
forms which are encountered in the determination of the relative extrema of a function (Chapter 3), the construction of variational principles (Chapter 7), and stability criteria (Chapters 7, 18). This discussion is restricted to the case where a is real. Reference 9 contains a definitive treatment of the underlying theory and computational procedures. 2—2.
SECOND-ORDER CHARACTERISTIC-VALUE PROBLEM
We know from Cramer's rule that nontrivial solutions of
2)x1 + a12x2
—
a21x1 + (a22 —
A)x2
—
0
=
0
(2—4)
are possible only if the determinant of the coefficient matrix vanishes, that is, when a11 —
a12
A
0
a22—).
a21
(2-5)
Expanding (2—5) results in the following equation (usually called the characteristic equation) for 1: 22
—
(a11 + a22)). + (a11a22 — a21a12) = 0
(2—6)
We let
= a11 + a22 = a11a22 — a12a21 = H and the characteristic equation reduces to 22
+ P2 =
—
0
(2—7)
(2—8)
The roots of (2—8) are the characteristic values of a. Denoting the roots by 22, the solution is
21,2 = (P1 ± When a is symmetrical, a12 = a21, and —
=
(2—9)
a22)2
(a11
+ 4(a12)2
Since this quantity is never negative, it follows that the characteristic values for a symmetrical second-order matrix are always real.
Example
2—i
[2 a={2
2
=2÷ 5=
7
P2 = (2)(5) — (2)(2) = 6 The characteristic equation for this matrix is 22
—
72
+6=
0
SECOND-ORDER CHARACTERISTIC-VALUE PROBLEM
SEC. 2—2.
49
Solving (a),
[1
—2
—!
131=0 A.1
2—
where
±j
i=
By definition, nontrivial solutions of (2—4) exist only when 2 = or 22. In what follows, we suppose the characteristic values are real. We consider
first the case where A
A.1. Equation (2—4) becomes (a11 —
A.1)x1
a21x1 +
+
a12x2
(a22 —
=
0
=
0
The second equation is related to the first by times the first eq.
second eq.
This follows from the fact that the coefficient matrix is singular. (Oii
—
%1)(a22 —
—
a12a21
=
0
Since only one equation isindcpendcnt and there are two unknowns, the soluAssuming* tiOn is not unique. We define as the solution for A = 0, the solution of the first equation is that a12
xi') = 1)
C1
= 012
where
c1 is an arbitrary constant. Continuing, we let
and take c1 such that = 1. This operation is called normalization, and the resulting column matrix, denoted by Q1, is referred to as the characteristic vector for
=
=
—_2112
+ L
By definition,
if a12
(2—10)
a12
Q1Q1 = 1
0, we work with the second equation.
J -
(2—11)
CHARACTERISTIC-VALUE PROBLEMS
50
Since
we see that
Qi is a solution of (2—4) for A =
aQ1 =
CHAP. 2
(2—12)
A1Q1
Following the same procedure for A =
A2,
we obtain
Q2 = c2
(2—13)
)
where
=
+
j
L
Also,
(2—14)
aQ2
=
It remains to discuss the case where A1 = A2. If a is symmetrical, the characteristic values will be equal only when a11 a22 and a12 = c121 = 0. Equation (2—4) takes the form (a11 — A)xi + (0)xz 0 (O)x1 + (a11
A)x2 = 0
These equations are linearly independent, and the two independent solutions {c1, 0} {0, c2}
The corresponding characteristic vectors are Q1
= {+1,0}
(2—15)
Q2 = {0, +1} If a is not symmetrical, there is only one independent nontrivial solution when the characteristic values are equal. It is of interest to examine the product, QfQ2. From (2—10) and (2—13),
we have Q1Q2
—
— —C1C2
/1 +
(a11
—
A1)(a11
— A2)
2 a12
-
Now, when a is symmetrical, the.right-hand term vanishes since a11 — A2
—(a22 — A1)
=
—
a11 —
we see that QrQ2 = 0. This result is also valid when the roots are equal. In general, QIQ2 0 when a is unsymmetrical. Two nth order column vectors U, V having the property that and
UTV=VTU=0
(2—16)
are said to be orthogonal. Using this terminology, Q1 and Q2 are orthogonal for the symmetrical case.
SEC. 2—2.
SECOND-ORDER CHARACTERISTIC-VALUE PROBLEM
Example 2—2
[2 a=[2
A1=+6
2
A2=+1
+6 are
The equations for A =
—4x1 + 2x2 = 0
=0
2x1 —
times the first equation. Solving the first equation,
We see that the second equation is we obtain
=
=
c1
= 2c1
Then, 2} and
the normalized solution is Q1
Repeating forA =
A2
= +1, we find
=
c2{1,
and
Q2 One can easily verify that
j = 1,2
= and
—
— —
Ii a=[t
8 3
The characteristic values and corresponding normalized solutions for this matrix are 1
Q2 =
We see that
{4,
—
l}
0. Actually,
QTQ2 =
[1
—2
—l
A1—_+i
A2=—i
We have included this example to illustrate the case where the characteristic values are
PROBLEMS
52
complex. The equations corresponding to 2 (1 — i)x1
CHAP. 2
are
—
2x2
0
(1 + i)x2 = 0
x5
Note that the second equation is (1 — i) times the first equation. The general solution is
Repeating for
2=
x(1t =
c1
=
c2
12, we find
1-3-i}
Now, we take c2 = c1. When the roots are complex, 12 is the complex conjugate of We determine c1 such that Then, xt2> is the complex conjugate of
= I Finally, the characteristic values and characteristic vectors are 21,2 =
Qt,2 = In general, the characteristic values are complex conjugate quantities when the elements of a are real. Also, the corresponding characteristic vectors are complex conjugates.
2—3.
SIMILARITY AND ORTHOGONAL TRANSFORMATIONS
The characteristic vectors for the relations:
aQ1 = aQ2 =
system satisfy the following (a)
22Q2
We can write (a) as
2j
Q2] = EQ1
(b)
Now, we let
q=[Qj
Q2] (2—17)
= We call q the normalized Column j of q contains the normalized solution for modal matrix* for a. With this notation, (b) takes the form
aq =
(2—18)
* This terminology has developed from dynamics, where the characteristic vectors define the normal modes of vibration for a discrete system.
AND ORTHOGONAL TRANSFORMATIONS
SEC. 2—3.
53
We have shown that the characteristic vectors are always linearly indepen-
dent when a is symmetrical. They are also independent when a is unsymmetrical, provided that 0 except for the case where a is 11.2. Then, unsymmetrical and the characteristic values are equal. If 0, q1 exists and we can express (2—18) as
q'aq =
(2—19)
The matrix operation, p is arbitrary, is called a similarity transformation. Equation (2—19) states that the similarity transformation, q1( )q, reduces a to a diagonal matrix whose elements are the characteristic values of a.
If a is symmetrical, the normalized characteristic vectors are orthogonal, that is, — —
— —
—
—
Also, by definition,
Using these properties, we see that q
[Qfl
q = [QTJ [Qi
Q2]
[1
0
[o
1
and it follows that qT
(2—20)
p'
A square matrix, say p, having the property that is called an = orthogonal matrix and the transformation, pT( )p, is called an orthogonal transformation. Note that an orthogonal transformation is also a similarity transformation. Then, the modal matrix for a symmetrical matrix is orthogonal and we can write qTaq = (2—21) Example
2—3
[2
2 5
+6
Q2{2,-1} —[0
q=[Qi
0
+1]
PROBLEMS
54
We verify that qT
CHAP. 2
= q' and qTaq = 21 [1
1 [1
1 [5
21
—lj[2 [2 21[1 [2
1
aq — —
q
T
aq
[1
I
21
[1
01
1 [6
2
[12
—1
—ij —
0
sj[o i
21[6 01
—ij[o
i[i
21[6
[6
21
_-1][12 —ii = [o
= 5L2
01
ij =
(2)
[1
8 3
= +5
—
01
Lo Since
a is not symmetrical, qT q
-1
(
+1}
Qi =
—ij Actually,
q
[,15/6 — [\/17/6 —
1—
j
[— One can easily verify that
[5
q
[1
01
—2
01
Lo —iJ
[1
1
q involves complex elements. Since the characteristic vectors are complex conjugates, they are linearly independent and q -' exists. We find q — 1, using the definition equation for the inverse (Equation (1—50)):
SEC. 2—4
THE nth-ORDER SYMMETRICAL CHARACTERISTICS
One can easily verify that q
2—4.
[+i
-1
55
01
0
THE nth-ORDER SYMMETRICAL CHARACTERISTIC-VALUE PROBLEM
The nth order symmetrical characteristic-value problem involves determining the characteristic values and corresponding nontrivial solutions for a11x1 + a12x2 + a12x1 + +
+ +
+
+
+
= Ax1 — AX2
(2—22)
We can write (2—22) as
ax = AX (a —
(2—23)
0
In what follows, we suppose a is real. For (2—23) to have a nontrivial solution, the coefficient matrix must be singular. a — AI,4
(2—24)
0
The expansion of the determinant is
=
+
+
0
where
= a11 + a22 ±
+
(2—25) -
is the sum of all the jth order minors that can be formed on the diagonal.* Letting 22,. . , denote the roots, and expressing the characteristic equation in factored form, we see that and
.
=
2522
+ 2523 + ... +
(2-26)
We summarize below the theoretical results for the real symmetrical case. The proofs are too detailed to be included here (see References 1 and 9): 1.
2.
The characteristic values are all real. 22,. . , The normalized characteristic vectors Q1, Q2,. . , Q,,, are orthogonal: .
.
QTQJ =
i,j =
1,2
* Minors having a diagonal pivot (e.g.. delete the kth row and column). They are generally called principal minors.
CHARACTERiSTIC-VALUE PROBLEMS
56 3.
CHAP. 2
a is diagonalized by the orthogonal transformation involving the normalized modal matrix. qTaq
where
= Example 2—4
5—2
0
a= —2
3
—.1
0
—1
1
Since a is symmetrical, its characteristic values are all real. We first determine /3k, $2, f33, using (2—25):
5+ $2 /33
3
+9
+1
+11 + 5 + 2 +18 = 5(2) — (—2)(—2) = +6
The characteristic equation is
182—6=0 and the approximate roots are 22
+0.42 +2.30 +6.28
To determine the characteristic solutions, we expand ax = 2x,
=
(5 — 2)x1
2x2
—x3 = —(3 — 2)x2
(l—,t)x3=
x2
Solving the first and third equations for x1 and x3 in terms of x2, the general solution is
j=l,2,3 Finally, the modal matrix (to 2-place accuracy) is
+0.22
(2)
—0.84
q = [Q1Q2Q3] = +0.50
+0.51 +0.68
+0.85
—0.52
—0.10
120
a= 2
1
0
0 .0
3
+0.54
QUADRATIC FORMS
SEC. 2—5.
The expansion of Ia —
213J
=
57
0 is
and the roots are
23=—i
22—3 Writing out ax = 2x, we have (1—2)x1
=0
+2x2 +(1 — 1)x2
2x1
(3 — 2)x3
When 2 =
3,
= =
0
(a)
0
(a) reduces to
+2x2 =
—2x3
0
—2x2 = 0
(b)
(0)x3=0 We see from (b) that (a —
213)
is of rank 1 when 2 =
xI=c1
x2=cl
3.
The general solution of (b) is
x3=cz
By specializing the constants, we can obtain two linearly independent solutions for the repeated root. Finally, the characteristic vectors for 22 = 3 are
Q2 = (0,01) When 2 = 23 = —1, (a) reduces to 2x3 + 2x2
=
0
2x1 + 2x2
0
4x3
=0
The general solution and characteristic vector for
are
=
and
-
0
o}
This example illustrates the case of a symmetrical matrix having two equal characteristic
values. The characteristic vectors corresponding to the repeated roots are linearly independent. This follows from the fact that a — 213 is of rank I for the repeated roots.
2-5. QUADRATIC FORMS The homogeneous second-degree function
F a quadratic form in
+ 2a12x1x2 +
x2. Using matrix notation, we can express
F as [a11
F=[xix2]t[a12
ajal
T
CHARACTERISTIC-VALUE PROBLEMS
58
CHAP. 2
In general, the function
=
F= where afk =
for]
(2—27)
j=1
k, is said to be a quadratic form in xj, x2, ..
If F =
x
.
, x,,.
x
positive definite a 0, we say that F is positive semidefinite. We define negative definite and negative semidefinite quadratic forms in a similar manner. A quadratic form is negative definite if F 0 for all x and F = 0 only when x = 0. The question as to whether a quadratic form a
we is zero for some x
for
is positive definite is quite important. For example, we will show that an equilibrium position for a discrete system is stable when a certain quadratic form is positive definite.
Consider the quadratic form
F= b1
0
b2
LXIX2 13
x2
(2—28)
1
When F involves only squares of the variables, it is said to be in canonical form.
According to the definition introduced above, F is positive definite when
b1 >0
>0
b2
It is positive semidefinite when b1
0
and at least one of the elements is zero. Now, to establish whether is positive definite, we first reduce a to a diagonal matrix by applying the transformation, q '()q, where q is the orthogonal normalized modal matrix for a. We write xTax = (xTq)(q_taq)(qlx)
= Then, letting
y=
qTx
x
qy
(2—29)
(a) reduces to a canonical form in y:
F = xTax =
(2—30)
It follows that F is positive definite with respect to y when all the characteristic values of a are positive. But y is uniquely related to x and y = 0 only when x = 0. Therefore, F is also positive definite with respect to x. The problem of establishing whether xTax is positive definite consists in determining whether all the characteristic values of a are positive.
QUADRATIC FORMS
SEC. 2—5.
We consider first the second-order symmetric matrix
[a11
cz12
Laiz
a22
Using (2—26), the characteristic values are related by
+
=aii -F a22
,t122 = 132
=
a11a22
=
—
aJ
We see from (a) that the conditions
132>0 are
equivalent to
,t2>0 Suppose we specify that
>
au
at —. a11a22
0
>
—
0
Since a1 > 0, it follows from the second requirement in (d) that a22 > Therefore, (d) is equivalent to (b). We let
= aj1j = A2 =
a11a12
a12a12
0.
a11
=
(2—31)
at
Then, a is positive definite when
132>0 (2—3 2)
A2>0
A1>O The quantities
and are called the invariants and discriminants of a. The above criteria also apply for the case. That is, one can show that a is positive definite when all its invariants are greater than zero.
>
131 > 0
where
is
0
•..
/3,,
>
0
(2—33)
the sum of all the jth-order principal minors. Equivalent conditions
can be expressed in terms of the discriminants. Let represent the determinant of the array consisting of the first j rows and columns.
A=
a11
a12
£112
a22
aU (2—34)
a2J
The conditions, A1 >
0
A2 >
0
...
are sufficient for a to be positive definite.* *
See
Ref. 1 for a detailed proof. Also see Prob. 2—15.
A,,
>
0
(2—35)
CHARACTERISTIC-VALUE PROBLEMS
60
CHAP. 2
Example 2—5
111
122 123
The discriminants are A1 = +1 = 2 — 1 = +1 = 1(6—4)— 1(3—2) + 1(2—2) = +1 Since all the discriminants are positive, this matrix is positive definite. The corresponding invariants are = 1 + 2 + 3 = +6
$2(2—1)+(3—1)+(64)
Since A2 is negative
+5
= A3 = +1
/13
=
—3),
1
1
1
—2
2
1
2
3
1
this matrix is nor positive definite.
Suppose b is obtained from a by an orthogonal transformation: b = pTap
p1ap
(2—36)
If a is symmetrical, b is also symmetrical:
bT =
pTaTp
pTap
(2—37)
Now, b and a have the same characteristic values.* This follows from —
=
—
a —
(2—38)
Then, if a is positive definite, b is also positive definite. In general, the positive definite character of a matrix is preserved under an orthogonal transformation, REFERENCES 1.
HILDSBRAND, F. B.: Methods of Applied Mathematics, Prentice-Hall, New York, 1952.
2. 3.
BODEWIG, E.: Matrix Calculus, Interscience Publishers, New York, 1956. SMiRNOV, V. I.: Linear Algebra, Addison-Wesley Publishing Co., Reading, Mass., 1964.
4. 5.
TURNBULL, H. W., and A. C. AITKEN: An Introduction to the Theory of Canonical Matrices, Dover Publications, New York. HADLEY, G.: Linear Algebra, Addison-Wesley Publishing Co., Reading, Mass., 1961.
* See
Prob. 2—5.
PROBLEMS 6.
7.
8. 9.
10.
11. 12.
13.
CRANDALL, S. H.: Engineering Analysis, McGraw-Hill, New York, 1956. NOBLE, B.: Applied Linear Algebra, Prentice-Hall, New York, 1969. FRAZER, R. A., W. J. DUNCAN and A. R. COLLAR: Elementary Matrices, Cambridge University Press, London, 1963. WILKINSON, 3. H.: The Algebraic Eigenvalue Problem, Oxford University Press, London, 1965. FADDEVA, V. N.: Computational Methods of Linear Algebra, Dover Publications, New York, 1953. RALSTON, A., and H. S. WILF: Mathematical Methods for Digital Computers, Vol. 2, Wiley, New York, 1967. FORSYTHE, G. E., and C. B. MALER: Computer Solution of Linear Algebraic Systems, Prentice-Hall, New York, 1967. with Band Synimetric PETERS, G., and J. H. WILKINSON: "Eigenvalues of AX = A and B," Comput. J., 12, 398—404, 1969.
PROBLEMS 2—1.
Consider the system
Ay = where A and B arc symmetrical nth-order matrices and is a scalar. Suppose B can be expressed as (see Prob. 1—25)
B = brb where b is nonsingular. Reduce (a) to the form
ax = where x 2—2.
let c1,
c2
by. Determine the expression for a in terms of A and b. Let x1, x2 be two nth-order column matrices or column vectors and be arbitrary scalars. If c1x_I
+ c2x2
0
only when c1 =
= 0, x1 and x2 are said to be linearly independent. It follows that x1 and x2 are linearly dependent when one is a scalar multiple of the other. and Q2 arc linearly independent when Using (2—10) and (2—13), show that 2—3.
Determine the characteristic values and the modal matrix for
(a)
[3 [2 [2 0
3
(b)
to s
0
[3
2
2 7
0
2—4. Following the procedure outlined in Prob. 2—I, determine the characteristic values and modal matrix for
+ 12Y2
l2Y1 +
=
CHARACTERISTIC-VALUE PROBLEMS
62 .2—5.
Suppose
CHAP. 2
that b is derived from a by a similarity transformation.
b = p'ap Then, lb —
=
t1,4
a
—
and it follows that b and a have the same characteristic equation. (a)
Deduce that 2(b) — '(a) k
k
a(b) — p(a) Yk Pk
=
1,
2,.. ., n
Demonstrate for [1
The fact that (b)
.
, /3,,
.
formation is quite useful. Show that
P[2[1
—21
1
are invariant under a similarity trans-
_. Q U') k
P
—1
(a) k
2—6. When a is symmetrical, we can write qTaq = Use this result to find the inverse of
in terms of q and
Express
[3 a=[2 2—7.
2
Positive integral powers of a square matrix, say a, arc defined as
a2 = a3 =
aa aa2
ar = If al # 0,
exists, and it follows from the definition that
= (a) (b)
2—8.
a''
Show that ar is symmetrical when a is symmetrical. is a characteristic Let 2L be a characteristic value of a. Show that value of ar and is the corresponding characteristic vector.
Hint: Start with = 2,Q, and premultiply by a. A linear combination of nonnegative integral powers of a is called a
polynomial function of a and written as P(a). For example, the third order polynomial has the form
P(a) =
c0a3
+ c1a2 + c2a + c3L
Note that P(a) is symmetrical when a is symmetrical.
PROBLEMS
Let F(1)
0 be the characteristic equation for a. When the characteristic
values of a are distinct, one can show that (see Ref: 1)
F(a) =
0
where 0 is an nth-order null matrix. That is, a satisfies its own characteristic equation. This result is known as the Cayley-Hamilton Theorem. (a) Verify this theorem for [2 1 2
(b)
Note: F(a) = Show that
a2
f31a
= '(a2 (c) 2—9.
+
/3212.
— /31a + /3213)
forn =
3
Establish a general expression for a —' using (2—25). Determine whether the following quadratic forms are positive definite.
(a) (b) 2—10.
F= F = 34 +
+ 4x1x2 +
+
4x1x2 + 6x1x3 — 8x2x3
—
Show that a necessary but not sufficient condition for a to be positive
definite is
a11>O,a22>O (Hint:
= Oforj 1,] = 1,2,.. .,n) ax = 0 has a nontrivial solution, say x1. What is. the
= 0, 2—11. If value of xfax1? Note that 2 0 is a characteristic value of a when a is singular. 2—12. Let C be a square matrix. Show that CTC is positive definite when 0 and positive sernidefinite when CI = 0. IC! (hint: Start with F = xT(CTC)x and let y = Cx. By definition, F can equal zero only when x 0 in order for the form to be positive definite.) 2—13. Consider the product CTaC, where a is positive definite and C is 0 and positive semisquare. Show that C1aC is positive definite when CI definite when CI = 0. Generalize this result for the multiple product, . .
.
.
.
2—14. Let a be an mth-order positive defInite matrix and let C be of order m x n. Consider the product, b = CTaC
Show that b is positive definite only when the rank of C is equal to n. What can we say about b when r(C) < n? 2—15. Consider the quadratic form a11
a12
x1
a12
a22
x2
:
:
a27,
x,,
CHARACTERISTIC-VALUE PROBLEMS
64
CHAP. 2
We partition a symmetrically, (pXl)
(pxp)
A121 f Xt
NzTvTl [A11
AT (qxp)
where
q=
n — p.
MV ("2
The expansion of F = XTaX has the form
F= Now, we take X2
A
(qxq)
XfA11X1 -i- 2XTA12A2 +
0 and denote the result by XTAIIX1
> 0 for arbitrary X1, A11 must be positive definite. Since 1A111 is equal For to the product of the characteristic values of A11, it follows that Ajj must be positive. (a)
By taking p =
1,
2,.. .
, n,
deduce that
p=1,2,...,n are necessary conditions for a to be positive definite. Note that it remains to show that they are also sufficient conditions. (b)
Discuss the case where
= 0.
2-46. Refer to Prob. t—25. Consider a to be symmetrical. (a) Deduce that one can always express a as the product of nonsingular lower and upper triangular matrices when a is positive definite. (b)
Suppose we take Show that a is positive definite when
j=1,2,...,n and positive semi-definite when
j=1,2,...,n and at least one of the diagonal elements of g is zero. Suppose we take g = hT. Then,
= and A
Show
— Ati.11 —..h2i.2
,..i.2 Upp
that the diagonal elements of b will always be real when a is
positive definite. 2—17. If a quasi-diagonal matrix, say a, is symmetrically partitioned, the submatrix A11 is also a quasi-diagonal matrix. Establish that
a=
i,j =
is positive definite only when A, (i = 1, 2,.
2,. . ., N
1, .
.
,
N) are positive definite.
PROBLEMS
65
Hint: Use the result of Prob, 1—23. Verify for
1100 2300 0021 0052
2—18. Suppose we express a as the product of two quasi-triangular matrices, for example, (pxp)
(nxn)
a
[G11
1)
= [G2,
(qxp)
where p +
q
(pxq)
1 [B11
G22j[O
B12
B22
= n. We take B11
1P
8221q
Show that the diagonal submatrices of g are nonsingular for arbitrary p when a is positive definite.
3
Relative Extrema for a Function 3—1.
RELATIVE EXTREMA FOR A FUNCTION OF ONE VARIABLE
Letf(x) be a function of x which is defined for the interval x1 x x2. If f(x) — f(a) 0 for all values of x in the total interval x1 x x2, except x a, we say the function has an absolute minimum at x a. If f(x) — f(a)> 0 for all values of x except x = a in the subinterval, x containing x = a, we say that f(a) is a relative minimum, that is, it is a minimum with respect to all other values of f(x) for the particular subinterval. Absolute and relative maxima are defined in a similar manner. The relative maximum and minimum values of a function are called relative extrema. One should note thatf(x) may have a number of relative extreme values in the total interval x1 x x2. As an illustration, consider the function shown in Fig. 3—1. The relative extrema are [(a), f(h), f(c), f(d). Using the notation introduced above, we say that f(b) is a relative minimum for the interval x fib. The absolute maximum and minimum values of f occur at x = a and x = d, respectively. f(x)
x1
x a
b
c
Fig. 3—i. Stationary points at points A, 8, C, and 0. 66
X2
SEC. 3—1. RELATIVE EXTREMA FOR A FUNCTION OF ONE VARIABLE
67
In general, values of x at which the slope changes sign correspond to relative extrema. To find the relative extrema for a continuous function, we first determine the points at which the first derivative vanishes. These points are called stationary points. We then test each stationary point to see if the slope changes sign. If the second derivative is positive (negative) the stationary point is a relative minimum (maximum). If the second derivative also vanishes, we must consider higher derivatives at the stationary point in order to determine whether
the slope actually changes sign. In this case, the third derivative must also vanish for the stationary point to be a relative extremum. Example
3—1
Setting the first derivative equal to zero,
x2 + 4x +
dx
I
=
0
2(x
+
2)
and solving for x, we obtain
x1,2 =
—2
±
The second derivative is d2f
=
2x
+4=
Thcn,x = x1 = —2 + corresponds to a relative maximum.
=
J(x) =
(x
—
a)3
x2
= —2—
+c
The first two derivatives are = 3(x
—
a)
Since both derivatives vanish at x =
a,
we must consider the third derivative: d3f dx3
6
The stationary point, x = a, is neither a relative minimum nor a relative maximum since the third derivative is finite. We could have also established this result by considering the expression for the slope. We see from (a) that the slope is positive on both sides of x = a. The general shape of this function is shown in Fig. E3—l.
RELATIVE EXTREMA FOR A FUNCTION
66
CHAP. 3 Fig. E3—1
f(x)
x
I
a
The sufficient condition for a stationary value to be a relative extremum (relative minimum (maximum) when d2f/dx2 > 0 (< 0)) follows from a consideration of the geometry of the f(x) vs. x curve in the vicinity of the stationary point. We can also establish the criteria for a relative extremum from the Taylor series expansion of f(x). Since this approach can he readily extended to functions of more than one independent variable we will describe it in detail. Suppose we know the value of f(x) at x = a and we want f(a + Ax) where Ax is some increment in x. If the first n + 1 derivatives off(x) are continuous in the interval, a x a + Ax, we can express f(a + Ax) as
f(a + Ax)
f(a) Ax +
=
where
remainder
(Ax)2
+
(Ax)" +
denotes the jth derivative of f(x) evaluated at x
(3.1) a, and the
is given by 1
(3-2)
where is an unknown number between a and a + Ax. Equation (3—1) is called
the Taylor series expansion* of f(x) about x =
a.
If f(x) is an eth-degree
polynomial, the (n + 1)th derivative vanishes for all x and the expansion will yield the exact value off(a + Ax) when n terms are retained. In all other cases, there will be some error, represented by due to truncating the series at n terms. Since depends on we can only establish bounds on The following example illustrates this point. See Ref. 1, Article 16—8.
SEC. 3—1. RELATIVE EXTREMA FOR A FUNCTION OF ONE VARIABLE
69
Example 3—2 We expand sin x in a Taylor series about x = 0 taking n = and noting that a = 0, we obtain sin Ax = Ax + R2
2.
Using (3—1) and (3—2),
The bounds on R2j are cos Ax <
R21
If we use (a) to find sin (0.2), the upper bound on the truncation error is
0.0013.
If Ax is small with respect to unity, the first term on the right-hand side of (3—1) is the dominant term in the expansion. Also, the second term is more nth terms. We refer to df/dx Ax as significant than the third, fourth the first-order increment in f(x) due to the increment, Ax. Similarly, we call 4d2f/dx2(Ax)2 the second-order increment, and so on. Now, f(a) is a relative minimum when f(a + Ax)— f(a) > 0 for all points in the neighborhood of Ax e, where a, that is, for all finite values of Ax in some interval, — and e are arbitrary small positive numbers. Considering Ax to be small, the first-order increment dominates and we can write
x
f(a + Ax) — f(a) =
Ax + (second- and higher-order terms) (3—3)
For f(a + Ax) — f(a) to be positive for both positive and negative values of Ax, the first order increment must vanish, that is, df(a)/dx must vanish. Note that this is a necessary but not sufficient condition for a relative minimum, if the first-order increment vanishes, the second-order increment will dominate: f(a + Ax) — f(a) =
(Ax)2
+ (third- and higher-order terms)
(3—4)
It follows from (3—4) that the second-order increment must be positive for > 0 to be satisfied. This requires d2f(a)/dx2 > 0. Finally, f(a + Ax) — the necessary and sufficient conditions for a relative minimum at x = a are df(a) dx
—
0
d2f(a)
dx2
lithe first two derivatives vanish at x = the dominant term in the expansion.
f(a + Ax) + f(a) =
a,
>
3
5
the third-order increment is now
(Ax)3 + (fourth- and higher-order terms)
(3—6)
Since the third-order increment depends on the sign of Ax, it must vanish for
RELATIVE EXTREMA FOR A FUNCTION
CHAP. 3
f(a) to be a relative extremum. The sufficient conditions for this case are as follows:
Relative Minimum d4f
d3f
dX4> (3—7)
Relative Maximum d4f
d3f
The notation used in the Taylor series expansion off(x) becomes somewhat cumbersome for more than one variable. In what follows, we introduce new notation which can be readily extended to the case of 11 variables. First, we to be the total increment in f(x) due to the increment, Z\x. define
41 = f(x +
Ax)
—
f(x)
(3—8)
This increment depends on Ax as well as x. Next, we define the differential operator, d, (3—9)
The result of operating onf(x) with d is called the first by df:
df=-1Ax =
df(x,Ax)
and is denoted (3—10)
The first differential off(x) is a function of two independent variables, namely, x and Ax. Iff(x) = x, then df/dx = 1 and
c/f = dx = Ax
(3—11)
One can use dx and Ax interchangeably; however, we will use Ax rather than dx.
Higher differentials of f(x) are defined by iteration. For example, the second differential is given by
d2f = d(df)
(3—12)
=
Since Ax is independent of x,
(Ax) =
0
and d2f reduces to
d2f =
(Ax)2
= d2f(x, Ax)
In forming the higher differentials, we take d(Ax) =
0.
(3—13)
SEC. 3—2.
FUNCTION OF n INDEPENDENT VARIABLES
Using differential notation, the Taylor series expansion (3—1) about x can be written as
(3-14)
The first differential represents the first-order increment in f(x) due to the increment, Ax. Similarly, the second differential is a measure of the secondorder increment, and so on. Then,f(x) is a stationary value when df = 0 for all permissible values of Ax. Also, the stationary point is a relative minimum (maximum) when d2f> 0 (<0) for all permissible values of Ax. The above criteria reduce to (3—5) when the differentials are expressed in terms of the derivatives. Rules for forming the differential of the sum or product of functions are listed below for reference. Problems 3—4 through 3—7 illustrate their application.
f= u(x) + v(x)
df=du+dv d2f = d(df)
15
3
d2u + d2v
—
f= u(x)v(x) df = u dv + v du 2 = ud2 v + 2dudv + vd 2u
(3—16
df
f = fly) where y = y(x) df dy
dy
3-2.
RELATIVE EXTREMA FOR A FUNCTION OF n INDEPENDENT VARIABLES
be a continuous function of n independent variables We define Af as the total increment in f due to increments in the independent variables (Ax1, Ax2 Let f(x1, (x1, x2
Af = f(x1 + Ax1, x2 + Ax2,. .
.
,
+
—
f(x1, x2,.. .
If Af> 0 (<0) for all points in the neighborhood of(x1, x2,.
,
(3—18)
we say . , that f(x1, x2 is a relative minimum (maximum). We establish criteria for a relative extremum by expanding f in an n-dimensional Taylor series. The procedure is identical to that followed in the one-dimensional case. Actually, we just have to extend the differential notation from one to n dimensions. .
RELATIVE EXTREMA FOR A FUNCTION
72
CHAP. 3
We define the n-dimensional differential operator as d
Ax1
+
(3-19)
=
+
Ax2 +
ox2
Ax2 where the increments are independent of (x1, x2,. The result obtained when d is applied to f is called the first
df
. ,
(3—20)
=
Higher differentials are defined by iteration. For example, the second differential has the form
d2f=d(df)= Since
(3—21)
are considered to be independent, (3—21) reduces to d2f
(3—22)
Axk k=i.
Now, we let
f(2)
r
1
j, k =
2, .
1,
,n
. .
(3—23)
LOXJOXkJ
Ax = and the expressions for the first two differentials simplify to
df = d2f =
AXTf(l)
(3—24)
AXTf(2) Ax
The Taylor series expansion forf about (x1, x2, terms of differentials, has the form
Af= df +
+
. .
+
We say that f(x1, x2 is stationary when df requirement is satisfied only when
=
,
when expressed in
+
(3-25)
.
0 for arbitrary Ax. This (3—26)
0
Equation (3—26) represents n scalar equations, namely,
=
0
j=
1,
2,. .
.
,n
(3—27)
The scalar equations corresponding to the stationary requirement are usually
SEC. 3—2.
called
FUNCTION OF n INDEPENDENT VARIABLES
73
the Euler equations for f, Note that the number of equations is equal
to the number of independent variables.
A stationary point corresponds to a relative minimum (maximum) of f when d2f is positive (negative) definite. It is called a neutral point when d2f is either positive or negative semidefinite and a saddle point when d2f is indifferent, i.e., the eigenvalues are both positive and negative. This terminology was originally introduced for the two dimensional case where it has geometrical significance.
To summarize, the solutions of the Euler equations correspond to points at which f is stationary. The classification of a stationary point is determined by evaluated at the point. the character (definite, semidefinite, indifferent) of We are interested in the extremum problem since it is closely related to the stability problem. The extremum problem is also related to certain other problems of interest, e.g., the characteristic-value problem. In the following examples, we illustrate various special forms of f which are encountered in member system analysis.
Example 3—3
f=
y,)
=
x2
df= 1
1
ek
k
1
j
Xk
1
Now,
>-—&Xk
k1
OXk
It follows that
df=
j=l
Repeating leads to
d2f
=
Consider the double sum,
f
>
k=t
The first differential (see Prob. 3—9) has the form
df =
(duJwJkvk +
dv1)
dwfkvk +
Introducing matrix notation,
u=
w= =
and letting
du
[w31]
v=
{v1}
RELATIVE EXTREMA FOR A FUNCTION
74
CHAP. 3
and so forth, we can write df as
df = =
d(urwv) duTwv
+ nTdWv + uTwdv
One operates on matrix products as if they were scalars, but the order must be preserved. As an illustration, consider
f= where
— x7c
a, c are constant and a is symmetrical. Noting that da =
dc
=
0
and dx
Ax,
the first two differentials are AxT(ax — c)
df
d2f =
AxTa Ax
Comparing (g) and (3—24), we see that fO)
ax —
=
c
a
The Euler equations are obtained by setting
equal to 0:
ax
c
The solution of (i) corresponds to a stationary value of (f). If a is positive definite, the stationary point is a relative minimum. One can visualize the problem of solving the system ax = c, where a is symmetrical from the point of view of finding the stationary — XTC. value of a polynomial having the form f =
Suppose
f
u/v. Using the fact that
ldu
3j7u\ \,vJ
UX1
:X3
1(6u
uôv
V
vax1
—
we can write
df =
=
—f dv)
We apply (b) to 1.
=
xx
where a is symmetrical, and obtain (see Prob. 3—5) 2 AXT
= —f---- (ax — 2x) (d)
(122 =
xx
Ax
A
AxT Ax — 2 dA AxTx)
Setting dA = 0 leads to the Euler equations for (c),
ax—Ax=0 which we recognize as the symmetrical
(e)
problem.
LAGRANGE MULTIPLIERS
SEC. 3—3.
75
The quotient xTax/xTx, where x is arbitrary and a is symmetrical, is called Rayleigh's
quotient. We have shown that the characteristic values of a are stationary values of Rayleigh's quotient. This property can he used to improve an initial estimate for a characteristic value. For a more detailed discussion, see Ref. 6 and Prob. 3—11.
3-3.
LAGRANGE MULTIPLIERS
Up to this point, we have considered only the case where the function is expressed in terms of independent variables. In what follows, we discuss how one can modify the procedure to handle the case where some of the variables are not independent. This modification is conveniently effected using Lagrange multipliers. Suppose f is expressed in terms of n variables, say x1, x2 some of which are not independent. The general stationary requirement is
df= >
(3—28)
j=1
for all arbitrary differentials of the independent variables. We use of
instead
to emphasize that some of the variables are dependent. In order to
establish the Euler equations, we must express df in terms of the differentials of the independent variables. Now, we suppose there are r relations between the variables, of the form =
g5(x1, x2
It =
0
1,
2, .
.
,r
(3—29)
One can consider these relations as constraint conditions on the variables. Actually, there are only n — r independent variables. We obtain r relations between the n differentials by operating on (3—29). Since = 0, it follows that 0. Then,
"a
k=1,2,...,r
j=t
(3—30)
Using (3—30), we can express r differentials in terms of the remaining n — r differentials. Finally, we reduce (3—28) to a sum involving the n — r indepen-
dent differentials. Equating the coefficients to zero leads to a system of n — r
equations which, together with the r constraint equations, are sufficient to determine the stationary points. Example We
3—4
illustrate the procedure for n =
2
and r =
1:
f= g(x1, x2) = 0
The first variation is ox1
Ox2
RELATIVE EXTREMA FOR A FUNCTION
76
CHAP. 3
Operating on g(x1, x2) we have
ax1
Now, we suppose ag/ax2 0. Solving (b) for dx2 (we replace dx1 by x1 is the independent variable.)
/ag\
dx2= —t—i-----itSx1 8x2/
and substituting in (a), we obtain
df = [PL ax1
—
8x1 ox2
ox2
Finally, the equations defining the stationary points arc
\ox1fOx2/ox2 g(x1, x2) = 0
ax1
To determine whether a stationary point actually corresponds to a relative extremum, we must investigate the behavior of the second differential. The general form of d2f for a function of two variables (which are not necessarily independcnt) is d2f = 2
=
2
2
dx1 +
k1
Of —i—
a quadratic form in the independent differential, d2x1
using (c), and noting
= 0, d2f
+ ax1ax.2
ox2
+
\0x1
+
ax2j Ox2
where
u=
Og Jag
The character of the stationary point is determined from the sign of the bracketed term.
An automatic procedure for handling constraint conditions involves the use of Lagrange multipliers. We first describe this procedure for the case of two variables and then generalize it for n variables and r restraints. The problem consists in determining the stationary values of f(x1, x2) subject to the constraint condition, g(x1, x2) = 0. We introduce the function H, defined by H(x1, x2, A.) = f(x1, x2) + Ag(x1, x2)
(3—31)
where A. is an unknown parameter, referred to as a Lagrange multiplier. We
LAGRANGE MULTIPLIERS
SEC. 3—3.
77
consider x1, x2 and ,% to be independent variables, and require H to be sta-
tionary. The Euler equations for H are OH
—
Og
+
Ox, Ox2
A
(3-32)
Ox2
Ox2
OH
—
Ox1 —
g(x1, x2) =
0
0. Then, solving the second equation in (3—32) for A, and substituting in the first equation, we obtain
We suppose Og/0x2
A=
(3-33)
0x2/ Ox2
and
=
0
g(x1, x2) =
0
—
Ox1
(3-34)
Equations (3—34) and (e) of the previous example are identical. We see that the Euler equations for II are the stationary conditions for f including the effect of constraints.
Example
3—5
f= g
We form H
f+
+ 2x1 + 7x2
+ =
0
2x1
+
=
— x2
+
+
2g,
H=
7x2
+
A(x1 — x2)
The stationary requirement for H treating x1, x2, and 2 as independent variables is
6x1 + 2 + 2 =
0
2
=
0
x2
=
0
4x2
+7— x1
—
Solving this system for x1, x2 and A we obtain 4x2
A
=
+7 = —9/10
This procedure can be readily generalized to the case of n variables and r constraints. The problem consists of determining the stationary values Of subject to the constraints gk(xl, x2,. . , = 0, where j(x1, x2, . . , , r. There will be r Lagrange multipliers for this case, and H has k = 1, 2, .
.
RELATIVE EXTREMA FOR A FUNCTION
78
CHAP. 3
the form
H=
f + k1
H(x1, x2
2k9k
.
.
(3—35)
,
The Euler equations for H are
+
0
..,n
(3—36)
= 1, 2,. . ., r
(3—37)
i = 1,2,.
k=1
9k = 0
k
We first solve r equations in (3—36) for the r Lagrange multipliers, and then determine the n coordinates of the stationary points from the remaining n — r equations in (3—36) and the r constraint equations (3—37). The use of Lagrange multipliers to introduce constraint conditions usually reduces the amount of algebra. REFERENCES 1.
THOMAS, G. B., JR., C'alculus and Analytical Geometry, Addison-Wesley Publishing
Co., Reading, Mass., 1953. COURANT; R., Differential and Integral Calculus, Vol. 1, Blackie, London, 1937. COURANT, R., Differential and Integral Calculus, Vol. 2, Interscience Publishers, New York, 1936. 4. HANCOCK, H., Theory of Maxima and Minima, Dover Publications, New York, 1960. 5. APOSTOL, T. M,, Mathematical Analysis, Addison-Wesley Publishing Co., Reading, Mass., 1957. 6. CRANDALL, S. H., Engineering Analysis, McGraw-Hill, New York, 1956. 7. HILDEBRAND, F. B., Methods of Applied Mathematics, Prentice-Hall, New York, 2. 3.
1952.
PROBLEMS 3—1.
Determine the relative extrema for
(a) (b) (c)
f(x) = f(x) = f(x) =
2x2 + —2x2 ax2 +
(e) (f) (g)
f(x) = f(x) =
(x
4x + 5 + 8x + 10 2bx + c
(d) f(x)=x3+2x2+x+10 f(x)=1x3+2x2+4x+15 —
4ax3
+ (x —
a)2
+ 4bx2 + cx + d
3—2. Expand cos x in a Taylor series about x = the upper and lower bounds on R3.
3—3.
n=
3.
Determine
Expand(1 + x)112inaTaylorseriesaboutx = Otakingn =
mine upper and lower bounds on R2. 3—4. Find df and d2f for
(a) f=x2+2x+5
(b) f=3x3+2x2+5x+6 (c) f=x2sinx (d)
0, taking
f=
cosywhcrey = x3
2.
Deter-
PROBLEMS
Let f =
3—5.
79
Show that
u(x)/v(x).
df =
(du
—
d2f =
f dv)
—
fd2v)
—
Let u1, u2, u3 be functions of x and f = f(u1, u2, u3). Determine df. Suppose f = u(x)w(y) where y = y(x). Determine expressions for df and d2f. Apply to 3—6. 3—7.
(a) u=x3—x (b) w=cosy (c) y=x2
Find the first two differentials for the following functions: + + + 5x1 — 4x2 + 6x1x2 + Consider f = uv, where
3—8.
(a) (b)
f= f' =
3—9.
u=
u(y1, .v2)
v
=
v(y1, Y2)
and = y2(x1, x2)
y1(x1, x2)
Yi
Show that
df =
u dv + v du
d(uv)
d2f =
ud2v
+ 2 du dv + vd2u
Note that the rule for forming the differential of a product is independent of whether the terms are functions of the independent variables (x1, x2) or of dependent variables. 3—10. Classify the stationary points for the following functions: (a) — 9x1 + 12x2 — 10 3xl + (b)
f f f f
3—11.
+ 6x1x2 6x1x2 + 2x1 + 6x1x2 + 34 — 3x1 = Consider Rayleigh's quotient, xTax
x is arbitrary. Since a is symmetrical, its characteristic vectors are linearly independent and we can express x as
x= (j = 1, 2,. . Show that
where Q3 (a)
.
,
n)
are the normalized characteristic vectors for a.
= j=j-
CHAP. 3
RELATIVE EXTREMA FOR A FUNCTION
80
(b) (c)
Suppose x differs only slightly from Qk. Then, ICjI Specialize (a) for this case. Hint: Factor out 2k and Use (b) to obtain an improved estimate for A.
<<
for j
k.
[3
a=[i x
The exact result is
2=1
{1, —3}
x={1,—2}
3—12. Using Lagrange multipliers, determine the stationary values for the following constrained functions: (a) g
+
x2
0
(b)
g1 =
x1 + x2 + X3
—
1
=
0
g2=x1—x2+2x3+2=0
3—13. Consider the problem of finding the stationary values of f = = xrarx subject to the constraint condition, = 1. Using (3—36) we write
H =f+ Ag = (a)
—
2(XTX
—1)
Show that the equations defining the stationary points off are
ax=Ax
xTx=1
Relate this problem to the characteristic value problem for a symmetrical matrix. 3—14. Supposef = a. Show that and g = I — xTax = 0 where aT the Euler equations for H have the form (b)
xTax=1 We see that the Lagrange multipliers are the reciprocals of the characteristic values of a. How are the multipliers related to the stationary values of f?
4
Differential Geometry of a Member Element The geometry of a member element is defined once the curve corresponding to the reference axis and the properties of the normal cross section (such as area, moments of inertia, etc.) are specified. In this chapter, we first discuss the. differential geometry of a space curve in considerable detail and then extend the results to a member element. Our primary objective is to introduce the
concept of a local reference frame for a member. 4—1.
PARAMETRIC REPRESENTATION OF A SPACE CURVE
A curve is defined as the locus of points whose position vector* is a function of a single parameter. We take an orthogonal cartesian reference frame having and X3 (see Fig. 4—1). Let F he the position Vector to a point directions X1, x3
X3(y)
i3
x2
X2(y) x1
Fig. 4—1. Cartesian reference frame with position vector ?(y). * The vector directed from the origin of a fixed reference frame to a point is called the position vector. A knowledge of vectors is assumed. For a review, see Ref. 1.
81
DIFFERENTiAL GEOMETRY OF A MEMBER ELEMENT
82
on the curve having coordinates
CHAP. 4
(j = 1, 2, 3) and let y be the parameter. We
can represent the curve by
=
3
an alternate representation is
Since F = i—i
=
(j =
1,2, 3)
(4—2)
Both forms are called the parametric representation of a space curve. Example
4—i
Consider a circle in the X1-X2 plane (Fig. E4—1A). We take y as the polar angle and The coordinates are
(1)
let a =
x1 = x2 = F =
and
a a
cos y sin y
+ asiny'12
(2) Consider the curve (Fig. E4—1B) defined by
=
a
cos y
x2 = bsiny X3 =
(4—3)
CY
where a, b, care constants. The projection on the X, -X2 plane is an ellipse having semiaxes a and b. The position vector for this curve has the form F
4—2.
= a cos
+ b sin Y12 + CYI3
ARC LENGTH
Figure 4—2 shows two neighboring points, P and Q, corresponding to y and
and The cartesian coordinates are length of the chord from P to Q is given by
y+
As Exy —* 0,
the chord length
+
(j
approaches the arc length,
1,
2, 3) and the
In the limit,
ds2
=
Noting that
dx1 =
dy
we can express ds as ds
+
+
+ dy
(4-4)
Fig. E4—1A
Fig. E4—1B
Q(+Ay)
I LISI
P(y) Fig. 4—2. Differentia' segment of a curve. 83
DIFFERENTIAL GEOMETRY OF A MEMBER ELEMENT
84
CHAP. 4
Finally, integrating (4—4) leads to 2
dx
s(y)
2
dx
dx
+
=
21/2 dy
(4—5)
+
We have defined ds such that s increases with increasing y. It is customary to call the sense of increasing s the positive sense of the curve. To simplify the expressions, we let dx
=
2
1/2
(4-6)
+ Then, the previous equations reduce to
ds =
dy (4—7)
One can visualize as a scale factor which converts dy into ds. Note that +1. x > 0. Also, if we take y = s, then Example
4—2
Consider the curve defined by (4—3). Using (4—6), the scale factor is
[a2 sin2 y +
cos2 y + c2]'12
a. One can always orient the axes such that this condition is satisfied.
We suppose that b
Then, we express
b2
as
= (b2 + c2)"2 [1 —
k2
sin2 y]"2
where
k2 = b2 — b2 +
2
c2
The arc length is given by s
dy =
(b2
+
[1 — k2 sin2 yJ112 dv
The integral for s is called an elliptic integral of the second kind and denoted by E(k, y). Then,
(b2 +
s
c2)'12
E(k, y)
Tables for E(k, y) as a function of k and y are contained in Ref. 3. When b = is called a circular helix and the relations reduce to
=
(a2
S = ny
+
const.
a,
the curve
SEC. 4—3. 4—3.
UNIT TANGENT VECTOR
85
UNIT TANGENT VECTOR
We consider again the neighboring points, P(y) and Q(y + shown in Figure 4—3. The corresponding position vectors are P(y + ky), and
approaches the tangent to the curve at P. Then, the unit tangent vector at P is given by* As L\y -+ 0,
-
t=
.
Jim
PQ d1 --=—ds
(4—8)
•
Using the chain rule, we can express I as — dP
— —
ds
dy
—
1
dy ds —
—
dF
(4—9)
dy
Since > 0, 1 always points in the positive direction of the curve, that is, in the direction of increasing s (or y). It follows that dP/dy is also a tangent vector and
— dy
/df dP\"2 \dy dy Equation (4—10) reduces to (4—6) when coordinates. +s
is expressed in
Q(y+6.y)
+
r(y) Fig. 4—3. Unit tangent vector at P(y).
* See Ref. 1, p. 401.
(4—10)
cartesian
DIFFERENTIAL GEOMETRY OF A MEMBER ELEMENT
86
CHAP. 4
Example 4—3 We determine the tangent vector for the curve defined by (4—3). The position vector is F
= a cos
+ b sin Y12 + cyi3
Differentiating P with respect to y, dP
dy
=
+ h cos Y12 +
—a sin
and using (4—9) and (4—10), we obtain
a = +[a2 sin2 y + b2 cos2 y + c2]"2
=1[—asinyT1 + bcosyi2 + c13] =' coast, and the angle between the t?ngent and the X3 When a b, a [a2 + direction is constant. A space curve having the property that the angle between the tangent and a fixed direction (X3 direction for this example) is constant is called a helix.*
4-4. PRINCIPAL NORMAL AND BINORMAL VECTORS Differentiating
= 1 with respect to y, we have -
dy
=
0
It follows from (a) that di/dy is orthogonal to f. The unit vector pointing in the direction of di/dy is called the principal normal vector and is usually denoted by ii. H=
ldt dy
where
d (1
(4..-lt) dF
The binormal vector, h, is defined by
b='?xh We see that b is also a unit vector and the three vectors.
(4-12) ñ,
b comprise a right-
handed mutually orthogonal system of unit vectors at a point on the curve. Note that the vectors are uniquely defined once y) is specified. The frame associated with b_and ii is called the moving trihedron and the planes determined by (1, ñ), (ii, b) and are referred to as the osculating normal, and rectifying planes (see Fig. 4—4). * See
Ref. 4, Chap.
1.
PRiNCIPAL NORMAL AND BINORMAL VECTORS
SEC. 4—4.
87
Normal plane
Rectifying
plane
Fig. 4—4. Definition of local planes.
Example 4—4 We determine fi and b for the circular helix. We have already found that
a — [a2 + and
=
sin VT1
+ a cos
+ c13]
Differentiating t with respect to y, we obtain
di
a — [cos ytj + Sm
—
Then, i dt fl
dt dy
— C05
— Sm
dy
The principal normal vector is parallel to the plane and points in the inward radial direction. It follows that the rectifying plane is orthogonal to the X1-X2 plane. We can determine b using the expansion for the vector product.
a
—asiny acosy
C
This reduces to
b
C. a
sin
The unit vectors are shown in Fig. E4—4.
C
— — cos
a
+
a £3
a
GEOMETRY OF A MEMBER ELEMENT
88
CHAP. 4 Fig, E4—4
4—5.
CURVATURE, TORSION, AND THE FRENET EQUATIONS
The derivative of the tangent vector with respect to arc length is called the curvature vector, K. K
dt
d2F
c/s
i/s2
ic/i
K
(4—13)
c/s2
Using (4—11), we can write —
ds
— Ku
(4—14)
0. The Note that K points in the same direction as Ft since we have taken K curvature has the dimension L1 and is a measure of the variation of the tangent vector with arc length. We let R be the reciprocal of the curvature:
R=
K1
iS)
In the case of a plane curve, R is the radius of the circle passing through three consecutive points* on the curve, and K = JdO/dsj where 6 is the angle between I and To show this, we express I in terms of 0 and then differentiate with respect to s. From Fig. 4—5, we have cos * See
+ Sifl 012
Ref. 4, p. 14, for a discussion of the terminokgy 'three consecutive points."
SEC. 4—5.
CURVATURE, TORSION, AND THE FRENET EQUATIONS
Then —
K
.
[—sin
+ cos 617]
dO a—
and
K
dO
1
ds
R
dO/ds
+ cos 612]
[— sin
In the case of a space curve, the tangents at two consecutive points, say P and Q, are in the osculating plane at F, that is, the plane determined by and ñ at. P. We can interpret R as the radius of the osculating circle at P. It should be noted
that the osculating plane will generally vary along the curve. x2
\ R
+ R
t i2
it
Fig. 4—S. Radius of curvature for a plane curve.
The binormal vector is normal to both and ñ and therefore is normal to the osculating plane. A measure of the variation of the osculating plane is given by db/ds. Since his a unit vector, db/ds is orthogonal to h. To determine whether db/ds involves we differentiate the orthogonality condition I b 0, with respect to s. -
db
-
ds
dl ds
But dl/ds Kñ and b ii = 0. Then, db/ds is also orthogonal to I and involves only ñ. We express db/ds as
db
= —tn
where r is called the torsion and has the dimension, L
(4—16)
DiFFERENTIAL GEOMETRY OF A MEMBER ELEMENT
90
CHAP. 4
It remains to develop an expression for a. Now, h is defined by
xn Differentiating with respect to s, we have db = di
This reduces to
db
diii
—=t x
ii =
0.
dñ
xn+t
Finally, using (4—16), the torsion is given by
l-dfl —
dñ
—b
——
ds
dy
(4—17)
Note that a can be positive or negative whereas K is always positive, according to our definition. The torsion is zero for a plane curve since the osculating plane coincides with the plane of the curve and b is constant.
Example
4—5
The unit vectors for a circular helix are
=
[—a
sin vij +
—cos
sin
—
b=
cos Yti + cT3]
a
yl' — ccosyi3 + at3]
where
a = (a2 +
c2)112
Then, a
K=-— adv
a
a +c
a
and 1—
a
dñ
c
dy
a
c 2
2—const
We have developed expressions for the rate of change of the tangent and binormal vectors. To complete the discussion, we consider the rate of change of the principal normal vector with respect to arc length. Since fi is a unit vector, dñ/ds is orthogonal to ñ. From (4—17), -
dñ
b— ds
a
SEC. 4—6.
GEOMETRICAL RELATIONS FOR A SPACE CURVE
91
To determine the component of dñ/ds in the I direction, we differentiate the
orthogonality relation, I n = 0. ds
(b)
ds
it follows from (a) and (b) that dñ
-
— = —I(t + tb us
(4—18)
The differentiation formulas for 1, ii, and b are called the Frenet equations. 4—6.
We
SUMMARY OF THE GEOMETRICAL RELATIONS FOR A SPACE CURVE
summarize the geometrical relations for a space curve: Orthogonal Unit Vectors t =
a=
ldi
di
= — = tangent vector exdy
thu
di
1
principal normal vector
—i-- i—
(4—19)
= I x ñ binormal vector di ds — dy
dy
Equations)
Di:fferenriation Formulas
lull
dl
— -— = Kn ds ady db 1db — = —— = —rn ds ctdy dñ -. 1 dñ —Kt ds ady
K= =
+tb
(4—20)
1 di — curvature a dy 1—dñ
—b
a
— = torsion dy
We use the orthogonal unit vectors (I, ñ, b) to define the local reference frame for a member element. This is discussed in the following sections. The Frenet
92
DIFFERENTIAL GEOMETRY OF A MEMBER ELEMENT
CHAP. 4
equations are utilized to establish the governing differential equations for a
member element.
4—7.
LOCAL REFERENCE FRAME FOR A MEMBER ELEMENT
The reference frame associated with ñ, and b at a point, say P, on a curve is uniquely defined once the curve is specified, that is, it is a property of the
curve. We refer to this frame as the natural frame at P. The components b) are actually the direction cosines for the natural of the unit vectors frame with respect to the basic cartesian frame which is defined by the orthogonal unit vectors (1k, 12, 13). We write the relations between the unit vectors as ft n
£12
=
133
11
t22 £32
(4—21)
12
e33
One can express* the direction cosines in terms of derivatives of the cartesian
coordinates (x1, x2, x3) by expanding (4—19). Since (1, b) are mutually or13) the direction cosines are related by thogonal unit vectors (as well as 1jm6m
=
j,
k
=
1,
2, 3
(4—22)
Equation (4—22) leads to the important result [ljk]T =
(4—23)
and we see that is an orthogonal matrix.f The results presented above arc applicable to an arbitrary continuous curve. Now, we consider the curve to be the reference axis for a member clement and take the positive tangent direction and two orthogonal directions in the normal plane as the directions for the local member frame. We denote the directions of the local frame by (Y1, Y2, 1'3) and the corresponding unit vectors by (t1,
2,
= 1) We will always take the positive tangent direction as the Y1 direction x and we work only with right handed systems t3). This notation is shown in Fig. 4-6. When the centroid of the normal cross-section coincides with the origin of the local frame (point P in Fig. 4—6) at every point, the reference axis is called the centroidal axis for the member. It is convenient, in this case, to take Y2, Y3 as the principal inertia directions for the cross section. In general, we can specify the orientation of the local frame with respect to the natural frame in terms of the angle between the principal normal direction and the I'2 direction. The unit vectors defining the local and natural frames * See
Prob. 4-5.
t See Prob. 4—6.
SEC. 4—7
are
LOCAL REFERENCE FRAME FOR A MEMBER ELEMENT
related by
93
-
tl — t2
=
COS 4)11
+ sin 4)b
(4—24)
çbn + cos 4th
Combining (4—21) and (4—24) and denoting the product of the two direction cosine matrices by the relation between the unit vectors for the local and basic
frames takes the concise form
t=
(4—25)
where
[
€12
€j3
£21cos4)+€31sin4)
€22cos4)+ €32sin4)
€23cosçb+ €33sinqS
[21sin4)+€31cos4)
—€22si+C32cos4)
—€23 sin 41+ £33cos41
Note that the elements of fi are the direction cosines for the local frame with respect to the basic frame. fJjk =
(4—26)
Xk)
'. We will utilize (4—25) in the next Since both frames are orthogonal, J1 chapter to establish the transformation law for the components of a vector. x3 Normat
Y1
Fig. 4—6. Definition of local reference frame for the normal cross section.
CHAP. 4
DIFFERENTIAL GEOMETRY OF A MEMBER ELEMENT
94
Example 4—6 We determine
for the circular helix. The natural frame is related to the basic frame.by
a.
a
c
a
a
a
sin y
0
c
a
—cosy —
——slay
I
=
— cos
y
—
C. —slay
b
—--cosy a
a
12
=
{Ik}
—
a
Using (4-.25)
a cos y a
a
—sinycosçb ——cosysin4
a
C.
+cosysm4 + —sinycos4
4—8.
a
sin y sin
— — cos
y cos
a
a
I. a
a
CURVILINEAR COORDINATES FOR A MEMBER ELEMENT
We take as curvilinear coordinates (yi, Yz' y3) for a point, say Q, the parameter
of the reference axis and the coordinates (Y2, of Q with respect to the orthogonal directions (Y2, 1'3) in the normal cross section (see Fig. 4—7). Let F(y1) the R(y1, Y2' Y3) be the position vector for Q(Yl, y3) and position vector for the reference axis. They are related by = r + Y2t2 + y3t3 where COS
+ cos4b
=
t3 =
+ Sifl (
We consider 4 to be a function of y1. Y2
y3
—
——
Y2
Fig. 4—7. Curvilinear coordinates for the cross section.
(4—27)
SEC. 4—8.
CURVILft4EAR COORDiNATES FOR A MEMBER ELEMENT
95
The curve through point Q corresponding to increasing Yj with Y2 and y3 held constant is called the parametric curve (or line) for yj. In general, there are three parametric curves through a point. We define as the unit tangent vector for the parametric curve through Q. By definition,
=
Ui
13R (4—28) aIR
= The differential arc length along the
curve is related to
aIR
by (4—29)
=
=
This notation is illustrated in Fig. 4—8. One can consider the vectors (or
to define a local reference frame at Q. x3
y2t2 +y3t3
x2
Fig. 4—8. Vectors defining the curvilinear directions.
Operating on (4—27), the partial derivatives of R are 0R
aR aR
— dy1 =
t2
=
t3
+
dt2 Y2
dy1
+ Y3
dy1
96
DIFFERENTIAL GEOMETRY OF A MEMBER ELEMENT
CHAP. 4
We see that
t2
g2
ü3=t3
1
(4—30)
g3=1
It remains to determine ü1 and g1. Now, dy1
=
=
Also, d12
(dñ
dy1
\dy1
dt3
=
+ dy1 —bj +
(dii
.
dçb
\dy1
.
j
1db '\dv1
dq5\
—
1db
+ b—)+ dy1j
\dy1
dy1 —11—— dy1
We use the Frenet equations to expand the derivatives of ñ and h. Then, cit2
dyj
dy1
I and finally, = cc(1
—
Ky'2)!1
Y2 COS 4)
+
/
+
d4)'\.
d4)\ _)(Y2t3
y3t2)
(4—31)
J73 sin 4)
We see from Fig. 4—9 that y'2 is the coordinate of the point with respect to the
principal normal direction. y3
\y3
Fig. 4—9.
of y.
Since 13R/ay1 (and therefore ii1) involve the reference frame defined and by iii, u2, will not be orthogonal. However, we can reduce it to an orthogonal
REFERENCES
system
97
by taking dy
=
(4—32)
which requires cer dy
=
(4—33)
150
When (4—32) is satisfied,
aR
=
— Ky'2)t1
and
= =
(4—34) cx(l —
In this case, the local frame at Q coincides with the frame at the centroid. One
should note that this simplification is practical only when ccc can be readily integrated.
Example
4—7
The parameters a and t are constant for a circular helix:
a=
(a2
+ c2)112
C
Then, C
at = — a
and integrating (4—33), we obtain tS
— Yo)
For this curve,
varies linearly with y (or arc length). The parameter g1 follows from (4—34).
hi =
ds1
= a(1
—
Ky2)
/
a
'\
cc-
x(l —
REFERENCES
3.
THOMAS, G. B., JR.: Analytical Geometry and Calculus, Addison-Wesley Publishing Co., Inc., Reading, Mass., 1953. HAY, U. 13.: Vector and Tensor Analysis, Dover Publications, New York, 1953. JA}INKE, E., and F. EMDE: Tables of Functions, Dover Publications, New York, 1943.
4.
STRUm, D. J.: Differential Geometry, Addision-Wesley Publishing Co., Reading,
1.
2.
Mass., 1950.
DIFFERENTIAL GEOMETRY OF A MEMBER ELEMENT
CHAP. 4
PROBLEMS 4—1.
(a) (b) (c) (d)
Il, b, ; K, x for the following curves: x2 = 3 sin y cos y x3 = 5y x1 3 cos y x2 = 6 sin y x3 = 5y + p313 + = x1 = cos y x2 = sin y
Determine
x1 =
3
x3 = cy where a, /3, c are real constants. 4—2. If 0, the curve lies in the plane. Then,r Oandb ±i3. The sign of b will depend on the relative orientation of ñ with respect to 1.
Suppose the equation defining the curve is expressed in the form
x3=0
x2=J(x1)
Equation (a) corresponds to taking x1 as the parameter for the curve. (a) Determine the expressions for 7, ñ, b, and K corresponding to this representation. Note that Let y and + f(x1)12 +
_Lf' (b)
.f"etc
Apply the results of (a) to
=
4a —
2 x1)
a and b are constants. This is the equation for a parabola symmetrical about x1 = b/2. Let 9 be the angle between and
where (c)
cos0 = I
•11.
= sec 0. Express t, h, !, and K in terms of 0. for the case where 02 is negligible with respect to unity. This approximation leads to Deduce that
(d)
Specialize
sin 0 cos 0 4—3.
tan 0
0
1
A curve is said to be shallow when 02 << 1. Show that (see (4—20)) Let K = l/R and t
di dh
— = —— 12 dy.
dñ
PROBLEMS 4—4.
99
The equations for an ellipse can be written as
x1 =
a
x2 =
cos y
b
sin y
or x22
x12
Determine 1, n for both parametric representations. Take x1 as the parameter for (b). Does y have any geometrical significance? 4—5. Show that dx,.
1k
C
dy
C2k
=
A
dy dyj j
1€21
0
°
€33j 4—6.
\ dy
42 €13
IA
dv2
[3 (d€ [,.=,. \ dyj J
€311 €32
72
Let
43]
42
Then, €1
[4k] €3
Using (4—22), show that
Ii' iT
—
i—i
Determine D for Prob. 4—la. Determine JI for Prob. 4—lb. 4-9. Specialize for the case where the reference axis is in the X1 — X2 plane. Note that b = €3313 where 431 = 1. When the reference axis is a plane curve and = 0, we call the member a "planar" member. as 4—10. We express the differentiation formulas for 4—7.
4—8.
dt — = at ds (a) (b)
Show that a is, in general, skewsymmetric for an orthogonal system of unit vectors, ie., c5Jk. Determine a. Suppose the reference axis is a plane curve but çb 0. The member is not planar in this case. Determine a.
5
Matrix Transformations for a Member Element 5—1.
ROTATION TRANSFORMATION
Suppose we know the scalar components of a vector with respect to a reference frame and we want to determine the components of the vector corresponding to a second reference frame. We can visualize the determination of the second set of components from the point of view of applying a transformation to the column matrix of initial components. We refer to this transformation as a rotation transformation. Also, we call the matrix which defines the transformation a rotation matrix. (j = 1, 2, 3 and n = 1, 2) be the directions and corresponding unit Let vectors for reference frame n. (See Fig. 5—1.) We will generally use a superscript to indicate the reference frame for directions, unit vectors, and scalar
a
Fig. 5—1. Directions for reference frames "1" and "2." 100
SEC. 5—1.
ROTATION TRANSFORMATION
components in this text. We consider a vector, a. The scalar components of
a with respect to frame n are
a
a is independent of the reference frame. Then,
a=
=
(a2)Ti2
To proceed further, we must relate the two reference frames. We write, the relations between the unit vectors as
i2 = Ili' where is the scalar component of with respect to The transformation matrix, is nonsingular when the unit vectors are linearly independent. Substituting for and equating the coefficients of i' leads to
a' = a2 Finally,
we let
R'2 R21 =
(53)
With this notation, the relations between the component matrices take the form
a2 = a1 = =
R21a2
The order of the superscripts on R corresponds to the direction of the trans-
formation. For example, R'2 is the rotation transformation matrix corresponding to a change from frame 1 to frame 2. We see that the transformation matrix for the scalar components of a vector is the inverse transpose of the transformation matrix governing the unit vectors for the reference frames.
Example 5—i We consider the two-dimensional case shown in Fig. E5-1. The relations between the unit vectors are = cos + sin = —sin + cos
We write (a) according to (5—2).
cos6 sinO
102
MATRIX TRANSFORMATIONS FOR A MEMBER ELEMENT
CHAP. 5
Then,
—sin4
R21
cos4
Lsinfl
R'2 = (fV)
1
= L—sjn 6
0 cos
+
cos U
o sin
and
Jafl
—
I
[
sin
cos
— ?J L — sin 0
cos oJ
When both frames are orthogonal, q5 = 0 and
ir Fig. E5—1
12
The result obtained in the preceding example can be readily extended to the
orthogonal reference frames. When both frames are orthogonal, the change in reference frames can be visualized as a rigid body rotation of one frame into the other, f3jk is the direction cosine for with respect to and the rotation transformation matrix is an orthogonal matrix: case of two
1 R'2 — — LI'jki
Pjk — COSt'X2 j,
A'k
(55)
In Sec. 4—7, we defined the orientation of the local frame (1k, 13, at a point on the reference axis of a member element with respect to the natural frame (1, ñ, b) at the point. This frame, in turn, was defined with respect to a fixed cartesian frame 12, 13). In order to distinguish between the three frames, we use superscripts p and p' for the local and natural frames at p and a superscript 1 for the basic cartesian frame: = = {Z, =
t2,
ii, b}
(5—6)
THREE-DIMENSIONAL FORCE TRANSFORMATIONS
SEC. 5—2.
103
With this notation, the relations between the unit vectors and the various
rotation matrices are: = R141
t"
= 1.
From(4—21),
2.
From (4—24),
(5—7)
—
1
=0 0
0
0
cos4; sin 4; —sin4; cos4;
3.
defined by (4—25). 5—2.
THREE-DIMENSIONAL FORCE TRANSFORMATIONS
The equilibrium analysis of a member element involves the determination of the internal force and moment vectors at a cross section due to external forces and moments acting on the member. We shall refer to both forces and moments as "forces." Also, we speak of the force and moment at a point, say P, as the "force system" at P. The relationship between the external force
system at P and the statically equivalent internal force system at Q hasa simple form when vector notation is used. Consider a force F and moment M acting at P shown in Fig. 5—2. The statically equivalent force and moment at Q are Feqrnv.
Mequiv = M +
--
XF
(5-8)
One can visualize (5—8) as a force transformation in which the force system
at P is transformed into the force system at Q. This transformation will be
Fcquiv.
Q
5—2. Equivalent force system.
linear if
is constant, that is, if the geometry of the element does not change
appreciably when the external loads are applied. We will write (5—8) in matrix form and treat force transformations as matrix transformations.
104
MATRIX TRANSFORMATIONS FOR A MEMBER ELEMENT
CHAP. 5
We develop first the matrix transformation associated with the moment of a force about a point. Let be a force vector acting at point P and MQ the moment vector at point Q corresponding to We will always indicate the point of application of a force or moment vector with a subscript. The relation
between MQ and
is
—
= QP x (5—9) We work with an orthogonal reference frame (frame 1) shown in Fig. 5—3 and write the component expansions as — v' ri PL..
1
._
'1
P
(5—10)
(jl)TM1
=
The scalar components of QP are 4,
Expanding the vector cross
—
product leads to =
(5—11)
0
(1
—
1\
(1.
(1
1
—
—(42
XQ2)
0
—
Note that matrix. One can interpret it as a forceis a into translation transformation matrix. The force at P is transformed by
F1P3
/
,1
'Ik
M1
i—-i
43
I' /
'i/
/ xP1
Xp3
Fig. 5—3. Notation for orthogonal reference frame.
— 2
SEC. 5—2.
THREE-DIMENSIONAL FORCE TRANSFORMATIONS
105
moment at Q. Note that the order of the subscripts for the translation transcorresponds to the order of the translation (from P formation matrix, and must be referred to the same frame, that is, the superto Q). Also, scripts must be equal. Up to this point, we have considered only one orthogonal reference frame. In general, there will be a local orthogonal reference frame associated with each point on the axis of the member, and these frames will coincide only when the member is prismatic. To handle the general case we must introduce rotation transformations which transform the components of F and M from the local frames to the basic frame (frame 1) and vice versa. We use a superscript p to indicate the local frame at point P and the rotation matrix corresponding to a transformation from the local frame at P to frame 1 is denoted by R'1. With this notation, = a
(5—12)
= and the general expression for
takes the form
=
(5—13)
We consider next the total force transformation. The statically equivalent force and moment at Q associated with a force and moment at P are given by —
+ QP x
MQ =
(a)
When all the vectors are referred to a common frame, say frame 1, the matrix transformation is 'Q)
We let
—
I
L3J
L"PQ
(b)
(5—14)
=
is called the force system at Q referred to frame 1. Using The 6 x 1 matrix this notation, (b) simplifies to d1 — When the force systems are referred to local frames, we mast first transform them to a common frame and then apply (5--15). Utilizing the general
matrix, I
and applying.
0
1
_
(5—16)
=
=
(a)
MATRIX TRANSFORMATIONS FOR A MEMBER ELEMENT
106
we
obtain
CHAP. 5
=
(5—17)
is applied Equation (5—17) states that when the matrix transformation we obtain its statical equivalent at Q. Actually, we could leave off the to appears subscripts and superscripts on when we write (5—17). However,
alone, we must include them. Note that the force transformation generally involves both translation and rotation. The order of the subscripts corresponds to the direction of the translation, e.g., from P to Q. Similarly the order of the superscripts defines the direction of the rotation or change in reference frames, e.g., from frame p to frame q. In general, the geometry of a member element is defined with respect to a basic we must determine reference frame which we take as frame 1. To evaluate from the geometrical relations for the member. We have and Rn', already discussed how one determines R1 in Secs. 4—7 and 5—f.
When the member is planar* and the geometry is fairly simple (such as a straight or circular member), we can take frame I parallel to one of the local frames. This eliminates one rotation transformation. For example, suppose we reduces to take frame I parallel to frame p. Then, R1 = 1 and —
[R
=
(S—18)
Similarly, if 1 and q are parallel,
-
-
I
—
When both p and q arc parallel to 1,
reduces
(5-19)
to
= 91i'PQ By transforming from P to Q and back to F, we obtain =
(5—20)
and it follows that (5—21)
is carried out in the order P S1, S1 are intermediate points, the transformais equal to the product of the intermediate transformation
If the transformation from P to
S2,...,
—÷
tion matrix,
Q, where
S2,
.
.
, S,,
matrices. —
* If the reference axis isa plane curve and the local frame coincides with the natural frame we say the member is planar.
(5—22)
=
0)
THREE-DIMENSIONAL FORCE TRANSFORMATIONS
SEC. 5—2.
where s1, s2, ..
.
, SN
107
are arbitrary reference frames. It is convenient to take a
common reference frame for the intermediate transformations.
Example
5—2
We consider the plane circular member shown in Fig. ES—2. We take frame 1 parallel to frame p. Then,
4
—
=
=
= {a sin 0, —a(1 — =
cos 0), O}
0
0
—a(1—cos6)
0
a(1 — cos 0)
0 a sin 0
0
cos6
0
= sin 0 0
—sinO cos 0
—a
Sin 6
0
0
The transformation matrix has the form — —
—
9 I
—
where 0
0
=0
0
+a(1 —cos6) —asinO
asin0
0
a(1 — cos 0)
Fig. E5—2
P p 2
t7
Example 5—3
As an illustration of the case where the geometry is defined with respect to a basic for a circular helix. The general
Cartesian frame, we consider the problem of finding
MATRIX TRANSFORMATIONS FOR A MEMBER ELEMENT
108
CHAP. 5
has the form
expansion for
[RPS
The parametric representation for a circular
= =
is given in Sec. 4—i:
cos y sin y
a a
= cy
(j = 1,2,3) are the cartesian coordinates with respect to the basic frame (frame 1). Let yp and y corresponding to points P and Q. The coordinate matrices for P and Q are
where
4 = {a cos Vp, a Sifl
Cyp)
= {a cos yQ, a sin YQ, cy0} Then, 0
=
C(yp
a(sin y,, — sin
— C(yp —
—a(sin
—
—a(cos
0
y0) sin yQ)
a(cos Yp
YQ)
,, — cos
0
C05 Y0)
To simplify the algebra, we suppose the local frame coincides with the natural frame at every point along the reference axis, that is, we take = 0. Using the results of Sec. 4—7, the rotation matrices reduce to a
a a
C
C. a
C
cc
cc
0
a2 + c2. Evaluating the product,
where cc2
(a\2
we obtain
/c\2
ac
a ——sin ij
I—I
cc
—i (1 — cos I
cc
I
c cc
/a\2
/c'\2 \ccJ
SEC. 5—3. THREE-DIMENSIONAL DISPLACEMENT TRANSFORMATIONS where
y0. Also,
= yp =
a2c —3-(2
ac —
a
— sinq)
I
a
—
cosn)(a5
—
c2)]
a
(IC
.
I
—
a
I
ac2
a —
—
c2)
a2
ci2c
a
a
—
a
a
a
a
a
a
Note that we can specialize the above general results for the case of a plane circular member (Example 5-2) by taking c = 0 and = 0.
5—3.
THREE-DIMENSIONAL DISPLACEMENT TRANSFORMATIONS
Let P and Q be two points on a rigid body. Suppose that the body experiences a translation and a rotation. We define Up and Ui,, as the translation and rotation* vectors for point P. The corresponding vectors for point Q are given by
UQ = Up + (Up X
(5—23)
= Wp
Equation (5—23) is valid only when Since PQ — QP and x PQ
is —
PQ x
negligible with respect to unity. an alternate form for is
UQ = Up + QP X
(5—24)
We define (5—25)
= as the displacement matrix for P referred to frame 1. The displaccment at Q resulting from the rigid body displacement at P is given by —
[13
XPQ1
L"l 3J
—
i
consider next the case where the local frames at P and Q do not coincide. The general relation between the displacements has the form
= 0
13
-
(5—27)
—
j
[o One can showt that alternate forms of (5—27) are
= The units
t See Prob. 5--7.
are
radians.
T
-
(yqp)T
d14
MATRIX TRANSFORMATIONS FOR A MEMBER ELEMENT
110
CHAP. 5
We see that the displacement transformation matrix is the inverse transpose of
the corresponding force transformation matrix. This result is quite useful. REFERENCES 1.
HALL, A. S., and R. W. WOODS-LEAD: Frame Analysis, 2d ed., Wiley, New York, 1967.
2. 3.
MORLCE, P. B.: Linear Structural Analysis, Ronald Press. New York, 1959. PESTEL, E., and LECKIE, F.: Matrix Methods in Elastomechanics, McGraw-Hill, New York, 1963. Livasr.Ey, R. K.: Matrix Methods of Structural Analysis, Pergamon Press, London,
4.
1964.
MARTLS.t, H. C.: Introduction to Matrix Methods of Structural Analysis, McGraw-Hill, New York, 1966.
5.
PROBLEMS Consider the two-dimensional {50, —100}, find a2.
5—1.
a1 =
cartesian reference frames shown. If Prob.
5—i
5—2. The orientation of two orthogonal frames is specified by the direction cosine table listed below.
1/2 1/2
1/2 1/2
\/2/2 (a)
Determine R12. Verify that (R12)T = (R12y1
o
PROBLEMS
(b) If a1 = {10, 5, 10}, find a2. (c) If a2 {5, 10, 10), find a'. 5—3. Consider two points, P and Q, having coordinates (6, 3,2) and (—5, 1,4)
with respect to frame 1. The direction cosine tables for the local reference frames are listed below. —,
—1
1/2
1/2
1/2
1/2
0
0
(a) (b) (c)
5--4.
—
1/2
1/2
1/2
1/2
12/2
Determine 91'),Q and
Determine Suppose
{l00, —50, 100, 20, —40, +60). Calculate Consider the planar member consisting of a circular segment and a
straight segment shown in the sketch below. Point P is at the center of the circle. Prob,. 5—4 S
t'i Q
Ic
(a) (b) (c) 5—5.
Determine by transforming directly from P to Q. Also find Determine by transforming from P to S and then from S to Q. Find corresponding to = {0, 0, 1, 0, 0, 0}. Consider the circular helix,
= (a)
Suppose 4(y) =
(b)
Suppose
=
2
0.
cos
+ 2 sin
Determine Determine
—y.
+ Take
13.
= ir/2, YQ =
-
MATRIX TRANSFORMATIONS FOR A MEMBER ELEMENT
112
5—6. —
= {1/2,
corresponding to
Refer to Problem 5—3. Determine 1/10, 0}. Verify that
1/4, 1/3, — 1/10,
Q_
Q
5—7.
rI IA
5—8.
P
P
Verify that (5—27) and (5—28) are equivalent forms. Note that •1
x2
CHAP. 5
ii
r
— I _T___
I
niT I
—c
ort
I
3 3J L" 1 Consider the plane member shown. The reference axis is defined by
= f(xj). Prob. 5—8 x1
(a)
(b)
Determine Note that the local frame at P coincides with the basic frame whereas the local frame at Q coincides with the natural frame at Q. Specialize part (a) for the case where —
and the x1 Prob. 4—2.
4a
coordinate of point Q
2 (x1b — Xj)
is
equal to h/4. Use the results of
Cl)
C
z
XCI)
6
Governing Equations for an Ideal Truss 6—i.
GENERAL
A system of bars* connected at their ends by frictionless hinges to joints and subjected only to forces applied at the joint centers is called an ideal truss.t The bars arc assumed to be weightless and so assembled that the line connecting the joint centers at the ends of each bar coincides with the centroidal
axis. Since the bars are weightless and the hinges are frictionless, it follows that
each bar is in a state of direct stress. There is only one force unknown associated with each bar, namely, the magnitude of the axial force; the direction of the force coincides with the line connecting the joint If the bars lie in one plane, the system is called a plane or two-dimensional truss. There are two displacement components associated with each joint of a plane truss. Similarly, a general system is called a space or three-dimensional truss, and there are three displacement components associated with each joint. We suppose there are in bars (members) and j joints. We define i as
= =
2 3
for a plane truss for a space truss
Using this notation, there are if displacement quantities associated with the] joints. In general, some of the joint-displacement components are prescribed. Let r be the number of prescribed displacement components (displacement restraints) and nd the total number of unknown joint displacements. It follows that = if — r (6—2) Corresponding to each joint displacement restraint is an unknown joint force A prismatic member is conventionally referred to as a bar in truss analysis. ± See Ref. 1.
115
116
GOVERNING EQUATIONS FOR AN iDEAL TRUSS
CHAP. 6
(reaction). We let flf be the total number of force unknowns. Then,
n1—m+r
(6—3)
Finally, the total number of unknowns, n, for an ideal truss is
fl—flj+flj—(j+tfl
(6—4)
The equilibrium equations for the bars have been used to establish the fact that the force in each bar has the direction of the line connecting the joint centers at the ends of the bar. There remains the equilibrium equations for the joints. Since each joint is subjected to a concurrent force system, there are iJ scalar force-equilibrium equations relating the bar forces, external joint loads, and direction cosines for the lines connecting the joint centers in the deformed state. In order to solve the problem, that is, to determine the bar forces, reactions, and joint displacements, m additional independent equations are required. These additional equations are referred to as the bar force—joint displacement relations and are obtained by combining the bar force—bar elongation relation and bar elongation—joint displacement relation for each of the in bars. In this chapter, we first derive the elongation—joint displacement relation for a single bar and then express the complete set of in relations as a single matrix equation. Thisprocedure is repeated for the bar force-elongation relations and the joint force-equilibrium equations. We then describe a procedure for introducing the joint-displacement restraints and summarize the governing equations. Finally, we briefly discuss the solvability of the governing equations for the linear case. In this case, the question of initial instability is directly related to the solvability.
In Chapter 7, we develop variational principles for an ideal truss. The two general procedures for solving the governing equations are described in Chapters 8 and 9. We refer to these procedures as the displacement and force methods. They are also called the stWhess and flexibility methods in some texts.
The basic concepts employed in formulating and solving the governing equations for an ideal truss are applicable, with slight extension, to a member system having moment resisting connections. Some authors start with the general system and then specialize the equations for the case of an ideal truss.
We prefer to proceed from the truss to the general system since the basic formulation techniques for the ideal truss can be more readily described. To adequately describe the formulation for a general system requires introducing a considerable amount of notation which tends to overpower the reader. 6—2.
ELONGATION—JOINT DISPLACEMENT RELATION FOR A BAR
We number the joints consecutively from 1 through j. It is convenient to refer the coordinates of a joint, the joint-displacement components, and the external joint load components to a common right-handed cartesian reference frame. Let (j = 1, 2, 3) be the axes and corresponding orthogonal unit
ELONGATION—JOINT DISPLACEMENT RELATION
SEC. 6—2.
vectors for the basic frame. The initial coordinates, displacement components,
and components of the resultant external force for joint k are denoted by
(j =
1,
2, 3) and the corresponding vectors are written as
rk = j= Uk = Ukl
1
(6—5)
Pk
The coordinates and position vector for joint k in the deformed state are 1k
=
+
11k
(6—6)
+ Uk
Figure 6—1 illustrates the notation associated with the joints.
14313
Deformed position
of joint k
//
//
x2
flkl
11k2
x1
Fig. 6—1. Notation for joints.
We number the bars from 1 through m and consider bar n to be connected to joints k and s. The centroidal axis of bar n coincides with the line connecting joints k and s. From Fig. 6—2 the initial length of bar n, denoted by = — is equal to the magnitude of the vector
=
(6—7)
116
GOVERNING EQUATIONS FOR AN IDEAL TRUSS
CHAP. 6
Since the basic frame is orthogonal, (6—7) reduces to
=
— xk)T(xs — xk)
(6—8)
— xkf)
= Before the orientation of the bar can be specified, a positive sense or direction must be selected. We take the positive sense for bar n to be from joint k 13
s
x.,3
I
/ / XkI
//
12
/
Xk2
Xg3
Fig. 6—2. Undeformed position of Bar n.
to joint s and define as the direction cosine for the positive sense of bar n direction: in the undeformed state with respect to the
=
1
(Ar
=
1
— XkJ)
(6—9)
It is convenient to list the direction cosines in a row matrix,
=L
—
Xk)
(6—10)
1, due to the orthogonality of the reference frame. Finally, Note that we let be the unit vector associated with the positive direction of bar n in the undeformed state. By definition, -
=
1
Ar = 12
The deformed position of bar n is shown in Fig. 6—3. The length and direction cosines for bar n are equal to the magnitude and direction cosines for the vector, the unit vector = — Pk. Let L12 + e12 be the deformed length,
ELONGATION—JOINT DISPLACEMENT RELATION
SEC. 6—2.
x3
Joints
Ap
x2
t?&2
x1
Fig. 6—3. Deformed position of Bar n.
associated with the positive direction in the deformed state, and the corresponding direction cosine matrix. These quantities are defined by +
(6—12)
eh)
—.
1
—
(6—13)
—
—
(6—14)
We consider first (6—12). Substituting for = A1 +
— Uk)
and noting (6—7), (6—11), we obtain, after dividing both sides by
(1 +
e
2
=
1
+
2.
1
—
Uk) +
The expression for the direction cosine, 1
=
[
[cx,,j
I +-;:-
We list the fl's in a row matrix,
— uk)T(US
ilk)
(6—15)
expands to
+
1
E
UkJ)
(6—16)
120
GOVERNING EQUATIONS FOR AN IDEAL TRUSS
[cm
uk)T]
+
CHAP. 6 (6—17)
By definition, is the change in length of bar n. Then, e,,/L7, is the extensional strain which is considerably less than unity for most engineering materials. For example, the strain is only for steel at a stress level of 3 x ksi. The relations simplify if we introduce the assumption of small strain, << I
Expanding the left-hand side of (6—15), and noting (6—18), we obtain e,,
—
(ii,, — uk)T(u$
Uk) +
— Uk)
(6—19)
The direction cosines for the deformed orientation reduce to
+
(6—20)
— Uk)
To simplify the expression for further, we need to interpret the quadratic terms. Using (6—20), we can write (6—19) as
= This
/
+
1
(u,
—
Uk)
form shows that the second-order terms arc related to the change in
orientation of the bar. If the initial geometry is such that the bar cannot experience a significant change in orientation, then we can neglect the nonlinear terms. We use the term linear geometry for this case. The linearized relations are — Uk)
(6—21)
We discuss this reduction in greater detail in Chapter 8. Since we are concerned in this chapter with the formulation of the governing equations, we will retain the nonlinear rotation terms. However, we will assume small strain, i.e., we work with (6—19), (6—20).
6—3.
GENERAL ELONGATION—JOINT DISPLACEMENT RELATION
We have derived expressions for the.direction cosines and elongation of a bar in terms of the initial coordinates and displacement components of the joints at the ends of the bar. By considering the truss as a system or network, the geometric relations for all the bars can be expressed as a single matrix equation. The relations for bar n, which is connected to joints s and k (positive direction
SEC. 6—3. GENERAL ELONGATION—JOINT DISPLACEMENT RELATION
121
from k to s) are summarized below for convenience:
= (x.
—
— Xk)
=
(x., — xk)
=
— Uk)
=
+
=
+
T
(ii, — uk)T
Uk)
Up to this point, we have considered joints s and k as coinciding with the. positive and negative ends of member n. Now we introduce new notation which
is more convenient for generalization of the geometric relations. Let n÷, n_ denote the joint numbers for the joints at the positive and negative ends of and k member n. The geometric relations take the form (we replace s by
bynJn(a)): =
—
x,,.)
—
=
=
(6—22)
—
= ci,,
+
=
+
ci,,
11)T
(u,,, —
u,,)
To proceed further, we must relate the bars and joints of the system, that is, we must specify the connectivity of the truss. The connectivity can be defined by a table having m rows and three columns. In the first column, we list the bar numbers in ascending order, and in the other two columns the corresponding numbers, and n, of the joints at the positive and negative ends of the members. This table is referred to as the branch-node incidence table in network theory.* For structural systems, a branch corresponds to a member and a node to a joint, and we shall refer to this table as the member-joint incidence table or simply as the connectivity table. It should be noted that the connectivity depends only on the numbering of the bars and joints, that is, it is independent of the initial geometry and distortion of the system.
Example
6—1
As an illustration, consider the two-dimensional truss shown. The positive directions of the bars are indicated by arrowheads and the bar numbers are encircled. The conneêtivity
See Ref. 8.
122
GOVERNING EQUATIONS FOR AN DEAL TRUSS
CHAP. 6
table (we list it horizontally to save space) for this numbering scheme takes the following
form: Bar,n
1
2
3
4
+Joint(n÷)
1
2
4
—Joint(n..)
2
3
5
0
3
8
9
10
11
3
1
2
4
5
6
5
6
2
3
5
6
7
5
1
2
6
4
5
Fig. E6—1
0
2
0 4
6
With the connectivity table, the evaluation of the initial length and direction cosines can be easily automated. The initial data consists of the j coordinate matrices, x1, x2 To compute and a,,, we first determine n÷ and n_ from the connectivity table and then use the first two equations of(6—22). For example, for bar 8, 8÷ 5, and 1, 8.. xs• —
x1 — x5 (x1 —
=
x)T(x * x5) —
x5)T
We define e and qj as the system elongation and joint-displacement matrices,
e=
{e1,
C2,...,
= {u1,u2
6—23
and express the m elongation-displacement relations as a single matrix equation
e = u
(6—24)
where d is of order m x ij. The elements in the uth row of d involve only Then, partitioning d into submatrices, of order 1 x 1, the elements of where k = 1, 2 in and £ = 1, 2 it follows that the only nonvanj, ishing submatrices for row n are the two subinatrices whose column number corresponds to the joint number at the positive or negative end of member it,
SEC. 6—3. GENERAL ELONGATION—JOINT DISPLACEMENT RELATION
namely,
n÷ and n.: =
+777
(6—25)
=
= Example
when
0
t'
n÷ orn
6—2
The .& matrix can be readily established by using the connectivity table. For row n, one puts +y,, at column at column n_, and null matrices at the other locations. The general form of the d matrix for the truss treated in Example 6—1 is listed below. We have also listed the elongations and joint displacement matrices to emphasize the significance of the rows and partitioned columns of .&. Uj
U2
U3
U4
U5
U6
0
0
0
0
0
0
0
e1
Ii
e2
0
12
e3
0
0
0
13
13
0
e4
0
0
0
0
14
14
e5
15
0
0
0
0
e6
0
is
0
0
—Ys
0
0
0
0
0
e5
is
0
0
0
e9
0
19
0
0
0
e10
0
1io
0
ho
0
e11
0
0
0
0 —19
0 0
The d matrix depends on both the geometry and the topology. It is of interest to express d in a form where these two effects are segregated. The form of (6—25) suggests that we list the y's in a quasi-diagonal matrix, 71
=
72
(6—26)
GOVERNING EQUATIONS FOR AN IDEAL TRUSS
124
CHAP. 6
and define C as k
C = [Ck(i
in
e= 1,2,...,j Cr,, -
*= + =0 Then,
1,2
(6—27)
= — Ij n÷ or n...
d = yC
(6—28)
The network terminology* for C is augmented branch-node incidence matrix. We shall refer to it simply as the connectivity matrix. Example 6—3 The connectivity matrix for Example 6—1 is listed below. The unit matrices are of order 2 since the system is two-dimensional.
Joint Numbers 1
2
1
12
2
+12
3
4
5
6
12
+12 12
Bar Numbers
+12 +12
12
+12
12
+12
+12
12
12
+12
One can consider row n of C to define the two joints associated with bar n. It follows that column k of C defines the bars associated with joint k. This association is usually See Prob. 6—6.
See also Ref. 8.
FORCE-ELONGATION RELATION FOR A BAR
SEC. 6—4.
referred
125
to as incidence. We say.that a joint is positive incident on a bar when it is at
the positive end of the bar. Similarly, a bar is positive incident on a joint when its positive
end is at the joint. For example, we see that joints 1 and 4 are incident on bar 5 and bars 3, 4, 6, 8, and 11 are incident on joint 5. We will use this property of the connectivity matrix later to generalize the joint force-equilibrium equations.
6-4. FORCE-ELONGATION RELATION FOR A BAR By definition, each bar of an ideal truss is prismatic and subjected only to axial load applied at the centroid of the end cross sections. It follows that the only nonvanishing stress component is the axial stress, a, and also, a is constant throughout the bar. We will consider each bar to be homogeneous but we will not require that all the bars be of the same material. The strain, s, will be constant when the bar is homogeneous and the force-elongation relation will be similar in form to the uniaxial stress-strain curve for the material. A typical a-c curve is shown in Fig. 6—4. The initial portion of the curve is essentially straight for engineering materials such as steel and aluminum. A material is said to be elastic when the stress-strain curve is unique, that is, when the curves corresponding to increasing and decreasing a coincide (OAB and BAO in Fig. 6—4). If the behavior for decreasing a is different, the material is said to be inelastic. For ductile materials, the unloading curve (BC) is essentially parallel to the initial curve.* Elastic behavior
B
C
0
Fig.
6—4.
Stress-strain curves for elastic and inelastic behavior.
We introduce the following notation:
A = cross sectional area F axial force, positive when tension = initial elongation, i.e., elongation not associated with stress *
A detailed discussion of the behavior of engineering materials is given in Chap. 5 of Ref. 2.
126
GOVERNING EQUATIONS FOR AN IDEAL TRUSS
CHAP. 6
Since the stress and strain are constant throughout the bar,
F= e = e0 =
Lg
(6—29)
Le0
We convert the a-c relation for the material to the force-elongation relation for the bar by applying (6—29).
C0
Fig. 6—5. Linear elastic behavior.
We consider first the case where the stress-strain relation is linear, as shown in Fig. 6—5. A material having this property is called Hookean. The initial and transformed relations are
a= F a
E(r
AE
=
L
—
(e — e0)
k(e — a0)
(6—30)
F + e0 = fF + e0
We call k, f the stiffness and flexibility factors for the bar. Physically, k is the force required per unit elongation and f, which is the inverse of k, is the elongation due to a unit force. We consider next the case where the stress-strain relation is approximated by a series of straight line segments. The material is said to be piecewise linear. Figure 6—6 shows this idealization for two segments. A superscript (j) is used to identify the modulus and limiting stress for segment j. The forceelongation relation will still be linear, but now we have to determine what
SEC. 6-4.
FORCE-ELONGATION RELATION FOR A BAR
127
Fig. 6—6. Piecewise linear approximation.
segment the deformation corresponds to and also whether the strain is increasing (loading) or decreasing (unloading). For unloading, the curve is assumed to be parallel to the initial segment.* The relations for the various possibilities are listed below. 1.
Loading or Unloading—initial Segment
=
F
F= 2.
k">(e
(6—3 1) —
Loading—Second Segment
F>l) < F F
= 3.
F>1>
,4a12>
F>2>
k'2>(e
Unloading—Second Segment
F
k>1>(e —
One can readily generalize these relations for the nth segment.f *
We
(6—32)
+ (f°> —
are neglecting the Bauschinger etlect. See Ref. 2, Sec. 5.9. or Ref. 3, Art. 74.
t See Prob. 6—8.
(6—33)
GOVERNING EQUATIONS FOR AN IDEAL TRUSS
128
CHAP. 6
Example 6—4 We consider a bilinear approximation, shown in Fig. E6—4. Fig. E6—4
40
41.7 30
(in./in.)
Taking
,4=lin.2
L=lQft=l2Oin we obtain
=
= 83.3 kips/in.
f"> =
1/k"> =
k>2> =
—.— = 41.7 kips/in. L
f"> =
24 x
F"> =
= 3okips
42>
L AE>2>
=
+ ([1)
—
Segment 1 Segment 2
120
12
x
in./kip
in./kip
— 0.36 in.
F
(83.3)(e — 120
F
(41.7)(e —
Suppose a force of 35 kips is applied and the bar is unloaded. The equivalent initial strain is (see Equation 6—33 and Fig. 6—6):
= =
—
+ (f>2) —
=
+ 0.06 in.
The procedure described above utilizes the segment stiffness, which can be interpreted as an average tangent stiffness for the segment. We have to modify the stiffness and equivalent initial elongation only when the limit of the seg-
SEC. 6-4.
FORCE-ELONGATION RELATION FOR A BAR
129
is reached. An alternate procedure is based on using the initial linear stiffness for all the segments. In what follows, we outline the initial st(ffness approach. ment
.4
I''I
/1
-eo,eq.
Fig. 6—7. Notation for the initial stiffness approach.
Consider Fig. 6—7. We write the force-elongation relation F—
= where e0,
eq
is
—
—
kU)(e
A
segment 2 as (6—34)
— eo,eq)
interpreted as the equivalent linear initial strain and is given by eo,eq = A
(6—35)
=
—
—
The equivalent initial strain, eoeq, depends on e, the actual strain. Since e in turn depends on F, one has to iterate on eoeq regardless of whether the segment limit has been exceeded. This disadvantage is offset somewhat by the use for all the segments. The notation introduced for the piecewise linear case is required in order to distinguish between the various segments and the two methods. Rather than continue with this detailed notation, which is too cumbersome, we will drop all the additional superscripts and write the force-deformation relations for bar n in the simple linear form of
= =
+
(6—36)
GOVERNING EQUATIONS FOR AN IDEAL TRUSS
130
CHAP. 6
where k, f, and e0 are defined by (6—31) through (6—35) for the physically
nonlinear case.
6—5.
GENERAL BAR FORCE—JOINT DISPLACEMENT RELATION
The force-deformation and deformation-displacement relations for bar n are given by (6—22) and (6—36). Combining these two relations leads to an expres-
sion for the bar force in terms of the displacement matrices for the joints at the ends of the bar. The two forms are: F,, =
=
— e0,
F0,,,
+
— k,,y,,u,,
(6—37)
F0,,, = —k,,e0,,,
and —
e0,, + f,,F,,
u,_) = e,,
(6—38)
We can express the force-displacement relations for the "m" bars as a single matrix equation by defining (6 39)
k1
k2
k=
km
and noting (6—24). The generalized forms of (6—37) and (6—38) are:
F=
k(e
—
e0)
=
F0
+
(6—40)
and
d°1I = 6—6.
e0
+ fF
(6—41)
JOINT FORCE-EQUILIBRIUM EQUATIONS
Let F,, be the axial force vector for bar n (see Fig. 6—8). The force vector has the direction of the unit vector, i,,, which defines the orientation of the bar in the deformed state. Now, = fi,,i. Then,
F,, =
(6—42)
When F,, is positive, the sense of F,, is the same as the positive sense for the bar. Continuing, we define F,,,, as the forces exerted by bar n on the joints at the positive and negative ends of the bar. From Fig. 6—8, = — F,, = — F,,fi,,i F,,,,
+F,, =
(6—43
JOINT FORCE-EQUILIBRIUM EQUATIONS
SEC. 6—6.
Joint n_
Fig. 6—8. Notation for barforce.
We consider next joint k. The external joint load vector is Pk, where For equilibrium, the resultant force vector must equal zero. Then, Pk =
- j+=k
Pk
—
The first summation involves the bars which are positive incident on joint k (positive end at joint k) and the second the bars which are negative incident. Using (6--43), the matrix equilibrium equation for joint k takes the form: Pk = Let
(6—44)
—
j+k
be the general external joint load matrix:
=
P2,
(if x
,
(6—45)
1)
We write the complete set of joint force-equilibrium equations as:
=
(6—46)
Note that the rows of pertain to the joints and the columns to the bars. We partition into submatrices of order i x 1. (if x
=
=
1,
2,.. ,j .
m)
k=
and
1,
2.
.
.
,m
(6—47)
Since a bar is incident only on two joints, there will be only two elements in any column of From (6—44), we see that, for column n, = =
=
0
when
(6—48) e
orn_
The matrix can be readily developed using the connectivity table. It will have the same form as dT with y, replaced by n,,. When the geometry is linear,
=
=
and
CHAP. 6
GOVERNING EQUATIONS FOR AN IDEAL TRUSS
132
Example 6—5 matrix for the truss of Example 6—1 has the following general form:
The
Bar Numbers 1
2
4
3
RT
a
7
8
10
9
11
+llç
1
Iz
6
5
+
nT
T 2
ftT
—plo
OF +p7
DT
-'
nT
P2
A
oT ±1J3
C
oT
OT
nT
C
J3
-I
oT +p4
øT
P6
oT +pU •r
I,
We could have also utilized the connectivity* matrix C to develop ft was pointed out in Example 6—3 that the elements of the kth column of C define the incidence of the bars on joint k. Using this property, we can write the generalized form of (6—44) as
where 0
-.
0
(rn x Lm) o
(6—49)
o
Finally, we have
= 6—7.
(6—50)
INTRODUCTION OF DISPLACEMENT RESTRAINTS; GOVERNING EQUATIONS
We have developed the following equations relating F, e,
and qj,
e = d°l1 = e0 + IF = and are the external joint-displacement and external joint-load matrices arranged in ascending order. Also, in our derivation, we where the elements of
have considered the components to be referred to a basic reference frame. Now, * See Sec. 6—3, Eq. 6—27.
SEC. 6—7.
INTRODUCTtON OF DISPLACEMENT RESTRAINTS
133
joint displacement restraints are imposed, there will be a reduction in the number of joint displacement unknowns and a corresponding increase in d, the number of force unknowns. This will require a rearrangement of and when
Let r be the number of displacement restraints and 11d the number of displace-
ment unknowns. There will be n4 prescribed joint loads and r unknown joint loads (usually called reactions) corresponding to the na unknown joint displacenients and the r known joint displacements. We let U1, U2 be the column matrices of unknown and prescribed joint displacement components and P1, P2 the corresponding prescribed and unknown joint load matrices. The rearranged system joint displacement and joint load matrices are written as U, P: (fld
x
1)
(r x 1) ><
—
-
6—51
1)
x 1)
+ V = 13
point out that the components contained in U (and P) may be referred to local reference frames at the various joints rather than to the basic frames. This is necessary when the restraint direction at a joint does not coincide with one of the directions of the basic frame. Finally, we let A and B be the transformation matrices associated with U and P. Then, (a) takes the form: We
= AU
e= P
e0 + fF
BF
We partition A, B consistent with the partitioning of U, P:
A2]
(mxr)
(fflxnd)
(6—52)
[Bil (nd x in) LB2i(r x m)
B
and write (b) in expanded form:
A1U1 + A2tJ2 =
e
= P2
e0
+ fF
(6—53)
B1F
(6—54)
B2F
(6—55)
Equation (6—53) represents equations relating the in unknown bar forces, the nd unknown displacements, and the r prescribed displacements. Equation
equations involving the in unknown bar forces and the prescribed joint loads. Lastly, Equation (6—55) represents r equations. for the r reactions in terms of the m bar forces. When the geometry is nonlinear, A and B involve the joint displacements. If the geometry is linear, A J3T, and (6—54) represents
= AT
j
= 1, 2
(6—56)
134
GOVERNING EQUATIONS FOR AN IDEAL TRUSS
CHAP. 6
We have introduced the displacement restraints into the formulation by with A, B. It remains to discuss how one determines A, B replacing d, In the following section, we treat the case of an arbitrary restraint from d,
direction. We also describe how one can represent the introduction of displacement restraints as a matrix transformation. 6—8.
ARBITRARY RESTRAINT DIRECTION
When all the restraint directions are parallel to the direction of the global reference frame, we obtain U from by simply rearranging the rows of such that the elements in the first rows are the unknown displacements and the last r rows contain the prescribed displacements. To obtain A, we perform the same operations on the columns of d, Finally, since P corresponds to U, we obtain B by operating on the rows of or alternately, by operating on the columns and then transposing the resulting matrix. When the restraint at a joint does not coincide with one of the directions of the basic frame, it is necessary first to transform the joint displacement and external load components from the basic frame to a local frame associated with the restraint at the joint. Suppose there is a displacement restraint at joint k. Let (j = 1, 2, 3) be the orthogonal directions for the local reference frame associated with the displacement restraint at joint k. Also, let and be the corresponding displacement and external joint load components. Finally, let R0k be the rotation transformation matrix for the local frame at joint k with respect to the basic frame (frame o). The components are related by:
k_
ok
k.. Pk—
ak
Uk—
Uk —
Pk
where ROk
= [cos
(6—58)
We have omitted the frame superscript (o) for quantities referred to the basic frame (ut, to simplify the notation. We define CU', as the system joint-displacement and -force matrices referred to the local joint reference frames,
= ... = as the system joint-rotation matrix, .
and
(6-59)
. .
R°1
=
R02 (6—60)
.
R0j
Then,
= =
(a)
ARBITRARY RESTRAINT DIRECTION
SEC. 6—8.
135
Operating on the initial equations with (a), b
(IP = leads to
=
(6—61)
The transformation to is the same as for the case where the restraint directions are parallel to the directions of the basic frame, that is, it will involve only a rearrangement of the rows of Similarly, we obtain A by rearranging the columns of .cifi. The steps are A2] -+
-* B LB2
Example 6—6 To obtain the submatrices in column k of
k of ri by R°"
T
we postmultiply
We can perform the same operation on
the submatrices in column
and then transpose the
resulting matrix or, alternately, we can premultiply the submatrices in row k of by R°". As an illustration, see the matrix for Example 6—5 on page 136. The matrix can be and replacing il,, by y,. determined by transposing
One can visualize the introduction of displacement restraints as a matrix transformation. We represent the operations U
and
P
(6—62)
as
U=D°lI and call D the displacement-restraint transformation matrix. When the restraint directions are parallel to the directions of the basic frame, D is a permutation matrix which rearranges the rows of We obtain D by applying the same row rearrangement to a unit matrix of order ij. Postmulti-
plication by Dr effects the same rearrangements on the columns. Also,* Dr D1. For the general case of arbitrary restraint directions, we first determine and then U. Now, = (a) The step,
—*
U, involves only a permutation of the rows of
U =
(6—63).
where H is the permutation matrix corresponding to the displacement restraints. *
See
Prob. I —36 for a discussion of permutation matrices.
pt
Pt
F1
—
F2
F3
F4
—
R04P1
F5 F6
for Example 6—6 F7
R°'pb
F8 F9
F10
—
ROSDTI
F11
Cl)
(I)
C
-1
m
C
> 2
:13
0
m
0 z C,)
—1
0 C
m
z z
m
0
C)
INITIAL INSTABILITY
SEC. 6—9.
137
Combining (a) and (6—63), we have
U= and it follows that
I) = Since both H and
HPII°'
(6—64)
are orthogonal matrices, D is also an orthogonal matrix.
Using (6—62),
A
dDT
then substituting for d, P.s, and D in terms of the geometrical, connectivity, local rotation matrices lead to and
B
( 6— 65 )
A=
Equation (6—65) is of interest since the various terms are isolated. However, one would not generate A, B with it. 6—9.
INITIAL INSTABILITY
The force equilibrium equations relating the prescribed external joint forces and the (internal) bar forces has been expressed as (see Equation 6—54):
P1=B1F I) and F is (m x 1). When the geometry is nonlinear, B1 depends on the joint displacements as well as on the initial geometry and
where P1 is
<
restraint directions. In this section, we are concerned with the behavior under an infinitesimal loading. Since the nonlinear terms depend on the load intensity, we they will be negligible in comparison to the linear terms for this case, take B1 as constant. Then, (a) represents linear equations in in unknowns. If these equations are inconsistent for an arbitrary infinitesimal loading, we say the system is initially unstable. When the geometry is linear, B1 is independent of the loading and the initial stability criterion is also applicable for a finite loading. This is not true for a nonlinear system. We treat stability under a finite loading in Chapter 7. Consider a set of j linear algebraic equations in k unknowns.
ax=c
(b)
In general, (b) can be solved only if a and [a c] have the same rank,* It follows
that the equations are consistent for an arbitrary right-hand side only when the rank of a is equal to], the total number of equations. Applying this condition *
See
Sec. 1—13; see also Prob. 1—45.
CHAP. 6
GOVERNING EQUATIONS FOR AN IDEAL TRUSS
138
to (a), we see that the truss is initially unstable when the rank of B1 is less than na. For the truss to be initially stable under an arbitrary loading, B1 must be That is, the number of bars must be at least This requires m of rank
equal to the number of unknown displacement components. Since the rank this condition is necessary but not sufficient for initial may still be less than stability. In order to determine whether a truss is initially stable, one must actually find the rank of B1. The following examples illustrate various cases of initial instability.
Example
6—7
The force-equilibrium equations for the accompanying sketch are: Fig. E6—7
x2
in =
4
na =
5
x1
F1 Pu
)
P21
F3
+1 B1
+1 +1
P22 P31
F4
—1
P12
—
F F2
+1
Row 3 is (— 1) times row 1. The equations are consistent only if P21 = — Pu Since m < we know the system is unstable for an arbitrary loading without actually finding r(B1).
INITIAL INSTAB1LITY
SEC. 6—9.
139
Example 6—8
We first develop the matrix for the truss shown in Fig. E6—8A and then specialize it for various restraint conditions. Fig. E6—8A
'1
J 0
M 4
3
I
F F
® Pit
F3
F5
F4
F6 —cosU
—1
© Piz ® Psi
F2
+1
+1
sinO
—a--—-— ±1
P22
® Psi
cosO .
sinO
+1
® P32
cosO —sinO
—1
© P41
—1
—1
—cosO —sinO
There are three relations between the rows (1)
row®+row®+row®= —row®
(2)
row® + row® + row
(3)
= —row®
(sin 8)(row ® + row ©) — cos U (row ®) = cos U (row ®)
GOVERNING EQUATIONS FOR AN IDEAL TRUSS
140
CHAP. 6
The first two relations correspond to the scalar force equilibrium conditions for the external
joint loads: Pkl
=
P11
Pk2 = P12
+ P21 + P31 + P41 =
0
+ P22 + P32 + P42 = 0
The third relation corresponds to the scalar moment equilibrium condition:
k1 Mk is the moment of the external force vector acting at joint k with respect to point
0, the origin of the basic frame. We obtain relation (3) by taking Oat joint 4. Equation (b) reduces to —d(p11 + P21) + b(p22
+ P32) =
0
Using ci
=L
b =
sin 8
L cos
0
we can write (c) as
cos 0p32
sin
O(pii + P21) — cos
which is relation 3.
We see that rows 2 and 5 arc independent. Thc remaining set (rows 1, 3, 4, 6, 7, 8) contains only three independent rows. Now, we obtain B1 from by first taking a linear combination of the rows (when the restraints are not parallel to the basic frame) and then deleting the rows corresponding to the joint forces associated with the prescribed joint displacements. Since has three linear dependent rows, it follows that we must introduce at least three restraints. Initial instability will occur if— 1. 2.
An insufficient number of restraints are introduced (n4 > 5). A sufficient number of restraints are introduced (/24 = 5) but the rows of B1 are not linearly independent. We say the restraints are not independent in this case. These cases are illustrated below.
Case
1
Fig. E6—8B 1
2
x2
m6 x1
INITIAL INSTABILITY
SEC. 6—9.
We obtain B1 by deleting rows 6 and 8 (corresponding to P32 and P42). The system is stable only when the applied joint loads satisfy the condition
Pu + P21 + P31 =
P41
Case 2 Fig. E6—8C
x2 rn = 6 —5
xl
We delete rows 4, 6, and 8. The number of restraints is sufficient (fld = 5) but the restraints are not independent since r(B1) < 5. Actually, r(81) = 4. To make the system stable, at least onc horizontal restraint must be introduced.
In Example 6—8, we showed that there are three relations between the rows for a two-dimensional truss. These relations correspond to the force- and moment-equilibrium conditions for the complete truss. To establish the relations for the three-dimensional case, we start with the equilibrium equations, (jxl) 3 o (2i—3)x 1
0
is the moment of with respect to an arbitrary moment center, 0. For convenience, we take 0 at the origin of the basic reference frame. Parwhere
titioning
(6—66)
where
is of order (i x m) and using the matrix notation introduced in
CHAP. 6
GOVERNING EQUATIONS FOR AN IDEAL TRUSS
142
Sec. 5—2
for the moment,* the equilibrium equations take the form PA,. = 0
(6—67)
=0
(6—68)
Equation (6—67) represents i relations between the rows of PA,
row q + row (q + I) + ... + row [i(j —
2)
+ q] = row [i(j
1)
+ q]
q=l,2,...i (6—69)
and (6—68) corresponds to (2i — 3) relations.
We have shown that there are at least 3(i — 1) relations between the tows of PA. Now, we obtain B by combining and rearranging the rows of PA. It follows that B will also have at least 3(i — 1) relations between its rows. Finally, we obtain by deleting the rows corresponding to the restraints. For the system to be initially stable, we must introduce at least 3(i — 1) restraints: r
no.
of restraints
3(i
— 1)
(6—70)
Note that this requirement is independent of the number of bars. Also, it is a necessary but not sufficient condition for initial stability. in. The number of restraints must also satisfy the necessary condition This requires r = ((j — — m) (6—71) Both (6—70) and (6—71) must be satisfied. Either condition may control r, depending on the arrangement of the bars. REFERENCES
C. H., and J. B. WILBUR: Elementary Structural Analysis, McGraw-Hill, New York, 1960. 2. CRANDALL, S. H,, and N. C. DAHL: An Introduction to the Mechanics of Solids, McGraw-Hill, New York, 1959. 3, TIMOSNENKO, S.: Strength of Materials, Part 2, Van Nostrand, New York. 1941. 4. TIMoSISENKO, S., and D. H. YOUNG: Theory of Structures, McGraw-Hill, New York, 1.
NORRIS,
5.
MCMINN, S. I.: Matrices jbr Structural Analysis, Wiley, New York, 1962. MARTIN, H. C.: Introduction to Matrix Methods of Structural Analysis, McGrawHill, New York, 1966. LIVIISLEY, R. K.: Matrix Methods of Structural Analysis, Pergamon Press, London,
1945. 6. 7.
1964. 8.
FENVES, S. J., and F. H. BRANIN: "Network-Topological Formulation of Structural Analysis," J. Struct. Div., ASCE, Vol. 89, No. ST4, pp. 483—514, 1963.
* See
Eq. 5—11.
PROBLEMS
143
PROBLEMS 6—1.
Determine in,J, r, and
for the following plane trusses: Prob.
6—1
(a)
(b)
6—2.
Suppose bar n is connected to joints s and k where Xk
(a)
= {l,
=
1,0] (ft)
{5, —5. —2] (ft)
Take the positive direction of bar n from k to s. Determine and
(b)
Suppose {1/10, 1/20, 1/l0}
Uk
=
—
1/10,
—
1/30}
(inches) (inches)
Note that the units of x and u must be consistent. Find 1k and Determine and ji,,, using the exact expressions (Equations 6—15, 6—17), the expressions specialized for the case of small strain (Equations 6—19, and 6—20), and the expressions for the linear geometric case (Equation 6—21). Compare the results for the three cases. 6—3. Discuss when the linear geometric relations are valid and develop the appropriate nonlinear elongation-displacement relations for the trusses shown. Assume no support movements. 6—4. Consider the truss shown: (a) Establish the connectivity table. (b)
List the initial direction cosines. Do we have to include nonlinear
(c)
geometric terms for this truss? Locate the nonzero submatrices in .sd, using the connectivity table. Determine the complete form of d.
GOVERNING EQUATIONS FOR AN IDEAL TRUSS
144
CHAP. 6
Prob. 6—3
2
.3
(a)
x2
3
X1
2
Ib)
x2
Prob. 6—4
I'
(d) (e) 6—5.
6—6.
Determine C.
Verify that d =
cxC.
Determine d for the three-dimensional truss shown. Consider the d-c network shown. The Junctions are generally called
nodes, and the line connecting two nodes is called a branch. The encircled numbers refer to the branches and the arrowheads indicate the positive sense (of the current) for each branch.
PROBLEMS
145
Let and n. (j = 1, 2,. . , 5) denote the potential at node j. Also, let denote the nodes at the positive and negative ends of branch n. The potential .
Prob. 6—5
x3
4 (0, 1,0)
xI
(1,0,0)
(1,1,0)
Prob. 6—6 3
0
0' drop for branch n, indicated by
is given by = v,,,
We define v and e as
v= e=
{v1,
v2,.. .
{e1, e2, .
.
.
,
,
v5}
= general node potential matrix = general branch potential difference matrix
and write the system of branch potential difference—node potential relations as
e=
.cjv
Determine d, using the branch-node connectivity table. Discuss how the truss
problem differs from the electrical network problem with respect to the, form of ad. How many independent columns does ad have? In network theory, ad is called the augmented branch node incidence matrix.
GOVERNING EQUATIONS FOR AN IDEAL TRUSS
146
6—7.
(a) (b) (c)
CHAP. 6
Take L 20 ft, A = 2 in2, and the a-s curve shown. Develop the piecewise linear force-elongation relations. Suppose a force of + 60 kips is applied and then removed. I)etermine the force-elongation relation for the inelastic case. Suppose the bar experiences a temperature increase of 1000 F. Determine the initial elongation. Consider the material to be aluminum. Prob. 6—7
6X
ksi
20 ksi
6—8.
Generalize Equation 6—32 for segmentj. Start with
+
e
and express eb0 in terms of quantities associated with segment (J — 1). 6—9. Generalize Equation 6—35 for segmcntj. 6—10.
Suppose
the stress-strain relation for initial loading is approximated,
as in the sketch, by
a=
E(s
—
be3)
Prob.
GA
Ee da
Et
6—10
PROBLEMS
147
Determine expressions for ES and E', the secant and tangent moduli. Determine expressions for k5 and kt. Suppose the material behaves inelastically for decreasing 4 Consider the unloading curve to be parallel to the initial tangent. Determine the force-elongation relation for AB. 6—li. Repeat Prob. 6—10, using the stress-strain relation (a)
(b) (c)
=
(u +
where E, c, and n are constants. 6—12.
For the accompanying sketch: Prob.
6—12
p
(a)
Locate the nonzero submatrices in
(b) Assemble for the linear geometric case. 6—13. Repeat Prob. 6—12 for the three-dimensional truss shown.
Prob.
6—13
LX2
I
6—14.
(a)
'I.
Consider the electrical network of Prob. 6—6. Let be the current in branch n. The positive sense of
is from, node
n_ to node n÷. Now, the total current flowing into a node must equal the total current flowing out of the node. This requirement leads to one equation for each node involving the branch currents incident on
GOVERNING EQUATIONS FOR AN IDEAL TRUSS
148
the
CHAP. 6
node. Let =
{i1, j2, .
.
.
= general branch-current matrix
,
Show that the complete system of node equations can be written as (Sxl) 0 (b)
where d is given in Prob. 6—6. How many independent equations does (a) represent?
(Hint: d has only four independent columns). When the resistance is linear, the current and potential drop for a branch are related by = e0
is the branch emf and R,, is the branch resistance. An
alternate form is i,, =
— e0 ,,)
Note the similarity between (b) and the linear elastic member forceelongation relation. Show that the complete system of branch curpotential relations can be written as
e=
= e0
= R1(e
—
+ Ri
e0) = R1dv
—
R1e0
Equations (a) and (c) are the governing unpartitioned equations for a linear-resistance d-c network. The partitioned equations are developed in Prob. 6—23. It should be noted that the network problem is onedimensional, that is, it does not involve geometry. The d matrix depends only on the topology (connectivity) of the system. Actually, d corresponds to the C matrix used in Sec. 6—3 with i = 1. 6—15. Refer to Prob. 6—12, Suppose u11, u42, is52 are prescribed. Identify B1 and B2. 6—16. Refer to Prob. 6—12. (a) (b)
Develop the general form of Suppose is21,
u42,
are prescribed. The orientation of the local
frame at joint 5 is shown in the sketch. Determine B1 and B2. x2
Prob.
6—16
PROBLEMS 6—17.
(a) (b)
Refer to Prob. 6—13
Develop the general form of Determine B1 and B2 corresponding to the following prescribed displacements: U11,
U12,
U3j,
U-33,
U23,
U13
The local frame at joint 2 is defined by the following direction cosine table.
x2
x1
x3 0
1/2
1/2
1/2
1/2
6—48. Consider the two-dimensional truss shown. The bars are of equal length and 0 is the center of the circumscribed circle. The restraint direction degrees counterclockwise from the tangent at each joint. Investigate the is initial stability of this system. Repeat for the case of four bars. Prob. 6—18
r (restraint direction) t (tangent)
11
= 13
I.
6—19.
Suppose na =
equations for P1
in.
Then, B1 is of order tn x m. The equilibrium
0 are (,nxrn)
B1
F=0
(mxl)
rnXl
If (a) has a nontrivial solution, the rank of B1 is less than m and the system is initially unstable (see Prob. 1—45). Rather than operate on B1, to determine r(B1), we can proceed as follows: (1) We take the force in some bar, say bar k, equal to C: Fk
=
C
GOVERNING EQUATIONS FOR AN DEAL TRUSS
150
(2)
CHAP. 6
Using the joint force-equilibrium equations, we express the remaining
bar forces in terms of C.
(3) The last equilibrium equation leads to an expression for Fk in terms of C. If this reduces to an identity, r(B1) < since a nontrivial solution for F exists. This procedure is called the zero load test. (a)
Apply this procedure to Prob. 6—18. Take F1 = C and determine F2, F3, and then F1 using the equilibrium condition (summation of forces normal to r must equal zero) for joints 1, 2, 3.
(b)
When n4 = m and the geometry is linear, the truss is said to be statically determinate. In this case, we can determine F, using only the equations of static equilibrium, since the system, P1 = B1F, is square. Do initial
elongations and support settlements introduce forces in the bars of a statically determinate truss? 6—20. Modify the zero load test for the case where na < in. Note that the general solution of B1F 0 involves m — r(B1) arbitrary constants. 6—21. Investigate the initial stability of the two-dimensional truss shown. Use the zero load test. Prob.
6—21
5
6—22. Investigate the initial stability of the system shown. The restraint directions are indicated by the slashed lines. Prob. 6—22
-j 4
-t
I 3
c
PROBLEMS 6—23. We generalize the results of Probs. 6—6. and 6—14 for a network having b branches and n nodes. Let
e v
branch potential duff, matrix = {e1, e2, -. = branch current matrix = {i1, j2 = node potential matrix = {v1, v2,. . . ,
.
,
The general relations are (1) node equations (n equations) (nxb)
&T
(bxl)
(nxl)
=
and (2) branch equations (b equations)
e = dv = e0 + Now, dT rows of
has
only n —
1
Ri
independent rows. One can easily show that the
are related by row k
row ii
= It follows that (a) represents only n —
independent equations, and one equation must be disregarded. Suppose we delete the last equation. This corresponds to deleting the last column of d (last row of dT). We partition d, (bxn)
1
bx(n—1)
bxl
d2]
and let d1 = A. The reduced system of node equations has the form ATj
=0
Note that AT corresponds to B1 for the truss problem. Equation (e) represents (n — 1) equations. Since v is of order n, one of the node potentials must be specified. That is, we can only determine the potential difference for the nodes with respect to an arbitrary node. We have deleted the last column of d which corresponds to node n. Therefore, we take as the reference potential. (a)
Let = {v1 —
v2 —
.
.
.
,
—
Show that
dv = AV (b)
Summarize the governing equations for the network. The operation
corresponds to introducing displacement restraints in the truss problem. Compare the necessary number of restraints required for the network and truss problems;
7
Variational Principles for an Ideal Truss 7-1.
GENERAL
The formulation of the governing equations for an ideal truss described in Chapter 6 involved three steps: 1.
2. 3.
The elbngation of a bar was related to the translations of the joints at the end of the bar. Next, the bar force was expressed in terms of the elongation and then in terms of the joint translations. Finally, the equilibrium conditions for the joints were enforced, resulting in equations relating the external joint loads and internal bar forces.
The system equations were obtained by generalizing the member forcedisplacement and joint force equilibrium equations and required defining only Later, in Chapter 10, we shall two additional transformation matrices follow essentially the same approach to establish the governing equations for an elastic solid. In this chapter, we develop two variational principles and illustrate their application to an ideal truss. The principle of virtual displacements is treated first. This principle is just an alternate statement of force equilibrium. Next, we discuss the principle of virtual forces and show that it is basically a geometrical compatibility relation. Both principles are then identified as the stationary requirements for certain functions. For this step, we utilize the material presented in Chapter 3, which treats relative extremas of a function. Finally, we discuss the question of stability of an elastic system and develop the stability criterion for an ideal truss. Why bother with variational principles when the derivation of the governing equations for an ideal truss is straightforward? Our objective in discussing them at this time is primarily to expose the ieader to this point of view. Also, we can illustrate these principles quite easily with the truss. Later, we shall 152
PRINCIPLE OF VIRTUAL DISPLACEMENTS
SEC. 7—2.
153
these principles, particularly the principle of virtual forces, to construct approximate formulations for a member. use
7-2. PRINCIPLE OF VIRTUAL DiSPLACEMENTS The principle of virtual displacements is basically an alternate statement of force equilibrium. We will establish its form by treating first a single particle and then extending the result to a system of particles interconnected with internal restraints. The principle utilizes the concept of incremental work and, for completeness, we review briefly the definition of work before starting with the derivation. Let v be the displacement of the point of application of a force F in the direction of F. The work done by F (see Fig. 7—1) is defined as
w
w0 +
JFdv
= W(v)
where v0 is an arbitrary reference displacement. Since W is a function of v, the increment in W due to an increment Ar can be expressed in terms of the differentials of W when F is a continuous function of yr f = dW + 4d2W + dW dW —Ar = dv
F
Au
(7—2)
d2W = d(dW)
refer to dW as the first-order work. Similarly, we call d2W the secondIf dF/dv is discontinuous, as in inelastic behavior, we must use the value of dF/dv corresponding to the sense of Av. This is illustrated in We
order work.
F
w—rv0
Fig. 7—1. Work integral for the one-dimensional force-displacement relation.
t Differential notation is introduced in Sec. 3—I.
VARIATIONAL PRINCIPLES FOR AN IDEAL TRUSS
154
CHAP 7
< 0. —k2 for of v when there is a reversal in the Note that W is not a single-valued function
> 0, and dF/du =
Fig. 7—2. We use dF/dv = +k1 for
F-v curve.
A
vs
V
Fig. 7—2. Work integral for direction-dependent force.
We consider first a single mass particle subjected to a system of forces (see Fig. 7—3), Let R be the resultant force vector. By definition, the particle is in equilibrium when R = 0. We visualize the particle experiencing a displacement increment Au from the initial position. The first-order work is
dW=R'Afi
(7—3)
If the initial position is an equilibrium position, dW 0 for arbitrary Au since 0. Therefore, an alternate statement of the equilibrium requirement is:
R=
The first-order work is zero for an arbitrary displacement of a particle from an equilibrium position.
74 —
is called a virtual displacement; this statement is the definition of the principle of virtual displacements. The incremental displacement
R+
Fig. 7—3. Virtual displacement of a single mass particle.
One can readily generalize (7—4) for the case of S particles. Let
first-order work associated with the forces acting on particle q and We consider the forces to be continuous functions of Au.
be the the
SEC. 7—2.
PRINCIPLE OF VIRTUAL DISPLACEMENTS
155
If particle q is in equilibrium,
corresponding virtual-displacement vector.
for arbitrary It follows that the scalar force-equilibrium equations for the system are equivalent to the general requirement,
=
0
dW
=
=
for arbitrary
0
Equation (7—5) is the definition of the principle of virtual work for a system of particles.
In general, some of the forces acting on the particles will be due to internal restraints. We define dW5 as the first-order work done by the external forces and dW1 as the work done by the internal restraint forces acting on the particles. Substituting for dW, (7—5) becomes
dW5 + dW1 =
0
for arbitrary
q=l,2
S
be the work done by the internal restraint forces acting on Now. let the restraints. We use the subscript D for this term since it involves the
/(Deformedl
F1
F1 I
.
El (Initial)
(Deformed) F1
F1
II
Fig. 7—4. Work done on the mass particles and internal restraints.
deformation of the restraints. The restraint force acting on a particle is equal in
magnitude, but opposite in sense, to the reaction of the particle on the restraint. Since the points of application coincide, it follows that
As an illustration, consider the simple system shown in Fig. 7—4. For this case, we have
dW,, =
—F1
Au1 + F1 Au2
dW1 = F1 Au1 —
F1 Au2
156
VARIATIONAL PRINCIPLES FOR AN IDEAL TRUSS
CHAP. 7
Using (b), we can write (a) as:
dWE =
for arbitrary
dWD
(7—6)
q=1,2,...,S
Also, the general principle of virtual displacements can be expressed as follows:
The first-order work done by the external forces is equal to the first-order work done by the internal forces acting on the restraints for any arbitrary virtual displacement of a system of particles from an equilibrium position.
We emphasize again that (7—6) is just an alternate statement of the force equilibrium conditions for the system. Some authors refer to (7—6) as the work equation. To apply the principle of virtual displacements to an ideal truss, we con-
sider the joints to be mass points and the bars to be internal restraints. We have defined and as the column matrices of external joint loads and corresponding joint displacements, Then,
= where &W contains the virtual joint displacements. The first-order work done
by the restraint forces acting on bar n due to the virtual displacements is
=
F,, deH
Generalizing (b), we have
dW0 = FraC
Finally, the work equation for an ideal truss has the form
Ml = FT de
for arbitrary MIt
(7—7)
The scalar force-equilibrium equations are obtained by substituting for de in terms of Mt. It is convenient to first establish the expression for the differential elongation of an individual bar and then assemble de. Operating on —
—
(
—
—
u,
U
and noting the definition of
de, =
(see (6—22)), we obtain
— u,)T]
+
=
—
(Au
—
Au,,)
Au,) (7—8)
1= J,,, F, de, = Wd(e,). We must use the rules for forming the differentials of a compound function since e, depends on the joint displacements. Using (3—17), we can write
dW4 = d2
dWd
de
de,
= F, de,
= d(F, de,,) =
de,
(de,)2
+ F, d2e,
PRINCIPLE OF VIRTUAL DISPLACEMENTS
SEC. 7—2.
157
The assembled form follows from (6—25). We just have to replace y,, with il,,:
MI
(7—9)
and requiring (a) to be satisfied for arbitrary
results in the joint force
de
Substituting for de in (7—7),
?,TMI = force equilibrium equations. For the geometrically linear case, e where d is constant and de = dM1 follows directly from e. We have treated the geometrically nonlinear
case here to show that the principle of virtual displacements leads to forceequilibrium equations which are consistent with the geometrical assumptions associated with the deformation-displacement relations. Example
7—1
We consider a rigid member subjected to a prescribed force, P. and reactions R,, R2, as in the diagram. There is no internal work since the body is rigid. Introducing the virtual
Fig.
b
displacements shown above, and evaluating the first-order work,
= R, Au1 + R2 Au2 —
dW = Now,
P
is not independent:
= Au,
(
d\
+ Au2
7d
—
Then,
dW =
+
—
Au2 {R2
—
—
Requiring (c) to be satisfied for arbitrary Au,, Au2 leads to
R2 = P which are the force and moment equilibrium equations, in that order.
=0
E7—1
158
VARIATIONAL PRINCIPLES FOR AN IDEAL TRUSS
CHAP. 7
Example 7—2 We consider the outside bars to be rigid (see sketch). To obtain the force equilibrium equation relating P and the internal bar forccs F1, F2, we introduce a virtual displacement, Au1, of the point of application of F:
dWE = PAu1 dW0 F1 de1 + F2 de2
The first-order increments in the elongations are de1 =
Au1 cos El
de2 =
—Au2 cos El
—Au1 cos U
where U defines the initial position. Then, equating dW5 and dW0, dWE =
P=
(1WD
(F1
—
for arbitrary Au1
Fjcos
0
The force in bar 3 does not appear explicitly in the equilibrium equation, (c). It is possible Fig. E7—2
Bars 3,4, 5, 6 are rigid to include F3 even though bar 3 is rigid by treating it as a Lagrange multiplier.f We
consider Au3 as independent in the work equation: P Au1 — (F1 cos 0)Au1 + (F, cos 0)Au2 = 0
(d)
Now, Au1 — Au2
=
0
(e)
Multiplying the constraint relation by —2, adding the result to (d), and collecting terms, we obtain Au1(P — F1 cos 0 — 2) + Au2(F2 cos 0 + 2) = 0 (f) Finally, we require (f) to be satisfied for arbitrary Au1 and Au2. The equilibrium equations See See, 3—3.
SEC. 7—3.
PRINCIPLE OF VIRTUAL FORCES
159
are
P=F1cosO+A F2cos6+2=O and we recognize 2 as the force in bar 3.
7—3.
PRINCJPLE OF VIRTUAL FORCES
The principle of virtual forces is basically an alternate statement of geometrical compatibility. We develop it here by operating on the elongation— joint displacement relations. Later, in Chapter 10, we generalize the principle. for a three-dimensional solid and describe an alternate derivation. We restrict this discussion to geometric linearity. The governing equations —
e
=
=
Now, we visualize a set of bar forces AF, and joint loads,
which satisfy
the force-equilibrium equations:
= A force system which satisfies the equations of static equilibrium is said to be statically permissible. Equation (b) relates the actual elongations and joint dissum over the bars, and placements. If we multiply the equation for Ck by note (c), we obtain the result AFTe = = which is the definition of the principle of virtual forces:
The actual elongations and joint displacements satisfy the condition AFTC —
=0
(7-10)
for any statically permissible system of bar forces and joint loads.
The principle of virtual forces is independent of material behavior but is restricted to the geometrically linear case. The statically permissible system (AF,
is called a virtual-force system.
To illustrate the application of this principle, we express cW and partitioned form,
in
where U2 contains the prescribed support movements. Using (a), (7—10) takes the form: AFTe — U2 = APr U1
VARIATIONAL PRINCIPLES FOR AN IDEAL TRUSS
160
CHAP. 7
If the elongations are known, we can determine the unknown displacements by specializing AP1. To determine a particular displacement component, say we generate a force system consisting of a unit value of PkJ and a set of bar forces and reactions which equilibrate Pkj = 1. = (7—11) = The internal bar forces and reactions are obtain from an equilibrium analysis of a statically determinate structure. Since only one element of is finite, APf U1
(l)ukj
and (b) reduces to Ukf = eTFJ;,k1_ U2TP2IPkJ1
(7—12)
The principle of virtual forces is also used to establish geometric compatibility relations required in the force method which is discussed in Chapters 9 and 17. We outline the approach here for completeness. One works with self-equilibrating virtual-force systems, i.e., statically permissible force systems which involve only bar forces and reactions. By definition, a self-equilibrating force system F*, f)* satisfies
B1F* =
P7 = P7 = B2F*
0
(7—13)
For this case, (b) reduces to —.
=
0
(7—14)
Equation (7—14) represents a restriction on the elongations and is called a geometric compatibility equation.
Example
7—3
The truss shown (Fig. E7—3A) has support movements and is subjccted to a loading which results in elongations (e1, e7) in the diagonal bars. We are coniidering the outside bars to be rigid.
Fig. E7—3A u,P
Bars 3,4, 5, 6 are rigid
SEC. 7—3.
PRINCIPLE OF VIRTUAL FORCES
To determine the translation, u, we select a statically determinate force system consisting of a unit force in the direction of u and a set of bar forces and reactions required to equilibrate the force. One possible choice is shown in Fig. E7.-3B. Evaluating (7—12) leads to u
=
cos 6
+
— tan
—
This truss is statically indeterminate to the first degree. A convenient choice of force redundant is one of the diagonal bar forces, say F2. The equation which determines F2 is Fig. E7—3B
0
derived from the gcometric compatibility relation, which, in turn, is obtained by taking a self-equilibrating force system consisting ofF2 = + I and a set of bar forces and reactions required for equilibrium. The forces are shown in Fig. E7--3C. Fig. E7—3C
— ens 6
0 Evaluating (7—14), we obtain
e1 + e2 = 0
To show that (a) represents a geometrical compatibility requirement, we note that the elongation-displacement relations for the diagonal bars are
ucos6
e2
= —ucosO
Specifying e1 determines u and also e2. We could have arrived at Equation (a) starting
from Equation (b) rather than (7—14). Flowever, (7—14) is more convenient since it does not involve any algebraic manipulation. We discuss this topic in depth later in Chapter 9.
VARIATIONAL PRINCIPLES FOR AN IDEAL TRUSS
162
7—4.
CHAP. 7
STRAIN ENERGY; PRINCIPLE OF STATIONARY POTENTIAL ENERGY
In this section, we specialize the principle of virtual displacements for elastic behavior and establish from it a variational principle for the joint displacements. We start with the general form developed in Sec. 7—2, FT
de =
gpT
for arbitrary to be arbitrary, i.e., unrestrained,
If we consider all the elements of
de =
A'W
and (a) leads to the complete set of force-equilibrium equations in unpartitioned form,
= can obtain the equation for P1 by rearranging (c) or by starting with the partitioned form of We
and noting that
U2 is prescribed. The reduced form is FT de
—
iW1 =
0
for arbitrary AU1
(7—15)
where now In what follows, we will work with (7—15).
Our objective is to interpret (7—15) as the stationary requirement for a function of U1. We consider F to be a function of e, where e = e(U1). The form of F = F(e) depends on the material behavior. We could express F in terms of U1 but it is more convenient to consider F as a compound function of e. The essential step involves defining a function, V1- = VT(e), according to FT dc =
(7—16)
With this definition, and letting
=
VT —
=
0
=
(7—17)
we can write (7—15) as
for arbitrary
(7—18)
We call the total strain energy function and the total potential energy. One should note that VT exists only when F is a continuous single-valued function of e. This requirement is satisfied when the material is elastic. Equation (7—18) states that the joint force-equilibrium equations (P1 = B1F) expressed in terms of the unknown displacements are the Euler equations for the
t See Secs. 6-4, 6—5.
STRAIN ENERGY
SEC. 7—4.
163
total potential energy. It follows that the actual displacements, i.e., the displacements which satisfy the equilibrium equations, correspond to a stationary value of 11,'
It remains to discuss how one generates the strain-energy function. By definition, dVT
and
= where is the strain energy for bar). Since we are considering pound function of e1, Equation (b) is equivalent to
to be a com(7—19)
= That is, the strain energy function for a bar has the property that its derivative with respect to the elongation is the bar force expressed in terms of the elongation. Finally, we can express as (7—20) = where e0 is the initial elongation, i.e., the elongation not associated with the force. Actually, the lower limit can be taken arbitrarily. This choice corresponds to taking as the area between the F-c curve and thee axis, as shown in Fig.7—5.
Fig. 7—5. Graphical representation of strain energy and complementary energy.
We consider the linearly elastic case. Using (6—30), F1
— e0,
Then
=
— e0,
)2
(7—21)
The total strain energy is obtained by summing over the bars. We can express VTas
=
VT
j=1
4(e
—
e0)Tk(e
—
eo)
(7—22)
VARIATIONAL PRNCIPLES FOR AN IDEAL TRUSS
164
CHAP. 7
Finally, we substitute for e in terms of U1, U2, using
e=
A1U1 + A2U2
(7—23)
When the geometry is linear, A1, A2 are constant and is a quadratic function. If the geometry is nonlinear, is a fourth degree function of the displacements. Up to this point, we have shown that the displacements defining an equilibrium position correspond to a stationary value of the potential energy function. To determine the character (relative maximum, relative minimum, indifferent,
neutral) of the stationary point, we must examine the behavior of the second differential, in the neighborhood of the stationary point. and noting that AP1 = 0 leads to Operating on d2
(7—24)
=
+
The next step involves expressing d2VT as a quadratic form in AU1. We restrict this discussion to linear behavior (both physical and geometrical). The general
nonlinear case is discussed in Sec. 17.6 When the geometry is linear, we can operate directly on (7—23) to generate the differentials of e, de
d2e =
A1AU1 0
since A1 is constant. When the material is linear,
dF=kde where k is a diagonal matrix containing the stiffness factors (AE/L) for the bars. Then, d2VT reduces to d2VT = dFT dc = deTk de AUT(ATkA1)AU1
If de
7 25
-
0 for all nontrivial AU1, d2VT is positive definite and the stationary
point is a relative minimum. This criterion is satisfied when the system is initially stable, since de =
0 for AU1
A1 AU1 =
0
0 would require that
(m equations in
unknowns)
have a nontrivial solution. But a nontrivial solution of (a) is possible only when for the geometrically linear case and r(B1) = r(A1) < ne,. However, A1 = when the system is initially stable. Therefore, it follows that the displacements
defining the equilibrium position for a stable linear system correspond to an absolute minimum value of the potential energy.
Example
7—4
We establish the total potential energy function for the truss considered in Example 7—2.
For convenience, we assume no initial elongation or support movement. The strain
COMPLEMENTARY ENERGY
SEC. 7—5.
165
energy is VT
=
+
Substituting for the elongations in terms of the displacement,
e2= —u2cosO= —u1cosO
e1 =u1cosO results in
=
+
cos2 0
=
+
cos2 0 — P1u1
and finally
The first differential of
is
= {{(k1 + k2)cos2 Ojuj Requiring
— P1}Au1
to be stationary leads to the Euler equation,
P1 =
[(k1 + k2)cos2 0]u1
which is just the force-equilibrium equation
P1 =
(F1
—
F2)cos 0
with the bar forces expressed in terms of the displacement using
F1 =
k1e1
The second differential of
=
ku1
cos 0
k2e2
=
—k2u1
cos 0
is
= and
F2 =
[(k1 + k,)cos2 0](Au1)2
we see that the solution, Ul
P1
= k1 + 0. Thc truss is initially unstable
corresponds to an absolute minimum value of H,, when 0 when 0 0.
7—5.
COMPLEMENTARY ENERGY; PRINCIPLE OF STATIONARY COMPLEMENTARY ENERGY
The principle of virtual forces can be transformed to a variational principle for the force redundants. We describe in this section how one effects the trans-
formation and utilize the principle later in Chapter 9. This discussion is restricted to linear geometry. We start with Equations (7—13) and (7—14), which we list below for convenience:
eTAF
=
—
where AF,
0
represent a self-equilibrating force- system, i.e., they satisfy the the following constraint relations: B1 iW = 0 B2 AF
166
VARIATIONAL PRINCIPLES FOR AN IDEAL TRUSS
CHAP. 7
Our objective is to establish a function of F, whose Euler equations are (a) and
(b). We cannot work only with (a) since F is not arbitrary but is constrained by the force-equilibrium equations,
P1 = We interpret
B1F
(fld
as the first differential of a function
= and call
equations in m variables)
eT
=
dV7
(7—26)
the complementary energy function for bar j. By definition,
=
(7—27)
That is, the complementary energy function for a bar has the property that its derivative with respect to the bar force is the elongation expressed in terms of the force. We express as dF1
(7—28)
This definition corresponds to taking Vj' as the area bounded by the F-e curve
and the F axis as shown in Fig. 7—5. Also, the strain and complementary energy functions are related by
=
+
(7—29)
When the material is linear elastic,
= 11
j—
=
e0,
+
1.1r2 rj T 2Jj' 3
—
+ 4FTfF
Next, we define II. as: 1-Ic =
=
7—31) —
the total complementary energy function. With these definitions, Equations (a), (b), and (c) can be interpreted as We call
0
subject to the constraint condition d(P1 — 81F)
=
0
We can combine (e) and (f) into a single equation by introducing Lagrange multipliers. Following the procedure described in Sec. 3—3, we add to (7—31) the joint force equilibrium equations and write the result as: + (P1 — where
.
,
(7—32)
contains the Lagrange multipliers. The Euler equa-
COMPLEMENTARY ENERGY
SEC. 7—5.
tions for
treating F and
167
as independent variables are
dfl, =
arbitrary
for AF,
0
+ B102
e(F) B1F = P1
(7—33)
We recognize the first equation in (7—33) as the member force-displacement relation, and it follows that = U1.
An alternate approach involves first solving the force-equilibrium equation, (d). There arena equations in m variables. Since B1 is of rank n4 when the system is initially stable, we can solve for na bar forces in terms of P1 and the remaining
bar forces. One can also work with a combination of bar forces and reactions as force unknowns. We let (rn —
q= m — X = {X1, X2
number of redundant forces Xq} = matrix of force redundants
(734)
and write the solution of the force-equilibrium equations as
F=
+ P2,0 + F0
P2
The force system corresponding to
=
is
(735)
self-equilibrating, i.e.,
for arbitrary X
0
We substitute for F in (7—3 1) and transform eT
=
—
Then,
UI AP2
AF —
(eTFx
to
(7—36)
UIP2,
and the Euler equations are eTFx
—
UIP2
x
0
(737)
Note that (7—37) is just a reduced form of (7—33). Also, we could have obtained this result by substituting directly in (a).
Up to this point, we have shown that the force redundants which satisfy the geometric compatibility equations correspond to a stationary value of the total complementary energy. To investigate the character of the stationary point, we evaluate the second differential. Operating on (g),
=
deTFx AX
d2 is positive definite with regard to AX, the stationary point is a relative minimum. This requirement is satisfied for the linear elastic case. To show this, we note that
de
= fAF = WXAX =
VARIATIONAL PRINCIPLES FOR AN IDEAL TRUSS
168
Since f contains only positive elements, AX provided that there does not exist a
CHAP. 7
is positive definite with regard to nontrivial solution of
AX =
=0
For (j) to have a nontrivial solution, there must be at least one relation between the columns of But this would correspond to taking force redundants which
are not independent, and the solution scheme would degenerate. Therefore, we can state that the actual force redundants correspond to an absolute minifor the linear elastic case. mum value of Example
7—5
We consider the truss treated in Example 7—3. It is statically indeterminate to the first degree with respect to the bars (statically detcrminate with respect to the reactions) and we take X = F2
The force influence matrices defined by (7—35) follow from the force results listed on the sketches: F0 =
0; 0; 0; 0; —tan 0; 0}
= {+1; +1; —cosU; —sin 0; —sin 0; —cos0} P20 = P{—l; —tanG; +(an0} =0
Assuming a bar is rigid is equivalent to setting f = mentary energy is due only to the diagonal bars: = yr + = ± e02F2 + We convert
0
for the bar. Then, the comple-
+ f2Fi')
to a function of X by substituting
F1 =
cos 0
± X
F2 = +X Finally,
has the form
=
+
e0,1
+
+
(eoi + e0,2 + f1
tan 0)P +
tan 0 —
+ 4(f1 + f2)X2
Differentiating (e) leads to dIlC =
{[eoi +
d211, = (f1 +
+
(fl
(g)
SEC. 7—6.
STABILITY CRITERIA
The Euler equation follows from (f):
e01 + C02 +f1
(f1
+ f2)X = 0
Comparing (h) with (a) of Example 7—3, we see that the Euler equation for geometric compatibility equation expressed in terms of the force redundant.
7—6.
is the
STABILITY CRITERIA
Section 6—9 dealt with initial stability, i.e., stability of a system under infinitesimal load. We showed there that initial stability is related to rigid body motion. A system is said to be initially unstable when the displacement restraints are insufficient to prevent rigid body motion. In this section, we develop criteria for stability of a system under finite loading. If a linear system is initially stable, it is also stable under a finite loading. However, a nonlinear (either physical or geometrical) system can become unstable under a finite load. We consider first a single mass particle subjected to a system of forces which are in equilibrium. Let Il be the displacement vector defining the equilibrium position. We introduce a differential displacement All, and let AW he the work done by the forces during the displacement All. if A W > 0, the particle energy is increased and motion would ensue. It follows that the equilibrium position (ll) is stable only when AW < 0 for arbitrary All. We consider next a system of particles interconnected by internal restraints. Let AWE be the incremental work done by the external forces and AW1 the incremental work done by the internal restraint forces acting on the particles. The total work, AW, is given by
=
AWE + AW1
The system is stable when A W < (I for all arbitrary permissible displacement increments, that is, for arbitrary increments of the variable displacements. Now, we let AW11 be the work done by the internal restraint forces acting on the restraints. Since —AW1, we can express the stability requirement as (7-38) One can interpret AW0 as the work required to deform the system to the alternate
position and as the actual work done on the system. When the behavior is continuous, we can express and AWE as Taylor series expansions in terms of the displacement increments (see (7—2)):
=
dW0
+ +
+ +
We have shown that the first-order work is zero at an equilibrium position: dWD — dWE
0
VARIATIONAL PRINCIPLES FOR AN IDEAL TRUSS
170
CHAP. 7
If we retain only the first two terms in (b), the general stability condition reduces to d2 WD — d2 W5
for all arbitrary permissible displacement increments
>0
(7—39)
Equation (7—39) is called the "classical stability criterion." Retaining only the first two differentials corresponds to considering only infinitesimal displacement increments. If (7—39) is satisfied, the equilibrium position is stable with respect to an infinitesimal disturbance. In order to determine whether it is stable with respect to a finite disturbance, one must use (7—38). If (7—40)
d2WD = d2WE
for a particular set of displacement increments, the equilibrium position is said position infinitesimally to be neutral, and there exists an alternate displaced from the first position. One can interpret (7—40) as the necessary condition for a bifurcation of equilibrium positions. To show this, suppose U and U represent the displacement components for the two possible equilibrium positions of a system where
Also, let R and We can express
represent the resultant forces corresponding to U and 0. as
= R + dR +
+
Now, the second-order work for the initial equilibrium position is given by d2W
If d2 W =
0
for some finite
d2W5
—
d2WD
=
txUT dR
it follows that
dR = R0AU =
0
The condition
=
0
is equivalent to (7—40). Finally, if we consider
to he infinitesimal,
R=R+dR and (7—40) implies R =
0.
To apply the classical stability criterion to an ideal truss, we note that the first-order work terms have the form dW5 =
P1
AU1
dWD =
where U2, P1 are prescribed. Operating on (a) yields d2W5
0
d2WD
>jFjd2ej +
(7-41)
STABILITY CRITERIA
SEC. 7—6.
and the stability criterion reduces to
stable neutral unstable
for arbitrary nontrivial AU1 d2 WD = 0 for a particular nontrivial AU1 (7—42) d2 WD < 0 for a particular nontrivial AU1 where d2WD is a quadratic form in AU1. We postpone discussing how one transforms (7—41) to a quadratic form in AU1 until the next chapter. When the material is elastic, we can identify (7—39) as the requirement that Fir, be a relative minimum. By definition, d2WD > 0
drIp = dVT
—
dWE
For elastic behavior, dVT = dWD
and it follows that (7—43) d2WD — d2WE = Finally, we can state: An equilibrium position for an elastic system is stable (neutral, un-
stable) if it corresponds to a relative minimum (neutral, indifferent) stationary point of the total potential energy. Example 7—6 The system shown in Fig. E7—6A consists of a rigid bar restrained by a linear elastic spring which can translate freely in the .x2 direction. Points A and A' denote the initial and deformed positions. We will first employ the principle of virtual displacements to establish the equilibrium relations and then investigate the stability of the system. Fig. E7—6A x1
j2
The first-order work terms are dWD = F de
=
P2 du2
(a)
VARIATIONAL PRINCIPLES FOR AN iDEAL TRUSS
172
CHAP. 7
where F, e are the spring force and extension. Since the bar is rigid, the system has only one
degree of freedom, i.e., only one displacement measure is required to define the configuration. It is convenient to take 0 as the displacement measure. The deformation-displacement relations follow from the sketch:
e=
= L(sin
u1
0 — sin
0)
0 — sin
and
de =
(cos 0)L
du, =
(sin 0)L
Using (a) and (d), the principle of virtual displacements takes the form dW0 — dWa
= {F cos 0 —
sin 0) (L AO) =
P2
for arbitrary AG
0
Finally, (e) leads to the equilibrium relation,
F cos 0 =
P2
sin 0
which is just the moment equilibrium condition with respect to point 0. We transform (1) to an equation for ()by substituting for F using (c). The result is sin 0
tan
0=
sin 00
—
Since the system is elastic, dW5
—
and (e) is equivalent to
=
for arbitrary AU
0
The potential energy function for this system has the form
=
P2u2
—
= 4kL2(sin 0 —
sin
0)
and (g) can be interpreted as
= of 00 are plotted in Fig. E7—6B. The result
0
for
0
consists of two curves, defined by
0=
0
cos 0 = P2/kl
for arbitrary P2/kL for (P2/kL)
1
To investigate the stability of an equilibrium position, we have to evaluate the secondorder work at the position. After some algebraic manipulation, we obtain =
d2W0
—
= k(L
AU)2
[cossU —P2/kL] cos 0
Let 0* represent a solution of(g). Applying (m) to 0* results in the following classification:
REFERENCES
stable
COS 0* >
neutral
cos3
unstable
cos3 0'
P2
P2
One can show that (n) is equivalent to stable
neutral
dP2
dP2
0
o
dO
unstable
dP2
A transition from stable to unstable equilibrium occurs at point A, the peak of the deflection curve. The solution for 0 is different in that its stable segment is the linear kL) corresponds to a branch point, solution and the neutral equilibrium point (P2 Both the linear and nonlinear branches are unstable. Fig. E7—6B
0
REFERENCES 1
2. 3. 4. 5. 6. 7.
WANG, C. T.: Applied Elasticity, McGraw-Hill, New York, 1953. LANGHAAR, H. L,: Energy Methods in Applied Mechanics, Wiley, New York, 1962. REISSNeR, E.: "On a Variational Theorem in Elasticity," J. Math. Phys., Vol. 29, pages 90-95, 1950. ARGYRIS, i. H., and S. KIsLseY: Energy Theorems and Structural Analysis, Butterworths, London, 1960. CI.IARLTON, T. M.: Energy Principles in Applied Statics, Blackie, London, 1959. HOFF, N. J.: The Anal vsis of Structures, Wiley, & New York, 1956. K.: Variational Methods in Elasticity and Plasticity, Pergamon Press, 1968.
CHAP. 7
VARIATIONAL PRINCIPLES FOR AN IDEAL TRUSS
174
PROBLEMS
Consider the two-dimensional symmetrical truss shown. Assume = 03 = 0. (a) Determine the first two differentials of e1 and ez by operating on the
7—1.
expanded expression (equation 6—19) for e. (b) (c)
When a b, we can neglect the nonlinear term involving u12 in the Specialize (a) for this case. expressions for e and When a b, we can neglect the nonlinear term involving u11 in the Specialize (a) for this case. expressions for e and Prob. 7—1
x2
T 21
3
7—2. Refer to the figure of Prob. 7—1. Assume = u3 = 0 and a> b. Using the principle of virtual displacements, determine the scalar force-equilibrium equations for joint 1. 7—3. Suppose a force F is expressed in terms of e,
F=
C1e
+ 4C2e3
where a is related to the independent variable u by a
(a)
u + 1u2
Determine the first two differentials of the work function, W = W(u), defined by W
(b)
F de
=
Suppose (a) applies for increasing e and
F=
C1(e
—
e decreasing from e*. Determine d2 W at a = e*. Refer to Prob. 6—23. The n — 1 independent node equations relating the branch currents are represented by 7—4.
ATI
U
Now, the branch potential differences, e, are related to the n — node potentials, V, by
e=
AV
1
independent
PROBLEMS
175
Deduce that the requirement, 1T de
=
for arbitrary
0
is equivalent to (a). Compare this principle with the principle of virtual displacements for an ideal truss. 7—5. Consider the two-dimensional truss shown. Assume u2 = = 0. (a)
(b)
Using (7—14), obtain a relation between the elongations and ü32. Take the virtual-force system as LxF2 and the necessary bar forces and reactions required to equilibrate AF2. Using (7—12), express u11, u12 in terms of e1, e3. Note that bar 2 is
not needed. One should always work with a statically determinate system when applying (7—12). Prob.
7—5
x2
2
7—6. Refer to Prob. 6—23. One can develop a variational principle similar to the principle of virtual forces by operating on the branch potential difference—node potential relations. Show that -
AiTe=0 for any permissible set of current increments. Note that the currents must satisfy the node equations ATi
0
Deduce Kirchhoff's law (the sum of the voltage drops around a closed loop must equal zero) by suitably specializing Lsi in (a). Illustrate for the circuit shown in Prob. 6—6, using branches 1, 2, 4, and 6. 7—7. By definition, the first differential of the strain-energy function due to an increment in U1 has the form
=
n1
dV.
=
F,, de,,
We work with expressed as a compound function of e = e(U) since it is more convenient than expressing V directly in terms of U1. One can also
VARIATIONAL PRINCIPLES FOR AN IDEAL TRUSS
176
write (a)
CHAP 7
as
Using (b), show that the system of if joint force-equilibrium equations expressed in terms of the joint displacements can be written as:
k=
ÔU(k
Equation c is called Castigliano's principle, part I. (b)
Show that an alternate form of (c) is
P(k= Note that (d) is just the expansion of (c). Rework P rob. 7—2, using (d).
7—8. Determine V(e), dv, and d2V for the case where the stress-strain relation has the form (see Prob. 6—10)
a=
—
be3)
7—9. Determine V*(F), dV*, and d2V* for the case where the stress-strain relation has the form
=
(a
+
ca3)
7—10. Show that (7—12) can be written as UkJ
= 0Pkj
is defined by (7—31). This result specialized for U2 = 0 is = called Castigliano's principle, part H. Apply it to Prob. 7—5, part b. Assume linear elastic material and f1 = = = f. 7—11. The current and potential drop for a linear resistance are related by where
ef
Inverting (a), we can express
e0,j
as a function of e1.
= (a)
+
—
which has the property that
Suppose we define a function,
= corresponding to (b).
Determine b
(b)
Let W
where h = total number of branches. Considering
the branch potential drops to be functions of the node potentials, deduce that the actual node potentials V correspond to a stationary value of W. Use the results of Prob. 7—4. The Euler equations for
PROBLEMS
177
W = W(V) are the node current equilibrium equations expressed in (c)
terms of the node potentials. Suppose we define a function
which has the property that
=
(d)
corresponding to (a).
Determine b
(d)
Let W* =
W7. Show that the Euler equations for
H=
iTe
=
—
1T(AV)
—
= H(i,V)
(e)
are the governing equations for a d-c network. Show that the actual currents correspond to a stationary value of One can either introduce the constraint condition, An = 0, in (e) or use the result of Prob. 7—6. 7—12. Investigate the stability of the system shown below. Take k, = aL2k5 (e)
P
Prob.
Linear translational restraint
Rigid rod
ICr
and
consider a to range from 0 to 6.
(Linear rotational restraint)
7—12
8
Displacement Method Ideal Truss 8—1.
GENERAL
The basic equations defining the behavior of an ideal truss consist of forceequilibrium equations and force-displacement relations. One can reduce the system to a set of equations involving only the unknown joint displacements by substituting the force-displacement relations into the force-equilibrium equations. This particular method of solution is called the displacement or method. Alternatively, one can, by eliminating the displacements, reduce the governing equations to a set of equations involving certain bar forces. The latter procedure is referred to as the ,fin'ce or flexibility method. We emphasize that these two methods are just alternate procedures for solving the same basic equations. The displacement method is easier to automate than the force method and has a wider range of application, However, it is a computer-based method, i.e., it is not suited for hand computation. In contrast, the force method is more suited to hand computation than to machine computation. In what follows, we first develop the equations for the displacement method by operating on the governing equations expressed in partitioned form. We then describe a procedure for assembling the necessary system matrices using
only the connectivity table. This procedure follows naturally if one first operates on the unpartitioned equations and then introduces the displacement restraints.
The remaining portion of the chapter is devoted to the treatment of nonlinear behavior. We outline an incremental analysis procedure, apply the classical stability criterion, and finally, discuss linearized stability analysis. 8—2.
The
OPERATION ON THE PARTITIONED EQUATIONS
governing partitioned equations for an ideal truss are developed in
Sec. 6—7. For convenience, we summarize these equations below. 178
SEC. 8—2.
OPERATION ON THE PARTITIONED EQUATIONS
= B1F P2 = B2F
179
eqs.)
P1
(r eqs.) (in eqs.)
F = F, + kA1U1 F, = k(—e0 + A2tJ2)
The unknowns are the in bar forces (F), the r reactions (P2), and the na joint
displacements (U1). One can consider F, to represent the initial bar forces, that is, the bar forces due to the initial elongations and support movements 0. The term kA1U1 represents the bar forces due to U1. When the with U1 material is linear elastic, k and e0 are constant. Also, = BT when the geometry is linear.
We obtain a set of equations relating the flj displacement unknowns, U1, by substituting for F in (a). The resulting matrix equation has the form (B1kA1)U1 =
—
B1F1
We solve (8—i) for U1, determine F from (e), and P2 from (b). The coefficient matrix for U1 is called the system stiffness matrix and written as K11 = B1kA1
(8—2)
as representing the initial joint forces due to the initial One can interpret elongations and support movements with U1 = 0. Then — B1F1 represents the net unbalanced joint forces. When the geometry is linear, K1 reduces to 1
K11 = B1kBT = AfkA1
If the material is linear, k is constant and positive definite for real materials. Then, the stiffness matrix for the linear case is posiLive definite when the system
is initially stable, that is, when r(B1) Conversely, if it is not positive definite, the system is initially unstable. If the material is nonlinear, k and e0 depend on e. We have employed a piecewise linear representation for the force-elongation curve which results in linear relations. However, one has to iterate when the limiting elongation for a segment is exceeded. The geometrically nonlinear case is more difficult since both A and B depend
on U1. One can iterate on (8—1), but this requires solving a nonsymmetrical system of equations. It is more efficient to transform (8—1) to a symmetrical system by transferring some nonlinear terms to the right-hand side. Nonlinear analysis procedures are treated in Sec. 8—4. Even when the behavior is completely linear, the procedure outlined above for generating the system matrices is not efficient for a large structure, since f See Prob. 2—14.
DISPLACEMENT METHOD: IDEAL TRUSS
180
CHAP. 8
it requires the multiplication of large sparse matrices. For example, one obtains the system stiffness matrix by evaluating the triple matrix product,
=AfkA1 One can take account of symmetry and the fact that k is diagonal, but A1 is generally quite sparse. Therefore, what is needed is a method of generating K which does not involve multiplication of large sparse matrices, A method which has proven to be extremely efficient is described in the next section. 8—3.
THE DIRECT STIFFNESS METHOD
We start with (6—37), the force-displacement relation for bar ii:
+
= F0,
= where n4, n. denote the joints at the positive and negative ends of barn. One can consider F0, as the bar force due to the initial elongation with the ends
= 0). Now, we let required to equilibrate the action of
be the external joint force matrices Noting (6—43), we see that
fixed (un,
= =
p,I_
Substituting for
(8—4)
(8—4) expands to
=
+
—
pn_ = One can interpret (b) as end action—joint displacement relations since the elements of ± are the components of the bar force with respect to the
basic frame. Continuing, we let (8—5)
Note that is of order i x i where I = 2 or 3 for a two or three-dimensional truss, respectively. When the geometry is linear, and is sym= y,, metrical. With this notation, (b) takes a more compact form, = =
+
—
—
+
(8—6)
We refer to as the bar stiffness matrix. Equation (8—6) defines the joint forces required for bar n. The total joint forces required are obtained by summing over the bars.
SEC. 8—3.
THE DIRECT STIFFNESS METHOD
We have defined p2, .
—
=
.
X 1) (ii x 1) (U
,
{u1, u2
as the general external joint force and joint displacement matrices. Now, we write the complete system of if joint force-equilibrium equations, expressed in terms of the displacements, as
=
+
(8—7)
We refer to if, which is of order if x as the unrestrained system stiffness matrix. The elements of are the required joint forces due to the initial elongations and represents the required joint forces due to the joint displacements.
We assemble if and in partitioned form, working with successive members. The contributions for member n follow directly from (8—6). (Partitioned Form is j x 1) in row
8—8
in row n.
if (Partitioned Form is j x j) +k, —
k,,
—ku
in row column in row column in row n_, column n
Example 8—1
The connectivity table and general form of if and
for the numbering shown in
Fig. E8—l are presented below: Fig. E8—l
0 4
3
DISPLACEMENT METHOD: IDEAL TRUSS
182
Bar
k1+k2
4
5
2
2
4
2
1
3
3
4
2
+joint
1
—joint
4
—k1
—k2
k2 + k3 + k5
—k2
—k3
k3 + k4
--k3
Pai ,10
P0.2
—k4
k1 + k4 + k0
—k4
—k5
—k1
U4
U3
U2
U1
P2
3
1
CHAP. 8
=
UT
L'
0.1PI —
L'
pT
0,212 r
L'
UT
t'
Po, 3
—
0.
3P3 —
Po, 4
—
0,
1I'I 1
ftT 0,2P2 nT 0,3P3
øT r0,SPS
UT
0,
4P4
uT 0, 41'4
L'
0. 5P5
Example 8—2 The external force matrix, involves and the displacement matrices for those joints connected to joint j by bars. Now, corresponds to row j and ii,, to column j of ir. By suitably numbering the joints, one can restrict the finite elements of X' to a zone about the diagonal. This is quite desirable from a computational point of view. Fig. E8—2
Sect. 1
0
ft® 2:
(71
fs
(-I -..- —--
/
\
3
%._ —--
L6 ',.
/
;
©
-.
Consider the structure shown. We group the vertical joints into sections. The equilibrium equations for section k involve only the joints in section k and the adjacent sections. For example, the equations for section 3 (which correspond to P6) will involve only the displacement matrices for sections 2, 3, 4. This suggests that we number the joints by section. The unpartitioned stiffness matrix corresponding to the above numbering scheme
THE DIRECT STIFFNESS METHOD
SEC. 8—3.
is listed below. Note that has the form of a quasi-tridiagonal band matrix when it is partitioned according to sections rather than individual joints. The submatrices for this truss are of order 4 x 4, U'
U2
k, +k2 —k, P2
U3
U6
U7
U8
I—k2
k,+k3 —k3
—k,
U4
—k4
+k4 —k2
—k3
k2+k3
—k5
—k6
—k7
I
+k6 —k4
—k,
k4+k51
—k8
k6+k9 —k9
—k6
—k10
+k,0 p6
—k7
k8+k7
—k9
—k8
—k1,
—k,2
+k9
I
+k,2 —k,0
JJ7
—k,,
k10+k,, —k,, 3
—k,,
p8
—k,3
k,2+k,,
The introduction of displacement restraints involves first transforming the partitioned elements and to local frames associated with the restraints, permuting the actual rows, and finally partitioning the actual rows. The steps are indicated below.
-+ U -+
We write the system of joint force-equilibrium equations referred to the local
joint frames as
=
+
(8—10)
The transformation la'vs for the submatrices of
= =
€,n= 1,2,...,j
and T
follow from (6—57). (8—11)
DISPLACEMENT METHOD: IDEAL TRUSS
184
The step,
—+
P,
CHAP. 8
involves only a rearrangement of the rows of
We
obtain the corresponding stiffness matrix, K, by performing the same operations on both the rows and columns of The rearranged system of equations is written as P = KU + P0 (8-12) Finally, we express (8—12) in partitioned form:
= P2 =
K11U1
+ K1202 + P0,1 + K2202 + P0,2
8
13
The first equation in (8—13) is identical to (8—1).
Example
8—3
It is of interest to express the partitioned elements of K in terms of the geometrical, We start with the general Unconnectivity, and displacement transformation partitioned equations(6—28), (6—40), and (6—44), (6—50):
F0 + kda/1 = F0 + kyC'W
F
Then, substituting for F in (a) and equating the result to (8—7) leads to
= =
The matrix, DTkY, is a quasi-diagonal matrix of order im. The diagonal submatrices arc of order i, and the submatrix at location n has the form, We have defined this product as k,,. Then, if we let
=
[ki I. r,T
2p272
k5
we can express
as
= CTk5c Carrying out (8—9) for n = 1, 2 m is the same as evaluating the triple matrix product. Obviously, (8—9) is more efficient than (f).
The introduction of displacement restraints can be represented as
P= 11
= =
D1dP
(g)
and
= DTU = DfU1 +
(h)
THE DIRECT STIFFNESS METHOD
SEC. 8—3.
185
Substituting (g) and (h) in (8—7) and equating the result to (8—13), we obtain
=
K,, =
=
P0.
t—12
DsCTDke0
—
In order to obtain (8—13), we must rearrange the rows and columns of
then partition. This operation is quite time-consuming. Also, it leads to rectangular submatrices. In what follows, we describe a procedure for introducing displacement restraints which avoids these difficulties. We start with the complete system of equations referred to the basic frame, and
(8—14)
We assemble
and
using (8—8) and (8—9). Then, we add to
the
external force matrices for those joints which are unrestrained. It remains to modify the rows and columns corresponding to joints which are either fully or partially restrained. Case A: Fit!! Restraint
is unknown. We replace the equation for Pq by
Suppose uq = Uq. Then
=
Uq
This involves the following operations on the submatrices of X and
On X. Set off diagonal matrix elements in row q and column q equal 1. to 0 and the diagonal matrix element equal to I,.
I=
0
(8—15)
= Ii 2.
On
Add terms in
due to
t
C
X(qUq (8—16)
j ease B: Partial Restraint—Local Frame
We suppose the rth element in
is
prescribed.
= prescribed = = unknown
186
DISPLACEMENT METHOD: IDEAL TRUSS
We have to delete the equation corresponding to
CHAP. 8
and replace it with
4= Step I —Assemblage of Basic Matrices
according to the following:
We assemble Eq, Gq, ui', 1.
Eq and Gq. We start with
G=O, and we set G,r
+1
2. us'. We start with an ith-order column vector having zero elements and we set the element in the rth row equal to Note that this matrix involves only the prescribed displacements (local frame) in their natural locations. We start with an ith-order column vector having zero elements and 3. we insert the values of the prescribed joint forces (local frame) in their natural locations. Note that the elements corresponding to the reactions are zero.
When the joint is fully restrained,
E=O, Suppose joint
G=11
5 is partially restrained, The data consist of:
The rotation matrix, R°5, defining the direction of the local frame at 5 with respect to the basic frame. (b) The direction (or directions) of the displacement restraint and the value (or values) of the prescribed displacement. (a)
direction r, (c)
The values of the prescribed joint forces:
j=1,...,i As an illustration, suppose r =
2.
Then, in (b), we read in
In (c), we read in —5
—5
Psi
The four basic matrices are (for r =
E5=
P53 2)
1
0
0
[0
0
0
0
0
0
Gs=IO
1
0
0
0
1
[0 0 0
THE DIRECT STIFFNESS METHOD
SEC. 8—3.
187
1°
In
—5
Step 2—Operation on Jr and 1.
Premultiply row q of it" and
by
= " N, q 2.
e
" N, q
'-'q 1%
Postmultiply column q of it" by —
— 1' 2' —
.
T11* and add to PPN.
= 3.
Postmultiply column q of it" by (Eq
irtq = 4.
1,
2, .
.
.
Add Gq to irqq
= it"qq + Gq 5.
and
Add
to
=
P'N,q + U
+
The operation on row q and column q are summarized below.
On Jr = X'qq =
.YV'eq(E9R0")T
+
(Eq R°").Y(qq(Eq
— —
1'
2
I
Oiz 2PN,
q
I?'N,q
=
R0q,
—
1, 2,. . . ,j
(8—18)
+
+0
When ir is symmetrical (this will be the case when the system is geometrically linear), we can work only with the submatrices on and above the diagonal. The contracted operations for, the symmetrical case are threefold:
=
—
it"eq(EqR°")T
€= 1,2,...,q.— 1
(8—19)
CHAP. 8
DISPLACEMENT METHOD: IDEAL TRUSS
188
Ttlq*
q—
q
it'qq =
+
+
(8—20)
+ Gq —
*
—
=
(8—21)
The operations outlined above are carried out for each restrained joint. Note that the modifications for joint q involve only row q and column q. We denote the mOdified system of equations by
=
(8—22)
will be Equation (8—22) represents if equations. The coefficient matrix nonsingular when K1 is nonsingular. To show this, we start with the first equation in (8—13) and an additional set of r dummy equations: 1
[K11
Olfuil
-fJ
f—P0,1 —
—
—
-
+
N
Equation (a) represents 1/equations. This system is transformed to (8—22) when to d?tJ, we permute U, They are related by (sec (6—63)) [1°?, J
U
= rVp
where H is a permutation matrix. It follows that
= HT[K11
and, since H is an orthogonal matrix,
=
(8—23)
IK11I
It is more convenient to work with (8—22) rather than (a) since the solution of (8—22) yields the joint displacement matrices listed in their natural order, that is, according to increasing joint number. Once ciii' is known, we convert the joint displacement matrices to the basic frame, using uq =
The bar forces are determined from
F,, = F0,,, + Next, we calculate F,,
and assemble
—
in partitioned form by summing the
THE DiRECT STIFFNESS METHOD
SEC. 8—3.
189
contribution for each member. For member n, we put (see (8—4)) in row n+ in row n_
+ FOIIPf —
Once is known, we convert the force matrix for each partially restrained joint to the local joint reference frame, using
= required to equilibrate the bar forces. This operation The final result is provides a static check on the solution in addition to furnishing the reactions. When the problem is geometrically nonlinear, y,, and depend on the joint displacements. In this case, it is generally more efficient to apply an incremental formulation rather than iterate on (8—22). Example
8—4
We illustrate these operations for the truss shown in Fig. E8—4. Fig. E8—4
/
/50
Ii
1.
Member-Joint Connectivity Table Bar(n)
2.
1
2
3
4
5
+joint(n+)
1
3
1
3
—joint(n...)
2
1
4
2
6
7
8
9
10
11
4
3
5
3
5
6
5
2
4
3
6
4
4
6
Assemblage of
We consider the geometry to be linear. Then, (8—9) results in
listed below.
=
and
=
Applying
DISPLACEMENT METHOD: DEAL TRUSS
190
N
2
1
k1+k2+k3
—k1
—
4
3
—k2
—k3
—k4
—k5
CHAP. 8
51
6
0
k1+k4+k5
2
—k1
3
—k2
---k4
k2+k4+k6 +k7+k8
—k6
—k7
-— k3
4
—k3
—k5
—k6
k3+k5±k6 +k9+k10
—k9
—k10
—k7
—k9
+k7+k9
—k11
5
+k11
0
—
.
—k8
6
—k10
—k11
+k8+k10 +k11
Note that i( 3.
is
symmetrical and quasi-tridiagonal. with submatrices of order 4 x 4.
Introduction of Joint Displacement Restraints
The original equations are =
=
where contains the external joint forces. We start with i?PN If joint q is un— restrained, we put in row q If joint q is fully restrained, we modify and according to (8—15) and (8—16). Finally, if joint q is partially restrained, we use (8—19) through (8—21). Since is symmetrical, we have to list only the submatrices on and above the diagonal. It is convenient to work with successive joint numbers. For this system,
joint 2 is fully restrained and joints 4, 6 are partially restrained. The basic matrices for joints 4,6 and the initial and final forms of.Yt', are listed below. Note that this procedure does not destroy the banding of the stiffness matrix. Joint 4
R°4
(u42 is prescribed)
2
ri
E4=[
[0
01
G4=[0
oj
= {O,ii42j Joint 1
ri E6=[0 — —
6
[
=
(t42 is prescribed) 1
ii ii
01
oJ =
0
INCREMENTAL FORMULATION
SEC. 8—4.
Initial matrices (ir and ("1)
(Us)
(U4)
)r33
ir34
=
ir22 ir4,5
.X44
Sym
Final matrices (ir* and (ui)
(u,)
(U4)
(U3)
o
ir,4E4
1,0
0
.K13
(Us)
U
.K34E4
J35
E4ir46(E6R°6)T
,
—
—
+
+ G4
+
ir56(E6R°6)T
—
(E6R°6)iq66(E6R°6)T
E6R°6(—
--
+ G6
8—4.
INCREMENTAL FORMULATION; CLASSICAL STABILITY CRITERION
Equations (8—13), (8—22) are valid for both linear and nonlinear behavior. However, it is more efficient with respect to computational effort to employ an incremental formulation when the system is nonlinear. With an incremental formulation, one applies the load in increments and determines the corresponding displacements. The total displacement is obtained by summing the displacement increments. An incremental loading procedure can
also be used with (8—13) but, in this case, one is working with total displacement
rather than with incremental displacement. In this section, we develop a set of equations relating the external load and the resulting incremental displacements. These equations are also nonlinear, but if one works with small
load increments, the equations can be linearized. Our approach will be similar to that followed previously. We first establish incremental member force-displacement relations and then apply the direct stiffness method to
DISPLACEMENT METHOD: IDEAL TRUSS
192
CHAP. S
the incremental system equations. We complete the section with a discussion of the classical stability criterion. We start with (8—4), which defines the external joint forces required to equilibrate the action of the force for bar n, generate
Pn*
=
=
p,,
Equations (a) are satisfied at an equilibrium position. We suppose an in-
cremental external load AP is applied and define AU as the resulting incremental
displacement for the new equilibrium position. Since F and depend on U, their values will change. Letting AF, AD be the total increments in F, D due to AU, and requiring (a) to be satisfied at both positions, leads to the following incremental force-equilibrium equations: =
Afif +
+
(8—24)
Ap,,.. =
To proceed further, we need to evaluate the increments in e and
D.
The exact
relations are given by (6—22):
= — ci,,
u,,_)
—
+
—
—
—
=
To allow for the possibility of retaining only certain nonlinear terms, we write (a) as fi,, —
= = =
u)Tg,
(u,,÷ —
—
u,,..) +
—
u,,)
(8—25)
Yh(ufl. — u,,_)
If all the nonlinear terms are retained, g,,
=
To neglect a particular displacement component, we delete the corresponding element in For geometrically linear behavior, = 0. Operating on (8—25),
we obtain dv,, =
Au,,)Tg,,
—
(8—26)
and Ae,, =
= d2e,,
+ Au,,)
—
(827)
—
It remains to evaluate AF,,. We allow for a piecewise linear material and employ the relationst developed in Sec. 6—4. For convenience, we drop all the t See (6—31), (6—32), and (6—33).
INCREMENTAL FORMULATION
SEC. 8—4.
193
notation pertaining to a segment and write the "generalized" incremental
expression in the simple form = k(Ae —
(8—28)
where k, are constant for a segment. They have to be changed if the limit is unknown, one of the segment is exceeded or the bar is unloading. Since has to iterate, taking the values of k, Ae0 corresponding to the initial equilibrium position as the first estimate. This is equivalent to using the tangent stiffness. The initial elongation, Ae0, is included to allow for an incremental temperature change. Substituting for (8—28) takes the form
At' _AC' — Finally, we substitute for
+
Q
Lw,, in
LXPn+ =
—
+ if
I
72 Ct
(8—29)
(8—24) and group the terms as follows:
&i,..) +
+
(8—30)
where
= Fag,, + =
(8—31)
=
+
+
-i-
We interpret k7 as the tangent stiffness matrix. The vector, L\p9, contains linear, quadratic, and cubic terms in We have included the subscript g to indicate that it is a nonlinear geometric term. We write the total set of incremental joint equilibrium equations as
+ M'0 +
=
(8-32)
is assembled using (8—9) and MPO + with (8—8). Note that is symmetrical. Finally, we introduce the displacement restraints by ap-
where
plying (8—19)—(8—21). The modified equations are
=
—
(8-33)
—
It is convenient to include the prescribed incremental support displacement terms in involves only the incremental temperature and so that the variable displacement increments. The contracted equations are K1,11 AU1 =
—
AP0,1
—
AP9,1
—
1(1,12
AU2
(8-34)
is symmetrical. We cannot solve (8—33) directly for
where K1,
since contains quadratic and cubic terms in MI. There are a number of techniques for solving nonlinear
algebraic equations. t We describe here the method of successive substitutions, t See Ref. 12.
CHAP. 8
DISPLACEMENT METHOD: IDEAL TRUSS
194
which is the easiest to implement, but its convergence rate is slower in com-
parison to most of the other methods. First, we note that and are independent of A'1/1. They depend only on the initial equilibrium position and the incremental loading. We combine and and write (8—33) as X7K MI1 =
(8-35)
—
Now, we let represent the nth estimate for LXa/IJ and determine the (n + 1)th estimate by solving )p* L\cW(n+
=
— &?P
(8—36)
The iteration involves only evaluation of and back-substitution once is transformed to a triangular matrix. The factor method is particularly convenient since X7 is symmetrical. With this method, STS
(8—37)
where S is an upper triangular matrix. We replace (8—36) with
= STQ = A9* S
(8—38) —
In linearized incremental analysis, we delete solution of
in (8—35) and take the
(8-39)
.
as the "actual" displacement increment. One can interpret this scheme as one cycle successive substitution. The solution degenerates when the tangent stiffness matrix becomes singular.
To investigate the behavior in the neighborhood of this point, we apply the classical stability criterion developed in Sec. 7—6. The appropriate form for a truss is given by (7—41):
±
for arbitrary AU1 with AU2 =
den)> 0
0 (a)
We have already evaluated the above terms. Using (8—26), (8—27), and (8—29) with Ae0 = 0,
+
—
—
Au,,)
(b)
and (a) can be written as d2WD
It follows that
ALT ICE,
AU1 > 0
for arbitrary AU1
(8—40)
must be positive definite for a stable equilibrium position.
¶ Iterative techniques are discussed in greater detail in Secs. 18—7, 18—8, 18—9.
_____ iNCREMENTAL FORMULATION
SEC. 8—4.
But K1,
and
are related by HT[Kt.11
(8—41)
where H is a nonsingular permutation matrix, which rearranges the elements of according to (8-42) =H Then, and K1 have the same definiteness Finally, we can classify the stability of an equilibrium position in terms of the determinant of the tangent stiffness matrix: D
Example
= 1K1,
=
D>O
stable neutral unstable
iii
D=0 D< 0
(8—43)
8—5
We illustrate the application of both the total (8—13) and incremental (8—34) formulations
to the truss shown in Fig. E8—5A. To simplify the analysis, we suppose the material is linearly elastic, k1 = k2 = k, and there are no initial elongations or support movement.
b <
Fig. E8—5A
x2
d_____
d
The initial direction cosines for the bars are —b]
—b]
The deformed geometric measures are defined by (8—25). They reduce to
+
fl = 1,2
= = cc, +
for this example. Since b cc d, we can neglect the nonlinear terms due to u11, i.e., we can take 1 [0 0 1
t See Prob. 2—13 for a proof. See Sec. 2—5.
196
DISPLACEMENT METHOD: IDEAL TRUSS
CHAP. 8
Using (c), 1 I
Pi
—b
+ u12]
12
—b
+
112
—b +
Continuing, the bar force-displacement relations are
=
n = 1,
= ky,,u1
2
Finally, the force-equilibrium equation for joint I follows by applying (8—6) to both bars.
=
(k1
+
k2)u1 = k(11r71 ± 11212)U1
Equations (e) and (1) expand to
Jusi 0
—
u12)(b
and
F1 =
—
F2 =
(h
—
—
(b
—
The diagonal form of the coefficient matrix is due to the fact that we neglected u11 in the expressions for y and This approximation uncouples the equations. Note that (g) is the first equation in (8—13) with U2 and P0 set to 0. Solving the first equationt in (g). we obtain
I (J\2
Pti
The corresponding bar forces are Pi
= This result is actually the solution for the linear geometric case. The expression for and the corresponding bar forces follow from the second equation
[2k —
P12
t Equation (g) is (8—1) with F1 =
0.
u12)(b
—
1
1u12)j 012
INCREMENTAL FORMULATION
SEC. 5—4. F1
=
=
F2
k
— — (b
L
2u12)u12
—
=
1
197 P12
2b—u12 L
We can write (k) as L2 U12
=—— 2k (b
1
u12)(b
—
and solve (m) by iteration. Alternatively, one can specify u12 and evaluate from (k). The latter approach works only when there is one variable. The solution is plotted in Fig. E8-5B.
Fig. E8—5B
P12 A 9
(I + 0 b
B
We describe next the generation of the incremental equations which follow from (8—26)— (8—32). Applying (8—26), (8—27) to (b)—(d) results in
= [o
= d[i,, = de,
=
d2e, =
a,1 Au11
Au1
=
+
+
(Au1 )2
n=l,2 We arc assuming no initial elongation. Then, k
=
k(de,,
+
The tangent stiffness matrix and incremental geometric load term are defined by (8—31). Using (n), we obtain + u12)
= Sym
+ u12)2 + (Au12)2
DISPLACEMENT METHOD: DEAL TRUSS
198
CHAP. 8
Finally, we assemble the incremental equilibrium equations for joint 1 using (8—30).
+ k, 2)Au1 + Ap1, i +
Ap1 =
o
?4(_b + u12)2 +
(r)
Al)9. 2
+ F2) 0 —
[3(—b + u12) + Au12]}
Note that (s) is (8—34). Also, the incremental equations are uncoupled. We restrict the analysis to only 112 loading. Setting F1 = F2 in (s) results in
+
+ u17)2)] Au12 =
[Au12 + 3(—b +
Ap12
(t)
where F is determined from (e). The coefficient of Au12 is the tangent stiffness with respect to u12. du12
=
+
L
+ u12)2)
L
(u)
Applying the classical stability criterion (8—43) to (t), we see that
> du12
stable neutral unstable
0
0
<
0
(v)
Points A, B are stability transition points and the segment A-B is unstable. If k = 0, the truss is neutral with respect to Au11. Now there is a discontinuity in k at F = — Feb, the pin-ended Euler load, when the material is linearly elastic:
k=0
(w)
,t2EI3
To determine whether the members buckle before point A is reached, we compare F4 With — Feb. Using (u),
F4 =
—
\2
AE(
AEb2
)=
(x)
Then, for system instability rather than member instability to occur, b must satisfy b
where p is the radius of gyration of the section.
=
(y)
INCREMENTAL FORMULATION
SEC. 8—4.
199
Lastly, we outline how one applies the method of successive substitution to (t). For convenience, we drop the subscripts and write (t) as
Ltu =
In the first step, we take Ap9 =
—
0.
=—
(aa)
The second estimate is determined from
=
(bb)
Generalizing (bb),
=
—
The convergence is illustrated in Fig. E8—5C. Case (b) shows how the scheme diverges Fig. ES—5C
—
(a)
Lip
Ib)
in the vicinity of a neutral point = 0). Convergence generally degenerates as and one has to resort to an alternate method.
0
DISPLACEMENT METHOD: IDEAL TRUSS
200
8—5.
CHAR 8
LINEARIZED STABILITY ANALYSIS
In the previous section, we illustrated the behavior of a geometrically nonlinear system. The analysis involves first solving the nonlinear equilibrium equations for the displacements and then applying the classical stability criterion to determine the stability of a particular equilibrium position. Once the nonlinear equilibrium equations are solved, the stability can be readily determined. Now, if a geometrically nonlinear system is loaded in such a way that it behaves as if it were geometrically linear, we can neglect the displacement that is, we can take = in the expression for terms in This approximation is quite convenient since we have only to solve the linear problem in order to apply the stability criterion. We refer to this procedure as linearized stability analysis.
According to (8—40), an equilibrium position is stable ineutral, unstable) when the tangent stiffness matrix is positive definite (positive semi-definite, intransform to and test different). We generate for positive definiteness. We have shown that and K,, u have the same definiteness property, i.e., K1, is positive definite if,K7 is positive definite. Working with i(7 rather than K,, avoids having to permute the rows and columns. In linearized stability analysis, we approximate k, with k,,
=
+ F,,g,,
(8—44)
The first term is the linear stiffness matrix. We interpret the second term as a geometric stiffness. The bar forces are determined from a linear analysis of the truss. If the loading is defined in terms of a single load parameter, we can write (8—44) as
F,, = k,,,, =
(8—45) k1,,,
+ ).k9,
The tangent stiffness matrix is generated by applying the Direct Stiffness Method to each term in (8—45). We express the actual and modified matrices as
=
K,,
and
=
K1,
,i + Kg, ii +
(8—46)
(8—47)
where K1 is the system stiffness matrix for linear behavior. It is symmetrical and positive definite when the system is initially stable. The geometrical stiffness, K9, is also symmetrical but it may not be positive definite. Equation (8—46) shows that the tangent stiffness matrix varies linearly with the load parameter. If the system is initially stable, K,, is positive definite for 2 = 0. As 2 is increased, a transition from stable to neutral equilibrium may occur at some load level, say 2cr To determine we note that neutral equilibrium (see (8—43)) corresponds to K,, = 0 which, in turn, can be interpreted as the existence of a 'non-trivial solution of K,,
=0
(a)
SEC. 8—5.
LINEARIZED STABILITY ANALYSIS
Substituting for K1,
transforms (a) to a characteristic value problem,
AU1 = and
201
—).K5,11 AU1
(8—48)
is the smallest eigenvaluet of (8—48). we can work with =
Since 1K1, itt
Mt'
=
0
1,1
Ault' =
Mlii
(8—49)
instead of (8—48). Both equations lead to the same value of (8—49) has r additional characteristic values equal to — r dummy equations. To show this, we substitute for note (8—42). 11T [ICe,
[o
0 1JAU51 U
Premultiplying by H =
AHT
— —
I
However, since we have added using (8—41) and
[Kg, ii
01 5Au1
[o
i,j
l,,AU2
(a) becomes
K1. ii AU1 = '2Kg. ii AU1
AU2 =
—2 AU2
(na eqs.)
(b)
(r eqs.)
(c)
The solution of (c) is
1
AU2C1
0
0
o
i
+C2 o
This solution must be disregardcd since AU2 is actually a null matrix.
Example
8—6
Consider the system shown. We suppose the bars are identical, the material is linearly elastic, and there is no support movement, The geometry change is negligible under a vertical load and we can use the linearized stability criterion. Working with the undeformcd geometry, we have
F1 =
F2
=
F
=
—
=
—
Matrix iteration (Ref. 1) is a convenient computational scheme for determining 2,,. We apply
jtto (—K9,,1)AU1 =
which satisfies the restrictions on the method,
AU1
DISPLACEMENT METHOD: IDEAL TRUSS
202
CHAP. &
Fig. £8—B
A
d <
12
We let k1 = k2 = k. The system stiffness matrices follow from (8—44) and (8—45).
=
+
+ k€ 2 = 0
(b)
— 21
and
=
k8,1
+ k9,2 =
(g5
—
+ g2)
(c)
It remains to determine g1 and g2 which are defined by
=
+ ufg,,
We neglect u12 in the general expression for approximate expressions for fi, and g, arc L
—b]
(d)
This is reasonable when d K< b. The
—b]
L
i[l
(e)
0 0 0
0
(d)2 = 11e,ii + AKçj,ji =
2k 0
(b)2
(g)
0
0
SEC. 8—5.
LINEARIZED STABILITY ANALYSIS
Neutral equilibrium (K1,
203
is semidefinite) occurs at 2kb
\\L)
L
\,LJ
Note that (g) has only one eigenvalue instead of two. This is a consequence of our using approximate expressions for Equations (e) instead of the exact expressions. At 2 = the system is neutral with respect to Au1 i.e., the buckling mode is antisymmetric. Neutral equilibrium also occurs when the bars either buckle or yield. The value of 2 for Euler buckling of the bars is 2b
eb
= — L
2b k2EIl
=
2AEb/irp'\2
L \L
LLL2J I
Comparing (h) and (i), we see that Euler buckling of the bars controls when
d> irp The exact expression for g, is
I,.
If we work with (k), Kg,jj =
(d)2 K1,
= K,,
=
+
2k
(b)2
—2
0
In this case, there are two characteristic values and therefore two critical values of 2. 1
= 2kb
2cr. 2
2kb
(d\2 (Ii)
C)
Acri
The second root corresponds to neutral equilibrium with respect to Au12. For this example, b, and the first root defines the critical load. It is of interest to compare 2cr 2 with the buckling load found in Example 8—5. There we considered d b and followed the nonlinear behavior up to the point at which the slope of the P12 — u12 curve vanished (neutral with respect to Au12): d
1(1
=
du12
=
0
(0)
mex
The linearized result is significantly higher than the true buckling load. In general, the linear buckling load is an upper bound. How close it is to the actual value will depend on the geometry and loading. When d << b, it is quite close, while it considerably overestimates the true load for d b.
CHAP. 8
DISPLACEMENT METHOD: IDEAL TRUSS
204
REFERENCES
C. FL, and J. B. WILCUR Elementary Structural Analysis, McGraw-Hill, New York, 1960. 2. HALL, A. S.. and R. W. W00DHEAD: Frame Analysis. Wiley, New York, 1967, 3. ARGYRIS, J. H. and S. Kelsey: Energy Theorems and StruciuralAnalysis, Butterworths, London, 1960. 4. LIVESLEY, R. K.: Matrix Methods of Structural Analysis, Pergamon Press, 1964. 5. DC VEUBEKE, B. F., Matrix Methods of Structural Analysis, Pergamon Press, 1964. 6. MARTIN, H. C. Introduction to Matris Methods of StructuralAnalysis, McGraw-Hill, New York, 1965. 7. ARGYSIS, J. H.: Recent Advances in Matrix Methods of Structural Analysis, Pergamon Press, 1964. 8. RUBINSTEIN, M. F.: Matric Computer Analvsis of Structures, Prentice-Hall, 1966. 9, PRZEMIENIECKI, J. S.: Theory of Matrix Structural Aizah'sis, McGraw-Hill, 1968. 10. THOMPSON, J. M. T. and A. C. WALKER: "The Nonlinear Perturbation Analysis of Discrete Structural Systems," Jut. J. Solids Structures, Vol. 4, 1968, pp. 757—768. 11. RUBINSTEIN, M. F.: Structural ,S'vsse,'ns—Statics. Dynamics, and Stability, PrenticeHall, 1970. 12. RALSTON, A.: A First Course in Numerical Analysis, McGraw-Hill, 1965. 1.
PROBLEMS 8—1.
Consider U2 and P1 to he prescribed and the behavior to be physically
linear. (a)
—
Express
= Vr VT
(b)
—
PTU1
in terms of U1, U2. Use —
=
Show that (8-4) are the Euler equations for dVT = FTde
(c) 8—2.
(a) (b)
eo)Tk(e
e0,1)2 =
de =
e0)
Note that
BfAT.11
Express as a quadratic form in AU1. Hint: Obtain d2e by operating on (7—8).
For the structure sketched: Determine K11.
Determine and F due to a temperature increase of 100°F for all the bars. Assume no support movements at joints 2, 3, 4. 8—3. For the structure sketched: Determine the displacements, bar forces, and reactions. 8—4.
Refer to Example 8—2. Suppose we number the joints as shown.
Develop the general form 8—5.
For the structure sketched, determine
of Example 8—2.
\EaJ For the structure sketched: Develop the general form of Indicate how you would obtain K11. 8—6.
PROBLEMS
205
Prob. 8—2
E=3X
ksi
Bar areas = 3
Coefficient of thermal expansion = 6
X 106/°F
Prob. 20'
10'
E=3X
I
8—3
ksi
Initial elongation of
=
Horizontal displacement of joint 2 =
in.
to the left
6 kips
I
x2
Prob. 8—4
0
©
807
0
605
206
DISPLACEMENT METHOD: IDEAL TRUSS
CHAP. 8
Prob. 8—5
20'
E constant for all bars
I
Bar
Area
I
3a
2
4a
3
3a
4
4a
5
2.5a
Y3
Prob. 8-6
8—7. Determine the load-deflection relation for the system shown. Consider the material to be linearly elastic and the bars to bc identical, Assume no initial elongation or support movement. 8—8. Investigate the elastic stability of the system shown. Assume the material is linearly elastic and no support movements. Use the linearized stability criterion and work with the exact expression for Rework the
problem, considering d b and using the corresponding approximate expression for 8—9. Determine the lowest critical load for the truss shown. Assume the material is linearly elastic and all bars have the same stiffness.
wp
Prob.8—7
0
d <
k1 —Ic2
_AE —-•y-
x1
H
Prob. 8—S
K2 2¼
Prob. 8—9
T 20'
H
DISPLACEMENT METHOD: IDEAL TRUSS
208
CHAP. 8
8—10. The governing equations for geometrically nonlinear behavior of a linearly elastic discrete system such as a truss are nonlinear algebraic equations containing up to third-degree displacement terms. We have expressed them as
+
=
±
=
contain linear displacement terms. This form is dictated by our choice of matrix notation. In order to expand (a), we must shift from matrix to indicial notation. For convenience, we employ the summation convention. If a subscript is repeated in a term, it is understood the term is summed over the range of the repeated subscript. An example is
where d,
(j = 1,
2
n)
We write the ith equilibrium equation for tile system as (this representation is suggested in Ref. 8-40):
=
+ KIJkU,, +
2P1
k, range over the total number of unknowns, U1 is the total value of the jth displacement unknown, 2 is a load parameter, defines the load distribution, and the K's are constants which can be interpreted as second-, third, and fourth-order tensors. The second-order tensor, is the linear
where i, j,
stiffness matrix. (a)
We generate the system tensors by superimposing the contribution of each bar. The first step involves converting the matrix expressions
=
p,,
(d)
p,, - = — p,,
where
k,,e,, + F0,,,
F,,
e,, =
'y,,(u,,÷ —
— u)T
+
=
—
+
over to indicial form. We drop the n subscript, define p and u as
(f)
(U,,J
l.Pn_j
and write (d) in the form
=
(k,1
+
+
+ Po,
PROBLEMS
209
Show that k11 =
=L
+
(Ii)
{::} = cu
(i)
= where c is defined by [1
Discuss how you would locate the appropriate addresses for the bar stiffness tensors in the system tensors. What symmetry properties do the k's exhibit? Do these properties also apply for the system tensors? (b) Develop the incremental equations relating Au, AA and compare with (8—30).
(c) 8—11.
Specialize the incremental equations for linearized stability analysis. For the structure sketched: Prob. 8—11
Linearly elastic material. No support movement or initial elongation. Ic1
(a)
=
=k
Determine the nonlinear incremental equilibrium equations at the equi0, P2 P2, Cr, the linearized critical load. Take Ap1 = 0 and solve for Ap2 as a function of Au1. Comment on how the system behaves when a small horizontal load, Pi ±ep2, is applied in addition to
librium position corresponding to Pi = (b)
9
Force Method Ideal Truss 9—i.
GENERAL
The basic equations for the linear geometric case have the form
P1 =
B1F
e = BfU1 +
e0 + fF
P2 = B2F where the elements of B1 and B2 are constants. Equation (a) represents linear equations relating the na prescribed joint forces and the in unknown bar forces.
that is, the rows of B1 must For the system to be initially stable, r(B1) = be linearly independent. This requires in In what follows, we consider the system is said to be statically determinate only stable systems. If in = since one can find the bar forces and reactions using only the equations of statics. The defect of (a) is equal to in — nd = q, and is called the degree of indeterminacy. One can solve (a) for na bar forces in terms of the applied forces and q bar forces. We refer to the system defined by the na bars as the primary structure and the q unknown forces as force redundants. In order to determine F, q additional equations relating the bar forces are required. These equations are called compatibility conditions and are obtained by operating on (b) which
represents m relations between the na unknown displacements and the bar forces.
The general procedure outlined above is called the Jbrce or flexibility method. This procedure is applicable only when the geometry is linear. In what follows, we first develop the governing equations for the force method by operating on (a)—(c). We then show how one can establish the compatibility equations using the principle of virtual forces and discuss the extremal character of the force redundants. Finally, we compare the force method for a truss with the mesh method for an electrical network. 210
SEC. 9—2.
9-2.
GOVERNING EQUATIONS—ALGEBRAIC APPROACH
211
GOVERNING EQUATIONS—ALGEBRAIC APPROACH
We consider the first columns of B1 to be linearly independent (if the system is initially stable, one can always renumber the bars such that this
condition is satisfied) and partition B1, B2 and F as follows:
=[B11
B1 x nil
ii
B12]
"a)
[B21
B2 (rxm)
(na x q)
I
B22]
(rxq)
(rxn4) (na x 1)
= F2 (qx 1)
The bars corresponding to F1 comprise the primary structure and F2 contains the q redundant bar forces. Using (9—1), the force-equilibrium equations ((a) and (c)) take the form B11F1 =
P2 =
P1 — B12F2
+ B22F2
B21F1
(ad eqs.)
(9—2)
(reqs.)
(9—3)
Since Bit! 0, we can solve (9—2) for F1, considering P1, —B12 as righthand sides. The complete set of q + 1 solutions is written as
F1 = F1,0 + Fl,F,F2
(9—4)
where F1.0 and F1, F2 satisfy
1111F10 = P1 B1IFI,F2 = —B12
Note that the kth column of F1, F2 contains the bar forces in the primary structure due to a unit value of the kth element in F2. Also, F1, contains the bar forces in the primary structure due to the applied joint loads, P1, with
F2 =
0.
The reactions follow from (9—3):
P2 = P2,o =
P2,0
+
B21F1,0
(9—6)
= B21FI,F2 + B22 We consider next (b). Partitioning e, e0, and f, (fl4X 1)
(nixl) e
=
c
e1
=
——— C2
j (q< 1) ii
L 0
f2 (q x q)
(9—7)
CHAP. 9
FORCE METHOD: IDEAL TRUSS
212
and using (9—1), the force-displacement relations expand to
= e1, o + f1F1 B11U1 + B21U2 = e2 = e2, + 12F2 Bf2U1 + Bf2U2
(ne, eqs.)
(9-8)
(q eqs.)
(9—9)
Once e1 is known, (9—8) can be solved for U1. We obtain the equation for F2 by eliminating U1 in (9—9). First (see (9—5))
we note that 12
BT 12 —
Then, premultiplying (9—8) by Ff,
\T_
T
1,F21 — — IF2
T 11
adding the result to (9—9), and using
(a), (9—6) leads to
P2, F2U2 =
=
ez
+ Ff
e2,0
(9—10)
+ f2F2 + Ff,FZ(CLO + f1F1)
(9—11)
The first form, (9—10), shows that the equations are actually restrictions on the elongations. One can interpret (9—10) as a compatibility condition, i.e., it must be satisfied in order for the bars tofit in the deJbrmed structure defined by U1. The second form, (9—11), follows when we express the elongations in terms of the bar forces. Finally, we substitute for F1 and write the result as 122F2
where
f122 .1
(9—12)
d2 C
—
T
U2 — —e2,0
t'T
1, F2t1' t,F2 r'T ( —
+
v
+ nT
f'r
1O...43
The coefficient matrix, f22, is called the flexibility matrix for F2. One can show that f22 is positive definite when the bar flexibility factors in f2 are all positive.t If the material is physically nonlinear, f. and e0,, depend on F1. Iteration
is minimized by applying the loading in increments and approximating the force-elongation relation with a piecewise linear representation. The incremental equations are similar in form to the total equations4 We just have to replace the force, displacement, and elongation terms with theit incremental values and interpret las a segmental (tangent) flexibility. At this point, we summarize the steps involved in the force method. I.
Determination of F1, 0, P2. 0, F1.
and P2, F2
We select a stable primary structure F1 and determine the bar forces and reactions due to P1 and a unit value of each force redundant. This step involves q + 1 force analyses on the primary structure. Note that we obtain the primary
structure by deleting q = ni — bars. The selection of a primary structure and solution of the force equilibrium equations can be completely t See Prob. 9—i.
X See Prob. 9—4.
§ We reduce
to an echelon matrix. See (1—61).
2.
APPROACH
GOVERNfNG
SEC. 9—2.
213
Deterininatio,z of F2, F,, aizdP2
We assemble d2, and solve f,,F, = d, for F,. Then, we determine F, and P2 by combining the q + basic solutions.
= F10 + = P2,0 + P2,F2F2
F1 P2
3.
Determination of U,
Once F1 is known, we can evaluate e,, e1
e1,0 +
f1F1
and then solve (9—8).
=
e,
for U,. If only a limited number of displacement components are desired, one can
determine these components without actually solving (9—8). To show this, we write U, as
U1—' —t
e1 —
21
—hT 111
2
We see from (9—15) that the kth column of Br,' contains the bar forces in the primary structure due to a unit value of the kth element in P1. Also, it follows from (9—6) that the kth column of B2 contains the reactions due to a unit value of the kth clement in P,. Now, we obtain the kth element in U1 (which corresponds to the kth element in P,) by multiplying the kth column of Br1' by the kth column of B21Br,' by Of. and adding the two scalars. Then, letting
F,pJk = F, P2.pj,, =
due to an unit value of PJJ, with F2 = 0 due to an unit value of with F2 0
P2
we can write the expression for 0,ik
Note
(9—14)
as
I
—
nT
fl
—
that one works with the statically determinant primary structure to
mine the displacements. Example
—
9—1
Step 1:
Determination of F,, o, P2,
F,, F2, and P2 F2
For the truss shown in Fig. E9—IA,
=
2
in
3
q=
1
We take F3 as the redundant b,tr force:
F1={F,,F2}
F2={F3}
The primary structure consists of bars I and 2. Note that all force analyses are performed on the primary structure. The forces and reactions corresponding to P, and F3 = + 1 can be readily obtained using the method ofjoints. The results are shown in Fig. E9—1B.
FORCE METHOD: DEAL TRUSS
214
CHAP. 9 Fig. E9—1A
2
3
10 kips
I
20 kips
(1) A1 = 1.Oin,2
A2 0.51n.2 A3 =0.5 in.2 (2) Material is linearly elastic. E = 3 )< i04 ksi for all bars. = 0. (3) e0,1 —1/16 in. eo,2
(4) u32 + 1/10th. u41 =—1/I5in. Fig. E9—1B 1/2
3.33
2.50 3/8
—20.83
10 kips
20 kips
We could have obtained the above results for F1 by solving B11F1 =
B12F'2
P1
which, for this system, has the form
[—.6 +61 (F1)
110)
[-.8 +.8flFj = Step 2:
[01
+ [1j{F3}
Determination of f22, d2, F1, and F2
Since only u32 and u41 are finite, we can contract U2 and P2,
= u41} = {u3,.
and write PT
fl
/D' 2
2
SEC. 9—2. The
GOVERNING EQUATIONS—ALGEBRAIC APPROACH
215
force matrices follow from step 1: {—2083, —4.17}
F1,0 F1, F2
(kips) (kips)
—
{
=
(kips)
Also, we are given that {eo,
e1, o
e0,
2} =
(inches)
T'6, 0}
{
e2,0 = {eo,3} = 0 (inches)
f2 and evaluate f22 and d2. It remains to assemble The flexibility factors are (in./kip) 12(25)
12(25)
=
3
12 =
x
13
=
12(20)
Then, x
= f2
=
[f3] =
0.8(2 x
Evaluating the various products in (9-43), (9—12) reduces to 1.3SF3 = —7.31
(a)
Solving (a), we obtain
F2 = {F3} = —5.27kips 1— 17.53kips
F1 = F1,0 +
=
0.S7kips
Equation (a) actually represents a restriction on the elongations. The original form of (a) follows from (9—10). — 8e1 — a
— —.
Equation (b) reduces to (a) when we substitute for the elongations in terms of the bar forces.
Step 3: Determination of the Displacements Suppose only u11 is desired. Using (9—15), )T
= Ff, p,1e1 — Now,
fr'2 — (1
UO'
= e1,0 +
(c)
1
15
f1F1
{-.-.24, —.018}
We apply a unit load at joint 1 in the X1 direction and determine the bar forces in the primary structure and the reactions (P32, P41) corresponding to the nonvanishing prescribed displacements 1
FORCE METHOD: IDEAL TRUSS
216
CHAP. 9
Substituting in (c), we obtain
a11 = +185
—
.033
= +15 in
If both displacement components are desired, we apply (9—15) twice. This is equivalent to solving (9—8).
9—3.
GOVERNING EQUATIONS—VARIATIONAL APPROACH
We obtained the elongation compatibility equations (9—10) by operating on the elongation-displacement equations. Alternatively, one can use the principle of virtual forces developed in Sec. 7—3. It is shown there (see Equation (7—14)) that the true elongations satisfy the condition,
AFTe
0
—
for any statically permissible system of virflial bar forces and reactions which satisfy the constraint condition, B1 AF =
=
AP1
0
Equation (b) states that the virtual bar forces cannot lead to increments in the prescribed jOint loads, i.e., they must be self-equilibrating. Now, using (9—4), (9—5), we can write
=
F
{Fio} +
where B1
and
=
B1
0
Then AF2
= satisfies (b) for arbitrary AF2. The reactions due to AF2 are obtained from (9—6):
A?2 =
B2
AF =
P2
F2
Substituting for AF and A?2, (a) expands to F2e1 + C2 —
0
F2U2)
Equation (h) must be satisfied for arbitrary AF2. Finally, it follows that FT,F,el + C2
P2 F2U2 =
0
(i)
Equation (i) is identical to (9—10). Note that the elongation compatibility
SEC. 9—4.
COMPARtSON OF THE FORCE AND MESH METHODS
217
equations are independent of the material behavior. If the material is physically
linear, (i) leads to a set of q linear equations in F2 when we substitute for the elongations in terms of the bar forces. We determine the displacements by applying the general form of the principle of virtual forces (see (7—10))
= APfU1
AFTe —
where the virtual forces satisfy the force-equilibrium equations,
= Since only F1 is required to equilibrate P1, we can take
= AF2 = 0 and (j) leads to
U1 =
—
Note that I
— 1)11
One can interpret the compatibility equations expressed in terms of F2 as the Euler cquations for the total complementary energy function, —
Pf02 =
This approach is discussed in sec. 7—5. We take X = F2 in (7—35). Then,
= P2,F, = and (7—37) coincides with (i). We have written the expanded form of (i) as f22F2 =
d2
Since (i) are the Euler equations for
= and it follows that
H _I T2 221 2 — c
—
d2) 2U2
for the linearly elastic case. One can show that the stationary point corresponds to a relative minimum value of H. when the tangent flexibility factors for the redundant bars are all positive. t
9-4. COMPARISON OF THE FORCE AND MESH METHODS
It is of interest to compare the force method for a truss with the procedure followed to find the currents in an electrical network. The latter involves the t See Prob. 9—8.
FORCE METHOD: IDEAL TRUSS
218
CHAP. 9
application of Kirchhoff's laws and is called the mesh method. Various phases of the electrical network formulation are discussed in Probs. 6—6, 6—14, and the
governing equations for a linear resistance d-c network are developed in Probs. 6—14, 6—23. We list the notation and governing equations for convenience:
b = number of branches number of nodes
n
N=n—1 M—b—N=b—n+1 = potential at node j with respect to the reference potential, = nodes at positive and negative ends of branch k = current in branch k, positive when directed from node k_ to node
k+, k
ek
e0,
= potential drop for branch k =
— Vk+
Vk.
k = emf for branch k = resistance for branch k
The governing equations expressed in matrix notation are (see Prob. 6—23): An = 0
(N eqs.)
e = AV =
e0
+ Ri
(b
eqs.)
(9—16) (9—17)
where i
e = {ej, e2,
.
.
,
(9—18)
v={V1,V2 R1
R2
R=
Rb
and A is obtained by deleting the last column of the branch-node connectivity matrix d. Note that d has only two entries1 in any row. For row k (Ic =
jlkk = +1 dkk+ = dkj
—1
=0
1
k+ ork.
(9—19)
1= 1,2,...,N
Actually, d is just the matrix equivalent of the branch-node connectivity table. Example
9—2
A network can be represented by a line drawing consisting of curves interconnected at various points. The curves and intersection points are conventionally called branches and nodes respectively. Each branch is terminated at two different nodes and no two branches have a point in common which is not a node. Also, two nodes are connected by at least one path. A collection of nodes and branches satisfying the above restrictions is called a linear connected graph, If each branch is assigned a direction, the graph is said to be
SEC. 9—4.
COMPARISON OF THE FORCE AND MESH METHODS
219
oriented. The connectivity relations for a network are topological properties of the corresponding oriented graph. Consider the oriented graph shown. We list the branch numbers vertically and the node numbers horizontally. We assemble d working with successive branches. Finally, we obtain A by deleting the last column (cot. 4) of d. Fig. E9—2
2
3
Node Branch 1
I
2
—1
+1
2
+1
3
+1
3
A —.1
4 —1
—1
N
Now, A has N linearly independent columns. Therefore, it is possible to solve (9—16) for N branch currents in terms of b — N = M branch currents. We suppose the branches are numbered such that the first N rows of A contain a nonvanishing determinant of order N and partition A, i after row N. (N x N)
(bxN) —
[ A1
[A2 (MXN) (Nx 1)
= JJi__ (Mx))
FORCE METHOD: IDEAL TRUSS
220
CHAP. 9
Introducing (9—20) in (9—16) leads to (9—21)
ATi1 =
0, we can solve for i1 in terms of i2. We write the solution of the Since tAil node equations as
i=
Ci2 (9—22)
[Cii.I'2
=
I
Note that C1 is of order N by M and is related to A1, A2 by
C1 =
(9—23)
It remains to determine a set of M equations for i2. One can express (9—17) in partitioned form and then eliminate V, or alternatively, one can use the variational principle developed in Prob. 7—6. Using the first approach, we write (9—17) as
= A1V = e2
(N eqs) (M eqs)
+ R1i1
e1
A2V = e20 + R212
(9—24)
Once i1 is known, we can find V from A1V
e1 =
+ R1i1
e1
(9—25)
Eliminating V from the second equation in (9—24) and using (9—23), we obtain
e2 +
CTe1
=
0
(9—26)
Equation (9—26) represents M equations relating the branch potential differences (voltages). Finally, substituting for in terms of leads to (R2 + CTR1C1)i2
—e2,0 — Cfe10
(9—27)
The coefficient matrix for i2 is positive definite when the branch resistances are positive. This will be the case for a real system. The essential step in the solution involves solving (9—21), that is, finding C1. Note that C1 corresponds to for the truss problem. Also, the branches comprising A1 (and i1) correspond to the primary structure. Although the equations for the truss and electrical network are similar in form, it should be noted that the network problem is one dimensional whereas the truss problem involves the geometry as well as the cpnnectivity of the system. One can assemble C1 using only the topological properties of the oriented graph which represents the network. To find the corresponding matrices (F1, 0 and for a truss, one must solve a system of linear equations. In what follows, we describe a procedure for assembling C1 directly from the oriented graph. A closed path containing only one repeated node that begins and ends at that node is called a mesh. One can represent a mesh by listing sequentially
the branches traversed. A tree is defined as a connected graph having no
SEC. 9—4.
COMPARISON OF THE FORCE AND MESH METHODS
221
Let hT be the number of branches in a tree connecting 11 nodes. One can easily show that meshes.
bT=n—l=N
(9—28)
We reduce a graph to a tree by removing a sufficient number of branches such that no meshes remain. The branches removed are generally called chords. The required number of chords is equal to b — = b — N = M. Now, we associate the branches comprising a tree with the rows of A1. Selecting a tree is equivalent to selecting N linearly independent rows in A. The M chords correspond to the redundant branches, that is, the rows of A2. Note that one can always number the branches such that the first N branches define a tree. Chord j and the unique path (in the tree) connecting the terminals of chord j define a mesh, say mesh j. We take the positive direction of mesh j (clockwise or counterclockwise) such that the mesh direction coincides with the positive direction for chord j. Now, the current is constwit in a niesh. Suppose branch r is contained in mesh j. Then, the current in branch r due to a unit value of is equal to + 1 (—1) if the positive directions of branch rand meshj coincide (are opposite in sense). We have expressed the solution of the node equations as I)
(NXM)
i1=C1
(Mxl) 12
Now, we take the elements of i2 as the chord (mesh) currents. Then i1 represents
the required branch currents in the tree. We assemble C1 working with the columns. The column corresponding to involves only those branches of the tree which are contained in mesh j. We enter (+1, — 1,0) in row k of this column if branch k is (positively, negatively, not) included in mesh j. Example 9—3 For the graph in example 9—2, N = n — 1 = 3 and b = 6. Then M = b — N = 3 and we must remove 3 branches to obtain a tree. We take branthes 4 , 5 , and 6 as the chords. The resulting tree is shown in Fig. E9—3. We indicate the chords by dashed lines. 2
1
Fig. E9—3
3
CHAP. 9
FORCE METHOD: IDEAL TRUSS
222
For this selection of a tree,
ij =
{i1, i2, 13}
i2
{i4, i5, i6}
The meshes associated with the chords follow directly from the sketch:
mesh4 mesh5 mesh6
To assemble C1 we list the branches of the tree vertically and the chord numbers horizontally. We work with successive columns, that is, successive chords. The resulting matrix is listed below. Note that C1 is just the matrix equivalent of(a).
4
5
6
0
+1
1
1
—
—l•
Branches
ofthe
2
+1
+1
0
3
0
—1
—1
tree 1
The matrices, A1 and A2, follow from Example 9—2:
0
+1 +1 +1
—1
0
0
0
0
—1
+1
0
—1
—!
A1=
0
A2=
0 0
—l
One can readily verify that
C1 = The matrix, C = {C1, Im}, is called thc branch-mesh incidence matrix. Using (9—23), we see that A and C have the property (N x Ill)
ATC
=
0
(9—29)
Also, we can express the compatibility equations, (9—26), as (SIx 1)
CTe
=
0
(9—30)
The rows of CT define the incidence of the meshes on the branches. Equation (9—30) states that the sum of the potential drops around each mesh must be zero and is just Kirchhoff's voltage law expressed in matrix form. The matrix
PROBLEMS
223
formulation of the network problem leads to the same system of equations
that one would obtain by applying Kirchhoff's current and voltage laws to the various nodes and meshes. This, of course, also applies to the truss problem. The two approaches differ only with respect to the assemblage of the governing equations. In the conventional approach, one assembles the equations individually. This involves repeated application of the basic laws. When the equations are expressed in matrix form, the steps reduce to a sequence of matrix multiplications. REFERENCES 1.
2. 3.
4. 5. 6.
C. H., and J. B. WILBUR: Elementa,'v Structural Analysis, McGraw-Hill, New York, 1960. HALL, A. S., and R. W. W000HEAD: Frame Analysis, Wiley. New York, 1967. MORICE, P. B.: Linear Structural Analysis, Ronald Press, New York, 1969. RUBINSTEIN, M. F.: Matrix computer Analyris of Structures, Prentice-Hall, 1966. PRZEMIENIECKI, J. S.: Theory of Ma/rLr Structural Analysis, McGraw-Hill, 1968. RUBINSTEIN, M. F.: Structural Systems—Statics, Dynamics, and Stability, PrenticeFlaIl, 1970.
7.
Di MAGGIO, F. D., and W. R. SNLLERS: "Network Analysis of Structures," Eng. Mech. Div.. A.S,C.E., Vol. 91, No. EM 3, June 1965.
8.
FF.NVE.S, S. .1., and F. H. BRANIN, JR.: "Network-Topological Formulation of Struc-
tural Analysis," J. Structures Div., A.S.C.E. Vol. 89, No. ST4, August 1963.
PROBLEMS 9—1. Show that the coefficient matrix f22 is positive definite for arbitrary rank of F1 P2 when is positive definite. Use the approach suggested in
Problems 2—12 through 2—14. 9—2. Solve the following system using the procedure outlined in Sec. 9—2. TakeX1 {xi,x2}
2
1
2
2
3
4
12
0 0 l.Y25
0
0
x1
0
0
x2
0
X3
00 1
0002 x4
3
+ 1
2
9—3. Consider a system of in equations in n unknowns, ax = c, where in> n. Suppose r(a) = a and the first a rows of a are linearly independent.
Let q =
m
— n.
(a)
Show that the consistency requirement. for the system leads to q
(b)
relations between the elements of c. Interpret (9—10) from this point of view.
CHAP. 9
FORCE METHOD: DEAL TRUSS
224
Develop an incremental "force" formulation starting with
9—4.
— B1
= Ae
B2
AF AF
Bf AU1 +
AU2 =
+ ft AF
represent the flexibility factor and incremental initial elongation where f', for the segment corresponding to the initial value of F. One has to modify bothft and if the limit of the segment is exceeded (see sec. 6—4 for a detailed treatment). Consider the case where the loading distribution is constant, i.e., where only the magnitude is increased. Let P1 1,/i where 2 is the load parameter and defines
the loading distribution. Discuss how you would organize the
computational scheme. Also discuss how you would account for either yielding
or buckling of a bar. Distinguish between a redundant bar and a bar in the primary structure. 9—5. Solve Prob. 8—3 with the force method: Take F3 as the force redundant. 9—6. Assemble the equations for F2 = (F8, F9, F10, F, for the truss shown. Prob, 9—6
x2
xl 15'
15'
(1) Material is linear elastic and the flexibility factors are equal.
= {u42 } (2) Only u42 is finite. Take (3) Only initial elongation for bar 4.
For the truss shown: Using (9—10), determine the elongation-compatibility relations. Takt as the redundant bars. bars ©, (b) Express u52 in terms of the elongations and support movements. 9—7.
(a)
9—8.
By definition (see (7—26) and (7—31))
= AFTe
—
PROBLEMS
225
Prob. 9—i
2
x2
x1
Then —
= AFT de
Express d2fl, as a quadratic form in AF2. Consider the material to be nonlinear elastic and establish criteria for the stationary point to be a relative minimum. 9—9. Consider the oriented linear graph shown.
Prob. 9—9
0 (a) (b) (c)
Determine A. Determine C. Verify that ATC = 0.
Part III ANALYSIS OF A MEMBER ELEMENT
10
Governing Equations for a Delormable Solid 10—1.
GENERAL
The formulation of the governing equations for the behavior of a deformable solid involves the following three steps: 1.
2.
Study of deformation. We analyze the change in shape of a differential volume element due to displacement of the body. The quantities required to specify the deformation (change in shape) are conventionally called strains. This step leads to a set of equations relating the strains and derivatives of the displacement components at a point. Note that the analysis of strain is purely a geometrical problem. Study of forces. We visualize the body to consist of a set of differential volume elements. The forces due to the interactions of adjacent volume elements are called internal forces. Also, the internal force per unit area acting on a differential area, say dAd, is defined as the stress vector,
this step, we analyze the state of stress at a point, that is, we investigate how the stress vector varies with orientation of the area element. We also apply the conditions of static equilibrium to the volume elements. This leads to a set of differential equations (called stress equilibrium equations) which must be satisfied at each point in the interior of the body and a set of algebraic equations (called stress boundary conditions) which must be satisfied at each point on the surface of the body. Note that the study of forces is purely an equilibrium problem. 3.
Relate forces and displacements. In this step, we first relate the stress and strain components at a point. The form of these equations depends on the material behavior (linear elastic, nonlinear elastic, inelastic, etc.). Substitution of the strain-displacement relations in the stress-strain
relations leads to a set of equations relating the stress and derivatives of the displacement components. We refer to this system as the stress-displacement relations. 229
GOVERNING EQUATIONS FOR A DEFORMABLE SOLID
230
CHAP. 10
The governing equations for a deformable solid consist of the stress equilib-
rium equations, stress-displacement relations, and the stress and displacement boundary conditions. In this chapter, we develop the governing equations for a linearly elastic solid following the steps outlined above. We also extend the variational principles developed in Chapter 7 for an ideal truss to a three-dimensional solid. In Chapter 11, we present St. Venarit's theory of torsion-fiexure of prismatic members and apply the theory to some simple cross sections. St. Venant's theory provides us with considerable insight as to the nature of the behavior and also as to how we can simplify the corresponding mathematical problem by introducing certain assumptions. The conventional engineering theory of
prismatic members is developed in Chapter 12 and a more refined theory for thin walled prismatic members which includes the effect of warping of the cross section is discussed in Chapter 13. In Chapter 14, we develop the engi-
neering theory for an arbitrary planar member. Finally, in Chapter 15, we present the engineering theory for an arbitrary space member. SUMMATION CONVENTION; CARTESIAN TENSORS
1O—2.
Let a and b.represent nth-order column matrices:
a= b=
{Oj, 02 10—1
{h1,b2
Their scalar (inner) product is defined as
aTb =
bTa
=
a1b1
+ a2h2 + ... +
=
To avoid having to write the summation sign, we introduce the convention that when an index is repeated in a term, it is understood the term is summed over the range of the index. According to this convention
(i = 1,2
ii)
(10—2)
and we write the scalar product as aTb = a1b1
(10—3)
The summation convention allows us to represent operations on multidimensional arrays in compact form. It is particularly convenient for formulation, i.e., establishing the governing equations. We illustrate its application below.
Example 1.
10—i
Consider the product of a rectangular matrix, a, and a column vector, x.
c=ax
aisrnx n
SUMMATION CONVENTION; CARTESIAN TENSORS
SEC. 10—2.
231
The typical term is
=>
c•
2.
(b)
Let a, b be square matrices, x a column vector, and f, g scalars defined by
frxTax g = xrbx
(c)
The matrix form of the product, fg, is fg =(xTax)(xTbx) One could expand (d) but it is more convenient to utilize (b) and
write (c)
as
f= g
=
bk(xkxe
fg =
alJbk,xlxfxkx(
= 3.
DIJk(XIXJXkX(
We return to part 1 The inner product of c is a scalar, H,
=
II
xT(aTa)x
Using (b),
H =
c-c1 = 0f50
The outer product is a second-order array, il,
d=
axxraT
ccT
and can be expressed as
=
= alkaf,xkx,.
= AIJk(XkXe
According to the summation convention,
=
d11
+ d22 +
= trace of d
Then, we can write (h) as
H=
d11
=
AIIk(XkX(
4. Let represent square second-order arrays. The inner product is defined as the sum of the products of corresponding elements:
Inner product
ç,) = =
+
+
+
+ g21e21 +
In order to represent this product as a matrix product, we must convert cki,
one-dimensional
(m)
ejj over
to
arrays.
Let represent a one-dimensional set of elements associated with an orthogonal reference frame having directions If the
GOVERNING EQUATIONS FOR A DEFORMABLE SOLID
232
CHAP. 10
corresponding set for a second reference frame is related to the first set by — —
k
=
(10—4)
cos 1, 2, 3
we
say that the elements of b comprise a first-order cartesian tensor. Noting
(5—5), we can write (10—4) as
=
(10—5)
and it follows that the set of orthogonal components of a vector are a first-order cartesian tensor. We know that the magnitude of a vector is invariant. Then, the sum of the squares of the elements of a first-order tensor is invariant. (10—6)
A second-order cartesian tensor is defined as a set of doubly subscripted elements which transform according to =
(10—7)
j. k.
,n.
1.2. 3
An alternate form is
=
(10—8)
The transformation (10—8) is orthogonal and the trace, sum of the principal second-order minors, and the determinant are invariant.t
= fl(2)
where
= = b12 L
021
7
022
+
b22 7
032
b23 7
033
+
b11
b13 1733
In the cases we encounter, b will be symmetrical.
10-3. ANALYSIS OF DEFORMATION; CARTESIAN STRAINS Let P denote an arbitrary point in the undeformed state of a body and the position vector for P with respect to 0, the origin of an orthogonal cartesian reference frame. The corresponding point and position vector in the deformed state are taken as F'; and the movement from P to P' is represented by the
displacement vector, fl. By definition, (10—10)
This notation is shown in Fig. 10—1. f See Prob. 2—5.
SEC. 10—3.
ANALYSIS OF DEFORMATION; CARTESIAN STRAINS
233
Excluding rigid body motion, the displacement from the initial undeformed position will be small for a solid, and it is reasonable to take the initial Cartesian
coordinates (xi) as the independent variables. This is known as the Lagrange
Undeformed
dp
F' (Deformed)
i3 2
112
Fig. 10—1. Geometric notation.
approach. Also, to simplify the derivation, we work with cartesian components for ü. Then, ii
=
We consider a differential line element at P represented by the vector dii. (See Fig. 10—1). The initial length and direction cosines are ds and using the subscript notation for partial differentiation.
=
/
=
We are
(10—12)
Since we are in the deformed state is The corresponding line arid we can write following the Lagrange approach, p =
=
=
(10—13)
The extensional strain, r, is defined as the relative change in length with respect
234
GOVERNING EQUATIONS FOR A DEFORMABLE SOLID
CHAP. 10
to the initial length.t = (1 +
(10— 14)
Using the dot product, (10—14) becomes (1
Finally, we write (a) as
e(1 + 4c) = ap.kejk (10—15) — = One can readily establish that (eJk) is a second-order symmetrical Cartesian tensor4 direction and letting Taking the line element to be initially parallel to the represent the extensional strain, we see that (no sum) = e0 (10—16) -1) = To interpret the off-diagonal terms, we consider 2 initially orthogonal line elements represented by (see Fig. 10—2) and having direction cosines
+
d r',
I
xa
— '/12 p.-
dp'1
x2
Fig. 10—2. Notation for shearing strain. t This is the definition of Lagrangian strain. In the Eulerian approach, the cartesian coordinates for the deformed state are taken as the independent variables, =
and the strain is defined as
=
(1 —
are also called the See Prob. 10—4. It is known as Green's strain tensor. The elements, components of finite strain. They relate the difference between the square of the initial and deformed lengths of the line element, i.e., an alternate definition of Cjk —
ds2 = 2eJkdxJdxk
SEC. 10—3.
ANALYSIS OF DEFORMATION; CARTESIAN STRAINS
235
We define as the angle between the lines in the deformed which is called the shearing strain, follows by state. The expression for taking the dot product of the deformed vectors.
(it
,
COS —
—
=
(1
J=
.
Y12 =
)
Substituting for k)dsf
(sum on k only)
+
and noting that the lines are initially orthogonal,
= (a)
takes the form
(1 +
+
=
(10—17)
shows that
Specializing (10—17) for lines parallel to X,, shearing strain.
(I +
=
+
=
2e13
is related to the (10—18)
Equations (10—15) and (10—17) are actually transformation laws for extensional and shearing strain. The state of strain is completely defined once the strain tensor is specified for a particular set of directions. To generalize these expressions, we consider two orthogonal frames defined by the unit vectors and (see Fig. 10—3), take the initial frame parallel to the global frame = ti), and let = 15 tk. With this notation:
+ (1 +
=
)
(10—19)
=
+
The strain measures (e, y) are small with respect to unity for engineering materials such as metals and concrete. For example, e for steel. Therefore, it is quite reasonable (aside from the fact that it simplifies the expressions) to assume r, y in the strain expressions. The relations for "small" strain are: 1
(10—20)
It remains to expand eJk. Now,
= Differentiating
+ ii
+
u,,Ji,,
with respect to S
OP
=
=
—
+ Un, j)l,n
236
and
GOVERNING EQUATIONS FOR A DEFORMABLE SOLID
substituting into the definition of eJk =
+
k + Uk.
Cl-lAP. 10
(Equation (10—15)) leads to (sum on m only)
4Um, ,u,,,, k
(10—21)
In order to simplify (10—21), we must establish the geometrical significance of the various terms. x3 t3
t2
/
t,3
X2
Fig. 10—3. Unit vectors defining transformation of orthogonal directions.
With this objective, we consider a line element initially parallel to the X1 axis. Figure 10—4 shows the initial and deformed positions, and the angles
which define the rotation of the line toward the X2, K3 directions. The geometrical relations of interest to us are 012,
sin
033 1 +
0j3 =
1421
sine12
(1 + 81)cos 0j3 )2
(1 +
+ uj 1
+
Also, by definition,
+ We solve (a), (b) for u2,
=
=
e11
U1,
1+
i+
i4,
i+
and 03, 03,1 1
(1 + 013
—
ANALYSIS OF DEFORMATION; CARTESIAN STRAINS
SEC. 10—3.
237
and then solve (c) for u1,
= A= 1
(1
+
{1
—
A}112
—
1
(10—23)
sin2 013 + cos2 013 sin2 012
Applying the binomial expansion, (1 — x)"2
= I
—
+
+
(10—24)
we can write (10—23) as
to (1 —
+
+
—
+
—
+
(10—25)
In what follows, we assume small strain and express the derivatives and extensional strain (see Equation (d)) as
1=
u3. 1 = 0(013) U1
a11 =
0(012,
n2
t/12, "13
1—
+
+
u1,
(f)
The various approximate theories are obtained by specializing (f). 'U3
dx1
dx1
1123 dx1
X1,u1
Fig. 10—4. Initial and deformed positions of a line element.
In the linear geometric case, the rotations are neglected with respect to strain. Formally, one sets 012 = 613 = 0 in (f) and the result is a linear relation between
strain and displacement,
-
a11
(g)
Note that, according to this approximation, the deformed orientation coincides
CHAP. 10
GOVERNING EQUATIONS FOR A DEFORMA8LE SOLID
238
with the initial orientation. The general relations for the linear geometric case
(small strain and infinitesimal rotation) are
= =
= =
en
(no sum)
(10—26)
+ ui,,
The next level of approximation is to consider 62 to be of the same order as strain.
02 = sin 0 cos 6
0(s) << 1
(10—27)
0 1
with respect to 1 in (f), but we must retain
We can neglect
1
and
1
since they are of 0(62).
u11 +
ci
+
(h)
The complete set of strain-displacement relations for small strain and smallfinite rotation are listed below for reference. =
= =
-
+ +
(no sum)
+
(10—28)
+ Uk iUk. j k
I
We utilize these expressions to develop a geometrically nonlinear formulation
for a member in Chapter 18. Lastly, if no restrictions are imposed on the magnitude of the rotations, one must use (10—21). The relations for finite rotation and small strain are = =
=
+
=
+
+
+
+ u11(l +
(no sum)
+ Uk.iUk,j
10—29
k
Note that the truss formulation presented in Chapter 6 allows for arbitrary magnitude of the rotations. We have shown that linear strain-displacement relations are based on the following restrictions: 1.
2.
The strains are negligible with respect to unity, and Products of the rotations are negligible with respect to the strains.
The first condition will always be satisfied for engineering materials such as metals, concrete, etc. Whether the second restriction is satisfied depends on the configuration of the body and the applied loading. If the body is massive in all three directions, the rotations are negligible with respect to the strains for an arbitrary loading. We have to include the nonlinear rotation terms in the strain displacement relations only if the body is thin (e.g., a thin plate or slender member) and the applied loading results in a significant change in the geometry. As an illustration, consider the simply supported member shown
SEC. 10—3.
ANALYSIS OF DEFORMATION; CARTESIAN STRAINS
239
in Fig. 10—5. We can neglect the change in geometry.if only a transverse loading
is applied (case 1). However, if both axial and transverse loads are applied (case 2), the change in geometry is no longer negligible and we must include the nonlinear rotation terms in the strain-displacement relations.
Case 2 (Q,P)
Case1 (Q)
Fig. 10—5. Example of linear and geometrically nonlinear behavior.
To treat a geometrically nonlinear problem, we must work with the deformed
geometry rather than the initial geometry. This can be defined by tracking the movement of a triad of line elements initially parallel to the global directions. We let be the initial set and the deformed set (see Fig. 10—6). By definition,
= = =
(no sum)
(no sum) (1
+
The unit vector pointing in the direction of dTh is denoted by
Using (a),
we can write vi
p.
=
for small strain Finally, we express
in terms of the unit vectors for the initial frame. /3jklk
(10—30)
+
I3jk —
+
for small strain
We will utilize (10—30) in the next section to establish the stress equilibrium equations for the geometrically nonlinear case. Equations (10—30) reduce to (10—31)
for the geometrically linear case and to + 13j1Jk
+
(no sum)
for the case of small strain and small-finite rotations.
[0—32
240
GOVERNING EQUATIONS FOR A DEFORMABLE SOLID
/
CHAP. 10
f
=dxji1
ii
Fig. 10—6. Initial and deformed geometries.
10—4.
ANALYSIS OF STRESS
The effects of the surroundings on a body such as contact pressure, gravitational attraction, etc.,result in internal forces. In this section, we establish the equilibrium conditions for the internal forces in a body. This step is generally called the analysis of stress. Consider a body subjected to some effect which results in internal forces. We pass a cutting plane through the deformed body and separate the two segments as shown in Fig. 10—7. We let in denote the outward normal direction for the internal face of body and refer to this face as the + in face. In general, the subscript, m, is used for quantities associated with the + m face. Now, we consider a differential area element AAm, and let A Fm be the resultant internal is defined as force vector acting on this element. The stress vector, hm
(10—33)
AA, -. I)
Note that has the units of force/area. Also, it depends on the orientation of the area element, i.e., on the direction of the outward normal. We do not allow for the possibility of the existence of a moment acting on a differential area element. One can include this effect by defining a vectort in addition to a stress vector. f See Ref. 6, p. 68.
ANALYSIS OF STRESS
SEC. 10—4.
We
241
consider next the corresponding area element in the —rn
face.
From
Newton's law,
= and it follows that
=
(1O—34)
—ö,,,
The stress vector has the same magnitude and line of action but it's sense is reversed.
Body
Body I
2
Note: Deformed state
Fig. 10—7. Notation for internal force.
In order to analyze the state of stress at a point, say Q, we need an expression for the stress vector associated with an arbitrary plane through Q. With this objective, we consider the tetrahedron shown in Fig; 10—8. The orientation of the arbitrary plane is defined by q, the outward normal direction. The outward normals for the other three faces are parallel to the reference axes (X1; j = 1, 2, 3). To simplify the notation, we use a subscript j for quantities associated with the X face, that is, the face whose outward normal points in the + direction. For example, we write
= =
=
(10—35)
= etc.
The force vectors acting at the centroids of the faces are shown in Fig. 10—8. The term M0 represents the change in due to translation from Q to the centroid. For equilibrium, the resultant force and moment vectors must vanish. In the
limit (as P
Q), the force system is concurrent and therefore we have to
CHAP. 10
GOVERNING EQUATIONS FOR A DEFORMABLE SOLID
242
consider only the force equilibrium condition. From Fig. 10—8, we have crq + &Tq
(a3 +
=
—
Now, AA1 is the projection of AAq on the X2-X3 plane. Noting that the projection of LxAq on a plane is equal to AAq times the scalar product of and the unit normal vector for the plane, and letting aqj be the direction cosine for the q direction with respect to the direction, we can write AA.
=
=
cos(q,
X1) =
(10—36)
Finally, in the limit, Equation (a) reduces to
(1037)
=
Once the stress vectors for three orthogonal planes at Q are known, we can determine the stress vector for an arbitrary plane through Q with (10—37). x2 +
i2
+
I
x3
+
2)LXA2
Fig. 10—8. Differential tetrahedral element.
Equation (10—37) is the transformation law for the Stress vector. The component of 5q in a particular direction is equal to the scalar product of 6q and a unit vector pointing in thedesired direction. Now, we express the stress vectors in terms of their components with respect to the coordinate axes (j = 1, 2, 3). j = 1,2,3 (10—38) = aqklk
ANALYSIS OF STRESS
SEC. 10—4.
243
Note that the first subscript on a stress component always refers to the flice,
and the second to the direction. For example. a12 acts on the X1 face and points in the X2 direction. The positive sense of the components for a negative The normal (a13) and in-plane (afk) comface is reversed since ponents are generally called normal and shearing stresses. This notation is illustrated in Fig. 10—9. x2
ta22
4
T for stress components.
FIg. 10—9. Notation
Substituting for the stress vectors in (10—37) results in
=
(10—39)
cxqjajk
The component of 5q with respect to an arbitrary direction, m,is determined from aqn,
Letting and noting (10—38), (a) expands to
We generalize (c) for two orthogonal frames specified by the unit vectors (see Fig. 10—3) where
t. = t3
Defining identifying
=
-ii
as the component acting on the 1 with
i,,,,
(10—40)
cJ3k:k
face
in the
direction and
(c) takes the form
= This result shows that the set
is a second-order cartesian tensor.
(10—41)
GOVERNING EQUATIONS FOR A DEFORMABLE SOUD
244
CHAP. 10
It remains to establish the equilibrium equations for a differential volume element. The equilibrium equations relate to the deformed state, i.e., we must consider a differential element on the deformed body. Since we have defined the stress components with respect to the global Cartesian directions, it is natural to work with a rectangular parallelepiped having sides parallel to the global directions. This is shown in Fig. 10—10. Point 0 is at the centroid of the element.
i3
+
+
(—
dfl3
dr13
di73
ant Fig. 10—10. Differential volume element in Eulerian representation.
The stress vectors are considered to be functions of the deformedt coordinates We obtain the forces acting on the faces by expanding the stress vectors
about 0 and retaining only the first two terms4 Letting b denote the external force per unit volume and enforcing the equilibrium conditions leads to (10-42) and
=
=
x
0
0
(10—43)
The scalar force equilibrium equations are obtained by expanding the vector equations using (10—38).
Force equilibrium
+
Moment equilibrium
=
=
0
k= k=
1
1,
2, 3
23
(10—44)
(10—45)
Moment equilibrium requires the shearing stress components to be symmetrical. Then, the stress tensor is symmetrical and there are only six independent stress measures for the three-dimensional case and three for the two-dimensional case. t We arc following the Eulerian approach here. Later we will shift back to the Lagrange approach Second- and higher-order terms will vanish in the limit, i.e.. when the element is shrunk to a point.
ANALYSIS OF STRESS
SEC. 10—4.
245
Equations (10—44) must be satisfied at each point in the interior of the body.
Also, at the boundary, the stress components must equilibrate the applied surface forces. We define and write
as
the outward normal vector at a point on the deformed surface
= The external force per unit deformed surface area is denoted by
(10—46)
(10—47) = pnjlj Applying (10—37) leads to the stress-boundary force-equilibrium relations:
Ph = (10—48) Pnj
i3nk0kj
f=
1,
2, 3
When p,0 is prescribed, i.e.,
(10—48) represent boundary conditions is a reaction. on the stress components. If is prescribed, Our derivation of strain-displacement relations employed the Lagrange approach, i.e., we considered the displacements (and strains) to be functions of the initial coordinates (xi). The analysis of stress described above is based on the Eulërian approach, where the deformed coordinates are taken as the independent variables. This poses a problem since the strain and stress measures are referred to different volume elements. Figure 10—Il shows the initial and
/
Initia'
Deformed
(I + €2)dx2 Lagrange
dX2[
(1 +e1)dxj dx1 4 U12
(I Eu'er drj1
(I
x2 Fig. 10—11. Comparison of Eulerian and Lagrangian representations for a volume element. f See Prob. 10—12.
GOVERNING EQUATIONS FOR A DEFORMABLE SOLID
246
CHAP. 10
deformed area elements corresponding to the two viewpoints. To be consistent
with the Lagrange strains, we must work with a nonorthogonal parallelepiped whose sides are parallel to the deformed line elements in the analysis of stress. Conversely, to be consistent with the Eulerian stresses, we have to refer the strain measures to nonorthogonal directions in the initial state. In the linear geometric case, we assume small strain and neglect the change in orientation due to rotation. The two approaches coalesce and we just have where is the direction cosine for the with to replace with and initial direction of the exterior normal. The linear equilibrium equations are:
CXj
+ hk
0 (10—49)
For the geometrically nonlinear case, we work with stress measures referred to the deformed directions (see Fig. 10—6) defined by the unit vectors, We define
as the stress vector per unit initial area acting on the face which
initially is normal to the direction, b* as the force per unit initial volume, and as the force per unit initial surface area. Figure .10—12 shows this notation for the two-dimensional case. The stress and force vectors are considered to be functions of the initial coordinates (xe). The equilibrium equations at an interior point are + b* =
(X1 (1
+
(10—50)
0
x
0
We express the body force and stress vectors as —
k
= The set,
(10-5 1)
+
is called the Kirchhoff stress tensor. Substituting for
using
(10—30), results in the following scalar equations, which correspond to (10—44) and (10—45):
-
+
b7
=
1,2,3
0
1,
2, 3
(10-52) (10—53)
The boundary equilibrium equations are obtained by expanding
= and have the form
*_
pnj
—
(10—54) k
—
ANALYSIS OF STRESS
SEC. 10—4.
These
247
equations apply for arbitrary strain and finite rotation. For small
strain, we neglect the change in dimensions and shape of the volume element. This assumption is introduced by taking (10—56)
Since the deformed unit vectors are orthogonal (toe 1), the Kirchhoff stresses now comprise a second-order cartesian tensor and they transform according p2
x2 1,, =
/
4
—oj dx2
p,,ds pn =
ij
dx2 dx1
c4 dx1
/ (1
e2)dx2
dx2
(1 + €1)dx1
dx
x1
Fig. 10—12. Definition of stress components in Lagrangian representation.
to (10—41).
The equations simplify somewhat if we assume small-finite
For infinitesimal
(linear geometry),
and
the equations reduce to (10—49), (10—50).
In what follows, we will work with the Kirchhoff stress components to keep the treatment general. However, we will assume small strain. t See Prob. 10—16.
GOVERNING EQUATIONS FOR A DEFORMABLE SOLID
248
10—5.
CHAP. 10
ELASTIC STRESS-STRAIN RELATIONS
A body is said to be elastic if it returns to its initial dimensions and shape when the applied forces are removed. The work done during the deformation process is independent of the order in which the body is deformed. We treat first an arbitrary elastic material and then specialize the results for a linearly elastic material. Our starting point is the first law of thermodynamics:
5W =
OVT
+ OQ
where OW = first-order work done by the forces acting on the body 0 VT = first-order change in the total strain energy (also called internal
energy)
= first-order change in the total heat content. When the deformation process is isothermal or adiabatic, OQ = 0, and (a) reduces to OW
Now, we apply (b) to a differential volume clement in the deformed state (see, e.g., Fig. 1Q—12). We define V as the strain energy per unit initial volume. In general, V is a function of the deformation measures.
=
V=
Y12'
.
.
.)
(10—57)
The material is said to be hyperelastic (Green-type) when V is a continuous function. This requires =
(10—58)
cejj ciek(
oek,
By definition,
=
OV(dx1 dx2 dx3) cW
=
(10—59)
cetj
is the first-order changet in due to an incremental displacement, M. Also, one can show that the first order work done by the force vectors where
acting on the element is OW =
.
+
+
.
+
h
Ltfl)dx1 dx2 dx3
dx2 dx3
(10—60)
Equating OVT and OW leads to the general form of the stress-strain relation for a Green-type material,
= (3•
tSeeProb. 10—11. See Prob. 10—18. The forces are in equilibrium. is. they satisfy (10—50).
(10—61)
ELASTIC STRESS-STRAIN RELATIONS
SEC. 10—5.
249
This definition applies for arbitrary strain. Once V is specified, we can obtain
expressions for the stresses in terms of the strains by differentiating V. Since V is continuous, the stress-strain relations must satisfy (10—58), which requires (10—62) 43e1j
In what follows, we restrict the discussion to small strain and a linearly elastic material, i.e., where the stress-strain relations are linear. We also shift from indicial notation to matrix notation, which is more convenient for this phase.
We list the stress and strain components in column matrices and drop the superscript k on the Kirchhoff stress components:
= = {e11, e22, e33, 2e12, 2e23, 2e31}
(10—63)
= With this notation, (10—64)
The matrix transformation laws are
=
(10—65)
Since ÔV is invariant under a transformation of reference frames, the transformation matrices are related by (10—66)
1
The total strain, e, is expressed as
a° ±
a
(10—67)
where a° contains the initial strains not associated with stress, e.g., strain due to a temperature increment, and A is called the material compliance matrix. We write the inverted relations as
= D(a
—
a°)
(10—68)
A'
where D = is the material rigidity matrix. Equation (10—62) requires D (and A) to be symmetrical. The elements of A are determined from material tests, and D is generated by inverting A. Substituting for in (10—64), we obtain the form of the strain energy density for the linear case,
V=
—
a°)TD(a
—
a°)
(10—69)
Since V > 0 for arbitrary (E — a°), D and A are positive definite matrices.
There are 21 material constants for a linearly elastic Green-type material. The number of independent constants is reduced if the material structure t See Prob. 10—6, 10—13.
GOVERNING EQUATIONS FOR A DEFORMABLE SOLID
250
CHAP. 10
In what follows, we describe the transition from an anisotropic material to an isotropic material. A material whose structure has three orthogonal axes of symmetry is called orthotropic. The structure of an orthotropic material appears identical after a 1800 rotation about a symmetry axis. To determine the number of independent constants for this case, we suppose X1, X2, X3 are axes of symmetry and consider a 180° rotation about X2. We use a prime superscript to indicate the rotated axes. From Fig. 10—13, exhibits
= —x1
= -x3 = x2 The stress and deformation quantities are related by (we replace 1 by — I and 3 by —3 in the shear terms)
= 1,2.3
= a12 = Y12 =
—a12 Y12
a23
—a23
=
Y23
a13 =
a13
Y13
Y13
Now, the stress-strain relations must be identical in form. We expand e = Acv', and substitute for using (b). Equating the expressions for a'
Fig. 10—13. Rotation of axes for symmetry with respect to the X2-X3 plane.
tA
material whose structure has no symmetry is said to be anisotropic.
ELASTIC STRESS-STRAIN RELATIONS
SEC. 10—5.
and
251
leads to the following relations between the elements of A,
= + a24a12 + a25a23 = a34a12 + a35a23 =
ti15a33
314(T12 —
—a24a12
—
—a34u12
—
a25a23
For (c) to be satisfied, the coefficients must vanish identically. This requires £434 = a15
=
0
a24 =
=
0
£435
a34a350
The symmetry conditions require We consider next the expansions for a46 = a56 = 0. By rotating 1800 about X1, we find = a36 = a45 0 a16 = A rotation about the X3 axis will not result in any additional conditions. Finally, when the strains are referred to the structural symmetry axes, the stress-strain relations for an orthotropic material reduce to a11
a12
0j3
a12
a22
a23
a1,
0 -
— £444 -—
0
0
0
0
Y31
0
0
a12
0
a23
a66
(733
(10-70)
We see that A is quasi-diagonal and involves 9 independent constants. There is no interaction between extension and shear. Also, the shearing effect is uncoupled, i.e., cr12 leads only to An alternate form of the orthotropic stress-strain relations is
AT +
a1 =
—
V32
1
—-----a33
E2
=
/13
1
AT +
(10—71) a33
1
Y12 =
!(733
—
Y23
=
—
— —i—-
1
1
Y31
=
where E4 are extensional moduli,
are shear moduli, Vjk are coupling coefficients, and AT is the temperature increment. The coupling terms are related by E2
E1
E3
E1
E3
E2
(10-72)
GOVERNING EQUATIONS FOR A DEFORMABLE SOLID
252
CHAP. 10
It is relatively straightforward to invert these relations:t One should note that (10—71) apply only when X, coincide with the material symmetry directions4
If the stress-strain relations are invariant for arbitrary directions in a plane, the material is said to be transversely orthotropic or isotropic with respect to the plane. We consider the case where the X1 direction is the preferred direction, i.e., where the material is isotropic with respect to the X2-X3 plane. By definition, A is invariant when we transform from X1-X2-X3 to This c, '—'12'-'31'--'
v32
v23
v
2(1 + v)
1
--
F and
the relations reduce to
AT +
=
(a22
—
= PAT +
+
= pAT +
— va27)
1
2(l+v)
1
Yi2
(10—74)
—
—
y23
Y31
There are five independent constants (F, v, E1, v1, G1).
Lastly, the material is called isotropic when the stress-strain relations are Invariant for arbitrary directions, For this case, A = A' for arbitrary The relations are obtained by specializing (10—74): = p AT +
(at,
+ akk))
—
(10—75)
2(1 + v) F
Note that now there are only two independent constants (F, v). The coupling coefficient, v,is called Poisson's ratio. The inverted form of (10—75) is written as
a= a0
= =
a0
+
+
IO--21.
Prob.
10—22.
+ (10—76)
+ 2G)pAT
t See Prob. 10—19 for the inverted form of (10—7 1). § See
+
SEC. 10—6.
where
PRINCIPLE OF VIRTUAL DISPLACEMENTS
253
G are called Lamé constants and are related to E, v by
G=
shear
modulus =
E 2(1
+ v)
yE A
— (1
(10—77)
+ v)(1 — 2v)
Since D must be positive definite, v is restricted to — 1 < v < 1/2. The limiting case where v = + 1/2 is discussed in Problem 10—24.
10—6.
PRINCIPLE OF VIRTUAL DISPLACEMENTS; PRINCIPLE OF. STATIONARY POTENTIAL ENERGY; CLASSICAL STABILITY CRITERIA
Chapter 7 dealt with variational principles for an ideal truss. For completeness, we derive here the 3-dimensional form of the principle of virtual
displacements, principle of stationary potential energy, and the classical stability criterion. The principle of virtual forces and stationary complementary energy are treated in the next section.
The principle of virtual displacements states that the Iirst-order work done is equal to the first oidcr work done by the internal forces acting on the restraints for an arbitrary virtual displacement of the body from an equilibrium position. f In the continuous case, the external loads and the internal forces are loading consists of body (b) and surface represented by the stress vectors. We follow the Lagrange approach, i.e., we work with Lagrange finite strain components (eJ,j, Kirchhoff stresses and external force measures per unit initial volume or area (b*, p*). This is consistent with our derivation of the equilibrium equations. Let Au denote the virtual displacement. The firstorder external work is by the external forces
= =
dx1 dx2 dx3 + JJj3* Au dx1 dx2 dx3 ± dfI
10—78
where fI is the initial surface area. The total internal deformation work is obtained by summing the first-order work done by the stress vectors acting on a differential volume element. *
= =
dx2dx3 dx1 dx2 dx3
(10—79)
Equating (a) and (b), we obtain the 3-dimensional form of the principle of See Sec. 7—2.
See Fig. 10—12. §
See (10—60).
GOVERNING EQUATIONS FOR A DEFORMABLE SOLID
254
CHAP. 10
virtual displacements, 5WD
=
dx1 dx2 dx3 = fJJh*
dx1 dx2 dx3 +
dx1 dx2 dx3 =
(10—80)
dx1 dx2 dx3 +
Requiring (10—80) to be satisfied for arbitrary (continuous) is equivalent to enforcing the equilibrium equations. To show this, we work with the vector form and utilize the following integration by parts formula: t
=
J
—
dx2dx3
(10-81)
is the direction cosine for the initial outward normal (n) with respect direction. Operating on the left-hand term and equating coefficients
where
to the
in the volume and surface integrals leads directly to (10—50) and (10—54).
The principle of virtual displacements applies for arbitrary loading (static or dynamic) and material behavior. When the behavior is elastic and the loading
is independent of time, it can be interpreted as a variational principle for the displacements. The essential steps required for the truss formulation are described in Sec. 7—4. Their extension to a continuous body is straightforward. When the behavior is elastic, = Letting
VT
denote the total strain energy, the left-hand side of (10—80) reduces to
fJJ öVdx1 dx2 dx3 =
dx1 dx2 dx3
We consider the surface area to consist of 2 zones as shown in Fig. 10—14.
+ where displacements are prescribed on on cd
U1
(10—82)
and surface force intensities arc prescribed on pni
Pni
on
The displacement variation, L\u1, is admissible if it is continuous and satisfies
=
0
on
(10—83)
We also consider the surface and body forces to be independent of the displacements. With these definitions, the principle of virtual displacements is transt See Prob. 10—25.
PRINCIPLE OF VIRTUAL DISPLACEMENTS
SEC. 10—6.
255
formed to
= fl,, =
for arbitrary admissible
0
cIx1 dx2 dx3
VT
(1084)
—
is the total potential energy functional. According to (10—84), the displacements defining an equilibrium position correspond to a stationary value of the total potential energy functional. Note that this result applies for arbitrary strain and finite rotations. The only restrictions are elastic behavior where
and static loading.
PH
Fig. 10—14. Classification of boundary zones.
Example 10—2 Direct methods of variational calculus such as Rayleigh-Ritz, Galerkin, weighted residuals, and others are applied to fl,, to determine approximate solutions for the displacements. In the Rayleigh-Ritz method, one expresses the displacements in terms of unknown parameters, q, and prescribed functions, x2, x3), U1
+
=
where
=
0
forj =
>
1.
2
The displacement boundary conditions on fd are called "essential" boundary conditions. to a function of the q's. When the material is linearly Substituting for transforms elastic, V is a quadratic function of the strains. Then, V will involve up to fourth-degree reduces terms for the geometrically nonlinear case. If the behavior is completely linear,
H, = q =
Const.
+ qTQ + . .
.
.
.
.
K is symmetrical
(3N x 1)
256
GOVERNING EQUATIONS FOR A DEFORMABLE SOLID
Finally, requiring
CHAP. 10
to be stationary for arbitrary c5q leads (for linear behavior) to
Kq = Q The strains are evaluated by operating on (a) and the stresses are determined from the
stress-strain relations.
Polynomials and trigonometric functions are generally used to construct the spatial distribution functions. The mathematical basis for direct methods is treated in numerous texts (see Refs. 9, 10).
The "classical" stability criterion for a stable equilibrium position ist —
o2WE
>
0
for arbitrary Ad
is the second-order work done by the external forces where = during the incremental displacement, Ad, and WD = ó(ö WD) is the second-
order work done by the internal forces acting on the restraints during the incremental deformation resulting from Ad. The form of the work terms for a continuous body are obtained by operating on (10—78) and = Sfl = = =
Ad dx1 dx2 dx3 + j(
Ad
Au1 dx1 dx2 dx3 + J$ Au, dx2 dx3 ie11 + dx2 dx3
(10—85)
If = Ô2WE for a particular Ad, the equilibrium position is neutral. The position is unstable if ö2 WD < o2 Note that öb, are itull vectors when the forces are prescribed.
For elastic behavior, the incremental deformation work is equal to the increment in strain energy
=
=
and (10—84) can be written as
>
0
for arbitrary Ad
(10—86)
It follows that a stable equilibrium position corresponds to a relative minimum
value of the total potential energy. Bifurcation (neutral equilibrium) occurs when = 0 for some Ad, say Ada. If the loading is prescribed, and ö2VT = 0 at bifurcation. The governing equations for bifurcation can be obtained by expanding This involves transforming the integrand of ö2WD = by applying (10—81). Since bifurcation corresponds to the existence of an alternate equilibrium position, it is more convenient to form the incremental equations directly. The equations for the case of linearly elastic material and prescribed external forces are listed below. f See Sec. 7—6 for a derivation of the classical stability criterion. See Probs. 10—11, 10—18.
PRINCIPLE OF VIRTUAL FORCES
SEC. 10—7.
L
Equilibrium Equation in the Interior
+ 2.
=
+
0
= 1,2,3
Stress-Boundary Force Equations on
+ 3.
257
+
Au1
0
J = 1, 2, 3
(10—87)
Stress-Strain Relations
= D 4.
Strain-Displacement Relations
=
3 + AUJ, +
=
Au1
10—7.
AIIm, j + Urn, j Am,
0
PRINCIPLE OF VIRTUAL FORCES; PRINCIPLE OF STATIONARY COMPLEMENTARY ENERGY
Let u1 be the actual displacements in a body due to some loading and the geometrically linear strain measures corresponding to u1. The strain and displacement measures are related by
=
+
u1=fl Once the strains are known, we can find the displacements by solving (a) and enforcing (b). The principle of virtual forces is basically a procedure for determining displacements without having to operate on (a). It applies only for linear geometry. We developed its form for an ideal truss in Sec. 7—3. We will follow the same approach here to establish the three-dimensional form. The essential step involves selecting a statically permissible force system, i.e., a force system which satisfies the linear equilibrium equations. For the continuous case, the force system consists of stresses, surface forces, on Static permissibility requires and reactions, on Ac31,3 = 0
on on
= =
(10—88)
If we multiply e13 by Ac13, integrate over the volume using (10—81), and note the static relations, we obtain the following identity,t Acr13 dx1 dx2 dx3 f See Prob. 10—26.
=
u1
0,,
+ $ Th
(10—89)
258
GOVERNING EQUATIONS FOR A DEFORMABLE SOLID
CHAP. 10
which is referred to as the principle of virtual forces (or stresses). This result is
applicable for arbitrary material behavior. 1-lowever, the geometry must be linear.
in the direction defined by is Suppose the translation at a point Q on desired (see Fig. 10—15). Let d0 be the displacement. We apply a unit force at Q in the tq direction and generate a statically permissible stress field. (1) 1q at
The integral on
point Q
and
Acr
reduces to (l)dQ, and it follows that dx1 dx2 dx1 —
=
(10—90)
A second application is in the force method, where one reduces the governing
equations (stress equilibrium and stress displacement) to a set of equations
Fig. 10—15. Notation for determination of the translation at point Q.
involving only force unknowns. We start by expressing the stress field in terms
of a prescribed distribution
and a "corrective" field + cit,
(10—91)
is a particular solution of the equilibrium equations which satisfies the boundary conditions on where
+
0
= and
Thu
on
(10-92)
satisfies
= =
0 0
on
(10—93)
on
Stress fields satisfying (10—93) are called seljequilibrating stress fields. For the
ideal truss, a-° corresponds to the forces in the primary structure due to the prescribed loading and ? represents the contribution of the force redundants.
PRINCIPLE OF VIRTUAL FORCES
SEC. 10—7.
259
The governing equations for the force redundants were obtained by enforcing
geometric compatibility, i.e., the bar elongations are constrained by the requirement that the deformed bar lengths fit in the assembled structure.
Geometric compatibility for a continuum requires the strains to lead to continuous displacements. One can establish the strain compatibility equations by operating on the strain- displacement relations. This approach is described
in Prob. 10—10. One can also obtain these equations with the principle of virtual forces by taking a self-equilibrating force system. Letting Aox, Apc denote the virtual stress system, (10--89) reduces to
dx1 dx2 dx3 =
(10—94)
The compatibility equations are determined by expressing in terms of stress functions and integrating the left-hand term by parts. We illustrate its application to the plane stress problem.
Example 10—3 If the stress components associated with the normal direction to a plane are zero, the stress state is called planar. We consider the case where
=
= 033 = 0. The
equilibrium equations and stress-boundary force relations reduce to + b1 = 0 2 + b2 = 0
+ 012 5 +
=
+
a,,2o21
+
•z,,2a22
The stress field, oi,, must satisfy (a) with h1 = h2 = 0 and also p,,1 = = 0 on We can satisfy the equilibrium equations by expressing in terms of a function, follows:t 033 = = = The boundary forces corresponding to
are
=
Pa
OS
where s is the arc length on the boundary (sense is from X1 —* X2).
Substituting for crC,
pC
in terms of i/i, (10-94) expands to + a2
if —
f
Ps
CS
There is no loss in generality by taking 22
f See Prob. 10—14.
11 — y52
+ a2,
=
0
— 712,
12)dx1
0 ,i
on S. Then, integrating (e) by parts, dx1 dx2
0
as
260
GOVERNING EQUATIONS FOR A DEFORMABLE SOLID
CHAP. 10
results in the strain compatibility equation,
and requiring (f) to be satisfied for arbitrary
+
0
—
which is actually a continuity requirement U1 122
We express (g) in terms of
+
211 — (u1 212 + 112 112) = 0
by substituting for the strains in terms of the stresses.t
The principle of virtual forces is also employed to generate approximate solutions for the stresses. It is convenient to shift over to matrix notation for this discussion, and we write (10—94) as dx1 dx2 dx3
ApC
if
We express the stress matrix in terms of prescribed stress states and unknown parameters, a1,
where
= =
+ +
+ (12(l)2
satisfies (10—92) and
(1
+
+ 04,,
'
1, 2,...,r) are self-equilibrating stress
states, i.e., They satisfy the homogenous equilibrium equations and boundary conditions on The corresponding surface forces arc p
p°
+
= p° + 0101 + 0209 + = p (i = 1, 2, = 0
(i =
Taking virtual-force systems corresponding to equations for the parameters. dx1 dx2 dx3 =
jjT9.
1
1, 2,
., r) results in r
1, 2,...,r
(10—97)
In order to proceed, we need to introduce the material properties. When the material is linearly elastic,
+ Ai'=
+
+
and the equations expand to
=
d1
f,j = d, =
i,j =
1,
2,....r
dx1 dx2 dx3 —
r$J1T(a°
(10—98)
+ A6°)dx1 dx2 dx3
One should note that (10—97) are weighted compatibility conditions. The true stresses must satisfy both equilibrium and compatibility throughout the t See Prob. 10—27.
PRINCIPLE OF VIRTUAL FORCES
SEC. 10—7.
261
the corrective stress field since it is required to correct the compatibility error due to For completeness, we describe here how one establishes a variational principle for Our starting point is (10—94) restricted to elastic behavior. We define = according to = c5V* = (10—99) domain. We call
and call V* the complementary energy density. The form of V* for a linearly elastic material is = (10—100) + By definition, V* complements V, i.e.,
V+
=
(10—101)
Then, letting
=
cjx1 dx2 dx3
(10—102)
we can write (10—94) as 0
(511. *
TiC
—
for arbitrary = $$
(10—103)
This form is called the principle of stationary complementary energy and shows
that the true stresses correspond to a stationary value of Since is linear in the second variation of reduces to (52fl
=
=
dx1 dx2 dx3
(10—104)
We shift over to matrix notation and express öe as
=
A,
(10— 105)
represents the tangent compliance matrix. Now, must be positive definite in order for the material to be stable.t Then, (5211. > 0 for arbitrary and we see that the solution actually corresponds to a relative minimum where
value of The approximate method described earlier can be applied to 11g. Substituting for given by (10—95) converts to a function of the stress parameters When the material is linearly elastic, (a1. a2, . , ar). .
.
H, =
—
ard + const
(10—106)
The equations for the stress parameters follow by requiring H. to be stationary for arbitrary (511, = — ci) = 0
fa=d A,
(10—107)
The classical stability criterion specialized for elastic material and linear geometry requires SCTD, & > 0 for arbitrary Sc which, in turn, requires D, to be positive definite. Since D1', it follows that A must be positive definite for a stable material.
262
GOVERNING EQUATIONS FOR A DEFORMABLE SOLID
CHAP. 10
Operating on
c52fl = LtaTfLui
(10—108)
and noting that ö211. > 0, we conclude that f is positive definite. REFERENCES I.
CRANDALL, S. J., and N. C. DAHL: An Introduction to the Mechanics of Solids, McGraw-Hill, New York, 1959.
2.
BISPLINGHOFF, R. L., MAR., J. W., and T. H. H. PlAN:
3. 4.
5. 6. 7. 8. 9. 10.
of Deformable Solids. Addison-Wesley, Reading, Mass., 1965. WANG, C. T.: Applied Elasticity, McGraw-I-jill, New York, 1953. TIMOSHENKO, S. J., and J. N. GooDiag: Theory of Elasticity, 3d ed., McGraw-Hill, New York, 1970. SOKOLNIKOFF, I. S.: Mathematical Theory of Elasticity, 2d ed. • McGraw-Hill. New York, 1956. FUNG, Y. C.: Foundations of Solid Mechanics, Prentice-Hall, 1965. LEKIINITSKU, S. G. Theory of Elasticity of an Anisotropic Elastic Body, Holden-Day, San Francisco, 1963. WAsmzu, K. Variational Methods in Elasticity and Plasticity, Pergamon Press, 1968. HLDEBRAND, F. B.: Methods of Applied Mathematics, Prentice-Hall, 1965. CRANDALL, S. J.: Engineering Analysis, McGraw-Hill. New York, 1956.
PROBLEMS 10—1.
Write out the expanded form of the following products. Consider
the repeated indices to range from 1 to 2. (a) (b)
+ u1, ± Urn,
10—2.
where
=
+ Urn. k) —
Let f be a continuous function of x1, x2, x3. Establish the trans-
formation laws for and (3Xk. 10—3. Establish the transformation law for tensors. 10—4. Prove that eJk =
Jbk where
are cartesian
— ôJk)
is a second-order cartesian tensor. Hint: Expand (3/3
P.
(If)
P.
10—5. Equations (10—19) are the strain transformation laws. Since is a symmetrical second-order cartesian tensor, there exists a particular set of directions, say Xi', for which is a diagonal array. What are the strain components for the frame? Consider a rectangular parallelepiped having sides dXy in the undeformed state. What is its deformed shape and relative change with respect to its initial volume? Specialize the expression for in volume, for small strain. Then determine for the initial (Xi) directions and small strain. Finally, show that r.., is invariant.
PROBLEMS
263
10—6.
(a)
Specialize (10—19) for small strain and write out the expressions for
(b)
Let
in terms of ei, 62, .
Develop the form of (c)
.
P13•
P12, P23, y31}. We can express the strain trans= formation (small strain) as = using the results of part a.
Evaluate TE in terms of cos 0, sin 0 for the rotation shown below. Comment on the transformation law for the out-of-plane shear strains P32. Prob. 10—6
x2
10—7.
Tn the Eulerian approach, the cartesian coordinates
for the
deformed state are taken to be the independent variables, i.e., =
Ui
Xj(f/k)
Almansi's strain tensor is defined as — (ds)2
=
2Efk
thik
Determine the expression for EJk in terms of the displacements. Compare the result with (10—21). 10—8.
Consider the case of two-dimensional deformation in the X1-X2 be the extensions in the a, b, c direc-
plane (83 = P13 = P23 = 0). Let 6b, tions defined below and let 6N = {8a, 6b,
We can write
= BE C= (a) (b)
Determine the general form of B. Determine for = 0, 9b Determine B1 for Oa = 0, 6h = 60°,
= 90c. = 120°. (d) Extend (a) to the three-dimensional case. Consider six directions having direction cosines GJ2, with respect to X1, X2, X3. Can we select the six directions arbitrarily? (c)
GOVERNING EQUATIONS FOR A DEFORMABLE SOLID
264
CHAP. 10 Prob. 10—8
xz
-a
For small strain, the volumetric strain is
10—9.
= Rather than work with
+
+
C3
= eti + &22 +
one can express it as the sum of two tensors,
=
+
is called the spherical strain tensor,
where
=
is the deviator strain tensor. and (a) Write out the expanded form for (b) Determine the first invariant of of ejj.
and and compare with the invariant
This question concerns strain compatibility equations. Show that
10—10.
(a)
+
=
+
CX,, cXj, — 8X,,,
(?X,,
where eflk =
ek,
=
1
(CII,,
+
—
2 \CXk
and k, in, n range from I to 3. This expression leads to six independent conditions, called geometric compatibility relations, on the strain measures. (b)
Show that for two-dimensional deformation in the X1-X2 plane = 0; this called plane strain) there is only one com= 813 = patibility equation, and it has the following form: 22 + 83 11 = Is
Y12. 12
the following strain state permissible?
=
+
82 = kx2 Y12 = 2kx1x2 k
= constant
PROBLEMS 10—11.
265
Equation (10—21) defines the strain measures due to displacements,
To analyze geometrically nonlinear behavior, one can employ an incremental formulation. Let represent the displacement increment and Ae1k the incremental strain. We write + = where
contains linear terms (Aug) and öeJk involves quadratic terms. The 5-symbol denotes the first-order change in a functional and is called the variational operator (see Ref. 8). We refer to 5e as the first variation of e. Determine the expressions for 10—12.
Let
i,, be the unit vector defining the initial orientation of the
differential line element d1,, at a point.
=
dsi,,
1,
=
The unit vector defining the orientation in the deformed state is = (1 +
= Determine the general expression for Then specialize it for small strain. 10—13. The several parts of this question concerns stress transformation. in terms of (a) Starting with (i0--41). write out the expressions for all, a22, . , = stress matrix. We express the (b) Let a22, a33, ai2, = stress transformation as a matrix product. .
a' = T,a Develop the form of T,, using the results of part a. (c)
Evaluate 1',, in terms of cos (9, sin (9 for the axes shown. Prob. 1O—13
x2
x,t
xl (d)
Plane stress refers to the case where a13 = with reduced stress and strain matrices,
a23
=
a33
=
0.
We work
{a11, a22,
Er
= and write the transformations in the same form as the three-dimensional case:
a' = a'
=
CHAP. 10
GOVERNING EQUATIONS FOR A DEFORMABLE SOLID
266
Evaluate T,. from part c above and T, from
Prob. 10—6. Verify that
13
10—14. This question develops a procedure for generating self-equilibrating stress fields. (a) Expand the linear equilibrium equations, (10—49) and (10—50). (b) Specialize the equilibrium equations for plane stress (a13 = a23 =
= 0). Suppose we express the two-dimensional stress components in terms of a function = as follows: a33
(c)
a11 = t1'.22 a22 = tI'. ii
—
a12
—
b1
dx1
fx7 b2
dx2
= —1//,12
The notation for body and surface forces is defined in the following sketch. Prob.
10—14
x2
x1
Verify that this definition satisfies the equilibrium equations in the interior. Show that the expressions for and P2 ifl terms of derivatives
with respect to x1, x2, and s are Pi
T t/"1
=
P2
10—15.
b1
—
dx1 b2
—
dx2
The mean stress, a,,,, is defined as am
Rather than work with
=
we
+
a22
+ a33)
can express it as the sum of two tensors, — —
L
aU
PROBLEMS
267
is called the spherical stress tensor,
where
=
óijOrn
and
is the deviator stress tensor. Write out the expanded forms for and Determine the first invariant of 10-46. Establish the stress-equilibrium equations for small-finite rotation and small strain. (a) (b)
10—17.
Starting with (10—52), (10—55) specialized for small strain, establish
Au, Ab*, and the incremental equilibrium equations in terms of Group according to linear and quadratic terms. Specialize these equations
for the case where the initial position is geometrically linear, i.e., where approximate with in the incremental equations. 10—18. Prove (10—60). Hint:
= /= 10—19.
can
+ P.k
Verify that the inverted form of(l0—71) is D(e —
where D11 = E1/C3
D12
D13
C4D11
D22 = E2/C1 + (C2/C1)D12 D23 = v32E2/C1 + (C2/C1)D13 D31 = E3 + v31D13 + v32D23
and
C1 = C2 =
1
—
v21
+ v31v32(E2/E3)
E2C1 C4 =
v31
+
"32
= 0) = = Consider 2 sets of orthogonal directions defined by the unit vectors The stress-strain relations for the two frames are
Specialize for plane strain 10—20.
and
=
+ (a°)' + A'&
Express A' in terms of A and Also determine D'. 1O--21. Consider the three-dimensional stress-strain relations defined by (10—71).
(a)
Specialize for plane stress
=
=
= 0).
GOVERMNG EQUATIONS FOR A DEFORMABLE SOUD
268
CHAP. 10
Let
(b)
a22, cri2}
C= C=
62, Y12}
Verify that D has the following form: V2t
0
I
1 G (1 —
n
=
E2
Assuming X1-X2 in the sketch are material symmetry directions, determine D' for the X'1-X'2 frame. Use the results of Prob. 10—13, 10—20. What relations between the properties are required in order for D' to be identical to D? Prob. 10—21
x2
xI 10—22. Verify (10—73). Start by requiring equal properties for the X2 and X3 directions. Then introduce a rotation about the X1 axis and consider the Isotropy in the X2-X3 plane requires expression for Y23
=7
I.
023
10—23. Verify that the directions of principal stress and strain coincide for an isotropic material. Is this also true for an orthotropic material? 10—24.
Equations (10—76) can be written as
+
a11 = a°&1 +
where
2Ge11
is the volumetric strain. Using the notation introduced in Probs. 10—9
and 10—15——
(a)
Show that
=
Ka,,
+ a0
PROBLEMS
where K is the bulk modulus = (E/3(1 (b)
269 —
2v)). Discuss the case where
Show that
= (c)
Verify that the strain-energy density can be written as V
—
= = (d)
—
+
+
for the isotropic case. Determine and When v = We must work with 7 stress measures ('u' Urn) = and the mean stress has to be determined from an equilibrium consideration. Summarize the governing equations for the incompressible case.
Prove (l0--81) for the two-dimensional case. Is this formula restricted to a specific direction of integration on the boundary? Does it apply for a multi-connected region, such as shown in the figure below? .10—25.
Prob.
10—26.
10—25
Verify Equation (10—89).
Refer to Example 10—3. Express (g) in terms of material to be orthotropic. 10—28. Verify that the stationary requirement 10—27.
=0
Consider the
for arbitrary
where
=
—
—
dx2
— —
= Kirchhoff stress = Lagrange strain =
—
+
+
1u,,,,
= complementary energy density (initial volume) = prescribed force measures (initial dimensions) leads to the complete set of, governing equations for an elastic solid, i.e., stress equilibrium equations 1. 2. stress-displacement relations 3. stress boundary conditions on 4. displacement boundary conditions on 5. expressions for the reaction surface forces on
GOVERNING EQUATIONS FOR A DEFORMABLE SOLID
270
CHAP. 10
This variational statement is called Reissner's principle (see Ref. 8).
Transform HR to by requiring the stresses to satisfy the stress displacement relations. Hint: Note (10—101). (b) Transform 11R to — by restricting the geometry to be linear = and (ui, + and requiring the stresses to satisfy the stress equilibrium equations and stress boundary conditions on Hint: Integrate by parts, using (10—8 1). (a)
10—29.
Interpret (10—90) as dQ ==
where PQ is a force applied at Q in the direction of the displacement measure, dQ.
11
St.
Venant Theory of of
Prismatic Members 11—1.
INTRODUCTION AND NOTATION
A body whose cross-sectional dimensions are small in comparison with its axial dimension is called a member. If the centroidal axis is straight and the the member shape and orientation of the normal cross section are is said to be prismatic. We define the member geometry with respect to a global reference frame (X1, X2, X3), as shown in Fig. 11—1. The X1 axis is taken to coincide with the centroidal axis and X2, X3 are taken as the principal inertia directions. We employ the following notation for the cross-sectional properties: A = if dx2 dx3 = dA 12 — Sj(x3)2 dA
13 = fl(x2)2 dA
Since X2, X3 pass through the centroid and are principal inertia directions, the centroidal coordinates and product of inertia vanish:
'23
jJx2x3 dA = 0
One can work with an arbitrary orientation of the reference axes, but this will
complicate the derivation. St. Venant's theory of torsion-flexure is restricted to linear behavior. It is an
exact linear formulation for a prismatic member subjected to a prescribed t The case where the cross-sectiona' shape is constant but the orientation varies along the centroidal axis is treated in Chapter 15. 271
TORSION-FLEXURE OF PRISMATIC MEMBERS
272
CHAP. it
distribution of surface forces applied on the end cross sections. Later, in
Chapter 13, we modify the St. Venant theory to account for displacement restraint at the ends and for geometric nonlinearity. x2
F3
Fig. 11—i. Notation for prismatic member.
The distribution of surface forces on a cross section is specified in terms of its statically equivalent force system at the centroid. Figure 11—1 shows the Stress components on a positive face. We define M.. as the force and moment vectors acting at the centroid which are statically equivalent to the distribution of stresses over the section. The components of F.., M÷ are called stress resultants and stress couples, respectively, and their definition equations are
F1 = ffcrij c/A M1 = M2 = M3 =
F2
c/A
F3 =
JJ(x2cr13 — x3c12)dA
JJx3crj1 dA
JJcc13 c/A
(11—3)
dA
The internal force and moment vectors acting on the negative face are denoted
byF_,M_. Since
F_ =
—F÷
M_ =
(11—4)
it follows that the positive sense of the stress resultants and couples for the negative face is opposite to that shown in Fig. 11 —1. We discuss next the pure-torsion case, i.e., where the end forces are statically equivalent to only M1. We then extend the formulation to account for flexure
SEC. 11—2.
THE PURE-TORSION PROBLEM
273
and treat torsional-flexural coupling. Finally, we describe an approximate procedure for determining the flexural shear stress distribution in thin-walled
sections. 11—2.
THE PURE-TORSION PROBLEM
Consider the prismatic member shown in Fig. 11—2. There are no boundary forces acting on the cylindrical surface. The boundary forces acting on the end
cross sections arc statically equivalent to just a twisting moment M1. Also, there is no restraint with respect to axial (out-of-plane) displacement at the ends.
The analysis of this member presents the pure-torsion problem. In what follows, we establish the governing equations for pure torsion, using the approach originally suggested by St. Venant.
0
Fig. 11—2. Prismatic member in pure torsion.
Rather than attempt to solve the three-dimensional problem directly, we impose the following conditions on the behavior and then determine what problem these conditions correspond to. 1.
2.
Each cross section is rigid with respect to deformation in its plane, i.e., = 723 = 0. = Each cross section experiences a rotation w5 about the X3 axist and an out-of-plane displacement u1.
These conditions lead to the following expansions for the in-place displacements: 112 = —C01X3 03
(11—5)
+W3.3.2
The corresponding linear strains are 13 = a3 = Li —Ui 1
U12 +
712 713
Y23
=
U5,
+ U3,
0
012 —
=
(11—6)
05,3 + X20)1, 1
t Problem 11—i treats the general case where the cross section rotates about an arbitrary point.
274
TORSION-FLEXURE OF PRISMATIC MEMBERS
CHAP. 11
Now, the strains must be independent of x1 since each cross section subjected to the same moment. This requires = const = k1 = u1(x2, x3)
is
We consider the left end to be fixed with respect to rotation and express co1,u1aS
=
k1x1
(11—8)
U.1 =
x3) defines the out-of-plane displacement (warping) of a cross = section. The strains and stresses corresponding to this postulated displacement behavior are = 0 = C3 Cl =
where
712
=
x3)
+ x2)
713
and
a11 =
a12 =
Gy12
U13 =
Go'13
=
a22
=
a33
O'23 = 0
=
2 — x3)
+ x2)
a12(x2, x3)
(11—10)
a1 3(x2, x3)
We are assuming that the material is and there are no initial strains. One step remains, namely, to satisfy the stress-equilibrium equations and stress boundary conditions on the cylindrical surface. The complete system of linear stress-equilibrium equations, (10—49), reduces to
=
U21,2 +
0
(11—11)
Substituting for the shearing stresses and noting that Gk1 is constant lead to the differential equation (11-12)
which must be satisfied at all points in the cross section. The exterior normal n for the cylindrical surface is perpendicular to the X1 direction. Then 0, and the stress boundary conditions, (10—49), reduce to Using (11—10), the boundary condition for 2 — x3)
=0
+
Pfli =
+
(11—13)
is
+ x2) =
0
(11—14) —
t Problem 11 —3 treats the orthotropic case.
(on S)
THE PURE-TORSION PROBLEM
SEC. 11—2.
275
The pure-torsion problem involves solving V2q5, = 0 subject to (11—14). Once
çb, is known, we determine the distribution of transverse shearing stresses from (11—10). Note that depends oniy on the shape of the cross section.
The shearing stress distribution must lead to no shearing stress resultants:
dA =
F2 =
F3 = J$a13 dA
This requires
0
0
JJ ('X3
J J OX2
To proceed further, we need certain integration formulas. We start with dA =
if
(IS
dA leads to
which is just a special case of (10—81). Applying (11—15) to
Green's theorem, JJV2VJdA
0X2
.1
If
11—16
ôn
=
is a harmonic function (i.e.,
+
0),
Green's theorem requires
dS =
0
Now, /, is a harmonic function. For the formulation to he consistent, (11—14) must satisfy (c). Usiiig (11—15), (c) transforms to
#(XH2x3
=
—
0
is specified on the boundary, we cannot apply (11—15) directly Since to (b). In this case, we use the fact that = 0 and write cxi
ax2
ax2j
ox3 \
ax31
(j=2,3)
Integrating (e),
(j=2,3) and then substituting for the normal derivative, verifies (h). The constant k1 is determined from the remaining boundary condition,
=
J$(x2c13
—
x3c12)dA
(11—17)
TORSION-FLEXURE OF PRISMATIC MEMBERS
276
CHAP. 11
We substitute for the shearing stresses and write the result as
Gk1J where J is a cross-sectional property,
+
if
— X3
dA
+ = At this point, we summarize the results for the pure-torsion problem. 1.
Displacements
= 02
U3 = W1X2
= k1x1
=
k1
2.
(if
Stresses M3
J
\(;X2
A'11
+ X2
0j3 = —H J 3.
(11—20)
Governing Equations
mA: on S:
—
It is possible to obtain the exact solution for for simple cross sections. The procedure outlined above is basically a displacement method. One can also use a force approach for this problem. We start by expressing the shearing stresses in terms of a stress function so that the stress-equilibrium equation (Equation 11—li) is identically satisfied. An appropriate definition is 012
X3
(11—21)
013
The shearing stresses for the 2,
v
directions, shown in Fig. 11—3, follow directly
from the definition equation 01A
cv
0lv = —
(11—22) CA.
SEC. 11-2.
THE PURE-TORSION PROBLEM
277
Taking S 900 counterclockwise from the exterior normal direction, and noting
that the stress boundary condition is
=
0,
lead to the boundary condition
fort/i,
= const on S
(11—23)
We establish the differential equation for t/i by requiring the warping function be continuous. First, we equate the expressions for a in terms of t/i and M1
a12 =
—
a13 =
x3)
+ x2)
=
Now, for continuity,
= Operating on (a), we obtain
= It is convenient to express t/.'
as
(11—24)
aret
The governing equations in terms of
a12 = =
M1 dt/ (11--25)
j
a 13
and
=
(mA)
—2
(on boundary S1)
tJi =
(11—26)
Substituting (11—25) in the definition equation forM1 leads to the following expression for J:
cr7 —
JJ \.
CX3J
that
Applying (10—8 1) to (a) and —
dS =
A1
=
area enclosed by the interior boundary curve, S1
=
C1
(b)
= const
t Equations (11—26) can be interpreted as the governing equations for an initially stretched membrane subjected to normal pressure. This interpretation is called the "membrane See Ref. 3.
The S direction is always taken such that n — S has the same sense as X2 — X3. Then, the + S direction for an interior boundary is opposite to the + S direction for an exterior boundary since the direction for n is reversed. This is the reason for the negative sign on the boundary integral.
278
we can write
TORSION-FLEXURE OF PRISMATIC MEMBERS
J=
dA +
CHAP. 11
(11—27)
= 0 on the exterior boundary. To determine the constants C, for the multiply connected case, we use the is continuous. This requires fact that where
Js ('IS
dS
(11—28)
0
for an arbitrary closed curve in the cross section,
x
x
x2
0
Fig. 11—3. Definition of n-s and A.-vdirections. x3
Fig. 11—4. Graphical representation of sector area.
THE PURE-TORSION PROBLEM
SEC. 11—2.
279
Consider the closed curve shown in Fig. 11 —4. The shearing strain
is
given by
Yis =
ct52y12
+ 0t53y13
Using (11—9), we can write (a) as
2+
Yis =
=
k1
3 — xacls2
+ (11-29)
+
where p is the projection of the radius vector on the outward normal.t The magnitude of p is equal to the perpendicular distance from the origin to the tangent. Integrating between points P, Q, we obtain = where APQ
=
r50 p
J
dS =
+ 2APQ)
—
(11—30)
area enclosed by the arc PQ and the radius vectors to P and Q. sector
Finally, taking P = 5dS = 2k1A5
(11—31)
where A5 denotes the area enclosed by the curve. Since
=
we can
write
(11-32)
2Gk1A5 =
Note that the +S direction for (11—32) is from .X2 toward X3. Also, this result is independent of the location of the origin. Instead of using (11—9), we could have started with the fact that the cross section rotates about the centroid. The displacement in the + S direction follows from Fig. 11—4:1 x
u,5 =
Substituting for
is)
=
w1p
+
s
k1x1p
(11—33)
in
Yss =
Us
(11—34)
and noting that Ut =
lead to (11—29). Using (11—22), we can write
=
M5
=
(11—35)
t This interpretation of p is valid only when S is directed from X2 to X3, i.e., counterclockwise for this case. See Prob. 11—14 for an alternate derivation.
§ This development applies for arbitrary choice of the +S direction. The sign of p is positive if a rotation about X1 produces a translation in the +S direction. Equation (11—29) is used to determine the warping distribution once the shearing stress distribution is known. See Prob. 11—4.
TORSION-FLEXURE OF PRISMATIC MEMBERS
280
Then, substituting for
CHAP. 11
in (11—32), we obtain
=
(11—36)
.3 s
where n is the outward normal, A5 is the area enclosed by S, and the + S sense is from X2 to X3. This result is valid for an arbitrary closed curve in the cross section. We employ (11—36) to determine the values of 17 at the interior boundaries of a multiply connected cross section. It is of interest to determine the energy functions associated with pure torsion. When the material is linearly elastic and there are no initial strains, the
strain and complementary energy densities are equal, i.e., V = We let
V dA
V
strain energy per unit length
(11—37)
The strain energy density is given by
+
V= Substituting for Y12' '/13,
V=
X3)
2
+(
+
x2)j 2
and integrating (b) over the cross section, we obtain
V= Since
=
V,
(11—38)
and M1 = GJk1, it follows that =
+
(11—39)
xl
WI
dx1
Fig. 11—5. Differential element for determination of the rotational work.
Instead of integrating the strain-energy density, we could have determined the work done by the moments acting on a differential element. Consider the element shown in Fig. 11—5. The boundary forces acting on a face are statically equivalent to just a torsional moment. Also, the cross sections are rigid in
THIN-WALLED OPEN CROSS SECTIONS
SEC. 11—S.
281
their plane and rotate about X1. The relative rotation of the faces is
/((01 + dw1 '\ —dx1 — dx1
and the first-order reduces to
=
dx1
,i
workdone by the external forces due to an increment in wj
5WE =
M1 ,Xk1 dx1
Now,
=
=
dx2dx3 = óVdx1
5jJ
for an elastic body. Then, expanding ö V.
and it follows that dk1
= M1 =
GJk1
V=
11—3.
APPROXIMATE SOLUTION OF THE TORSION PROBLEM FOR THIN-WALLED OPEN CROSS SECTIONS
We consider first the rectangular cross section shown in Fig. 11—6. The exact solution for this problem is contained in numerous texts (e.g., see Art. 5—3 of Ref. 1) and thcrefore we will only summarize the results obtained. x3
I
dl 2
d, 2
HFigS 11—6. Notation for rectangular section.
282
TORSION-FLEXURE OF PRISMATIC MEMBERS
CHAP. 11
occurs at x2 = ± t/2, x3 =
When t d, the maximum shearing stress (points 5, 6). The exact expressions are
0
dt3
J = K1— (11-41) =
K2t
where
K1 =
192 (t'\ 1
I
tanh
— =
K2 =
8
1
1
1
(2n+1)2 cosh
—
A,,
2n+1 Id Values of K1, K2 for d/t ranging from 1 to 10 are tabulated below: d/t 1
K1
K2
0.422
0.675 .930 .985 .997 .999 1.000
3
.687 .789
4
.843
2
5
.873
10
0.936
If t d, we say the cross section is thin. The approximate solution for a thin rectangle is J 4dt3 (113
2—-—x2
=
2Gk1x2
(11-42) x2x3
(t)2 take d/t = in the exact solution.) The shearing stress across the thickness and
(We
M1
varies linearly
3M1
A view of the warped cross section is shown in Fig. 11—7.
Since the stress function approach is quite convenient for the analysis of thin-walled cross sections, we illustrate its application to a thin rectangular
THIN-WALLED OPEN CROSS SECTIONS
SEC. 11—3.
283
section. Later, we shall extend the results obtained for this case to an arbitrary thin walled open cross section. The governing equations for a simply connected cross section are summarized below for convenience (see (11—26), cross
(11—27)):
=
—2
(in A)
0
(on the boundary)
= J =
(1A
where the S direction is 900 counterclockwise from the is direction.t Since t is small and a12, the shearing Stress component in the thickness direction, must
_:k,1,
Fig. 11—7. Warping function for a rectangular cross section.
= 0 at all points independent of x3. The
vanish on the boundary faces, it is reasonable to assume
in the cross section. This corresponds to taking equations reduce to d2
=
—2
Solving (b), we obtain
-
J
=
dx2
=
M1 = ——---—-- = 2——x2
J
t This applies for X3 counterclockwise from X2. The general requirement is the n — S sense must coincide with the X2-X3 sense.
284
TORSION-FLEXURE OF PRISMATIC MEMBERS
The expression for (x3
CHAP. 11
developed above must be corrected near the ends
± d/2) since it does not satisfy the boundary condition, ti
This will lead to a12 0 near the ends, but will have a negligible effect on J Actually, the moment due to the approximate linear expansion for and is equal to only one half the applied moment: I't/2
x2a13 dx, =
d
P4
/1
dt3) =
J
The corrective stress system (a12) carries M1/2. This is reasonable since, even is small in comparison to amax, its moment arm is large. though
-f-s
t(s)
Fig. 11—8. Notation for thin-walled open cross section.
We consider next the arbitrary thin-walled open cross section shown in Fig. 11 —8. The S curve defines the centerline (bisects the thickness) and the n direction is normal to S. We assume = 0 and take = —n2 + t2/4. This corresponds to using the solution for the thin rectangle and is reasonable when S is a smooth curve. The resulting expressions for I and are
J=
4
t3 dS
(11—43)
M1
a15, ma, =
= Gkitrnax
THIN-WALLED OPEN CROSS SECTIONS
SEC. 11—3.
The
285
results for a single thin rectangle are also applied to a cross section
consisting of thin rectangular elements. Let of element i. We take J as
t1 denote the length and thickness
J=
(11—44)
Asan illustration, consider the symmetrical section shown in Fig. 11—9. Applying (11—44), we obtain 3 1'ff + w4v ..i
The
3
maximum shearing stress in the center zone of an element is taken as M1
= —7t1 = Gk1t1
(11—45)
In general, there is a stress concentration at a reentrant corner (e.g., point A in Fig. 11—9) which depends on the ratio of fillet radius to thickness. For the case bi
+
(Iw
I Fig. 11—9. Symmetrical wide-flange section.
of an angle having equal flange thicknesses, the formulat
=
\
+
(1146)
4rf/
is the fillet radius and 0rn is given by (14—45), gives good results for where rf/t < 0.3. The stress increase can be significant for small values of rf/t. For example, for Tf = 0,lt. Numerical procedures such as finite differences or the finite element must be resorted to in order to obtain exact solutions for irregular sections. -
t See Ref. 2 and Appendix of Ref. 9. See Ref. 4.
CHAP. 11
TORSION-FLEXURE OF PRISMATIC MEMBERS
286
11—4.
APPROXIMATE SOLUTION OF THE TORSION PROBLEM FOR THIN-WALLED CLOSED CROSS SECTIONS
The stress function method is generally used to analyze thin-walled closed cross sections. For convenience, the governing equations are summarized below (see (11—26), (11—27), (11—36)):
(in A)
—2
(on the exterior boundary) (on the interior boundary, S,) = area enclosed by
— ci J
—
dA +
S
and +S sense from X2 toward X3,)
—
£
j
on
dS
=
—2A5
We consider first the single cell shown in Fig. 11—10. The curve defines the centerline. Since there is an interior boundary, we have to add a term n S.
E
Sect. E-E
Fig. 11—10. Single closed cell.
involving C1 to the approximate expression for We take as + tz
where
used for the open section.
2n\
(11—47)
represents the contribution of the interior boundary. This expression
SEC. 11—4.
satisfies
THIN-WALLED CLOSED CROSS SECTIONS
287
the one-dimensional compatibility equation and boundary conditions, 2
=
atn— +t/2 n = — t/2
at
C1
(a)
and is a reasonable approximation when S is a smooth curve. Differentiating b
(fl
and substituting (b) in the expressions for the shearing stress components lead to 0
M1 /
+ C1\ 7)
(11—48)
cr?5 + The tangential shearing stress varies linearly over the thickness and its average We let q be the shear stress resultant per unit length along S, value is positive when pointing in the + S direction, 1/2
q
=
(11—49)
cr15
J —1/2
and call q the shear flow. Substituting for a
we find (11—50)
q
The additional shearing stress due to the interior boundary (i.e., closed cell) corresponds to a constant shear flow around the cell. One can readily verifyt that the distribution, q = const, is statically equivalent to only a torsional moment,
given by
=
(11—51)
The torsional constant is determined from
J Substituting for
dA + 2C1A1
M1/Gk1
(a)
using (11—47), we obtain
j = Jo + 4
(11—52)
=
t3 dS
Equation (a) was established by substituting for the shearing stresses in terms in the definition equation forM1 and then transforming the integrand. We could have arrived at (11—52) by first expressing the total torsional moment as
M1 = See Prob. 11-5.
+
(11—53)
288
TORSION-FLEXURE OF PRfSMATIC MEMBERS
CHAP. 11
is due to the closure. Next,
where MI is the open section contribution and
we write
M1 = GkIJ
it'll =
Gk1J°
J=
+ JC
= Gk1J'
(11—54)
Then, J0
(11—55)
and it follows that Jc
Jo
(11-56)
Finally, using (11—5 1), we can express JC JC
as
= Mu/(M1/J) =
(11—57)
This result shows that we should work with a modified shear flow, C
q/(M1/J)
(11—58)
rather than with the actual shear flow. Note that C C1 for the single cell. It remains to determine C1 by enforcing continuity of the warping function on the centerline curve. Applying (11—32) to
we have
=
(11—59)
Substituting for
q/t
M1C1
=
leads
(1160) One should note that C1 is a property of the cross section. Once C1 is known, we can evaluate .J from (11—52) and the shearing stress from
M1 (
(11—61)
+ -i—,)
Example 11—1 Consider the rectangular section shown. The thickness is constant and a, b are centerline properties are dimensions. The various CdS
Cl =
t See Prob. 11—6.
2(a+h)
SEC. 11-4.
THIN-WALLED CLOSED CROSS SECTIONS
289
We express J as
=
+
For this section,
1 (r'Y /
J°
We consider a >
b.
Then, Jo
J'
(t'Y
= 01— \\h
The section is said to be thin-walled when
c< b.
In this case, it is reasonable to neglect
Jo vs.
Fig. Eli—i
b H r+tb
+s, q
I
-
The strcss follows from (11—61),
M1C1/ t2'\ = —-———ii ± —-I = J C11
t\,
(I
± —s—
where, for this section, t2
/
h'\t
(t
If the section is thin-walled, we can neglect the contribution of = q/t =
We
i.e., we can take
M1
consider next the section shown in Fig. 11—li. Rather than work with
it is more convenient to work with the shear flows for the segments. We
number the closed cells consecutively and take the + S sense to coincide with the X2-X3 sense. The +S sense for the open segments is arbitrary. We define q3 as the shear flow for çellj and write (11—62)
Note that is the value of on the interior boundary of cell j and the shear flow is constant along a segment. The total shear flow distribution is obtained
290
TORSION-FLEXURE OF PRISMATIC MEMBERS
CHAP. 11
q2. S2
x3
Fig. 11—11. Cross section consisting of closed cells and open segments; A, and A,, are centerline areas.
by superimposing the individual cell flows. Then, the shear flow in the segment common to cells i andj is the difference between qj and q1. The sign depends on the sense of S.
q=
— q2
=
— C2)
q=q2—q1
for S1
(11—63)
forS2
The shearing stress is assumed to vary linearly ovcr the thickness. For convenience, we drop the subscripts on and write the limiting values as cr = ±a° + Cr" where M1 /Cnet
cr=—1-t
(11—64)
It remains to determine C1, C2, and J. We have shown (see (11—55)) that
J=
Jo + Jc
(a)
and
=
(h)
We determine J° from (11—65) segments
THIN-WALLED CLOSED CROSS SECTIONS
SEC. 11—4.
291
Substituting for
MI =
2qjA1 + 2q2A2
+ C2A2) in (b) leads to
+ A2C2)
-
(11—66)
The constants are obtained by enforcing continuity of on the centerline of each cell. This can also be interpreted as requiring each cell to have the same twist deformation, k14
I = 1,2
(11-67)
Substituting for q in terms of C and letting
=
C
dS
JS, t
a22
Cads
dS
C
=
=
.J52
=
—I
Jc
t
where a12 involves the segment common to cells 1, 2, the continuity equations
take the following form:
+ a12C2 = a12C1 + a22C2 =
(11—69) 2A2
C2, then determine f with (11—66), We solve this system of equations for and finally evaluate the stresses with (11—64). We can represent the governing equations in compact form by introducing matrix notation. The form of the equations suggests that we define
c
A
=
a
[a11
a121
a22j
(11—70)
With this notation, JC
= 2ATC (11—7.!)
aC
2A
Substituting for A in the expression for JC
and noting that JC is
CTaC
positive, we conclude that a must be positive definite,
The complementary energy per unit length along the centroidal axis is defined by (11—39), 11
We apply (11—51) to each cell. See (11—32).
2
TORSION-FLEXURE OF PRISMATIC MEMBERS
292
Since
CHAP. 11
ais varies linearly over the thickness, the open and closed stress dis-
tributions are uncoupled, i.e., we can write
=
+
toq
where
=
2G
(11-72) aq)
i
—
It is reasonable to neglect the open contribution when the section is thin-walled.
Examp'e 11—2 The open-section torsional constant for the section shown is
=
± 2(b
+ htfl
+d+
Applying (11 —68) to this section, we obtain
=
hd
A2=hb a11 =
1(h + 2d) + 11
t2
6
012
=
a22
= tl
+ 2b) +
t2
and the following equations for C1. C2 and i.
+
+2
C2
=
2
dt1
—
c1 +
±2
+
C2 = 2bt1
—
J=
Jo + Jo
Finally, the shear stress intensities in the various segments are
=
M1 (C1 (k— + M1 /C1 — C2
J
t2
M1 (C2
=7 =
M1
t3
+ t1
+t2
/
(a)
____ _____________ TORSION-FLEXURE WITH UNRESTRAINED WARPING
SEC. 11—5.
293
Fig. E11—2
tl
t3
03
1
Ii I
I
032 A2
h
X3
a'
M1
When
d=
b.
Cl =
C2
=
—s 2bt3
1 + 213
and the section functions as a single cell with respect to shear flow.
11—5.
TORSION-FLEXIJRE WITH UNRESTRAINED WARPING
Consider the prismatic member shown in Fig. 11—12. There are no boundary
forces acting on the cylindrical surface. The distribution of boundary forces x2
x2
—_____
if
xI
I +S
P2
Fig. 11—12. Prismatic member in shear loading.
on the cross section at x1 = L is statically equivalent to a single force P212, acting at the centroid. Also, the end cross sections are not restrained against displacement. In what follows, we describe St. warping, i.e., Venant's torsion-flexure formulation for this problem. Later, in Chapter 13, we shall modify the theory to include restraint against warping.
294
TORSION-FLEXURE OF PRISMATIC MEMBERS
CHAP. il
We start by postulating expansions for the stresses. The stress resultants and couples required for equilibrium at x1 are
=A4,r0 F2 = P2 M3 = P2(L
—
x1)
Introducing (a) in the definition equations for the stress resultants and couples leads to the following conditions on the stresses:
dA = JJx3a13 c/A = dA = P2(L — x1) Jj712 c/A =
0
P2
$fcrj3 c/A
=
0
S$(x2a13 — x3cr12)dA
—,
0
The expansion,
=
M3 13
=
P2 —
13
satisfies the first three conditions (i.e., F1,
M3) identically since
JJx2 c/A = jJx2x3 dA = 0
dA =
13
The last three conditions (i.c., F2, F3, M1) require a12, a13 to be independent of x1. This suggests that we consider the following postulated stress behavior:
cru =
——--—x2 =
a1 2
a1 2(x2, x3)
13
P2 —
x1)x2
13
(11 —73)
a13 = a13(x2, X3) a22
0
a33
Introducing (11—73) in the stress-equilibrium equations and stress boundary conditions for the cylindrical surface leads to
a21,2 + a31,3 + 2a21 +
P2 13
0
(mA)
=0
(on S)
(11—74)
At this point, we can either introduce a stress function or express (11—74) in terms of a warping function. We will describe the latter approach first. The displacements can be found by integrating the stress-displacement relations. We suppose the material is linearly elastic, isotropic with respect to the X2-X3 plane, and orthotropic with respect to the axial direction. This is a convenient way of keeping track of the coupling between axial and in-plane
SEC. 11—5.
TORSION-FLEXIJRE WITH UNRESTRAINED WARPING
295
deformation. Substituting for the stresses in (10—74), we obtain
=
=
u1
I
P2
E1
E113 V1
= u2,2 =
'Y12
Y23
= U2,3 + U3,2
v1P2 —7—(L
LI3
I
=
xj)x2
—.
x1)x2
—
xj)x2
= function ofx2, x3
1
U1 3 + U3, 1 =
Y13
—(L El3
=
= u1, 2 + u2,
—
V1P2
=
v1
63 = u33 =
(L
= function ofx2, x3
0
Integrating the first three equations leads to u1
=
(Lx1
E1 13
v1P2
(L
—
u3 = ——- (L
—
= v1P2
El3
1
+ f1(x2, x3)
—
+ .f2(xj, x3) x1)x2x3
+
(b)
x2)
The functions f1, f2, f3 are determined by substituting (b) in the last three equations. We omit the details and just list the resulting expressions, which involve seven constants:
= f2 =
C1
+ C5x2 + C6x3 +
C2
—
v1P2 —
=
x3)
+ C4x3 — k1x1x3
C5x1
Xi.
+ —--7-(L —
(c)
2E,iJ 3
C3 — C5x1 — C4x2
+ k1x1x2
The constants C1, C2, ..., C6 are associated with rigid body motion and k3 is associated with the twist deformation.t We consider the following displacement boundary conditions: 1.
The origin is fixed:
u1=u2=u3==0 2.
at(0,0,0)
A line element oh the centroidal axis at the origin is fixed:
= fSee Eq. (11—5).
u3,1
=
0
at(0,0,O)
CHAP. Ii
TORSION-FLEXURE OF PRISMATIC MEMBERS
296 3.
A line element on the X2 axis at the origin is fixed with respect to rotation in the X2-X3 plane: at (0, 0, 0) u2,3 = 0
These conditions correspond to the "fixed-end" case and are sufficient to eliminate the rigid body terms. The final displacement expressions are (Lxj —
u1
=
+ 4)(x2, x3)
+
—
—
(11—75)
—
—
Vj
2
El3
-
— x1)x2x3 + k1x1x2
One step remains, namely, to satisfy the equilibrium equation and boundary condition. The transverse shearing stresses are given by 1
41,2 — k1x3
1
(7j3
+
+ k1x2 —
4),
v1P2
2L13 v1P2
LI3
—.
x2) (11—76)
x2x3
Substituting for the stresses in (11—74), we obtain the following differential equation and boundary condition for 4):
l\
P2f'2v1
+
(mA)
-
The form of the above equations suggests that we express 4) 4)
= kjq5t
+
—
+
1
12) +
v1P2 C
+
- 77
as
(11-78)
where
is the warping function for pure torsion and 4)2. and 4)2d are harmonic functions which define the warping due to flexure. Substituting for 4) leads to the following boundary conditions for 4)2. and 2
2
)
2
One can show, by using (11—15), that cn dS
=
0
=
0
+
(11—79)
TORSION-FLEXURE WITH UNRESTRAINED WARPING
SEC. 11—5.
297
therefore the formulation is consistent. Terms involving vj/E are due to in-plane deformation, i.e., deformation in the plane of the cross section, and setting v1/E = 0 corresponds to assuming the cross section is rigid. Then, defines the flexural warping for a rigid cross section and represents the correction due to in-plane deformation. The shearing stress is obtained by substituting for in (11—76). We write the result as (j = 2,3) (11—SO) + 01j,r + 01j = and
where crU, is the pure-torsion distribution and butions corresponding to and 42d:
d are
r,
flexural distri-
=
2
2 2
P2
x2x3)
The pure torsion distribution is statically equivalent to only a torsional mo-
ment,
=
G1k1J.
One can show thatt dA
J$a12,a dA
dA — 0 dA 0
P2
=
0
(11—82)
Note that the shear stress due to in-plane deformation does not contribute to P2. The total torsional moment consists of a pure torsion term and two flexural terms, M1 = G1k1J +
+ X24)2r 3)dA
S2r
2
S2d
+ X242d,3
X34)2a,2)dA
and depend only on the shape of the cross section, it follows that and S2d are properties of the cross section. For convenience, we let
Since
X3
(1184)
1
13\
£.
J
and (11—83) reduces to
= Now, —
—
(11—85)
is the statically equivalent torsional moment at the centroid due
tSeeProb. Il—lO.
298
TORSION-FLEXURE OF PRISMATIC MEMBERS
CHAP. 11
to the fiexural shear stress distribution. Then,
defines the location of the resultant of the flexural shear stress distribution with respect to the centroid. The twist deformation is determined from
+
k1
(11—86)
where M1 is the applied torsional moment with respect to the centroid. If P2 is applied at the centroid, M1 = 0, and k1 =
= 0. Suppose P2 has an eccentricity e3. In this case (see Fig. 11—13), M1 = —e3P2, and
The cross section will twist unless
k1
P2 —
e3)
For Ilexure alone to occur, e3 must equal x3
Fig. 11—13. Notation for eccentric load.
Now, to Whether twist occurs depends on the relative eccentricity, e3 — find x3, one must determine S2. and S2d. This involves solving two secondorder partial differential equations. Exact solutions can be obtained for simple cross sections. In the section following, we present the exact solution for a rectangular cross section. If the section is irregular, one must resort to such numerical procedures as finite differences to solve the equations. In Sec. 11—7, we describe an approximate procedure for determining the flexural shear stress distribution in thin walled cross sections. Suppose the cross section is symmetrical with respect to the X2 axis. Then, is an odd function of x3. The form of the is an even function of x3 and boundary conditions (11—79) requires and to be even functions of x3
SEC. 11—5.
TORSION-FLEXURE WITH UNRESTRAINED WARPING
299
this case. Finally, it Ibilows thatt S2,. = 0 and S2d = 0. Generalizing this result, we can state: The resultant of the shear stress distribution due to fiexure in the is an axis of symmetry direction passes through the centroid when for the cross section. for
x3
Shear center
Fig. 11—14. Coordinates of the shear center.
We consider next the case where the member is subjected to P2, P3 and at the right end (see Fig. 11—14). The governing equations for the P3 loading can be obtained by transforming the equations for the P2 case according to >
X3
U2 —* U3 13
(3
ox2
Ox3
.-
U3—> —U2
—---+-—
—--*-----13
(3
Ox3
Ox2
U12
U13
—a12
13
'*
12
Two additional flexural warping functions must be determined. The expres-
sions defining the flexural shear stress distributions due to P3 are P3
413r. 2 -r 12
cr12 r
3
r
d
'2
=
= t
is even in x>,
((/33r, 3
(11—87)
.v1G1 P3 Vj D
x2x31!
2
L.
12
i —i---
+
+ (,b3d, 3]
is odd in x3, and S2r, Ssd involve only integrals of odd functions of
300
TORSION-FLEXURE OF PRISMATIC MEMBERS
where q53r,
are
CHAP. 11
harmonic functionst satisfying the following boundary
conditions:
= 2
(11—88)
(X2 +
=
2
Note that the distribution due to
leads to no shearing stress resultants. Finally, the total normal stress is given by M2
M3
12
13
(P3
P2
'\
13
J
(11—89)
Superimposing the shearing stresses and evaluating the torsional moment,
we obtain
M1 = where
defines
G1k1J
+
—
(11—90)
the location of the resultant of the flexural shear stress distri-
bution due to P3. One can interpret X2, x3 as the coordinates of a point, called the shear center. The required twist follows from (11—90): k1
(M1 +
=
—
Since (see Fig. 11—14)
M1 + P2x3 — P3x2 = the applied moment with respect to the shear center = MT
(11—91)
we can write (a) as
k1 =
(11—92)
To determine the twist deformation (and the resulting torsional stresses), one must work with the torsional moment with respect to the shear center, not the centroid. For no twist, the applied force must pass through the shear center. In general, the shear center lies on an axis of symmetry. lithe cross section is completely symmetrical, the shear center coincides with the centroid. It is of interest to determine the complementary energy associated with Then torsion-flexure. The only finite stress components are 012, aild V* reduces to =
+
+
dA
(a)
follows directly by substituting (11 —89) and using The contribution from the definition equations for 13. t The total flexural warping function for P3 is P3
(
I
—
'\
+
v,P3(
1
+ —4
SEC. 11—5.
TORSION-FLEXURE WITH UNRESTRAINED WARPING
=
1
M2
M2
(11-93)
+ Now, the total shearing stress is the sum of three terms: a pure torsional distribution due to MT the flexural distribution due to F2 the fiexural distribution due to F3
1.
2. 3.
Each of the flexural distributions can be further subdivided into— dr.
the
distribution corresponding to a rigid cross section (defined
by 2.
dId,
the distribution associated with in-plane deformation of the cross
section (defined by 4)jd)
We combine the flexural distributions and express the total stress as C33
d12,t + C12,, + C12.d = 013.1 + C13r + 013,d
where the various terms are defined by (11—81) and (11—87). For example,
r=
F2
F3
• 2
13
2
13
The complementary energy due to pure torsion follows from (11—38) and (11—92):
+
a
as
C12r
F3_
F2
—
2
3
r+
JJ =
1 See Prob. 11—il.
r) and integrating over the cross section, we obtaint F2
+
+
Jj
3
+
2FF
+
F2
(1196)
dA
2+
3413r, 3)dA
=
CIA
JJ
TORSION-FLEXURE OF PRISMATIC MEMBERS
302
The
CHAP. 11
coupling term, I/A23, vanishes when the cross section has an axis of
symmetry. We consider next the coupling between or., and JJ(a12,to12,r +
=
±
2
+ x2)
+
=
+
dA
+
MT ""F2 Jj 13
—
2)
+
F3
+
+
—
(11—97)
dA = 0
12
The remaining terms involve a,a, the shearing stress distribution due to inplane deformation of the cross section, We will not attempt to expand these terms since we are interested primarily in the rigid cross section case. Summarizing, the complementary energy for flexure-torsion with unrestrained warping is given by 1
M2
M2
M2
FF
F2
I
F2
(11-98)
+ terms involving v1/E — We introduce the assumption of negligible M1 + in-plane deformation by setting v1/E = 0. Similarly, we introduce the assumption of negligible warping due to flexure by setting 1/A1 = (&3r 1/A2 1/A23 = 0. In Sec. 11 —7, we develop an approximate procedure, called the engineering theory, for determining the flexural shear stress distribution, which is based upon integrating the stress-equilibrium equation directly. This approach is
where
similar to the torsional stress analysis procedure described in the previous section. Since the shear stress distribution is statically indeterminate when the cross section is closed, the force redundants have to he determined by requiring the warping function to be continuous. For pure torsion, continuity requires (see (11—32))
= 2G1k1A5
where the integration is carried out in the X2-X3 sense around S. and
is the
area enclosed by S. To establish the continuity conditions for flexure, we operate on (11 —81) and (11
There are four requirements:
j (aisd)F2 dS dS =
2,3
2v1G1P2
X3
dA
2v1G1P3 rr ii X2 dA El2 ,jj 4,
(11—99)
SEC. 11—6.
EXACT FLEXURAL SHEAR STRESS DISTRIBUTION
303
In the engineering theory of flex ural shear stress distribution, the cross section
is considered to be rigid, i.e., the distribution due to in-plane deformation is neglected. The consistent continuity condition on the flexural shearing stress
4sajsdSO
(11—100)
One can take the + S direction as either clockwise or counterclockwise. By coincides with the + S direction.
definition, the positive sense for 11—6.
EXACT FLEXURAL SHEAR STRESS DiSTRIBUTION FOR A RECTANGULAR CROSS SECTION
We consider the problem of determining the exact shear stress distribution due to F2 for the rectangular cross section shown in Fig. 11—15. For con-
venience, we first list the governing equations: x2
rd3
'2
—
dt3
A = dt
d
r Fig. 11—15. Notation for rectangular cross section. 1.
Warping functions +
= an
= =
1
2
+ 2
)
+
TORSION-FLEXURE OF PRISMATIC MEMBERS
304 2.
CHAP. 11
Shearing stresses 012
-
= F2
3) +
= T13 Determination
+
+ v1G1E2
+ xi)]
x2x3)
of
The boundary conditions for
are
l(d'\2
d
=0
atx3 =
We can take the solution as .i.
=
112
X2
The corresponding stresses and warping function are 2r
012,r
= 4)2r
13
12 X2
—
13 (11—101)
= 0
0j3,r One can readily show that
F2
Finally, we evaluate 1/A2 using (11—96):
=
(11—102)
Determinatio,, of The boundary conditions for /2d are
1(d2
2=
+
'\
at
x2 =
d
atx3 = Now, the form of (a) suggests that we express q52d
=
—
as
— f(x2, x3)
(b)
EXACT FLEXURAL SHEAR STRESS D!STRJBUTON
SEC. 11—6.
305
f is an harmonic function. The shearing stresses and boundary conditions expressed in terms of f are
where
=
v1G1 P2
2
—f 2)
VLGI F2
=
f3=0 It remains to solve V2f = 0 subject to (d). Since the cross section is symmetricaL f must be an even function of x3 and an odd function of x2. We express f as
f=
B0x2
+
sinh
B,, cos
(2nxx2)
This expansion satisfies V2f = 0 and the boundary condition at x3 = ± t/2. The remaining boundary condition requires B0 +
B,,
(2nn
co sh
nicd\
—)
2nicx3
c os——-
=
< x3 <
x32
(f)
in a Fourier cosine series and equating coefficients leads to
Expanding
B0
B,, =
t2
(±..Y mr1
mtd cosh ——
The final expressions for the shearing stresses are
=
v1G1 P2
1/t\2
cosh
I
—
t
.a
2nxx2\ 1 nnd
cosh— t
[ cJl3,d =
sinh
v1G1 F2
E
n2
13
[ This system is statically equivalent to zero.
—-— t mid
cosh— t
/J
1
2nirx2
.
2n7tx3
sin ——t
(11—103)
j
TORSION-FLEXURE OF PRISMATIC MEMBERS
306
CHAP. 11
To investigate the error involved in assuming the cross section is rigid, we • occurs at x2 0: note that the maximum value of F2 ci
2,
Specializing
d
for x2
2
0,
(512, d)x20 =
2nnx3
v1G1 F2 d2
C,, COS
13 '1
t
where 1 1
2,,
cosh 2,,
Now, C,, decreases rapidly with n. Retaining only the first term in (b) leads to the following error estimate, 4
/
1
1
I5lz,di
I
2,.
cosh
Results for a representative range of d/t and isotropic material are listed below.
They show that it is reasonable to neglect the corrective stress system for a rectangular cross section. The error decreases as the section becomes thinner, i.e., as d/t becomes large with respect to unity: d/t 2 1
11—7.
1512,4/512.r
0.024 0.092 0.122
ENGINEEffiNG THEORY OF FLEXURAL SHEAR STRESS DISTRiBUTION IN THiN-WALLED CROSS SECTIONS
The "exact" solution of the flexure problem involves solving four secondorder partial differential equations. If one assumes the cross section is rigid with respect to in-plane deformation, only two equations have to be solved. Even in this case, solutions can be found for only simple cross sections. When the cross section is irregular, one must resort to a numerical procedure such as finite differences or, alternatively, introduce simplifying assumptions as to the stress distribution. In what follows, we describe the latter approach for a thinwalled cross section. The resulting theory is generally called the engineering
theory of shear stress. We apply the engineering theory to typical cross sections
SEC. 11—7.
and
ENGINEERING THEORY OF FLEXURAL SHEAR STRESS
307
also illustrate the determination of the shear center and the energy
efficients, 1/Ai (j 2, 3). Figure 11—16 shows a segment defined by cutting planes at x1 and x1 + dx1. Since the cross section is thin-walled, it is reasonable to assume that the normal Also, we work stress, Oh, is constant through the thickness and to neglect
qA
/
xl thin-wafled segment.
Fig. 11—16.
Integrating the axial force-equiwith the shear flow, q, rather than with librium equation, — + —(a11t) 0 C2S
with respect to S, we obtain the following expression for q, ('S
(11—104) JSA
Equation (11—104) is the starting point for the engineering theory of shear over the cross section is known, stress distribution. Once the variation of have shown that the normal stress varies linearly we can evaluate q. Now, we the member is subjected to a constant shear (F2, F3 over the cross section when
TORSION-FLEXURE OF PRISMATIC MEMBERS
308
constant)
CHAP. 11
and the end sections can warp freely. Noting that the member is
prismatic, the derivative of
for this case is — X3 —
x2dM3
dM2
13 dx1
12 dx1 3
'2
X3 +
X2
13
12
and (11—104) expands to
,' i'S 12!i'S x2tdS—-7-H x3tdS q=qA—-—I 13 j54 JSA
The integrals represent the moment of the segmental area with respect to X2, X3 and are generally denoted by Q2, Q3:
rs Q2
=j
x3tdS = Q2(S,SA) (11—105)
x2t c/S = Q3(S, SA)
=
With this notation, (b) simplifies to F2
q=
—
F3 — —-Q2
13
(11—106)
Equation (11—106) defines the shear flow distribution for the case of negligible
restraint against warping, i.e., for a linear variation in normal stress. Note that q is positive when pointing in the + S direction. We consider first the open section shown in Fig. 11—17. The end faces are unstressed, i.e.,
=
=0
Taking the origin for S at A, (11—106) reduces to
q= Q3
F2
=J5x2tdS
F3
f-Q2
=$x3tdS
Q2
We determine Q2, Q3 and then combine according to (11—107). The shearing stress distribution corresponding to F2,
q= satisfies
ffai2 c/A = dA =
F2
0
identically. To show this, we expand =
+
(11-107)
SEC. 11—7.
ENGINEERING THEORY OF FLEXURAL SHEAR STRESS
309
x3 —
B
I
Shear center
X2
Centroid
Fig. 11—17. Flexural shear flow—open segment. and
evaluate the shear stress resultants:
Jjui2dA
=
J
dS
Equation (b) requires rsa J0
= =0
Now,
Integrating (e) by parts and noting that X2, X3 are principal centroidal axes,t
we obtain
=
—
J"
dS =
13
= x2x3t dS = 0 — J The shear stress distribution predicted by (a) is statically equivalent to a we evaluate the force F212. To determine the location of its line of moment with respect to a convenient moment center. By applying the same argument, one can show that the shear flow corresponding to F3 is statically t See Eq. (11—2). 11—12.
TORSJON-FLEXURE OF PRISMATIC MEMBERS
310
CHAP. 11
The intersection of the lines of action of the two resultants is the shear center for the cross section (see Fig. 11—17). equivalent to a force,
Example
11—3
Consider the thin rectangular section shown. We take + S in the + X2 direction. Then, The various terms are + q points in the +X2 direction and q/t =
=
j -d(2
tx2
dx2 = 2
q=
F2/'d2
q
4
—
tF2 (d2
F2
—
2
This result coincides with the solution for
obtained in Sec. 11—6. Actually, the engineering theory is exact for a rigid cross section, i.e., for r1/E = 0.
x2
I
Fig. E11—3
+
Example 11—4 We determine the distribution of q corresponding to F2 for the symmetrical section of Fig. Eli —4A. Only two segments, AB and BC, have to be considered since 1Q31 is
symmetrical.
Segment AB = q
=
According to our definition, + q points in the + S direction (from A to B). Since q is negative for this segment, it actually acts in the negative S direction (from B to A).
____________________________
SEC. 11-7.
ENGINEERING THEORY OF FLEXURAL SHEAR STRESS
311
Pg. E11—4A
IC
d
tw
tf
It H
I,
Segment BC
We measure S from B to C. Then,
=
+
q=
—
[hh1t1 +
—
xi)]
Note that the actual sense of q is from C to B. The distribution and sense of q are shown in Fig. E1I—413.
It is of interest to evaluate A2. Specializing (11-96) for a thin-walled section,
if
dA
dS
5
=
fl
and substituting for q yields 1
=
2dS
1
j
We let
= area of the web = A1 = total flange area = A2 = kA,.
2b1t1
The resulting expression for kis
lÀ,. I
3Af\
2A F
6A1
id, l/bf\2
TORSION-FLEXURE OF PRISMATIC MEMBERS
312
x2
CHAP. 11 Fig. E11—48
.4
— X2)
x3
4.
This factor is quite close to unity. For example, taking as typical, for a wide-flange section,
= 2ç (If = 3, 1..
we find
.4f =
k=
0.95
The shearing stress corresponding to F3 varies parabolically in the flanges and is zero in the web. Each flange carries half the shear and
61
1
A3 —
5
1
A1 — 5 hr1
Examp'e 11—5
Cross-Sectional Properties This section (Fig. Eli —5A) is symmetrical with respect to X2. The shift in the centroid from the center of the web due to the difference in flange areas is b2t2 —
(a)
b,t1 + b2t2 + We neglect the contribution of the web in '2 since it involves 12
+ ('2)2 =
+
(b)
SEC. 11—7.
ENGINEERING THEORY OF FLEXURAL SHEAR STRESS
313
Determination of Taking S as shown in the sketch, we have
=
t[(b)2 t2
=0
[,"b2\2
—X32
(c)
since X2 is an axis of symmetry. Fig. E11—5A x3
I I
Centroid
Distribution of q Corresponding to The shear flow corresponding to F3 is obtained by applying
q=
p3 Q2
and is shown in Fig. El 1—SB. The shear stress vanishes in the web and varies parabolically in each flange. Integrating the shear flow over each flange, we obtain
=
F3
Then, the distribution is statically equivalent to F373 acting at a distance e from the left flange, where e
Since
=
R2d R
=
(12)2 12
X2 is an axis of symmetry, the shear center is located at the intersection of R and X2,
CHAP. 11
TORSION-FLEXURE OF PRISMATIC MEMBERS
314
Fig. E11--58
R = F3 q
1'3t2frbf\2 12
2
X3
R1
The coordinates of the shear center with respect to the centroid are 0
,c3
= e —(1 +
=d
(I
—
L12
2
Torsional Shear Stress to
The flexural shear stress distribution is statically equivalent to a torsional moment equal with respect to the centroid. We have defined M1 as the required torsional moment
with respect to the centroid. Then, the moment which must be balanced by torsion is — F3; the required torsional moment with respect to the shear center. Using the approximate theory developed in Sec. 1 1 —3, the maximum torsional shear stress in a segment is = M ti
where .1 =
+
+
We consider next the closed cross section shown in Fig. 11 —18. We take the origin for S at some arbitrary point and apply (11-106) to the segment Se-S:
q= =
F2
F3
13
12
q1 — -1—Q3 —
(11— 108)
Q3 =
where q1 is the shear flow at P. The shear flow distribution is statically indeter-
minate since q1 in unknown. We have previously shown that q
q1 = con-
SEC. 11—7.
ENGINEERING THEORY OF FLEXURAL SHEAR STRESS
315
stant is statically equivalent to only a torsional moment equal to
The second and third terms are statically equivalent to F212 and F313. The constant q1 is determined by applying the continuity requirement to the
centerline curve. Since the engineering theory corresponds to assuming the cross section is rigid with respect to in-plane deformation, we use (11—100).
P
Fig. 11—18. Notation for closed cell.
The flexural shear stress distribution must satisfy 0
for an arbitrary closed C
(11—109)
Substituting for q. dS
F2 C
f
dS
(IS
2
and considering separately the distributions corresponding to F2 and F3, we obtain + q =
= T (B2 —
(B3 — Q2)
Q3)
QdS B
27 —
QdS B
dS
(11-110)
I
Each distribution satisfies (11—109) identically. Also, the distribution is statically equivalent to a force Fyi,, located Y4 units from the centroid. Note that q = B3 leads only to a torsional moment equal to f One can interpret (11-109) as requiring the flexural shear stress distribution to lead to no twist deformation, See Prob. 11 —14 for the more general expression, which allows for a variable shear modulus.
TORSION-FLEXURE OF PRISMATIC MEMBERS
CHAP. 11
The general expression for 1/Ai follows from (11— 96):
j = 2,3
==
(11—111)
Substituting for 1
=
and
we
2BJQk
—
'k
+
k;j, k
(j
2,3)
noting that .1
obtain 2dS
dS
(11—112)
QkT
42j
which applies for an arbitrary single cell.
Example 11—6 We illustrate the determination of for the square section of Fig. convenient to take P at the midpoint since the centerline is symmetrical.
Eli —6A.
It is
Fig. E11—6A
H
2z
H
Cross-Sectional Properties
= =
a2
(a3\
= (at)(a/2) 5a1
it
fdS
a
= 3.5t
a2
+ —
10
3
= 4a3t
SEC. 11—7.
ENGINEERING THEORY OF FLEXURAL SHEAR STRESS
317
Determination of We start at P and work counterclockwise around the centerline. The resulting distribuarc shown in Fig. Eli —6B. Note that + Q2 corresponds tion and actual sense of q due to to a negative i.e., clockwise, q. Fig. E11—6B
Evaluation of B3 By definition,
"
1
B3=
dS
rds.
it
I
Using the above results, and noting that the area of a parabola is equal to (2/3) (base) x (height), we obtain
I
dS
+
a3
B3 =
of Flexural Shear Flov for F3 The shear flow is given by q—
F3((
F3 —
12
Q2)
a
4Q2 —
3at
(+ sense clockwise). The two distributions are plotted in Fig. Ell—6C. To locate the line of action of the resultant, we sum moments about the midpoint (0 in the sketch:
(M)0 =
I
/a\
4F3 /a\ +
=
The resultant acts e units to the right of 0, where
e=
19
19
a
aF
_ CHAP. 11
TORSION-FLEXURE OF PRISMATIC MEMBERS
318
Fig. Eli —6C
t
j
F3/21
I
q — a 1.21
I F3
30
i F3
6a
F3f 4 Q2
Finally, the coordinates of the shear center with respect to the ccntroid (which is A units to the right of 0) are x2
= e — LI = +
16
X3 = 0
Torsional Shear The shear flow for pure torsion is due to Mr, the torsional moment with respect to the shear center. For this section,
Mr
5L2F3 +
M1 —
We apply the theory developed in Sec. II —4. One just has to replace M1 with MT in
SEC. 11—7.
ENGtNEERING THEORY OF FLEXURAL SHEAR STRESS
Equation (11—61):
—
±
Ci = 4at
J=
I C1
+
Determination of 1/A3 Applying (11—112), we find 1
-
1
2
dS
I-
B
dS'\
—
1.276
I) - 3at
Note that 3at is the total web area.
We consider next the analysis of a two-cell section and include open segments
for generality. There is one redundant shear flow for each cell. We select a convenient point in each cell and take the shear how at the point as the redundant for the cell. This is illustrated in Fig. 11—19: qj represents the shear flow redundant for cell j and the + S sense coincides with the X2-X3 sense to be consistent with the pure-torsion analysis. The + S sense for the open segments is inward from the free edge. For convenience, we drop the CL (centerline) subscript on S and A. +5, q
x3
x2
Fig. 11—19. Notation for
section.
TORSION-FLEXURE OF PRISMATIC MEMBERS
320
CHAP. 11
The total shear flow is the sum of q0, the open cross-section distribution
(q1 =
q2
=
0),
and
q0 +
q
(11—113)
We determine q0 by applying (11—107) to the various segments. The redundant shear-flow distribution is the same as for pure torsion (see Fig. 11—11). Finally, we obtain a system of equations relating q1, q2 to F2, F3 by applying the continuity requirement to each centerline, t
j = 1,2
q—--= 0
(11—114)
where q is positive if it points in the + S direction. Using the aJk notation defined by (11—68), the equations take the following form:
a11q1 + a12q2 = a12q1 + a22q2 =
11—115
t
The shear flows q2, for pure torsion are related by (we multiply (11—71) by MT/J and note (11—62))
+ a12q2,1 =
2A1
(11--1l6) a12q1,1 + a22q2,1
2A2
MT
Thus, the complete shear stress analysis involves solving aq = b for three different right-hand sides. The equations developed above can he readily generalized.
Example
11—7
We determine the flexural shear stress distribution corresponding to F3 for the section shown in Fig. E11—7A. We locate and P2 at the midpoints to take advantage of
symmetry.
Cross-Sectional Properties A1 =
202
A2
a2
= (a3t\ '2
[ a21 + 2[(3at)_4_j
Ga
= —
a22
=
4a
7
012 =
a
See Prob. 1!-- 14 for the more general expression, which allows for a variable shear modulus.
_______a _______ SEC. 11—7.
ENGINEERING THEORY OF FLEXURAL SHEAR STRESS
321
Fig. Eli—lA
I
x2
+S,q 2a
a
Distribution of This system (Fig. E1l—7B) is statically equivalent to a moment 2a2(2q1 +
Distribution of q0 Due to F3 We apply —--i02 '2 to the various segments starting at points P1. P2. The resulting distribution is shown in Fig. Eli —7C.
Deter,niiuztion of q1.
q2
= C
.322
dS
q0—=+ t
2F3
The equations for q1 and q2 are 6q1 — q2
=
1 F3
7a
2 F3 —q1 + 4q2 = ——
70
Solving (a), we find
q1 =
2F3 —
q2 = +
a 11 F3
161
The total distribution is obtained by adding qR and q0 algebraically.
TORSION-FLEXURE OF PRISMATIC MEMBERS
322
CHAP. 11 Fig. El 1 —7B
q2
4
q2a
2q1a
q1a
q1
Fig. Eli —7C
P3
I
14 a
Location of the Shear Center Taking moments about the midpoint of the left web, and letting e be the distance to the action of the resultant, we obtain
line
2a2(2q1 + q2) + (2a)
M( + e
=
(2
+
32\
a=
+ (3a)
F3) =
eF3
1.61a
The shear center is located on the K2 axis and
=
c —
(2a
—
= +O.055a
REFERENCES 1.
2,
WANG, C. T.: Applied Elasticity, McGraw-Hill, New York. 1953. TIMOSHENKO, S. J., and J. N. Oooo!ETt: Theory of Elasticity, 3d ed., McGraw-Hill, New York, 1970.
PROBLEMS 3.
Dmt
HARTOG,
323
J. P.: Advanced Strength of Materials, McGraw-Hill, New York,
1952. 4.
5.
HERRMANN, L. R.: "Elastic Torsional Analysis of Irregular Shapes," I. Eng. Mech. Div., A.S.C.E., December 1965. SOKOLNIKOFF, LS.: Mathematical Theory of Elasticity. McGraw-Hill, New York, 1956.
6. 7.
8. 9.
TIMOSHFNKO, S. J.: Strength of Materials, Part 2, Van Nostrand, New York, 1941. CEiRNICA, J. N.: Strength of Materials, I-Jolt, Rinchart, New York, 1966. DABROWSKI, R.: GekrUmmte diinnwan.dige Trëger ("Curved Thin-walled Girders"), Springer-Verlag, Berlin, 1968. KOLLBRUNNER, C. F., and K. Basler: Torsion in Structures, Springer-Verlag, Berlin, 1969.
10. 11.
VLASOV, V. Z.: Thin- Walled Elastic Beams, Israel Program for Scientific Translations, Jerusalem, 1961. ODEN, J. T.: Mechanics of Elastic Structures, McGraw-Hill, New York, 1967.
PROBLEMS
il—i. The pure-torsion formulation presented in Sec. 11—2 considers the cross section to rotate about the centroid, i.e., it takes U2
wj =
= W1X3 = +(01x2
a1 =
Suppose we consider the cross-section to rotate about an arbitrary point The general form of (a) is
wj = k1x1 + c1 a1 = k147
a2 = —wj(x3 — U3
(a) (b)
= +w1(x2 —
Starting with Equation (h), derive the expressions for governing equations for What form do the equations take if we write
=
a13 and the
+
+ C2
Do the torsional shearing stress distribution and torsional constant J depend on the center of twist? 11—2. Show that .1 can be expressed as
Hint:
= = I,,
+
2)
)2 —
—
+
3)2]dA
dA =
0
3)2]dA
Compare this result with the solution for a circular cross section and comment on the relative efficiency of circular vs. noncircular cross section for torsion. 11—3. Derive the governing differential equation and boundary condition for for the case where the material is orthotropic and the material symmetry axes coincide with the X1, X2, X3 directions.
TORSION-FLEXURE OF PRISMATIC MEMBERS
324
11—4.
CHAP. 11
along an arbitrary curve S
The variation in the warping function
is obtained by integrating (11—29),
=
+
k1
(a)
is the perpendicular distance from the center of twist to the tangent. One selects a positive sense for S. The sign of p is positive when a rotation about the center of twist results in translation in the +.S direction. We express
where as
lM1
1
(b)
and (a) reduces to
+ a15 Determine the variation of sections shown.
(c)
along the centerline for the two thin-walled open Prob. 11—4
x3
tf
Center
/
of twist —
Center of twist
4
b/2
b/2 (a)
11—5.
(bi
Verify that the distribution. q = const, satisfies
F2 = ja12 = F3 = dA = x =
= dS = = dS
0 0
for the closed cross section sketched. 11—6.
Refer to Prob. 11—4. To apply Equation (c) to the centerline of a
closed cell, we note that (see (11—50)) —
a15 =
J
Jq
C
= k—— = —
(a)
PROBLEMS
325
Prob. 11—5 +
=
£3
x2
Then, Ct
+-C
Integrating (b) leads to the distribution of Apply (h) to the section shown. = 0 at point P. Discuss the case where a = b.
Take
Prob. 11-6
T
11—7.
p
.
Determine the torsional shear stress distribution and torsional
constant J for the section shown. Specialize for t a. 11—8. Determine the equations for (j = 1, 2, 3) and J for the section shown. Generalize for a section consisting of "n." cells. I 1—9. Determine the distribution of torsional shear stress, the torsional constant J, and the distribution of the warping function for the section shown. Take = 0 on the symmetry axis and use the results presented in Prob. 11—6. 11—10. 11—11.
Verify Equation (11—82). Utilize (11—15).
The flexural warping function
satisfy
mA on S
TORS!ON-FLEXURE OF PRISMA1]C MEMBERS
326
CHAP. 11
Prob. 11—7
0 +S1
—
I
+S2
t
H
Prob. 11—8
t
t
I :f
T
t—.
0 Si
t
t
t
0
I
-
0
t
t
a—
a
a
I
t
Prob.11—9
I Utilizing the following integration formula,
if (fr,
212,2 +
3f2, 3)dx2 dx3
dS
—
JJII V2f2 dx2 dx3
where 11,12 are arbitrary functions, verify Equation (11—96). 11—42. Refer to Fig. 11—17. Starting with (11—107), derive the expressions for the coordinates of the shear center in terms of the cross-sectional parameters.
_a PROBLEMS
327
11 —13. Determine the flexural shear flow distributions due to F2, F3 and locate the shear center for the five thin-walled sections shown.
Prob.
11—13
I-
T
R
d/2
+
I
x3
(b) (a)
2/
2/
2/
I I H-a
+
-j (dl
(c)
I :L 1
=
Ca
(el
TORSION-FLEXURE OF PRISMATIC MEMBERS
328
CHAP. 11
11—14. We established the expression for the twist deformation (Equation (11—31) by requiring the torsional warping function to be continuous. One can also obtain this result by applying the principle of virtual forces to the segment shown as part of the accompanying figure.
Prob.
11—14
Arbitrary closed curve 44—
-44
wi
M'11
LsM1
w1 +
(a)
(b)
2G(2E)
x3 GtE)
G(E)
X21
2G(2E)
a-
F Ic)
dx1
PROBLEMS
329
The general principle states that (a) x+dx = (if AbTu dA)dx1 + if ApT for a statically permissible force system. Now, we select a force system acting on the end faces which is statically equivalent to only a torsional moment M1. If we consider the cross section to be rigid, the right-hand side of (a) reduces to AM1o,1 1 dx1,and we can write
(J$eT
dA)dx1
Next, we select an arbitrary closed curve, S (part b of figure), and consider the region defined by S and the differential thickness dn. We specialize the virtualstress system such that AG = 0 outside this domain and only is finite inside the domain. Finally, using (11 -51), we can write dn(&riz)
=
if
and Equation (b) reduces to k1
The derivations presented in the text arc based on a constant shear modulus G throughout the section, so we replace (d) with Gk1
=
If G is a variable, say G = fG* (where f = f(x2, x3)), we have to work with G*k1 =
Also, we define the torsional constant J according to
G*k1J Consider a thin-walled section comprising discrete elements having different
material properties. Develop the expressions for the torsional and fiexural shear flow distributions accounting for variable G and E. Determine the normal stress distribution from the stress-strain relation. Assume a linear variation in extensional strain and evaluate the coefficients of the strain expansion
from the definition equations for F1, Al2, and M3. Apply your formulation to the section shown in part c of the figure.
Engineering Theory of Prismatic Members INTRODUCTION
12—1. St.
Venant's theory of fiexure-torsion is restricted to the case where—-
1.
2.
There are no surface forces applied to the cylindrical surface. The end cross sections can warp freely.
The warping function consists of a term due to flexure (ç&j) and a term due to Since is independent of x1, the linear expansion pure torsion M2
F1
M3
a11 =—;—+——X3 ——---Xz 13 '2
is the exact solutiont for
The total shearing stress is given by
=
+ crj
(12—2)
and represents the is the pure-torsion distribution (due to flexural distribution (due to We generally determine by applying the engineering theory of shear stress distribution, which assumes that the cross section is rigid with respect to in-plane deformation. Using (12—1) leads to the where
following expression for the flexural shear flow (see (11—106)):
=
—
—
(12—3)
The warping function will depend on x1 if forces are applied to the cylindrical
surface or the ends are restrained with respect to warping. A term due to variable warping must be added to the linear expansion for This leads to an additional term in the expression for the flexural shear flow. Since (12—1) t A linear variation of normal stress is exact for a homogeneous beam. Composite beams (e.g.. a sandwich beam) are treated by assuming a linear variation in extensional strain and obtaining the distributions of from the stress-strain relation. See Probs. 1—14 and 12—1. 330
FORCE-EQUILIBRIUM EQUATIONS
SEC. 12—2.
331
the definition equations for F1, M2, M3 identically, the normal stress correction is self-equilibrating; i.e., it is statically equivalent to zero. Also, the shear flow correction is statically equivalent to only a torsional moment since (12—3) satisfies the definition equations for F2, F3 identically. In the engineering theory of members, we neglect the effect of variable warping on the normal and shearing stress; i.e., we use the stress distribution predicted by the St. Venant theory, which is based on cons tant warping and no warping restraint at the ends. In what follows, we develop the governing equations for the engineering theory and illustrate the two general solution procedures. This formulation is restricted to the linear geometric case. In the next chapter, we present a more refined theory which accounts for warping restraint, and investigate the error involved in the engineering theory. satisfies
12—2.
FORCE.EQUILIBRIUM EQUATIONS
In the engineering theory, we take the stress resultants and couples referred to the centroid as force quantities, and determine the stresses using (12—1), (12—3), and the pure-torsional distribution due to MT. To establish the forceequilibrium equations, we consider the differential element shown in Fig. 12--i. The statically equivalent external force and moment vectors per unit F—dxi/2
.-I.
clxl/2___H —
dx1
'
+
4L
dF+ dx1
+ dx1
dx1
Fig. 12—1. Differential element for equilibrium analysis.
length along X1 are denoted by b,
Summing forces and moments about 0
leads to the following vector equilibrium equations (note that F = dx1
dM÷ dx1
—
+ m+
+ —
=o
x F+) =
-
0
—
ENGINEERING THEORY OF
332
MEMBERS
CHAP. 12
We obtain the scalar equilibrium equations by introducing the component expansions and equating the coefficients of the unit vectors to zero. The resulting system uncouples into four sets of equations that arc associated with stretching, flexure in the X1-X2 plane, flexure in the X1-X3 plane, and twist.
Stretching dF1
Flexure in X1-X2
=
+
b1
+
b2
0
+ F2 =
0
--s- + b3 =
0
dx1
0
Plane
dF2 dx3 dx1
+
m3
(12—4)
Flexure in X1 -X3 Plane dx1
dM2 dx1
Twist
+ ni2 —
F3
= 0
dM1 —— + m1 = 0
ax
This uncoupling is characteristic only of prismatic members the equilibrium equations for an arbitrary curved member are generally coupled, as we shall show in Chapter 15. The fiexure equilibrium equations can be reduced by solving for the shear force in terms of the bending moment, and then substituting in the remaining equations. We list the results below for future reference. Flexure in X1 -A'2 Plane dM3
F2 d2it.13 2
dx1 Flexure
—
dx1
+
din3
— b2
dx1
ui3
= 0 (1 25)
in X1 -A'3 Plane
dM2 dx1
d2M2 dx12
din2
+
in2
+ -— + b3 = 0 dx1
FORCE-DISPLACEMENT RELATIONS
SEC. 12—3.
333
Note that the shearing force is known once the bending moment variation is
determined. The statically equivalent external force and moment components acting on the end cross sections are called end forces. We generally use a bar superscript to indicate an end action in this text. Also, we use A, B to denote the negative and positive end points (see Fig. 12—2) and take the positive sense of an end x2 MA2
2
x3
—L Fig. 12—2. Notation and positive direction for end forces.
to coincide with the corrcsponding coordinate axis. The end forces are related to the stress resultants and couples by force
= FAJ = MAJ = A
(J
1
2 3)
(12—6)
minus sign is required at A, since it is a negative face.
12—3.
We
FORCE-DISPLACEMENT RELATIONS; PRINCIPLE OF VIRTUAL FORCES
started by selecting the stress resultants and stress couples as force
parameters. Applying the equilibrium conditions to a differential element results in a set of six differential equations relating the six force parameters. To complete the formulation, we must select a set of displacement parameters and relate the force and displacement parameters. These equations are generally called force-displacement relations. Since we have six equilibrium equations, we must introduce six displacement parameters in order for the formulation to be consistent. Now, the force parameters are actually the statically equivalent forces and moments acting at the centroid. This suggests that we take as displacement
ENGINEERING THEORY OF PRISMATIC MEMBERS
334
parameters
the equivalent rigid body translations and rotations of the cross
section at the centroid. We define i? and
as
= equivalent rigid body translation vector at the centroid
= =
CHAP. 12
equivalent rigid body rotation vector
(12—7)
By equivalent displacements, we mean (force intensity) (displacement) dA =
fl
+M
(.12—8)
Note that (12—7) corresponds to a linear distribution of displacements over the cross section, whereas the actual distribution is nonlinear, owing to shear deformation. In this approach, we are allowing for an average shear deformation determined such that the energy is invariant.
We establish the force-displacement relations by applying the principle of virtual forces to the differential element shown in Fig. 12—3. The virtual-force +
dx1
dx1
I
dudXl
dx1 2 2
Fig. 12—3. Statically permissible force system.
is statically permissible; that is, it satisfies the one-dimensional equilibrium equations system
ö
dx1 dx1
+
=
x
O
Specializing the principle of virtual forces for the one-dimensional elastic case, we can write dV* dx1 =
AP1
represents a displacement quantity, and is the external force quantity corresponding to The term dV* is the first-order change in the onedimensional complementary energy density due to increments in the stress resultants and couples. where
FORCE-DISPLACEMENT RELATfONS
SEC. 12—3.
335
Evaluating the right-hand side of (b), we have
+ A]\+
=
+
dx1
Using the second equation in (a), (c) takes the form
+
+
x
Finally, evaluating the products, we obtain
= [AF1u1,1 + /XF2(u2, +
— co3)
+ zXM2a2
+ AF3(u3, 1 +
j]dx1
+
(02)
(12—9)
Continuing, we expand dV*:
dV* =
+
\CFJ 3
=
+ 1
and k1 are one-dimensional measures. Equating (12—9) and (12—10) leads to the following relation between the deformation measures and the displacements: The quantities
Cl
=
av*
= u1,
=
—— = 112. j —. (03
e2
k2
aF2
(123
=
WL
cM2
(02, 1
(12—Il)
23
We see that— 1.
2. 3. 4.
e1 is the average extensional strain. e2, e3 are average transverse shear deformations. k1 is a twist deformation. k2, k3 are average bending deformation measures (relative rotations of the cross section about X2, X3).
Once the form of V" is specified, we can evaluate the partial derivatives. In what follows, we suppose that the material is linearly elastic. We allow for the
possibility of an initial extensional strain, but no initial shear strain. The general expression for
=
is
+
+
+
denotes the initial extensional strain. Now, for unrestrained where torsion-Ilexure is given by (11—98). Since we are using the engineering theory
MEMBERS
ENGINEERING THEORY OF
336
CHAP. 12
of shear stress distribution, it is inconsistent to retain terms involving in-plane F1/A, and neglecting deformation, i.e., v1/E. Adding terms due to
the coupling between F2, F3 leads to
= Fie? +
+
+
2AE
2GA3
+
+
+
+
LF2
2GA2
(12—12)
+
where
M1 + F2x3 —
MT e?
=
$5 x2e1 cIA
We take (12—12) as the definition of the one-dimensional linearly elastic com-
plementary energy density for the engineering theory. One can interpret as "weighted" or equivalent initial strain measures. Differentiating (12—12) with respect to the stress resultants and couples, and substituting in (12—11), we obtain the following force-displacement relations: e?,
F1 F2
F3
+
MT
(03
k2
=k2
X2 = U3 1 + (02
k3
= k3 +
U21
M2
(12—13)
= (03,
To interpret the coupling between the shear and twist deformations, we note (see Fig. 12—4) that U2
X3W1
U3
defines the centroidal displacements due to a rigid body rotation about the shear center. Comparing (a) with (12—13), we see that the cross section twists about the shear center, not the centroid. This result is a consequence of neglecting the in-plane deformation terms in i.e., of using (12—12). Instead of working with centroidal quantities (M1, u2, u3), we could have
started with M1. and the translations of the shear center. This presupposes that the cross section rotates about the shear center. We replace u2, u3 (see Fig. 12—4) by U2
U3 =
+ (01X3 t01X2
(12—14)
SEC. 12—3.
FORCE-DtSPLACEMENT
337
where um, U53 denote the translations of the shear center. The terms involving
F2, F3, Al1 in (12—9) transform to
1+
Then, taking
1
— w3)
+
1
+ w2)
(a)
as an independent force parameter, we obtain
=
a)1,1
F2 US2,
F3
(12— 15)
1
053 1 + (02
Since the section twists about the shear center, it is more convenient to work and the translations of the shear center. Once 052, 053, and w1 are
with
x2
Fig. 12—4. Translations of the centroid and the shear center.
known, we can determine 02, 03 from (12—14). We list the uncoupled sets of force-displacement relations below for future reference. Stretching
e? +
F1
01:1
Flexure in X1 -X2 Plane
F2 GA2
+
El3
= =
1 — (03
ENGINEERING THEORY OF PRISMATIC MEMBERS
338
Flexure in X1-X3 Plane
(12—16) F3
+
U53
k2
CHAP. 12
+
=
I
Twist About the Shear Center
MT
= Wi,1
The development presented above is restricted to an elastic material. Now, the principle of virtual forces applies for an arbitrary material. Instead of first specializing it for the elastic case, we could have started with its general form (see (10—94)),
=
dA]
[ss
AP1
(12—17)
where represents the actual strain matrix, and denotes a system of statically permissible stresses due to the external force system, AP1. We express the integral as .
if
j=1
A
and determine
using
as defined by the engineering theory. For example,
taking AF1 A
AM2
AM3
+ ——
—
13
12
leads to
e1 =
if
k2
if x2c1 dA if
k3
dA
Once the extensional strain distribution is known, we can evaluate (b). Using (12—18), the one-dimensional principle of virtual forces takes the form
+
AM1)]dxj =
AP1
(12—19)
The virtual-force system must satisfy the one-dimensional equilibrium equations (12—4). One should note that (12—19) is applicable for an arbitrary material.
When the material is elastic, the bracketed term is equal to dV*, and we can write it as dV's
dx1 =
>d1 AP1
(12—20)
SEC. 12—4.
SUMMARY OF THE GOVERNING EQUATIONS
339
The expanded form for the linearly elastic case is
J
[(eq +
AF1 +
+
AF2
+
AF3
(12-21)
+
+ El2) AM2 +
+
El) AM3] dx1 =
We use (12—21) in the force method discussed in Sec. 12—6.
SUMMARY OF THE GOVERNING EQUATIONS
12—4.
At this point, we summarize the governing equations for the linear engineering theory of prismatic members. We list the equations according to the different modes of deformation (stretching. flexure, etc.). The boundary conditions reduce to either a force or the corresponding displacement is prescribed at each end. Stretching (F1, u1) F1
I + b1 =
0 (12—22)
F1 F1
or u1 prescribed at x1 =
0,
L
Flexure in X1-X2 Plane (F2, M3, U2, 0)3)
+ b2 = 0 M31 + m3 + F2 = F2,
F2
=
U2
0
i—
2
M u2
(12—23)
+
(03,1
or F2 prescribed at x1 =
0,
M3 or 0)3 prescribed at .x1 =
L
0, L
Flexure in the X1-X3 Plane (F3, M2, U3,
F3,1 + b3 M2, + F3
=
u3
0 —
1+
F3
=
0
0)3
M
u3 or F3 prescribed at x1 = 0, L 0)3 or M2 prescribed at x1 = 0, L
(12—24)
ENGINEERING THEORY OF PRISMATIC MEMBERS
340
CHAP. 12
Twist About the Shear Center (MT, (01, u2, u3)
+
MT. 1
MT
0
01T
= or a1 prescribed at x1 =
m5. =
12-5.
(12—25)
+ b2x'3 — b3g2
X3W1
U2 U3
m1
0, L
=
DISPLACEMENT METHOD OF SOLUTION—PRISMATIC MEMBER
The displacement method involves integrating the governing differential equations and leads to expressions for the force and displacement parameters as functions of x1. When the applied external loads are independent of the displacements, we can integrate the force-equilibrium equations directly and then find the displacements from the force-displacement relations. If the applied load depends on the displacements (e.g., a beam on an elastic foundation), we must first express the equilibrium equations in terms of the displacement parameters. This.problem is more difficult, since it requires solving a differential equation rather than just successive integration. The following examples illustrate the application of the displacement method to a prismatic member.
Example
12—1
We consider the case where b2 = coast (Fig. This loading will produce flexure in the X1-X2 plane and also twist about the shear center if the shear center does not lie on
the X2 axis. We solve the two uncoupled problems, superimpose the results, and then apply the boundary conditions. Flexure in X1 -A'2 Plane We start with the force-equilibrium equations,
F21 M3,1 =
—F2
Integrating (a), and noting that b2 = coast, we have
F2 =
—
b2x1
For convenience, we use subscripts A, B for quantities associated with x1 =
=
etc.
FAJ
With this notation, (c) simplifies to
F2 =
FA2
—
b2x1
Substituting for F2 in (b), and integrating, we obtain
= MAI
—
XIFA2
+
0,
L:
SEC. 12—5.
DISPLACEMENT METHOD OF SOLUTION
We consider next the force-displacement relations,
M3 £03, 3
Uz, i
F2
+
—.
GA2
Integrating (g) and then (h), we obtain £03 =
WA3
+ — (x1MA3
—
4 \
4 \2E13/
GA
12Ff)
The general flexual solution (for b2 = const) is given by (e), (f), and (I). Fig. E12—1
x2 b3 13
Shear center
Centroid L
H
Twist About the Shear Center The applied torsional moment with respect to thc shear center is
mr = Substituting for mr in the governing equations,
= (01,1
MT
=
and integrating, we obtain
MT = MAT
=
—
b2y3x1
+
The additional centroidal displacements due to twist are U2 =
U3 =
X3W1
ENGINEERING THEORY OF PRISMATIC MEMBERS
342
CHAP. 12
Cantilever Case We suppose that the left end is fixed, and the right end is free. The boundary conditions are
=0
=
UA2
FBZ
= M53 = MBT
Specializing the general solution for these boundary conditions requires
= An
b2L
11. i2
A3
2
MAT =
b2;L
and the final expressions reduce to
F2 = b2(L M3 =
x1)
— Lx1 +
= U2
—
— x1)
=
+ b2Lx1
+
-
+
()
U3 = b2
4L 4
a)1
It is of interest to compare the deflections due to bending and shear deformation. Evaluating u2 at x1 = L, we have I
U52
h2L
LI lh2L2
UB2
E
= =
13
GL2A2
an illustration, we consider a rectangular cross section and isotropic material with v = 0.3 (d = depth): As
613d2
13
A2
5
A
10
=
By definition, d/L is small with respect to unity for a member element and, therefore, it is
SEC. 12—5.
DISPLACEMENT METHOD OF SOLUTION
343
reasonable to neglect transverse shear deformation with respect to bending deformation
for the isotropic case.t Formally, one sets 1/A2 = 0.
Fixed-End Case We consider next the case where both ends are fixed. The boundary conditions are 0.42
= W.43 = (ISA! = 0
=
=
=0
Specializing (h), (i), and (k) for this case, we obtain b2L
FAZ=---2
M.43 =
b2L2 12
b2T3L
MAT =
The final expressions are
IL2
M3 =
b2
MT =
b2T3
+ —
u2
=
U3
= —X20)1
(03
2)
+
b2IL2 = El =
Lx1
b2y3
— x1
Xj
L
+
+
b2 —
2L4 + xi)
(u)
xi)
xi)
Example 12—2 We consider a member (Fig. E12—2) restrained at the left end, and subjected only to forces applied at the right end. We allow for the possibility of support movement at A. The expressions for the translations and rotations at B in tcrms of the end actions at B and support movement at A are called member force-displacement relations. We can
obtain these relations for a prismatic member by direct integration of the force-displacement
t For shear deformation to be significant with respect to bending deformation, G/E must be of the same order as l/A,L2 where A, is the shear area. This is not possible for the isotropic case. However, it may be satisfied for a sandwich beam having a soft core. See Prob. 12—1.
CHAP. 12
ENGINEERING THEORY OF PRISMATIC MEMBERS
344
relations. In the next section, we illustrate an alternative approach, which utilizes the
principle of virtual forces4' Fig. E12—2
x2
M1, WB 1
FA3 UB 3
MA3/
/
3
x3
The boundary conditions at x1 = L are = =
(a)
Integrating the force-equilibrium equations and applying (a) lead to the following expressions for the stress resultants and couples:
(j= 1,2,3)
= MT
M2 =
M52
M3
MB3 + (L
(L
—
x1)F53
—
Using (b), the force-displacement relations take the form
+ (L — xi)F5]
C03,j = U3,1 =
+
+
—
= El2 U3 j = — 0)2 +
_
M5.,—
Col,i = t See Prob. 12—11.
(b)
SEC. 12—5.
DISPLACEMENT METHOD OF SOLUTION
345
Integrating (c) and setting x1 = L, we obtain UAI +
UB1
L_
L_M83 +
03.43 +
(.053
UA2 + LWA3 +
+
(052
U83 =
=
co,0
L2
M82 —
UA3 — LWA2 —
L
+
L2
+
LX3_ Mjjr +
—
+
/L
+
L3\..
L2_
L2
—
IL
+
L3\_
MET
Finally, we replace MET by
MET =
+ X31B2
and write the equations in matrix form:
L
US'
L U82
+
L3
+ L
L3
GJ L
L (082
L2
(053
+
UA2
+
L
U.43
LOA2,
(.0,42, (.0A3}
(f)
The coefficient matrix is called the member "flexibility" matrix and is generally denoted by fB.
We obtain expressions for the end forces in terms of the end displacements by inverting f. The final relations are listed below for future reference:
ENGINEERING THEORY OF PRISMATIC MEMBERS
346
CHAP. 12
AE (um
=
— UA1)
FB2
+
—
WA3)
(u53 — UA3)
rGJ MB1
+
12E
—
L3
+ COA2)
-1
+
—
+ (2 —
(053
—
—
+
— UA3)
(co81 —
WAI)
a2)
WA!)
j_ WA2
UA2) +
(4 +
WA!)
— WA1)
— UA3)
=
+
+ C°A2)
+ (0A3)
+ (4 + a2) M83
(co82
(082 — UA2) +
+ MB2
+ +
=
(coB! —
—
—
+ (2
(1)4j)
a3)_L_coA3
where a2
12E13
12E12 a3
=
=
1+03 introduce the assumption of negligible transverse shear deformation by setting 03 = a3 = 0 The end forces at A and B are related by We
(j= 1,2,3) MA! = —!V151
+ LF53
MA2 MA3
We list only the expressions for MA2
=
—M53 — LF52
M43: —
+ (4 + a2) MA3
0A3)
+
—
WA2 + (2 — 02)
—T-(uB2 — UA2) +
+ (4 + a3)
WA!)
C0A3
—
L2
+ (2 —
a3)
WA!)
SEC. 12—5.
DISPLACEMENT METHOD OF SOLUTION
347
Example 12—3 We consider next the case where the applied loads depend on the displacements. To simplify the discussion, we suppose the shear center is on the X2 axis and the member is loaded only in the plane. The member will experience only flexure in the X1-X2 plane under these conditions. The governing equations are given by (12—23):
F2 =
—M3,1
—
(b)
in3
M3
(03
An alternate form of(a) is M3
+ 1723,
b2
0
Once M3 is known, we can, using (b), find F3. Now, we solve (d) for 03 and substitute in (c): F2
F,1
(03,1
+
p2,11
Then, M3
+
E13(u3
b2 —
and
F2 =
El2 (02
—m3 —
+
b2,
—
Finally, we substitute forM3 in (e) and obtain a fourth-order differential equation involving 02 and the load terms: d4u2
+
d2 (
'\
b2 —
I
+
(din3
\
— h2)
=
0
The problem reduces to solving (i) and satisfying the boundary conditions: F2 or 02 prescribed or (03 prescribed)
Neglecting transverse shear deformation simplifies the equations somewhat. The resulting equations are (we set 1/GA2 = 0) (03
j
= E13(u2, 11 —
F3 = d4u3
—in3
—
d2
E13(u2 1
—
+
(din3
—
'\
— b2) = 0
ENGINEERING THEORY OF PRISMATIC MEMBERS
348
CHAP. 12
As an illustration, consider the case of linear restraint against translation of the centroid,
e.g., a beam on a linearly elastic foundation. The distributed loading consists of two terms, one due to the applied external loading and the other due to the restraint force. We write
=
q — ku3
where q denotes the external distributed load and k is the stiffness factor for the restraint.
We suppose rn3 = = 0, k is constant, and transverse shear deformation is negligible. Specializing (k) for this case, we have (03 = U2
M3 = E13u2,
F2 = d4u2 F3
—E13u2
k
+
q
=
or u2 prescribed 1
0L
M3 or (03 prescrihedJ
The general solution of (n) is sin A.x1 + C2 cos Ax1) +
+
/
sin Ax1 + C4 cos Ax1)
k
represents the particular solution due to q. Enforcement of the boundary conditions at x 0, L leads to the equations relating the four integration constants. increases with increasing x. The function e_2x decays with increasing x, whereas where u2
For Ax > 0. If the member length L is greater than 2(3/A) = 2Lb (we interpret 3, Lb as the width of the boundary layer), we can approximate the solution by the following: 0
x1
<
=
Lb:
+
sin Ax1 + C2 cos Ax1)
LB<xl
L:
u2 =
+
sin Ax1 + C4 cos Ax1)
The constants (C1, C3) are determined from the boundary conditions at x1 = (C3, C4) from the conditions at x1 = L. Note that C3 and C4 must be of order
u2isfiniteatx1 =
L.
Application .1 The boundary conditions at x1 =
0
(Fig. E12—3A) are
u2 = M3 =
0
E13u2,11
=
0
Since q is constant, the particular solution follows directly from (11),
= q/k The complete solution is U2
=
(1
cos Ax1)
0
and since
FORCE METHOD OF SOLUTION
SEC. 12—6.
349
Fig. E12—3A
q = const
//////////// ///////////////////////////////////////// X1 x2
Application 2 The boundary conditions at x1 =
(Fig. E12—3B) are
0
U2,1 =
0
F2 = —E13u2,111 = —P12 and the solution is U2 =
PA Ax1
+ sin Ax1)
The four basic functions encountered are
=
Ax + sin Ax)
=
sin Ax =
—
(12—26)
= e
cos Ax =
—
Their values over the range from Ax = 0 to Ax =
5
are presented in Table 12—1. Fig. E12—3B
////////////////////j///
x1
x2
12—6.
FORCE METHOD OF SOLUTION
In the force method, we apply the principle of virtual forces to determine the displacement at a point and also to establish the equations relating the force redundants for a statically indeterminate member. We start with the onedimensional form of the principle of virtual forces developed in Sec. 12—3 (see Equation 12—19):
+
=
d1 AP1
ENGINEERING THEORY OF PRISMATIC MEMBERS
350
CHAP. 12
Table 12—i
Numerical Values of the ,tX
iJi1
0.0 0.2 0.4 0.6 0.8
1.000 0.965 0.878 0.763 0.635
0.000 0.163
1.0
0.508 0.390 0.285 0.196 0.123
0.310
Functions AX
1.000 0.640 0.356 0.143 —0.009
1.000 0.802 0.617 0.453 0.313
0 0.2 0.4 0.6
0.281 0.243 0.202 0.161
—0.111 —0.172 —0.201 —0.208 —0.199
0.199 0.109 0.042
1.0 1.2
2.6 2.8
0.067 0.024 —0.006 —0.025 —0.037
0.123 0.090 0.061 0.038 0.020
3.0 3.2 3.4 3.6 3.8
—0.042 —0,043 —0.041 —0.037 —0.031
1.2
1.4 1.6
1.8
2.0 2.2 2.4
4.0 4.2 4.4 4.6 4.8 5.0
'
0.261
0.310 0,322
0.8
1.4
--0.006
1.6
—0.038
1.8
—0.179 —0.155 —0.128 —0.102 —0.078
—0.056 —0.065 —0.067 —0.064 —0.057
2.0 2.2 2.4 2.6
0.007 —0.002 —0.009 —0.012 —0.014
—0.056 —0.038 —0.024 —0,012 —0.004
—0.049 —0.041 —0.032 —0.024 —0.018
3.0 3.2 3.4 3.6 3.8
—0.026 —0.020 —0.016 —0.011 —0.008
—0.014 —0.013 —0.012 —0.010 —0.008
0.002 0.006 0.008 0.009 0.009
—-0.012
—0.007 —0.004 —0.001 0.001
4.0 4.2 4.4 4.6 4.8
—0.005
—0.007
0.008
0.002
5.0
2.8
are the actual one-dimensional deformation measures; d, represents a displacement quantity; AP, is an external virtual force applied in the direction of
where e3,
The relations between the deformation measures and the internal forces depend on the material properties and the assumed stress expansions. The appropriate
relations for the linear elastic engineering theory are given by (12—13). If a displacement is prescribed, the corresponding force is actually a reaction. We use AR,. to denote a prescribed displacement and the corresponding reaction increment, and write (a) as d,. AR,. d, AP, (12—27) + AM3)]dx,, — where d, represents an unknown displacement quantity. To determine the displacement at some point, say Q, in the direction defined by the unit vector we apply a virtual force APQIq, and generate the necessary internal forces and reactions required for equilibrium using the one-dimensional
force-equilibrium equations. We express the required virtual-force system as = AP0 (12—28) = MJ,QAPQ ARk==Rk,QAPQ
SEC. 12—6.
FORCE METHOD OF SOLUTION
Introducing (12—28) in (12—27) and canceling dQ =
—
leads to
+
Q
351
+ k3M1, Q)]dxj
(12—29)
This expression is applicable for an arbitrary material, but is restricted to the linear geometric case. Since the only requirement on the virtual force system is that it be statically permissible, one can always work with a statically determinate virtual force system. The expanded form of (12—29) for the linearly elastic case follows from (12—21): dQ
+
=
(F2\ +
+
[(e? +
$ +
Fj,Q
i'F3\ +
El21
+ +
M1.
M3\
-JM3,Q 11dx1
J..J3/
where
if
(12—30)
J
dA
=7±JJx2s?dA Finally, we can express (12—29) for the elastic case in terms of V*: L
k
dQ
—
JxLt'AQ
(12—31)
ORQ
This form follows from (12—20) and applies for an arbitrary elastic material.
Example
12—4
We consider the channel member shown in Fig. E12--4A. We suppose that the material is linearly elastic and that there is no support movement. We will determine the vertical
Fig. E12—4A
x2 I)
Centroid Q
Shear center
ENGINEERING THEORY OF PRISMATIC MEMBERS
352
CHAP. 12
displacement of the web at point Q due to— 1.
2.
the concentrated force P a temperature increase AT, given by
AT =
a1x1
+ a2x1x2 + a3x1x3
Force System Due to P Applying the equilibrium conditions to the segment shown in Fig. E12—4B leads to
.F2 = = +Pe M3 = —P(L — x1) F1 = F3 = M2 0 Fig. E12—4B •1
.
M3
(
P
I
Fe
I
F2,
Shear center axis
System We take dQ positive when downward, i.e., in the — X, direction. To be consistent, we must apply a unit downward force at Q. The required internal forces follow from Fig. E12—4C:
F2,0 = Mr,0
—1 C
/L
M3,Q =
2
F5,0 = F3,0
(p. =o 0
\
(b)
— x1)
M2,Q
(j=
0
1,2,3)
Fig. E12—4C
I
S
F2,Q
/
Shear center axis
FORCE METHOD OF SOLUTION
SEC. 12—6.
353
hzitial Deformations The initial extensional strain due to the temperature increase is
=
=
a
+ a2x1x, + a3x1x3)
(d)
The equivalent one-dimensional initial deformations are e? = .1
d,4
JJ = ±.
Jf
(e)
dA = aa3x1 dA = —aa2x,
=
Deter,ninatio,, of (IQ Substituting for the forces and initial deformations in (12—30), we obtain ('Lii I
dQJ
P
r
Pc2
P
1
IL
\)
(f)
L
5 1))
c2L
cxa,L2
=
Example 12—5 When the material is nonlinear, we must use (12—29) rather than (12—30). To illustrate the nonlinear case, we determine the vertical displacement due to P at the right end Fig. E12—5
x2
xi P
"I
Centroid (and shear
center)
of the member shown in Fig. E12—5. We suppose that transverse shear deformation is negligible, and take the relation between k3 and M3 as k3
= a1M3 +
(a)
354
ENGINEERING THEORY OF PRISMATIC MEMBERS
Noting that only F2, and M3,0 are finite, and letting e2 = reduces to L
d0
0,
CHAP. 12
the general expression for
k3M30dx1
Now,
M3 = —P(L M3,Q = —(L
—
—
x1)
x1)
Then,
k3 = —Pa1(L
—
x1)
—
P3a3(L
—
x1)3
Substituting for k3 in (b), we obtain dQ =
+ P3a3
Pa1
We describe next the application of the principle of virtual forces in the analysis of a statically indeterminate member. We suppose that the member is statically indeterminate to the rth degree. The first step involves selecting r force quantities, Z1, Z2,. , Z.. These quantities may be either internal forces or reactions, and are generally called force redundwns. Using the force-equilibrium equations, we express the internal forces and reactions in terms of the prescribed external forces and the force redundants. .
.
=
+
= M3,0
=
Mf,kZk
+
R1,
(12—32)
kZk
The member corresponding to = Z2 = ''' = Z,. = 0 is conventionally called the primary structure. Note that all the force analyses are carried out on the primary structure. The set (F3, 0, M3, R1, represents the internal forces and reactions for the primary structure due to the prescribed external forces. Also, (F1, k, M3, k, R1, k) represents the forces and reactions for the primary structure due to a unit value of Zk. One must select the force resultants such that the resulting primary structure is stable, Once the force redundants are known, we can find the total forces from (12—32). It remains to establish a system of r equations relating the force redundants. With this objective, we consider the virtual-force system consisting of AZ,, and the corresponding internal forces and reactions,
= F3,,,AZ,, AM3 = MJ,,,AZ,, AR, =
FORCE METHOD OF SOLUTION
SEC. 12—6.
355
This system is statically permissible. Substituting (a) in (12—27), and noting
=
that
we obtain
0,
k + kJMi.k)] dx1 =
(12—33)
k
Taking k = 1, 2, ..., r results in a set of r relating the actual deformations. One can interpret these equations as compatibility conditions, since they represent restrictions on the deformations. To proceed further, we must express the deformations in terms of In what follows, we suppose that the material is linearly elastic. The compatibility conditions for the linearly elastic case are given by
J
[(eq +
+
+
+
+
+
+
+
(12—34)
dx1
A more compact form, which is valid for an arbitrary elastic material, is C
(7R Ic
(k = i,2,..
The final step involves substituting for resulting equations as (k
.fIcJZJ = is,,
j= where
= fjk
+
=J
M1 using (12—32). We write the
=
1
+
1,
2. .
+
.
.
, r)
(12—36)
j
[(eq + F1.o)F +
1
+
+
—
(1235)
1
rr1 =>
.,r)
Ic
+
dx1
+
+
+
+
The various terms in (12—36) have geometrical significance. Using (12—30), is the displacement of the primary structure in the direction of we see that Z1 due to a unit value of ZIc. Since fik = fkJ, it is also equal to the displacement in the direction of Zk due to a unit value of Z3. Generalizing this result, we can write
=
(12—37)
ENGINEERING THEORY OF PRISMATIC MEMBERS
356
CHAP. 12
corresponds to arbitrary points, and i.e., i has the same direction and sense. Equation (12—37) is called Maxwell's law of reciprocal deflections, and follows directly from (12—30). The term Ak is the actual displacement of the point of application of Zk. minus the displacement of the primary structure in the direction of Zk due to support movement, initial strain, and the prescribed external forces. If we take Zk as an internal force quantity (stress resultant or stress couple), Ak represents a relative displacement (translation or rotation) of adjacent cross sections. One can interpret (12—36) as a superposition of the displacements due to the various effects. They are generally called superposition equations in elementary texts.t If the material is physically nonlinear, (12—36) are not applicable, and one must start with (12—33). The approach is basically the same as for the linear case. However, the final equations will be nonlinear. The following examples illustrate some of the details involved in applying the force method to statically indeterminate prismatic members. where i, j are
Example
12—6
This loading (Fig. E12—6A) will produce flexure in the plane and twist about the shear center; i.e., only F2, M3 and are finite. The member is indeterminate to the first degree. We will take the reaction at B as the force redundant.
Fig. E12—6A x2
x2
X3 Shear
center
Primary Structure One can select the positive sense of the reactions arbitrarily. (See Fig. El 2—6B.) We work with the twisting moment with respect to the shear center. The reactions are related to the internal forces by
=
Z1
R2
—
R3
= = +[MT]x,=o
R4
t See, for example, Art. 13—2 in Ref. 3.
FORCE METHOD OF SOLUTION
SEC. 12—6.
357 E12—68
x2
x2
R3,d3
R2, d2
ZI = 0
Force System Due to Prescribed External Forces
R1
Fig. E12—6C
q
—__ lB Mr,o F2,0
Shear center axis -.——p-*'
F2,0 =
MTo = M3,0 = F1,0
=
qe(L
— x1)
R1,0
x1)
R2,0
—
q —
F3,0
b
= M2,0 =
= 0 = qL
R3,0 =
x1)2 0
I qe
(b)
qL2
R4,0 = qeL
Force System Due to Z1 = + Fig. E12—60 M3,1
Bt
e
F2,1
Shear center axis
ENGINEERING THEORY OF PRISMATIC MEMBERS
358
CHAP. 12
F2,1=+1 MT,j =
—e
R2,1 =
—1
R4•1 =
—e
M3,1=+(L—x1) F1,
=
F3,
1
= M2, 1 = 0
Equation for Z1 We suppose that the member is linearly elastic. Specializing (12—36) for this problem,
f 11Z1 =
A1 1)2
=
+
1)2
)2]dX
+
(d)
= and then substituting for the forces and evaluating the resulting integrals, we obtain L
fit
Le2
L3
(e)
(L — x1)dx1 —
The value of Z1 for no initial strain or support movement is
z1 = 8
Final Forces The total forces are obtained by superimposing the forces due to the prescribed external system and the redundants:
F2=F20+Z1F2,1 = —q(L—x1)+Z1 MT
qe(L — x1)
M3 =
—
—
eZ1
(L — (g)
= qL
R3 =
=
—
Z1
L2
e(qL
LZ1 —
Z1)
FORCE METHOD OF SOLUTION
SEC. 12—6.
359
Example 12—7 This loading (Fig. E12—7A) will produce only flexure in the X1-X2 plane. We suppose the material is physically nonlinear and take the expression for k3 as
k3 =
+ a1M3 +
(a)
To simplify the analysis, we neglect transverse shear deformation. Fig. E12—7A
q
V.
f /// / F
Primary Structure
=
Z1
R7
R3 =
=
(b)
Fig. E12—78
x1
R2, d2
z1 =0
R1,2j
Force System Due to Prescribed External Forces (see Example 12—6) F2,0 = —q(L — x1) M3,0 = R1,0 =
0
R2,0 = qL
—
R3,0 =
qL2
ENGINEERING THEORY OF PRISMATIC MEMBERS
360
CHAP. 12
Due to Z1 = + 1 (see Example 12—6)
Force
F2,
= L
M3,
+1
—
R21=—1
R1,3=+1 R3,1 = —L
Compatibility Equation Since the material is nonlinear, we must use (12—33). Neglecting the transverse shear deformation term (e2), the compatibility condition reduces to
dx1 =
J We substitute for k3 using (a):
J
(ajM3 +
—
dx1 =
JL
Now, 1W'3
= M3,
+ Z1M3,
=
x1)2
—
+ Z1(L — xt)
Introducing (g) in (f), we obtain the following cubic equation for Z1:
z?
+
(asLs) +
+ = —
+
— x1)dx1
—
+
For the physically linear case,
030 and (h) reduces to
=
Example
+
—
—
—
fLko(L
—
xi)dxi]
12—8
The member shown (Fig. E12—8A) is fixed at both ends. We consider the case where the material is linearly elastic, and there are no support movements or initial strains. We take the end actions at B referred to the shear Center as the force redundants.
= = Z3 = Z2
MTB
The forces acting on the primary structure are shown in Fig. E12—8B.
Initial Force System F2,0 =
=
P
MT,0
P(a
—
x1)
361
FORCE METHOD OF SOLUTION
SEC. 12—6.
Fig. E12—8A
tP x2 Shear
b
a
•
L
Fig. E12—8B
z3
x3
Fig. E12—8C
M30 MTO
(
Shear center axis
P A
Px3 F2,0
_________________ ENGINEERING THEORY OF PRISMATIC MEMBERS
362
CHAP. 12
Z = +1 Fig. E12—8D Al31
'(
ti
F2,1
Shear center axis
L—x1 '-I
F2,1=+1
M31=L—x1
(c)
—0
Z2
= +1 Fig. E12—8E M3,2
(
M72
M3,2 = +1
F22 =
MT,2
0
z3 = +1 Fig. E12—8F Al33
(I
//// Shear center axis
I. —x1
=+
1
F3,
M3,
=
0
(e)
FORCE METHOD OF SOLUTION
SEC. 12—6.
363
Goinpatibility Equations The compatibility equations for this problem have the form
fkj =
i:
f
=
= 1,2,3)
(k
JkJZJ =
+
+
+
+
Substituting for the various forces and evaluating the resulting integrals lead to the following equations:
I) \ +l—1Z2 /l——+-—-—1Z1 L 7 V \GA2 3E13)
=
/L\ Z2 =
(L2\ Z1
[a [GA2 Pa2
(a3
I
a2b
El3 \ 3
2
(g)
+ '\GJJ
GJ
Finally, solving (g), we obtain 6E13
=
1±
—P
2b1 PICA L2GAZ
Z3__Pox3
Z3
Application Suppose the member is subjected to the distributed loading shown in Fig. E12—8G. We can determine the force redundants by substituting for P, a, and b in (h),
P = q dx1 a= b=L
—
x1
and integrating the resulting expressions. The general solution is CL (
=
2
+
2
z2 Z3
X3 —
j
x1q dx1
—
x1)
+
6E13 —
1)
ENGINEERING THEORY OF PRISMATIC MEMBERS
364
CHAP. 12
where
C=
1
+
12E13
L2GA2
As an illustration, we consider the case where q is
q
const in (j),
we obtain qL z1 = — — 2
z2 =
12
Fig. E12—80 x2
q(xj)
H
L
H
REFERENCES 1.
2.
TIMOSHaNKO, S. J. : Advanced Strength of Materials, Van Nostrand, New York, 1941. HETENY!, M.: Beams on Elastic Foundation, University of Michigan Press, Ann Arbor, 1946.
3.
4. 5. 6.
7. 8.
Elementary Structural Analysis, McGraw-Hill, Noiuus, C. H., and J. B. New York, 1960. ASPLUND, S. 0.: Structural Mechanics: Classical and Matrix Methods, Prentice-Hall, 1966. DEN HARTOG, J. P.: Advanced Strength of Materials, McGraw-Hill, New York, 1952. ODEN, J. T.: Mechanics of Elastic Structures, McGraw-Hill, New York, 1967.
Geac, J. M. and WEAVER, W.: Analysis of Framed Structures, Van Nostrand, 1965. MARTIN, H. C.: Introduction to Matrix Methods of StructuralAnalysis, McGraw-Hill, New York, 1966.
PROBLEMS
12-1. The accompanying sketch shows a sandwich beam consisting of a core and symmetrical face plates. The distribution of normal stress over the depth is determined by assuming a linear variation for the extensional strain:
PROBLEMS
365
= —x2k3
in the definition equation for M3:
We relate k3 to M3 by substituting for
M3 =
x2a11 dA
+ Ef13,f)k3
M3
To simplify the notation, we drop the subscript and write (b) as iVI
where (EI)equiv is the equivalent homogeneous flexural rigidity. Prob.
12—1
x2
012 A*
f M3
I The shearing stress distribution is determined by applying the engineering
theory developed in Sec. 11—7. Integrating the axial force-equilibrium equation
over the area A* and assuming
is
constant over the width, we obtain
JJ(aii,i + a21,2 +
a31,
3)dA = o (d)
ba12 =
cru, dA
Then, substituting for cru, M
= —(Ek3)x2 (
and noting that F2
—M3,
(—Ex2)
I)equiv
(d) becomes
a12 =
J'J x2E
dA
(e)
ENGINEERING THEORY OF PRISMATIC MEMBERS
366
(a)
(b)
CHAP. 12
Apply Equations (e) and (f) to the given section. The flange thickness is small with respect to the core depth for a typical
and G. are
beam. Also, the core material is relatively soft, i.e., small with respect to Ef. Specialize part a for determine the equivalent shear rigidity (17*)
(c)
1
dA
= J)
0 and tf/h 1. Also which is defined as
2
The member force-deformation relations are F Y2
= M3
k
= (EI)equjv
Refer to Example 12—1. Specialize Equation (q) for this section and discuss when transverse shear deformation has to be considered. 12—2. Using the displacement method, determine the complete solution
for the problem presented in the accompanying sketch. Comment on the influence of transverse shear deformation. Prob.
12—2
q = const
F
b
x1
12—3.
For the problem sketched, determine the complete solution by the
displacement method. 12—4. Determine the solution for the cases sketched. Express the solution in terms of the functions defined by (12—26).
PROBLEMS
367
Prob. 12—3
x2
q
const
Shear
HeH
R
Prob. 12—4
(0)
/////////////////////////////7/////////////////// Ib) Jr
(c)
12—5. The formulation for the beam on an elastic foundation is based on a continuous distribution of stiffness; i.e., we wrote b2 =
—ku2
(a)
Note that k has units of force/(Iength)2,
We can apply it to the system of discrete restraints diagrammed in part a of the accompanying sketch, provided that restraint spacing c is small in
ENGINEERING THEORY OF PRISMATIC MEMBERS
368
CHAP. 12
comparison to characteristic length (boundary layer) Lb, which we have taken as 3
3
2
(k/4E1)'14
A reasonable upper limit on c is
c< Letting k4 denote the discrete stiffness, we determine the equivalent distributed stiffness k from
k=
kd/c
Evaluate Lb with (b), and then check c with (c). Prob. 1
J
>
r
12—5
(
.
J
J
J ...—
r
r +
C
r
+
(a)
III C+C
L,E,11
a/2
/7/7
/7/7
7
(b)
Consider the beam of part b, supported by cross members which are fixed at their ends. Following the approach outlined above, determine the distribution of force applied to the cross members due to the concentrated load, P.
PROBLEMS
369
Evaluate this distribution for
L=64ft
a=241t
c—lit
12—6. Refer to Example 12—3. The governing equation for a prismatic beam on a linearly elastic foundation with transverse shear deformation included is obtained by setting b2 = q — ku2 in (i). For convenience, we drop
the subscripts: d4u
—
k d2u +
k
u=
/
1
d2 7 —
We let
q —
+
&112
and (a) takes the form d4
dx
Note that
,12
4
u—q—
dx 2
is dimensionless and A has units of 1/length. The homogeneous
solution is
cos bx + C2 sin bx) +
u
cos bx + C4 sin bx)
where
a=
2(1
+
=
2(1
—
b
= 0. Determine the expression for the boundary layer length (e3 0). Determine the solution for the loading shown. Assume L large with respect to Lb. The boundary conditions at x 0 are 0) = 0 P
To specialize (d) for negligible transverse shear deformation, we set (a) (b)
F2
Investigate the variation of Mmax and
Umax
with
to vary
Consider
from 0 to 1. Prob. P
/////////)///////////// ////////////)//////// a
L
X
12—6
ENGINEERING THEORY OF
370
MEMBERS
CHAP, 12
12—7. Refer to the sketch for Prob. 12—3. Determine the reaction R and centroidal displacements at x1 L/2 due to a concentrated force Pi2 applied to the web at x1 L/2. Employ the force method. 12—8. Refer to Example 12—7. Assuming Equation (h) is solved for Z1, discuss how you would determine the translation u2 at x1 = L/2.
12—9. Consider the four-span beam shown. Assume linearly elastic behavior, the shear center coincides with the centroid, and planar loading. (a) Compare the following choices for the force redundants with respect to computational effort: 1. reactions at the interior supports 2. bending moments at the interior supports
(b)
Discuss how you would employ Maxwell's law of reciprocal detlections
to generate influence lines for the redundants due to a concentrated force moving from left to right. Prob.
12—9
12—10. Consider a linearly elastic member fixed at both ends and subjected to a temperature increase
Determine the end actions and displacements (translations and rotations) at mid-span. 12—11. Consider a linearly elastic member fixed at the left end (A) and subjected to forces acting at the right end (B) and support movement at A. Determine the expressions for the displacements at B in terms of the support movement at A and end forces at B with the force method. Compare this approach with that followed in Example 12—2.
13
Restrained of. a 13—1.
Prismatic Member
INTRODUCTION
The engineering theory of prismatic members developed in Chapter 12 is based on the assumption that the effect of variable warping of the cross section on the normal and shearing stresses is negligible, i.e., the stress distributions predicted by the St. Venant theory, which is valid only for constant warping
and no warping restraint at the ends, are used. We also assume the cross section is rigid with respect to in-plane deformation. This leads to the result that the cross section twists about the shear center, a fixed point in the cross section. Torsion and flexure are uncoupled when one works with the torsional moment about the shear center rather than the centroid. The complete set of governing equations for the engineering theory arc summarized in Sec. 12—4. Variable warping or warping restraint at the ends of the member leads to additional normal and shearing stresses. Since the St. Venant normal stress distribution satisfies the definition equations for F'1, M2, M3 identically, the additional normal stress, must be statically equivalent to zero, i.e., it must satisfy
dA =
dA =
dA =
0
(13—1)
The St. Venant flexural shear flow distribution is obtained by applying the engineering theory developed in Sec. 11—7. This distribution is statically equiva-
lent to F2, F3 acting at the shear center. It follows that the additional shear due to warping restraint must be statically equivalent stresses, and to only a torsional moment: Sfri2 dA dA
0
=
(13—2)
0
To account for warping restraint, one must modify the torsion relations. We will still assume the cross section is rigid with respect to in-plane deformation. 371
RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13
372
En what follows, we develop the governing equations for restrained torsion. We start by introducing displacement expansions and apply the principle of virtual displacements to establish the force parameters and force-equilibrium equations for the geometrically linear case. We discuss next two procedures for establishing the force-displacement relations. The first method is a puredisplacement approach, i.e., it takes the stresses as determined from the strain
(displacement) expansions. The second method is similar to what we employed for the engineering theory. We introduce expansions for the stresses in terms
of the force parameters and apply the principle of virtual forces. This corresponds to a mixed formulation, since we are actually working with expansions
for both displacements and stresses. Solutions of the governing equations for the linear mixed formulation are obtained and applied to thin-walled open and closed cross sections. Finally, we derive the governing equations for geomettrically nonlinear restrained torsion. 13—2.
DISPLACEMENT EXPANSIONS; EQUILIBRIUM EQUATIONS
The principle of virtual displacementst states that
JJJaT ös d(vol.) =
d(vol.) + JJJJT
SSSbT
d(surface area)
is identically satisfied for arbitrary displacement, L\u, when the stresses (r) are in equilibrium with the applied body (b) and surface (p) forces. We obtain a system of one-dimensional force-equilibrium equations by introducing expansions for the displacements over the cross section in terms of one-dimensional displacement parameters. This leads to force quantities consistent with the displacement parameters chosen. We use the same notation as in Chapters 11, 12. The X1 axis coincides with the centroid; X2, X3 are principal inertia axes; and x3 are the coordinates of the shear center. We assume the cross section is rigid with respect to in-plane deformation, work with the translations of the shear center, and take the displacement expansions (see Fig. 13—1) as U1 =
U1
+
W3X2 +
W2X3
w1(x3
U2
—
U3 =
+ w1(x2 — x2)
—
(13—3)
where 4 is a prescribed function of x2, x3, and— 2.
u1, ;2, u53 are the rigid body translations of the cross section. W2, £03 are the rigid body rotations of the cross section about the shear center and the X2, X3 axes.
3.
f is a parameter definining the warping of the cross section. The
1.
variation over the cross section is defined by Note that all seven parameters are functions only of x1. For pure torsion t See
Sec. 10—6.
SEC. 13—2. DISPLACEMENT EXPANSIONS; EQUILIBRIUM EQUATIONS
373
the St. Venant theory developed in Chapter 11), one sets f = co1, i = const and For unrestrained variable torsion (i.e., the engineering theory = developed in Chapter 12), one sets f = 0. Since there are seven displacement parameters, application of the principle of virtual displacements will result in seven equilibrium equations. (i.e.,
x3
Shear center
— —:: — —e
U52
I
I
I
x2
Centroid
Fig. 13—1. Notation for displacement measures.
The strain expansioust corresponding to (13—3) are 6j Ytz
= =
+f
u1, 1 + (02.
(03, 1x2
us2, 1 — C03 — COj,
1(x3 — x3)
us3,
1
+ (02 + cot. 1(x2 —
+ +
2
(13—4)
3
Using (13—4), the left-hand side of (a) expands to öe d(voL)
Au1,
+ F2(Au,2, 1 — Aw3)
+ F3(Au33 + Aw2) + M2 Aw2, + M3 Ac)3,1 A1 + MR Af]dxi + MT Aw5,1 + where the two additional force parameters are defined by
= MR =
dA
+
(13—5)
Note that Mç1, has units of (force) (length)2 and MR has units of moment. The quantity Mci, is called the biinornent. t This derivation is restricted to linear geometry. The nonlinear strain expansions are detived in Sec. 13—9.
RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13
374
To reduce the right-hand side of (a), we refer the transverse loading to the shear center, The additional load terms are m4,
dS = distributed bimornent = external bimoment at an end section (x1 =
=
4picb
=
$5pjcb dA
0, L)
(136)
Then SJJbT Au d(vol.) + SSPT Au d(surface area)
=
Au1 + b2 1xu32 + b3 Au,3 ± mr + rn2 Aw2 + m3 Aw3 + m# Af]dx1 + JF1 Au1 + F2 Au,2 + F3Au,3 + MTLXWI + M2Aw2 + M3 Aw3 +
The definitions of
are the same as for the engineering
mj, mj', F,,,
theory. Finally, we equate (b), (c) and require the relation to be satisfied for arbitrary
variations of the displacement parameters. This step involves first integrating (b) by parts to eliminate the derivatives and then equating the coefficients of the displacement parameters. The resulting equilibrium equations and boundary conditions are as follows: Equilibrium Equations F1
+ b1
0
F2,
+ b2
0
F3,
+ b3 =
0
1
MTI+m-I-=O M2,
1
— F3
+ rn2 =
0
M3,1 + F2 + m3 = 0 M4, — MR + 0
Boundary Conditions at x1 u1
=
(13—7)
0
u1
or
Us2
F1 = F2 =
—F1
F3
—F3
US3
or or
w1=w1
or
Mr=—MT
or or or
M2 =
w3=co3
f=J
Boundary conditions at x1 = L These are the same as for x1 = For example:
f=J
0
with the minus sign replaced with a plus sign.
or
SEC. 13—3.
DISPLACEMENT MODEL
375
We recognize the first six equations as the governing equations for the engineering theory. The additional equation,
O<xj
f=f
or
is due to warping restraint. Also, we see that one specifies either f or the bimoment at the ends of the member. The condition f = 7' applies when the
end cross section is restrained with respect to warping. If the end cross section is free to warp, the boundary condition is = ± M4, (+ for x1 = L). To interpret the equation relating MR and the bimoment, we consider the definition for MR, MR = + Integrating (e) by parts leads to
=
SS4(ail,2 + a13, 3)dA Utilizing the axial stress equilibrium equation, a12, 2 + c713,
3
+ a11,
0
1
we can write MR
=
+ JJç'au,j dA
We see that (h) corresponds to the axial equilibrium equations weighted with
respect to
+ a13,3 +
$$(a12,2
— x,,3a13)çdS = 0
+ 4(PI —
(i) i
+
—
MR
=
0
In most cases, there is no surface loading on S. i.e., Pi = 0 on the cylindrical boundary. We will discuss the determination of stresses in a later section. We simply point out here that MR involves only the additional shear stresses due correspond to to warping restraint since the St. Venant shearing
1
3—3.
FORCE-DISPLACEMENT RELATIONS—DISPLACEMENT MODEL
To establish the relation between force parameters and the displacement parameters, we consider (13-4) to define the actual (as well as virtual) strain distribution and apply the stress-strain relations. We also consider the material to be isotropic and suppose there is no initial strain. The stress expansions are au a12
1
=
Gy12
=
+ — 0)3
a13 = Gy13 = G[u,3, 1 +
t M5 =
torsion.
=
0 for
+ X3) + f4, 2] + cot, 1(x2 — + f4, 3] i
cot, 1(x3
St. Venant (pure) torsion. We neglect
and
(13—8)
for unrestrained variab'e
376
RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13
denotes the effective modulus. Although our displacement expansions correspond to plane strain (&2 = = 0), the in-plane stresses vanish on the boundary. Therefore, it seems more reasonable to use the extensional stressstrain relations for plane stress. In what follows, we will take = Young's modulus, F. Consider the expression for The term involving is due to warping of the cross section. This additional stress must satisfy (13—1), which, in turn, requires 4 to satisfy the following orthogonality conditions :t where
dA =
dA
dA =
0
(13—9)
Assuming (13—9) is satisfied, and noting that X2, X3 are principal centroidal axes, the expressions for F1, M2, M3, and the reduce to: EAu1,1 M2
E12w2,1 E13a,3 1
M3
13—10
j
where cIA
We have included the subscript r on E to keep track of the normal stress due to warping restraint. Inverting (13—10) and then substituting in the expression for lead to F1 'Yii
+
M2
13
F2
1 — (03
=
+
(13—Il)
and MR expand to
The expressions for F2, F3,
F3
M3 —
+
i) + fS2
+ fS3
+
A(u33, i
(13—12)
1
11c01,1 —
=
+
—
+
+
+
where
Si = polar moment of inertia = —
= t F1 =
M2 =
12
x3q5,2)dA
+
= 0 for c11 due to warping restraint.
+
+ 13
DISPLACEMENT MODEL
SEC. 13—3.
377
Also, the expressions for the shearing stresses canbe written as a12 —
+G
1
f
+
3
—
The essential step is the selection of which, to this point,must satisfy only the ortliogonality conditions (13—9). To gain some insight as to a suitable form let us reexamine the St. Venant theory of unrestrained torsion. We suppose the section twists about an arbitrary point instead .of about the centroid as in Sec. 11—2. The displacement expansions are
for
u2 = —w1(x3
coj(x2
U3
—
= where
i=
M1/GJ = const. Operating on (a) leads to
=0 a12 =
(x3 — x3)]
2
7
3
+
The equation and boundary condition for equation and boundary condition,
follow from the axial equilibrium
mA —
We can express
on S
—
as
=
C
+
—
x'2x3
+
where C is also an arbitrary constant. The boundary condition and expressions
for tile stresses become —
a12 = M1 cr13
Since
3
—
x3)
+
x2)
-
depends only on the cross section, it follows that the stress distribution
RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13
378
torsional constant are independent of the center of twist. Also, one can showt that = 0 SJcbt,2dA = 2)dA 3)2]dA 3 — $S[(4)t. 2) + and
are evaluated by requiring
Suppose we take 4' = The constants (C, 4' to satisfy (13—9), and we obtain
C= = =
dA
Now, one can shows that the equations for are identical to the equations for the coordinates of the shear center when the cross section is considered to be rigid with respect to in-plane deformation. That is, the warping function for unrestrained torsion about the shear center is orthogonal with respect to 1, x2, x3.
Summarizing, we have shown that C — T3x2
4)
+ x2x3 +
(13—14)
is a permissible warping function. The cross-sectional properties and forcedisplacement relations corresponding to this choice for 4' are listed below: Properties
S2 =
= 14,
14,
'13—15 2)
+
3)2]dA
Shear Stresses
a12 =
F2
+ G(—x3co1,1
+ (13—16)
= t See Sec. 11—2 and Prob. 11—2. See Prob. 13—1.
+ G(x2w1,1 + f4)t,
SEC. 13—4.
RESTRAINED TORSION—DISPLACEMENT MODEL
379
Force-Displacement Relations
MT = MR =
G11w1, i
w1, us2, j
F3 j
We
+
—
—
+ .T2F3
—
(13—17)
— w1,
W3
+ W2 + x2(f
Wi,
introduce the assumption of negligible restraint against warping by
setting Er
to MR =
0.
0. Then, M4, 0, and thc seventh equilibrium equation reduces Specializing (13—17) for this case, we obtain f — to1,
=
MT =
(x3F2
—
—,
x21'3)
(13—18)
GJW1,1
and
F2(1
F3( x2x3
\J1
us3,1 = to2 +
F2(
G\
F3!'! + -- +
(13—19) 14,
The shearing stress distributions due to F2, F3 do not satisfy the stress boundary condition 0 on S +
However, one can show that they satisfy
+
=0
for arbitrary F2, F3. Equations (13—19) are similar in form to the results obtained in Chapter 12, which were based on shear stress expansions satisfying (a) identically on the boundary.
Finally, we point out that torsion and flexure are uncoupled only when warping restraint is neglected (F. = 0). Equations (13—17) show that restrained torsion results in translation of the shear center. We will return to this point in the next section. 13—4.
SOLUTION FOR RESTRAINED TORSION—DISPLACEMENT MODEL
To obtain an indication of the effect of warping restraint, we apply the theory developed in the previous section to a cantilever member having a
rectangular cross section. (See Fig. 13—2). The left end (x1 =
0)
is fixed with
380
RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13
x3
T 2b
/
A
Sect. A-A
Fig. 13—2. Restrained torsion-cantilever with rectangular cross section.
to both rotation and warping while the right end (x1 = L) is free to warp. The boundary conditions are respect
x1=O
x1=L
M1=M
(a)
For convenience, we list the governing equations for restrained torsion: Equilibrium Equations (See (13—7))
M1,1+m1=O
(b) (c)
Force-Displacement Relations (See (13—10) and (13—12). F2 = F3 = 0)
= M1 = =
G11w11
+ +
Note that
(d)
Boundary Conditions (for this example)
At xj= 0, (e)
Atx1
= M
1,1=0 Xj
We start with (b). Integrating (b) and enforcing the boundary condition at = L leads to (13—20) M1 = M
RESTRAINED TORSION—DISPLACEMENT MODEL
SEC. 13—4.
381
Next, we combine (c) and (d):
=M
G11co11 + i +
Solving (f) for w1,
=
(f) (g)
t
(h)
G111, and then substituting in (g) lead to
(i) —
where 2 is defined as
[i,,
G
13
- 21
Note that has units of(1/length)2. The solution of(i) and (h) which satisfies the boundary conditions (e) is (we drop the subscript on x for convenience)
f=
{l
=
—
cosh
x
±
+ tanh [sinh
sinh
+ (1 — cosh
tL]}
(13—22)
=
+ '1 The rate of decay of the exponential terms depends on we can take tanh )L 1, and the solution reduces to I,fr
For
)L >
2.5,
—
As
a point of interest, the St. Venant solution is dw1
M (J)
We see that is a measure of the length, Lh, of the interval in which warping restraint is significant. We refer to Lb as the characteristic length or boundary layer. By definition, 0
(13—24)
In what follows, we shall take (13—25)
The results obtained show that is the key parameter. Now, depends on the assumed warping function. If the ratio G/ET and on terms derived from
RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13
382
the warping functiowt for unrestrained torsion defined by
we take 4)
(13—14), the various coefficients are related by
J
(13—26)
—
J
—
At this point, we restrict the discussion to a rectangular section (see Fig. 13—2) We evaluate the various integrals defined by (13—15) and write and 4) = the results as
J=
(13—27)
= K1a3b =
where the K's are dimensionless functions of b/a. With these definitions, the expression for takes the form 1 /G'\112 K22= --—
K
-
The coefficients are tabulated in Table 13—1. We see that
is essentially
constant. Assuming E 2.6G and K1 3.2, we find 2/b and Lb 2b. The influence of warping restraint is confined to a region of the order of the depth. Although this result was derived for a rectangular cross section, we will show later that it is typical of solid and also thin-walled closed cross sections. Table
13—1
b /<4
1
2 3
10
2.25 3.66 4.21 4.99
.0311 .165 .283 .425
.156 .450
.683 .964
3.36 3.16 3.23 3.32
We consider next the problem of locating the center of twist. We utilize the and large solution corresponding to 4)
M1 fC =
=
0
(13—29)
,,
for a rectangular section and
—e reduces to
MIXEIJ FORMULATION
SEC. 13—5.
383
The translations of the shear center follow from (13—17):
Us31
= x3(f — = —x2(f —
(13—30)
By definition, the translations are zero at the center of twist. Setting i12 =
denote the coordinates of the center of twist
in (13—3) and letting lead to 0
X2 = gx2 x
)
see that the center of twist approaches the shear center as x increases. The maximum difference occurs at x = 0 and the minimum at x = L. We
1
=—
=
1—
For unrestrained warping, E, = 13—5.
0,
=
2L!1
oc,
and g =
1.
FORCE-DISPLACEMENT RELATIONS—MIXED FORMULATION
We first review briefly the basic variational principles for the three-dimensional formulation. The principle of virtual displacements requires
5e d(vol.) = JJjbT Au d(vol.) +
Au d(surface area)
to be satisfied for arbitrary Au and leads to the stress-equilibrium equations and stress-boundary force relations. Note that ôg is a function of Au and is obtained using the strain-displacement relations, The stress-strain relations can be represented as since, by definition of the complementary energy density,
=
av*
=
combining (a) and (b), we obtain a variational principle which leads to both sets of equations. The stationary requirement, By
— liT11 —
V*)d(vol.)
—
considering i, u as independent quantities, g called Reissner's principle.t
d(surface area)] = E(u),
and
0
(13—33)
b prescribed, is
See Ref. 11 and Prob. 10—28. Reissner's principle applies for arbitrary geometry and elastic material. This discussion is restricted to linear geometry. The nonlinear case is treated in Sec. 13—9.
RESTRAINED TORSION-FLEXURE OF PRTSMATIC MEMBER CHAP. 13
384
The essential point to recognize is that Reissner's principle allows one to work with and u as independent quantities. In a displacement formulation (Sec. 13—3), we take as a function of u, using the stress-displacement relations — V* reduces to V, the strain-energy density. In a = and = mixed formulation we start by introducing expansions for the displacements. The Euler equations for the displacement parameters are obtained by expanding (a). This step leads to the definition of force parameters and forceequilibrium equations. We then generate expansions for the stresses in terms
of the force-parameters from an equilibrium consideration. The relations between the force and displacement parameters are obtained from the second stationary requirement: Tx1
[f$(CT
—
öV*)dA]dxj =
0
(13—34)
The first step was carried out in Sec. 13—2 and the expanded form of 5$
dA
is given by (b) of Sec. 13—2. Letting represent the complementary energy per unit length along X1, and using (13—4), the stationary requirement on the stresses (Equation 13—34) expands to
+ +
1 — W3) + öF3(u,3, i
+
+
+ co2) + +
i —
=o
13—35)
In order to proceed further, we must express in terms of the force parameters (F1, F2, ..., MR). Equating the coefficients of each force variation to zero results in the force-displacement relations. Instead of applying (13—34), one can also obtain (13—35) by applying the principle of virtual forces to a differential element. We followed this approach in Chapter 12 and, since it is of interest, we outline the additional steps required for restrained torsion. One starts with (see Fig. 13—3) 5V* dx1 =
op
+ [$$uT ISp
(a)
The boundary forces are the stress components acting on the end faces. Taking u according to (13—3) and considering only MR. we have $5
oiiTudA = ±SJISrsTudA
= ±(ISMTWI + ISM4f) where the plus sign applies for a positive face. The virtual-force system must be statically permissible, i.e., it must satisfy the one-dimensional equilibrium equations. This requires const
= Then,
=
dxi{f,i
= dx1{f 0M4 + .[ISMR +
+ w1,1 1 ISMT}
(d)
MIXED FORMULATION
SEC. 13—5.
385
The first procedure (based on (13—34)) is more convenient since it avoids introducing the equilibrium equations. However, one has to have the straindisplacement relations. In certain cases, e.g., a curved member, it is relatively
6Mg, +
—aM,5
WI
o,1+w11dx1
f
f + f,1 dxj Fig. 13—3. Virtual force system.
to establish the force-equilibrium equations by applying the equilibrium conditions to a differential element. We obtain the force-displacement relations by applying the second procedure (principle of virtual forces) without having to introduce strain expansions.t In what follows, we consider the material to be homogeneous, linearly elastic easy
and isotropic. To simplify the treatment, we also suppose there is no initial strain. The complementary energy density is =
if
dA +
+
if
(13-36)
It remains to introduce expansions for the stress components in terms of the force parameters such that the definition equations for the force parameters are identically satisfied. Considering first the normal stress, we can F1
where
M2
M3
(a)
satisfies the orthogonality conditions: § dA
=
JJx2çb c/A
=
c/A = 0
(b)
Note that we have imposed a restriction on q5. The complementary energy due to c11 expands to —,
1
I
t The approach based on the principle of virtual forces is not applicable for the geometrically nonlinear case. See 1). Problem § F1 = = M3 = 0 for
treats the case of a nonhomogeneous material. due to warping restraint.
386
RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13
Finally, substituting for
in (13—35), we obtain F1
W3, I
f,
L.13
—
These expansions coincide with the corresponding relations obtained with the displacement model (see (13—10)).
The shearing stress distribution must satisfy thc definition equations for F2, F3, MT and MR identically. We can obtain suitable expansions by adding
a term due to warping restraint to the results for unrestrained torsion and fiexure. We write
+ + ('If = the flcxural distribution due to F2, F3;
(13—37)
is the unrestrained is torsion distribution; and is the distribution due to restrained torsion. Since we are assuming no in-plane deformation, the fiexural distribution for a thin-walled section can he obtained by applying the engineering theory
where
developed in Sec. 11—7. For a solid section, we utilize the results of Sec. 11—5,
taking v = 0. The shear stress distribution for unrestrained torsion is treated in Sees. 11—2 through 11—4. Since the restrained-torsion distribution is statically equivalent to a torsional moment, we have to distinguish between the unrestrained and restrained torsional moments: MT =
+
= =
(13—38)
It remains to determine We follow the same approach as in the engineering theory of flexural shear stress, i.e., we utilize the axial equilibrium equations and stress boundary condition:
mA
('12,2 +
+
on S
0
Differentiating the expression for au and noting the equilibrium equations, we obtain ('11,1 =
Since
F2 13
+
F3
+
MR
12
satisfies (a) for arbitrary F2, F3 and a" corresponds to
follows that
r is
=
0,
due to MR: MR
y r ('12.2 + ('13.3 =
+
4 0
(mA) —-
(on 5)
it
MIXED FORMULAflON
SEC. 13—5.
and boundary condition on a' ensure thatt
The orthogonality conditions on
dA =
=
dA =
0
We solve (13—39) and then evaluate
= J$[—(x3
387
—
0
from (c)
x3)a12 + (X2 -.-
Noting (13—40), we see that
Finally, we write (c1 as
= where
(13—41)
With this definition,
is a cross-sectional property which depends on
MT =
+
(13—42)
When the cross section is thin-walled, we neglect
= = We take
and
flow qr
(13 —40)
and (a) reduces to
d
at a free edge
0
to be constant over the thickness t and work with the shear Equation (d) becomes
1çô1
= qV
=
0
(13—43)
at a free edge
and boundary condition on
The orthogonality conditions on
=
(IS — 0
=
0
ensure that (13—44)
and equating to (13—41). by evaluating Finally, we determine We consider next the complementary energy density. We write the expanded form of the shear contribution as
+
V*,hear
JJ
+
+
+
+
(134)
in Sec. 11—5. For convenience, these have evaluated and results are summarized below (See Equation 11—98) /z'2 I 13 I. 112 Vf = + + We
23
3
(a) LI
11*
V uf f See Prob. 13—2.
2GJ —
RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13
388
The coupling term, 1/A23, vanishes when the section has an axis of symmetry.
= 0 is a consequence of our assuming the cross section is rigid with respect to in-plane deformation. We evaluate using (13—39) ((13—43) for the thin-walled case), and write the results as Also
=
+
if
(13—46)
where Cr is a dimensionless factor which depends on q5. The coupling between unrestrained and restrained torsion is expressed as
=
+
if
(13-47)
It is obvious that = 0 for a thin-walled open section since is an odd function of n whereas a' is constant over the, thickness. We will show later that it is possible to make vanish for a closed section by specializing the homogeneous solution of (13—43). Therefore, in what follows, we will take Cur
0
Finally, we write the coupling bctwcen flex ural and restrained torsion as
+
=
(13—48)
=
1
+ X2rF3Mg)
where Xjr have units of length. if X2 is an axis of symmetry,
= 0 since is symmetrical and a' is antisymmetrical with respect to the X2 axis. We substitute for in (13—35). replace Mr with + Ccj,MR, and equate and <5MR. The resulting force-displacement
the coefficients of oF2, OF3,
relations are I
U521 — 0)3
Us3,1 + 0)2
Wi.j
"F2
x3,
F3
+ —-- + —
= GA23
A3
+
J
(13—49)
=
I+I =
C,
MR +
1
(x3,F2 + x2,F3)
The corresponding relations for the displacement model are given by (13—12). Up to this point, we have required to satisfy the orthogonality relations and also determined a' such that there is no energy coupling between au and (C,1, = 0). If, in addition, we take ISC If" — X3X2 — — 5) — -t-- X2X3 -r —
SEC. 13—6.
RESTRAINED TORSION—MIXED FORMULATION
Cçt, = +1 MrT=+MR
389
(13—50
Note that
is the warping function for unrestrained torsion about the shear center. We discuss the determination of 4> in Secs. 13—7 and 13—8. One neglects shear deformations due to flexure by setting (13-51)
Similarly, we neglect shear deformation due to restrained torsion by setting Cr = X2r = X3r
0
(13—52)
This assumption leads to the center of twist coinciding with the shear center and 1 =
One now has to determine
(13—53)
from the equilibrium relation, + in4,
is known, it is more convenient to work with
If
tsr — —
AS
—
AK:'
In what follows, we outline the solution procedure for restrained torsion and list results for various loadings. We then discuss the application to open and closed cross section. 13—6.
SOLUTION FOR RESTRAINED TORSION—MIXED FORMULATION
We suppose only torsional loading is applied. The force-displacement relations are obtained by setting F2, F3, co2, w3 equal to zero and C,1, +1 in (13—49). For convenience, we summarize the governing equations below.
Equilibrium Equations MT. 1
+ 112T = 0
MT— —
Force-Displacement Relations (4> E,.l4,j
AK
1
= GJw1,1
GJ
+f)
f See Prob. 13—3. We include the minus sign so that C1
will be positive.
390
RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13
Boundary Conditions
or or
MT
prescribed at each end
Translations of the Shear renter u52, 1
=
=
We start by integrating (a): MT
dx1 =
C1 —
C1
+
(13—54)
Substituting (c) in (b) and (13—54) leads to the governing equations for w1 and f:
+
(1 + Cr)O)i,t
GJ(w1,1 + f) = 0 After some manipulation, (f) becomes C1x1
Eric,,
f
—
=
(C1
I
C
+
where ,12 is defined ast
cc = 1
22
C
+ Cr
(1355)
ErI#
Equation (g) corresponds to (h), (i) of Sec. 13—4. The general solution for f and has the following form:
f Wi
C3 cosh 2x + C4 sinh 2x —
+
+ C2 +
= —
(13—56)
sinh 2x + C4 cosh 2x) —
is the particular solution due to on x1 for convenience. t The corresponding paramater for the
We have dropped the subscript formulation is 2 (see (13—21)).
SEC. 13—6.
RESTRAINED TORSION—MIXED FORMULATION
391
The significance of A has been discussed in Sec. 13—4. We should expect, on the basis of the results obtained there, that AL will be large with respect to unity for a closed section. We will return to the evaluation of A in the next section. In the examples below, we list for future reference the solution for various loading and boundary conditions.
Example
13—1
Cantilever—Concentrated Moment Fig.
E13—1
X1
The boundary conditions (Fig. E13—l) are
x=O
cv1=f=O
x=L
f. = =
Starting with (13—54), we set
0
0
and C1 = M. The remaining constants are deter-
mined from
atx=0 atx=L
w1—f——0
and the final solution ist
M[
coshA(L—x)
GJ
cosh AL
[x —
WI
M
r LA
{sinh AL — sinh A(L — x)}
—c 1 sinh ).(L — x) cosh ,,L.
(13—57)
[ cosh
M
—
t The corresponding solution based on the displacement model is given by (13—22), (13—26). The expressions for! differ by a minus sign. This is due to our choice of We took in the displacement model and = in the mixed model.
_______
RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13
392
Note that C, 1 when the complementary energy term due to the restrained torsion shear stress (o') is neglected. The translations of the shear center are obtained by integrating
=
u,2,, = and requiring u,2, 0s3 to
vanish at x = U,2 =
I
u=j Let
0.
We write the result as =
X3.U
X2,.U
M
(13-58)
M1dx=x—co1
denote the coordinates and translations of the center of twist. By definition,
=
u,2
x3) = 0 —X2)=0 —
—
=u,3 + Substituting for
and w1, we obtain t X2 —
g
—1
+ x
The limiting
— gx3,
—
(13—59)
c
—
——f—-— [sinh ).L — sinh A. cosh ,.L
of g occur at x =
— x1]
0, L.
(13—60) 1
1
+
k) is an axis of symmetry for the cross section. Also, x2. = = 0 if we neglect shear deformation due to the restrained shear stress and, in this case, the center of twist coincides with the shear center throughout the length. Note that Xi,. = 0 if X,, (j
x3,.
Example 13—2 We consider next the case where warping is restrained at both ends; the left end (x =
0)
is fixed and the right end rotates a specified amount w under the action of a torsional moment. The boundary conditions are
x=0 x=L
co1=f=0 ce1=w
[=0
f See (13—31), (13—32) for the displacement model solution. There is no twist or translation at x = 0. We determine g(O) by applying L'I-IOspital's rule to (13—59).
RESTRAINED TORSION—MIXED FORMULATION
SEC. 13—6.
393
To simplify the analysis, we suppose there is no distributed load. Starting with the
general solution,
MT = C1
f = C3 cosh Ax + C4 sinh =
+ C2 —
—
{C3 sinh Ax + C4 cosh Ax}
and enforcing the boundary conditions leads to the following relations:
11—c
C3=—
GJ = cosh AL
=
s = sinh AL
4
C1L
C, [2(1 — c)11
+
C1(
/1—c'\
Ax
=
sinh Ax —
(0 1
(1361)
+ (01
=
+
= ci {i = Mç1,
—
(1.
—
[cosh Ax +
cosh Ax) — sinh Ax]}
sirih Ax]}
—
= ErI#A
{sinh Ax +
(L__f) cosh
Ax}
We write the relation between the end rotation, (0, and the end moment M, as M
L,5
= where L,ff denotes the effective length:
=L
r
2C
(c— 1'\1 (13-62)
= L(1 — cC3)
The following table shows the variation of with AL. For AL > 4, C2 2/AL. Note = 1 if transverse shear deformation due to restrained torsion is neglected. that AL
C3
0.5 1
0.98 .924
2
.76
3
.60
4
.48
RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13
394
Example 13—3
Uniform Distributed Moment-Symmetrical Supports The general solution for
MT =
for convenience) is:
m
C1 — ?flX
C4.
C3
f= —coshAx +
C1
flIX
+
—
C1
=
m /x2
x + C2
c
+
—
(a) (C3
sirth Ax + C4 cosh Ax)
—
We consider the boundary conditions to be identical at both ends and measure x from the midpoint (Fig. E13—3). Symmetry requires
MT=O}
atx=O
(b)
and (a) reduces to
MT =
[=
sinh Ax +
= C2 —
(13-63)
x
+
cosh xx
—
We treat first the case where the end section is fixed with respect to both rotation and warping. Requiring (13—63) to satisfy
f=
at x = L/2
=0
(a)
results in
I=
{x
sinh Ax}
—
mL2 fi[ CO1
/x"t21
U—
(cosh ,,x —
j+
c)
MT = —mx (
= ,nL1—
x
+
C
(13—64)
)
AL 1
c=
AL
s=
.
AL
The solution represents an upper bound. A lower bound is obtained by allowing the section to wrap, i.e., by taking
APPLICATION TO THIN-WALLED OPEN CROSS SECTIONS
SEC. 13—7.
395
Fig. E13—3
—x1 H
and the result is {x
f=
—
sinh
xc
mL2 (1 [ —
=
c=
13—7.
C, —
c)
—mx (
=
j+
(13—65)
C
—x+fsinhAx
IC,
cosh
I
cosh Ax —
AL
APPLICATION TO THIN-WALLED OPEN CROSS SECTIONS
In what follows, we apply the mixed formulation theory to a wide flange section and also to a channel section. We first determine the cross-sectional and then obtain general expressions for properties corresponding to = — the stresses in terms of dimensionless geometric parameters. Before discussing
the individual sections, we briefly outline the procedure for an arbitrary section.
Consider the arbitrary segment shown in Fig. 13—4. We select a positive sense for S and an arbitrary origin (point P). The unrestrained torsion warping function is obtained by applying (11—29) to the centerline curve and requiring the section to rotate about the shear center.t
is positive when translation in the + S direction rotates the position vector about the + X1 direction. The unrestrained torsional shear flow is zero By definition, Ic1 = We work with q" rather than and mixed sections where one generates q" in terms of
to facilitate treatment of closed
396
RESTRAfNED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13
=
for an open section. Then, taking
—
sP
and integrating leads to (13—67)
pscclS
Note that one can select the sense of S arbitrarily. Also, varies linearly with S when the segment is straight. The constant is evaluated by enforcing the —* F1 = 0), orthogonality condition dS
=0
If the section has an axis of symmetry, = 0, if we take P on the symmetry ?v.t2 = M3 axis. The remaining orthogonality conditions (a'j1 0),
$4x2tdS =
0
are identically satisfied by definition of the shear center. t x3
Shear center
IPsc
x2
Fig. 13—4. Notation for determination of the warping function.
When the section has branches, we apply (13—67) to each branch. One has only to require continuity of 4) at the junction point. As an illustration, consider the section shown in Fig. 13—5. The distribution of 4) for the three branches is given by A— B 4)p+$gpscdS 4)
B—C B—D
b4)B+JopscdS
We are taking the origin at B for branches B — tSeeProb. 13—1.
C
and B — D.
SEC. 13—7.
APPLICATION TO THIN-WALLED OPEN CROSS SECTIONS
The shear flow due to
397
is obtained by integrating (13—43) and noting
(13—50). Forconvenience, we let qr
=
(13—68) 145
With this notation, the resulting expression simplifies to
+ J
S
+
q5tdS =
(13—69)
We start at a free edge and work inward. A +q points in the +S direction corresponds to _qr, i.e. qr acting in the —S (see Fig. 13—5). Then, a + direction, If the section has an axis of symmetry, is an odd function with respect to the axis and
is an even function.
x3
S.q
p S,q
Fig. 13—5. Example of a section with branches.
Once
and
are known, we can evaluate 1w,, and
with (13—10), (13—46):
= JJqYdA = 542tdS Cr =
if
+
In order to evaluate XZr, X3r, we need the fiexural shear stress distributions. and write We let q(J) be the distribution due to qU)
j=2 j=3
(13—71)
'(=3 k= 2
RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13
398
The coupling terms are defined by (13—-48), which reduces to
I
j
X3r +
q q
X21.
—
Substituting for qr and qf results in
for a thin-walled section with 4) = —
3dS
J I X3r
If X2 is an axis of symmetry,
=
Y
(13-72) —
t
is an even function of x3, qt2t is an odd function,
and X3r = 0. By analogy, x2, = 0 if X3 is an axis of symmetry. apply for an arbitrary thinThe definition equations for Cr, Ia,, x2r, and
walled section. When the section is closed, we have only to modify the equaand We will discuss this further in the next section. tions for 4) Example
13—4
Symmetrical I Section The I section shown (Fig. El 3—4A) has two axes of symmetry; it follows that the shear center coincides with the centroid and the warping function is odd with respect to K2, X3. Fig. E13—4A
x3
1, Applying (13—67), we obtain
q5=O q5 =
forweb S
for flange
Note that the sense of S is reversed for the bottom flange.
SEC. 13—7.
APPLICATION TO THIN-WALLED OPEN CROSS SECTIONS
399
The shear flow vanishes at S = ± b/2. Applying (13—69) and starting from pt. A, we find
= The distributions of
(25)2]
= S-b/2
and q' are shown in Fig. E13—4B, where the arrows indicate the
sense of q' for + Ms-. Fig. E13—48 b2ht
Plot of qr
Plot
We express the cross-sectional properties in terms of Ii, t, and a shape factor
= b/h 3 =
=
=
ht3
+
th5
(t)2
8(1 + = The dimensionless parameters occurring in the solution of the differential equations for the mixed formulations are and AL (see (13—55)). Using (c) and assuming a value of 1/3 for Poisson's ratio, we write
[3(1 + =
I
AL
=
400
RESTRAINED TORSION-FLEXURE OF PR!SMATtC MEMBER CHAP. 13
The coefficients
are tabulated below:
0.75
2.4 2.66
4.22
0.50
3.2
6.93
3
1. The warping parameter, ),L, depends Since (t/h)2 << 1 and 0(1), we see that on t//i as well as L/h. This is the essential difference between open and closed cross sections. For the solid section, we found that AL = 0(L/h) and, since L/h is generally large in com-
parison to unity, the influence of restrained warping is Iocalized.f The value of AL for an open section is O(L//,) 00/1,) and the effect of warping restraint is no longer confined to a region on the order of the depth at the end but extends further into the interior.
We consider next the determination of the stresses due to restrained warping. The general expressions are M4,
q
r
dTts
=
7
Using the distribution for çb and qr shown above, the maximum values of normal and shear
stress are =
6
=
The shearing stress due to unrestrained torsion is obtained from 3
To gain some insight as to the relative magnitude of the various strcsses, we consider a member fully restrained at one end and subjected to a torsional moment M at the other end. This problem is solved in Example 13—1. The maximum values of the moments are tanh
atx = 0
AL
= C5M
J
We substitute for the moments in (f), (g) and write the results in terms
t We defined the boundary layer length,
(sec (13—24). (13—25)) as 0
Lh
4
L
;.L
the maximum
SEC. 13—7. shear
APPLICATION TO THIN-WALLED OPEN CROSS SECTIONS
stress for unrestrained torsion: tanh
,,, =
(f)) Mt —
The variation of these coefficients with b/h is shown below: b 1?
Since
and
1
2
0.75 0.50
2.11 2.31
1.5 1.67 2
are of 0(1), it follows that
=
0(d)
The additional shearing stress (at) is small in comparison to the unrestrained valu Therefore, it is reasonable to neglect the terms in the complementary energy density due to ic., to take C, = 1 for an open section. We will show in the next section 0 and that this assumption is not valid for a closed section.
Example
13—5
Channel Section
We consider next the channel section shown in Fig. El3—SA. Since X2 is an axis of = x3, = 0. The expressions for the location of the centroid, shear center,
symmetry,
Fig. E13—5A S
Shear center
x2
H°
RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13
402
and 12 are
=
b
I+
e=b
=be
I+
th3 b h
The dimensionless coefficient? is essentially constant, as the following table shows: b
?
h
0.429 0.409 0.375
1.00 0.75
0.50
We determine by applying (13—67) to the three segments. Taking S as indicated above, and noting that is odd with respect to X2, we obtain:
Segment 1—2 6 Psc
=
=
hh(
- -S
bh(
2S\
Segment 2—3
The distribution is plotted in Fig. E13—5B. Since? < 1/2, the maximum value of q5 occurs
at point I (and 4). We generate next the distribution of and using (b):
starting at point 1 (since q =
Segment 1—2 S
152
bin
Segment 2—3
/ = The distribution of
'\
+
is plotted in Fig. El3--5C.
+
s2
0
at that point)
SEC. 13—7.
APPLICATION TO THIN-WALLED OPEN CROSS SECTIONS
403
Fig. E13—5B
—(1— e)
—e —
Distribution of Fig. E13—SC
D2® D1
)+Mi D2
D2
I
0
0
Distribution of qr/1242t The expressions lbr J, example:
and AL are written in the same form as for the previous
(1 ±
I=
=
I —h5 c
(t'\2
+
±
+
+
f=
=
Cs = AL
=
(t\2
+
(t)
=
404
RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13
The following table shows the variation
and
with b/h for G/E. = 3/8. i.e., Poisson's
ratio equal to 1/3. Note that the comments made for the wide-flange section also apply to the channel section. b
c =— h
1
2.33
2.55
0.75 0.50
2.65 3.4
3.39
5.24
In order to evaluate X2r, we need the flexural shear stress distribution due to F3. Applying (11—106) leads to
Segment 1—2 4(3)
Segment 2—3 4(3) =
—
— S)
The distribution is plotted in Fig. E13—SD; the arrows indicate the sense of q for a +F3. Fig. E13—5D
+
I
I
t÷F3
t
—1
Distribution of
lb/it
/2
SEC. 13—8.
Substituting for
THIN-WALLED CLOSED CROSS SECTIONS
405
and the cross-sectional constants in (13—72) leads to
(t\2
= =
+
(1 +
+
The coefficient is of order unity, as the following table shows:
1
0.926
0.5
1.03
In Example 13—1, we determined expressions for the coordinates of the center of twist in terms of .'c,, and It is of interest to evaluate these expressions for this cross section.
The coordinates at x
0(sec (13—59), (13—60)) are
=
0
= X2
——i—1
Substituting for
+
and evaluating
we obtain X2
=
—
=
1
0.5
13—8.
0.476 0.625
0.836 0.485
APPLICATION TO THIN-WALLED CLOSED CROSS SECTIONS
We treat first a single closed cell and then generalize the procedure for multicell sections. Consider the section shown in Fig. 13—6. The +S direction is from X2 toward X3 (corresponding to a rotation about the +X1 direction). Using the results developed in Sec. 11—4, the shear flow for unrestrained tor-
sion is
q =-1-C
2A
406
RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13
where A is the area enclosed by the centerline curve. The shearing stress varies
linearly over the thickness,
=
+
C'\
+
—) =
but the open-section term has a zero resultant. x3 q
S
x2
Fig. 13—6. Notation for single closed cell. —
Substituting for qU in (13—66), taking 4 P lead to
and integrating from point
CS
= We determine
+
dS — C
(13—73)
by enforcing
=
0
The two additional orthogonality conditions 4x2gbtdS =
0
are identically satisfied by definition of the shear center. t The shear flow due to is defined by (13—69),
q=—-7--q 14 + Q4 t Noting that x2t =
dQs/ds, we can write
#x24.t = We merely have to identify this term as the moment of the flexural shear stress about the shear center. See Prob. 11-12.
SEC. 13—8.
where
is
THIN-WALLED CLOSED CROSS SECTIONS
407
indeterminate. Our formulation is based on no energy coupling
between qU and
i.e., we require (see (13 --47))
=
Noting that
(13—74)
is constant for a single cell, and using (e), we obtain
f =
dS
(13—75)
——
The flexural shear flow distributions for F2, F3 are generated with (11—110). We merely point out here that there is no energy coupling between qU and quqf
0
(f)
One can interpret (13—74) and (f) as requiring qr to lead to no twist deforma0. We have expressed the fiexural shear flows as (see (13—71)):
tion, i.e., w1
ft — qjij=q
(J) —
=
FJ_0)
2
k
j=3 k=2
Finally, the definition equations for the cross-sectional properties have the same form as for the open-section: Eq. 13—70
Cr
Eq. 13—72 X2 is an axis of symmetry. Then, is an odd function of x3. If we take the origin for S (point p) on the X2 axis, = 0. Also, is an even function of x3 and = 0. In what follows, we illustrate the application of the procedure to a rectangular cross section.
Example
13—6
Rectangular Section—C'onsta,,t Thickness Applying (13—73) and taking q5 = 0 at point
shown in (Fig. E13—6A) leads to
ci + b
fa — b\
The distribution is plotted in Fig. E13—6B. Note that
section of constant thickness does not warp.
=
0
when a =
b,
i.e., a square
408
RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13 Fig. E13-.6A
T
- x2
Centroid
2a—
Fig. E13—6B
(a—b +
b
Distribution of
We determine Q4, by integrating (a),
= at
— b'\ S2 (
= (Q4 + and evaluate
for segment 1—2
+ bJ — 2
/ a+b1\
(a — b\
for segment 2—3
with (13—75): dS
.11
a—b
SEC. 13—8.
THIN-WALLED CLOSED CROSS SECTIONS
The distribution of
follows from (b), (c), 2
j(
409
2a\)
1(S\2\
2a(S
j/
2a
2 \a+b corresponds to q' acting in the clockwise and is plotted in Fig. E13—6C. Note that (— S) direction for + Ms-. Also, D is negative for b > a. Fig. E13—6C
x3
b
T 2a b
q'/D
qr(+
D-
2
We introduce a shape factor (, depth width
b
a
410
RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13 The resulting relations are
and express the various coefficients in terms of a, t, and
J= =
(neglecting
16a3t
4a5t [2(1
the contribution oft)
—
LATh
+
4
+
+
5(1
9(1
(G'\ /
I
\) 1/2 L
L
= x2, = x3,.
The variation of C,,
0
and b
= --
with b/a is shown in the table below;
.'
L;f for
C,
\
a
2 3
cc 10.43 4.41
G
3
E
8
0
0.98
0.0877 0.185
1.27
1.39
We found
(g)
=0
T)
for an open section. Our results for the single cell indicate that
=0
IL
C,>> 1 C,
1
for a closed section. We obtained a similar result for using the displacement-model formulation for a solid section. Since is due to the restrained shearing stress (q'), we see that shear deformation due to q' cannot be neglected for a closed cross section. We discuss next the determination of the normal and shearing stresses due to warping. The general expressions are
°isq q' t
tie
THIN-WALLED CLOSED CROSS SECTIONS
SEC. 13—8.
411
The maximum normal stress occurs at point 2 while the maximum shear stress can occur
at either points I or 3. We consider the same problem as was treated in Example 13—4, i.e., a member fully restrained at one end and subjected to a torsional moment M at the other end. We express the stresses in terms of ag,, the maximum shear stress for unrestrained torsion,
M(
+
= J
C
which reduces to
MC = since
7
M =
we are considering the section to be thin-walled. The maximum stresses are 2
0$
,nax,I
=
i
tanh 2.L
= [3C,1112
—
The variation of
and
2
S.C
with height/width is shown below. We are taking Poisson's
ratio equal to 1/3.
=
h/a
c,
(point 2)
1
0
2
1.04
3
—1.51
(point 1)
(point 3)
0 —0.35 —0.46
+0.44 +0.65
0
For large tanh I and we see that both the normal and shear stress are of the order of the unrestrained-torsion stress. In the open section case, we found the restrainedtorsion shear stress to be of the order of (thickness/depth) times the unrestrained shear stress.
To illustrate the procedure for a multicell section. we consider the section shown in Fig. 13—7. The unrestrained-torsion analysis for this section is treated in Sec. 11—4 (see Fig. 11—11). For convenience, we summarize the essential results here.
We nttmber the cells consecutively and take the +S sense from X2 to X3 for the closed segments and inward for the open segments. The total shear flow is obtained by superimposing the individual ccli flows
q' = qU =
0
for an exterior (open) segment constant for an interior segment
We let —
—
(U
WIT
RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13
412 e
"2
+
q,S
q1 ,S1
Fig. 13—7. Notation for mixed cross section.
The constants C1, C2 are determined by requiring each cell to have the same twist deformation, w1,
Enforcing (11—67),t
=
=
for each cell leads to 2A
where a, A are defined as
f
dS
—
,Jsj t
= a21
dS
=
A = {A1, A2}
The warping function is generated by applying (13—6): 4'
=
(13—76)
a
= Psc —
7
We start at point P1 in cell 1 and integrate around the centerline, enforcing continuity of 4, at the junction points b, c, and d. For example, at b, we require
t See also (11—32).
SEC. 13—8.
413
THIN-WALLED CLOSED CROSS SECTIONS
which leads to a relation between
and 4),,,:
Jb = 4)e +
j
Psc
+
dS =
dS —
Repeating for points C and d results in the distribution of 4) expressed in One can easily verify that 4) is continuous, i.e., & determined terms of determined from segment cdcL. Finally, we from segment ca is equal to evaluate
by enforcing
JJ4)dA=J4)tdS=O where the integral extends over the total centerline. Note that
0 if P1
is taken on an axis of symmetry. The shear flow for restrained torsion is obtained with (13—69): a
=
as
The steps are the same as for the flexural shear determination discussed in Sec. 11—7. We take the shear flow at points P1. P2 as the redundants, =
J = 1, 2
(13—77)
and express the shear flow as
+ ii.
(13—78)
is due to The distribution, has the same form as We just have to replace C with C'S. We generate by integrating (i) around the centerline, and enforcing equilibrium at the junction points. For example, at point b (see Fig. 13—7), where Zj0 is the open section distribution and
+ = Note that = 0 at points P1, P2. e andf The redundant shear flows are evaluated by requiring no energy coupling between qU and qr which is equivalent to requiring qr to lead to no twist deformation, j. Noting (c), we can write
= Finally, substituting for
j = 1,2
0
we obtain aCr = B 1'
f See footnote on page 385.
(13—79)
—
dS
(13—80)
RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13
414
Once 4) and zir are known, the cross-sectional properties (1 , Ci., X2r, xar) can be evaluated. Also we can readily generalize the above approach for an
n-cell section. 1
3—9.
GOVERNING EQUATIONS—GEOMETRICALLY NONLINEAR RESTRAINED TORSION
In this section, we establish the governing equations for geometrically nonlinear restrained torsion by applying Reissner's principle. This approach is a mixed formulation, i.e., one introduces expansions for both stresses and dis-
placements. The linear case was treated in Sec. 13—5. To extend the formulation
into the geometrically nonlinear realm is straightforward. One has only to introduce the appropriate nonlinear strain-displacement relations. Our starting point is the stationary requirement t —
V*)d(vol.)
d(surface area)]
—
=
0
V*(o.),
a, are independent variables, C and b are where e(u), prescribed. We take the displacement expansions according to (13—3) and use the strain-
displacement relations for small strain and small finite rotations4 U1
= U1 + C02X3 —
U3 =
+ w1(x2
= Yiz = Y13
W3X2
+
+
/4)
x3)
cn1(x3
112
—
(13—81
+
+ U2,1 + U3,3U3,2 l.1j3 + U31 + 1)2
The in-plane strain measures (62, 63, Y23) are of 0(w2), which is negligible
according to the assumption of sinai! finite rotations. Actually we assume O'22 =
1723
0, i.e., plane Stress. Substituting for the displacements and
noting the definition equations for the force parameters, the first term in (a) expands to d(vol.) =
1
1
1)]
— W3 + + F2[u,2. i + F3[u53 S + — t+ + M2{w2, 1 — co1, 1(u52 i + x3w1, + M3[w3, i — w1, i — T2w1. 1)] M0f1 + MRf + 1
+
i+
MQW1W1, 1}dxj
f See Eqs. 13—33 and corresponding footnote. We are working with Kirchhoff Stress and Lagrangian strain here. See Sec. 10—3. Eq. 10—28. Tile displacement expansions assume small-finite rotation, i.e., sin w and cos w 1. To be consistent, we must use (10—28).
GOVERNING EQUATIONS
SEC. 13—9.
415
where the two additional force parameters are = ÷ MQ = $J(x2c12 + x3a13)dA
The terms involving the external forces have the same form as for the linear case, but we list them again here for convenience (see (13—6)):
JJcbTu d(vol.) + jJpTu d(surface area)
+
+ + mrwi + in2w2 + m3oj3 + rn4f)dxi + F3u33 + MTO1 + M2(02 + M3co3 +
(13—83)
+ F1u1 + where the end forces (the barred quantities) are defined as previously, for example,
=
etc.
(5Jp1
It remains to introduce expansions for the stresses in terms of the independent
force parameters and to expand V*. In the linear case, there are 8 force M3, and M4, MR. Two additional force measures (Me, MQ) are present for the nonlinear case but they can be related to the previous force measures. We proceed as follows. We use the stress expansions employed for the linear case with = They are summarized below for convenience measures. F1
(see Sec. 13--5): F1
a11
±
A
M2
M3
—1—-x3 — -T—X2
13
±
+
+ &ij
= —
+ +
MT
Il
_.-
where 4,, f, q, h2 and h3 are functions of x2, x3. Introducing (a) in the definition
equations for
and MQ leads to
=
f11F1
= $2
+ fl2M2 + fl3M3 +
if if
=
$3 =
if
/34, 4,
+
= +
(13-84)
+
RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13
416
andt
+ (k =
+ +xlh3k)dA
+
MQ —
$J(x2h2k
2, 3)
(13—85)
=
+ = Certain coefficients vanish if the cross section has an axis of symmetry4 One can readily verify that fi1F1
(13—86)
0
MQ
when the section is doubly symmetric. For generality, we will retain all the terms here.
The complementary energy density function has the same form as for the linear case: —
=
1
—----'
2Ek.,A
+
+ —-——
13)
'2
1 + ——'
+
+
+
+
((Mw +
+
+ X2rF3)
We have shown that it is quite reasonable to neglect transverse shear deformation due to warping (C. X2r X3r = 0) for a thin-walled open section. Substituting Equations (13—82)—(13—87) in Reissner's functional and re-
quiring it to be stationary with respect to the seven displacement and eight force measures leads to the following governing equations: Equilibrium Equations
F1,1 + b1 = 0 +
j+
+ (1 +
+
+
—
1
= 0 + m2 = M3,1 + F2 + m3 =
+
—
See Prob. 13—12.
1M2} + b2
+
M2, 1 — F3
t See Prob. 13—il.
— w1
1 1
1
w1F3
+ F3 + w1F2 — wi,1M3} + b3 =
—
(1 +
—
+
0 0
+
2J32w1,
0 0
SEC. 13—9.
GOVERNING EQUATIONS
417
where
Relations
=
1FF2
1FF2
+
+
G
M
1
X3r
F3
+
1 + Wj,
1+
1+
—
w3
+ wi[u53, 1 — Wj, 1/33]
1 MrTJ =
+
(02
+
1+
+
=
(13—88)
=
(02,1
+ (0l,j(—US2,j + /32(01,1)
+ +
+ /33(01,1) j
[CrM?+ X3rF2 +
= I' +
Conditions (+ for x1 = L,
—
+
for x1 = 0)
+ T3w1,1) + F2 — wjF3 (01,1M2 = ±F2 — x2w1, 1M3 + F3 + w1F2 — ±F3 1 + 7J1w1, 1
+ (1 + + M2(—u52,j + + (02 prescribed or M2 = ±M2 (03 prescribed or M3 = ± M3 prescribed or = ± + (0j(172F2 +
i)
—
= 1.
u1 prescribed or F1 = prescribed or prescribed or wi prescribed or
+
=
=
Boundarp
i — x2uS3,
+ (I + 1+
+
= ±MT
f
These equations simplify considerably when the cross section is symmetric and transverse shear deformation is neglected.1' We discuss the general solution of (13—88) in Chapter 18. The following example treats one of the cases, a member subjected to an axial force and torsional moment. t See Prob. 13—13.
418
RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER
CHAP. 13
Example 13—7 We consider a prismatic member (see Fig. E13—7A) having a doubly symmetric cross section, fully restrained at one end and loaded by an axial force P and torsional moment M. We are interested here in evaluating the influence of axial force on the torsional behavior. The linear solution (with no axial force) was derived in Example 13—i.
Fig. E13—7A
x2
P
M
L
F
Equilibrium Equations (symmetrical cross section and no distributed load) = F1. d dx1
0
i) =
(M1 +
0
Force-Displacement Relations = GJw11
= ErI,ji F1 =
i
+
=
0
Boundary conditions
.xi=O xj = L
F1
P
M1 + J3tF1w1,1 = Al
Integrating the last two equations in (a) and noting the boundary conditions, lead to
F1 = M1 + /31F1w1,
The first equilibrium equation takes the form 1,11
2
const
= corlst
=P =M
GOVERNING EQUATIONS
SEC. 13—9.
419
where P11 7;:ij
GJA
2GJ
i±P I+
+ F)
This expression reduces to Equation (g) of Sec. 13—6 when P = 0. Once f is known, we can determine the rotation by integrating (d), which expands to + F
f
M—
+ when we substitute for M1 using (b). The general solution is,
f= [GJ
(i
C1
M
cosli ,ux + C2 smh
+ +
=
C3
+ Mx {i +
—
{C1
sinh px + C2 cosh
(We drop the subscript on x1 for convenience.) Finally, specializing (g) for these particular boundary conditions result in
f= wi =
{
—1 + cosh
—
sinh
tanh
{sinh jtx + (1 — cosh
— —
These equations reduce to (13-57) when P = 0. A tensile force (P > 0) increases the torsional stiffness whereas a compressive force (P < 0) decreases the stiffness. Equation (h) shows that the limiting value of P is 1. We let F, represent the critical axial force and the corresponding axial stress
11
(;J
to be less than the yield stress, (J/11) must be small with respect to unity. As an illustration, consider the section shown in Fig. E13—7B. The various coefficients (see Example 13—4) are In order for
J=
+
RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13
420
Fig. E13—78
x3
X2
and
r,,
(t'\2(
G
REFERENCES 1.
2. 3.
4. 5. 6.
T., and N. B. CHRISTENSEN: "Methods of Analysis of Torsion with Variable Twist," J. Aero. Sci., pp. 110—124, April 1944. TIMOSHENKO, S. J:: "Theory of Bending, Torsion and Buckling of Thin-Walled Members of Open Cross Section," J. Franklin Inst., pp.559—609, 1945. VON KARMAN, T.. and W. C. CHIEN: "Torsion with Variable Twist," J. Aero. Sci., Vol. 13, No. 10, pp. 503—510, October 1946. BENSCOTER, S. U.: "Secondary Stresses in Thin-Walled Beams with Closed Cross Sections," NACA—TN 2529, Washington, D. C., 1951. BENSCOTER, S. U.: "A Theory of Torsion Bending for Multiceil Beams," J. Appi. Mech., Vol 21, No. 1, 1954. VON
VLASOV, V. Z.: Thin Walled Elastic Brains, israel Program for Scientific Translations,
Office of Technical Services, IJ.S. Dept. of Commerce, Washington. D.C. 1961. HEILIG, R.: "Der Schuberverformungseinfiuss auf die Wölbkrafttorsion Von Stilben mit offenern Profil," Der Stahlbau, April 1961. 8. HEILIG, R,: "l3eitrag zur Theorie der Kastentrhger beliehiger Der Stahlbau, December 1961. 9. J. T.: Mechanics of Elastic Structures, McGraw-Hill. New York, 1967. 7.
10.
KOLLORUNNER, C. F., and K. BASLER: Torsion in Structures, Springer-Verlag. Berlin, 1969.
Ii.
K.: Variational Methods in
and Plasticity, Pergarnon Press.
1968. 12,
MAISEL, B. I.: "Review of Literature Related to the Analysis and Design of ThinWalled Beams," Technical Report 440, Cement and Concrete Association, London, July 1970.
PROBLEMS 13.
14. 15. 16. 17.
18.
421
DABROWSKi, R.: "Gekrüinmte dUnnwandige Trager," Springer-Verlag, Berlin, 1968. GALAMBOS, T. V.: Structural Members and Fiames, Prentice-Hall, 1968. BLEICH, F.: Buckling Strength of Metal Structures, McGraw-Hill, New York, 1952. BURGERMEISTER, G., and H. STEin': Srabilitar Theorie, Part 1. Akademie Verlag, Berlin, 1957.
CHILVER, A. H.: Thin- Walled Structures, Chatto and Windus, London, 1967. REISSER, E.: "Note on Torsion with Variable Twist." .J. AppI. Mech., Vol. 23, No. 2, pp. 315—316, June 1956,
PROBLEMS 13—1.
The shear stress distribution due to
= where
is given by (see (11—95)) F2
F2 (733
2
13
13
'
3
are fiexural warping functions which satisfy
=
— x2
(in A)
(onS) This result applies when the cross section is assumed to be rigid with respect to in-plane deformation. The coordinate of the shear center is defined by
=
if
X3
3
X3
where
is the St. Venant torsional warping function. Hint: See Prob. 11—11
and Equation (11—97). 13—2. 13—3.
(a)
Verify (13—40) and (13—44). This problem reviews the subject of the chapter in two aspects.
No coupling between the unrestrained and restrained torsional distribution requires 0 +
The unrestrained torsional shear stress distribution for twist about the shear center (see Sec. 13—3, Equation (b)) is given by ,f U
IVIT O'12
=
=
—
— X3 + X3]
+ x2
—
—
x2]
The restrained torsional shear stress distribution is determined from (13—39). Verify that
=
MR when ç& =
and (a) is enforced.
RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13
422
(b)
When the cross section is thin-walled, (a) and (b) take the form
•fquqr_ = o
is the perpendicular distance from the shear center to the tangent at the centerline, Equation (d) follows from (11—29) and where
Prob. 11—4. We determine qf from (13—43). Finally, the force parameters for the thin-walled case are defined as
= MR = Verify that 1.
2. 3.
= MR when Open section Closed section Mixed section
=
dS Jqrc&
dS
Consider the following cases:
I and compare vs. Mu. Also evaluate L and compare with the unrestrained value. 13—5. Refer to Examples 12—2 and 13—2. Discuss how you would modify the member force-displacement relations developed in Example 12—2 to account for restrained torsion. Consider 1, X3r = 0, and—— (a) warping restrained at both ends (b) warping restrained only at x L 13—6. Refer to Example 13—2. Determine the translations of the shear center. Consider the cross section fixed at x 0. Discuss how the solution has to be modified when the cross section at x = L is restrained against translation. 13—7. Starting with the force-deformation relations based on the mixed formulation (13—49), derive the member force-displacement relations (see Example 12—2). Consider no warping at the end sections and take = + 1. Specialize for— (a) symmetrical cross section (b) no shear deformation due to restrained torsion and flexure—arbitrary cross section,
13—4.
Specialize (13—57) for .L >
at x
13—S.
Consider a thin-walled section comprising discrete elements of
material properties (F, G). Discuss how the displacement and mixed formulations haveto be modified to account for variable material properties. Note: The unrestrained torsion and flexural stress distributions are treated in Prob. 11—14 and 12—1.
Determine the distribution of qr, and expressions for Cr, for the cross sections shown in parts a and b and part e—d of the accompanying sketch (four different sets of data). and qr for the section shown. 13—10. Determine 13—11. Using the fiexural shear distributions listed in Prob. 13—1, show that 13—9.
-
'12 =
423
PROBLEMS
Prob. 13—9
I
Ii
T F— 0.75k (b)
I I
I
2k
/z
+
See part c.
(d)
'i—H
(c)
Prob. 13-10 t
0
0 ç1s2
t I
H
a
I
RESTRAINED TOIRSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13
424
Hint: One can write
22
13 •JJ
Also show that
(x2 V q52r +
113 —
13—12. Specialize Equations (13—84) and (13—85) for the case where the cross section is symmetrical with respect to the X2 axis. Utilize
x3)H0(x2, x3)dA =
0
where He is an even function and H,, an odd function of x3. Evaluate the coefficients for the channel section of Example 13—5. Finally, specialize the equations for a doubly symmetric section. 13—13. Specialize (13—88) for a doubly symmetrical cross Section. Then specialize further for negligible transverse shear deformation due to flexure and warping. The symmetry reductions are X2
=0
=
X2r
X3r = 0
i/A23=O 'li 0
= Consider the two following problems involving doubly symmetric cross section. (a) Establish "linearized" incremental equations by operating on (13—88) !72
13—14.
and retaining only linear terms in the displacement increments. (b)
Specialize for a doubly symmetric cross section (see Prob. 13—12). Consider the case where the cross section is doubly symmetric and the initial state is pure compression (F1 —P). Determine the critical
load with respect to torsional buckling for the following boundary conditions: 1.
co1 =
2.
==
f
0
at x =
=0
at x
0,
L
(restrained warping)
(unrestrained warping)
0, L
Neutral equilibrium (buckling) is defined as the existence of a nontrivial solution of the linearized incremental equations for the same external load. One sets
F1 = U2
—P
U3 = W1 = (02 = (03 = f
0
and determines the value of P for which a nontrivial solution which satisfies the boundary conditions is possible. Employ the notation introduced in Example 13—7.
Determine the form of V, the strain energy density function (strain energy per unit length along the centroidal axis), expressed in terms of displace13—15.
ments. Assume no initial strain but allow for geometric nonlinearity. Note that V = V* when there is no initial strain.
14
Planar Deformation of a Planar Member 14—i.
INTRODUCTION: GEOMETRICAL RELATIONS
A member is said to be planar if— 1.
The centroidal axis is a plane curve.
2.
The plane containing the centroidal axis also contains one of the
3.
principal inertia axes for the cross section. The shear center axis coincides with or is parallel to the axis. However, the present discussion will be limited to the case where the shear center axis lies in the plane containing the centroidal axis.
We consider the centroidal axis to he defined with respect to a global reference frame having directions X1 and K2. '[his is shown in Fig. 14—i. The orthogonal unit vectors defining the orientation of the local frame (Y1, Y2) at a point are x 12 = where points in the positive tangent direction and denoted by 13. Item 2 requires Y2 to be a principal inertia axis for the cross section. x2
Yl n r2
tl
S B
A
n i2
x1
ii
Fig. 14—i.
notation for planecurve. 425
426
PLANAR DEFORMATION OF A PLANAR MEMBER
CHAP. 14
By definition, t
=
=
dx1
+
dx2
Since we are taking t2 according to 11 x t2 = -
t2
dx2
+
=
(14-1)
13, it
follows that
dx5
(142)
The differentiation formulas for the unit vectors are dt1
1
(14-3)
where dt1
1
d2x2 dx1
d2x1 dx2
—
According to this definition, R is negative when d11/dS points in the negative t2 direction, e.g., for segment AB in Fig. 14—1. One could take t2 = ii, the unit normal vector defined by -
1
d11
(14-4) ciS
x 12 = 13 but this choice is inconvenient when there is a reversal in curvature. Also, this definition degenerates at an inflection point, i.e., when dt/dS = O. If the sense of the curvature is constant, one can always orient the X1-X2 frame so that coincides with ñ, to avoid working with a negative R. To complete the geometrical treatment, we consider the general parametric representation for the curve defining the centroidal axis,
rather than according to
x1 = x1(y) x2 = x2(y)
(j45
where y is a parameter. The differential arc length is related to dy by dS
+
d. -
2
2
+ (p)]
1/2
dv
=
dy
(14—6)
According to this definition, the +S sense coincides with the direction of t We summarize here for convenience the essential geometric relations for a plane curve which are developed in Chapter 4.
SEC. 14—2.
FORCE-EQUILIBRIUM EQUATIONS
and 1/R in terms of y are
increasing y. Using (14—6), the expressions for
-
t2 = — ( —-
R
1j
if
-
1
dx2
I 7dx1
t1 = — (
427
+— dy
dx2.. ——-—--
dy
+
dx1
dy
(14_
if._ (It1 = -( t2 dy — —
1
( k\
d2x1 dx2 dv2
dy
+
d2x2 dx1
dy2 dy
A planar member subjected to in-plane forces plane for our notation) will experience oniy in-plane deformation. In what follows, we develop the governing equations for planar deformation of an arbitrary planar member. This formulation is restricted to the linear geometric case. The two basic solution procedures, namely, the displacement and force methods, are described and applied to a circular member.
We also present a simplified formulation (Marguerre's equations) which is valid for a shallow member. Finally, we include a discussion of numerical integration techniques, since one must resort to numerical integration when the cross section is not constant. 14—2.
FORCE-EQUILIBRIUM EQUATIONS
The notation associated with a positive normal cross section, i.e., a cross
section whose outward normal points in the + S direction, is shown in Fig. 14—2.
We use the same notation as for the prismatic case, except that now the vector - I'3
dA
012
Centroidal axis
Fig. 14—2. Force and moment components acting on a positive cross section.
PLANAR DEFORMATION OF A PLANAR MEMBER
428
CHAP. 14
components are with respect to the local frame (Y1, Y2, Y3) rather than the
basic frame (X1, X2, X3). The cross-sectional properties are defined by
Since
if dy2 dy2
A
=
13
= JJ(y2)2 dii
=
if dii
(14—8)
'2 = .iJ(Y3)2
Y2, Y3 pass through the centroid and are principal directions, it follows
that
dA = flY3 dA =
dii =
SSY2Y3
0
(14—9)
When the member is planar (X1-X2 plane) and is subjected to a planar loading, F3
0
M2
M1
(14—10)
in this case, we work with reduced expressions for F÷ and M÷ (see Fig. 14—3) and drop the subscript on M3:
=
M+ =
M3t3
+ F212' = Mt3
(14-11)
Note that 13 is constant for a planar member. x2
) = t1 x t2
x1
Fig. 14—3. Force and moment components in planar behavior,
To establish the force-equilibrium equations, we consider the differential volume element shown in Fig. 14—4. We define b and as the statically equivalent external force and moment vectors per unit arc length acting at the centroid.
For equilibrium, the resultant force and moment vectors must vanish. These conditions lead to the following vector differential equilibrium equations: dS
—
dM÷
+
,_.
+ —
=
in + r1 x F+ =
o
-
0
(14—12)
SEC. 14—3.
We
expand b and
PRINCIPLE OF VIRTUAL FORCES
429
in terms of the unit vectors for the local frame:
b= + = mt3
b212
(14—13)
Introducing the component expansions in (14—12), and using the differentiation formulas for the unit vectors (14—3), lead to the following scalar differential equilibrium equations: dF1
—
F2
+ b1 = 0
(14-14)
dM
+
+m
0
that the force-equilibrium equations are coupled due to the curvature. The moment equilibrium equation has the same form as for the prismatic case. dS
r(S)
Fig. 14—4. Differential element for equilibrium analysis.
The positive sense of the end forces is shown in Fig. 14—5. We work with components referred to the local frame at each end. The end forces are related to the stress resultants and stress couples by
= = Mj52 = MA= —MISA
14—3.
(14-15)
j=1,2
FORCE-DISPLACEMENT RELATIONS; PRINCIPLE OF VIRTUAL FORCES
We establish the force-displacement relations by applying the principal of virtual forces to a differential element. The procedure is the same as for the
CHAP. 14
PLANAR DEFORMATION OF A PLANAR MEMBER
430
prismatic case described in Sec. 12—3, except that now we work with displacement components referred to the local frame at each point. We define ü and as
=
=
=
centroid. = equivalent rigid-body rotation vector
rigid-body translation vector at the (14—16)
For planar deformation, only u1, u2 and 0J3 are finite, and the terms involving u3, co1, and w2 can be deleted: u1t1 + U2T2 — C03t3
(14—17)
Wt3
The positive sense of the displacement components is shown in Fig. 14—6.
F41
Fig. 14—5. Convention for end forces.
x2
x1
FIg. 14—6. Definition of displacement measures. t "Equivalence" refers to work. See (12—8).
PRINCIPLE OF VIRTUAL FORCES
SEC. 14—3.
431
as the complementary energy per unit arc length. For planar
We define
deformation, = (F1, F2, M). One determines by taking expansions for the stresses in terms of F1, F2, M, substituting in the complementary energy density, and integrating with respect to the cross-sectional coordinates Y2, y3. We will discuss the determination of later.
Specializing the three-dimensional principle of virtual forces for the onedimensional elastic case, and writing
=
cF1
=
e1
AF1
+
0F2
AF2
AF1 + e2 AF2 +
k
+
cM
AM (14—18)
AM
lead to the one-dimensional form Ss(ei AF1
+ e2 AF2 +
k
AM)dS
AP1
(14—19)
where is a displacement measure and is the force measure corresponding to d1. The virtual-force system (AF1, AF2, AM, AP1) must be statically permis-
sible, i.e., it must satisfy the one-dimensional equilibrium equations.
(
Fig. 14—7. Virtual force system
We apply (14—19) to the differential element shown in Fig. 14—7. The virtual
force system must satisfy the force-equilibrium equations (14—17), dS
Evaluating
AF÷ =
0
(a)
AP1,
=
+AM÷
+ (b)
+ AF2
= {AF1 —
+
—
+
dS
432
PLANAR DEFORMATION OF A PLANAR MEMBER
CHAP. 14
and then substituting in (14-49) results in the following relations between the
force and displacement parameters: cj
eV*
du1
U2
du2
u1
(14-20)
dw dS
k
We interpret e1 as an average extension, e, as an average transverse shear deformation, and k as a bending deformation. Actually, k is the relative rotation of adjacent cross sections. In what follows, we discuss the determination of Consider the differential volume element shown in Fig. 14—8. The vector defining the arc QQ1 is QQ1
=
ar2
dy =
+
di2
+
dt-\
dv
Noting that —
dy (112
dv
—7k-ti
=
o
for a planar member, (a) can be written as dS2
By definition,
=
=
—
— = is the complementary energy per unit length along the
centroidal axis. Substituting for dS2 in the general definition, we obtain dS
dS2 dv2 dv Y2,Y3
(14—2 1)
if
.-
In general, V* = V" (ô11, We select suitable expansions for the stress components in terms of F1, F2, M, expand V*, and integrate over the cross section. The only restriction on the stress expansions is that they satisfy the definition equations for the stress resultants and couples identically:
= F2 $5c12 dA SSa13 dA = JJy3aii dA = 0 —ify2ci1 dA = M
dA
J$(y2a13 — y3a12)dA = 0
0
433
PRINCIPLE OF VURTUAL FORCES
SEC. 14—3.
The most convenient choice for iH is the linear expansion,t M
(14—22)
—
where I 13. A logical choice for (when the cross section is thin-walled) is the distribution predicted by the engineering theory of flexural shear stress distribution described in Sec. 11—7: a11 = 1q(F2)
q=
F2t/i
(14—23)
where t denotes the local thickness, and q is the flexural shear flow due to F2. Both expansions satisfy (a). x2
r +1)212 +Y33
r2
r1ty +dy)
Y2
it
Fig. 14—8. Differential volume element.
In what follows, we consider the material to be linearly elastic. The complementary energy density is given by 11*.....
—
0
2
a12
2
where c? is the initial extensional strain. Substituting (a) in (14—21) and taking
the stresses according to (14—22), (14—23) results in the following expression f This applies for a homogeneous beam. Composite beams are more conveniently treated with the approach described in the next section.
PLANAR DEFORMATION OF A PLANAR MEMBER
434
CHAP. 14
for V*:
= e?Fi + k°M +
+
+
+
2GA2*
(14-24)
where
=
(i
if
dA
-
55
I
—
\\
—
R}
1 and If the section is symmetrical with respect to the 1'3 axis, 1* The deformation-force relations correspoiiding to this choice for —ei e,
M
F1
F2
du2
=
=
+
-w
M
F1
A2.
u2
dr,1
U1
= are
(14-25)
dw
Note that the axial force and moment are coupled, due to the curvature. Inverting (14—25) leads to expressions for the forces in terms of the deformations:
F1 = M— —
LA
e1)
—
—
k
R(l
)
EI*
+
Ô)(el —
R(1
/
k°
(14-26)
F2
We observe that
I —
where p is the radius of gyration and d is the depth of the cross section, For example, 1
d2
AR2 = i2R2 for a rectangular cross section. Then, is of the order of (d/R2) and can be neglected when (dIR)2 1. A curved member is said to be thin when O(d/R) 1.
We set ö =
0
1
and thick when O(d/R)2
for a thick member. The thinness assumption is introduced
PRINCIPLE OF VIRTUAL DISPLACEMENTS
SEC. 14—4.
435
neglecting y2/R with respect to unity in the expression for the differential arc length, i.e., by taking by
dS
14 27
-
Assuming a curved member to be thin is equivalent to using the expression
for V* developed for a prismatic member. The approximate form of (14—25) for a thin member is F1
dii1
Li2
(14—28)
i—k° To complete the treatment of the linear elastic case, we list the expanded forms of the principle of virtual forces for thick and thin members. Note that these expressions are based on a linear variation in normal stress over the cross
section. Thick Member
Cit0
+
F1
+
M'\ AF1
+
F2
/XF2
(14—29)
+ (ko +
+
AM} dS —
Thin
J
14—4.
+
+
/ M'\ + (\kO + h-i)
1
dS =
(14—30) d1 AP1
FORCE-DISPLACEMENT RELATIONS—DISPLACEMENT EXPANSION APPROACH; PRINCIPLE OF VIRTUAL DISPLACEMENTS
In the variational procedure for establishing one-dimensional force-displacement relations, it is not necessary to analyze the deformation, i.e., to determine
the strains at a point. One has only to introduce suitable expansions for the stress components in terms of the one-dimensional force parameters. Ndw, we• can also establish force-displacement relations by starting with expansions for the displacement components in terms of one-dimensional displacement parameters and determining the corresponding strain distribution. We express the
PLANAR OEFORMA11ON OF A PLANAR MEMBER
436
CHAP. 14
stresses in
terms of the displacement parameters using the stress-strain relations, and then substitute the stress expansions in the definition equations for F1, F2, and M. The effect of transverse shear deformation is usually neglected in this
approach. To determine the strain distribution, we must first analyze the deformation at a point. This step is described in detail below. Figure 14—9 shows the initial position of two orthogonal line elements, QQ1 and QQ2, at a point (y, Y2' y3). The vectors defining these elements are QQ1
dy2
QQ2 =
a2 =
I
dy2 t2
(14—31)
a —
We use a prime superscript to denote qua Iltities associated with the deformed position of the member, which is shown in Fig. 14—10; for example:
?'=
= position vector to point P(y) in the deformed position (point P'). tangent vector to the deformed centroidal axis. = position vector to Q(y, Y2, y3) in the deformed position (point Q'). x2 Q2 Q1
Pj(y +dy) P(y) axis
x1
Fig. 14—9. Initial geometry for orthogonal curvihnear line elements.
SEC. 14—4.
PRINCIPLE OF VIRTUAL DISPLACEMENTS
437
From Fig. 14—10, and noting (14—3 1): —
=
P'P'1 =
/
0))
=
+
or2
/
&Y2
\.
C))J c,u2
—
dy
(14—32)
dy2
The analysis of strain consists of determining the extensions and change in
angle between the line elements. We denote the extensional strains by
(j =
1,
2) and the shearing strain by Y12 The general expressions are
'—12 3—
(1-3)
Sin Y12
Now, we restrict this discussion to small strain, Substituting for the deformed
vectors and neglecting strains with respect to unity, (14—33) expands to
Istj,
au2 c'y
+
2(a2)"
—.
ô,V
1 ('U2
2
"
tj -'
Y12
a2
t2
-±
cc2
The nonlinear terms arc associated with the rotation of the tangent vector. Neglecting these terms corresponds to neglecting the difference between the deformed and undeformed geometry, i.e.. to assuming linear geometry. The next step involves introducing an expansion for in terms of y2. We express ü2 as a linear function of ü
wv211
(14—35)
where co = w(y) and U
U1t1 + U2t2 = 1kv)
(14—36)
is the displacement vector for a point on the centroidal axis. Equation (14—35) implies that a normal cross section remains a plane after deformation. One can interpret co as the rotation of the cross section in the direction from toward t2. This notation is illustrated in Fig. 14—1 1. In what follows, we consider only linear geometry. Substituting for ü2, taking y = S, and evaluating the derivatives lead to the following strain expansions:
PLANAR DEFORMATION OF A PLANAR MEMBER
438
=
—
= du,
y2k)
u2 —
1
e2
=
CHAP. 14
+
= I61IY20 (14—37)
0)
doi
The vanishing of c2 is due to our choice for ü2. One could include an addiThis would give tional linear term, = $ and, additional terms in the x2 Q2
x1
Fig. 14—10. Deformed geometry for orthogonal curvilinear line elements. u2t2
(u1 —Wy2)tl
Centroidal axis
UI tl
Fig. 14—11. Displacement expansion.
PRINCIPLE OF VIRTUAL DISPLACEMENTS
SEC. 14—4.
expressions for
439
that the assumption that a normal cross
and Y12• Note
section remains plane does not lead to a linear variation in extensional strain over the depth when the member is curved. We introduce the assumption of negligible transverse deformation by setting e2 = 0. The resulting expressions for (0 and k in terms of u1 and u2 are e2
=
0
du2
u1
+
dS
R
(14—38)
dIui
—
dS
—
dS2
When transverse shear deformation is neglected, one must determine F2 using the moment-equilibrium equation. The next step involves expressing F1, F,, and M in terms of the one-dimensional deformation parameters e1e2 and k. In what follows, we consider the material to be linearly elastic and take the stress-strain relations for c12 as:
=
= Gy12
E(c1
Substituting for r1, Y12, using (14—37),
=
F 1 —y2/R
— y,k) —
———--—-(e1
Fe1
(14—39)
and then evaluating F1, F2, and M, we obtain
F2
d
= Ge2 if =
(14-40)
+ Ekjj
—Fe1
+
The various integrals can be expressed in terms of only one integral by using the identity 1
1 — y2/R
and noting that Y is a
—
f The relation for
1
axis: $5Y2 dA
member.
y2/R
1-F
is exact only when
=
0
= (733
11
We generally neglect
for a
440
PLANAR DEFORMATION OF A PLANAR MEMBER
CHAP. 14
One can easily show that
ri
I'
c/A
dA
=
L JJ
(14—41)
1
- y2/R
For completeness, we list the inverted form of (14—40),
=
+
M
+
F2
k
= k° +
F1
lvi
+
where
=
+
= A(1 4—42
e? =
if
k° =
(i
c/A —
if
-
dA
The expressions for e1 are identical with the result (see (14—25)) obtained with
the variational approach. However, the result for k differs in the coefficient for M. This difference (1' or F') is due to the nonlinear expansion used for Example 14—1
We determine I' for the rectangular cross section shown in Fig. F.14—1.
I' =
11 1
y2/R
=h
J—a;2 1 — y2/R
=—R2bd+R3bln To obtain a more tractable form, we expand the log terms, using
(1+x'\
I
PRINCIPLE OF VIRTUAL DISPLACEMENTS
SEC. 14—4.
441
This series converges for xI < 1. Then
=
In
d
d3
+
3(d\4
3(d\2
I
ii
+
+
+
and
I' =
+
{
3
d
2
3
d
+ ..
+
Fig. E14—1 H
I
Y3
The relations listed above involve exactintegrals. Now, when the member is thick, we neglect (y2/R)2 with respect to unity. This assumption is introduced by taking
1 —y2/R
=
1
+
+ ...
+
+
and I':
in the expansions for
Co
—.--..e2
Y2
- yJR
=i{i
JJ
dA
+
442
PLANAR DEFORMATiON OF A PLANAR MEMBER
CHAP. 14
To be consistent, we must also neglect 1'/AR2 with respect to unity in the expression for A'2 and I". When the member is thin, we neglect y2/R with respect to unity.
1—y2/R —
y2k
(14—44)
—
at2
It is of interest to establish the one-dimensional form of the principle of virtual displacements corresponding to the linear displacement expansion used in this development. The general three-dimensional form for an orthogonal coordinate system is (see Sec. 10—6):
SJJ(aii
+
)d(vol.) =
+ a12 öy12
represents an external force quantity and d1 is the displacement quantity We consider only and Viz to be finitc, and express the corresponding to differential volume in terms of the cross-sectional coordinates Y2' Y3 and arc length along the ccntroidal axes (see Fig. 14—9): where
d(vol.) =
dS2
(i
dy2 dv3
dS
d7 dy3
—
Then (a) reduces to (a11
(i
+ a12
dA] dS =
(14—45)
—
We take (14—45) as the form of the principle of virtual displacements for planar
deformation.
The strains corresponding to a linear expansion for displacements and linear geometry are defined by (14—37), which are listed below for convenience: ci
—
Y12
—
y2/R
du1
= k
du2
e2
U2
+
u1
— U)
do dS
Substituting for e1, Y12 and using the definition equations for F1, F2, and M,
SEC. 14—4.
we
PRINCIPLE OF VIRTUAL DISPLACEMENTS
443
obtain
Ad1 (14—46) + F2 + M ök]dS = Js[Fi This result depends only on the strain expansions, i.e., (c). One can apply it for the geometrically nonlinear case, provided that (cS) are taken as defining
the strain distribution over the cross section. We use the principle of virtual displacements to establish consistent forceequilibrium equations. One starts with one-dimensional deformation-displacement relations, substitutes in (14—46), and integrates the left-hand side by parts. Equating coefficients of the displacement parameters leads to a set of force equilibrium equations and boundary conditions that are consistent with the geometrical assumptions introduced in establishing the deformation-displacement relations. The following example illustrates this application.
Example
14—2
The assumption of negligible transverse shear deformation is introduced by setting e2
equal to zero. This leads to an expression for the rotation. w, in terms of the translation components,
=
(In2
+
and the relations for negligible transverse shear deformation reduce to
hEFt ôe1 + M ök]dS
d (du2
1< =
u1
=
+
Substituting for Aw and the strain variations, d
= Au1 —id
=
(5k
Aui
d2
Au2 1
AU2
—
+
Au2
d
I
and integrating by parts, the left- and right-hand sides of (b) expand to
j 54
[F1
+ M (5k]dS
=
/ F1 /
—
I
\
+
F1 + I
+j
Rj M\
Rj r
J
Au1
dM —
dS
dM
Au2
Au1 — —-- Au2
1Aui
[—
dS
dF1
1
—
+M
d uS
An2
d
+M—
dM1
dS
+ Au2
[ [—
F1
dS
+
PLANAR DEFORMATION OF A PLANAR MEMBER
444
CHAP. 14
and
+
(b1 +
Ad,
=
dS
+ (p22 + ma) Au32 +
+ (PB! +
+ (p41 +
(b2
MA)
+ (r42
A
A
014) AUA2 + M4 A
The consistent equilibrium equations and boundary conditions for negligible transverse shear deformation follow by equating corresponding coefficients of the displacement variations in (e) and (f):
S4<S<S3 +
dF1
+
d2M
F1
+-
1dM
—
+ b1 +
01
0
drn
+
=0
—
s—sn u1
prescribed or
u2
prescribed or
do2
F1
—
p42
prescribed or
M=
—MA
UI
prescribed or
F1 =
F21
U2
prescribed or
=
— m
S=
du2
prescribed or
—F32
—
in
M
One can obtain (g) by solving the last equation in (14—14) for F2 and substituting in the first two equations. Suppose we neglect u1/R in the expression for w: do2 CD
d2u2
k
This assumptiont is generally referred to as Mush tori's approximation. The equilibrium
equations for the tangential direction reduce to dF1
t See Ref. 5.
CARTESIAN FORMULATION
SEC. 14—5.
445
The other equilibrium equation and the boundary conditions are not changed. Using (h) instead of (a) eliminates the shear term, F2/R, in the tangential force-equilibrium
equation.
14—5.
CARTESIAN FORMULATION
We consider the case where the equation defining the centroidal axis has the form x2 = f(x1). The geometrical relations for this parametric representation are obtained by taking y x1 in (14—7). They are summarized belowt for convenience and the notation is shown in Fig. 14—12:
dS = dx1
r [
7df\21112
ylx1jj. f'df'\ + I—)
= i[ I
-
I
t2
[ / df \
1
cos0 1 j
-
-
(14—47)
+ '2
= t1 X t2 = 13 d2f
I
ci:
In the previous formulation, we worked with displacement components and external force components referred to the local frame. An alternate approach, originally suggested by involves working with components referred to the basic frame rather than the local frame. The resulting expressions differ, and it is therefore of interest to describe this approach in detail. We start with the determination of the force-equilibrium equations. Consider the differential clement shown in Fig. 14—13. The vector, equilibrium
equations are dF+ dx1 See Prob. 14—1. See Ref. 6.
+
x dx1
-
-
ax1
(14—48)
+
=
0
PLANAR DEFORMATION OF A PLANAR MEMBER
446
CHAP. 14
x2 Y2
YI
X2
dx1
x1 'I
Fig. 14—12. Notation for Cartesian formulation.
dx1
2
pN2 12 1
F1t1
lj Fig. 14—13. Differential element for equilibrium analysis.
CARTESIAN FORMULATION
SEC. 14—5.
447
h are the external applied force and moment vectors per unit projected length, i.e., per unit x1. They are related to b and (see Fig. 14—4) by
where fl,
dx1 = b dS = (cth)dx1 hdx1 = iñdS = (cthi)dx1
(1449)
Substituting for the force and moment vectors,
= F1t1 + F2t2 = N171 + N212 it—hi3 + P = N1 = F1 cos 0 — F2 sin 0 N2 = F1 sin 0 + F2 cos 0
(14—50)
the equilibrium equations expand to dN1 —dx1
dx1
d = ——(F1 cosP — dx1
=
(F1
dx1
—1(dM
'\
——-(—-- + hJ=' \dx1 ,i We
sin 0) =
sin0 + F2 cos 0)
F2
(14-51)
P2
—N1 sm0 + N2cosO
restrict this treatment to an elastic material and establish the force-
displacement relations, using the principle of virtual forces,
dx1 =
[e1
AF1 + e2 AF2 + k
dx1 =
d1
(a)
where V* V* (F1, F2 M) is the complementary energy per unit arc length. Consider the differential element shown in Fig. 14—14. The virtual-force system is statically permissible, i.e., it satisfies the force-equilibrium equations identically:
=
dx1
+ J1 x
dx1
Expanding
o
=
d1
x
and then substituting for the displacement and rotation vectors, O= (0
v111
+ 1)212
(013 = (0t3
(14—52)
PLANAR DEFORMATION OF A PLANAR MEMBER
448
we
CHAP. 14
obtain
=
+
+
(d)
dx1; Finally, substituting for N1, N2, in terms of F1, F2 and equating coefficients of the force increments result in dx1
oV
dx1
do1
do2
cos2 0-— + sinOcosO—dx1
.
e2
a v*
=
—sin 0 COS
do1
,
U
--
do2
+cosO——w dx1
(14—53)
dw
k=
= ——— cos 0
dx1
The member is said to be shallow when 02 << 1. One introduces this assumption by setting 4f cos U
Sm 0
1
tan 0 =
(14—54)
in (14—50), which relate the cartesian and local forces.
1)2
2 V1
Fig. 14—14. Virtual force system.
Marguerre's equations are obtained by assuming the member is shallow and, in addition, neglecting the contribution of F2 in the expression for N1.
DISPLACEMENT METHOD OF SOLUTION
SEC. 14—6.
449
Marguerre starts with
N,
F1
N2
F2 +
and the resulting equations are
dF, —— + Pi = dx, dF2
dx,
+
d 7
(F1
dx, dM
——
F2
dx1
=
dx,j
J
+ P2 =
0
— in (14—55)
dv'
e, = e2
df\
0
d.x,
df dv2 + —— dx, dx,
dv2
0F2
= —- — co
dx, dw
k
OM
dx,
One step remains, namely, to establish the boundary conditions. The general conditions are v1 or N, v2 or N2 prescribed at each end (14—56) M or w We obtain the appropriate boundary conditions for the various cases considered above by substituting for N,, N2 and ox For example, the boundary conditions for the Marguerre formulation are
w
14—6.
The
or
F2 +
or
M
F, prescribed at each end
(14-57)
DISPLACEMENT METHOD OF SOLUTION—CIRCULAR MEMBER
displacement method involves solving the system of governing dif-
ferential equations which, for the planar case, consist of three force-equilibrium
equations and three force-displacement equations. If the applied loads are independent of the displacements, we can first solve the force equilibrium equations and then integrate the force-displacement relations. This method is quite straightforward for the prismatic case since stretching and flexure are uncoupled. However, it is usually quite difficult to apply when the member is
PLANAR DEFORMATION OF A PLANAR MEMBER
450
CHAP. 14
curved (except when it is circular) or the cross section varies. In what follows,
we illustrate the application of the displacement method to a circular member having a constant cross section, starting with— 1.
2.
the exact equations (based on stress expansions) for a thick member Marguerre's equations for a thin member
The results obtained for this simple geometry provide us with some insight as to the relative importance of transverse shear deformation and stretching deformation versus bending deformation. When the centroidal axis is a circular segment, R = const, and the equations simplify somewhat. It is convenient to take the polar angle 8 as the independent
variable in this case. We list the governing equations below for convenience and summarize the notation in Fig. 14—15:
/
dM
dF1
—
=
RF1
R2b2
m
(14—58)
—
1dM F1
= ej + Eq. (14—25)
M
+
= k°
+
1 f'du1 —
+ ui)
e2
k=
=
F1
M
+
=
—
0)
(14—59)
Idw
Solution of the Force-Equilibrium Equations
We consider the external forces to be independent of the displacements. Integrating the first equilibrium equation, we have
RF1 = —M
—
R2
j
(b1
+
+
where C1 is an integration constant. Substituting for F1 in the second equation
results in a second-order differential equation for M:
+ M
C1 + R2[b2
—
+
The general solution of (b) is
M=
C1
+ C2cosO + C3sinO +
(14—60)
DISPLACEMENT METHOD OF SOLUTION
SEC. 14—6.
451
denotes the particular solution due to the external distributed loading and C2, C3 are constants. Once M is known, we find F1 using (a) and F2 from the moment equilibrium equation. The resulting expressions are
where
F1 =
j(b1
cosO + C3 sinO + Mr)— R
±
;)dU (14—61)
sine + c3 cos 9 +
F2 =
dS
— in
RdO
F
F1
Fig. 14—15. Notation for circular member.
Integration of the Force-Displacement Relations
We start with (14—59) written in a slightly rearrangcd form: du1
du2
— u2
+ u1
RF2
+ Rco
= Rk° + To determine u1 and u2,
+ RF1)
Re? +
[M +
(RF1)]
transform the first two equations to
= u2 + Re? +
(M + RF1)
(14—62)
PLANAR DEFORMATION OF A PLANAR MEMBER
452
CHAP. 14
and d2u2
+
=
1/1
+
=
Re?
± R2k°
+ RF1)
—
+ a1R2 —
(14—63)
Re?
I a1 =
—
We have previously shownt that
is of the order of (d/R)2. It is reasonable
to neglect with respect to 1 but we will retain it in order to keep track of the influence of extensional deformation. We solve (14—63) for u2, determine u1 from (14—62), and w from the second equation in (a), F2
1
W
(du,
'\
(14-64)
This leads to three additional integration constants. The six constants are determined by enforcing the three boundary conditions at each end. Various loading conditions are treated in the following examples.
Example
14—3
Consider a member (Fig. El4—3) fixed at the negative end (A) and subjected only
at the right end (B). The boundary conditions for this case are
to
F2=M=O u1 =u2=w=O
F1=Fa1;
atO=011
atO=O
Specializing the force solution for no external distributed loading and enforcing the boundary conditions at B, we obtain
F1 = F2 =
FRI Sifl(08
M
RF51(l — cos(08
F81
cos(08 — 0) 8) —
0))
To simplify the analysis, we suppose there is no initial deformation. Using (b). the form — F81
R'
'I'
where
EI*
=
f See Sec. 14—3, Eq. (14—26).
[a1
— 02
COS(08
(d\'
0)1
takes
Note that
453
DISPLACEMENT METHOD OF SOLUTION
SEC. 14—6.
is associated with transverse shear deformation. Substituting for tJ' in (14—63)
and integrating, we obtain
c4
cos
0 + c5
[ai +
sin o +
0
—
0)]
The solution for u1 follows from (14—62):
=
sin 0 — C5 cos 0 + C6
{o +
+
[o
—
0)
+ sin(66 — 0)]}
Next, we determine w using (14—64), U) =
C6
+
{O
+
sm(OB — O)}
Finally, the constants are found by enforcing the displacement boundary conditions at 0 = 0:
F R3
C4 = C5 =
(05 —
sin
C6 = To determine the relative importance of stretching and shear deformation versus bending deformation, we evaluate the displacements at 0 and write the resulting expressions Fig. E14—3 A
FBI
Constant cross section
PLANAR DEFORMATION OF A PLANAR MEMBER
454
CHAP. 14
in the following form:
WE = UBI
=
—
+ b1 Oe)
sin
—2 sin 63 + 4sin 03 COS 4Sin
— COS 0B
± b2
+ b3 &)
—
sin
(I)
— Slfl 63 b2
—
—4 03
+ 2 sin 03 — 4 sin 03 COS 013
63 +
sin 03 Co
4(03 — sin_03 cos 03)
b
T 4sin2 63
b—
-
—
I + COS 63
The coefficients (b1,..., b4) are of order unity or less when is not small with respect to unity, i.e., when the segment is not shallow. Also, öe and ó, are of order (d/R)2. It follows that the displacements due to stretching and shear deformation are of order (dIR)2 times the displacement due to bending deformation for a nonshallow member. To investigate the shallow case, we replace the trigometric terms in (i) by their Taylor series expansions,
/
sin 0 =
02
+
—
—
02
cos 6 = I — sin0cos6
0(1
+
—
+
—
—
and neglect
with respect to unity. The resulting expressions are P8152 1 1
U3j
P31S3 fOfl 032
=
+
Now, 1*
El"
(d'12
=
EI*
[1
1*
SEC. 14—6.
DISPLACEMENT METHOD OF SOLUTION
455
For example,
1 (d\2 AS2 EI*
E (d\2
=
lOG
=
(d'\2
0,26
for a rectangular section and v = 0.3. Since (cl/S)2 << I for a member, we can neglect the transverse shcar terms in UBL, UB2 and the stretching term in co8. However, we must retain the stretching term in u51 since it is of the same order as the bending term. The appropriate expression for is PatS3 In sum, we have shown that the percentage of error due to neglecting stretching and transverse shear deformation is of the order of (d/R)2 for a nonshallow circular member. If the member is shallow < I 5°), we catnwt neglect stretching deformation. Actually, the stretching term dominates when the member is quite shallow. The error due to neglecting transverse shear deformation for the shallow case is still only of the order of (d/R)2.
Example
14—4
The internal force distributions due to
acting on the cantilever member shown
in Fig. E14—4 are given by F1
sin(011 —
F2 = F52
cos(05
M
R sin(02
0)
0) 0)
We suppose the member is not shallow and neglect stretching and shear deformation. The force-displacement relations reduce to (we set A = = in (14—59)) du1
= Re?
—
dO du2
+
u1
dw dO
= Rw RM
= Rk° + EJ*
Eliminating u1 from the first two equations, we obtain 12u
+ u2 = du1
=
R2k° u2 I
—
Re?
+
R2
M
+ Re? (du2
We determine u2, then u1, and finally oi. Note that (c) corresponds to (14—62), (14—63) and (14—64) with A = A2 = cc. The final expressions (for no initial deformation or support
PLANAR DEFORMATION OF A PLANAR MEMBER
456
CHAP. 14
movement) are
= {(O COS(OB — 0) — sin
02
= 0) =
Example
0 cos
cos
F52R {
0
0 Sjfl(OB — 0) + COS(OB
0)}
COS(OB — 0) — cos
14—5
We analyze the shallow parabolic member shown in Fig. E14—5 using Marguerre's equations. We consider the member to be thin and neglect transverse shear deformation. Taking f = and = rn = 0, the governing equations (see (14—55) and (14—57)) reduce to
dx1
d2M
al'1 — P2 = 0
dx1
dM
F2
=
e?
F1
+
AL
dv1
= -— + ax1 dx1
dx1
dv2 Cl)
dx1
k= v1,
k
M
+
=
d2v2
v2, w prescribed at x1
N2 =
—
dx1
0
+ ax1F1 = 0 at x1 = I.
M =0 Integrating (a) and using the boundary conditions at x1 = L, we obtain
M= F2 = We suppose e? = k° = relation,
0
—
p2(L
—
xi)2 x1)
—
—
— x?)
ax1N51
to simplify the discussion. Integrating the moment-curvature
=M =
x1)2
CINBI(L2
4)
SEC. 14—6.
DISPLACEMENT METHOD OF SOLUTION
457
Fig. E14—4
B
l.A const
Fig. E14—5
Al
—,.- F1 a= L2
(h/L)2<(1
t P2 =
COflSt
B NB!
j
PLANAR DEFORMATfON OF A PLANAR MEMBER
458
and noting that v2
dv2/dxj =
0
Ely2 =
at x1 =
lead to the solution for v2,
0
+
—
CHAP. 14
—
—
The axial displacement is determined by integrating the extensional strain displacement
relation, dv1
=
-
-
F1
=
dv2 —
+
÷
+
-
We express the last term in (g) as NB1X1
AE
1
+
6 ) 1) [VT)
Now, a
2h,/L2
Then
a2L4(A'\
—
2(h'\2
6
and
we see that this term dominatcs when h is larger with respect to the cross-sectional
depth.
14—7.
FORCE METHOD OF SOLUTION
Our starting point is the principle of virtual forces restricted to planar deformation, Js(ei AF1 + e2 AF2 + k AM)dS — AR1 = d1 (14—65)
represents a support movement, and AR1 is the corresponding reaction increment. The relations where the virtual-force system is statically permissible,
between the deformation measures (e1, e2, k) and the internal forces (F1, F2, M)
depend on the material properties and on whether one employs stress or displacement expansions. This discussion is limited to a linearly elastic material but one should note that (14--65) is valid for arbitrary material. For convenience, we list the force-deformation relations below. The notation for internal force quantities is shown in Fig. 14—3.
Arbitrary Linearly Elastic Menther — e1
e2 =
k=
+
F1
+
M
F2
+
(14—66)
F1
AER.
M
+ -= El
FORCE METHOD OF SOLUTION
SEC. 14—7.
459
k°, A2, and] are defined by(14—24) for the stress-expansion approach and (14—42) for the displacement-expansion approach. Thin Linearly Elastic Member
e1 =
e?
+
F1
(14-67)
where A2, a?, k° are the same as for a prismatic member. When the member is nor shallow, it is reasonable to neglect stretching and transverse shear deformation. As shown in ExampLe 14—3, this approximation
introduces a percentage error of O(cl/R)2. Formally, one sets A = A2 = If the member is shallow, we can still neglect transverse shear deformation but we must include stretching deformation. The basic steps involved in applying the force method to a curved member arc the same as for the prismatic case. However, the algebra is usually more complicated, due to the geometry. We will discuss first the determination of the displacement at a point. To determine the displacement at Q in the direction defined by tQ, we apply an external virtual force, APQ7Q, generate a statically determinate system of internal forces and reactions corresponding to = FJ,QAPQ AM = M AP0 ARk =
Rk.
Q
(j =
1,2)
(14-68)
APQ
and substitute in (14—65): dQ = SS(eIFI,Q + e2F2Q + kM,Q)dS —
(14—69)
This expression is valid for an arbitrary material. We set e2 = 0 if transverse a? if stretching deformation is
shear deformation is negligible and a1 = negligible.
Example 14—6 We consider the thin linearly elastic circular segment shown in Fig. E14—6A. We suppose the member is not shallow and neglect stretching and franslerse shear deformation. The reactions are the end forces at A for this example, and (14—69) expands to
=
+ (k0 +
+ OA1FAIQ + UA2FA2.Q + WAMA,Q
In what follows, we illustrate tile application of (a)
(a)
PLANAR DEFORMATION OF A PLANAR MEMBER
460
CHAP. 14 Fig. E14—6A
U2L
Displacements at B
Expressions
= AF51. The internal virtual-force system corresponds 0 as tlie independent variable =
To determine u81, we take
to F81 = ±1. It is convenient to work with
rather than 0. The force-influence coefficients (F10, F2,Q, M.Q) follow directly from Fig. E14—6B:
F10 =
=
COSq
F20 = =
(b)
cos
R(1
Substituting (b) in (a) results in the following general expression for um
=
{e? cos
R
+ R (ko +
(1 — cos
Jo3
+
cos
sin
+
— cos
Once the loading is specified, we can evaluate the integral. Terms involving the support displacements define the rigid body displacement at B. leads to expression for u82 and Taking = SF82, We list them below for future reference: "
u82 = R
j
Jo COB =
W.4 + R
f3
(d)
Elj
(. (ko +
Solution for a Concentrated Loading at an Arbitrary Interior Point We consider an arbitrary force vector, in Fig. E14—6C.
and moment, Mc, applied at point C as shown
FORCE METHOD OF SOLUTION
SEC. 14—7.
461
Fig. E14—6B
ill Ffi2 ,t482
RI
— eQS
Fig. El 4—SC
A.
_______________
PLANAR DEFORMATION OF A PLANAR MEMBER
462
CHAP. 14
+
= =
The expressions for the displacements at B due to an external loading are obtained by specializing (c) and (d) for no initial deformation or support movement and noting that
M=O M=
RPc1[1
—
iic)] +
sin(1i —
ilc) +
ivIc
The solution for constant I is
Pc1R3(
.
+
+ =
PC2R( MCR2
.
.
+ Sill 1k +
Sifl 0c — sin
=
+
— cos
(Oc + Sin
PciR3I
Or
1
cos 1k —
.
sin
Oc sin
—
1
—
PC2R3 /1
+ —-—- (ens 1k —
—
sin
Sill 0c cos
sin
Sin
OR + cos
+ —--h-—
I
.
1k +
1
sm Oc sin
sin 9c cos
cos
Sin Oc) +
R2Pc2
cos Oc) +
If we take point C to coincide with B, = 0 and = OR. The resulting equations relate the displacement at B due to forces applied at B in the directions of the local frame at B and can be interpreted as member force-deformation relations. It is convenient to express these relations in matrix form:
— 2 sin
= R2[l — cos
+ 4sin 08 cos 08]
=
We call
Symmetrical
OR
—
sin OR]
F81
OR] —
sin °B cos OR]
R[l —
cos 98]
F82
the member flexibility matrix.
We describe next the application of the principle of virtual forces in the analysis of a statically indeterminate planar member. Let the member be indeterminate to the rth degree and let Z1, . , Z, represent the force redundants. Using the equilibrium equations, we express the internal forces and reactions in .
FORCE METHOD OF SOLUTION
SEC. 14—7.
terms
463
of the applied loads and the force redundants:
F1 = F10 + k-=
F2 =
1
F2,0 + (14—70)
M = M0 + k= 1
R1 = R10 +
k1
R1.kZk
(which is statically Substituting the virtual force system corresponding to permissible and in (14—65) and letting j range from 1 to r lead to the compatibility equations relating the actual deformations:
+
=
+ kM,1)dS
0
(a)
l,...,r
1=
When the material is linearly elastic, the compatibility equations take the form
(j =
k1
...,r)
(14—71)
where
=fkj =
f
+
+
+
=
—
+
Wesetl = I,A2 =
+ Fl.kM,f)
+
+
+ Fl,OM.J) +
±
Oforathinmember.
,42,and 1/AR
Note that fik is the displacement of the primary structure in the direction of is the actual displacement of the point of Z1 due to a unit value of Zk. Also, minUs the displacement of the primary structure in the direcapplication of tion of Z, due to support movement, initial deformation, and the prescribed external forces.
Example
14—7
Consider the symmetrical closed ring shown in Fig. E14—7. From symmetry,
at6r=0 F2 =
0
j
(a)
464
PLANAR DEFORMATION OF A PLANAR MEMBER
CHAP. 14
We take
the moment at 0 = 0 as the force redundant. To simplify the algebra, we suppose the member is thin and neglect stretching and shear deformation. The compatibility equations reduces to f11Z1 =
fii = &= due to a unit value of Z1 and is the relative rotation is the relative Note that rotation (X) due to the applied load. Equation (h) states that the net relative rotation must vanish. E14—7
fm/F
M
F1
1'
Now,
M
=
R(t — cos 0)
We consider I to be constant. Then, (b) reduces to
1
dS
JM21 cIS
=
PR1
TI\1
2\ ——)
—
(1 —
—J
cos 8)dO
/PR\
FORCE METHOD OF SOLUTION
SEC. 14—7.
465
Because of symmetry, we need to integrate over only a quarter of the ring. Finally, the total moment is
The axial and shear force variations are given by F1
F2 =
When the equation defining the centroidal axis is expressed in the form x2 = f(x1), it is more convenient to work with force and displacement tities referred to the basic frame rather than to the local frame, i.e., to use the cartesian formulation developed in Sec. (14—5). The cartesian notation is summarized in Fig. 14—16.
xl N2, F2
——--'-
i2
it
Ag. 14—16. Notation for Cartesian formulation.
The geometrical quantities and relations between the internal force components are
tan 6 = dS =
F1 = F2 =
dx1
cos 6
cos 6 + N2 sin U —N1 sin (9 + N2 cos 0
N1
(14—72)
PLANAR DEFORMATION OF A PLANAR MEMBER
CHAP. 14
first find N1, N2 and then determine F1, F2. To obtain the equations for the Cartesian case, we just have to replace dS by dx1/cos U in the general expressions ((14—69) and (14—71)). In what follows, we suppose the member is thin and linearly elastic. When the member is not shallow, we can neglect the stretching and transverse shear deformation terms. The equations for this case reduce to: We
Displacement at Point Q dQ
=
+ (ke +
L
(14-73)
C'oinpatibility Equations
\dx1
= j =
(14—74) —
5[e?Fi,i
the terms involving F1
We can
Plo =
+ (ko +
in terms of N1
cos
and N2
since
0 N10j + sin 0
Then, + ,f'N2, (
j'
(14—75)
One must generally resort to numerical integration in order to evaluate the integrals, due to the presence of the term 1/cos 9. When the member is shallow, 02 1, and we can approximate (14—72) with cos 0 sm 0 cls
1
tan 0 = dx1
(14—76)
F1
N1 +f'N2
F2
—f'N1 + N2
We cannot neglect the stretching deformation term in this case. However, it is reasonable to take F1 N1. We also introduced this assumption in the development of Marguerre's equations. The equations for the shallow case with negligible transverse shear deformation and F1 N1 have the forms listed below:
Displacement at Point Q dQ
=J
[(ei° +
N1,
+ (ko +
M,
dx1 —
R1,
(14—77)
FORCE METHOD OF SOLUTION
SEC. 14—7.
467
Compatibility Equation
>j1k4 = fjk
=
+
dx1
(14—78)
Jx,
= —
f
+
+ (k0 +
Example 14—8
Consider the two-hinged arch shown in Fig. E14—8A. We work with reaction components referred to the basic frame and take the horizontal reaction at B as the force redundant. Fig. E14—8A
x2,
P
Primary We must carry out two force analyses on the primary structure (Fig. E14—SB), one for the external forces (condition Z1 = 0) and the other for Z1 1. The results are displayed in Figs. E14—8C and D, respectively. Fig. E14—8B V2
= Z1
zI
R242
CHAP. 14
PLANAR DEFORMATION OF A PLANAR MEMBER
468
Fig. E14—8C N2
KM 0
Fig. E14—8D
H
(+)
'VI,'
ii
(+) N2,,
M,1
Compatibility Equation We suppose the member is not shallow. The compatibility equations for Z1 follow from (14—74):
I J11—I El
dx t cosO
+ f'N2,1) +
= Jo L
(ko
+
FORCE METHOD OF SOLUTION
SEC. 14—7.
469
Using the results listed above, the various terms in (a) expand to
f
— h
= .ft0E1
±
=
s:
+ f'N2
+ k°
cost?
L
+
dx1
JL[(
JLIJLI(h)(h) +
+
+
k0(f_
!L
LIcosO
L
(+P(xi
- a))dxi
Once the integrals are evaluated, we can determine Z1 from (c)
Finally, the total forces are obtained by superposition of the two loadings:
M = M•0 + Z1M1 R. = + Z1R,
=
(d) 1,
2, 3
R4 =
To evaluate the vertical displacement at point Q, we apply a unit vertical load at Q on the prinwry structure and determine the required internal forces and reactions plotted in Fig. E14—8E.
Fig. F—
XQ
1/L
Q
(i_IL) FQ=+l
PLANAR DEFORMATION OF A PLANAR MEMBER
470
CHAP. 14
Applying (14—73), we obtain VQ2
=
+
VA2
—
(13B2
VA2)
+
—
('XQ
—
I'
M\
CL
dx1
x1Ik°+—1—---—---—x1
I
\ 1L
El)cosO XQ1 M'\ dx1 ( +— x1(k°+-—J-—— EljcosO L jof \ Jo
/
dx1
(k°+—'l—-— El) cosO
I
A numerical procedure for evaluating these integrals is described in the next section.
Example
14—9
The symmetrical nonshallow two-hinged parabolic arch shown in Fig. E14—9A is subjected to a uniform load per unit horizontal length, that is, per unit x1. The equation for the centroidal axis is
4h( where h is the elevation at mid-span
= L/2). We take the horizontal reaction at the
right end as the force redundant and consider only bending deformation. Figures E14—9B and C carry through an analysis parallel to that of the preceding example,
Deternzination of Z1 and Total Forces The equation for Z1 follows from (14—74): 1L
Jo
EJcosO
pL2
ElcosO
Note that this result is valid for an arbitrary variation of El. Finally, the total forces are N1 — N10 + Z1N1 N2
=
= N20 + Z1N21 = —
M=M,0+Z1M,1 =0 Since M =
0, the deformed shape of the arch coincides with the initial shape when axial deformation is neglected. It follows that (c) also apply for the fixed nonshollolv case.
When the arch is shallow, the effect of axial deformation cannot be neglected. The expression for Z1 follows from (14—78): —
+
FORCE METHOD OF SOLUTION
SEC. 14—7.
471
Fig. E14—9A
p = Coflst
B
xi
Pnmary structure
R2,d2
Fig. E14—9B
N2
xI L
N5,00
M0
pL(
Force System Due to p
Xj\ )
pL
PLANAR DEFORMATION OF A PLANAR MEMBER
472
CHAP. 14
If E is constant, (d) reduces to JL dx1
pL2
1L1
8h
—
+
—
pL2
—
8h
1
I
+
jL1
The parameter ö is a measure of the influence of axial deformation. As an illustration, we
consider A and I to be constant and evaluate ö for this geometry. The result is
I
15(p'\2
— 8 Ah2
—
8
where p is the radius of gyration for the cross section. Fig.
A'2
N11 = +1
M,1 = +f
N2,1 = 0
Force System Due to Z1 = + I One should note that (e) applies only for the shallow case, in, for (f')2 K< 1. Now, 4/1
/
2x1
For the assumption of shallowness to be valid, 16(h/L)2 must he small with respect to unity. The total forces for the shallow case are
=
N1 =
M=
p1?
f(
pL2
ô
1
\=
PL( —
SEC. 14—8.
NUMERICAL INTEGRATION PROCEDURES
473
It is of interest to determine the rotation at B. The "Q" loading consists of a unit moment applied at B to the primary structure (see Fig. E14—9D). Applying (14—77) (note that Fig. E14—9D
PQ
+l
the stretching terms vanish since Nj,Q = 1L
M
0),
we obtain
P/
x1
j
\
/
,
When El is constant, (i) reduces to
pL3(
(5
2
Since
14—8.
0, the results for the fixed end shallow case will differ slightly from (h).
NUMERICAL INTEGRATION PROCEDURES
One of the steps in the force method involves evaluating certain integrals which depend on the member geometry and the cross-sectional properties. Closed-form solutions can be obtained for only simple geometries, and one usually must resort to a numerical integration procedure. In what follows, we describe two proceduresi which can be conveniently automated and illustrate their application in deflection computations.
We consider the problem of evaluating
J t See Ref. 8 for a more detailed treatment of numerical integration schemes.
(14—79)
__________________ PLANAR DEFORMATION OF A PLANAR MEMBER
474
where f(x) is a reasonably smooth function in the interval XA divide the total interval into n equal segments, of length h:
CHAP. 14
x
— XA
h=
We
(14—80)
If f(x) is discontinuous, we work with subintervals and use a different spacing x,, represent the for each subinterval. For convenience, we let x0, x1, the coordinates of the equally spaced points on the x axis, and f0, corresponding values of the function. This notation is shown in Figure 14—17. .
12
,
1;,
a
11
X2
X1
XO
.
B X,,
Fig. 14—17. Coordinate discretization for numerical integration.
The simplest approach consists in approximating the actual curve by a set
of straight lines connecting approximation,
etc., as shown in Fig. 14—18. With this ('xk
EXJk1,k =
f(x)dx dx =
=
h
+
(14—81)
+
If only the total integral is desired, we use,
J
f(x)dx
+ L)
=
(14—82)
+
which is called the trapezoidal rule. A more accurate formula is obtained by approximating the curve connecting
three consecutive points with a second-degree polynomial, as shown in Fig. 14—19. This leads to AJk,k+2
=
fdx
h
+ 4fk+1 +
(14—83)
J
Jk÷2= Jk + To apply (14—83), we must take an even number of segments, that is, n must
be an even integer. If the values of J at odd points are also desired, they can
NUMERICAL INTEGRATION PROCEDURES
SEC. 14—8.
be
475
determined using h
= Finally, one can express
12
Jo
{Sfk +
(14—84)
— Jk+2]
as
in =
+
+ 4(11 + .13 +
+ (14—85)
+ 2(f2 +f4 + Equation (14—85) is called Simpson's rule.
f
N fk+1
fk
fk—1
h
S
Xk_1
x
Xk
Fig. 14—18. Linear approximation.
I
S Xk
fk÷2
fk+1
fk
S
Xkf.1
Fig. 14—19. Parabolic approximation.
________________M CHAP. 14
PLANAR DEFORMATION OF A PLANAR MEMBER
476
Example 14—10 Consider the problem of determining the vertical displacement at Q for the straight member of Fig. E14—lO. We suppose shear deformation is negligible. The deflection due
Fig. E14—10
XQ
M
PQ
+1
XQ(1t)
to bending deformation (we consider the material to be linear elastic) is given by dQ
MQ dx
J
where M is the actual moment and M0 is due to the "Q" loading. Substituting for M, expands to
(a)
=
/ 1L M
CXQ
— J
M
'\
M
M
+ J
—
NUMERICAL INTEGRATION PROCEDURES
SEC. 14—8.
477
To evaluate (b), we divide the total length into ii equal segments of length h, number the points from 0 to n, and let
M
—dx J0 El Cx M x—dx El Jo I
I
With this notation, (b) takes the form = Xk
+ "k
—
Xkjk
If, in determining we also evaluate the integrals the interior points, then we can readily determine the displacement distribution using cd).
Example
14—11
Consider the simply supported nonshallow arch shown. We suppose there is some distribution of It'! and we want to determine the vertical deflection at Q. Considering Fig. E14—11
A
ill
El L
only bending deformation, dQ is given by dQ
=
M,
1W
ds
1W,
J
dx
PLANAR DEFORMATION OF A PLANAR MEMBER
478
CHAP. 14
Now, the distribution of M
is the same as for the straight member, Then, the procedure followed in Example 14—10 is also applicable here. We just have to replace M/EI with M/EI cos 0 in Equation (c) of Example 14—10.
REFERENCES 1.
TIMOSHENKO, S. 1.: Advanced Strength of Materials, Van Nostrand, New York, 1941.
2.
Boan, S. F., and J. J. GENNARO: Advanced Structural Analysis, Van Nostrand, New York, 1959.
3.
REISSNER, E.: "Variational Considerations for Elastic Beams and Shells," J. Lag. Mech. Div., A.S.C.E, Vol. 88, No. EM 1, February 1962.
4.
MARTIN, H. C.: introduction to Matrix lvi etliods of Structural Analysis, McGraw—Hill, New York, 1966. MUSRTARI, K, M., and K. Z. (IALIMOV: "Nonlinear Theory of Thin Elastic Shells,"
5. 6. 7.
8.
Israel Program for Scientific Translations. Jerusalem, 1962 MARGUERRE, K.: "Zur Theoric der gekriimmten Platte grosser Formanderung," Proc. 5th mt. Congress App!. Mccli. pp. 93—101. 1938. Onai'i, J. 1.: Mechanics of Elastic Structures. McGraw-Hill, New York, 1967. HILDEBRAND, F. J.: introduction to Numerical Analysis, McGraw-Hill, New York, 1956.
PROBLEMS
Specialize (14—7) for the case where Yi = x1. Let x2 = f(x1) and 14—1. let 9 be the angle from X1 to Y1 as shown below. Evaluate the various terms for a parabola
f=
+
Finally, specialize the relations for a shallow curve, i.e., where Prob.14—1 32
14—2. Evaluate 1* and the sketch. 14—3. Verify (14—34).
14—4. 14—5.
(see Equation 14-24) for the section defined by
Verify (14—41) and (14—42).
Discuss the difference between the deformation-force relations based
on stress and displacement expansions (Equations (14—25) and (14—42)). Illustrate for the rectangular section treated in Example 14—i. Which set of relations would you employ?
PROBLEMS
479
2t
I
Prob.
14—2
Prob.
14—6
T b=O.75d
d
I 14—6.
Evaluate I' and 1" for the symmetrical section shown.
h=O.75d
t=d120
14—7. Consider a circular sandwich member comprised of three layers, as shown. The core layer is soft (E 0). and the face thickness is small in comparison to the depth d). Establish force-deformation relations based
on strain expansions (see (14—37)). Prob. 14—i
I
II I
Starting with (14—34) and (14—35), derive a set of nonlinear strain 14—8. displacement relations for a thin member. Assume small finite rotation, and
PLANAR DEFORMATION OF A PLANAR MEMBER
480
CHAP. 14
linearize the expressions with respect to Yz' i.e., take
=
e1 —
Y12
Determine the corresponding force-equilibrium equations with the principle of virtual displacements. 14—9. Refer to Fig. 14—10 and Equation (14—31). If we neglect transverse is orthogonal to t'1 and we can write shear deformation,
=
(1 +
—
tj
ldF' -
= fl1t1 + fl2t2
= —/32t1 + f31t2 1 + e1 .-, di'2 I + e1
dt'1
=
R'
= (a)
Verify that
+
= (1 +
can be expressed as 1
(i+e1
c
Cl
I
- y,k}
Also determine e1 and R' for small strain. Express ü in terms of th initial tangent vectors,
ü= and take y (b)
S (i.e.,
U1t1
+ U2t2
1).
Derive the force-equilibrium equations, starting with the vector equations (see (14—12) and Fig. 14—4),
dS
dM÷
+m +
+
=
0
—
x F÷ = 0
and expanding the force vectors in terms of components referred to the deformed frame: = F11'I + = Mi3 (c)
h=
b11'1
+
Assume small strain. Derive the force-equilibrium equations with the principle of virtual displacements. Take the strain distribution according to Equation (b).
________ PROBLEMS
(d)
481
Derive the nonlinear deformation-displacement and equilibrium equations for the cartesian formulation. Refer the translations and loading to the basic frame, i.e., take
= P
+ V212
Pi'i + P2t2
Specialize the equations for the case of a shallow member. 14—10. The accompanying sketch applies to both phases of this
problem.
Prob. 14—10
h2 = const
(a)
(b)
Determine the complete solution for the circular member shown. Utilize symmetry at point A = co = F2 = 0) and work with (14—58), (14—59). Discuss the effect of neglecting extensional and shear deformation, i.e., setting (1/A) (1/42) = 0. Repeat (a), using Mushtari's equations for a thin member with no
transverse shear deformation, which are developed in Example 14—2. Show that Mushtari's approximation (u1 << du2/dO) is valid when the segment is shallow. 14—11. The sketch presents the information relevant to the problem: Prob. 14—11 P2 =
cOnSt
2
L
L
x2
'j
L2
PLANAR DEFORMATION OF A PLANAR MEMBER
482
CHAP. 14
Apply the Cartesian formulation to the symmetrical parabolic arch
(a)
shown. Consider the member to be thin and neglect transverse shear deformation. (b) Specialize (a) for negligible extensional deformation (set 1/A = 0).
Specialize (a) for the shallow case and investigate the validity of
(c)
Marquerre's approximation. 14-42. Refer to Example 14—6. Determine UB2 due to a uniform distributed constant. loading, b2 Determine the displacement measures at B (see sketch). Consider 14—13. only bending deformation. Note: It may be more convenient to integrate the governing equations rather than apply (14—69). Prob. 14—13
A
14—14.
Solve two problems with the information sketched: Prob. 14—14
(a)
Determine the fixed end forces and radial displacement at point B with the force method. Consider only bending deformation and utilize symmetry at B.
(b)
Generalize for an arbitrarily located radial force.
PROBLEMS 14—15.
(a) (b)
483
Refer to Example 14—7.
Determine the radial displacement at B defined in Fig. E14—7. Determine the force solution for the loading shown. Prob. 14—15
P
P 14—16.
The sketch defines a thin parabolic two-hinged arch. Prob. 14—16
x2
Determine the horizontal reaction at B due to the concentrated load. Consider the arch to be nonshallow. (h) Utilize the results of(a) to obtain the solution for a distributed loading (a)
(c)
p2(x) per unit x1. Determine the reactions resulting from a uniform temperature increase, T.
(d)
Suppose the horizontal support at B is replaced by a prismatic member extending from A to B. Assume the connections are frictionless hinges. Repeat parts a and c.
484
14—17.
PLANAR DEFORMATION OF A PLANAR MEMBER
CHAP. 14
Consider the arbitrary two-hinged arch shown. Discuss how you
Prob.
14—17
would generate the influence line for the horizontal reaction. Utilize the results contained in Examples 14—10 and 14—11.
15
Engineering Theory of an Arbitrary Member 15—i.
INTRODUCTION; GEOMETRICAL RELATIONS
In the first part of this chapter, we establish the governing equations for a member whose centroidal axis is an arbitrary space curve. The formulation is restricted to linear geometry and negligible warping and is referred to as the theory. Examples illustrating the application of the displacement and force methods are presented. Next, we outline a restrained warping formulation and apply it to a planar circular member. Lastly, we cast the force method for the engineering theory in matrix form and develop the member force-displacement relations which are required for the analysis of a system of member elements. The geometrical relations for a member are derived in Chapter 4. For convenience, we summarize the differentiation formulas here. Figure 15—1 shows the natural and local frames. They are related by —
cos çbñ
+ sin çt'b
= —sin4iii + where = we obtain
Differentiating (a) and using the Frenet equations (4—20), dt
= at
0
0
t
(15-1) 0
Note that a is skew-symmetric. 485
1t3
486
ENGINEERING THEORY OF AN ARBITRARY MEMBER
CHAP. 15
n
I
b
Local reference directions
Fig. 15-i. Natural and local reference frames for a member element.
of a point, say Q, are taken as S and Y2, y3. The curvilinear Letting K be the position vector to Q (see Fig. 15—2),
R=
+ y2t2(S) +
(15—2)
and differentiating, we find 13R
3K
= (t
-
y2a12
—
—
y3a23t2 + y2a23t3
— = t2 —
(15—3)
3K 3y3
The differential volume at Q is d(vol.)
Y2a12 — y3a13)dS dy2 dy3
(1
=
(i
E
(15—4)
dS dy2 dy3
is the coordinate of Q in the normal (11) direction and the radius of curvature. Also, where
3R
(1
..
Y3
3R
f See Sec. 4—8.
y2a23 =
/1 Y2
+
= 1/K
is
+
dcb
(15—5)
INTRODUCTION; GEOMETRICAL RELATIONS
SEC.
and the local vectors at Q are orthogonal when a23
487
0, which requires
a23 =0 (15—6)
d/
It is reasonable to neglect y/R terms with respect to unity when the member is thin, i.e., when the cross-sectional dimensions are small in comparison to
ay2
—
Fig. 15—2. Curvilinear directions.
and
We express d4,/dS as dq5
(15—7)
dS
where L is the total arc length and is the total increment in The nonorthogonality due to can be neglected when the member is only slightly twisted, i.e., when <<
1
(15—8)
where b is a typical cross-sectional dimension. In what follows, we will assume the member is thin, (15—8) is satisfied, and defines the orientation of the principal inertia directions.
ENGINEERING THEORY OF AN ARBITRARY MEMBER
488
CHAP. 15
Example 15—i The curvature and torsion for a circular helix are derived in Example 4—5: 1
R,
(H'\2
R
1
where R is the radius of the base circle and H is the rise in one full revolution. The helix is thin when b/R c< 1, where b is a typical cross-sectional dimension.
Example
15—2
By definition, a member is planar if r = 0 and the normal direction (Il) is an axis of symmetry for the cross section. We take the centroidal axis to be in the X1-X2 plane and define the sense oft2 according tot2 x 13 = Theangle is constant and equal to
either 0° (12
or 180° (12 = —ii). Only a12 is finite for a planar member:
a13=a23=0 Example
15—3
Consider the case where the centroidal axis is straight and
The member is said to he naturally twisted. Only 023
=
a13
=
(Ic!)
T= If bk <
15—2.
1,
is
varies linearly with finite for this case:
S.
0
const
IC
we can assume aR/aS is orthogonal to F2,
FORCE-EQUILIBRIUM EQUATIONS
To establish the force-equilibrium equations, we consider the differential element shown in Fig. 15—3. We use the same notation as for the planar case. The vector equilibrium equations follow from the requirement that the resultant force and moment vectors must vanish: dS
_. + rn
—
+ t1 x F÷ =
-
0
We express the force and moment vectors in terms of components referred to the local frame, = FTt = (15—10)
=
m't
FORCE-EQUILIBRIUM EQUATIONS
SEC. 15—2.
489
bc/S +
+
c/S
dS2
2
inc/S c/S
+
2
Fig. 15—3. Differential element for equilibrium analysis.
where
F=
F2, F3} etc. The vector derivatives are dFT = -------t + 'Tat
dF÷
dS
tIS
—
dS
dS
(a)
+
Also,
= FJ3
x
—
Substituting in (15—9), and noting that librium equations:
_F3,F2}Tt
=
{O,
=
—-a,
(b)
lead to the following equi-
aF + b =
0
dM
+F0J
dF3 dM1
+
— a12F2
—a33F3 + b1 =
+
a12F1
—
+
a13F1
+ a23F2 + b3 =
0
+ m1 =
0
a23F3
a12M2 — a13M3
a12M1
—
0
+ b2 = 0.
a23M3 + m2 — F3
(15—11)
0
When the member is planar, a13 = a23 = 0 and the equations uncouple naturally into two systems, one associated with in-plane loading (b1, b2, rn3,
490
ENGINEERING THEORY OF AN ARBITRARY MEMBER
CHAP. 15
F2, M3) and the other with out-of-plane loading (b3, in1, m2, F3, M1, M2). The in-plane equations coincide with (14—14) when we set a12 = 1/R and the out-of-plane equations take the form F1,
dF3
dM1 dM2
15—3.
+
1
1.
+ m2
+
in1
=
0
—
F3
=
0
(15—12)
RELATIONS—NEGLIGIBLE WARPING RESTRAINT; PRINCIPLE OF VIRTUAL FORCES
We consider the material to be elastic and define V" as the complementary energy per unit arc length. Since we are neglecting warping restraint, is a function only of F and M. We let
Iav*
-
.
(15 13)
— 1, 2, 3
k = {kJ and write the one-dimensional principle of virtual forces as
dS =
Js(eT
AF + kT AM)dS
(15—14)
dS It
(0
Fig. 15—4. Virtual force system.
NEGLIGIBLE WARPING RESTRAINT
SEC. 15—3.
491
Now, we apply the principle of virtual forces to the element shown in Fig. 15—4. We define ü and ii5 as
= uTt = equivalent rigid body translation (15—15) vector at the centroid = = equivalent rigid body rotation vector
=
The virtual system satisfies the equilibrium equations (15—5) identically and therefore is statically permissible. Evaluating APi,
an +
AFT
+ AMT
- aw) dS
and substituting in (15—14) lead to the following force-displacement relations:
10
du
e
—
an
+ (15—16)
k
dO)
=
e1 = C2
=
e3 =
aY*
=
k2
du2
a v*
(113
=
=
k3 =
du3
dw1
= DV"
=
—
Cj2
aV* k1
du1
(*
dw2
=
=
d0)3
—
+ a12u1 —
—
+
+
+ 023u2
— 0120)2
— (1130)3
+
0120)1 —
023C03
+
+
0230)2
Once is specified, the left-hand terms can be expanded. The form. of depends on the material properties, the particular stress expansions selected, and the member geometry. In what follows, we consider the material to be linearly elastic and approximation with the complementary energy function
ENGINEERiNG THEORY OF AN ARBITRARY MEMBER
492
CHAP. 15
for the prismatic case, which is developed in Sec. 1 2—3:
=
+
+
+
+ (15-17)
+
+
+
+ where
= torsional moment with respect to the shear center coordinates of the shear center with respect to the centroid
MT = M1 + Y2,
F3y2
—
=
dA
if
= —j-'-
Note that (lST-i7) is based on
dA
if
a linear expansion for the normal stress, M3
M-,
F1
=7+
13
and using the shear stress distribution predicted by the engineering theory,
+ = is the where is the unrestrained torsional distribution due to MT and flexural distribution due to F2, In addition to these approximations, we are also neglecting the effect of curvature, i.e., we are considering the member to be thin. The approximate force -displacement relations for a linearly elastic thin curve member are F1
+ F2
k1 =
k3 =
F3
MT
MT
=
+
du1 —
=
+
= e3
=
dü1
=
du2 du3
—
+ a12u1
—
C21U3 —
-
+ a13u1 + a21tc2 + (03 15
a12co2 — a13w3
M2
dw2
M3
dw3
UI3
U,)
—a23co3
+ a13co1 + 1223(02
18
CIRCULAR PLANAR MEMBER
SEC. 15—4.
493
When the member is planar, the shear center is on the Y2 axist and there is no coupling between in-plane (u1, U2, and out-of-plane (u3, (02) displacements. That is, an out-of-plane loading will produce only out-of-plane displacements. The approximate force-displacement relations for out-of-plane deformation for a thin planar member are
e3 =
F3
+
= dw1
MT
k2
du3
MT — I
(15-19)
M2
= k2
+
1
= M1 — y2F3. Note that flexure and twist are coupled, due to the curvature, even when the shear center coincides with the centroid. where
DiSPLACEMENT METHOD—CIRCULAR PLANAR MEMBER
15—4.
Since the displacement method involves integrating the governing differential equations, its application is restricted to simple geometries. In what follows, we apply the displacement method to a circular planar member subjected to out-of-plane loading. We suppose the cross section is constant and the shear center coincides with the centroid. It is convenient to take the polar angle e as the independent variable. The governing equations are summarized below
and the notation is defined in Fig. 15—5. Equilibrium Equations (sec (15—12))
+ —
dM2
Rh3
0
M2 + R,n1
=0
+Rrn2—RF3=O
Force-Displacement Relations (see (15—19)) e3
ldu3
F3
=
=
k1
k2 =
I /dw1
M1
k2o
+
+
M2
=
l/dw2
+
t The shear center axis lies in the plane containing the centroidal axis, which, by definition, is a plane of symmetry for the cross Section.
ENGiNEERING THEORY OF AN ARBiTRARY MEMBER
494
CHAP. 15
Boundary Conditions
F3 M1
M2 The
or or or
u3
prescribed at each end (pts. A, B)
solution of the equilibrium equations is quite straightforward. We
integrate the first equation directly:
F3=C1—RJ0b3d0
(15—20)
The remaining two equations can be transformed to (d)
=
dM1
+ Rin1
(e)
We solve (d) for M1 and determine M2 from (e). The resulting expressions are
M1 = C2cosO + C3 sinG + M2 = —C2sinO + C3cosO + where M1
,,
is
+ Rm1
the particular solution of (d). P3
x2
i313
F3 B
the X1 —X2 plane. are axes of symmetry.
Fig. 15—5. Notation for circular member.
CIRCULAR PLANAR MEMBER
SEC. 15—4.
The
495
solution of the force-displacement relations is also straightforward.
First, we transform (b) to d2co1
+ w1 =
Rk2
R2 —
in1
+
R
(1 +
RM5
- Rw2
= where
(f)
is a dimensionless parameter, (15—22)
=
which is an indicator for torsional deformation. Solving the first equation for and then determining W2 and u3 from the second and third equations lead to
= C4cosO + C5 sin fJ + wi,, =
—C4
sin U + C5cosU +
C6 —
+j
RM1
d
dO
+
is the particular solution for wi. The complete solution involves six integration constants which are determined by enforcing the boundary conditions. The following examples illustrate the application of the above equations.
where
Example
15—4
The member shown is Sxed at A and subjected to a uniform distributed loading. Taking
Fig. E15—4 b3 = const
h3
= coast in (15—20), we obtain
F3 =
C1 — Rb30
(a)
ENGINEERING THEORY OF AN ARBITRARY MEMBER
496
CHAP. 15
The equation for M3 reduces to
+
M1
= RF3 = RC1
—
R250
= RF3 =
—
R2b30
Then,
and the solution for M3 and M2 follows from (15—21),
=
C2
cos 0 + C3 sin 0 + RF3
—C2.sinO + C3cosO —
M2
R2b3
The boundary conditions at B require
F3=M1=M2=O
Replacing
—
0
by
= C2 =
R5308
C3
R2b3 cos
0=08
at
—R2h3 sin
the final solution is
F3 = =
Rb30
R2b3[q
M2
—
sin
?q]
.—R2h3[1 — cos
Example 15-5 can be determined by applying the The force system due to the end action, librium conditions directly to the segment shown in Fig. E15—SA. This leads to
F3 = = FB3R(l
M2 = We
F83R[l •—
cos
—
sin
—
=—
sin(Oa
—
0)]
(a)
9)
suppose there is no initial deformation. Using (a), the equation for w3 becomes
+ w1 =
(1
+
—
(b)
0)
The particular solution of (b) is (1
+
[0 cos(OB
—
9)]
Using the above results and specializing (15—23) for this support condition lead to the following expressions for the displacements: WAI COS 0 +
+
sin 0 C05
+
sin 0 —
9
cos(08 —
CIRCULAR PLANAR MEMBER
SEC. 15—4.
I
cosO +
SlflO +
1
.
Sm 0
Osln(Oa
—
(d)
—
—
UA3 + RthAI(l — cos RFB3 IF I
133
+ cosO]
.
.
497
+
£1) — Rö5A2 Sifl
0
C1j sin P
COS
SW
1L cos(05 — 0)
0
+ c, +
+
sln(OR —
and define the rigid body displacements due to support movement. Also, terms involving c1 are due to twist deformation. The rotations and Terms involving
Fig. E15—5A F3 M2 A
I?
translation at B are listed below:
+
COS
+
sin 0a
R2P83 f COS
+
sin
+ +
033 -. 1]
+ Roi41(l — cos 1—c3
+
c1] sin
—
—--—- sin 2
1
cos
+ —h--— 13)33
±
Rw42 Sm 0)3
+
sin
C05
El2
3
1—c,
Th
— 2c1
sin
498
ENGINEERING THEORY OF AN ARBITRARY MEMBER
CHAP. 15
To investigate the relative importance of the various deformation terms, we consider the rectangular cross section shown in Fig. El 5—5)3. The properties aret
61 _6
1
I
5A — 5d2d3
A3
J =
(for
d3)
Then,
(d3\2 El2
—
GA3R2 —
The values of 4k and
for d3/d2 =
E
[1 (di'\21
G [io k,i2) J 1,
2, 3 and v =
0.3
are tabulated below:
= E12/GJ d3/d2
4k
(Ibr v
1
1,69
1.54
2
2.75
3.8
3
3.16
7.4
0.3)
Fig. E15—5B
T
1(3
t The torsional constant for a rectangular cross Section is developed in Sec. 11—3.
SEC. 15—5.
FORCE METHOD—EXAMPLES
499
Since (d2/R)2 << 1, we see that it is reasonable to neglect transverse shear deformation. In general, we cannot neglect twist deformation when the member is not shallow. For the shallow case, we can neglect in the expressions for UB3.
Example 15—6 Consider a closed circular ring (Fig. El 5—6) subjected to a uniformly distributed twisting 0 and M1, M2 are constant. Then, using (15—16), we find
moment. From symmetry, F3
M1 =
0
1142
Rm1
The displacements follow from (15—18) U3 = (13; — 0
RM2 WI
=
R2m1
= Fig. E15—6
x2
b3 =
m2
0
const
15—5.
FORCE METHOD—EXAMPLES
In this section, we illustrate the application of the principle of virtual forces to curved members. The steps involved are the same as for the prismatic or planar case and therefore we will not reiterate them here. We restrict this discussion to the case where the material is linearly elastic, the member is thin
and slightly twisted, and warping is neglected. The general form of the expression for the displacement at an arbitrary point and the compatibility equations corresponding to these restrictions (see (15—14), (15—17)) follow.
ENGINEERING THEORY OF AN ARBITRARY MEMBER
Displacement
at Point Q
= —i
+ j [(ei°
+
+
+
CHAP. 15
(k20
Fj,Q + (a-)
+
(15—24)
+
+
+
Compatibility Equations
Zr = force redundants
Z1, Z2
= F30 + = M3,0 +
kA
R1 = R1,0 + (k
.fkJZJ 3=
kZk
=
r)
1, 2
(15—25)
1
where
+
=
+ ETM31M3,kldS
JMT.JMT.k + +
—
+
+
= M1 + y3F2 —
+
+
+
+
v2F3
The reduced form for out-of-plane deformation is obtained by setting F1 = F2 = 0. + Example 15—7 Consider the nonprisinatic member shown below. The centroidal axis is straight but the orientations of the principal inertia axes vary. We take X1 to coincide with the ceniroidal axis and X2, X3 to coincide with the principal inertia directions at the left end (point A). The principal inertia directions are defined by the unit vectors t2, t3.
+ sin
= cos
=
çbt3
+ cos
—sin
at
x1=O
FORCE METHOD—EXAMPLES
SEC. 15—5.
Now, we consider the problem of determining the translations of the centroid at B due to the loadingshown in Fig. E15—7A. It is convenient to work with translationcomponents (v52, v53) referred to the basic frame, i.e., the X2, X3 directions. We suppose that the shear Fig. E15—7A
x2
x1 axts
center coincides with the centroid and cializing (15—24), and noting that M1 = the displacement expression reduces to dQ =
0
shear deformation is negligible. Spefor a transverse load applied at the centroid,
11L(i +
(b)
Force Systems The moment vectors acting on a positive cross section due to P2, P3 applied at B (Fig. E15—7B) are
—
P2(L
= To find M2, M3, we must These follow from Fig. E15—7C:
—
— P3(L
—
x1)12
the components of It,! with respect to the local frame.
For P3.
M2 = P2(L
—
P2(L
—
M3
==
x1)cos
q5
502
CHAP. 15
ENGINEERING THEORY OF AN ARBITRARY MEMBER
Fig. E15—7B
1'2(L—x1)i3
/B
L-x,
'I
P3
—P3(L
Fig. E15—7C M2r2
——P2(L—x1)z1
M3t3
For
M2 = M3 = +P3(L
Determination of
—
1)cos 4)
—
x1)sin4)
(C )
Due to P2
The virtual-force system for we obtain
corresponds to P2 = +
1.
Introducing (d) in (b), (f)
Determination of CR3 Due to P2 The virtual-force system for VB3
corresponds to P3 = + 1. Using (e) leads to
FORCE METHOD—EXAMPLES
SEC. 15—5.
503
Example 15—8 We rework Example 15—6 with the force method. Using symmetry, we see that
M1 =
0
=
Suppose the rotation w1 in the direction of in1 is desired. The virtual loading for this displacement is rn1 = + 1. Starting with
and substituting for M2, we obtain TO
COl
R2
=
Example 15—9 Consider the closed ring shown. Only M1 and M2 arc finite for this loading. Also, the behavior is symmetrical with respect to X1 and we have to analyze only one half the ring.
Fig. E15—9 x2
42 44
12,J are constant T
We take the torsional moment at 0 = M1
T
0
as the force redundant. The moment distributions
504
CHAP. 15
ENGINEERING THEORY OF AN ARBITRARY MEMBER
Specializing (l5—25) for this problem,
f11Z1 =
A1
[M1 0M1
A1 = —2R
0M2,
+
dO
El2
,,/2
f11
=
2R
1
J
dO
+
and then substituting for M1, M2, A1
'
=
12
[sin2
-
=
cos 0
E12) sin
0
and it follows that Z1 = 0. We could have arrived at this result by noting that the behavior is also symmetrical with respect to X2. This requires M2 to be an even function of 0. The virtual-force system for WAI is T = + 1. Using (15—24) and (a) leads to
r/T sin 0\sin 0
2wAi = 2RJ
fT cos O'\ cos 01 +
RTa[l
I
COAl
Example 15-10 We analyze the planar circular mcmber shown in Fig. El 5— bA. The loading is out-ofplane, and only F3, M1, and M2 are finite. To simplify the algebra, we consider the shear center to coincide with the centroid and neglect transverse shear deformation. It is convenient to take the reaction at B as the force redundant. Fig. E15—1OA A P13
C
(Displacement restraint
iriX1 direction at B)
B
FORCE METHOD—EXAMPLES
SEC. 15—5.
505
Primary Structure The primary structure is defined in Fig. El 5—lOB:
B1 = d1 — R4
=
R2
Z1
=
= —MAI
—MA2
d2 = d4 =
UA3
F53
=
E15—1OB
M1 ,
xl Force Analyses The force solutions for the loadings shown in Fig. El 5— [OC are:
For F: F'3,0 = +P
M1, = PR[l M2,0 =
ForZ1 = +1:
COS(ii —
—PR sin(lJ — 'Ic)
llc)] (b)
'Ic
F3,1 = +1 = R(1 — cos M21 —Rsini1
ii)
(c)
506
ENGINEERING THEORY OF AN ARBITRARY MEMBER
CHAP. 15 Fig. E15—1OC
C
=+i B
Compatibility Equation (15—25) IL,j =
fit
!ii = +
R
+
M2
Substituting for the internal force and reactions, we obtain the following expressions and
for
R3 [(1 +
(1 —
Sin
—
Sill
COS
— —
+ R2
UA3 + RrOA2 sin
sin(05
{o. [i +
—
—
—
cos
—
sin
O)dG
cos(05
—
Or)]
—
sin
Oc
—
+ sin
=
El2 GJ
Note that we could have determined A1 and
using the results of Example 15—5.
RESTRAINED WARPING FORMULATION
SEC. 15—6.
15-6.
507
RESTRAINED WARPING FORMULATION
In what follows, we consider the member to be thin and slightly twisted. Referring to Fig. 15—2, these restrictions lead to -
dR
tl
(15—26)
dS dr2 dy3
d(vol.)
Therefore, in analyzing the strain at Q (S, y3), we can treat the differential line elements as if they were orthogonal. Thc approach followed for the prismatic case is also applicable here. One has only to work with stress and strain measures referred to the local frame 12, 13) rather than the global frame. Our formulation is based on Reissner's principle (13—33): V*)d(vol.)
bTIi
d(surface area)] = — r, ü = independent quantities = e(iI) p, b = prescribed forces = = complementary energy density
0
We introduce expansions for fl, in terms of one-dimensional displacement and force measures (functions of S) and integrate over the cross section. The force-equilibrium equations follow from the stationary requirement with respect to displacement measures. We start with the strain measures, = One can show thatt Y12. S1
au
ti tj
-.
+ t2 +
(3U
(15—27)
-
(3
is the displacement vector for Q (S, Yi' Y2). We use the same displacement expansion as for the prismatic case: where
+ u212 + 113(3 U1 + W2y3 — WIY2
U1
lAsZ —
IA2
=
+
—
co1(y2
± f4i (15—28)
y3)
— Y2)
Y3)
Expanding
= t See Prob. 15—5.
+
+ cr13y13)dy2dy3
ENGINEERING THEORY OF AN ARBITRARY MEMBER leads
CHAP. 15
to
= F1e1 + F2e2 + F3e3 + MTkI
dy2
+ M3k3 + MRf + Mj,. e1, e2,. .
, k3 (defined by (15—16)) = SSai14) dy2 dy3 + a134))]dy2dy3 MR = JJ[ci2(4),2 + a12çb) + .
(i5—29)
The equilibrium equations consist of(15—11) and the equation due to warping restraint, (15—30)
MR
which can be interpreted as the stress equilibrium equation for the weighted with respect to 4).
direction
Now, we use the stress expansion developed for the prismatic case. The
derivation is discussed in Sec. 13—5, so we only list the essential results here. The normal stress is expressed as F1
=
+
M2 '2
M3 Y3 —
+ -r
'3
'1)
(15—31)
the St. Venant warping function referred to the shear center. 4) = — We write the transverse shear stress distribution as where
+
±
(15—32)
= are functions of
The corresponding complementary energy function
+
= $SV* dy2 dy3 =
M32)
+
+ + 2F2F3
+
+
+
(15—33)
+
+
+
Also, (15—32) satisfies (see(13—50))
JS(a124),2 + cr134),3)dy2dy3
Finally, noting (b), we express MR as
MR = (1 +
+ b2F2 + b3F3
where the b's involve the curvature (a32, a1 3). A23
Y3r = Y2r
(15—34)
If the cross section is symmetrical, br =
0
RESTRAINED WARPING FORMULATION
SEC. 15—6.
509
and b2, b3 are due to self-equilibrating stress distributions.t It is reasonable, in this case, to take b2 0 and compute the shear coefficients (A2, A3) based on the primary fiexural shear stress distributions. Expanding the stationary requirement with respect to force measures yields the force-displacement relations, ej
a
=
av* k2
av* k3
=
e2 + b2f =
=
e3 +
aF2
b3f =
aV*
(15—35)
i3V*
, k3 are defined by (15—16). The corresponding unrestrained where e1, e2, warping relations arc (15—18). .
Example
15—11
To investigate the influence of warping restraint, we consider a planar circular member having a doubly svrnnietricai cross section (Fig. his—il), clampedat one end and subjected Fig. E15—11
NM
to a torsional moment at the other end. We neglect transverse shear deformation due to restrained torsion. The governing equations for this loading (See Sec. 15—4) follow.
Equilibrium Equations dM1
dM2 dO
M1 t See Prob. 15—6.
= =
R dO + M'j
ENGINEERING THEORY OF AN ARBITRARY MEMBER
CHAP. 15
Force-Displacement Relations
M2 =
E12k2
El2
+
f
El4, R
=
(10
(b)
1 (dco1
f= = GJk1
Boundary ('onditions
(01C02J0
0—U
1v11 =M
M2 =
0
One can write the equilibrium solution directly from the sketch: M
M1
M2 = Al sin(05 —
— 0)
0)
We substitute for the moments in the force-displacement relations.
M1 =
—
GJk1
=
0)
—
1
k1
—
I
lvi
'\
(dco2
+ Wi)
=
— 0)
and solve for k1, and then Wi. The resulting expressions are G.J
=
ME
GJ + =
—
GJ —
0) — cos
—
siiih
tanh
1
{o cos(05 — 0)
+
El2
.,.
2 2 =———
0
{sinh
sin 0 cos
cos(05 — 0)
—
+
0
and
sin 0 cos
+ cos ü tanli
cosh
tanh
Warping restraint is neglected by setting Er =
—
=
SEC. 15—7.
COMPLETE END RESTRAINT
The rotation at B is
=
(RM\
—
+
COS 0B)
l\ El2)
K=
1
+
1
Sin 0B
—
tanh 704
cos2
[
—
If we set On
On
and let (g) reduces to (13—57), the prismatic solution. The influence of warping restraint depends on 2 and Values of K vs. 2 for = ir/4, it/2 are tabulated below: 1
for
=
+
for
= it/4
= ir/2
1
0.179
0.51)0
5
0.786
0.96
10
0.907
0.99
We showed in Chapter 13 that 2
0
(open section)
(j) 2 =0
(closed section)
where t is the wall thickness and h is a depth measure. Since 2 = RA and R/li for a thin curved member, the influence of warping restraint is not as significant as for the 1
prismatic case.
15—7.
MEMBER FORCE-DISPLACEMENT RELATIONS—COMPLETE END RESTRAINT
-
In the analysis of a member system, one needs the relations between the for
and displacements at the ends of the member. For a truss, these equations
512
ENGINEERING THEORY OF AN ARBITRARY MEMBER
CHAP. 15
to a single relation between the bar force and the elongation. Matrix notation is particularly convenient for this derivation so we start by expressing the principle of virtual forces and the complementary energy density in terms of generalized force and deformation matrices. Referring back to Sec. 15—3, we define reduce
a v*
(1536) aM1
and write the principle of virtual forces as dS
Ss
dT
dS
J
(15—37)
Note that we are working with M1, not MT. We use the complementary energy function for a thin slightly twisted member with negligible warping restraint (i.e., (15—17)). With the above notation,
+
Eq. (15—17)
where
g=
(15-38)
gfm
1
gf
A2G
+
Y2Y3
(7
Sym
o
GJ I
+
ojo 1
J'3
El2 0
Sym
0
The force-deformation relation implied by (15—38) Is
+
(15—39)
We will use these general expressions for planar and out-of-plane deformation as well as for the arbitrary case. One has only to delete the rows and columns
COMPLETE END RESTRMNT
SEC. 15—7.
513
of g corresponding to the zero force measures. For example, {F1F2M3}
=
0
(15—40)
El for planar loading applied to a planar member. Finally, we substitute fort in (15 —37) and distinguish between prescribed and unknown displacements. The principle of virtual forces expands to
J5(t° +
dS — dT
dT
(15—41)
contains prescribed displacements and R are the corresponding reactions; d contains unknown displacements and are forces corresponding to d. The virtual-force system must satisfy the force-equilibrium equations, (15—11). It is more convenient to generate and R with the equilibrium equations for a finite segment rather than attempt to solve (15—il). Consider the arbitrary member shown in Fig. 15—6. Each end is completely restrained against displacement. The positive sense of S is from A toward B. where
B
13
Basic member frame
Fig. 15—6. Arbitrary curved member.
514
ENGINEERING THEORY OF AN ARBITRARY MEMBER
CHAP. 15
suppose the geometry of the member is defined with respect to a basic frame which we refer to as frame ii, and take the end forces at B as the force redundants. Then, the primary structure consists of the member cantilevered We
from A.
Throughout the remaining portion of the chapter, we will employ the notation for force and displacement transformations that is developed in Chapter 5. A superscript n is used to denote a quantity referred to (lie basic frame. When no frame superscript is used, it is understood the quantity is referred to the local frame. For example, represents the internal force matrix at point Q referred to the local frame at Q. Note that acts on the positive face. The force matrix for the negative face is — The end forces at A, B are denoted and are related to the internal force matrices by by
=
=
15
=—
—
42
Also, the displacement matrix at point Q is written as 0//Q
= {u1,
(Oj, W2, c03}Q
U2, U3
{u}
(15—43)
For this system, and are prescribed. We determine for the primary structure, i.e. the member cantilevered from A, due to displacement of A, temperature, loads applied along the member, and the end forces at B and then equate it to the actual The virtual-force system is
Lw = AR = =
= =
Also,
d= d
Introducing (a), (b) in (15—41), we obtain
+
+ Il
+ Next, we express
+
as
+
(15—44)
is the internal force matrix at Q due to the prescribed external loading applied to the member cantilevered from A. Finally, substituting for where
COMPLETE END RESTRAINT
SEC. 15—7.
515
leads to
=
+
[SB
.
T(18 +
0)dS (15—45)
[JB
+ The first term is due to rigid body motion of the member about A whereas the second and third terms are due to deformation of the member. We define as the member deformation matrix:
=
—
to rigid body
(1546)
notion aI,out .4
By definition, We also define
is
equal to the sum of the second and third terms in (15—45).
JSB
0)dS = initial deformation matrix
+
S-4
(15—47)
member flexibility matrix
fll
and (15—45) reduces to —
=
+
(15—48)
Equation is the force-displacement relation for an arbitrary member with complete end restraint. It is analogous to the force-elongation relation for the ideal truss element that we developed in Chapter 6. The member flexibility matrix, f", is a natural property of the member since it depends only on the geometry and material properties. For simple members such as a prismatic member or a planar circular member with constant cross section, one can obtain the explicit form off. When the geometry is complex, one must generally resort to numerical integration such as described in Sec. 14—8 in order to determine f and This problem is discussed in the next section. Finally, we point out that the general definitions off, are also valid for in-plane or out-of-plane deformation of a planar member. One simply has to use the appropriate forms for the various matrices. Up to this point, we have considered only a simple member. Now suppose the actual member consists of a set of members rigidly connected to each other and the flexibility matrix for each member is known. We can obtain the total flexibility matrix by compounding the flexibility matrices for the individual elements. To illustrate the procedure, we consider two members, AA1 and A1B, shown in Fig. 15—7.
The matrix, f", contains the displacements at B due to the end forces at B with A fixed: Now, suppose point A1 is fixed. Then, the displacement at B due to the tion of member A1B is A1B
516
ENGINEERING THEORY OF AN ARBITRARY MEMBER
CHAP. 15
is the flexibility matrix for member A1B referred to frame n. The additional displacement at B due to movement of A1 is
where
—
• B,' displaccment
at 4 —
' BA,
'Ai
It remains to determine
B
Fig. 15—7. Segmented member.
The force system at A1 due to the end forces at B is given by A,
CF-n B
—
and the resulting deformation of member AA1 is —
A, mernberAA, —
ffl
A,
— ffl
AA,'
(TI? It
Finally, we have = ffl
+
=
(15—49)
The end forces at B are found by inverting (15—48):
=
(fn)_ 1
—
1
=
member stiffness matrix •,J/'fl
—
1550
+
—
—
The first term is due to external load applied along the member and represents
the initial (or fixed-end) forces at B. For convenience, let (15—51)
The second and third terms are the end forces at B due to end displacement at B, A. Once is known, we can evaluate the interior force matrix at a point using (15—44),
=
0
+
(a)
Thus, the analysis of a completely restrained member reduces to a set of matrix multiplications once the member stiffness and initial deformation matrices are established.
SEC. 15—8.
GENERATION OF MEMBER MATRICES
517
When analyzing a system of members by the displacement method, expres-
sions for the end forces in terms of the end displacements are required. In addition to (15—50), we need an expression for
= Substituting for
—
Now,
0
leads to
+
i
——
A,0
—
(
— BA
)
B.i
represents the initial end forces. In order to express the equations where in a more compact form, we let
in KBB in k AB
ii
inwn.T
— —
—
n
BA
L
BA
jn
—
BA
BA
With this notation, the force-displacement relations simplify to
= = Note that only
15-8.
and
+
+ +
are
(15—54)
+
required in order to evaluate
and
GENERATION OF MEMBER MATRICES
The member flexibility matrix is defined by
= Noting that
g014)dS A
= and letting
=
(15—55)
we can write ffl
(15—56)
If numerical integration is used, the values of the integral at intermediate points along the centroidal axis as well as the total integral can he determined in the same operation. This is desirable since, as we shall show later, the intermediate values can be utilized to evaluate the initial deformation matrix. We consider next the initial deformation matrix:
= We transform 4',g, and
+
0)dS
from the local frame to the basic frame, using (15—55)
518
ENGINEERING THEORY OF AN ARBITRARY MEMBER
and
CHAP. 15
= =
The contributions of temperature and external load are T,O)dS 0)dS
(15—57) (15—58)
Suppose there is an external force system applied at an intermediate point, denote the force and moment matrices and the total force matrix:
say C. Let
(15—59)
= Normally, the external force quantities are referred to the basic frame for the member, i.e., frame n. The initial force matrix at Q due to this loading is given by 0
Sc
SA
_IIU
c
07fl Q.0 —
(15—60)
Writing —
or" or"
CD
and introducing the above relations in (15—59) result in (
= The bracketed term is an intermediate value of the integral defining
Finally,
we let (15—61)
= With this notation, (f) simplifies to
(15—62)
Also, (15—63)
The determination of the member flexibility matrix reduces to evaluating J defined by (15—61). One can work with unpartitioned matrices, i.e., g, but it is more convenient to express the integrand in partitioned form. The partitioning is consistent with the partitioning of into F, M. Since the formulation is applicable for arbitrary deformation, it is desirable to maintain this generality when expanding in partitioned form. Therefore, we define a as the row order of F and /3 as the row order of M. (15—64)
Continuing, we partition
and g symmetrically, consistent with (15—64),
GENERATION OF MEMBER MATRICES
SEC. 15—8.
519
and simplify the notation somewhat: 0
=
(15—65)
= (fixfj) —
gQ—
() g12
()
g11 E
g12
i
g22
The translation and rotation transformation matrices are developed in Secs. 5—1, 5—2 and the form of g for a thin curved member is given by (15—38).
The local ilexihility matrix is defined by (15—55). Using the above notation, the expressions for the submatrices are
= (15—66)
g22 —
g22 are diagonal matrices when the shear center Note that g12 = 0 and coincides with the centroid. If, in addition, axial and shear deformation are neglected, g11 = We let
0.
=
(15—67)
=
The submatrices follow from (15—65):
+
+ '1112 '1122
= =
+
XHQgZ2XBQ (15—68)
-F-
Next, we partition J consistent with'1':
j— —
'I' dS — I
Jp22j
—
(15—69)
x
JP
= F"
is
Finally, we partition f": -
r
ffl
ffl
=
(15-70)
520
CHAP. 15
ENGINEERING THEORY OF AN ARBITRARY MEMBER
The initial deformation matrix due to an arbitrary loading at point C can be determined with (15—62). Its partitioned form is
= 1V01 —
r cli
C.12 Ii
I
BC
2V'BC
L"c. 12 —-
lYe)
n
C,12
—
rT
22J
C —
'C
(flxfl)
denote the initial translation and rotation matrices. We write The member stiffness matrix, is obtained by inverting
where
= (r)'
=I
[ku
(15—72)
k22j I
(11 )< /1)
for convenience)
One can easily show that (we drop the frame superscript on —(
k,'11 — k ii —
12 22
\1 12) (15—73)
çT i
—If
1n
22k $ —
Once is known, the stiffness matriccs and can be generated. Expanding (15—53) leads to the following partitioned forms: knB
—
in KBA
I
—
AA
—I
rinIi
I
I
nvn,T llJtB,!
T
ml 12
I= A
[k';1
]
I
rnii I
I
L
I
1n.!T 12
I
(15—74
A
—
15—9. MEMBER MATRICES—PRISMATIC MEMBER
In Chapter 12, we developed the governing equations for a prismatic member
and presented a number of examples which illustrate the displacement and force methods of solution. Actually, we obtained the complete set of forcedisplacement relations and also the initial end forces for concentrated and uniform loading. Now, in this section, we generate the member flexibility matrix using the matrix formulation. We also list for future reference the various member stiffness matrices. The notation is summarized in Fig. 15—8. For convenience, we drop the frame reference superscript n, since the basic frame coincides with the local frame, i.e., = I. The positive sense of a displacement, external force, or end forces coincides with the positive sense of the corresponding coordinate axis.
Starting with (15—66), we have since R$ are identity matrices. = from (15—68). Once XBQ is assembled we can determine the submatriees of
_________ ______ MEMBER MATRICES—PRISMATIC MEMBER
SEC. 15—9.
521
i2
MB!, WB i
UB!
I.
/
X1 is centroiclal axis. X2, X3 are principal inertia directions.
Fig. 15—8. Summary of notation for a prismatic member.
Now,
X5Q=
o
o
0
0
0
—(L—.xQ1) 0
(L—xQt)
o
(a)
Then, using g defined by (15—38), we obtain
L/AE
0
/1 L
0
1) +
+
+
0
L2
GJ
2E12
L/GJ Sym
2E13 +
LT2
=
V
0 -—
10
+
0
0
.
\-2x3
I)
Sym
f12= GJ
L
0 0
LIEu
(15—75)
ENGINEERING THEORY OF AN ARBITRARY MEMBER
522
CHAP. 15
The submatrices of k are generated with (1 5—73), (1 5— 74) and are listed below 03 0:
for reference. Transverse shear deformation is neglected by setting a2 =
a2 =
12E12
a3
'2 2
b1 = AE
+
k11=
1+03
-
12E
0
L
= = —I'---
1 + a2 GJ
12E11
+
0
L3
Sym 0
0
0
6E1
0
k12 =
L2
(15—76)
0
L2
L2
(4 +
k22
0
Fl*
(4 +
Sym 0
0
—
0
0
L L3
L2
10
(change sign of(2, 3) and (3, 2) in k12) —
b1
B=
L2
L2
0
1! L2
0
MEMBER
SEC. 15—9.
MEMBER
523
—6E!!Ye2
b
V
L2
FI*
=
(4
+ a3)_L_
(change sign of(1, 2), (1, 3) in k22)
Finally, the fixed end forces due to a concentrated transverse force and a uniform transverse loading are summarized below.
concentrated Force
a3 =
12E13
MB3
—
(15—77) FB2
—
+
MBI = MA1
= — F112
MA3 =
+ FB2 +
—L
Force a2
12E12 — GA1L2
÷
—
+
FB3
(15—78)
+ FB3) MAI
—MB1 —
MA2
_PC3 — FR3
L
+
—
MB2)
ENGINEERING THEORY OF AN ARBITRARY MEMBER
524
CHAP. 15
Concentrated Torque
MA1
= =
—
(15—79
Tc1(1
—
Uniformly Distributed Load, b2
-
F82 =
-
b2L
FA2
h2L2
—
M83 = —MA3 =
MAI =
M81
(15—80)
0
Uniformly Distributed Load, b3 —
F83 = 1q43 = —
b3L. b3L2
—
M82 =
= =0
=
(15—81)
——u—
Uniformly Distributed Torque, rn M81
15—10.
MAI =
(15—82)
MEMBER MATRICES—THIN PLANAR CIRCULAR MEMBER
In this section, we generate the flexibility and initial deformation matrices for a thin planar circular member, of constant cross section, using matrix operations. We include extensional and transverse shear deformation for the
sake of generality. Some of the relations have already been obtained as illustrative examples of the force and displacement methods. In particular, the reader should review Example 14—6, which treats planar deformation, and Examples 15—4, 15—5, 15—10 for out-of-plane deformation.
The notation is summarized in Fig. 15—9. By definition, Y2 and Y3 are principal inertia axes and p3 = 0, i.e., the shear center lies in the plane containing the centroidal axis. It is convenient to take the basic frame (frame n) to be parallel to the local frame at B. The three-dimensional forms of and
are cos
=
= sin
— sin
01
cos
0
17] =
=
= rRbq °
I
0 1
(15—83)
SEC. 15—ID.
THIN PLANAR CIRCULAR MEMBER
525
jO
0
0
I
=
R sin
R sin ,fl for planar deformation and
R0=1 —
[—R(1
—
cos
SQ
for out-of-plane deformation. Since the complete flexibility matrix is desired, it is just as convenient to work with subinatrices of order 3 as to consider
separately the planar and out-of-plane cases.
x;
Centroidal axis b
B2
F;3,U;,
Mw Fig. 15—9. Summary of notation for a planar circular member.
We
consider the member to be thin and use the local flexibility matrix
defined by (15—38). Expanding (15—66), (15—68) leads to the member flexibility
matrix.
ENGINEERiNG THEORY OF AN ARBITRARY MEMBER
526
13
as =
ae
El2 Ct
Cs
= a1
=
=
CHAP. 15
El3 El7
ae + a7
a2 = ae —
c1 =
+
=
—
(15—84)
(
Y2\ —
—\2 Y2
\
Symmetrica' -
±
sin
I
—
+ az)i} R3
- cosO5
sin2 05(1 2
+ a2)}
+r!j) a2)sin
cos
0
0
+ c41 — 2c3
R2
0
0
0
0
05 — C2 Sifl
— sin
1,12
— T— t212t) —
—cos05)
+ 0
+ c2 sin
Ib22
cos f15}
COS
SEC. 15-10.
THIN PLANAR CIRCULAR MEMBER
527
We consider next the determination of the initial deformation matrix due to an arbitrary concentrltted load at an interior point, C. Now, the flexibility matrix for the segment AC referred to the local frame at C, which we denote is known. We just have to change to and superscript b to c in by
When the external load is referred to the local frame at C, the displacement at C is given by ('IC
The displacement at B due to rigid body motion about C is T011hc.
TrI &
—
"B —
BC
BC
3
Finally, we can write
=
=
U
j
1, — b V0 — UB —..
T c
+
Tçc. T .4C.i21
c C
/1
+
(15—85)
T
r
1'
—
v-I,. T
AC,11
c
(
4C,121 C
Ii
Tcc
\
C
The uncoupled expressions follow. Planar Loading
/
=
= M3 +
I'
Slfl
1
R3 I
+ -'-'--—
—
cog
I
+
cos
1+a1
0c ± -—-—
+ Sifl
1:13
=
2
sin ic
R
5jfl
(15—86)
ç
— —s-—-L
Sill 'Ic
1+02.
.
1
2
+
—
j
+ 00,3
—
o4
I
+
'Ic
1+02 Cos
+
—
+
1+a1
+
WB3
{cos 'Ic — cos OB}Tc3 —
sin
+ (1
COS
+
RO
ENGINEERING THEORY OF AN ARBITRARY MEMBER
528
CHAP. 15
Out-of-Plane Loading V0.
=
=
R3 (
+ c3(— sin
—
R21
+ —+
COS
'ic + c4) —
sin
°B + sin
cos 'ic + C2 —
£12 (
+
RI
— C2
+
+
—
c3(1
cos (15—87)
Cos 'ic — 5jfl
=
CZS1flGCS1flOB
COS 'ic + c3(sin 0B — sin 'ic)
+
2=
6c COS 0J3 + C3 Sifl
j
R2
+ c2 sin
Sifl 0c cos 0B
)
R21
1=
C2
+ C2 Sifl °c
o5}Tc2
Sjfl 'ic + c3(cos
£12
—
COS
'ic)
)
SIfl Oc Sin
i-i—
E12
)[CIOC sin 'ic + c2 sin
sin
cos 'ic + c2 sin 0c cos
When the loading is symmetrical, one can utilize symmetry to determine the fixed end forces. The most convenient choice of unknowns is the internal forces at the midpoint, i.e., 6 = OB/2; F1 and M3 are unknown for the planar case and
only M2 is unknown for the out-of-plane case. Explicit expressions for the fixed end forces due to various loading conditions arc listed below. Planar Loading Fig. 15—10 defines the notation for the planar case.
We consider two loadings: a concentrated radial force P applied at C, and a uniform distributed radial load b2 applied per unit arc length over the entire segment. The basic frame is chosen to utilize symmetry. We determine the axial force and moment at C from the symmetry conditions u1 = w3 = 0. CASE 1—CONCENTRATED RADIAL FORCE P
THIN PLANAR CIRCULAR MEMBER
SEC. 15—10.
529
P
1;.fl
B!
MA
R
Fig. 15—10. Notation for planar loading.
1+Q2
SiflO! — CL
(1 + a1\ CL
I
cos cc) + —--—— SIfl 2 2
sin2 cc
+
—
I
(1 + a2\
sin
CL
cos
cc
=
'
— 131 —
—
=
RP ci —-i7771
Al
— COS
(
(15—88) —
— 77 —
+
Cl
1'
—
'A2 PR I
—
MA
1
— co.s
SIfl
+ 1,/I
(sin
COS
CASE 2—UNIFORM DISTRIBUTED RADIAL LOAD b2
=
0
—Rb2(1
sin
(1 + ai) 2
sin2 —
cc
cc
+
(1 + a2'\
sin cc cos
cc
ENGINEERING THEORY OF AN ARBITRARY MEMBER
530
(15—89)
Mc=R2b2aecb(_1 = =
=
—
—Rb2 Sill
CHAP. 15
a4)
C(
— cos
= —MA = R2b2aecb
Out-of-Plane Loading
Figure 15—11 defines the notation for the out-of-plane case. We consider four loadings: a concentrated force P, and a couple T—both applied at C; a uniform distributed force h3; and a uniform distributed couple in1. CD2
Tile bending moment at C is obtained using the symmetry condition = 0. F3
Mt
1
R
Fig. 15—li. Notation for out-of-plane loading. CASE 1—CONCENTRATED FORCE P
=
P
=0 PR 2
+ c3(l — cosa)
FLEXIBILITY MATRIX—CIRCULAR HELIX
SEC. 15—11.
PR
MB1 = MA1
-
=
—
--i-- (1
-
cos
—
531
cx)
PR.
=
—
----- sin
cx
=
FB =
CASE 2—CONCENTRATED TORQUE T
=
0
= T
=
c2sin2cx
2 cxc1 + c2 sin cx cos cx
(15—91)
=
FB=FA=0
-
CASE 3—UNIFORM DISTRIBUTED LOAD h3
0 —
= =
R 2 b3
c1(sincx
—
cx)+ c2 sin cx(i — coscx)+
+ = R2b3(sin
= =FA=—PRcx
cx —
—
cx
—
cxcoscx)
COS
C2
(15—92)
cos cx)
R2b3(cx sin cx —
1
+ cos cx)
CASE 4—UNIFORM DISTRIBUTED COUPLE m1
= =
0
cj(a — sin cx) + c2 sin cx(cos cx —1)
m1R——--——-—--
cxc1 + C2
SJfl cx cos cx
(15—93)
—m1R sin cx
MA2 = 15—11.
—
rn1R(1
—
cos
cx)
FLEXIBILITY MATRIX—CIRCULAR HELIX
In this section, we develop the flexibility matrix for a member whose centroidal
axis is a circular helix. The notation is shown in Fig. 15—12. The principal inertia direction, Y2, is considered to coincide with the normal direction, i.e., the inward radial direction, at each point. We also suppose the properties are constant. For convenience, we summarize the geometrical
CHAP. 15
ENGINEERING THEORY OF AN ARBITRARY MEMBER
532
relations:
x3 =
R cos 0
x2
R sin 0
x3=C8 dS
a dO
a=
[R2 + C2)112 = constant
=
+ Rcos 012 + Cl3)
fl
t2 = -—COS
b=
t3
sin 012
—
=
—
Ccos 072 + R13)
R.
a
Rc, = R,3 =
=
C.
C ——-cos 0
—sin 0 0
—
R(sin
0)
0 —
R
a
— C(OB —
= C(OB — 0)
IC
—cosO a
—---sinO
a
R(sin —
R(cos
sin 0)
—
R(cos
sin 0) —
cos 0)
0
The steps involve only algebraic operations and integration. We first determine using (15—66), then from (15—68), and finally with (15—70). In what follows, we assume the shear center coincides with the centroid and neglect
extensional and transverse shear deformation. With these restrictions, = = g12 = () 1
GJ
g2= El3
and the expressions for
reduce to
= 'I'12
'V22
The flexibility matrix for a constant cross section is given below. t See Examples 4—6 and 5—3.
SEC. 15—li.
FLEXIBILITY MATRIX—CIRCULAR HELIX
j
Centroidal axis
Yi
Xi", X', X31 —directions of basic frame Y2, Y3—principal inertia directions
Fig. 15—12. Notation for circular helix.
Notation—-Din,ensionless Para,neters
= C12
a3
=
R2E12 R2
+
a5
=
a7= a9
1+a1 a4+a6 2
a6+3a4 2
C2l2
C2 (E12\ 13
RC[12 Ra3
+
ij T
El2 El2
R2 [12
a6
=
a8= a10
1—a1
2
a6—3a4 2
533
ENGINEERING THEORY OF AN ARBITRARY MEMBER
534
CHAP. 15
Elements of f71
[fit
Sym
1
f22
I
f33j
= C2a( +
a
+ (1
R2a2cz 1
+ Sin208)
El3 f
a6
02B
a8
RCr1
f31 =
+
= R2a2cc 1
+El = El2 + a8 f33 =
R2a
—
/1
+
a10
COS
05 + COS OB)}
cos0,1)
a4 sin2 05 + a9OssinOn}
— 2a4OB
cos
2
sin 08) + a1005 COS
0B + a4 sin
{(ai + a5)05 —
—
COS
—
sin 08 Cos
+
sin
0B —
1 + cos
+ (a1 — a4)(i
C2x
OB}
2sinO8 +
—
0B + a4(OB
0B(08 sin
El3
—
+ 2a4 sin 08 + a10 sin
COS
2a1
a6 sin
sin
cos
(15—94)
Elements of 1161
[114
f261 Lf34
f35
f36j
sin 08 + a705 + a8 sin
f14 =
COS
02 8 + a8 sin2
f16 =
El2
—
(
f24 —
=
—
08}
(
El3
+ a8 sin2 0B f
Sffi 0B C05
<08
sin
a4(1 — cos
1 + cosoB}
PARTIAL END RESTRAINT
SEC. 15—12.
Ca3cJ
f26= =
535
.
sin OB
— 0B
=
a1(1 —
COS
cos
f36 = —
Elements
of
{a508 + a6 sin
El2
15—12.
—
Sym a
sin 2
=
COS 013j
cos08)
{a508
aa3
— a6
sin 08 cos OB} aa2
.
112
MEMBER FORCE-DISPLACEMENT RELATIONS—PARTIAL END RESTRAINT
In Sec. 15—7, we considered an arbitrary member which is completely restrained at both ends. This led to the definition of the member flexibility matrix and a set of equations relating the end forces and the end displacements. Now, when the member is only partially restrained, there is a reduction in the number of member force unknowns. For example, if there is no restraint against rotation at B, M8 = 0, and there are only a unknowns (where a is the order of F8), the rotation at B has no effect on the end forces. To handle the case of partial restraint, we first determine the compatibility equations corresponding to the reduced set of force unknowns. Inverting these equations and using the equilibrium relations for the end forces results in force-displacement relations which are consistent with the displacement releases.
Let Z denote the force redundants. Normally, one would work with the primary structure corresponding to Z = 0. However, suppose we first express the force at a point, say Q, in terms of the end forces at B, using, as a primary structure, the member cantilevered from A:
B
Next, using the primary system corresponding to Z = 0, we express
(a
in terms
of the applied external load and the force redundants:
= EZ + G
(15-95)
536
ENGINEERING THEORY OF AN ARBITRARY MEMBER
CHAP. 15
0) due to the applied external if Z contains only end forces at B. Now, the principle of virtual forces requires
The elements of G are the end forces at B (for Z
loads. Note that G =
0
=
+
TO/fl
+
for any self-equilibrating virtual-force system. Taking the system due to AZ
results in the compatibility equations for Z. It is convenient to work first with the virtual force system due to Equation (b) reduces to T(,fçfl +
T(O//,l
= ffl are the initial deformation and flexibility matrices for thefull end using (15—95), and requiring the resulting restraint case. Substituting for expression to he satisfied for arbitrary AZ, we obtain
where
(ETIftE)Z + ET(
+
=
=
(15—96)
are the displacements of the supports at B, A. It should be noted that We suppose Z is of order q x 1, i.e., there are q force redundants. Also, we let t be the row order of (and 0?t).
(F
(15—97)
= ([3 x 1)
With this notation,
E is i x q
Gisi x 1 and (15—96) represents q equations. For convenience, we let
(q x q)
=
+ rG
(15—98 )
1)
(1
and the member force-deformation relations take the form
trZ = —
—
ETJc)
n
z) Tcjjgn
'A
—
' O,Z
(
—
)
We refer to 1. as the reduced flexibility matrix since, in general, q < i. Actually,
is the flexibility matrix for Z and it is positive definite since E must be of rank
q, i.e., the force systems corresponding to the redundants must be linearly independent. Note that one can determine directly by working with the primary system corresponding to Z = 0. This is the normal approach. The approach that we have followed is convenient when the member flexibility matrix is known.
PARTIAL END RESTRAINT
SEC. 15—12.
537
At this point, we summarize the force-displacement relations for partial end restraint: Z = member force matrix
= EZ + G
0,z
= 0 — = reduced flexibility matrix (q x q) — 'I/'' — p
f.Z
,)
f,.
ETrE (15—100)
=
—
Note that, for complete end restraint, '7
—.
14'
B
G=0
15—101
=
We will use (15—100) in Chapter 17 when we develop the formulation for a member system. Continuing, we let
kr =
(15—102)
The force redundants are obtained by inverting (15—99):
Z=
(15-103)
—
—
Substituting for Z, the end forces at B are given by
= We defined
+G
—
—
(e)
as the effective member stiffness matrix: = EkYET (
LET
15—104)
In general, k is singular when q < i, since E is only of rank q. Equation (e)
takes the form
+
= = =
—
+G
(15—105)
+ (I, —
The end forces at A are determined from (a):
=
+
—
—
A,0
— ar"
'BA
(
iLl
Finally, we write the relations in the generalized form B,i
=
I +
I n'
RB
L
+
BA
'A
ENGINEERING THEORY OF AN
538
MEMBER
where
CHAP. 15
(15—107)
= k"AA
—
— ar"
—
n
BA BA
BA e
T BA
Comparing (15—107) with (15—53), the corresponding expressions for the complete restraint case, we see that one has only to replace by in the partitioned forms for and The equation for is different, however, due to the presence of the G term.
Example
15—12
Suppose there is no restraint and generate E, G with (15—95).
against rotation at B. Then,
0. We take Z
- Fl +G —
For this case, G =
0.
The reduced flexibility and stiffness matrices follow from (15—98),
(15—102),
= and the effective stiffness matrix follows from (15—104): rc". —
1
I
o
Finally, the force-displacement relations are (see (15—99)): rt,
ii
—
,i
it
vn, T
it
Note that premultiplication of
by Er eliminates 9", the relative rotation at B. There is no compatibility requirement for the end rotations in this case; i.e., the support rotation at B, which we have defined as does not introduce any member deformation.
REFERENCES I. 2. 3.
4. 5.
REISSNER, F.: "Variational Considerations for Elastic Beams and Shells." J. Eng. Mech. Div. A.S.C.E., Vol. 88, No. EMI, February 1962. HALL, A. S. and R. W. W000HEAD: Frame Analysis, Wiley, New York, 1967. GEaR, J. M. and W. WEAVER: Analysis of Framed Structures, Van Nostrand, New York, 1965. RUEINSTRJN, M. F.: Matrix Computer Analysis of Structures, Prentice-Hall, 1966. LIVESLEY, R. K.: Matrix Methods of Structural Analysis, Pcrgamon Press, London, 1964.
6.
DAEROWSKJ, R.: Gekriimmte dünnwandige Trüger (Curved thin-walled beams), Springer-Verlag, Berlin, 1968.
PROBLEMS
539
V. Z.: Thin Walled Elastic Beams, Israel Program for Scientific Transla-
7.
tions, OfiIce of Technical Services, U.S. Dept. of Commerce, Washington, D.C., 1961. 8.
BAZANT, Z. P.:
"Nonuniform Torsion of Thin-walled Bars of Variable Section,"
International Association for Bridge and Structural Engineering Publications, Zurich, Vol. 25, 1965, pp. 17—39.
PROBLEMS 15—1.
Refer to Example 15—5. Determine c, for a typical wide-flange section
and a square single cell. Comment on the relative importance of torsional deformation vs. bending deformation (i.e., terms involving in Equation (e)). Distinguish between deep and shallow members. 15—2. Refer to Example 15—7. Consider a rectangular cross section and
I/p j3\
varying linearly with x1, as shown in the sketch. Evaluate VB2 /(
/ \3E12J
vB3/
3E12
and
for a range of 4 and a/b. Prob. 15—2
x2
'2 b
x3
13
y3
15—3. Determine the reaction at B and translation (in the direction of F) at C for the member sketched. Neglect transverse shear deformation.
P
/
Prob. 15—3
Shear center
I— Vertical restraint at B
ENGINEERING THEORY OF AN ARBITRARY MEMBER
540
CHAP. 15
15—4. Repeat Prob. 15—3, considering complete fixity at B. Utilize symmetry with respect to point C. Derive (15—27). Start with the definitions for the strain measures 15—5.
(see Fig. 1 5—2),
(1 + r,)
DR
S[fl Y12 =
=
D15
Df)
DS
Dy2
E3f)
DS Dy2
neglect second-order terms, and note (15—26). 15—6.
Summarize the governing equations for restrained torsion. Evaluate
b2 and b3 (see (15—34)) for a symmetrical wide-flange section and a symmetrical
rectangular closed cell. Comment on whether one can neglect these terms. — 15—7. Refer to Example 15—11. Specialize the solution (Equations f) for = 1L 1. Verify that (g) reduces to the prismatic solution, (13—57), when
—*
0.
Consider a member comprising of three segments. Assuming the flexibility matrices for the segments are known, determine an expression for the member flexibility matrix in terms of the segmental flexibility matrices. 15—8.
Generalize for n segments. 15—9. Discuss how you would apply the numerical integration schemes described in Sec. 14—8 to evaluate defined by (15—69). 15—10. 15—11.
Verify (15—73) and (15—74).
Determine the fixed end forces for the member shown, using (15—77)
and (15—79). X2
15—12. 15—13.
Solve Prob. 15—3 using (15—84) and (15—87). Verify (15—90) and (15—91). Apply them to Prob. 15—4.
Prob. 15—11
PROBLEMS
541
15—14. Starting with (15—87), develop expressions for the initial deformations due to an aribitrary distributed loading, h3 = b3(O). Specialize for b3 =
constant and verify(15—92). 15—15.
Using the geometric relations and flexibility matrix for a circular
helix (constant cross section; Y2 coincides with the normal direction) developed in Sec. 15—11:
1)evelop a matrix equation for the displacements at B due to a loading referred to the global frame and applied at flint: See (15—85). (b) Evaluate for the loading and geometry shown. (a)
Prob. 15—15 Y3, b
I I
Y2,fl
P
——
— ir/2 c
G
=R/2 =E/2
x1 15—16. Determine the reduced member flexibility matrix for no restraint against rotation at an interior point P.
15—17.
For the planar member shown, determine E and G corresponding to
Z =
MB M4}
Then specialize for rotation releases at A, B and determine kg.
Part IV ANALYSiS OF A MEMBER SYSTEM
16
Direct Stiffness Method Linear System 1 6—i.
We
INTRODUCTION
consider a system comprised of in members which are connected at j
joints. We suppose the geometry of the assembled system is defined with
respect to a global framet and use a superscript o to indicate quantities referred to the global frame. The external force and displacement matrices for joint k are denoted by (p0)
(ax!) (16—1)
(xxi)
(jOt)
(fbi) where c is the number of translation (force) components, /1 is the number of rotation (moment) components, and i + fi. Note that = 2, /3 = 1 for a
planar system subjected to in-plane loading and 1, /3 = 2 for a planar /3 = 3. system subjected to out-of-plane loading. For an arbitrary system, In what follows, we assume the material is linearly elastic and the geometry is linear, i.e., we neglect the change in geometry due to deformation. The governing equations consist of joint force-equilibrium equations and member force-displacement relations. We have already developed the member forcedisplacement relations in Chapter 15, so that it remains only to establish the joint force-equilibrium equations. In this chapter, we apply the direct stiffness method, which consists in assembling the system stiffness and initial force matrices by superimposing the. contribution of each member. the next chapter, we present the general formulation for a linear member system and obtain the equations corresponding to the force and displacement solution by t By global
frame, we mean a fixed Cartesian frame.
545
546
DIRECT STIFFNESS METHOD—LINEAR SYSTEM
CHAP. 16
matrix operations. Finally, in Chapter 18, we extend the direct stiffness method
to include geometrical nonlinearity. 16—2.
MEMBER FORCE-DISPLACEMENT RELATIONS
In developing the relations between the end forces and end displacements for a member, we considered the member geometry and loading to be referred to a basic member frame (frame n) and used A, B to denote the negative and positive ends of the member. The general relations were written (see (15—107)) as
=
+ +
+ +
Note that (a) also applies when there is only partial end restraint or internal releases.
Now, we define n_ as the joints at the positive and negative ends of member n. Replacing B by n, n the member frame take the form
= •
+ +
+ +
where —
\T
'
We transform the force and displacement quantities from the member frame to the global frame for the system by applying
= =
(16—2)
to (b). This step is necessary since we are working with joint forces and displacements referred to the global frame. The final expressions arc:
=
+ +
=
+ +
(16—3)
where the global member stiffness and initial force matrices are generated with k
—
=
TI
n
rzou 16—4
Once the displacements are known, we evaluate using (16—3) and then transform to the member frame. Since the initial end force and stiffness matrices are generated in partitioned form, it is natural to express (16—4) in partitioned form. Using the notation
SYSTEM EQUILIBRIUM EQUATIONS
SEC. 16—3.
547
introduced in Section 1 5—8, we write (o
plion
=
ip
=
t.MoJ
(16—5)
(flxl)
()O,22
(Sxo)
Expanding (16—4) leads to on —
o
12 —
k 1
21 =
on,T fi
ii
11
L
00. 12
on, Ti n
—
0
n
— Don,
on on —
21
"(')(),22
on
o,T ( )( ), 12
(16—6)
P
t 0,1
= depends on is a natural property of the member whereas Note that the orientation of the member frame with respect to the global frame. The operations defined by (16--6) can be considered as the element matrix generation phase. The member force-displacement relations satisfy the equilibrium conditions
for the member and compatibility between the restrained end displacements and the corresponding joint displacements. Actually, the equilibrium condi= tions were used to determine Compatibility is satisfied by setting = When there is only partial restraint at an end, there will and be displacement discontinuities. For example, if there is a rotation release at the positive end, will not be equal to the end rotation matrix. We have treated'r partial end restraint by defining an effective member stiffness matrix k0. In the derivation of k0, we consider °1IA to be the displacements of the supports (i.e., the joints) and enforce continuity of only the restrained end displacements. 16—3.
SYSTEM EQIJILIBRIIJM EQUATIONS
The equilibrium equations for joint k arc obtained by summing the end
forces for the members incident on k:
t See Sec. 16—12.
DIRECT STIFFNESS METHOD—LINEAR SYSTEM
548
CHAP. 16
depends on and the displacements of those joints which are In general, connected to joint k. We define as the total (or system) external joint force and joint displacement matrices: (if x 1) (if <
=
16—7
and write the complete set of if joint force-equilibrium equations as
=
+
91'o
(16-8)
contains the joint forces required to equilibrate the initial end forces We have dropped the reference frame superscript for convenience. The most efficient way to assemble and is to work with submatrices of order 1, the natural partition size, and superimpose the contributions of each member which follow directly from (16—3). This operation requires no matrix multiplications. The terms due to member n are listed below. where
(Partitioned Form Is] x 1):
In
in row n÷ in row
(16—9)
111 X (Partitioned Form Isj x
oT
in row column n+ column n_ in row in row n_, column
1
inrown...,columnn_ Since 16—4.
is
symmetrical, only the upper or lower half has to be stored.
INTRODUCTION OF JOINT DISPLACEMENT RESTRAINTS
In this section, we extend the procedure described in Sec. 8—3 for introducing
joint translation restraints in the formulation for an ideal truss to an arbitrary member system. Actually, only the notation for the joint force and joint displacement matrices has to be changed. The governing equations are: =
—
= 1
1
=
+
=
(16—11)
+
The stiffness and initial force matrices are assembled using (16—9) and (16—10). It remains to introduce the prescribed external forces and displaceis prescribed, and we just add ment restraints. If joint q is unrestrained,
SEC. 16—4.
INTRODUCTION OF JOINT DISPLACEMENT RESTRAINTS
549
is unknown. We replace q is completely restrained, to — the matrix equation for with the matrix identity, Finally, ifjoint q is partially restrained, some of the elements in 19q are unknown.
In this case, we replace the scalar equations for the unknown reactions by scalar identities.
We suppose joint q is partially restrained and, for generality, consider the translation and rotation restraint directions to he arbitrarily orientated with respect to the basic frame. We define X'1, .. , as the orthogonal directions .
for the translational restraint frame and
as the orthogonal directions
for the rotational restraint frame. Quantities referred to the restraint frames are indicated with primes and a single superscript is used for the total matrix: =
(flxj)
L°"i
Now,
(16—12)
P,q = T" —
(16—13 /3
q
We define 9/°" as the total rotation transformation matrix: 1
0
=
I
I
L0
(16—14)
With this notation, the transformation laws take the form 9/oqçpr (
The modification requires two operations. First, we transform (16—11) to This is accomplished by premultiplying row q of
with
16—15 ) in PPO
and postmultiplying column q of with 9/°" In the second step, we replace the equations corresponding to the unknown elements in with identities. This operation can also be represented in matrix form. Suppose the rth element in is prescribed. We assemble four matrices, Eq, and as follows: 1.
EqandGq
We start with
E=I,
G=O,
and set
+1
CHAP. 16
DIRECT STIFFNESS METHOID—.-LtNEAR SYSTEM
550
2.
We start with an ith-order column matrix having zero elements and set the element in row r equal to the prescribed displacement.
We
start with an ith-order column matrix having zero elements and enter
the values of the prescribed forces and moments referred to the restraint frames. Note that element r is zero. Premultiplying transformed row q of 3r, t?P0 with Eq reduces the rth equation to 0 = 0. Then, adding Gq to Eqirqq and to — + q introduces the identity for the rth element and includes the prescribed external forces We also operate on the qth column of in to preserve syrnnwtry and include the terms due to prescribed displacements in The complete set of operations for joint q are listed below: 1.
€=1,2,...,q—l T)a/e;
T)F
2.
9N, q
+
q
= 3.
+
T)Eq + Gq
(l616)
€=q+i,q+2,...,j = =
—
The operations defined by (16—16) are carried out for each joint, working with successive joint members. We represent the modified equations as
=
(16—17)
The superscript J is placed on % to indicate that the joint displacement matrices are referred to the local joint restraint frames, which may not coincide with the global frame. Again we point out that the primary advantage of this modifica-
tion procedure is that no row or column rearrangement is required. Solving (16—17) yields the joint displacements (local restraint frame) listed in their natural order, i.e., according to increasing joint number. The modified stiffness matrix,
will be positive definite when the system is stable.
Once QgJ is known, we transform the displacements from the restraint frames
to the global frame, using (16—15), and evaluate the member end forces from (16—3). Next, we the total external force matrix, The contribution
SEC. 16—4.
INTRODUCTION OF JOINT DISPLACEMENT RESTRAINTS
of member n
551
is
(16—IS)
in row fl...
Finally, we transform the external joint forces from the global frame to the local restraint frames. This step determines the reactions and also provides a statics check on the solution. Example Suppose
16—1
joint q is completely restrained. Then, "li
for E, G are
and
= 0.. The forms
Gqlj
EqOi and (16—16) reduces to
1.
€=L2,...,q—l —
—
,t,. q
2.
j
3.
Oi
= Example 16—2 Suppose
joint q is completely restrained against translation. Then, the translation are prescribed. The appropriate matrices for
matrix and external moment matrix this case are
r0
r1
01 Gg
=
=
}
= Example 16—3 We consider the case where joint q is restrained with respect to translation in one direc-
tion and there is no restraint against rotation. This corresponds to a "roller" support. We take to coincide with the restraint direction and X'2, as mutually orthogonal directions comprising a right-handed system. The translation, is prescribed. The prescribed forces are P52,
and
552
DIRECT STIFFNESS METHOD—LINEAR SYSTEM
We first assemble
CHAP. 16
From (16—14),
= where
rRoa' I
[
13
r,s
=
1,2,3
= The forms of E, G,
are
and
0
=
=
We specialize the results for a planar system subjected to planar loading. In order for only planar deformation to occur, the translation restraint direction must lie in the plane plane. It is convenient to select the orientation of the system, which we take as the of X'2 such that X3 coincides with X",. 'Ihe specialized forms are
=
r,s =
= [S,.,] 0 Eq
1,2
1
=
Gq
(d)
= ---H 0
Finally, we consider the case of a planar system subjected to an out-of-plane loading. direction in order for only The translational restraint direction must be parallel to the and arc prescribed. The out-of-plane deformation to occur. For this case, specialized forms are Cq
Eq
=
=
qe;
= Note that (e) is obtained by settinge =
= 1,
=
2
in (a) of Example 16—2.
REFEREt'JCES
REFERENCES R. K.: Matrix Methods of Structural Analysis, Pergamon Press, London,
1.
1964. 2. 3.
4.
MARTIN, H. C.: Introduction to Matrix Methods of StructuralAnalysis, McGraw-Hill, New York, 1965. RUBINSTEIN, M. F.: Matrix computer Analysis of Structures, Prentice-Hall, New York, 1966.
Gere, I. M. and W. Weaver: Analysis of Framed Structures, Van Nostrand, New York, 1965.
17
General Formulation Linear System 17-1.
INTRODUCTION
We consider a system comprising m linear elastic members interconnected at j joints. We suppose there are i degrees of freedom per joint (i.e., the joint displacement and force matrices are of order i x 1) and the geometry and joint quantities are referred to a global frame. Also, we neglect geometry change due to deformation. In the previous chapter, we applied the direct stillness method, which is actually a displacement method, to this system. Now, in this chapter, we first develop the governing matrix equations and then deduce the equations corresponding to the force and displacement solution procedures.
We also establish variational principles for the force and displacement methods. Finally, we discuss how one can introduce member deformation constraints in the displacement method. Since the basic steps involved in the member system formulation are the same as for the ideal truss formulation, we recommend that
the reader review Chapters 6 through 9 before starting this chapter. Let r be the number of prescribed joint displacements. Then, the total number is of joint displacement unknowns, na = if
—
r
(17—1)
The total number of force unknowns, flf, is equal to r (the reactions corresponding to the prescribed displacements) plus qT, the total number of member force unknowns:
nf=?+qT=r+(qj +q2+ '.. +qm)
(17—2)
represents the number of force unknowns for member n. By definition, is equal to the number of force quantities that have to be specified in order to be able to determine the total internal force matrix at an arbitrary point. If the member is fully restrained at each end, q1, = i. For partial restraint, q,, is equal to i minus the number of independent force releases. Note that when the member is pinned at both ends, = 1 since there are only five independent moment releases. where
554
MEMBER EQUATIONS
SEC. 17—2.
555
There are qT equations relating the member forces and the joint displacements.
Also, there are ij equilibrium equations relating the external joint forces and the member forces. The formulation is consistent, i.e., the number of equations is equal to the number of unknowns. If flf if, the system is said to be statically determinate since the force unknowns can be determined using only the equi-
librium equations. The difference, flf — ii, is generally called the degree of static indeterminacy, and represents the order of the final system of equations for the force method. For the displacement method, the final system ofequations arc of order In what follows, we first establish the member force—joint displacement relations by generalizing the results of Sec. 15—12. Then, we assemble the joint force-equilibrium equations. Finally, we introduce the joint displacement restraints. 17—2.
MEMBER EQUATIONS
The reduced member equations were developed in Sec. 15—12. For convenience, we summarize the notation and equations below (see (15—100)):
Z=
member force matrix (q, x 1)
(ixi)
—
member flexibility matrix (i x i) = ETfnE reduced member flexibility matrix (q,, x f, — = member deformation matrix (i x i) = + f"G = initial member deformation matrix (i x i) = = — )
f"
These equations include the effect of partial end restraint, internal force releases,
and reductions due to symmetry or antisymmetry. We can also use (a) for complete end restraint by setting F = and G = 0. Now, we introduce new notation which is more convenient. First, we note that G contains the end forces at B due to the external member loads acting are the on the primary structure defined by Z = 0. Also, — — end forces at A. Then we write
=G
(17—3) —
is a compatibility Next, we note that the equation relating Z and K", requirement. The term fZ + ETeC?OZ is the relative deformation in the positive sense of Z due to the member loads and the member redundants, Z, whereas is the relative deformation in the negative sense of Z due to support (joint) movement. The net relative deformation must be zero for continuity.
GENERAL FORMULATION—LINEAR SYSTEM
556
CHAP, 17
Then, we define
= reduced member deformation matrix (q,, x 1) = = — reduced initial member deformation matrix x 1) = ETI/'O z = + f"G)
17—4
With this notation, the member equations take the form — B
R,o
(175) — + frZ = We generalize the relations for member n by setting
A=n_
B—n÷
Z=Z.
E=E,, =
(17-6)
"r,
n
= fr.
the joint quantities are referred to the global frame, we must transform the end forces and displacements from the member frame (frame n) to the Since
global frame (frame o), using
= The final equations follow. Member Forces—End Forces
= =
TE)Z +
17 7
+
Member Forces—Joint Displacements (q,, ,t
n+ (
—
17—8)
The force translation transformation matrix, 2C, is a second-order tensor, i.e., it transforms according tot
q
and it T T
t See Sec. 10—2.
SYSTEM FORCE-DISPLACEMENT RELATtONS
SEC. 17—3.
and 'r.
Using (17—JO), we can express
557
as.
TF)Z
—
=
.
—
since it is a natural property of the member whereas We prefer to work with depends on the selection of the global frame.
SYSTEM FORCE-DISPLACEMENT RELATIONS
17—3.
force-deformation relations for member n. Equation (17—8) represents the By defining general flexibility and deformation matrices, we can express the member force-deformation relations as a single matrix complete set of equation. We let
Z
x 1)
total member force matrix = {Z1, Z2, , Z,,,} .
.
.
x 1) = total reduced member deformation matrix = 2 1, m} = total reduced initial member deformation matrix (q,. x 1) —
ro,
I
f
,
'17—12
ro, mJ
total reduced member flexibility matrix
x
fr, 2
Note that f is quasi-diagonal, symmetrical, and positive definite. With this force-deformation relations are given by notation, the -V =
-V,,
+ fZ
(17—13)
It remains to generalize the deformation-displacement relations. We define as the total joint displacement matrix referred to the global frame. (ij x 1)
=
°/4,
.
. .
(17—14)
,
and express 'V as
=
(17—15)
The partitioned form is
d12
...
-
1
(17-16)
558
GENERAL FORMULATION—LINEAR SYSTEM
CHAP. 17
Row n of d corresponds to member n. The submatrices in row n are of order
x i. Now, we see from (17—8) that there are only two non-zero elements in row n and they are at columns n_. The assembly of d is defined by
d 17
= sins = 0
17
fl_
S
ii =
1,
2,
. .
.
, in
It is of interest to express si in factored form. First, we define the following matrices: (inz x 1)
411
,..., 411,,,_ }
_,
=
(im
x 1)
(mi x un)
E
E=
(im x q1)
([7—18)
Em
(zinxirn)
(un
x im)
Using this notation,the expression for "V takes the form
= =
ET9II(Q71+ —
(a)
T411)
using member-joint connectivity matrices for and negative (C_) ends:
Next, we relate 411÷, 411_ to
the positive
411÷
C÷%
17
- 19
SYSTEM EQU)L)BRIUM EQUATIONS
SEC. 17—4.
559
Note that rows a of C ÷, C correspond to member a. There is only one nonzero —
element in a row. For row n, we enter n
in column n÷ of of C_. Finally, combining (a) and (17—19), we have
= and it follows that
and column
17 20
—
d. = =
—
For an ideal truss, (17—21),reduces to (see Equation 6—28)
=
—
where
contains the direction cosines for the bars.
17—4.
SYSTEM EQUILIBRIUM EQUATIONS
We have used the member force-equilibrium equations in developing the member force-displacement relations, so it remains only to satisfy equilibrium of the joints. There arc i equations for each joint, and a total of if equations. The expressions for the end forces in terms of the member forces are given by (17—7). Assembling the joint force.equilihrium equations involves only summing at each joint the end forces incident on the joint. as the total external joint force matrix referred to the global We define frame: (17—22) = 1)
and
as the initial (Z = 0) joint force matrix: (ijx j)
=
1'
(17—23)
2
are the joint forces due to external forces acting on the We express the complete set of equations as
The elements of
members with Z
0.
+ Z1
p1,1
,
P1 =
Z2
+
Zm
I
We assemble
and
(17-24)
working with successive members. The contribution
of member a follows from (17—7):
in row
GENERAL FORMULATION—LINEAR SYSTEM
CHAP. 17
Column n TE n_n —
p171sn
= =
S
s=
Tdyn
— —
(17—26)
TE
0
fl+,
Comparing (17—26) with (17—17), we see that
=
(17—27)
We let ——
1'
1,,
2,
(17—28)
= Then, we can express
as
+ C!1..O
= 17—5.
(17—29)
INTRODUCTION OF JOINT DISPLACEMENT RESTRAINTS; GOVERNING EQUATIONS
The governing equations for the unrestrained system are
(17-30)
Now, we suppose r joint displacements are prescribed. We rearrange and that the prescribed displacements arc last. We also rearrange -+
—*
cu1) (U2)
U = P
so
(rxl)
=
(17—31)
Pf (r
x
where U2, P1, and P1 are prescribed. We use B, A to represent the rearranged
forms
d:
x r)
(qr
A2]
[liii [ I
B=
_J
=
[An [ A2
(17—32)
j
fr'qr)
JOINT DISPLACEMENT RESTRAINTS
SEC. 17—5.
561
Finally, we write the equations for the restrai;ied system as
Pi =
+ B1Z P1 + AfZ P2 = 2 + B2Z P1, 2 + = fZ + 1/,, A1U1 + A2U2
(na eqs.) (r eqs.)
(17—33) (17—34)
17 35
-
The unknowns are the
member forces (Z), the
displacements (U1), and
the r reactions (P2).
If the restraints are parallel to the directions of the global frame, the transformation of d to A (or to B) involve only a permutation of the columns of (rows The same permutation is applied to the rows Suppose joint q is partially restrained and the restraint directions do not coincide with the global frame directions. We first transform the force and dis-
d
placement matrices for joint q from the global frame to the restraint frame, using A
O/Iq
(17—36)
= row q of
by
T
and premultiplying We write the transformed equations as
This step involves postmultiplying column q of d by
= J)1 + 4L
(1737)
+ fZ =
where the superscript J indicates that joint forces and displacements are referred
to local restraint frames. The final equations are obtained by permuting the the rows of and then partitioning. columns of d2 (rows of The transformation of a71 to U can be expressed as a matrix product, U = DQ/ = where
(17—38)
contains the rotation matrices for the joint restraint frames,
=
...
(17—39)
and H is the row permutation matrix. One can generate H by starting with I and permuting the rows according to the new listing of the joint displacements, i.e., with the prescribed displacements last. Now, D is an orthogonal matrix, (17—40)
Then,
= DTU
P=
GENERAL FORMULATION—LINEAR SYSTEM
562
CHAP. 17
and it follows that
P, — A = B
(17—41)
AT =
The partitioned forms are obtained by partitioning D:
D=
[D 1
(n
LD2J
(r
x (j) .,
(17—42)
Finally, we can write
A1 = A2 = P1 = 2 =
= BT
= Bf
(17-43)
1
To determine the requirement for initial stability, wc consider (17—33),
B1Z =
P2
-.
which represents nj equations in unknowns. For the equations to he consistent for an arbitrary loading, the rank of 'B1 must equal n,,. Therefore, the stability requirement for the system is r(B1) = ;(A1)
Since B1 is of order
=
(17—44)
a necessary hut not sufficient condition for stability
x
(17—45)
Equation (17—44) is the stability requirement for a geometrically linear system. It is also the initial stability requirement for a geometrically nonlinear system.
In the next chapter, we develop the stability criteria for a geometrically nonlinear system subjected to a finite loading. 17—6.
NETWORK FORMULATION
In the formulation presented in the previous articles, we worked with the actual joint displacements and external joint forces referred to the global frame.
The governing equations are given by (17—30), which we list below for convenience:
=
+
where
Tc)
—
=
+
One assembles d, 9PI, using(17—17), (17—25), which are actually the expansions
of(b). By introducing new joint variables, we can express d in terms of only one
NETWORK FORMULATION
SEC. 17—6.
563
C. The rule (17—17) for assembling d still applies except that now Let Y denote some arbitrary point. Suppose we express the actual force and
connectivity matrix, C.4 —
displacement matrices for joint k in terms of their equivalents at point Y. We define
= statically equivalent force at Y due to the actual force matrix at joint k. (1746) = displacement at Y due to rigid body motion about joint k. The actual and equivalent quantities are related by r)r°
—
kY
Y, k
k
(17—47) k
where 0 kY
=
I"
0
—
—
0 Xy2;
4i)
plane
(17-48)
0
— Xy1
t planar
We could operate on (b), but it is more convenient to start with (17—11): —
'Kr,, = Now, by definition,
= Substituting for
using (17—47), and noting that
we obtain 'Kr,,, =
(17—48)
The remaining steps are the same as followed previously. We let 'Wy = {0/t°y,
1,
'Wy
2
(17—49)
and write 'K =
The generation of
(17—50)
follows from (17—48). For now n,
= =
0
= (q,, x 1)
S
s=
1, 2
j
(17—51)
564
To express
GENERAL FORMULATION—UNEAR SYSTEM
CHAP. 17
in factored form, we let 1+Y
=
(irn
x irn)
(17—52)
+ Y
Then, C_)
=
(17—53)
We transform the joint forces, using (17—47), and write the resulting equations
as
+
= =
(17-54)
+ fZ =
To relate corresponding terms in (a) and (17—54'). we generalize (17—47): (17—55)
=
(if x i/) •
It follows that
,,
—
g,YI =
jY
r (17—56)
reduces to (17--53) when (d) and (c) are introduced. The formulation developed above can be interpreted as a network formulation since the connectivity term appears seperately in the factored form of d. A simplified version which does not allow for member force releases has been presented by Fenves and Branin (see Ref. 1). The only operational advantage of not working with the actual joint quantities is in the generation øf d,m÷ and d,,,. This advantage is trivial compared to the additional operations required ., to introduce the displacement restraints, and finally, to to generate to once the solution is obtained. Another serious disadvantage transform is that the equations tend to become ill-condii.ioned. The expression for
Fenves and Branin's primary objective was to show that the governing equations for a member system can be cast in a form such that geometrical and topological effects are separated, i.e., a network formulation. DiMaggio and Spillars (Ref. 2) have also presented a network formulation for a rigid jointed member system. Actually their formulation is a special case of our first formulation. It is not, strictly speaking, a true network formulation since connectivity is not completely separated from geometry (see (17—21)). The only way that one can separate connectivity from geometry is to redefine the joint variables. Note
565
DISPLACEMENT METHOD
SEC. 17—7.
that the ideal truss is an exception. Connectivity and geometry are naturally
uncoupled for this system. Whether one interprets the governing equations for a member system from a network viewpoint is of academic interest only. In the displacement method,
the equations reduce to the equations for the direct stiffness method. The only possible advantage of the network interpretation is in the force method. There one can use certain concepts of the mesh methodt to select a primary structure, provided that there are no member force releases or partial joint restraints. However, the selection of a primary structure for a rigid-jointed frame having fixed supports is quite simple, and even this advantage is debatable. 17—7.
DISPLACEMENT METHOD
The governing equations are given by (17—33), (17—34), and (17—35). Once the member forces are known, we can find the reactions from (17—34). Now, we start by solving (17—35) for Z in ternis of the displacements,
Z = Z1 + kA1UE + kA2U2
17-57)
where
= initial member force matrix (q1 x 1)
= k=
(17—58)
= reduced member stiffness matrix
x
Note that k is quasi-diagonal, symmetrical, and positive definite. The matrix,
Z1, contains the initial member forces due to external loads acting on the members and initial deformation resulting from fabrication errors or temperature changes. We substitute for Z in (17—33) and write the result as
=
P0
+
+ K12U2
(17—59)
where
= K12 = ATkA2 = +
(nd x lid)
x r) (lid
X 1)
(17—60)
The elements of P0 are the joint forces due to the initial end forces. Since A1 is of rank (when the system is stable) and k is positive definite, it f See Sec. 9—5. See Prob. 2—18.
566
GENERAL FORMULATION—LINEAR SYSTEM
CHAP. 17
that K1.1 is positive definite. Conversely, jfK1 is not positive definite, the system is unstable. The joint displacements are determined by solving (17—59) and the member forces are obtained by back substitution in (17—57). Operating on the restrained equations, as we have done above, is not efficient
since the various coefficient matrices must be generated by matrix multiplication. By first manipulating the unrestrained equations and then introducing
the displacement restraints, one can avoid any matrix multiplication. This procedure corresponds to the direct stiffness method. Operating on (17—30), we obtain Z = Z1
and
+ kd°ll
(17-61)
+ c/TZ +
= =
(17—62)
+
Equation (17—62) is identical to (16—8). The of (16—9), (16—10) when we introduce the factored forms of
reduces to
d, Z1.
First, we review the definitions of the member stiffness matrices, The effective member stiffness matrix (see (16—104)) has been defined as k to the global frame and applying (16- 107) reads to T n e.H
ke.n
= and L0 .._1O fl+n+ — e,fl
takes the form
Now, substituting for d using (17—21), the expression for
=
—
where
=
(17—63)
Finally, we expand (d):
= +
+
T)C
+
—
One can easily show that (17—64) reduces to (16—10) when the properties of
C... are taken into account. The initial end actions for member n
=
+ T, ,,
t See Eqs. (17—7), (17—8), and (17—il).
TE)( — kr, n'17ro. n) TEn(_kr.
n)
SEC. 17—8.
FORCE METHOD
=
+
+ The general form
567
d, and 4, the expression for
Using the factored forms for form
takes the
(17—65)
—
defined according to (16—9) is
=
+
(17—66)
Substituting (e) in (17—65) results in (17—66).
In Sec. 16—4, we presented a procedure for introducing joint displacement restraints and represented the modified equations as
=
(ii eqs.)
Now, (f) consists of (17—59) plus r relations for the prescribed displacements. We obtain (f) by starting with
=
[0
IrJ
U2
L
and permuting the rows and columns. This operation can be represented in terms of the permutation matrix, 11, defined by
u=[IdllJ = HTP Then,
= is positive definite when K11 is positive definite, i.e., when It follows that the system is stable.
17—8.
FORCE METHOD
We start with the governing equations for the restrained system:
B1Z =
P1,
P1
= 'V0 + fZ
BfU1 + =
2
+ 82Z
(fld eqs.)
(a)
(qr eqs.) (r eqs.)
(b) (c)
unknowns where Also, equations in Equation (a) represents The system is statically determinate when n4. We let B1 is of rank be the degree of static indeterminacy, i.e., the number of member force redundants: (17—67) = —
568
GENERAL FORMULATION—LINEAR SYSTEM
CHAP. 17
we can solve (a) for member forces in terms of the net prescribed joint forces (P1 — P1 i) and r member forces. The compatibility Since B1 is of rank
equations for the member force redundants are obtained by eliminating U1 from (b). This is possible since (b) represents qT equations whereas U1 is only x 1. In the next section, we specialize the principle of virtual forces of order for a member system and utilize it to establish the compatibility equations. We suppose the first 0d columns of B1 are linearly independent. If the system is initially stable, the member force matrix Z can always be rearranged so that this condition is satisfied. We partition Z after row
1z,)
Z
(17—68)
(ZR)
1)
1)
The elements of ZR are the force redundants for the system. We refer to the system obtained by setting ZR = 0 as the primary system. Continuing, we partition B1 and B2 consistent with (17—68): (fld 0 qj-)
x fld)
("s
=[
B1
'Jo,)
BIR
B2
69
17 —
(r
(i
0
(r °qo,)
"d>
The equilibrium equations take the form
=
P1
—
P1,2 +
BIRZR + B2RZR
(17—70)
KZR
(17—72)
(17—71)
We write the solution of (17—70) as
+
ZP
The force influence matrices can be expressed as ("a
1)
Zr,,, =
(B1pY'(P1
—
ZPR = Actually, the solution procedure can be completely automated.( The complete solution for is
but it is not necessary to determine
xqo,)
z
(17-74)
Note that the member forces due to B1Z = 0. Finally, we substitute for
are self-equilibrating, i.e., they satisfy in the expression for P2 and write the
result as
P2 = t See Sec. 9—2.
P2,0
+
P2 RZR
(17—75)
FORCE METHOD
SEC. 17—8.
where
569
(rXI)
=
P1,2 + J37PZP,0
'17—76)
B2R + B2pZp,ft (r X
It remains to determine Zft. Equation (b) represents qr equations in unknowns, U1. Since + = qft, there are excess equations. We partition (b) consistent with the partioning of Z,
rDT 1
rDT 1 1
+
I.e'
1
1
1
2
+
—
*ftftJ o,R) L PR V R) and obtain the following two sets of equations relating to U1 and ZR:
L"2RJ
BfPU1 +
'V'p
=
BfftU, +
+ = 1'O.J( +
+ + fRRZR
ft
(fld eqs.)
(17—77)
eqs.)
(d)
The joint displacements can he determined from (17—77) once Zft is known.
Eliminating U1 from (d) leads to RU2
I/R + ZP +
+
+ fftftZft ±
+ fpftZR)
17—78
Equation (17—78) represents the compatibility equations for the force redun-
dants. Finally, we substitute for equations as
using (17—72) and write the resulting
fZRZR = A
(17—79)
where >< qg)
—f
— 'RR
7T c
7T 4• 7 1-q', R'PP'—P. ft
17T —
A
=
+
P2, ftU2 —
o) —
p
+
These equations are similar in form to the corresponding equations for the ideal truss developed in Sec. 9—2. The flexibility matrix, fzR, can be expressed as
lil [Zp
= —
[ZP
'RRJ J [zp,R1T f[ZP.R1
j
ft
(17—81)
j
Now, f is positive definite for a deformable system. Then, it follows that fzR is also positive definite. In a later article, we consider the case where certain member deformations may be prescribed. Once the preliminary force analyses have been carried out, the remaining steps are straightforward. We generate A, solve for Zft. and then determine
GENERAL FORMULATION—LINEAR SYSTEM
570
Zr,,
CHAP. 17
P2 by back substitution. If the displacements are also desired, they can be
determined by solving (17—77).
The final number of equations for the force method is usually smaller than for the displacement method VS. lid). However, the force method requires considerably more operations to generate the equations. The force method can be completely automated, but not as conveniently as the direct stiffness method. Also, automating the preliminary force analyses requires solving an additional set of nd equations. Another disadvantage of the force method is that the compatibility equations tend to he ill-conditioned unless one is careful in selecting force redundants. 17—9.
VARIATIONAL PRINCIPLES
In Chapter 7, we developed variational principles for the displacement and force formulations for an ideal truss, Now, in this section, we develop the corresponding variational principles for a member system. The extension is quite straightforward since the governing equations are almost identical in form.
We start with the force-equilibrium equations,t
P = P, + ATZ The partitioned form is
= P,,1 + AfZ = P1,2 +
To interpret (a) as a stationary rcquircment, we consider the deformationdisplacement relation,
"K = AU = A1U1+ A2U2 The first differential of "K due to an increment in U is
d"K = A AU =
A1
AU1 + A2 AU,
(17-82)
Then, the requirement that
PTAU =
+ ZTd.K
(17-83)
be satisfied for arbitrary AU is equivalent to (a). If we consider U, to be prescribed, (17—83) results in only (b). We refer to (17—83) as the principle of virtual
displacements for a member system. In the displacement method, we substitute for Z in the joint force-equilibrium equations, using — "1/',,) Z= = (AU —
form of (17—83) suggests that we define a scalar quantity, V = V(U), having the property dV = = dV(U) (17—84) The
t We work with the governing equations for the restrained system. See (17—33), (17—34), (17—35).
VARIATIONAL PRINCIPLES
SEC. 17—9.
One
571
can interpret V as the strain energy function for the members. For the
linear case, V ôan be expressed as
V= =
—
17 85
—
—
Continuing, we define the potential energy function,
V+
—
as
PTU
(17-86)
The Euler equations for H,, are the unpartitioned joint force-equilibrium equations expressed in terms of U. Finally, we introduce the joint displacement constraint condition, U2 =02, by writing (17—86) as
H,, =
V
+
1U1 +
—
—
—
(17-87)
U2)
where U1, U2, and P2 are variables. The Euler equations for (17—87) are the partitioned equilibrium equations (Equations (h), (c)) expressed in terms of the displacements with U2 set equal to 02, i.e., they are the governing equations for the displacement formulation presented in Sec. 17—7.
If only the equations for P1 are desired, we set U2 = 02 in (17—87),
=
V
+
(17—88)
1U1 —
where
V=
+ A202 —
+ A202 —
(17—89)
The Euler equation for (17—88) is (17—59), and the second differential has the form
i2rr — /tITTIATI A \AIT — 1)
'4
(17—90)
= AUrK11 AU1
Since K11 is positive definite, we can state that the displacements defining the equilibrium position correspond to a minimum value of defined by (17—88) or (17—87).
We consider next the force-method formulation. We let AP, AZ be a statically permissible virtual-force system. By definition,
= ATAZ = BAZ
(17-91)
Premultiplying both sides of (d) with AZT and introducing (17—91) leads to the principle of virtual forces, APTU = (17-92) Note that (17—92) is valid only for a statically permissible virtual-force system, i.e., one which satisfies (17—91).
The compatibility equations follow directly from the principle of virtual forces by requiring the virtual-force system to be self-equilibrating. If AZ satisfies
then (17--92) reduces to
-
AP1
=
B1
=
AZ =
0
(17-93) (17-94)
CHAP. 17
GENERAL FORMULATION—LiNEAR SYSTEM
572
This result is valid for an arbitrary self-equilibrating virtual-force system. The
formulation presented in the previous section corresponds to taking AZ
[ZP,Rl
=
A ZR
(17-95)
AP2 = We define the member complementary energy function, V's' = V*(Z), such that dV* = (17—96) For the linear case, and
= 4ZTfZ + ZT.K0
(17—97)
We also define the total complementary energy function,
=
as (17—93)
—
The deformation compatibility equations, (17—94), can he interpreted as the stationary requirement for 11. subject to the following constraints on Z, P2:
= =
—
+ B2Z
The constraint conditions are the joint force-equilibrium equations. Operating on (g), and noting that P, 2 are prescribed, lead to the constraint conditions on the force variations B1 AZ = AP2 =
0 B2 AZ
Note that (h) require the virtual-force system to be statically permissible and self-equilibrating.
In the previous section, we expressed Z, P2 as Z
+
=
P20 +
P2RZR
This representation satisfies (g) and (h) identically for arbitrary AZR, Sub-
stituting for Z, P2 in (17-98) and expanding V* using (17-97), we obtain
=
+ —
ZR
+
ZR]
(17-99)
2. RU2 + const
The Euler equations for (17—99) are (17—79), and the second differential has the form
=
AZR
(17—100)
SEC. 17—10.
MEMBER DEFORMATION CONSTRAINTS
573
is positive definite, it follows that the true forces, i.e., the forces that satisfy compatibility as well as equilibrium, correspond to a minimum value Since
of
Instead of developing separate principles for the displacements and force redundants, we could have started with a general variational principle whose Euler equations are the complete set of governing equations. One can easily show that the stationary requirement for
rIR = ZT(8TU1 + IT
T
considering
1
— Pfu1
—
IT
(
—
7—10
Z, U1, U2, and P2 as variables, lead to the partitioned joint
force-equilibrium equations and the member force-joint displacement relations. This principle is a specialized form of Rcissner's principle. We obtain (17—87) from (17—101) by introducing the force-displacement
relations as a constraint condition on Z, k(BfU1 +
Z
=
—
—
and noting that, by definition,
ZT(BTIJ1 ±
—
V" = y
Introducing the joint force-equilibrium equations as constraint conditions reduces 11a to —11. as defined by (17—98). 17—10.
INTRODUCTION OF MEMBER DEFORMATION CONSTRAINTS
Suppose a member is assumed to he either completely or partially restrained with respect to deformation due to force. The rigidity assumption is introduced by setting the corresponding deformation parameters equal to zero in the local flexibility matrix, g. For example, if axial extension is to be neglected, we set 1/AE = 0. For complete rigidity, we set g Now, in what follows, we discuss the case where neglecting member deformation parameters causes the mmher flexibility matrix t. to be singular, This happens, for example, when axial extension is neglected for a straight member. The rank is decreasedt by I and the axial force-deformation relation degenerates to
=
—
= v,,
+
-
(a)
is the initial axial deformation due to temperature and fabrication error. Note that now the axial force has to be determined from the equilibrium equations. For rigidity, = 0, and the force-displacement relations where
(see (17—5)) degenerate to (b) See (16—75).
GENERAL FORMULATION—LINEAR SYSTEM
574
CHAP. 17
One can interpret (a), (b) as either member deformation constraints or as con-
straint conditions on the joint displacements. In general, the decrease in rank of the system flexibility matrix f is equal to the number of constraint conditions. We consider first the force method. The governing equations are given by fZrZR = A
(qR eqs.)
where
= "Zr
L
Suppose these are c deformation constraints. Then, f is of rank order to solve (c), must be nonsingular, i.e., it must be of rank requires qT —
In This
— C.
(17—402)
C
That is, there must be at least
unconstrained member deformations. This condition is necessary but not sufficient as we will illustrate below. Aside from insuring that the flexibility matrix is of rank there is no difliculty involved in introducing member deformation constraints in the force formulation. Example
17—1
Consider the ideal truss shown. For this system. qT
4
q4 = 2 We take the forces in bars 3, 4 as the redundants: cF1)
IF3
(F2J
ff4
Then,
Zr5[0ri
0 1
and
1
1
0
01 0
= We can specify that, at most, two bars are rigid. No difficulty is encountered if only one bar is rigid. However, we cannot specify that bars 1, 3 or 2, 4 are rigid.
We consider next the displacement formulation. Since f is singular, k does not exist and, therefore, we cannot invert the complete set of force-displacement
SEC. 17—10.
MEMBER DEFORMATION CONSTRAINTS
575 Fig. E17—1
0
0 relations, i.e., (17—57) are not applicable. In what follows, we first develop the
appropriate equations by manipulating the original set of governing equations. We then show how the equations can be deduced from the variational principle for displacements. The governing equations are
P1 =
P1
1
+ AfZ
eqs.)
+ fZ = A1U1 + A2U2
(qT eqs.)
Now, we suppose there are c deformation constraints and the elements of are listed such that the last c elements are the prescribed deformations. We partition '/7' and Z as follows:
=
1
z
(cx1)
(17—103)
=
contains the constrained member deformations and Z. the corresponding member forces. We use subscripts c, u to indicate quantities associated with the constrained and unconstrained deformations. Continuing, we partition
where
A
A1 (q-r
(cx I'd)
'ia) —
(cxr) •)
(qT
f
(17—104)
(qr—e)xi
05 (cx I)
xl)
=
fT
=
-
0 (cxc)
GENERAL FORMULATION—LINEAR SYSTEM
576
CHAP. 17
The deformation constraints are introduced by setting
= 0. Note that, in order for f to be singular, there must be no coupling between and i.e., f must have the form shown above. Using this notation, the governing equations take the form + AfCZC + + = + A2aU2 = A1,U1 + A2,02
P1
= =
(17—105)
(17106) (17—107)
Equation (17—107) represents c constraint conditions on the unknown joint displacements, U1. The rank of A1, is equal to the number of independent constraint equations. One can easily demonstrate that c independent constraint
conditions are required in order to be able to analyze the system for an arbitrary loading.
Example
17—2
Suppose we specify that bars 1, 3 are rigid for the system considered in Example 17—I.
The constraint equations arc (we take
= {e1, e1}) — U21 = = —u11 +
—
e3 =
For (a) to be consistent, we must have er,., + e30 =
+
U41
Even if(b) is satisfied, we cannot find the forces in bars 1, 3 due to Pi 1.
In what follows, we assume A1, is of rank c. We solve (17—106) for 4 and substitute in (17—105). This is permissible since is nonsingular. The resulting equations are
4= =
(17—108)
+
—
and
+ AfCZC = P1
A1,U1 =
—
—
—
—
A2,02
(17—109)
(17—110)
Now, the coefficient matrix, is nonsingular only when the structure obtained by deleting the restraint forces (4) is stable. By suitably redefining 4, we can transform (17—109) such that the coefficient matrix is always nonsingular. Suppose we write
Z.
where 4
represents
=4+ =4+
—
+ A2,02 —
the new force variable and
(17—ill)
is an arbitrary symmetrical
MEMBER DEFORMATION CONSTRAINTS
SEC. 17—10.
in (17—109), we obtain
positive definite matrix of order c. Substituting for
P1 =
P1,1
+
577
+ 1 ([A1,,
[k,,
T
By defining
=
1
rf
kcj =
L
rk L
(a)
A2,,1
11 (17-112)
and noting (17—104), we can write (a) as
Pi
+ (Afk'A1)U1 +
+ ATk'(A2tY2 —
Using the notation introduction in Sec. 17—7 (see (17—60)), we let ATk'A,,
Kr,
= =
(17—113)
±
Finally, the governing equations take the form Z
(17-114)
—
= = =
= K11U1 +
—
—
— K1202 = H1 = H2
(17—115)
(17—116)
Since A1 must he of rank for stability and we have required to be positive definite, it follows that K1 is positive definite. Also, the solution for U1 must satisfy (17—116) and we see from (17—111) that is equal to the actual constraint force matrix, for arbitrary k'.
with Z. deleted, have the same form as the The expressions for Z and unconstrained expressions (17—57) and (17—59). Now, it is not necessary to
rearrange Z such that the constraint forces arc last. One can work with the natural member force listing, Z = {Z1, .
.
.
, Z,,,}
and take arbitrary values for the member deformation parameters that are to be negelected. We obtain K11 and Fl1 by first generating using the
direct stiffness method and then deleting the rows and columns corresponding to the prescribed displacements. The constrained deformations, 1/v, can be listed arbitrarily. It is only necessary to specify the locations of the constraint forces (elements of in. the natural member force listing. Once the ments and constraint forces are known, we can determine the force matrix for member n by first evaluating (see (17—8) and (17—11))
Z =k'r, —
r, fl
/
(O/g° '.
n+
ro. ,,j —
" n
n -—
GENERAL FORMULATION—LINEAR SYSTEM
578
CHAP. 17
where k.,, is the modified stiffness matrix, and then adding the constraint forces in the appropriate locations. In what follows, we describe two procedures for
solving (17—115) and (17—116).
In the first method, we solve (17—115) for U1,
U1 =
(17—118)
—
and then substitute in (17—116),
=
— 112
(17—119)
The coefficient matrix for is positive definite since K1 is positive definite is of rank c. Note that, with this procedure, we must invert an ndth and order matrix and also solve a set of c equations. For the unconstrained case, we have to solve only equations. Example
17—3
We suppose bar 2 (Fig. E17—3) is rigid. The constraint equation is e2
= U, =
To simplify the example, we consider only the effect of joint forces. Using the notation
introduced above, the various matrices for this example are U1
{u1,u2}
P1
Pz}
4= 4=
= e1 e2
F1
F2
U2,
= =
c=1
11,1=2 (P1
k,,
are null matrices.) Fig. E17—3
0 Bar
We start by assembling A1,
= e2 1
is rigid
SEC. 17—10.
MEMBER DEFORMATION CONSTRAINTS
579
and then partition according to (17-104):
Ii 1]
Note that we cannot invert (17—109), since Af,,k,,A1,, is singular. Now, we assume an arbitrary value for the stiffness of bar 2, —
a is an arbitrary positive constant, and assemble K1 0
=
kJ
L
K11
i:
LU
k1•[I'
a 1
= Afk'A1 —
The governing equations (17—114), (17—115), and (17-116) reduce to
=
[u-.]
+
K11U1 +
(h)
=
P1
(i)
=
0
(j)
The solution follows from (17—118), (17—119). We just have to take
H2 =
H1 =
=
0
(k)
The inverse of K11 is
1 [1+2a —ii
+Ij
I
(1k1[-l
Then
.L[1
(1)
+11
= and (17—119) reduces to
=
ak1
ak1
F'2—p2—p1 Substituting for F'2 in (17—118), we obtain U1
= 2p' Id1
U2 = 0
Finally, we substitute for
U1, u2
in (h):
F1 = F2
=
F'2
=
P2 — P1
(n)
GENERAL FORMULATION—LINEAR SYSTEM
CHAP. 17
one can start with
Instead of first solving (17—115) for U1 in terms of (17—116),
=
=
—
H2
which represents c relations between the displacements. Since A is of rank c, we can express c displacements in terms of rid — c displacements, i.e., there are only nd — c independent displacements. We suppose the first c columns of are linearly independent. Since is of rank c, we can always permute the columns such that this requirement is satisfied. We let (17—120)
— c
and partition
U1: (cx 1)
U1
(nxl)
(cxnd)
(Cxc)
(17—121)
The elements of U are the independent displacements. By definition, is nonsingular. Then, solving (a) for the constrained displacements, we have
=
2U
—
(17—122)
Finally, we express U1 as
U1 = BU + H3
(17-123)
where a)
(cx
=
f fl
I
L
(axx)
I
-J
(cxl)
H3 = (
0
(ax!)
j
Note that B is of rank n and (17125)
0
H2 can be completely automated using the The generation of B, H3 from same procedure as employed in the force method to select the primary structure. We consider next the joint force-equilibrium equations, (17—115),
= H1
K11U1 +
(fld eqs.)
Substituting for U1 leads to (K11B)U +
=
-- K11H3
114
from (b) by premultiplying by BT and noting (17—125). The resulting system of n equations for U is We eliminate
(BTK11B)U = BTH4
(n eqs.)
(17—126)
Since B is of rank n, the coefficient matrix is positive definite. One can interpret
MEMBER DEFORMATION CONSTRAINTS
SEC. 17—10.
BTK1
581
1B as the reduced system stiffness matrix. We solve for U and then
evaluate U1 from (17—123). It remains to determine the restraint forces, We consider again Eq. (a). Assuming U.1 is known, we can write
=
=
K11U1
—
(n0 eqs.)
115
(17—127)
The matrix, 115, is the difference between the external applied force, P1, and the joint force due to member force with the constraint forces deleted, i.e., 115 =
— P1,
1
(c)
— ATZ
where (see (17—114))
Z
—
(d)
= k'(A1U1 + A2U2 —
We determine Z using the member force-displacement relations and assemble P1 + AfZ by the direct stiffness method. Now has c independent rows.
In determining B, we permuted the columns
such that the first c columns
are linearly independent. We apply the same permutation to (17—127) and partition after row C: [ATe.
i
(17—128)
H5 2
Considering the first c equations, we have (17—129)
1
Since
is nonsingular, we can solve (17—129) for
We obtain the final
member forces by adding the elements of Z defined by (17—114) and (d).
In this approach, we have to invert a matrix of order c and solve a system equations. Although the final number of equations is less than in the first approach, there is more preliminary computation (generation of B) and the procedure cannot be automated as easily. of no — c
Example
17—4
For this example (Fig. E17—4),
c—4
n=1
n42
021 —
e10 e,0
e3
U22 — 032
e30
e4
031
144j
e4,
(a)
The constraint conditions are e1
=
e2
U12
=
=
(b)
CHAP. 17
GENERAL FORMULATION—LINEAR SYSTEM
582
Note that (b) corresponds to (17—107). The form corresponding to (17—116) is
+1
1e10 +
U12(
+1
—1
Je,0 e30 + u32f
+1 +1
..e4,o + U41J U31
I
1-
A1,
U1
FL2
Columns 2,4, 5, and either I or 3 comprise a linearly independent set. Then, we can take either u1 or u21 as U. It is convenient to take U = u1 We permute the columns according to 1
1 —*5
2—.
1
3-.2 The rearranged form of U1 is
U1 = {u12,u21,u22u31 = (U, U}
u11}
We determine U,by applying (17—122). This step is simple for this example since
I.
Finally, we assemble U1 defined by (e) and then permute the rows to obtain the initial Fig. E17—4 2
x2
ears
1,2,3,4 are rigid
listing of U1. The final result is U11
+1 0
U21
= +1 {u11} +
0
e1,,,
-I—
1142
e2,0
U22
0
e30
u31
0
e4,0 + U4j
1'
B
+ 1132
I 113
MEMBER DEFORMATION CONSTRMNTS
SEC. 17—10.
583
The constraint forces are determined from (17—127), which for this example has the form
+1 +1
I
+1
[
F4
H5,5
I H,
I
I
=
permute the rows of (g) according to (d) and consider only the first four equations. The resulting equations correspond to (17—129). We
It is of interest to derive the equations for the constrained case by suitably specializing the variational principle for displacements. We start with the unconstrained form of
developed in Sec. 17—9,
=
V + !5L1U1 —
where
V= "K = A1U1 + A202
Now, the displacements are constrained by
=
+
= Then, V reduces to
)Tk('K —
V= +
We obtain the appropriate form of
by substituting for V using (d) and = 0:
introducing the constraint condition, "Kr —
=
V
+
—
+
—
(17-430)
The elements of 4 are Lagrange multipliers. One can easily show that the stationary requirement for (17—130) considering U1 and 4 as independent variables leads to (17—109) and (17—110). Since = v", we can add the term
'(1"'
—
to (d). Taking
V=
—
—
(17—131)
in (17—130) leads to (17—115) and (17—116).
In the second approach, we substitute
U1=BU+H3
(f)
GENERAL FORMULATION—LINEAR SYSTEM
584
CHAP. 17
in (a) and (17—131):
= V
V +
+ H3)
— —
—
(17—132)
A1BU + A1H3 + A2U2 The variation of
considering U as the independent variable is
= AUT[BT(P1
1
P1)
+
+ BTATkr(ASH3 + A2U2 —
=
—
(g)
BTH4]
Requiring to be stationary for arbitrary AU results in (.17—126). Note that we could have used the reduced form for V, i.e., equation (d). Also, we still have to determine the constraint forces. REFERENCES 1.
S. J., and F. H. BRANIN, JR., "Network-Topological Formulation of Structural Analysis," J. Structural Div., A.S.C.E., Vol. 89, No, ST4, August, 1963, pp. FENVES,
483—514. 2.
3.
DIMAGOT0, F. L., and W. R. SPILLARS. "Network Analysis of Structures," .1. Eng. Mech, Div., 4.S.C.E., Vol 91, No. EM3, June, 1965, pp. 169—188. ARGYRIS, J. H;, "The Matrix Analysis of Structures with Cut-Outs and Modifications," Proc. Ninth International congress App!. Mech., Vol. ô, 1957, pp. 131—142.
18
Analysis of Geometrically Nonlinear Systems 18-1.
INTRODUCTION
In this chapter, we extend the displacement formulation to include geometric nonlinearity. The derivation is restricted to small rotation, i.e., where squares of rotations are negligible with respect to unity. We also consider the material to be linearly elastic and the member to be prismatic. The first phase involves developing appropriate member force-displacement relations by integrating the governing equations derived in Sec. 13—9. We treat first planar deformation, since the equations for this case are easily integrated and it reveals the essential nonlinear effects. The three-dimensional problem is more formidable and one has to introduce numerous approximations in order to generate an explicit solution. We will briefly sketch out the solution strategy and then present a linearized solution applicable for doubly symmetric crosssections.
The direct stiffness method is employed to assemble the system equations. This phase is essentially the same as for the linear case. However, the governing equations are now nonlinear.
Next, we described two iterative procedures for solving a set of nonlinear algebraic equations, successive substitution and Newton-Raphson iteration. These methods are applied to the system equations and the appropriate rerelations are developed. Finally, we utilize the classical stability criterion to investigate the stability of an equilibrium position.
18-2.
MEMBER EQUATIONS—PLANAR DEFORMATION
-
Figure shows the initial and deformed positions of the member. The centroidal axis initially coincides with the X1 direction and X2 is an axis of symmetry for the cross section. We work with displacements (u1, u2, co3), 585
586
ANALYSIS OF GEOMETRICALLY NONUNEAR SYSTEMS
CHAP. 18
distributed external force (b2), and end forces (F1, F2. M3) referred to the
initial (X1-X2-X3) member frame. The rotation of the chord is denoted by p3 and is related to the end displacements by — U,42
L
The governing equations follow from (13—88). For convenience, we drop the subscript on x1, and M3, w3, 13. Also, we consider h1 = rn3 = 0.
Deformed position 1182
b2 dxi
x1 ,
Centroidai exis
Fig. 18—1. Notation for p'anar bending.
Equilibrium Equations
=0 (F1u2,
+ F2) + b2
0
(a)
F2 = Force-Displacement Relatio,is F1
F2
= ULx +
=
U2.
=
(0,
2
M
— CO
(b)
MEMBER EQUATIONS—PLANAR DEFORMATION
SEC. 18—2.
587
Boundary Conditions
Forx =
0:
=
or or or
UAI
U2 =
w=
WA3
1F110 = —FAI
iF2 +
Mb =
Forx = U1
or or
=
=
IF2 +
or
=
+FB1
FIlL
(d)
F112
Ml,.
Integrating (a) leads to
F1 =
P
F2 + Pu2 = — C3P
C2Px + Jx(Jx b2 dx)dx
—
where C2, C3 are integration constants. We include the factor P so that the dimensions are consistent. The axial displacement is determined from the first equation in (a),
PL 11111 — UAI
=
('1.
1
j
—
(u2
2
dx
(18—2)
Combining the remaining two equations in (a), we obtain
M = El
/
+
P\ u2
ri
+
Finally, the governing equation for u2 follows from the third equation in (e),
+ C3)+
+ where
—
(18—3)
2__ El
The solutions for u2 and M arc
=
C4
cos px + C5 sin
(i
co
+
+
C2
C4
+ C2x + C3 + U2b sin j.tx + C5 cos px)
b2 dx +
+
(i
+
(18—4) x
-
where U2b denotes the particular solution due to b2. If b2 is constant, U2b
b(EI
—
1/
(18—5)
ANALYSIS OF GEOMETRICALLY NONLINEAR SYSTEMS
588
CHAP. 18
Enforcing the boundary conditions on u2, w at x =
0, L leads to four linear equations relating (C2 C5). When the coefficient matrix is singular, the member is said to have buckled. In what follows, we exclude member buckling. We also neglect transverse shear deformation since its effect is small for a homogeneous cross section. We consider the case where the end displacements are prescribed. The net displacements are
u = (u — CD' = (a5 Evaluating (18—4) with A2 =
C2 = C3 =
oc,
1.
(18—6)
U2bX)X_OL
we obtain
— jzC5
C4
—
—
C
— — 1
C5 =
—
sin 1iL
1
2(1
(18—7)
1— COS/LL
.
1tL —
—
—
D=
,u sin —
—
Note that D 4 0 as
This defines the upper limit on P, i.e., the member
buckling load: (18_8)
PJrnax
The end forces can be obtained with (c—e). We omit the algebraic details since they are obvious and list the final form below. MA3 =
+
+
=
+
+
MB3
—
+
—
—
=
where
D=
UA2)]
—
UA2)]
1
2
— Unz)]
C0A3 —
('Lj
P1.
(u2,
2
dx =
(u52 — uA2)
—
+
P
PL
— erL
—j 2(1 — cos
—
—
Dç62 =
=
uA2)]
—
+
+ FB2 =
—
—
+
pL sin pL
1iL cos
sin iL) cos 1zL)
(18—9)
MEMBER EQUATIONS—PLANAR DEFORMATION
SEC. 18—2.
The
589
functions were introduced by Livesley (Ref. 7), and are plotted in Fig.
—+ 27t. The initial end forces depend on 18 —2. They degenerate rapidly as the transverse loading, b2. If b2 is constant,
bL A2 —
52 —
bL2 —
1
(18—10)
-
1
—
= In order to evaluate the stiffness coefficients, P has to be known. If one end, say B, is unrestrained with respect to axial displacement, there is no difficulty since is now prescribed. The relative displacement is determined from
UB1 =
PL
+
UA1
Le.
—
('L
dx =
=
er( jiL, UA2, U52, WA, w5)
—— j
+ Dg54 = C5
—
w43)
2
+
L
sin
=
—
(U52
WAS) +
= =
(UB2 —
j
(18—11)
—
— WA3
sin jiL cos jiL) + 2(1 — cos (1
—
cos jiL)
if
—
(1
—
eos
jiL'\
4
+
-
sin jiL cos jiL
+ We call Cr the relative end shortening due to rotation. However, when both axial displacements are prescribed, we have to resort to iteration in order to evaluate P since e, is a nonlinear function of P. The simplest iterative scheme is
=
p(i+ 1)
=
(u51 — UAI)
+
and convergence is rapid when jiL is not close to 2it.
(18—12) -
for the incremental end forces due to increments in the end displacements are needed in the procedure and also for Expressions
stability analysis. If jiL is not close to 2ic, we can assume the stability functions
ANALYSIS OF GEOMETRICALLY NONLINEAR SYSTEMS
590
CHAP. 18
I
+02
pL —2
—4 —6
Oi
—8
—10
A Fig. 18—2. Plot of the 0 functions. are
constant and equal to their values at the initial position, when operating
on (18—9). The resulting expressions are
+
(IMA3
Aco,43 + c&2
+
dMB3 = dMç3 +
+
dP42
41
112
112
—
(18—13)
——h-——- dP
—
—
(Au,32 —
—
+ AWA3 —
P — ,JL'i
AU42)]
—
42
dFB1
dP =
.41
1 42
=
dP
dF41 = —
—dP
Au41)
+ AEder
SEC. 18—3.
where
MEMBER EQUATIONS—ARBITRARY DEFORMATION
591
the incremental initial end forces are due to loading, Lxb2. We can
obtain an estimate for
by assuming Au2
is constant. AuA2)
Au12.x dx
(18—14)
The coefficients in (18—13) arc tangent stiffnesses. They are not exact since we have assumed and Au2, constant. To obtain the exact coefficients, we have to add El3 F
,
+
— 1LA2
j
—
,
1
4)3] dCuL)
(18-15) d(,uL) =
to dM4 and similar terms to dM5, .
—
.
,
2b13
dP
The derivatives of the stability
,
functions are listed below for reference: 2(1iL)2 sin
—
D
=
- 41 =
+
+
-
(18—16)
pL)
—
We also have to use the exact expression for
der =
ae
tie.
+ ——Au52 +
cIUA2
+ -—--
+ —--—A(1zL)
(WA
in the equation for dP. An improvement on (18—14) is obtained by operating on (18—11), and assuming ,tL is constant. 18—3.
MEMBER EQUATIONS—ARBITRARY DEFORMATION
The positive sense of the end forces for the th case is shown in Fig. 18—3. Note that the force and displacement measures are referred to the fixed member frame. The governing equations for small rotations were
derived in Sec. 13—9. They are nonlinear, and one must resort to an approximate method such as the Galerkin scheme,t in which the displacement measures are expressed in terms of prescribed functions (of x) and parameters. The problem
is transformed into a set of nonlinear algebraic equations relating the eters. Some applications of this technique are presented in Ref. 5. t This method is outlined in Sec. 1O—6.
ANALYSIS OF GEOMETRICALLY NONLINEAR SYSTEMS
592
CHAP. 18
M52, WB2
x2
!152
I
//
//
x1
P2
Note:
The centroidal axis coincides with X1, X2 and X3 are principal inertia directions.
Fig. 18—3. Notation for three-dimensional behavior.
If we consider b1 = 0, the axial force F1 is constant along the member and the nonlinear terms involve and coupling terms such as co1M2;
Neglecting these terms results in linearized equations, called the Kap pus equations. Their form is:
Equilibrium Equations
P
F1
+ x3w1,1) +
dx1
[P(u,3
1+
1 + rn-1- +
d
+ 711w1
—
+
=0 0
M, + F2 + in3 =
0
=
0
M4,1 11
0
=
1142
=
0
b2
+ F3] + b3 =
i
dx1
—
F2]+
+
1
—
F3
+
+
(18—18)
MEMBER EQUATIONS—ARBITRARY DEFORMATION
SEC. 18—3.
593
Force-Displacement Relations
=
u1,
i
+
1
+ u53,
1
I U52,1 —
U53,
CO3
+
+ "F2
F3
..— +
GA2
+ Ui2 =
w1,
1/F2
X3r
A23
J
F3
x2
+
GA23
i—
A3
+
M2
J
M3
(Dz,ij7f + C01,t
+ X3rF7 ± X2rF3)
Boundary Goiidif ions (± for x = L,
—
x=
for
0)
P
+ M!1 + M2 = ±M2
P(u521
+
P(u,3
—
•t
=
+ F2 = F3 = ±F3 = =
i) +
+ 711w1
—
M3 = ±M3
To interpret the linearization, we consider (13—81). If one neglects the nonlinear terms in the shearing strains,
takes
Y12
u12 + 112.1
Y13
u13 + U3,1
-
the extensional strain as a1
u1
+
+ zfu2, i + U3,
+
1
and assumes
+ + +
= = =
0
0 0
one obtains (13—81). Equations are exact when the section is doubly symmetric, Assumptions (a) and (b) are reasonable if is small w.r. to u2 and However, they introduce considerable error when co1 is the dominant u3, term. This has been demonstrated by Black (Ref. 5). When the cross section is doubly symmetric, 1
=
= X2r = X3r =
A23
=0 (18—19)
1
—
= r2
ANALYSIS OF GEOMETRICALLY NONLINEAR SYSTEMS
594
(r
CHAP. 18
is the radius of gyration with respect to the centroid) and the problem
uncouples to— plane Flexure in 1. 2, Flexure in X1-X3 plane 3. Restrained torsion
We have already determined the solution for fiexure in the X1-X2 plane. If we introduce a subscript for /L and Cu2)2
P
P
=
El2
=
(18—20)
cbjCu3L)
and then replace (02 U3 —* —U2
—4
F2
F3
F3
—F2
(18—21)
M'2 -
in (18—9), (18—13), we obtain the member relations for flexure in the X1-X3
frame. For example, +
+
MA3
U.42)]
—
(18—22)
=
+
E12
+
+
U43)]
and
F42 = F43 =
+
[_WB2
W42
—
U43)]
—
— U.43)
The expressions for the axial end forces expands to
=P P =
AE
r2
=
—
J'Al = + AE(Cri + er2 + er3)
U41)
Ci-'
j
CL
1
dx1
dx1
er2
(18—23)
J er3
=
1
2L
f
(u3
dx1
where is obtained from er2 by applying (18—21). We generate the restrained torsion solution following the procedure described in Example 13—7. If the joints are moment resisting (i.e., rigid), it is reasonable
to assume no warping, which requires f = 0 at x = 0, L. The corresponding solution is summarized below:
________1+P SEC. 18—3.
r2P P=— GJ
-
_
MEMBER EQUATIONS—ARBITRARY DEFORMATION
1+P
GJ U
595
+ '>
Erlcb 1 +
GJ
MB1
<1)A1)
MA1
—MB1
[iiL(1 + Cr(1 + P))
sinh 1uL
+
—
(18—24)
2
1+
(
GJ(1 + P)
1 1—coshuL [ + —-——--—(1 — — sinh,uL /1(1 + Cr(1 + P))[ We neglect shear deformation due to restrained torsion by setting C,.
0.
If warping restraint is neglected, 0
En4,
c%)
I
(18—25)
+P
At this point, we summarize the member force-displacement relations for a doubly symmetric cross section. For convenience, we introduce matrix notation: = {F1P2F3M1M2M3}8 {u1u2u3w1w2w3}B
etc.
(18—26)
= where
+ ÷
+ kBJl%A + +
—
contains nonlinear terms due to chord rotation and end shortening
{AE(eni + er2 + e,.3);
uA2);
—
uA3);
0; 0;
contains
the initial end forces due to member loads; and AE L
El3
El3
El2
El2 'P33
GJ
Sym
El3
ANALYSIS OF GEOMETRICALLY NONLINEAR SYSTEMS
596
CHAP. 18
AE L
El3 El2
.
El2
k44 =
GJ
El3
Operating on
leads
to the incremental equations. i.e., the three-
dimensional form of (18—13). Assuming the stability functions are constant and taking dP —
—
+
UA2)(AU82
+ r2(w81 —
—
—
iiA3)(AuB3
—
AuA3)
(18—27)
AWAI)}
we obtain
+ (knE +
=
+ (knA
—
+ (kBA —
Mi11 + (kAA + kr)L\%A
(18 —28)
SEC. 18-4.
SOLUTION TECHNIQUES; STABILITY ANALYSIS
597
where k, is the incremental stiffness matrix due to rotation, 0
P3 2
f.J3
+
AE
kr
Pz
rp1
0
P2P3
r 2 p1p3
0
0
—r2p1p2
0
0
(r2p1)2
0
0
0
0
LFB1.
+
L
TAT
Symmetrical
0 p3
=
—
uA2)
p2 =
— UA3)
Pi =
—
WA1)
1fF is close to the member buckling load, one mustinclude additional terms due to the variation in the stability functions and use the exact expression for der
Kappas's equations have also been solved explicitly for a monosymmetric section with warping and shear deformation neglected. Since the equations are linear, one can write down the general solution for an arbitrary cross section. It will involve twelve integration constants which are evaluated by enforcing the
displacement boundary conditions. The algebra is untractable unless one introduces symmetry restrictions. 18—4.
SOLUTION TECHNIQUES; STABILITY ANALYSIS
In this section, we present the mathematical background for two solution techniques, successive substitution and Newton-Raphson iteration, and then apply them to the governing equations for a nonlinear member system. Consider the problem of solving the nonlinear equation
= Let
(18—29)
0
represent one of the roots. By definition, =
0
(18—30)
In the method of successive substitution,t (18—29) is rewritten in an equivalent form,
x=
g(x)
(18—31)
and successive estimates of the solution arc determined, using q(k)
—
represents the kth estimate. The exact solution satisfies
where
t See Ref. 9.
x=g
(18—32)
ANALYSIS OF GEOMETRICALLY NONLINEAR SYSTEMS
598
Then,
=
Expanding
CHAP. 18
—
in a Taylor series about
=
g(k)
+ g(T —
+
+
—
and retaining only the first two terms lead to the convergence measure —
where
is between
(18—33)
—
and T.
In the Newton-Raphson method,t
=
is expanded in a Taylor series about
Ax +
+
+
=0
where Ax is the exact correction to
Ax =
—
An estimate for Ax is obtained by neglecting second- and higher-order terms:
=
(18—34)
+
The convergence measure for this method can be obtained by combining (a) and (18—34), and has the form —
(18—35)
—
Note that the Newton-Raphson method has second-order convergence whereas successive substitution has only first-order convergence. We consider next a set of n nonlinear equations:
=
'I' =
An exact solution is denoted by In successive substitution,
0
x2,...,
=
(18—36)
Also,
= is rearranged to
ax =
(18—37)
c — g
where a, c are constant, g = g(x), and the recurrence relation is taken as =c
—
(18—38)
The exact solution satisfies
a5 =
c —
Then, 1))
t See Ref. 9.
=
— g(k))
SEC. 18—4.
SOLUTION TECHNIQUES; STABILITY ANALYSIS
599
Expanding g in a Taylor series about
=
=
g(k)
+
+
—
=
9i,i
91,2
91n
92,1
92,2
92,n
L0xrJ '
and retaining only the first two terms results in the convergence measure
=
(x —
(18—39)
a
For convergence, the norm of
where lies between xk and less than unity.
'g. must be
The generalized Newton-Raphson method consists in first expanding about
=
+
=
+
0
where
= [T'—j
=
—
=
(18—40)
= Neglecting the second differential leads to the recurrence relation
= =
( 18—41 )
+
The corresponding convergence measure is
=
—
(18—42)
Let us now apply these solution techniques to the structural problem. The governing equations are the nodal equations referred to the qlobcil system frame,
=0
1?e —
(1843)
contains the external nodal forces and — is the summation of the member end forces incident on node i. One first has to rotate the member end forces, (18—26), from the member frame to the global frame using = where
k° = is constant. We introduce the displacement restraints and write the final equations as
In our formulation, the member frame is fixed, i.e.
-
m
e
Pm
P1
+
+ KU
1844
ANALYSIS OF GEOMETRICALLY NONLINEAR SYSTEMS
600
CHAP. 18
depend on the axial forces while Pr depends on both the axial force and the member rigid body chord rotation, if the axial forces are small in comparison to the member buckling loads, we can replace K with Note that K and
K1, the linear stiffness matrix. Applying successive substitution, we write
KU =
—
and iterate on U, holding K constant during the iteration: P1 — (18—45) = We employ (18—45) together with an incremental loading scheme since K is actually a variable. The steps are outlined here:
1.
Apply the first load increment,
2.
Update K using the axial forces corresponding to Pe(l)• Then apply
and solve for U(I), using K
K1.
Pe(2) anditerate on
(ni_pe(1)
K (1) 3.
_p(n—1) e(2)
r
Continue for successive load increments.
A convenient convergence criterion is the relative change in the Euclidean norm, N, of the nodal displacements.
N= '\
—
.
Jabs
(a specified value)
e
1
(18—46)
va]uc
This scheme is particularly efficient when the member axial forces are small
with respect to the Euler loads since, in this case, we can take K = K1 during the entire solution phase. In the Newton-Raphson procedure, we operate on Vi according to (18—41): =—
4,(fl)
Now, Pe is prescribed so that
=
due to
+ dPr + K LW +
= where
(18—47)
—
denotes the tangent stiffness matrix. The iteration cycle is AU(n)
= =
—
+
(
18—48 )
iterate on (18—48) for successive load increments. This scheme is more has to be updated for each cycle. However, its convergence rate is more rapid than direct substitution. If we assume the stability functions We
expensive since
SEC. 18—4.
are
SOLUTION TECHNIQUES; STABILITY ANALYSIS
601
due to AU, the tangent stiffness matrix reduces to
constant in forming
dI(
dP,
0
0
(18—49)
K + Kr
where K. is generated with (18—28). We include the incremental member loads at the start of the iteration cycle. Rather than update at each cycle, in fixed for a limited number of cycles. This is called mod (fled one can hold
Newton-Raphson. The convergence rate is lower than for regular NewtonRaphson but higher than successive substitution. We consider next the question of stability. According to the classical stability criterion,t an equilibrium position is classified as: stable neutral unstable
>0
—
0
d2W,,, — d2We d2 W,,
d2
(18—50)
<0
is the second-order work done by the external forces during a where d2 displacement increment AU, and is the second-order work done by the member end forces acting on the members. With our notation, Pe)TAU
d2w,
=
(d
(18—51)
AU
= AUTK, AU
and the criteria transform to
/
(AU)TK, AU —
/
\T
< 0
Pe) AU = 0 >0
stable neutral unstable
(18—52)
The most frequent case is Pe prescribed, and for a constant loading, the tangent stiffness matrix must be posil.ive definite,
To detect instability, we keep track of the sign of the determinant of the tangent stiffness matrix during the iteration. The sign is obtained at no cost (i.e., no additional computation) if Gauss elimination or the factor method are used to solve the correction equation, (18—48). When the determinant changes sign, we have passed through a stability transition. Another indication of the existence of a bifurcation point (K1 singular) is the degeneration of the convergence rate for Newton-Raphson. The correction tends to diverge and oscillate in sign and one has to employ a higher iterative scheme. Finally, we consider the special case where the loading does not produce significant chord rotation. A typical example is shown in Fig. 18—4. Both the t See Sees. 7—6 and 10—6.
______________________ 602
ANALYSIS OF GEOMETRICALLY NONLINEAR SYSTEMS
CHAP. 18
frame and loading are symmetrical and the displacement is due only to short-
ening of the columns. To investigate the stability of this structure, we deletet the rotation terms in K, and write K
The member axial K is due to a unit value of the load parameter forces are determined from a linear analysis. Then, the bifurcation problem reduces to determining the value of 2 for which a nontrivial solution of (K + 2K;)AU
0
(18—54)
exists. This is a nonlinear eigenvalue problem, since K = K(2). 12X
I
I
Fig. 18—4. Example of structure and loading for which linearized stability analysis is applicable.
In linearized stability analysis, K is assumed to be K1 and one solves
K, AU =
—2K AU
(18-55)
Both K, and K; are symmetrical. Also, K1 is positive definite, Usually, only the lowest critical load is of interest, and this can be obtained by applying inverse iterations to (—K;)Au 1
(18—56)
REFERENCES 1.
2. 3.
M. GERE: Theory of Elastic Stability, 2d ed., McGraw-Hill, New York, 1961. KOLLBRUNNSR, C. F., and M. MEIsTER: Knicken, Biegedriilknicken, Kippen. 2d ed., Springer-Verlag, Berlin, 1961. BLEICH, F.: Buckling Strength of Metal Structures, McGraw-Hill, New York, 1952.
TIMOSHENKO, S. P., and J.
= 0 in (18—28). t Set Pi P2 = See Refs. 11 and 12 of Chapter 2.
REFERENCES 4:
5. 6. 7.
603
G.. and F!. STEUP: Stabilidhsrheorie, Part 1, Akademie-Verlag. Berlin, 1957, CFJtLVER, A. H., ed.: Thin- Walled Structures, Chatto & Windus, London, 1967. VLASOV, V. Z.: Thin Walled Elastic Beams, Israel for Scientific Translations, Office of Technical Services: U.S. Dept. of Commerce, Washington, D.C., 1961. LIVESLItY, R. K.: Matrix Methods of Structural Anal vsis, Pergamon Press, London, BLYRGERMEISTEa,
1964. 8. 9.
AROYRIS, J. H.: Recent Advances in Matrix Methods of Structural Analysis, Pergamon Press, London, 1964. HILDEBRAND, F. B.: Introduction to Numerical Analysis, McGraw-Hill, New York, 1956.
10.
11.
GALAMBOS, T. V.: Structural Members and Frames, Prentice Hall, 1968. BRUSU, D. and B. ALMROTH: Buckling of Bars, Plates, and Shells, McGraw-Hill, New York, 1975.
index
Associative multiplication, 8
Augmented branch-node trix, 124, 222
incidence ma-
Augmented matrix, 33 Axial deformation, influence on bending
of planar member, 472 Bar stiffness matrix, 180 Bifurcation; Neutral equilibrium Bimoment, MqS, 373
Branch-node incidence table, 121, 145 Cç1,
Constraint conditions treated. with La-
grange multipliers, 76, 80 Curved member definition of thin and thick, 434 thin, 487 slightly twisted, 487
Defect, of a system of linear algebraic equations, 31 Deformation
for out-of-plane loading of a circular member, transverse shear, twist, and bending, 498
C,,r—coefficients appearing in complementary energy expression Cr,
for restrained torsion, 387, 388, 416 Canonical form, 58 Cartesian formulation, principle of vir-
tual forces for a planar member, 465 Castigliano's principles, 176 Cayley—Hamilton Theorem, 63
Center of twist, 383, 389 Characteristic values of a matrix, 46 Chord rotation. p. 586 Circular helix, definition equation, 84, 86 Circular segment out-of-plane loading, 504 restrained warping solution, 509 Classical stability criterion continuum, 256 member system, 603 truss, 170 Closed ring, out-of-plane loading, 503 Cofactor, 19 Column matrix, 4
for
planar member,
stretching
and
transverse shear vs. bending, 454 Deformation constraints force method, 573 displacement method, 576 variational approach, 583 Deformed geometry, vector orientation, 239
Degree of statical indeterminacy member, 555, 567 truss, 210 Determinant, 16, 37, 39 Diagonal matrix, 10 Differential notation for a function, 70, 72, 79
Direction cosine matrix for a bar, 119 Discriminant, 40, 59 Distributive multiplication, 8 Echelon matrix, 29 Effective shear area, cross-sectional properties, 302 Elastic behavior, 125, 248 End shortening due to geometrically nonlinear behavior, 589
Column vector, 4 Complementary energy continuum, 261 member system, 572 planar curved member, 434
Engineering theory of a member, basic
restrained torsion, 385; 387, 388
assumptions, 330, 485 Equivalence, of matrices, 27 Equivalent rigid body displacements, 334, 414, 430 Euler equations for a function, 73 Eulerian strain, 234
unrestrained torsion-flexure, 301 Conformable matrices, 8, 35 Connectivity matrix, member system, 563 Connectivity table for a truss, 121, 143.
Consistency, of a set of linear algebraic equations, 31, 44 605
INDEX
606
First law of thermodynamics, 248 Fixed end forces prismatic member, 523 thin planar circular member, 528 Flexibility matrix arbitrary curved member, 515 circular helix, 534 planar member, 462 prismatic member, 345, 521 thin planar circular member, 526 Flexural warping functions, 296, 300/n Frenet equations, 91
Gauss's integration by parts formula. 254 Geometric compatibility equation arbitrary member, 499 continuum, 259, 264 member system, 569 planar member, 463, 466 prismatic member, 355 truss, 160, 212, 216, 223 unrestrained torsion, 279, 315 Geometric stiffness matrix for a bar, 200 Geometrically nonlinear restrained torsion solution, 595 Green's strain tensor, 234 Hookean material, 126, 249 Hyperelastic material, 248 Incremental system stiffness matrix member system, 601 truss, 193 Inelastic behavior, 125 Initial stability member system, 562 truss, 137 Invariants of a matrix, 59, 62 Isotropic material, 252 Kappus equations, 592 Kronecker delta notation, 11
Lagrange multipliers, 76, 80, 583 Lagrangian strain, 234 Lamé constants, 253 Laplace expansion for a determinant, 20, 38
Linear connected graph, 218 Linear geometry, 120, 143, 237 Linearized stability analysis, 602 Local member reference frame, 92 Marguerre equations, 449, 456 Material compliance matrix, 249 Material rigidity matrix, 249 Matrix iteration, computational method, 201
Maxwell's law of reciprocal defiections. 356
Member, definition, 271 Member buckling, 588 Member force displacement relations. 537, 546, 556 Member on an elastic foundation, 384, 369 Mesh, network, 220
Minor, of a square array, 19 Modal matrix, 52 Modified Neuton-Raphson iteration, 601 Moment, MR, Mushtari's equations, 444
Natural member reference frame, 92 Negative definite, 58 Negligible transverse shear deformation, planar member, 443, 454, 498 Network, topological, 220 Neutral equilibrium, 170, 256, 601 Newton-Raphson iteration, member system, 598 Normalization of a vector, 49 Null matrix, 4 One-dimensional deformation 335, 338, 432
measures,
arbitrary member, 491 Orthogonal matrices and trnasformations, 50, 53 Orthotropic material, 250, 251
Permutation matrix, 42, 135 Permutation of a set of integers, 16, 37 Piecewise linear material, 126, 146 Plane curve, 98, 425 Poisson's ratio, 252 Positive definite matrix, 58, 63 Positive semi-definite matrix, 58 Postmultiplication, matrix, S Potential energy function, member system, 571
Premuftiplication, matrix, 8 Primary structure member system, 568 planar member, 463 prismatic member, 354 truss, 211 Principle minors, 55 Principle of virtual displacements member system, 570 planar member, 442 Principle of virtual forces arbitrary member, 490, 492, 512 member systens, 571 planar member, 435, 458 prismatic member, 338, 351
Quadratic forms, 57
INDEX
modification for partial end restraint, 535
Quasi-diagonal matrix, 15, 38 Quasi-triangular matrix, 39
Radius of gyration, 434 Rank of a matrix, 27, 42, 43 Rayleigh's quotient, 75, 79 Reissner's principle continuum, 270 member, 383, 414 member system, 573 Relative minimum or maximum value of a function, relative extrema, 66 Restrained torsion solution, member linear geometry, 391 nonlinear geometry, 595
607
prismatic
Restrained torsion stress distribution and cross-sectional parameters channel section, 401 multicell section, 411 symmetrical I section, 398 thin rectangular cell, 407
Rigid body displacement transformation, 109
Rotation transformation matrix, 101, 232 Row matrix, 4
prismatic member, geometrically nonlinear behavior, 588, 595 prismatic member, linear geometry, 522 Strain and complementary energy for pure torsion, 280 Strain energy density. 248
Stress and strain component trnasformations. 249 Stress components Eulerian, 242 Kirchhoff, 246 Stress function, torsion, 276 Stress resultants and stress couples, 272 Stress vector, 240 Stress vector transformation, 242 Submatrices (matrix partitioning), 12, 36 Successive substitution, iterative method member system, 597 truss, 193 Summary of system equations, force equiibrium and force displacement, 561 Symmetrical matrix, Il, 35 System stiffness matrix member system, 548, 550, 565 truss, 179, 180, 188, 206
Self-equilibrating force systems, 160, 211, 258
member systems, 568 Shallow member, assumptions, 448 Shear center, 297, 300, 309, 378, 389 Shear flow, 287 Shear flow distribution for unrestrained torsiOn, 308 Similarity transformation, 53, 62 Simpson's rule, 475 Singular matrix, 22 Skew symmetrical matrix, 11 Small strain, 120, 235 Small-finite rotation approximation, 238 Square matrix, 4
Stability of an equilibrium position, 171, 195
Stability functions (4), prismatic member, 589
Statically equivalent force system, 103, 106
Tangent stiffness matrix for a bar, 193 prismatic member, 590, 596 Tensor invariants, 232 Torsion solution, rectangle, 281 Torsional constant, J, 276, 278, 323 Torsional warping function, 274, 377 Transverse orthotropic material, 252 Transverse shear deformation planar member, 454, 498 prismatic member, 355
irapezoidal rule, 474 Tree, network, 220 Triangular matrix, 12 Two-hinged arch solutions, 467, 470 Unit matrix, 10
Stationary values of a function, 67, 79
Variable warping parameter, f, for restrained torsion, 372 Vector, definition (mechanics), 4/n
Stiffness matrix arbitrary curved member, 516, 520
Work done by a force, definition, 153, 156
Statically permissible force system, 159, 216, 257
