Vibration Analysis for Electronic Equipment

VIBRATION ANALYSIS FOR ELECTRONIC EQUIPMENT THIRD EDITION

Dave S. Steinberg Steinberg & Associates and University of California, Los Angeles

A WILEY-INTERSCIENCE PUBLICATION

JOHN WILEY & SONS, INC. New York

Chichester

Weinheim

Brisbane

Singapore

Toronto

This book is printed on acid-free paper. @ Copyright G2000 by John Wiley & Sons, Inc. All rights reserved. Published simultaneously in Canada. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 750-4744. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 605 Third Avenue, New York, NY 10158-0012, (212) 850-6011, fax (212) 850-6008, E-Mail: [email protected]. Library of Congress Cataloging in Publication Data:

Steinberg, Dave S. Vibration analysis for electronic equipment/Dave S. Steinberg.--3rd ed. p. cm “ A Wiley-Interscience publication.” ISBN 0-471-37685-X 1. Electronic apparatus and appliances-Vibration. I. Title. TK7870S8218 2000 621.381--dc21 Printed in the United States of America 109876

99-056617

To my wife, Annette, and to my two daughters, Con and Stacie

Electronic equipment continues its relentless expansion into virtually every area associated with commercial, industrial, and military applications throughout the entire world. Exotic technology has become commonplace in medicine, entertainment, communication, travel, transportation, manufacturing, education, and commerce. The internet has dramatically and forever changed the way many companies run their businesses, without the need for large offices and a large staff of people to perform clerical duties, largely due to the extensive use of the fast, small, powerful personal computers. This has resulted in improved profit margins in companies that embraced the new technologies, and disasters for companies that resisted the tidal wave of changes. Extensive changes were made in this third edition to reflect the changes that are taking place in the world of electronics. One of the big changes is the dramatic reduction in the costs of manufacturing the electronic equipment due to improved automation. This has resulted in the use of more electronics with improved performance over mechanical functions in a wide variety of products such as cameras, automobile controls for braking, ignition. air conditioners, transmission shifting, and combustion control, computers, computer printers, FAX machines, and televisions. One of the most dramatic changes has occurred in the military area, where high costs have forced the Department of Defense to scrap many of the military specifications in favor of best commercial practice electronic hardware for their new sophisticated equipment. Lower costs often lead to lower quality, which can result in a reduced reliability. This may only be an inconvenience when your television set fails or your automobile will not start. This can be a disaster in a military vehicle if the guns do not fire, or if the navigation system fails, or if the communications system fails. Therefore a new chapter was added that investigates and evaluates the effects of manufacturing methods and tolerances on the reliability of the electronic hardware. Another new chapter was added to show methods for improving the ruggedness of commercial hardware for improved reliability in harsh military environments. An additional chapter was added that relates to a better understanding of the bending distortion of the circuit board at its resonant condition, and how it effects the forces, stresses, and fatigue life in the component lead wires and solder joints. A new chapter was also added in the area of environmental stress screening, to provide a better understanding of how various combinations of xvii

XViii

PREFACE

thermal cycling and vibration can affect the accumulated damage and the amount of fatigue life used up by various screens. New equations are derived that show how to evaluate the effectiveness of a lead wire strain relief to reduce the dynamic forces and stresses in the lead wires and solder joints, and the resulting increase in the fatigue life. A simpler method for evaluating the fatigue life in random vibration is shown, which has the same accuracy as the three-band technique with much less work. A quick way of estimating the effectiveness of stiffening ribs on plug-in PCBs is shown, which is very simple to use and easy to apply. A new added section discusses a quick and easy method for evaluating the first three natural frequencies for flat plates and PCBs that are rectangular, octagonal, triangular, and round, with 16 different edge and point supports for each shape. A new and more accurate method for calculating the expected transmissibilty Q values for beams, circuit boards, and chassis is presented that relates to the natural frequency and the input acceleration G level. The data for this method are based on extensive vibration tests which showed that very low input acceleration G levels produced very low stresses and damping, which resulted in very high Q values. At the opposite end, high input G levels produced high stresses and damping, which resulted in low Q values. Many detailed sample problems are shown throughout this textbook, to demonstrate methods of analysis and evaluation that are based on 40 years of analysis and testing many different types of electronic systems. A section on vibration testing case histories was added to provide additional knowledge and insights for improving the testing results. Field failures in electronic equipment hardware compiled by the U.S. Air Force over a period of about 20 years show that about 40% of these failures are related to connectors, 30% to interconnects, and 20% to component parts. Failures in these areas can be due to handling, vibration, shock, and thermal cycling. Field failures related to operating environments show that about 55% of the failures are due to high temperatures and temperature cycling, 20% of the failures are related to vibration and shock, and 20% are due to humidity. Since high temperatures and temperature cycling events appear to be the major environmental causes of electronic failures, more information in these areas is very important. These subjects are not addressed in this book in great detail. More information is available on high temperatures and temperature cycling forces, stresses, and fatigue life effects [54]. Details on electronic components, electrical lead wires, solder joints, and plated through-holes for surface mounted and through-hole components, due to differences in the thermal coefficients of expansion (TCE), can be found in my book entitled Cooling Techniques for Electronic Equipment, Second Edition published by John Wiley & Sons. DAVES. STEINBERG Westlake Village, California September, 1999

DLIST OF SYMBOLS

A

Area (in.2), amplification g (dimensionless ratio) a length (in.) H ASIC Application-specific integrated circuit h Length (in.) Hz B Fatigue exponent, width (in.) I b Dynamic constant C C Length (in.) Lri C G Center of Gravity Distance from neutral axis J C K to outer fiber, length (in.), damping coefficient (lb . s/in.) Critical damping (lb .s/in.) cc D Dissipation energy KIN (lb .in./s) diameter (in.), plate bending stiffness Kt factor (lb . in.) Decibel dB KO D,, Plate torsional stiffness factor (Ib * in.) d L Diameter (in.), length (in.) DIP Dual inline package LCCC E Modulus of elasticity (lb/in.2) A4 e MT Bolted efficiency factor (%) ESS Environmental stress screen M , F Force (lb) Frequency (Hz) f f" Natural frequency (Hz) Rotational natural Mo fr MS frequency (Hz) FEM Finite element method Shear modulus (lb/in.2) G rn N with subscript: acceleration in gravity units (dimensionless)

Acceleration of gravity, 386 in./s2 Horizontal force Ob), drop heightr (in. or ft) Height (in.), thickness (in.) Frequency (cycles/s) Moment of inertia (area) (in.4) Mass moment of inertia (Ib in: s2) Torsional form factor (in.4> Linear spring rate (Ib in.), stiffness ratio (dimensionless), buckling form factor dimensionless) Kinetic energy Theoretical stressconcentration factor (dimensionless) Angular spring rate (Ib .in./rad) Length (in.) Leadless ceramic chip carrier Bending moment (lb . in.) Total moment (lb in.) Bending moment per unit length along the X axis (in: lb/inJ, bending moment at point X(lb. in.) Momentum (lb .s) Margin of safety (dimensionless) mass (lb s2/in.) Number of fatigue cycles to fail xix

XX

LIST OF SYMBOLS

Number of positive zero crossings, (Hz) n Actual number of fatigue cycles NS Number of sweeps through a resonance P Force (lb) P. PSD Power spectral density

N,+

wd W

X X X

x

z

Greek Symbols

(G*/HZ)

PCB d'

P

Q

4

R

Rn

RC

R, RMS RSS r sb

se

S SMD TCE T t

U V

W

Printed circuit board Dynamic force Unit load (lb/in.) Transmissibility (dimensionless ratio) Shear flow (lb/in.), dynamic pressure (lb/in.) Radius (in.), reaction (lb) Stress ratio (dimensionless), sweep rate (octave/min) Fatigue-cycle ratio (dimensionless) Damping ratio (dimensionless) Frequency ratio (dimensionless) Root mean square Root sum square radius (in.), relative position factor Bending stress (lb/in.2> Endurance limit stress (lb/in.*) Stress (lb/in.*) Surface mounted device Thermal coefficient of expansion (in./in./'C) Kinetic energy Ob-in.), torque (in/. lb) Time (min), temperature, Thickness (in.) Strain energy (lb in.), work (lb in.) Velocity (in./s>, force (lb) Weight (lb)

Dynamic load (lb) Unit load (lb/in.) Displacement along X axis, Xi, E;, Zcoordinate axes First derivative, velocity Second derivative, acceleration Displacement (in.)

CY

s 0 P

P

4 A

R

a"

Angle (degrees, radians), thermal coefficient of expansion (in./in./"C) Displacement (in.) Angular displacement (radians) Poisson's ratio (dimensionless) Density (lb/in.), mass per unit area (lb * in.^) Phase angle (degrees) Difference Angular velocity (rad/s) Natural frequency (rad/s) Subscripts

av b C

d e eq in m ax n 0

out st su t tu tY U

Y ST S T

Average Bending Critical Dynamic, desired Endurance Equivalent Input Maximum Natural Maximum Output or response Static Shear ultimate Tension Tensile ultimate Tensile yield Ultimate Yield Shear tearout Shear Torsion

-

CONTENTS

Preface

xvii

List of Symbols

xix

1. Introduction

1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13

Vibration Sources Definitions Vibration Representation Degrees of Freedom Vibration Modes Vibration Nodes Coupled Modes Fasteners Electronic Equipment for Airplanes and Missiles Electronic Equipment for Ships and Submarines Electronic Equipment for Automobiles, Trucks, and Trains Electronics for Oil Drilling Equipment Electronics for Computers, Communication, and Entertainment

2. Vibrations of Simple Electronic Systems

2.1

2.2 2.3 2.4 2.5 2.6

Single Spring-Mass System Without Damping Sample Problem-Natural Frequency of a Cantilever Beam Single-Degree-of-Freedom Torsional Systems Sample Problem-Natural Frequency of a Torsion System Springs in Series and Parallel Sample Problem-Resonant Frequency of a Spring System Relation of Frequency and Acceleration to Displacement Sample Problem-Natural Frequency and Stress in a Beam Forced Vibrations with Viscous Damping Transmissibility as a Function of Frequency Sample Problem-Relating the Resonant Frequency to the Dynamic Displacement

1

1 2 3 3 5 5 6 7 10 13 15 16

16 17

17 19

21 22 23 25 26 27 30 34 34 vii

Viii

CONTENTS

2.7

Multiple Spring-Mass Systems Without Damping Sample Problem-Resonant Frequency of a System

3 Component Lead Wire and Solder Joint Vibration Fatigue Life

3.1 3.2

3.3 3.4 3.5

3.6

Introduction Vibration Problems with Components Mounted High Above the PCB Sample Problem-Vibration Fatigue Life in the Wires of a TO-5 Transistor Vibration Fatigue Life in Solder Joints of a TO-5 Transistor Recommendations to Fix the Wire Vibration Problem Dynamic Forces Developed in Transformer Wires During Vibration Sample Problem-Dynamic Forces and Fatigue Life in Transformer Lead Wires Relative Displacements Between PCB and Component Produce Lead Wire Strain Sample Problem-Effects of PCB Displacement on Hybrid Reliability

4. Beam Structures for Electronic Subassemblies

4.1 4.2

4.3

Natural Frequency of a Uniform Beam Sample Problem-Natural Frequencies of Beams Nonuniform Cross Sections Sample Problem-Natural Frequency of a Box with Nonuniform Sections Composite Beams

5. Component Lead Wires as Bents, Frames, and Arcs 5.1 5.2 5.3 5.4 5.5 5.6 5.7

Electronic Components Mounted on Circuit Boards Bent with a Lateral Load-Hinged Ends Strain Energy-Bent with Hinged Ends Strain Energy-Bent with Fixed Ends Strain Energy-Circular Arc with Hinged Ends Strain Energy-Circular Arc with Fixed Ends Strain Energy-Circular Arcs for Lead Wire Strain Relief Sample Problem-Adding an Offset in a Wire to Increase the Fatigue Life

36 37 39 39 39 40

43 45 46 46 49 50 56

56 60 64 68 69

75 75 77 80 83 90 92 94 97

CONTENTS

6 Printed Circuit Boards and Flat Plates

6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10 6.11 6.12 6.13

Various Types of Printed Circuit Boards Changes in Circuit Board Edge Conditions Estimating the Transmissibility of a Printed Circuit Board Natural Frequency Using a Trigonometric Series Natural Frequency Using a Polynomial Series Sample Problem-Resonant Frequency of a PCB Natural Frequency Equations Derived Using the Rayleigh Method Dynamic Stresses in the Circuit Board Sample Problem-Vibration Stresses in a PCB Ribs on Printed Circuit Boards Ribs Fastened to Circuit Boards with Screws Printed Circuit Boards With Ribs in Two Directions Proper Use of Ribs to Stiffen Plates and Circuit Boards Quick Way to Estimate the Required Rib Spacing for Circuit Boards Natural Frequencies for Different PCB Shapes with Different Supports Sample Problem-Natural Frequency of a Triangular PCB with Three Point Supports

7. Octave Rule, Snubbing, and Damping to Increase the PCB Fatigue Life

ix

103 103 106 108 111 116 120 122 127 131 132 137 141 141 142 144 149 150

7.1

Dynamic Coupling Between the PCBs and Their Support Structures 7.2 Effects of Loose Edge Guides on Plug-in Type PCBs 7.3 Description of Dynamic Computer Study for the Octave Rule 7.4 The Forward Octave Rule Always Works 7.5 The Reverse Octave Rule Must Have Lightweight PCBs Sample Problem-Vibration Problems with Relays Mounted on PCBs 7.6 Proposed Corrective Action for Relays 7.7 Using Snubbers to Reduce PCB Displacements and Stresses Sample Problem-Adding Snubbers to Improve PCB Reliability 7.8 Controlling the PCB Transmissibility with Damping 7.9 Properties of Material Damping 7.10 Constrained Layer Damping with Viscoelastic Materials 7.11 Why Stiffening Ribs on PCBs are Often Better than Damping 7.12 Problems with PCB Viscoelastic Dampers

150 154 154 155 155 156 157 159 161 162 162 163 164 164

X

CONTENTS

8. Preventing Sinusoidal Vibration Failures in Electronic Equipment 8.1 8.2

8.3 8.4

8.5 8.6

8.7

8.8

Introduction Estimating the Vibration Fatigue Life Sample Problem-Qualification Test for an Electronic System Electronic Component Lead Wire Strain Relief Designing PCBs for Sinusoidal Vibration Environments Sample Problem-Determining Desired PCB Resonant Frequency How Location and Orientation of Component on PCB Affect Life How Wedge Clamps Affect the PCB Resonant Frequency Sample Problem-Resonant Frequency of PCB with Side Wedge Clamps Effects of Loose PCB Side Edge Guides Sample Problem-Resonant Frequency of PCB with Loose Edge Guides Sine Sweep Through a Resonance Sample Problem-Fatigue Cycles Accumulated During a Sine Sweep

9. Designing Electronics for Random Vibration

9.1 9.2 9.3 9.4 9.5

Introduction Basic Failure Modes in Random Vibration Characteristics of Random Vibration Differences Between Sinusoidal and Random Vibrations Random Vibration Input Curves Sample Problem-Determining the Input RMS Acceleration Level 9.6 Random Vibration Units 9.7 Shaped Random Vibration Input Curves Sample Problem-Input RMS Accelerations for Sloped PSD Curves 9.8 Relation Between Decibels and Slope 9.9 Integration Method for Obtaining the Area Under a PSD Curve 9.10 Finding Points on the PSD Curve Sample Problem-Finding PSD Values 9.11 Using Basic Logarithms to Find Points on the PSD Curve 9.12 Probability Distribution Functions

166

166 167 168 169 171 174 175 177 179 182 185 185 187 188

188 188 189 190 192 193 193 194 195 197 198 200 200 20 1 202

CONTENTS

9.13 Gaussian or Normal Distribution Curve 9.14 Correlating Random Vibration Failures Using the Three-Band Technique 9.15 Rayleigh Distribution Function 9.16 Response of a Single-Degree-of-Freedom System to Random Vibration Sample Problem-Estimating the Random Vibration Fatigue Life 9.17 How PCBs Respond to Random Vibration 9.18 Designing PCBs for Random Vibration Environments Sample Problem-Finding the Desired PCB Resonant Frequency 9.19 Effects of Relative Motion on Component Fatigue Life Sample Problem-Component Fatigue Life 9.20 It’s the Input PSD that Counts, Not the Input RMS Acceleration 9.21 Connector Wear and Surface Fretting Corrosion Sample Problem-Determining Approximate Connector Fatigue Life 9.22 Multiple-Degree-of-Freedom Systems 9.23 Octave Rule for Random Vibration Sample Problem-Response of Chassis and PCB to Random Vibration Sample Problem-Dynamic Analysis of an Electronic Chassis 9.24 Determining the Number of Positive Zero Crossings Sample Problem-Determining the Number of Positive Zero Crossings

10 Acoustic Noise Effects on Electronics 10.1 Introduction Sample Problem-Determining the Sound Pressure Level 10.2 Microphonic Effects in Electronic Equipment 10.3 Methods for Generating Acoustic Noise Tests 10.4 One-Third Octave Bandwidth 10.5 Determining the Sound Pressure Spectral Density 10.6 Sound Pressure Response to Acoustic Noise Excitation Sample Problem-Fatigue Life of a Sheet-Metal Panel Exposed to Acoustic Noise 10.7 Determining the Sound Acceleration Spectral Density Sample Problem-Alternate Method of Acoustic Noise Analysis

xi

202 204 205 206 208 214 215 218 220 221 222 223 224 224 225 226 229 23 1 233

234 234 234 235 236 238 238 239 240 245 246

Xii

CONTENTS

11. Designing Electronics for Shock Environments 11.1 11.2 11.3 11.4

11.5 11.6 11.7 11.8

11.9

11.10 11.11

11.12 11.13 11.14 11.1s 11.16 11.17

11.18

11.19 11.20

11.21

Introduction Specifying the Shock Environment Pulse Shock Half-Sine Shock Pulse for Zero Rebound and Full Rebound Sample Problem-Half-Sine Shock-Pulse Drop Test Response of Electronic Structures to Shock Pulses Response of a Simple System to Various Shock Pulses How PCBs Respond to Shock Pulses Determining the Desired PCB Resonant Frequency for Shock Sample Problem-Response of a PCB to a Half-Sine Shock Pulse Response of PCB to Other Shock Pulses Sample Problem-Shock Response of a Transformer Mounting Bracket Equivalent Shock Pulse Sample Problem-Shipping Crate for an Electronic Box Low Values of the Frequency Ratio R Sample Problem-Shock Amplification for Low Frequency Ratio R Shock Isolators Sample Problem-Heat Developed in an Isolator Information Required for Shock Isolators Sample Problem-Selecting a Set of Shock Isolators Ringing Effects in Systems with Light Damping How Two-Degree-of-Freedom Systems Respond to Shock The Octave Rule for Shock Velocity Shock Sample Problem-Designing a Cabinet for Velocity Shock Nonlinear Velocity Shock Sample Problem-Cushioning Material for a Sensitive Electronic Box Shock Response Spectrum How Chassis and PCBs Respond to Shock Sample Problem-Shock Response Spectrum Analysis for Chassis and PCB How Pyrotechnic Shock Can Affect Electronic Components Sample Problem-Resonant Frequency of a Hybrid Die Bond Wire

248

248 249 25 1 252 253 257 258 260 260 262 264 265 269 269 274 274 275 276 277 278 28 1 282 284 285 285 286 288 288 29 1

292 296 298

CONTENTS

12. Design and Analysis of Electronic Boxes 12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8 12.9 12.10 12.11 12.12 12.13

Introduction Different Types of Mounts Preliminary Dynamic Analysis Bolted Covers Coupled Modes Dynamic Loads in a Chassis Bending Stresses in the Chassis Buckling Stress Ratio for Bending Torsional Stresses in the Chassis Buckling Stress Ratio for Shear Margin of Safety for Buckling Center-of-Gravity Mount Simpler Method for Obtaining Dynamic Forces and Stresses on a Chassis

13. Effects of Manufacturing Methods on the Reliability of Electronics 13.1 13.2

13.3 13.4 13.5 13.6 13.7 13.8 13.9

Introduction Typical Tolerances in Electronic Components and Lead Wires Sample Problem-Effects of PCB Tolerances on Frequency and Fatigue Life Problems Associated with Tolerances on PCB Thickness Effects of Poor Bonding Methods on Structural Stiffness Soldering Small Axial Leaded Components on ThroughHole PCBs Areas Where Poor Manufacturing Methods Have Been Known to Cause Problems Avionic Integrity Program and Automotive Integrity Program (AVIP) The Basic Philosophy for Performing an AVIP Analysis Different Perspectives of Reliability

14. Vibration Fixtures and Vibration Testing

14.1 14.2 14.3 14.4 14.5 14.6

Vibration Simulation Equipment Mounting the Vibration Machine Vibration Test Fixtures Basic Fixture Design Considerations Effective Spring Rates for Bolts Bolt Preload Torque

Xiii

300 300 300 303 305 308 311 316 318 320 324 325 326 328

330 330 331 332 333 334 335 336 338 340 343

346 346 347 347 348 350 352

XiV

CONTENTS

Sample Problem-Determining Desired Bolt Torque Rocking Modes and Overturning Moments Oil-Film Slider Tables Vibration Fixture Counterweights A Summary for Good Fixture Design Suspension Systems Mechanical Fuses Distinguishing Bending Modes from Rocking Modes Push-Bar Couplings Slider Plate Longitudinal Resonance Acceleration Force Capability of Shaker Positioning the Servo-Control Accelerometer More Accurate Method for Estimating the Transmissibility Q in Structures Sample Problem-Transmissibility Expected for a Plug-in PCB

353 353 355 356 357 357 358 359 360 364 365 366

Vibration Testing Case Histories 14.19 Cross-Coupling Effects in Vibration Test Fixtures 14.20 Progressive Vibration Shear Failures in Bolted Structures 14.21 Vibration Push-Bar Couplers with Bolts Loaded in Shear 14.22 Bolting PCB Centers Together to Improve Their Vibration Fatigue Life 14.23 Vibration Failures Caused by Careless Manufacturing Methods 14.24 Alleged Vibration Failure that was Really Caused by Dropping a Large Chassis 14.25 Methods for Increasing the Vibration and Shock Capability on Existing Systems

369 369 370 371

14.7 14.8 14.9 14.10 14.11 14.12 14.13 14.14 14.15 14.16 14.17 14.18

15. Environmental Stress Screening for Electronic Equipment (ESSEE) 15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8 15.9

Introduction Environmental Stress Screening Philosophy Screening Environments Things an Acceptable Screen Are Expected to D o Things an Acceptable Screen Are Not Expected to D o To Screen or Not to Screen, That is the Problem Preparations Prior to the Start of a Screening Program Combined Thermal Cycling, Random Vibration, and Electrical Operation Separate Thermal Cycling, Random Vibration, and Electrical Operation

367 368

373 375 376 377

379 379 379 381 383 383 384 384 387 389

CONTENTS

15.10 Importance of the Screening Environment Sequence 15.11 How Damage Can Be Developed in a Thermal Cycling Screen 15.12 Estimating the Amount of Fatigue Life Used Up in a Random Vibration Screen Sample Problem-Fatigue Life Used Up in Vibration and Thermal Cycling Screen

XV

389

390 392 395

Bibliography

401

Index

405

-

CHAPTER 1

Introduction

1.I

VIBRATION SOURCES

Electronic equipment can be subjected to many different forms of vibration over wide frequency ranges and acceleration levels. It is probably safe to say that all electronic equipment will be subjected to some type of vibration at some time in its life. If the vibration is not due to an active association with some sort of a machine or a moving vehicle, then it may be due to transporting the equipment from the manufacturer to the customer. Vibrations encountered during transportation and handling can produce many different types of failures in electronic equipment unless the proper consideration are given to the mechanical design of the electronic structure and the shipping containers. Vibration is usually considered to be an undesirable condition, and in most cases it is. However, there are many applications where vibration is deliberately imposed to improve a function. Some sophisticated applications of vibrations are in the use of ultrasonics to clean medical instruments, measure wall thickness, and find flaws in castings. Vibration is also used in sorting rocks of different sizes by passing them over vibrating screens, which have groups of graduated holes. When the first jet airplanes were introduced, standard instruments from airplanes with piston engines and propellers were used. Theres instruments had a tendency to stick when they were used on the jet airplanes. The difference was due to the lower-frequency vibration levels developed in the piston engines. In order to make these same instruments work in the first jet airplanes, small vibrators had to be mounted on the instrument panels. Mechanical vibrations can have many different sources. In household products such as blenders and washing machines, the vibrations are due to the unbalance created by rotating and tumbling masses. In vehicles such as automobiles, trucks, and trains most of the vibration is due to rough surfaces over which these vehicles travel. In ships and submarines the vibration is due to the engines and to buffeting by the water. In airplanes, missiles, and rockets the vibration is due to jet and rocket engines and to aerodynamic buffeting. Most of the vibration in a missile, during subsonic flight, is due to 1

2

INTRODUCTION

+ Displacement ~

Period

-

Displacement

FIGURE 1.1. Simple harmonic motion.

the sound field developed by the rocket engines [1,6].* This is due to the extreme turbulence of the jet exhaust downstream from the rocket engine. 1.2

DEFINITIONS

Vibration, in a broad sense, is taken to mean an oscillating motion where some structure or body moves back and forth. If the motion repeats itself, with all of the individual characteristics, after a certain period of time, it is called periodic motion. This motion can be quite complex, but as long as it repeats itself, it is still periodic. If continuous motion never seems to repeat itself, it is called random motion. Simple harmonic motion is the simplest form of periodic motion, and it is usually represented by a continuous sine wave on a plot of displacement versus time, as shown in Fig. 1.1. The reciprocal of the period is known as the frequency of the vibration and is measured in cycles per second, or hertz (Hz), in honor of the German who first experimented with radiowaves. The maximum displacement is called the amplitude of the vibration. Crede [2,3] defines shock as a transient condition where the equilibrium of a system is disrupted by a sudden applied force or increment of force, or by a sudden change in the direction or magnitude of a velocity vector. Shock is not transmitted easily in the light airframe structures generally used in airplanes and missiles. Impact forces often result in a transient type of vibration, which is influenced by the natural frequencies of the airframe. Only steady-state linear vibrations are considered in this book. Linear vibrations occur with linear elastic media where the displacements are directly proportional to the applied force. If the force is doubled, the displacement is doubled. Only stresses up to the elastic limit of any given material are considered. Higher stresses generally result in permanent deformations, which extend into the plastic deformation region and are not considered here. *Numbers in brackets [ ] refer to bibliography entries in the back of the book.

DEGREES OF FREEDOM

3

FIGURE 1.2. Rotating vector simulating a single-degree-of-freedom system.

1.3 VIBRATION REPRESENTATION

A rotating vector can be used to describe the simple harmonic vibration of a single mass suspended on a coil spring (Fig. 1.2). The vector Yo rotates counterclockwise with a uniform angular velocity of R rad/s. The projection of the vector on the vertical axis represents the instantaneous displacement Y of the mass as it vibrates up and down. This can be written as

When the vector Yo rotates through one revolution, it rotates through an angle of 360°, which is 277- radians, for one complete cycle. The angular velocity is measured in radians per second, and the frequency f is measured in cycles per second. This leads to the relation

1.4

DEGREES OF FREEDOM

A vibrating system requires some coordinates to describe the positions of the elements in the system. If there is only one element in the system that is restricted to move along only one axis, and only one dimension is required to locate the position of the element at any instant of time with respect to some initial starting point, then it is a single-degree-of-freedom system. The same is true for a torsional system. If there is only one element that is restricted to rotate about only one axis so that only one dimension is required to locate the position of the element at any instant of time with respect to some initial starting point, it is a single-degree-of-freedom system. Some samples of systems with a single degree of freedom are shown in Fig. 1.3.

4

INTRODUCTION

Spring and mass

Rod and disk

Pendulum

(a)

(b)

(Ci

FIGURE 1.3. Single-degree-of-freedomsystems.

X+

F

Y

(a)

f6)

(di

(C)

FIGURE 1.4. Two-degree-of-freedomsystems.

A two-degree-of-freedom system requires two coordinates to describe the positions of the elements. Some samples of systems with two degrees of freedom are shown in Fig. 1.4. In rigid-body mechanics, a single body can have six degrees of freedom. These are translation along each of its three mutually perpendicular X , Y , and Z axes, and rotation about each of them, as shown in Fig. 1.5. A typical beam can have an infinite number of degrees of freedom since it can bend in an infinite number of shapes, or modes, as shown in Fig. 1.6.

FIGURE 1.5. A single mass with six degrees

of freedom.

+Z

-Y

VIBRATION NODES

5

_ _ - - -FIGURE 1.6. A simply supported beam showing several degrees of freedom.

k

....

Node point

,/--------

‘.._--

”

First harmonic mode

Second harmonic mode

(a)

(bi

FIGURE 1.7. First and second harmonic modes for a simply supported beam.

1.5 VIBRATION MODES

A standard manner in which a particular system can vibrate is known as a vibration mode. Each vibration mode is associated with a particular natural frequency and represents a degree of freedom. A single-degree-of-freedom system will have only one vibration mode and only one resonant frequency. The six-degree-of-freedom system shown in Fig. 1.5 can have six vibration modes and six resonant frequencies. The simply supported beam shown in Fig. 1.6 can have an infinite number of vibration modes; this is the same as saying this beam can have an infinite number of different shapes for each of its resonances, which are also infinite. The fundamental resonant mode of a vibrating system is usually called the natural frequency or the resonant frequency of the system. Sometimes it is called the first harmonic mode of the system. For example, a simply supported uniform beam vibrating at its fundamental resonant frequency has the . this beam is mode shape of a half sine wave as shown in Fig. 1 . 7 ~When vibrating at its second natural frequency, in its second harmonic mode, it has the mode shape of the full sine wave shown in Fig. 1.7b. The first harmonic mode of a system, with the lowest natural frequency, is the fundamental resonant mode; this often has the greatest displacement amplitudes and usually the greatest stresses. The second harmonic mode, or second resonance, usually has a smaller displacement than the first harmonic mode, so the stresses are usually smaller. The displacements continue to decrease for the higher resonant modes.

1.6 VIBRATION NODES

Vibration nodes are unsupported points on a vibrating body that have zero displacements. Nodes are generally associated with bending or torsion modes. At the first harmonic bending resonant mode in a beam there are no node

6

INTRODUCTION

First mode

Second mode

Third mode

Fourth mode

FIGURE 1.8. First four harmonic modes of a circular membrane.

First

mode

Second mode

Third

mode

FIGURE 1.9. First three harmonic modes of a square plate.

points. At the second harmonic mode there is one node point, the third harmonic mode has two node points, and so on. Figure 1 . 7 ~shows the bending mode of a vibrating beam with no node points, and Fig. 1.7b shows a beam with one node point. Vibrating plates can have straight-line bending nodes and circular nodes. The first four harmonic modes of a circular membrane are shown in Fig. 1.8. The plus ( + ) signs show positive displacements and the minus ( - > signs show negative displacements. The dashed lines represent nodal locations of zero displacement. The first three harmonic bending modes of a square plate with free edges are shown in Fig. 1.9.

1.7

COUPLED MODES

In a system with two or more degrees of freedom, the vibration mode of one degree often influences the vibration mode of the other degree. For example, in Fig. 1 . 4 if~ mass 2 is held rigidly while mass 1 is displaced in the vertical direction, mass 1 will oscillate up and down. Now if mass 2 is released, the motion of mass 1 will act upon mass 2 so mass 2 will begin to oscillate up and down. Because the motion of mass 1 has a direct effect on the motion of mass 2, these two vibration modes are defined as being coupled. In coupled modes, the vibration in one mode cannot occur independently of the vibration in the other mode. Coupled modes can occur in translation, rotation, and combinations of translation and rotation for systems with more than one degree of freedom. For coupled modes in translation and rotation, it is often possible to

FASTENERS

7

Y A

L

3

I I 1

I

> r

L+L+

FIGURE 1.10. A single mass with two springs.

determine whether the system is coupled or uncoupled by making a simple test. Apply a steady load to the body of the system at its center of gravity (CG), in a specific direction. If the body moves in the direction of the applied load without rotation, then the translational mode is not coupled with the rotational mode for motion along the direction of the applied load. Consider, for example, the mass with two springs as shown in Fig. 1.10. If a steady load is applied to the CG along the X axis, the mass will not only translate along the X axis, but it will also tend to rotate at the same time. This test shows that for vibration along the X axis, the translational mode is coupled with the rotational mode. If a steady load is applied to the C G along the Y axis, the mass will translate along the Y axis without rotation. Also, if a moment is applied to the mass about the CG, then rotation will take place without translation. These tests show that for vibration along the Y axis the translational mode is decoupled from the rotational mode.

1.8 FASTENERS

Many different types of fasteners are used in electronic equipment. These include screws, nuts, rivets, and clips. Fasteners are responsible for a very large percentage of field failures and, in most shock tests, they are the largest single source of failure. Although fasteners have been used in very large quantities in electronic equipment for many years, most of the applications are based on static installation considerations. The fastening techniques, which are generally chosen for ease of installation and low cost, are usually not satisfactory for severe shock and vibration environments. A contributing factor is that fasteners are such small items and their use so universal that their application tends to be semiautomatic without regard to their strength. This is particularly the case for machine screws. When large screws are used, it is often because someone has had some association with the automotive or aircraft industry, where very small screws are seldom used.

8

INTRODUCTION

Because fasteners play an important part in the overall reliability of the electronic equipment, extra consideration should be given in their selection as follows [ 5 ] : 1. Select the proper type of fastener (screws, rivets, etc.), considering such trade-offs as environment, strength, maintenance, and cost. 2. Select the correct fastener size and location based on dynamic loads and geometry of the structure. 3. Select the correct locking device for screws and nuts. 4. Select the right installation technique.

Most electronics manufacturers choose screws and rivets on a production basis. If the designer uses screws, the size and locking device are selected according to the tolerances as well as for ease in assembly. In installation, the designer depends on production personnel to use their own judgment in the proper installation of fasteners. The results may be that the wrong fasteners are used, the wrong sizes are used, the wrong lockwashers are used, and the wrong installation torques are used. It has been found, generally speaking, that cold-driven shank-expanding rivets are very satisfactory and should be used more frequently in electronic assemblies. It has been found, too, that when screws are used, they should be of larger sizes than those customarily considered and they should be made of better materials. Many locking devices commonly used are unsatisfactory. Some are often the source of many severe problems. It has also been found that a mechanic’s judgment in tightening a machine screw is usually faulty. Many failures in electronic equipment have resulted because bolts have come loose. Consider what can happen to a sensitive electronic chassis when a large transformer comes loose and rattles around during vibration. Although many investigations have been made with high-speed films and strain gages to study the loosening action of screws and nuts, the mechanism by which this occurs is still not well established. It appears, however, that the bolt stretches slightly under the action of a dynamic load, so that the interface friction forces in the area of the bolt are suddenly and sharply reduced. Since the threads in the screw or nut tend to return to their original shape during the small time increments, the small geometric changes become a driving force, which tends to loosen the screw and nut. Some specific recommendations can be made to improve the quality of the fasteners in electronic equipment: 1. Steel screws should be used in all screw fastenings. The steel should have the minimum properties of SAE 1010. 2. The screws should be tightened by a torque device, which can be preset to the required value.

FASTENERS

9

TABLE 1.1 ~~

Screw Size

Torque

Screw

Torque

(in: lb)

Size

(in: lb)

2-64 4-40 6-32 8-32 10-24 10-32

3-3.5 5-6.5 10-12 20-24 22-27 34-42

12-24 12-28 $20 -28 i-16 f -24

45-56 50-64 65-80

85-100 250-320 330-415

3. The tightening torque should be 6 0 4 0 % of the torque required to twist the head off the screw. These torque values for steel screws are given in Table 1.1 [lll. 4. The screw head should permit a positive nonslipping grip for the driving device and should also withstand the driving forces. A slottedhex-head machine screw seems to be, for general purposes, the most satisfactory. 5 . In all applications involving through-holes, locknuts should be used instead of lockwashers. Most of the standard steel locknuts are satisfactory. 6. Blind-tapped holes should be avoided, if possible. When they are necessary, locking devices such as lockwashers should be used under the screw head in order to prevent the screw from backing out during vibration. The tightening torque should be increased by an amount equal to the resisting frictional torque of the locking device. 7 . The fasteners for a unit should be distributed so that the failure of one fastener will not free the unit or cause it to malfunction. Even for very small components there should be a minimum of two fasteners. Lockwashers and screw-locking inserts should be used with care in electronic systems that will be used in the zero-G environment of outer space. These two devices create a binding friction in the screw by biting into the metal during installation. This action will very often shave metal particles from the screw. If these particles are not removed, they can float around in a zero-G environment and create electrical problems. An actual count was made of the metal particles after inserting 24 screws with external-star-type lockwashers, and a total of about 1000 metal particles were counted. The crossed-recess screwdriver slot (Phillips head) will also tend to be cut by the action of the screwdriver. When the screw is seated and the screwdriver is twisted, the screwdriver will very often twist out of the slot and shave small bits of metal out of the screw head.

10

INTRODUCTION

In order to avoid shaving small bits of metal from the screws with devices that bit into the metal, many electronic firms use liquids such as Loctite and Glyptal, which bind the screws. Some firms use nylon inserts in the screws. Nylon inserts are usually good for about a dozen insertion and removal cycles before the nylon cold flows and reduces the binding torque. NASA will not permit the use of volatile products in the area of displays and optical devices for their spacecraft. These products tend to outgas in a vacuum and then deposit themselves in a thin film; this can coat optical lenses and interfere with the operation of sensitive optical units such as cameras and telescopes. 1.9 ELECTRONIC EQUIPMENT FOR AIRPLANES AND MISSILES

Electronic boxes used in airplanes and missiles often have odd shapes that permit them to make maximum use of the volume available in tight spaces. Since volume and weight are generally quite critical, the electronic boxes usually have a high packaging density. This value normally ranges from about 0.03 to about 0.04 lb/in.3, depending on the severity of the environmental requirements. The weight of a typical electronic box will range from about 10 to about 80 lb. The vibration frequency spectrum for airplanes will vary from about 3 to about 1000 Hz, with acceleration levels that can range from about 1 G to about 5 G peak. The highest accelerations appear to occur in the vertical direction in the frequency range of about 100-400 Hz. The lowest accelerations appear to occur in the longitudinal direction, with maximum levels of about 1 G in the same frequency range. For helicopters, the frequency spectrum will vary from about 3 to about 500 Hz and acceleration will range from about 0.5 to about 4 G. The highest accelerations appear to occur in the vertical direction at frequencies near 500 Hz. The displacement at low frequency are very large, with values of about 0.20-in. double amplitude at about 10 Hz. Missiles have the highest frequency range in this group, with values that generally go up to 5000 Hz [7]. The lower frequency limit is about 3 Hz, and this appears to be due to bending modes in the airframe structure. Acceleration levels range from about 5 to about 30 G peak, with the maximum levels occurring during power-plant ignition at frequencies above 1000 Hz. The vibration environment in supersonic airplanes and missiles is actually more random in nature than it is periodic. However, sinusoidal vibration tests are still being used to evaluate and to qualify electronic equipment that will be used in these vehicles. Because the forcing frequencies in airplanes and missiles are so high, it is virtually impossible to design resonance-free electronic systems for these environments. Of course, it is always possible to completely encapsulate an entire electronic box with some expanding rigid type of foam, which could

ELECTRONIC EQUIPMENT FOR AIRPLANES AND MISSILES

11

drive the resonant frequency well above 1000 Hz (possibly to 2000 Hz) for a small box. This is generally considered to be impractical, however, because it becomes too expensive to maintain, troubleshoot, and repair such a system. The obvious conclusion is that the forcing frequencies present in airplanes and missiles will excite many resonant modes in every electronic box. What becomes equally obvious is that extra care must be taken in the design and analysis of an electronic system, or it can literally shake itself to pieces. The electronic support structure must be dynamically tuned with respect to the electronic components to prevent coincident resonances that can lead to rapid fatigue failures. The first thought that comes to the mind of an experienced mechanical design engineer, when confronted with a severe vibration specification, is to mount the electronic equipment on vibration isolators. There is no doubt that a set of isolators, properly designed, can control shock and vibration. There are four major factors that must be considered when isolation mounts are being discussed. 1. Sway space must be provided all around the electronic equipment to keep it from colliding with other objects. If volume is scarce, it might be more practical to pack more electronics into the same volume by using a larger electronic box with hard mounts. 2. Cold plates are being used more and more in electronic structures to remove the heat dissipated by the electronic equipment. If isolators are used, flexible couplings must be provided between the airframe structure and the electronic box to take care of the large displacements developed by the isolators. Reliable flexible couplings must be used because cooling effectiveness may be sharply reduced if a coupling fails. 3. Electrical wire cables and harnesses must be used to connect the typical electronic box to the main electrical system in the airplane or missile. If isolators are used, these cables and harnesses will be forced through large amplitudes because of the sway space required by the isolators. Special precautions must be taken to prevent fatigue failures in the cables and harnesses. 4. A good vibration isolator is often a poor shock isolator, and a good shock isolator is often a poor vibration isolator. The proper design must be incorporated into the isolator to satisfy both the vibration and the shock requirements.

Cold-plate designs for airborne electronic equipment have become more and more sophisticated with the use of air and liquids. Air heat exchangers make extensive use of multiple fins, wavy fins, split fins, and pin fins in order to improve the heat-transfer characteristics. The multiple-fin heat exchangers are usually dip-brazed aluminum with as many as 22 fins per inch. These fins may be only 0.006 in. thick, but the large number of fins in a typical heat

12

INTRODUCTION

exchanger makes it quite rigid for its weight. Airplanes make extensive use of electronic equipment where a cooling-air heat exchanger is built right into the electronic support structure. The heat exchanger is riveted, brazed, or cemented to the major structural members in the electronic box, so that the heat exchanger itself becomes a major load-carrying member of the system. The cooling air for the cold plate is usually taken from the compressor on the jet engine that powers the airplane. This air must be conditioned before it can be used for cooling, because the air temperature from the first stage is usually greater than 300°F. Liquid-cooled cold plates are usually used to cool electronic equipment on spacecraft or very-high-flying research airplanes. The cooling liquid can be a mixture of ethylene glycol and water, or some other liquids such as FC-75, a fluorocarbon or Coolanol 45, a silicone fluid. The liquid-cooled cold plates are usually made part of the spacecraft airframe structure instead of the electronic box structure. Then, when the electronic box must be removed from the spacecraft, it is not necessary to disconnect fluid lines, which can become quite messy. Since the cold plate stays with the airframe, the heat dissipated by the electronic box is usually transferred to the cold plate through a flat interface on the mounting surface of the electronic box that makes intimate contact with the cold plate. The trend in commercial and military electronics is toward the line replaceable unit (LRU), with which it is possible to replace a defective electronic box right on the flight line in a matter of minutes. This is accomplished by providing all of the required interface connections, both mechanical and electrical, at the back end of the electronic box. The box becomes similar to a printed circuit board (PCB) that can be plugged into its receptacle. If there are several large electrical connectors on the back end of the electronic box, it may be quite difficult to insert the box and engage the electrical connectors properly. Some connectors may require a force of 0.50 lb/pin for proper engagement. When there are 8 connectors, each with 100 pins, there is a total of 800 pins, which will require a 400-lb force to engage them. The plug-in electronic box is usually engaged and locked into position by some mechanism at the front of the box. Since the connectors are at the rear of the box, this means the force required to engage the connectors must pass through the box. When this type of electronic box is subjected to vibration, in many instances the vibration loads must be added to the installation loads to determine the total load acting on the structure. There has been an attempt to standardize electronic equipment used in military and commercial airplanes by establishing certain sizes for modular electronic units. These modular units are then mounted in a standard air transport rack (ATR), which provides rear-located dowel pins and connectors and a quick-release fastener at the front.

ELECTRONIC EQUIPMENT FOR SHIPS AND SUBMARINES

13

1. l o ELECTRONIC EQUIPMENT FOR SHIPS AND SUBMARINES

Ships and submarines will generally make use of console cabinets to support their electronic equipment, since there is usually more room available and weight is not very critical. The electronic components are usually mounted on panels and in sliding drawers. Panels are generally used to support dials, gages, manual controls, and test points. Only small masses are mounted on panels, because they are fastened to a frame or rack in the cabinet and they cannot withstand large dynamic loads. Drawers are often used to support the more massive electronic components such as those normally used in power supplies. The drawers are mounted on telescoping slides to provide access to the equipment. For safety, the drawers usually lock in the open and closed positions. For convenience, the drawers may also tilt to improve access in tight spaces. The vibration frequency spectrum for ships and submarines varies from about 1 to about 50 Hz, but the most common range is from about 12 to about 33 Hz. The maximum acceleration level in this range is about 1 G and appears to be due to vibrations set up by the engines and propellers. In military ships, shock is an important factor, due generally to various explosions, which can do extensive damage to electronic equipment, unless proper consideration is given to the design and installation. For example, it is not desirable to have a very rigid structure supporting the electronics, because a very rigid structure may not deform enough to absorb much stain energy. Theoretically, any structure that does not deform when it is subjected to an impact load will receive an infinite acceleration. A large displacement is desirable, since it can substantially reduce acceleration levels. Either this displacement must be confined to the structure or shock mounts must be used. In either case, provisions must be made in the design and installation to make sure parts will not collide and equipment will not break loose [36]. If shock isolators are used, they should be designed to deflect enough to absorb the shock energy without transmitting excessive loads to the electronic equipment. The shock mounts should have a minimum resonant frequency of about 25 Hz [2,51. Ideally, the resonant frequency of the electronic components should be at least twice that of the shock mounts, but never below 60 Hz. If the electronic components have resonances substantially below 60 Hz, it might bring them into the range of the most common vibration forcing frequencies, which, as previously mentioned, are as high as 33 Hz. If this should happen, the electronic components would be driven continuously near their resonance and could suffer fatigue failures. When the forcing frequency of the ship’s structure is near its higher frequency limit (around 25 Hz), the resonant frequency of the electronic equipment cabinet will be excited, since the shock isolators also have their resonance at 25 Hz. This condition should not impose high stresses on the electronic components mounted in the cabinet if the component resonance is

14

INTRODUCTION

twice that of the isolators. The cabinet support structure will, however, have to withstand the dynamic vibration loads. These loads will be determined by the amplification characteristics of the shock isolators during vibration. Shock isolators are available that provide a vibration amplification of about 3 for the conditions described above. Since the general vibration acceleration input level in these frequency ranges is normally quite low, an amplification factor of 3 for the isolators does not result in high stresses in the equipment cabinet. It is generally not desirable to increase the resonant frequency of the electronic equipment as high as possible. If the equipment is very stiff, it may be good for the vibration condition but poor for the shock condition. A very high spring rate may result in very high shock stresses because of the high acceleration loads. Lee [5] recommends a maximum resonant frequency of about 100 Hz and a maximum acceleration of 200 G on the electronic components. On tall narrow cabinets the load-carrying isolators should be at the base and stabilizing isolators should be at the top. A rigid structure must be used to support the stabilizing isolators at the top. If there is excessive deflection in the top support structure, it can change the characteristics of the entire system because of severe rocking modes. If shock isolators are not used, the shock energy must be absorbed by deflections in the electronic equipment cabinet and in the structure of the ship supporting the cabinet. In this case the natural frequency of the assembly, which consists of the cabinet and the ship’s structure, should be about 60 Hz. The natural frequency of the electronic components mounted on the equipment cabinet should be twice that of the assembly, or about 120 Hz. The ship’s structure must provide a good part of the deflection required to attenuate the shock force, or the dynamic stresses in the equipment cabinet may be high enough to cause structural failures. When the vibration forcing frequency in the ship’s structure is near its normal maximum limit of 33 Hz, the dynamic loads in the equipment cabinet will not be amplified to any great extent since the resonant frequency of the cabinet, at 60 Hz, is almost twice the forcing frequency. Furthermore, the electronic components mounted in the equipment cabinet have a still higher resonant frequency (120 Hz), so their dynamic vibration loads should be relatively small. The materials that are best suited for shock are ductile materials with a high yield point, a high ultimate strength, and a high percentage elongation. In general, metals that are mechanically formed are more desirable than cast metals, which have a relatively low percentage elongation. Since acceleration forces become progressively smaller as they propagate into the interior of the equipment cabinet, the electronic components that can withstand the highest G forces should be mounted near the exterior of the cabinet. Electronic components that have their own rigid structures

ELECTRONIC EQUIPMENT FOR AUTOMOBILES, TRUCKS, AND TRAINS

15

should be used to lend additional strength to the outer structural elements in the cabinet. Electronic components that cannot withstand high G forces should be mounted at the maximum elastic distance from the application points of the shock load. This will usually be at the center of the cabinet. The load path for each mass element in the system should be examined closely to determine the path the load will take as it passes through the structure. For example, a large transformer should be mounted close to a major structural support in order to reduce the length of the load path. This will result in smaller deflections and stresses.

1.1 1 ELECTRONIC EQUIPMENT FOR AUTOMOBILES, TRUCKS, AND TRAINS

Electronic equipment for automobiles and trucks has grown very rapidly in the past few years. Electronics are being used in their antilocking brake systems, fuel-air mixture control, radios, ignition systems, air conditioners, heaters, automatic transmission shifting points, rear view mirrors, door locks, instruments, global positioning systems (GPSs), windows, sun roofs, cruise control, air bags, television, telephones, and many other features, with more being added every year. Anticollision systems are being developed that will automatically engage brakes when vehicles are dangerously close to each other at high speeds. There are plans for computerized car trains that will link several automobiles together going to similar areas. Antitheft systems such as Lojack are available that can be activated by radio to send out a silent alarm to police when an automobile is stolen. Automobile and truck electronics must operate in harsh environments that include wide temperature swings, high under-hood operating temperatures of 140"C, high humidity with condensation, high-velocity splashing water, and subfreezing temperatures with only moderately high vibration and shock levels for the automobiles. Trucks are often required to travel over unpaved country roads for deliveries in outlying areas. Test data on a 2.5-ton truck instrumented with several accelerometers showed acceleration levels between about 15 and 19 G peak at speeds of 10-15 mph, at frequencies between about 15 and 40 Hz. At truck speeds above about 35 mph over rough roads the frequency appears to be random in nature. Diesel electric trains have been making use of electronics on their drive wheels to sense when the wheels are just reaching the slipping point on the tracks. A large engine may have six pairs of driving wheels. The electronics can sense when any one pair of drive wheels will slip. Sand is automatically dispensed to increase friction and the electric motor driving the one set of drive wheels is automatically slowed slightly to prevent slopping on the

16

INTRODUCTION

tracks. A single diesel electric engine with this new feature can go up steeper grades and pull three times as many cars as previous diesel electric trains. 1.12 ELECTRONICS FOR OIL DRILLING EQUIPMENT

Oil drilling equipment probably has the most severe thermal and vibration requirements for electronic systems. Deep drilling at 30,000 feet can encounter temperatures as high as 200°C. The electronics are usually located in a 6-foot long tube section just above the cutters. The coolant is the 200°C mud washing over the electronics tube section. Sometimes the cutters get stuck in a rock section while drilling at 30,000 feet. The turntable at the surface keeps turning while the cutters are stationary. The long drilling pipe winds up and when it breaks loose it generates random vibration acceleration levels as high as 30 G RMS. This means that the electronics section in an oil drilling rig must be very rugged to withstand the high vibration levels and the high temperatures and still provide a high reliability. When an electronic failure occurs, it is a very expensive and time-consuming task to remove 30,000 feet of pipe, replace the electronics section, and drop 30,000 feet of pipe down the hole again. 1. I 3 ELECTRONICS FOR COMPUTERS, COMMUNICATION, AND ENTERTAINMENT

According to many people in the three fields of computers, communication, and entertainment, they are rapidly converging into one giant industry with the internet. Sometimes this is called bandwidth, because fiber optic cables are being added, which can carry a very wide range of frequencies that are required to provide services for television, radio, data processing, stock trading, banking, internet, telephone, security, and more energy efficient homes and offices.

-

CHAPTER 2

Vibrations of Simple Electronic Systems

2.1

SINGLE SPRING-MASS SYSTEM WITHOUT DAMPING

The natural frequency of many single-degree-of-freedom systems can be determined by evaluating the characteristics of the strain energy and the kinetic energy of each system. Considering a single spring-mass system, for example, if there is no energy lost, then the maximum kinetic energy of the mass must be equal to the maximum strain energy in the spring, if the spring mass is negligible. All real systems have some damping. If a mass is suspended on a real coil spring and the spring is stretched and released, the mass will vibrate up and down. This free vibration may continue for a long time, but eventually all free vibrations die out and the mass stops vibrating. This is because damping in the spring dissipates a little energy with each cycle and eventually the mass stops moving. If there is no damping, the mass will theoretically keep on vibrating up and down with the same amplitude and frequency forever. In many systems, the damping is so small that it has very little effect on the natural frequency. Under these circumstances the natural frequency of the damped system can conveniently be approximated by the natural frequency of the undamped system. The maximum kinetic energy of a vibrating spring-mass system with no damping is at the point of maximum velocity. This occurs as the vibrating mass passes through the zero-displacement point, as shown in Fig. 2.1. From elementary physics the maximum kinetic energy T of the vibrating mass is

To = i m V 2 (2.1) The instantaneous tangential velocity I/ can be expressed in terms of rotational velocity as shown in Fig. 1.3:

v=

Yon (2.2) Substituting Eq. 2.2 into Eq. 2.1, the kinetic energy of the system becomes To = ;my,Zn*

(2.3) 17

18

VIBRATIONS OF SIMPLE ELECTRONIC SYSTEMS

T

,

+Displacement

t

< -<

rJ-1

;- - _ - - - - - --

Maximum strain energy

J Time

'Maximum velocity -Displacement

FIGURE 2.1. A vibrating spring-mass system.

The maximum strain energy can be determined from the work done on the spring by the mass as it stretches and compresses the spring during vibration. Since the spring is linear, the deflection is directly proportional to the force as shown in Fig. 2.2. The area under the curve represents the work done on the spring. Since the work is equal to the strain energy, the strain energy becomes

U" = +P,Y"

(2.4)

The spring rate K can be defined in terms of the maximum load P, and the maximum deflection Yo as K = P,/Y,, so

Po = KY,,

(2.5)

Substituting Eq. 2.5 into Eq. 2.4, the maximum strain energy becomes

uo= ;KY; If damping is assumed to be zero, then at resonance the kinetic energy will be equal to the strain energy: i,y2flz 2

=' K y 2

(I

2

0

[E) 1/ 2

Q" =

SINGLE SPRING-MASS

SYSTEM WITHOUT DAMPING

I9

This is the natural frequency in radians per second. Using Eq. 1.2, the natural frequency in cycles per second (Hz) is

The natural frequency equation can be written in a slightly different form by considering that the static deflection a,, of the spring is due to the action of the weight W acting on the spring. The spring rate can then be written as

Expressing the mass in terms of weight W and acceleration of gravity g , we have

Substituting Eqs. 2.8 and 2.9 into Eq. 2.7 gives another relation for the natural frequency: (2.10)

Frequency of a Cantilever Beam

Sample Problem-Natural

Determine the natural frequency of the cantilevered beam with an end mass, as shown in Fig. 2.3, in the vertical direction. Solution. When the weight of the aluminum beam is small compared to the weight of the end mass, it can be ignored without too much error. The cantilever beam can then be analyzed as a concentrated load on a massless beam, which retains its elastic properties. The static deflection at the end of

1

A

y

q

g

O 50

-1rL3 B-r Sect A A

1.4 -100

-

in

---+

FIGURE 2.3. Cantilever beam with an end mass.

20

VIBRATIONS OF SIMPLE ELECTRONIC SYSTEMS

this beam can be determined with the use of a structural handbook: (2.11) where W = 10.0 Ib (end weight) L = 10.0 in. (length) E = 10.5 x l o 6 lb/in.’ (modulus of elasticity, aluminum) I = bh3/12 = (0.50)(3.0)3/12 = 1.125 in.4 (moment of inertia) (10.0)(10.0)~ =

3(10.5

X

106)(1.125)

=

0.282

in.

X

(2.12)

Substitute Eq. 2.12 into Eq. 2.10 and note that the acceleration of gravity, g , is 386 in./s2. The natural frequency becomes 1 =

187 HZ

(2.13)

The natural frequency of the cantilever beam can also be determined from the spring rate, K , of the beam using Eq. 2.11: (2.14) Using the values previously calculated, the spring rate becomes

K=

3( 10.5 x l o 6 )(1.125)

( 10.0)3

= 3.54 X

l o 4 lb/in.

Also,

w 10 m=-=-= g 386

0.0259 Ib . s’/in.

Substituting into Eq. 2.7, the natural frequency becomes

”’-[-)

1

K

27r m

f,

=

1/2

1

3.54 x l o 4 0.0259

=z(

187 HZ

The results are exactly the same as Eq. 2.13.

1

1/ ?

(2.15)

SINGLE-DEGREE-OF-FREEDOM TORSIONAL SYSTEMS

2.2

21

SINGLE-DEGREE-OF-FREEDOM TORSIONAL SYSTEMS

The energy method is convenient for determining the natural frequency of a torsional system with one degree of freedom, as shown in Fig. 2.4. The torsional system is similar to the spring-mass system shown in Fig. 2.1, except that the spring action is due to the twisting of the rod and the inertia is due to the moment of inertia of the disk about an axis perpendicular to the plane of the disk. Assume the rod mass is small. The maximum kinetic energy of the oscillating disk is

KIN = ;zme2 where I

(2.16)

= mass

moment of inertia of disk 6 = rotational angular velocity

The maximum angular velocity for the oscillating system moving through the angle Bo is

io= eon

(2.17)

Substituting Eq. 2.17 into Eq. 2.16, the maximum kinetic energy becomes (assuming a small rod mass)

(2.18) The maximum strain energy can be determined from the work done on the twisting rod by the disk as it oscillates. Since the spring rate of the rod is linear, the angular deflection will be directly proportional to the torque T applied, as shown in Fig. 2.5. The area under the curve represents the work

22

VIBRATIONS OF SIMPLE ELECTRONIC SYSTEMS

done on the rod. Since the work is equal to the strain energy, the strain energy becomes

u, = +T,O,

(2.19)

The torsional spring rate K , can be defined in terms of the torque To and the angular displacement On as follows:

K 0 = -TO

(2.20a)

To = K , Bo

(2.20b)

On

so

Substituting Eq. 2.20 into Eq. 2.19, the maximum strain energy becomes

U, = i K , O i

(2.21)

If damping is assumed to be zero, then at resonance the kinetic energy (Eq. 2.18) must equal the strain energy (Eq. 2.21):

(2.22) The natural frequency (in Hz) becomes

j""('= f

Sample Problem-Natural

1/2

2 T I,,

(2.23)

Frequency of a Torsion System

Determine the natural frequency of the torsional system shown in Fig. 2.6.

Aluminum

{-X7" 'i FIGURE 2.6. A single-degree-of-freedom torsional system.

o&

L-

f

R =' 6 0 1" = 0 50 in

23

SPRINGS IN SERIES AND PARALLEL

Solution. The torsional spring rate K , of the rod can be determined using its angular displacement 8 under the action of an external torque To: TOL e= GJ

TO K O =-

and

e

so

where L = 10.0 in. (length of rod) G = 4.0 X l o 6 1b/ine2 (shear modulus of aluminum) J = r d 4 / 3 2 = ~ ( 1 . 0 ) ~ / 3=20.0981 in.4 (polar moment of inertia)

(4.0 X 106)(0.0981) K, =

10.0

= 3.92 X

IO4 in:lb/rad

The mass moment of inertia of the aluminum disk must be taken about the axis perpendicular to the plane of the disk:

I,

’

WR

=-

2g

where W = ~(6.0)~(0.50)/(0.10l b / i r ~ . ~=)5.65 lb (disk weight) R = 6.0-in. disk radius g = 386 in./s2 (gravity)

I,

(5.65) (6.0)2 =

2( 386)

= 0.264

1b.in:s’

The torsional resonant frequency can now be determined from Eq. 2.23:

’=

I

%[

3.92 x 104 0.264

1

1,/2

=

61.5 HZ

(2.24)

2.3 SPRINGS IN SERIES AND PARALLEL If a mass is suspended on two different springs in such a manner that the load path is directly from the mass through one spring, and then through the second spring before it reaches the support, the springs are said to be in series. The series spring, therefore, implies a series load path where the load must first pass through one of the springs before it can pass through the

24

VIBRATIONS OF SIMPLE ELECTRONIC SYSTEMS

Tension a n d compression

-

Bending

Torsion

FIGURE 2.7. Springs in series.

other spring. Cutting either spring would completely destroy the system. Some samples of springs in series are shown in Fig. 2.7. Springs in series can be combined into one equivalent spring using the relation

1

1 -

Ke,

K,

1

1

K?

K3

+-+-+

(2.25)

If a mass is suspended on two different springs in such a manner that the load path from the mass to the support is split between the springs, then the springs are said to be in parallel. The parallel path then permits the load to pass through one spring without passing through the other. Cutting one spring would still permit the other spring to carry the load. Some samples of springs in parallel are shown in Fig. 2.8. Springs in parallel can be combined into one equivalent spring using the relation

K,,

=K ,

+ K , + K , + ...

(2.26)

Kz = 10001b/in.

-

SPRINGSIN SERIES AND PARALLEL

25

< < K3 = 1000 Ib/in. 6

1 <

5

3

> K4 = 800 Ib/in. >

FIGURE 2.9. A combination of series and parallel springs.

Frequency of a Spring System

Sample Problem-Resonant

Determine the natural frequency of a mass supported by several springs in series and parallel combinations as shown in Fig. 2.9, by obtaining the equivalent spring rate for the system. Solution. Springs K, and K, are in parallel, so they can be combined using Eq. 2.26:

K,

= K,

+ K, = 1000 + 1000

Now there are three springs-K,, combined using Eq. 2.25: 1

--

K,,

_-

1

K,

=

2000 lb/in. series, so they can be

K,, and K,-in

1

1

1

1

K,

K,

4000

2000

+-+-=-+-

+- 1

800

The system now consists of one spring and one mass (Fig. 2.10). The natural frequency can be determined from Eqs. 2.7 and 2.9 as follows:

f,,= 49.5 HZ

3; K,,= 500 >

1

Ib/in.

W = 2 Ib

FIGURE 2.10. A single spring-mass system.

(2.27)

26

VIBRATIONS OF SIMPLE ELECTRONIC SYSTEMS

2.4 RELATION OF FREQUENCY AND ACCELERATION TO DISPLACEMENT

Many important approximations can be related to the dynamic displacements developed during resonant conditions. For example, if the geometry of a specific structure can be defined, then dynamic bending moments and bending stresses can be calculated from the dynamic displacements. Since displacements are very seldom measured during vibration tests, because optical measurements are often difficult to make, accelerometers are generally used. This permits test data to be taken in terms of frequency and acceleration G forces. The relation of frequency and acceleration to the displacement must therefore be determined to use the test data. Consider a rotating vector that is used to describe the simple harmonic motion of a single mass suspended by a spring. The vertical displacement Y of the mass can be determined by the projection of the vector Yo on the vertical axis as shown in Fig. 2.11. The vertical displacement can be represented by the equation

Y = Yosin nt

(2.28)

The velocity is the first derivative, I/= Y =

dY -=

dt

Szy, cos nr

The acceleration is the second derivative.

A

..

=

Y=

~

d2Y = dt’

-a2yosin f i t

The negative sign indicates that acceleration acts in the direction opposite to displacement.

27

RELATION OF FREQUENCY AND ACCELERATION TO DISPLACEMENT

The maximum acceleration will occur when sin Rt is 1: A,,,

= R2Y0

(2.29)

The acceleration in gravity units (G) can be determined by dividing the maximum acceleration by the acceleration of gravity: g = 32.2 ft/s2

= 386 in./s2

Also, changing radiants to cycles per second, we have fl=2.rrf Substituting into Eq. 2.29,

G = -A,,, g

- 4X2f2YJ 386 (2.30)

Note that the displacement Yo is the single-amplitude displacement. Sample Problem-Natural

Frequency and Stress in a Beam

Determine the resonant frequency, dynamic displacement, and maximum dynamic stress in the bracket with a transformer, as shown in Fig. 2.12, for a 5.0-G peak sinusoidal vibration input. Solution. Consider the transformer as a concentrated mass at the center of a massless beam. Since small screws are used to support the beam, the ends are considered to be simply supported. The natural frequency of a simply supported beam with a concentrated center mass can be determined from the static displacement as shown in Fig. 2.13, using Eq. 2.10. The static displacement can be obtained from a handbook: 6,,

W Transformer Screw

WL3 48 EI

(2.31)

=-

Bracket

L ---

0503 Sect A A FIGURE 2.12. A transformer mounted o n a bracket.

28

VIBRATIONS OF SIMPLE ELECTRONIC SYSTEMS

W=201b

-- --

FIGURE 2.13. A simply supported beam with a concen-

trated load.

__

~

-8O---.--.

where W = 2.0 Ib (transformer weight) L = 8.0 in. (length) E = 10.5 x l o 6 (aluminum-beam modulus of elasticity) I = bh3/12 = (1.2)(0.50)'/12 = 0.0125 i n 4 (moment of inertia) g = 386 in./s2 (2.0) ( 8.0)3 =

48(10.5

X

106)(0.0125)

=

1.62 X

in.

Substitute into Eq. 2.10 for the natural frequency:

f,

= 246

Hz (simply supported ends)

(2.32)

Equation 2.30 can be used to determine the dynamic deflection at the center of the transformer bracket, using the 246-Hz resonance: 9.8G 9,8Gi,Q yo=-- -

f2

(2.33)

f 2

where G,, = 5.0 G (input acceleration) Q = 30 (transmissibility at resonance) f , = 246 Hz (simply supported beam)

Yo=

9.8(5 .O) (30) =

(246)'

0.0243 in.

(2.34)

The approximate transmissibility for a beam-type structure can be obtained from the resonant frequency. For an electronic subassembly, experience has shown that a value equal to two times the square root of the resonant frequency, or about 30, gives reasonable results.

RELATION OF FREQUENCY AND ACCELERATION TO DISPLACEMENT

29

The dynamic load acting on the bracket can be approximated by letting Eq. 2.34 equal Eq. 2.31 and solving for the dynamic load wd:

w, =

w,=

48 EW, ~

L3 48(10.5 X 106)(0.0125)(0.0243)

(8.0)3

W, = 300 lb

(dynamic load)

(2.35)

If the acceleration loads were assumed to act directly on the transformer, the dynamic load would be ?t$ = WG,,Q

(2.36)

where W = 2.0 lb (transformer weight) Gin= 5.0 G (peak input) Q = 30 (transmissibility at resonance) Thus W, = (2.0) ( 5 .O) (30)

= 300 Ib

(2.37)

This result is exactly the same as that shown in Eq. 2.35, which was obtained from the dynamic deflection of a simply supported beam with a concentrated load. The dynamic bending stress in the simply supported transformer bracket can be determined from the geometry of the structure, using the dynamic loading as shown in Fig. 2.14. The maximum dynamic bending moment

Lr7

Wd = 300 Ib

4.0

Jlransfcyy

bracket

8.0

R = 150

R = 150

_L50_t.-150-~-

Shear diagram

FIGURE 2.14. Shear and bending-moment diagram for a beam

30

VIBRATIONS OF SIMPLE ELECTRONIC SYSTEMS

occurs at the center of the bracket: =150(4) =600in:lb

(2.38)

The maximum dynamic bending stress can be determined from the standard bending-stress equation:

s, =

MC r

(2.39)

where M = 600 in: lb (bending moment) c = h/2 = 0.50/2 = 0.25 in. (ref. Sect. A.A. Fig. 2.12) I = bh3/12 = (1.2)(0.50)'/12 = 0.0125 ine4 Hence

s, =

(600)(0.25) 0.0125

=

12,000 lb/in.2

(2.40)

A stress concentration factor must be used when several million stress reversals are expected. When only a few thousand stress cycles are expected in a ductile material, then a stress concentration factor is not critical. 2.5

FORCED VIBRATIONS WITH VISCOUS DAMPING

When a harmonic shaking force acts on a spring-mass system with damping, the resulting forced vibration will also be harmonic. The final frequency of the mass will be the same as that of the shaking force, because the initial transient vibrations will eventually be dissipated by the damping. Consider the harmonic shaking force Po cos R t acting on a damped spring-mass system. A free-body force diagram is shown in Fig. 2.15. The differential equation of motion with an external force exciting the system is mY+

C Y + Ky=

P" cos at

(2.41)

The solution of this equation consists of a complementary function plus a particular function. The complementary solution is the free vibrations. These will die out because of the damping. The particular solution can be taken in the form Y=Y,cos(Rt-

e)

(2.42)

FORCED VIBRATIONS WITH VISCOUS DAMPING

t

1

Po cos Ltt

31

Po cos nt

FIGURE 2.15. Forced vibration forces acting on a simple system.

The maximum displacement, Yo,can be expressed in terms of the maximum impressed force, Po, as follows:

Yo=

PO

[ ( K - mQ2)' + c 2 R 2 ]'''

(2.43)

Dividing the numerator and denominator of the above equation by K and substituting Eq. 2.6 leads to the following:

Po -

--

K

(2.44:)

Let Y,,= P o / K , where Y,, is the deflection of the system due to the maximum dynamic input load acting as a static load. For additional simplification, let

R R,=-

a"

C

and

R , = -cc

(2.45)

This leads to the general amplification (not transmissibility) equation A = -YO =

1

(2.46)

A plot of the dynamic amplification ratio A against the frequency ratio R , is shown in Fig. 2.16. This is not a force-transmissibility curve. A force-transmissibility curve is shown in Fig. 2.17.

32

VIBRATIONS OF SIMPLE ELECTRONIC SYSTEMS

10 8

6 4

2

10 08

06 04

02 01

0

10

0.5

15

R"

=

25

20

Forcing frequency

30

f

--

f,

Natural frequency

FIGURE 2.16. A dynamic amplification curve for a simple system.

10

a 6 4

02 01

0

05

10

R"

15

25

20

30

Forcing frequency - f Natural frequency f,

_-

=

FIGURE 2.17. A transmissibility curve for a simple system.

33

FORCED VIBRATIONS WITH VISCOUS DAMPING

The instantaneous magnitude of the force experienced by the support is the vector sum of the spring force and the damper force. These two forces have a 90” phase angle between them, so the force becomes

Fo = Yo(K 2

+~~fl’)’~’

Substituting Eq. 2.43 into the above expression,

Fo =

+C2fl2)li2 [ ( K - mn’)2 + C W ] 1/ 2 Po( K 2

The transmissibility Q, which represents the ratio of the maximum output force F,, to the maximum input force Po, can be determined by dividing the numerator and denominator by K and substituting Eq. 2.6 into the above equation:

By substituting Eq. 2.45 into Eq. 2.47, the transmissibility Q becomes

1 + (2R,Rc)2

(2.48)

(1 - R $ ) ’ + ( 2 R , R c ) 2

A plot of the force transmissibility Q against the frequency ratio R , is shown in Fig. 2.17. This curve is very similar to the amplification curve shown in Fig. 2.16. In the resonant area, where R , = 1.0, the two curves are almost identical for systems with a low damping ratio. For lightly damped systems, where R: is small enough to be assumed zero, the transmissibility expression reduces to

1 e=--1-Rk

(2.49)

34

VIBRATIONS OF SIMPLE ELECTRONIC SYSTEMS

A very convenient relation can be obtained from Eq. 2.48 by considering the transmissibility at resonance, where R , = 1.0: 1

+ (2RC)' W C l 2

For lightly damped systems, the ratio R: is very small compared to the value of 1 in the numerator; thus the transmissibility equation can be reduced to Q=-

1

(2.50)

2Rc 2.6

TRANSMISSIBILITY AS A FUNCTION OF FREQUENCY

The transmissibility of a system can be related to the resonant frequency and damping as follows: c=

/E Q2- 1

(2.51)

For a lightly damped system, where Q is greater than about 10, the relation can be solved approximately for the transmissibility:

Q=

(Km)"l

(2.52)

When the spring rate K is increased, the resonant frequency is increased. This decreases the dynamic displacement and the stresses, so the damping c is decreased, which increases the transmissibility Q. When the mass m is increased, the resonant frequency is decreased. This increases the dynamic displacement and the stresses, so the damping is increased, which decreases the transmissibility. Sample Problem-Relating the Resonant Frequency to the Dynamic Displacement

The natural frequency of a printed circuit board (PCB) is 100 Hz, with a transmissibility of 10 at the center of the circuit board. It is desirable to keep the maximum dynamic deflection at the center of the circuit board to 0.020 in. (single amplitude), in order to keep from overstressing the electrical lead wires on the resistors, capacitors, and diodes mounted on the circuit board. What should the natural frequency of the circuit board be for a 5-G peak sinusoidal vibration input?

TRANSMISSIBILITY AS A FUNCTION OF FREQUENCY

35

Solution. The PCB can be approximated as a single-degree-of-freedom system when it is vibrating at its fundamental resonant mode. Therefore Eq. 2.33 can be used to estimate the single-amplitude displacement for the 100-Hz resonant frequency:

Yo=

9.8Gi, Q

f2

where Gin= 5.0-G input (peak) Q = 10 (transmissibility at resonance) f, = 100 Hz (resonant frequency)

Yo=

9.8( 5 .O) (10) (

= 0.0490

in.

(single amplitude) (2.53)

This displacement is much too high. If the resonant frequency is increased, the displacement will decrease very rapidly as shown by Eq. 2.33. However, Eq. 2.52 shows that the transmissibility will also increase as the resonant frequency increases. Extensive test data on PCBs, with various edge restraints, indicate that many epoxy fiberglass circuit boards, with closely spaced electronic component parts, have a transmissibility that can be approximated by the relation

Q=fl

(2.54)

Substitute into Eq. 2.33 and solve for the resonant frequency:

( 2 3)

To obtain a single-amplitude displacement of 0.020 in. with a 5-G peak vibration input, the circuit board resonant frequency must be increased to the following: (2.56)

36 2.7

VIBRATIONS OF SIMPLE ELECTRONIC SYSTEMS

MULTIPLE SPRING-MASS

SYSTEMS WITHOUT DAMPING

The fundamental resonant frequency of a multiple spring-mass system, without damping, can be determined by considering the strain energy and the kinetic energy of the system. If there is no energy lost, then the total kinetic energy of the masses must be equal to the total strain energy of the springs. Consider the multiple spring-mass system as shown in Fig. 2.18. The total kinetic energy of the system is the sum of the individual kinetic energies of each mass as shown by Eq. 2.3:

T,,,,,

=

$m,Y:02 + i2 m 2 Y’O’ 2

+ $m,Y;’02

This can be written as

(2.57)

The total strain energy of the system is the sum of the individual strain energies of each spring, as shown by Eq. 2.4:

This can be written as

( 2S 8 )

FIGURE 2.18. A multiple spring-mass system without damping.

rF1 yj

MULTIPLE SPRING-MASS

SYSTEMS WITHOUT DAMPING

37

Since the total kinetic energy must equal the total strain energy, set Eq. 2.57 equal to Eq. 2.58 and solve for the natural frequency:

(2.59)

Note that the deflections in the above equation are the total deflections. For example, in Fig. 2.18 the total deflection of mass 3 must include the deflections of masses 1 and 2, and the total deflection of mass 2 must include the deflection of mass 1. Also, the effective weight of mass 1 on spring 1 must include all three weights. Sample Problem-Resonant Frequency of a System

Calculate the fundamental resonant frequency of the two spring-mass system shown in Fig. 2.19. Solution. The static deflection of weight 1 will be influenced by weight 2, since weight 2 is attached to weight 1; thus both weights will act on spring 3 . We have

Yl =

wl+ w,=--2 + 4 Kl

1000

- 0.006 in

(2.60)

The static deflection of weight 2 must include the static deflection of weight 1, since weight 2 is attached to weight 1:

Yz

1

1

w2

Yl + - = 0.006 K,

4

+ -- 0.00733 in. 3000

(2.61)

2.0 Ib

. 5

3 K2 = 3000 Ib/in. 4.0 Ib

FIGURE 2.19. A two-degree-of-freedom system without damping.

38

VIBRATIONS OF SIMPLE ELECTRONIC SYSTEMS

FIGURE 2.20. A free-body force diagram for a two-mass system.

Substitute into Eq. 2.59:

[(2)(0.006)

1

f f,

=-[

+ (4)(0.00733)](386)

(2)(0.006)2 + (4)(0.00733)*

2~

= 37.3

(2.62)

HZ

The static deflections of weights 1 and 2 in Fig. 2.19 could also have been determined by the simultaneous solution of the deflection equations for each mass. A free-body force diagram of the system is shown in Fig. 2.20. Consider the forces in the vertical direction: For weight 1,

- K , Y , + W , + K 2 ( Y 2 - Y Y =i )O -(

K , + K 2 ) Y ,+ K 2 Y 2+ W , = O

(2.63)

For weight 2,

- K , ( Y 2 - Y,)+ w,= 0 -KzYz

+ K,Y, + Wz = 0

(2.64)

Substituting values for K , , K 2 , W,, and W, into the above equations and solving,

+2 = 0

+

- 4000Y1 3000Y2 3000Y, - 3000Y,+ - IOOOY,

4= 0

+6=0

-

h

Y , = -= 0.006 in 1000

Yz =

3000(0.006) 3000

+4

(2.65) = 0.00733

in.

(2.66)

These results are exactly the same as those shown by Eqs. 2.60 and 2.61.

Component Lead Wire and Solder Joint Vibration Fatigue Life

3.1

INTRODUCTION

Electronic components are available in a large variety of types, sizes, and materials for leaded and leadless mounting and for through-hole or surface mounted applications. Most of these components will end up on printed circuit boards (PCBs) for use in everything electronic from cameras and washing machines to telephones and space shuttles. Some types of electronic equipment will be used in homes and offices that are air conditioned with very quiet environments. Other types of equipment will be used in military programs with high vibration and shock levels. It does not matter what the electronic equipment is doing, or what functions it performs, or how the equipment is used-in all cases the buyers or the consumers want a low cost, reliable, easy to use piece of equipment. Sometimes they get it and sometimes they don’t. Luck often seems to play a large part in the quality of electronic products. With good luck they last for years. With bad luck they last for weeks. A satisfied customer is a repeat customer. A quality product will usually attract repeat customers, even when the price of the product is slightly higher than the price of the competition. Most electronic failures are mechanical in nature. Many of these mechanical failures occur in the component lead wires and solder joints. Extensive military testing experience over a period of many years has shown that about 80% of the electromechanical failures are due to some type of thermal condition and about 20% of the failures are due to some form of vibration and shock. This chapter is devoted to the examination and evaluation of sinusoidal vibration, and its fatigue effects on the electrical lead wires and solder joints on different types of components and PCBs [25]. VIBRATION PROBLEMS WITH COMPONENTS MOUNTED HIGH ABOVE THE PCB

3.2

Electronic components are usually attached to PCBs with the use of solder. Different attachment methods can be used including dip soldering, wave 39

40

COMPONENT LEAD WIRE AND SOLDER JOINT VIBRATION FATIGUE LIFE

soldering, and infrared, vapor phase, oven, and even hand soldering. Eutectic solder is a mixture of 63% tin and 37% lead. It has a melting temperature of about 183°C. A wave soldering machine for through-hole devices uses a solder temperature of about 230°C to get the proper wicking action up the plated through-holes with the lead wires. Most semiconductor electronic components are rated below a temperature of about 200°C. Some types of capacitors are only rated at 85°C ambient. When these components are wave soldered to a PCB, the heat from the solder can be conducted up the metal lead wires directly into the silicon semiconductor chips, causing them to malfunction. The cure for this problem is often to raise the components high above the PCB. Longer lead wires increase the length of the heat flow path to the temperature-sensitive chips. This increases the thermal resistance so the temperature rise to the chip is reduced. The thermal problem is solved but it creates a vibration problem as shown below. Sample Problem-Vibration TO-5 Transistor

Fatigue Life in the Wires of a

A TO-5 transistor is mounted on the top of a 0.25-in. high plastic tube, which is sitting on the top surface of a PCB. The purpose of the high mounting is to reduce the excessive heat that is conducted up the three wires during the wave soldering operation, causing electrical malfunctions in the transistor. The longer wires reduce the temperature rise, which solves the thermal problem. Will the tall transistor wires survive a qualification dwell test requirement of 30 min with a 4.0-G peak sinusoidal vibration input level up to a frequency of 2000 Hz?

Solution Find the Response of the Transistor and the Wire Fatigue Life. The approximate natural frequency of the tall transistor can be obtained by considering it to be similar to a cantilevered beam with a concentrated end mass, using the static displacement natural frequency relationship:

2%-

(see Eq. 2.10)

where W = 1.0 g = 0.0022 lb (transistor weight) L = 0.25-in. (length of three lead wires) E = 20 x lo6 lb/in.* (kovar modulus of elasticity)

I

=

%-dJ/64= %-(0.016)'/64 = 3.22 X (one wire moment of inertia)

in.4 (3.1)

VIBRATION PROBLEMS WITH COMPONENTS MOUNTED HIGH ABOVE THE PCB

g

= 386

in./s2 (acceleration of gravity)

WL3

Sst =

S,,

41

(0.0022) (0.25)

E + (3)(20 X 106)(3 X 3.22 X l o p 9 )

= 5.93 X

f,,= % .2

lo-’ in.

(see Eq. 2.11)

(displacement for three wires)

/-

5 . 9 3 x lo-’

= 406

Hz (natural frequency)

(3.2) (3.3)

The dynamic force Pd acting on the transistor can be obtained from the acceleration and the mass of the transistor. The acceleration depends on the transmissibility expected at the natural frequency of the transistor. A good approximation for a beam-type structure is about two times the square root of the natural frequency. However, since the transistor is sitting on (and not cemented to) the top of a plastic tube, the interface slapping and impacting of the transistor on the tube during vibration will tend to reduce the expected transmissibility by about 50%. A transmissibility equal to the square root of the natural frequency will therefore be used here:

Pd = WGinQ (see Eq. 2.36) where W = 0.0022 lb (transistor weight) Gin= 4.0 peak (acceleration input level) = 20 (approximate transmissibility expected) Q=

Pd= (0.0022) (4.0) (20)

= 0.176

Ib

(dynamic force on transistor) (3.4)

The dynamic bending stress in the three lead wires can be obtained from the standard bending stress equation shown below. Stress concentrations due to cuts, scrapes, and scratches in the wires must be considered when many thousand stress reversal cycles are expected. The stress concentration can be used in the bending stress equation below, or it can be used to define the slope of the S-N fatigue curve for the kovar lead wire material. The stress concentration factor is normally used only once in either location. In this problem the stress concentration factor will not be used in the bending stress equation below. The stress concentration will be used in the kovar S-N fatigue curve, which is shown in Fig. 3.1. S,

where M

MC = -

I

(see Eq. 2.39)

= PdL = (0.176)(0.25) = 0.044

lb *in.(dynamic bending moment (3.5) c = d / 2 = 0.016/2 = 0.008 in. (wire radius)

42

COMPONENT LEAD WIRE AND SOLDER JOINT VIBRATION FATIGUE LIFE

Kovar wire

I

1

N , cycles to fail

FIGURE 3.1. S-N log-log fatigue curve for kovar wire, with a stress concentration factor of 2.

I

=

(3X3.22 x = 9.66 x wires) (see Eq. 3.1)

s, =

in.4 (moment of inertia, three

(0.044) (0.008)

9.66 x 10-9

= 36,439

lb/in.'

(3.6)

The fatigue life expected for the transistor kovar lead wires can be obtained from the S-N fatigue curve shown in Fig. 3.1. This is a log-log plot of the stress S and the number of cycles N to fail. Fatigue tests are run to failure on many similar closely machined samples. The results show there is a lot of scatter in the failure data. The best average straight line is drawn through the scattered test points to represent the typical fatigue properties of the material. The slope of the fatigue curve is adjusted to include a stress concentration factor of 2.0. This results in a fatigue exponent slope of b = 6.4. Every point on the straight sloped line represents a failure. The equation for every point on the sloped line is shown below: N,SP

=

N2Sf

(3.7)

Then

where N2 = 1000 cycles to fail at 84,000 lb/in.' (kovar wire reference point) S, = 84,000 lb/in.* (ultimate tensile strength of kovar lead wire)

VIBRATION FATIGUE LIFE IN SOLDER JOINTS OF A TO-5 TRANSISTOR

43

S , = 36,439 lb/in.’ (wire bending stress) (see Eq. 3.6) b = 6.4 (slope of kovar fatigue curve with stress concentration of 2)

N,

=

Wire life

=

84,000 6 . 4 (1000) 36,439

]

(

= 2.096 X

2.096 x l o 5 cycles to fail (406 cycles/s) (60 s/min)

l o 5 wire cycles to fail (3.8)

= 8.6

minutes to fail

(3.9)

3.3 VIBRATION FATIGUE LIFE IN SOLDER JOINTS OF A TO-5 TRANSISTOR Vibration forces generate overturning moments in the three lead wires. These moments result in shear tearout stresses in the solder joints, which can be obtained from the equation shown below. Microscopic examinations of failed solder joints have been made. This shows the solder appears to fail someplace between the outer diameter of the wire and the inner diameter of the plated through-hole. Therefore, an average of the wire diameter and the plated through-hole diameter is used to find the average of the solder joint [121.

S,,

=

A4 (solder shear tearout stress) hA

(3.10)

where A4 = 0.044 lb in. (see Eq. 3.5) h = 0.062 in. (solder height assumed equal to PCB thickness) A = (3>.rrd2/4 = ( 3 ) ~ ( 0 . 0 2 2 ) ~ /= 4 0.00114 i n 2 (average area of three solder joints)

”‘

=

0.044 (0.062) (0.00114)

=

622.5 Ib/in.*

(3.10a)

The vibration fatigue life for solder can be obtained from the log-log fatigue curve shown in Fig. 3.2, and with the use of Eq. 3.7. Very rapid fatigue cycles are considered here since the vibration frequency is so fast that the solder will not have a chance to creep and relax the stresses. Solder creep is common in thermal cycling environments, where the stress cycles are often very slow. This allows the solder to creep at high temperatures and relax a large percentage of any high stresses that are developed. Note that if forced

44

COMPONENT LEAD WIRE AND SOLDER JOINT VIBRATION FATIGUE LIFE

63/37 Tinilead solder

I

h’,cycles to f a i l FIGURE 3.2. S-N log-log vibration and thermal cycling fatigue curve for 63/37 tin/lead solder.

cooling is used immediately after the soldering operation it will increase the solder fatigue life.

1000 cycles to fail at 6500 Ib/in.2 (solder vibration life reference) S , = 6500 lb/in.l (solder ultimate shear reference) S , = 622.5 Ib/in.* (see Eq. (3.10a) b = 4.0 (slope of solder fatigue curve for vibration rapid cycles)

where N ,

=

N,

=

Solder life

=

6500 (1000) 622.5

[

‘

=

1.19 X lo7 solder cycles to fail (3.11)

1.19 x IO’ cycles to fail (406 cycles/s) (3600 s/h)

=

8.14 hours to fail

(3.12)

Comparing the solder fatigue life in Eq. 3.12 with the wire fatigue life of 8.6 minutes in Eq. 3.9 shows the wires are far more critical than the solder joints in vibration. Sinusoidal vibration tests were run using the same 4.0-G peak input level to establish the approximate fatigue life of the lead wires. The test results

RECOMMENDATIONS TO FIX THE WIRE VIBRATION PROBLEM

45

showed the wires failed after several minutes of vibration. The solder joints showed no visible cracks using a 35 power microscope. The vibration analysis and the vibration tests showed the 0.25-in. high transistor wires are critical. They will not be able to survive a 30-minute sinusoidal vibration qualification test using an input acceleration level of 4.0 G peak. Several changes were recommended to improve the fatigue life of the lead wires.

3.4 RECOMMENDATIONS TO FIX THE WIRE VIBRATION PROBLEM 1. Remove the 0.25-in. tall tube spacer and bring the transistor down to a height of about 0.060 in. above the top surface of the PCB. Attach heat sink clips to each of the transistor lead wires just before the wave soldering operation. This will thermally short-circuit the heat flow into the clips and reduce the high temperatures produced in the transistor. Remove the heat sink clips after the wave soldering operation. Experience with this approach has been very good in reducing hot spot temperatures. The reduced lead wire length will drive the transistor natural frequency above the 2000-Hz range, so there will be no vibration problems. This fix is labor intensive because it requires hand labor to add and to remove the heat sink clips. It is only cost effective for a production run of several hundred units. This fix would not be acceptable for production runs that involve several million units, as in the communication and automobile business. 2. Cement the 0.25-in. tall tube spacer to the PCB and to the transistor body to increase the stiffness of the assembly. This should drive the transistor natural frequency above the highest forcing frequency range of 2000 Hz, so there will be no vibration problems. This fix can be automated for high production runs. It may still not be acceptable for the super high communication and automobile production runs because their costs are calculated down to 0.001 penny. 3. Automate an operation that dimples the wires to automatically position the transistor 0.025 in. above the PCB for the wave solder operation. After soldering, automate another operation that adds a measured amount of viscous material, such as RTV, that encloses the three wires and fills most of the 0.025-in. space under the transistor. It is not necessary to fill the entire area under the transistor to substantially increase the damping and easily reduce the transmissibility more than 60%. This will reduce the wire bending stresses more than 60% and increase the wire fatigue life to about 50 hours. A fatigue life of only 30 minutes is required, plus some additional time to account for the usual false starts in most qualification test programs. This fix will solve the vibration problem with a slight increase in the production costs.

46

COMPONENT LEAD WIRE AND SOLDER JOINT VIBRATION FATIGUE LIFE

3.5 DYNAMIC FORCES DEVELOPED IN TRANSFORMER WIRES DURING VIBRATION

Transformers are made with copper and iron so they are often quite heavy. When these components are mounted on PCBs that are exposed to vibration, high dynamic forces and stresses can be produced in the lead wires and solder joints. Heavy components must be carefully mounted, properly restrained, and adequately supported to keep the lead wires and solder joints from failing in severe vibration and shock environments. Transformers are often in the shape of cylinders or cubes. Surface mounted transformers have the lead wires extending from the sides so they can be bent down 90" to meet the PCB. For through-hole mounting, straight wires are typically used, where they extend from the bottom surface of the component directly into the PCB. When transformers are mounted on through-hole PCBs that are exposed to vibration, high acceleration forces can be generated in the lead wires and solder joints when the natural frequency of the PCB is excited. The arching of the PCB under these conditions produces axial forces and bending forces in the lead wires. When the lead wires are not spaced very far apart, there will be very little bending in the wires, so almost all of the inertia force produced in the transformer wires will be carried in direct tension. Copper wires are typically used for the transformer core winding and for the external leads. The fatigue properties of the copper wire must be known to find the approximate fatigue life expected in vibration conditions.

Sample Problem-Dynamic Transformer Lead Wires

Forces and Fatigue Life in

A 0.80-lb transformer, with four 0.028-in. diameter copper wires, is mounted near the center of a 0.062-in. thick plug-in PCB that has a natural frequency of 225 Hz. The system is required to operate in a 3.0-G peak sinusoidal vibration environment. Find the following: (a) Dynamic force in the wires and solder joints. (b) Dynamic stresses in the lead wires and solder joints. (c) Approximate fatigue life expected in the lead wires and solder joints. Solution Use the Square Root of the Natural Frequency to Find the PCB Transmissibility. (a>The dynamic force Pd can be obtained from the following equation: Pd = WG,,Q

(see Eq. 2.36)

47

DYNAMIC FORCES DEVELOPED IN TRANSFORMER WIRES DURING VIBRATION

where W = 0.80 lb (transformer weight) Gin= 3.0 (peak sine input) Q= = \/225 = 15 (estimated PCB transmissibility)

a

Pd

=

(0.80)(3.0)(15)

=

36.0 lb

( 3.13)

(b) The dynamic tensile stresses in the wire can be obtained from a handbook, as shown below:

s,=

Pd

-

where Pd = 36.0 lb (shown above) d = 0.028 in. (copper wire diameter) A , = n d 2 / 4 = [ ~ ( 0 . 0 2 8 > ~ /X4 4] wires copper wires) 36.0

sw=---= 0.00246

(3.14)

A,

= 0.00246

i n 2 (area of four

14,634 lb/in.* (wire tensile stress) (3.15)

The dynamic shear stress in the solder joint can be obtained from Eq. 3.14 with a slight change as shown below. Test data show that solder joints appear to fail at some point between the wire and the plated through-hole. Therefore, the average solder joint diameter is obtained by using the average of the lead wire diameter plus the plated through-hole diameter. The thickness of the PCB,without any solder fillets, is normally used for the solder shear area. This is a conservative estimate since visual inspections show solder fillets of about 0.010 in. on the top and bottom surfaces of the PCB for through-hole mounted components. Pd

ss= A,

(3.16)

where Pd= 36.0 lb (dynamic load) (see Eq. 3.13) d = (0.028 + 0.040)/2 = 0.034 in. (average solder joint diameter) A s = (4)(n)(0.034)(0.062) = 0.0265 ine2(area of four solder joints) 36.0

s,=-0.0265

- 1358 lb/in.2

(solder shear stress) (3.17)

(c) The approximate fatigue life of the lead wire can be obtained from the log-log S-N fatigue curve for the copper lead wire shown in Fig. 3.3. This curve includes a stress concentration factor of 2.0, to account for cuts,

48

COMPONENT LEAD WIRE AND SOLDER JOINT VIBRATION FATIGUE LIFE

Copper wire

108

103 iV, cycles to f a i l

FIGURE 3.3. S-N log-log fatigue curve for copper wire, with a stress concentration factor of 2.

scrapes, and scratches normally found in component lead wires. The fatigue damage relation is used to find the number of stress cycles required to produce a fatigue failure, as well as the expected fatigue life.

N,S,b = N,S,b (see Eq. 3.7) where N , = 1000 stress cycles to fail at 45,000 lb/in., (copper reference point) S , = 45,000 lb/in., (stress reference point) b = 6.4 (reference slope of fatigue curve for copper lead wire with k = 2) S , = 14,634 lb/in.’ (wire stress) (see Eq. 3.15) Nl

=

Wire life

=

(1000)

[

~

45,000 14,634

1

6.4

=

1.33 x l o 6 wire cycles to fail (3.18)

1.33 x lo6 cycles to fail

(225 cycles/s) (3600 s/h)

=

1.64 hours

(3.19)

The approximate fatigue life of the solder can be obtained from the log-log fatigue curve for solder shown in Fig. 3.2. Solder creep is not considered here because the stress cycles are very fast, so the solder does not have a chance to relax any internal stresses. The fatigue damage equation (Eq. 3.7) is used to find the number of stress cycles to fail, as well as the expected fatigue life.

DISPLACEMENTS BETWEEN PCB AND COMPONENT PRODUCE LEAD WIRE STRAIN

49

where N2 = 1000 stress cycles to fail at 6500 Ib/in2 (solder reference point) S , = 6500 lb/in.* (stress reference point) SI = 1358 Ib/in.* (solder shear) (see Eq. 3.17) b = 4.0 (reference slope of fatigue curve for solder vibration)

[

6500

1

N,= (1000) Solder life =

1358

= 5.25 X

5.25 x 10' cycles to fail (225 cycles/s) (3600 s/h)

IO5 solder cycles to fail (3.20) = 0.65

hour

(3.21)

Comparing Eq. 3.19 (for the wire life) and Eq. 3.21 (for the solder life) shows the solder is more critical than the lead wire in this case. 3.6 RELATIVE DISPLACEMENTS BETWEEN PCB AND COMPONENT PRODUCE LEAD WIRE STRAIN

Plug-in PCBs can produce large dynamic displacements in high vibration and shock environments, when their natural frequencies are excited. This often results in large relative displacements between the PCB and any large electronic components mounted on the PCB. The body of a large component may be thicker and stiffer than the PCB that supports the component. When the component is attached to the PCB with many lead wires in a through-hole or surface mounted arrangement, the relative displacement between the PCB and the component can generate a significant amount of strain in the lead wires, as shown in Fig. 3.4. The lead wires that are soldered to the PCB will

50

COMPONENT LEAD WIRE AND SOLDER JOINT VIBRATION FATIGUE LIFE

stay perpendicular to the PCB when it bends away from the component body. The lead wires will also stay perpendicular to the component body. This motion will generate an axial force and a bending moment in the lead wires. When the length of the component body is small compared to the length of the PCB, the axial force in the lead wire will be much greater than the bending moment in the lead wire. The bending moment in the wire can then be ignored without introducing a large error. Finite element methods (FEMs) of analysis show that a long component body will bend much more than the lead wires will stretch. Also, when the component body is stiff and when it has many relatively stiff wires, such as a dual inline package (DIP), the PCB will be stiffened in the local area of the component. This will tend to reduce the relative displacement between the component and the PCB. In the following analysis only the component body bending and the lead wire stretching will be considered. The change in the PCB curvature is too complex to consider here for hand calculations. Approximations mentioned above will significantly reduce the amount of calculation time-and still give reasonable results that are conservative-and can be used to estimate the fatigue life. A trigonometric expression can be used to describe the PCB displacement at any point. The relative displacement ( 6 ) between the component and the PCB will be the sum of the component body bending deflection and the lead wires stretching deflection. The component considered in the analysis is a ceramic hybrid with kovar lead wires spaced on 0.050-in. centers. It becomes too complex to examine the forces in every wire so only four end wires are assumed to carry the dynamic loads produced by the relative displacements. Chapter 8 has a section that relates to finding the minimum desired PCB natural frequency to achieve a 10 million cycle fatigue life with sinusoidal vibration, based on the component type, the PCB geometry, and the expected environment, This information was based on testing many different types of PCBs with different types of components until they failed. The data were used with FEMs to cross check the results analytically. The following sample problem starts with the minimum desired PCB natural frequency equation to show how the approximate fatigue life in the lead wires and solder joints can be established using approximate hand calculation methods. Sample Problem-Effects

of PCB Displacement on Hybrid Reliability

A ceramic hybrid 1.5 in. long, with 0.025-in. diameter kovar lead wires, is through-hole mounted 0.050 in. above the top surface and at the center of a 0.062-in. thick epoxy fiberglass plug-in PCB, parallel to the 9.0-in. edge, as shown in Fig. 3.4. The PCB is required to operate in a 5.0-G peak sinusoidal vibration environment. Simplify the analysis by assuming only four wires at each end of the component carry the dynamic load. Find the following:

(a> Minimum desired PCB natural frequency for a 10 million cycle fatigue life.

DISPLACEMENTS BETWEEN PCB AND COMPONENT PRODUCE LEAD WIRE STRAIN

51

(b) Relative dynamic displacement expected between the PCB and the component. (c) Dynamic forces and stresses in the lead wires and solder joints. (d) Approximate fatigue life expected in the lead wires and solder joints. (e) Bending stress in the ceramic component body.

Solution. (a) The minimum desired PCB natural frequency equation is shown below. -

[

9.8GinChr&

)2’3

0.00022B

(see Eq. 8.13)

where Gin= 5.0 (peak sine vibration) C = 1.26 (component type for hybrid with pins extending from bottom surface) h = 0.062 in. (thickness of PCB) r = 1.0 (relative position factor for component at center of PCB) L = 1.5 in. (length of component) B = 9.0 in. (length of PCB parallel to length of component)

fd =

[

(9.8) (5 .O) ( 1.26) (0.062) ( 1.O) (0.00022)(9.0)

=

177 Hz (3.22)

(b) The relative dynamic displacement difference ( 6 ) between the component and the PCB can be obtained from the dynamic displacement curve, which is trigonometric, as shown below. 7TX

2,= Znsin-

and

U

6= Z ,

(3.23)

- 2,

Then 6 = Z , 1 -sini

)

(3.24)

7aT x

where Gin= 5.0 (peak sine vibration) f , = 177 Hz (to achieve 10 million cycle fatigue life) = 13.3 (approximate PCB transmissibility expected) Q= =

fi

2,= 9.8Gi, Q/(fnI2 = (9.8)(5.0)(13.3)/(177)2

= 0.02080 in.

(3.25)

x = 3.75 in. (to edge of component) u

=

9.0 in. (length of PCB)

1

.rr(3*75) 9.0

= 0.00071

in.

(3.26)

52

COMPONENT LEAD WIRE AND SOLDER JOINT VIBRATION FATIGUE LIFE

This represents the deflection between the PCB and the edge of the component. This also represents the bending deflection of the component and the stretching deflection of the four lead wires assumed to carry the dynamic force. (c) There is not enough information at this point to find the forces and stresses in the lead wires and solder joints. Additional information must be developed. There are two unknowns-the deflection of the component and the deflection of the lead wires-so two equations must be generated to solve the problem. The first equation says that the force in the component must be the same as the force in the lead wires. The second equation says that the sum of both deflections must equal 0.00071 in. Start with the deflection of the component body (Z,) as a cantilevered beam with an end load generated by the four lead wires. PC L3, z,= 3% 1,

where E , I, L,

so

Pc

=

3ECICZC L3,

(3.27)

= 40 X

lo6 1b/in2 (ceramic component modulus of elasticity) = bh3/12 = (0.562)(0.170)'/12 = 2.301 X l o p 4 = 0.75 in. (half the full length of the component) (3)(40 x 106)(2.301X 10-4)Zc P- =

'L

( 0.75)

= h 545 .- X

10JZc (3.28)

Find the axial deflection of the four lead wires (Z,) assuming pure tension only. 2,

where d, A, E, L,.

=

PWL, ( 4 wires) AWE,

so

P,

4A,EwZ, =

(3.29)

L,

in. (wire diameter) = 7r(d,)2/4 = ~(0.025)'/4 = 0.000491 in.2 (area of one wire) = 20 x 10 Ib/ine2 (kovar wire modulus of elasticity) = 0.050 2(0.025) 2(0.025) = 0.150 in. (includes two wire diameters into the component and two wire diameters into PCB for equivalent length) = 0.025

+

P,

+

(4)( 0.000491)( 20 x 1 0 6 ) Z , =

0.150

= 2.618

x 105ZW (3.30)

For equilibrium conditions the bending force in the component must equal the axial force in the four wires so Eq. 3.28 must equal Eq. 3.30. 6.545 x 104Z,

= 2.618

x 105Zz, then

Z,

=

4.002,

(3.31)

DISPLACEMENTS B E W E E N PCB AND COMPONENT PRODUCE LEAD WIRE STRAIN

53

The relative displacement (6) between the component and the PCB will be equal to the bending displacement of the component and the axial displacement of the four wires. 6=Z,+Z, where S Z,

= 0.00071 = 4.002,

(3.32)

in. (see Eq. 3.26) (see above)

Then 0.00071 = 4.002,

+Z,

= 5.00Z,

so Z,

=

1.420 X

in.

(expected wire displacement)

(3.33)

The resulting dynamic force in one lead wire can be obtained from Eq. 3.29 considering one wire. This will be the same force acting on the solder joint of the lead wire.

P,

A,E,Z, =

L

-

,

(0.000491)(20 X 106)(1.420X 0.15

= 9.296

lb (3.34)

(d) The dynamic tensile stress expected in the lead wire can be obtained as follows:

P,

sW -- - A=,

9.296 0.000491

=

18,933 lb/in.,

(3.35)

The approximate fatigue life expected for the kovar lead wire can be obtained from Eq. 3.7 and the kovar fatigue curve shown in Fig. 3.1, using the information shown below.

where

N,= 1000 stress cycles (reference point at 84,000 lb/in.*) = 84,000 lb/in.2 (reference point) S , = 18,933 lb/in.* (tensile stress shown above) b = 6.4 (slope of fatigue curve with a stress concentration of 2.0 to account for cuts, scrapes, and scratches in the lead wires)

S,

Nl

=

84,000 6.4 (1000) 18,933)

[

=

13.83 X 106 wire cycles to fail (3.36)

54

COMPONENT LEAD WIRE AND SOLDER JOINT VIBRATION FATIGUE LIFE

This is close to the 10 million cycle fatigue life typically expected for components in a sinusoidal vibration environment when the PCB natural frequency is obtained using Eq. 8.13. The approximate wire fatigue life for a sinusoidal vibration resonant dwell test condition can be obtained as shown below. Life

13.83 X l o 6 cycles to fail =

(177 cycles/s) (3600 s/h)

= 21.7

hours to fail

(3.37)

The dynamic shear stress expected in the solder joint can be obtained using the following information.

& = - ps -

ps --

rrdh

where P, d h A,

A,

= 9.296

lb (same as Eq. 3.34) = (0.025 + 0.040)/2 = 0.0325 in. (average solder joint diameter) = 0.062 + 2(0.015) = 0.092 in. (solder joint height including fillets) = n-dh = n-(0.0325)(0.092) = 0.00939 in.* (solder joint shear area)

Ps 9.296 S S -- - = A, 0.00939 ~

=

990 lb/in.2

(3.38)

The approximate fatigue life for the solder joint can be obtained using Eq. 3.7 and the vibration portion of the solder fatigue curve shown in Fig. 3.2, with the following data:

where N , = 1000 cycles to fail at 6500 1b/im2 (reference point) S , = 6500 1 b / h 2 (stress reference point) S , = 990 1 b / h 2 (shear stress in solder) b = 4.0 (slope of solder fatigue curve for vibration) N,

=

[ 7;: il

( 1000) -

=

1.86 x l o 6 solder cycle to fail (3.39)

The approximate solder joint time to fail in a sinusoidal vibration condition will be as follows: Life

1.86 x l o 6 cycles to fail =

(177 cycles/s) (3600 s/h)

= 2.92

hours to fail

(3.40)

DISPLACEMENTS BETWEEN PCB AND COMPONENT PRODUCE LEAD WIRE STRAIN

55

The solder joint with this geometry and application is more critical than the lead wire. (e) The bending stress in the ceramic hybrid body can be obtained from the standard bending stress equation with the following information:

where P = 9.296 lb (in one wire, but four wires carry the load) (see Eq. 3.34) L = 0.75 in. (half of the component length) M = PL = (4)(9.296)(0.75) = 27.88 lb . in. c = 0.17/2 = 0.085 in. (half the component thickness) in.4 (moment of inertia) (see Eq. 3.28) I = 2.301 X s b =

(27.88)(0.085) 2.301 X

=

10,300 lb/in.’

(3.41)

Ceramics have a compressive strength of about 250,000 1b/ins2 but the tensile strength is only about 25,000-30,000 lb/in.2, so the ceramic component body appears to have a good safety factor.

-

CHAPTER 4

Beam Structures for Electronic Subassemblies

4.1 NATURAL FREQUENCY OF A UNIFORM BEAM The natural frequency of a uniform beam can usually be determined by considering the strain energy and the kinetic energy of the vibrating beam. There are several different ways of expressing these quantities. Some methods are approximate, while others are exact. The exact methods usually require more work, and the approximate methods are often satisfactory for engineering solutions, so the latter will be given more consideration. Boundary conditions are very important in exact methods as well as in approximate methods. For approximate methods, if the geometric boundary conditions are satisfied, the resulting natural frequency equations will usually be quite accurate for the fundamental resonant frequency. The geometric boundary conditions consist only of the slope and the deflection. However, these boundary conditions must be met by the type of deflection curve that is used to describe the vibrating beam. The technique of assuming a deflection curve for a vibrating system, from which the natural frequency equation is determined, is called the Rayleigh method, after Lord Rayleigh [28], who first proposed it. If the deflection curve happens to be the exact curve for the boundary conditions, then the frequency equation will also be exact. A trigonometric function can be used to describe the deflection curve of the simply supported beam shown in Fig. 4.1. Consider the deflection equation

TX

Y = Yosin L This curve satisfies the geometric boundary conditions of slope and deflection as follows: Deflections:

x = 0, x =L , X=L/2,

56

Y=O Y=O Y=Y,

NATURAL FREQUENCY OF A UNIFORM BEAM

57

I yo

FIGURE 4.1. Deflection curve for a simply supported beam.

Slopes:

If then

X=O,

8=-Y0 L

X=L,

8=--Y LO

x = -2 ’

8=0

77

The strain energy can be obtained from the general bending moment equation d2Y E I y = M dX

as

From Eq. 4.1,

dY 7~ “iX _- -Yo cosdX L L d2Y “i2 - - -Yo dX L2

--

($1

2

T4 =

“iX sin L

“iX

F Y i sin’-- L

(4.4)

58

BEAM STRUCTURES FOR ELECTRONIC SUBASSEMBLIES

Substituting Eq. 4.4 into Eq. 4.3 for the strain energy of the beam, we find

U=

(4.5)

The kinetic energy of the vibrating beam will be as follows:

At resonance, the kinetic energy is equal to the strain energy, so Eq. 4.6 is equal to Eq. 4.5:

EIli Y i 4 L3

-

w O2Y;L

4g li 'EIg

=-

="i"i wL4

f

1/ 2

(4.7)

21i w L 4

This results in L e exact frequency equation because L e exact deflection curve was used in Eq. 4.1. A polynomial can also be used to describe the deflection curve for a simply supported beam, shown in Fig. 4.2. Consider the deflection equation

w (lb/in )

[La-* I

E -x

FIGURE 4.2. A simply supported beam with the Y axis at the center.

NATURAL FREQUENCY OF A UNIFORM BEAM

59

This curve satisfies the geometric boundary conditions of slope and deflection as follows: Deflections:

x =0 ,

Y=Yo

X=a, X=-a,

Y=O Y=O

e = -d=Y- -

2Y0X

Slopes:

dX

so

x =0 ,

a*

o=o

From Eq. 4.8,

--

dY

2Y0X

dX

a*

d2Y dX

2Y0 a*

--

(4.9) For the strain energy, substituting Eq. 4.9 into Eq. 4.3,

2 EIY: u= 2 EIY,, ( a ) = ___ ~

a'

a3

(4.10)

The kinetic energy can be obtained by using Eq. 4.8 as follows:

(4.11)

60

BEAM STRUCTURES FOR ELECTRONIC SUBASSEMBLIES

At resonance, the kinetic energy must equal the strain energy, so Eq. 4.11 must equal Eq. 4.10:

2EIY;

--

U3

Since u

-

4wR2Y,'u 15g

=L/2,

(4.12) Comparing these results with the exact frequency equation (Eq. 4.7) shows the polynomial results in a frequency that is about 11.1% too high. In fact, the Rayleigh method always results in a frequency value that is greater than the true value. unless the exact deflection curve is used. Sample Problem-Natural

Frequencies of Beams

Determine the approximate resonant frequency of the beam in Fig. 4.3 when it is made of aluminum and when it is made of steel, and vibration is in the vertical direction. Solution. If the beam is made of aluminum, = 10.5 x 106 1 b / h 2 (modulus of elasticity) b = 0.50 in. (beam width) h = 1.0 in. (height of beam) I = b h 3 / 1 2 = (0..50)(1.0)'/12 = 0.0417 in.4 (moment of inertia) g = 386 in./s2 (gravity) L = 10.0 in. (length) p = 0.10 1 b / h 3 (density of aluminum) W = b h L p = (0.50)(1.0)(10.0)(0.10)= 0.50 lb (weight of beam) w = W / L = 0.50/10.0 = 0.05 lb/in.

E

Section A A

FIGURE 4.3. A uniform beam with simply supported ends.

NATURAL FREQUENCY OF A UNIFORM BEAM

61

Substituting into Eq. 4.12,

fn= fn =

i

10.96 (10.5 X 106)(0.0417)(386)

(0.05) ( 10.0)'

1013 Hz

(4.13)

Now assume the beam is made of steel: 29 x lo6 lb/in.* (modulus of elasticity) p = 0.283 lb/in.3 (density) W = (0.50)(1.0)(10.0>(0.283)= 1.415 lb (weight) w = W / L = 1.415/10.0 = 0.1415 lb/in.

E

=

Substituting into Eq. 4.12,

fn= fn =

i

10.96 (29 x 106)(0.0417)(386) (0.141) ( 10.0)'

1005 HZ

(4.14)

Comparing Eqs. 4.13 and 4.14 shows the natural frequency of the steel beam is almost the same as that of the aluminum beam, even though the steel beam is almost three times heavier. A closer examination of the natural frequency equation shows that it depends on the ratio of the modulus of elasticity to the density. These two factors can be examined for the beam shown in Fig. 4.3, considering several different materials:

E

6.5 x lo6

100 x l o 6 in.

Magnesium:

-P

Aluminum:

-P

E

10.5 X l o 6 = 105 x l o 6 in. 0.10

Steel :

-=

E

29.0

Beryllium :

0.065

=

P

lo6 0.283

E

42.0 x lo6

P

0.068

_ --

X

=

102 x IO6 in.

= 619

x l o 6 in.

(4.15)

The only two items that will change in the natural frequency equation are the modulus of elasticity and the density. An examination of the ratios shown above indicates that the natural frequency of the beam will be about the

62

BEAM STRUCTURES FOR ELECTRONIC SUBASSEMBLIES

same for magnesium, aluminum, and steel. A beryllium beam, however, would have a much higher natural frequency. The Rayleigh method can be used to determine the fundamental resonant frequency for beams with various end conditions. This method can be used with a trigonometric or a polynomial expression as long as the geometric boundary conditions of slope and deflection are satisfied. Some typical deflections for different boundary conditions are shown, along with the resulting natural frequency equations, in Fig. 4.4. The resonant frequencies of beams with different edge conditions are shown in Fig. 4.5. CASE

1

Cantilever Beam- Uniform Load

Trigonometric

Polynomial

Y

Y - L -

-x

(

Y = Y,, l-cos-

CASE

2

2L

-

-X

Y

=

Y,,

(;I2

Fixed-Fixed Beam- Uniform Load

Trigonometric

Y

Polynomial

Y

FIGURE 4.4. Displacement curves and natural frequency equations.

fn =

C

J@$

Where C = Modal constant

BEAM TYPE

Cantilever

t-

$JL

k C = 56

Simply supported ends or hinged-hinged

MODE 3

MODE 2

MODE 1

,644

C = 351 500

r====7 4-7

&

C = 982 ___ ___~

Fixed ends

=

C = 6.28

m C

F ree-free

1.57

=

224

+ I-+C = 1 4 1

I

19.2

=

500

~

C

=

_

252

_

Fixed-hinged

=

C

=

c =192

9.82

132

277

723

868

I , , i

356

C

=

b (' =

C = 394

F & C = 166

245

('=

795

446

1<-

=

853

I,

7 95

zlY\8 C = 166

,

409 511, 774 910 0602i7, -

C = 318

982

616

H inged-free

_

500

/

C

* 400 800

359 .641

356

776

C

333 667

w * L C

MODE 5

MODE 4

C

=

284

JJLI 235

(' =

A

-

('=475 429 810

v=---=2 3 8 1 6/91

<- =

43 3

381 763

707

28 4

FIGURE 4.5. Resonant frequencies of uniform beams [51].

c

=

43 3

64 4.2

BEAM STRUCTURES FOR ELECTRONIC SUBASSEMBLIES

NONUNIFORM CROSS SECTIONS

Electronic equipment in close quarters must make use of every cubic inch of volume. When there is a little extra space, it is usually located in an area that is difficult to reach. Often structural modifications must be made to accommodate equipment that seems to grow larger with every redesign. This usually requires notching, reinforcing, or moving major load-carrying members, The net result is a chassis or a structure that supports all of the electronics at the expense of a cross section with notches and cutouts along its length. In order to determine the resonant frequency of this type of structure, it is necessary to consider the effects of nonuniform cross sections. If there are many different cross sections that must be considered, most methods of analysis can become very long and time consuming. However, there is a trade-off that can be made, which reduces the amount of work at the expense of some accuracy. This trade-off involves the calculation of an equivalent moment of inertia for a uniform structure that will have approximately the same stiffness as the nonuniform structure. The loss of accuracy is only about 5-1096 in the natural frequency for many cases, which is well worth the amount of work that can be saved. Consider a cantilever beam with two different cross sections that form two steps (Fig. 4.6). Castigliano’s strain energy theorem can be used to determine the deflection under the load at the free end of the beam. The general deflection equation is

The deflection for the step beam shown in Fig. 4.6 is (4.16)

FIGURE 4.6. A cantilever beam with two different cross sections.

NONUNIFORM CROSS SECTIONS

65

The bending moment at point 1 is

M,

= PX,

The bending moment at point 2 is

M,

=P("

+X,)

Substituting into Eq. 4.16,

P a 6=--/X EI, o

3EI,

"2 + 2 ax,+ x2') dX, ;d X , + -Ib( o

P

(4.17)

EI,

Let

L "=b=2

(4.18)

PL3 6s-+24

[

E;*

I

(4.19)

The deflection of a uniform cantilever beam with a concentrated load at the end can be obtained from any structural handbook: (4.20) An average moment of inertia I,, is used in the above expression so that the average moment of inertia can be determined by forcing the deflections in Eqs. 4.19 and 4.20 to be equal:

PL3

--

3EIa,

-+&I

PL3 24

EI,

Solving for the average moment of inertia, we have (4.21)

66

BEAM STRUCTURES FOR ELECTRONIC SUBASSEMBLIES

The average moment of inertia for the two-step beam in Fig. 4.6 can be determined for two different moments of inertia. Let

I , = 2 in.4 and

I 2 = 4 in.'

(4.22)

Substituting Eq. 4.22 into Eq. 4.21, (4.23) Now let us see if there is some other way that an average moment of inertia might be determined, so as to give results similar to Eq. 4.23 without going through so much work. Consider a method for averaging the moment of inertia as a function of the length of each section for the stepped beam shown in Fig. 4.6: aI, + b12 I' = (4.24) a' a+b Substituting Eqs. 4.18 and 4.22 into Eq. 4.24, (4.25) Comparing Eq. 4.25 with Eq. 4.23 indicates the approximate averaging method shown in Eq. 4.24 is only about 15.5% lower than the true value. Since the natural frequency varies as the square root of the moment of inertia, the error in the natural frequency will be only about 8.1% too low. Some other proportions can be examined. For example, consider the proportion a=2b so a = $ L and b = S L (4.26) Substituting Eqs. 4.26 and 4.17, (4.27) To get the same deflection as a uniform cantilever beam with a concentrated load, force Eq. 4.27 to equal Eq. 4.20:

PL'

PL'

3EIa,

S1E

-=-

(

(

191, + 8 1 2

I,I,

I , I2 = 27 191, + 81,)

I

NONUNIFORM CROSS SECTIONS

67

Substituting Eq. 4.22 into the above,

I,,

= 27

8 (j8+32/

- = 3.08 in.4

(4.28)

This is the correct value for the average moment of inertia. Compute the approximate average value with the method shown by Eq. 4.24. This can be done with Eqs. 4.26 and 4.22 as follows: (4.29) Comparing Eq. 4.29 with Eq. 4.28 indicates the approximate averaging method shown by Eq. 4.24 is only about 13.6% lower than the true value. There is a second method of averaging the moment of inertia as a function of the length of each section for the stepped beam shown in Fig. 4.6. Consider the following: (4.30)

Substituting Eqs. 4.18 and 4.22 into Eq. 4.30,

-L+ - L I"

=

L/2

-+2

L/2

= 2.66

in.4

(4.31)

4

Comparing this with the correct value of 3.55 in.4 shown by Eq. 4.23 indicates the approximate averaging method shown by Eq. 4.30 is about 25% lower than the true value when a = b. The resonant frequency would be 13.4% lower. When a = 2b (as shown by Eq. 4.26 along with Eq. 4.22), Eq. 4.30 becomes (4.32)

- 2+ -

4

Comparing this with the correct value of 3.08 i n 4 shown by Eq. 4.28 indicates the approximate averaging method shown by Eq. 4.30 is about 22% lower than the true value when a = 26. The resonant frequency would be about 11.7% lower.

68

BEAM STRUCTURES FOR ELECTRONIC SUBASSEMBLIES

There are, of course, many other methods that can be used to determine an average moment of inertia for a stepped beam. It is generally desirable to choose one that is conservative, so that it will produce a moment of inertia that is slightly lower than the true value. This will then result in a calculated natural frequency that is slightly lower than the true value. This is preferred because most electronic structures have many bolted and riveted joints, which tend to reduce the stiffness and therefore the natural frequency. Sample Problem-Natural Frequency of a Box with Nonuniform Sections

An electronic box has mounting flanges at each end and a cross section that has four different moments of inertia along its length (Fig. 4.7). It is required to determine the approximate resonant frequency during vibration in the vertical direction (along the Y axis) if

I,

= 3.5

in.4, Ib = 2.5 in.',

I,

= 4.2

in.4 Id = 5.5 in.4

Solution. The approximate average moment of inertia for an equivalent uniform beam can be determined with an equation similar to Eqs. 4.24 and 4.30. Using Eq. 4.24 first, we have a[,

+ bIb + CI, + dId

I'a\

=

I'a b

=

I:,

= 3.96

a+b+c+d (4.5)(3.5) + ( 6 . 5 ) ( 2 . 5 ) + (3.0)(4.2) 4.5 + 6.5

(4.33)

+ (7.0)(5.5)

+ 3.0 + 7.0

in.'

Approximating the chassis as a simply supported uniform beam, the natural frequency is (see Eq. 4.7) (4.34)

Mounting flange -,

FIGURE 4.7. A n electronic chassis with four different cross sections.

COMPOSITE BEAMS

69

where E = 10.5 X lo6 lb/in.’ (aluminum) I = I:” = 3.96 in.4 g = 386 in./s2 (gravity) W = wL = 28.5 Ib (weight) L = 21.0 in. (length)

so we have T

fn =

f,

i

= 387

(10.5 X 106)(3.96)(3.86 x l o 2 ) (28.5) (21 .O)

Hz

(4.35)

If there is some question concerning the integrity of several structural members that are part of a riveted assembly and there is a fear of some relative motion in this assembly, there could be a reduction in the stiffness of the electronic box, which would reduce the natural frequency. A conservative approximation of the natural frequency could then be made by using an approximate average moment of inertia that is known to be relatively low. Under these circumstances it might be desirable to use an expression similar to Eq. 4.30:

1’’ a =v

a+b+c+d a b c d

-+-+-+Io

I”

=

av

I:”

Ib

IC

(4.36)

Id

+

4.5 + 6.5 3.0 + 7.0 4.5 6.5 3.0 7.0 -+ - + - + 3.5 2.5 4.2 5.5

= 3.57

in.4

Substituting this value into Eq. 4.34, fn = 370 Hz

(4.37)

For this particular electronic box, there is not much difference in the resonant frequency for the two different methods used in determining the approximate average moment of inertia.

4.3 COMPOSITE BEAMS Composite laminations are often used in electronic boxes because of electrical, thermal, and vibration requirements. Sometimes a printed circuit board (PCB) is mounted so close to a metal bulkhead that it is possible for bare

70

BEAM STRUCTURES FOR ELECTRONIC SUBASSEMBLIES

metal lead wires on the circuit board to contact the bulkhead during vibration. Short circuits can result, and they may damage the electronic equipment. In order to prevent this, a thin strip of epoxy fiberglass, 0.005 in. thick, may be cemented to the bulkhead. Epoxy fiberglass is hard and tough and can resist the pounding of many sharp points during a resonant condition. The bulkhead, of course, then becomes a composite laimination. PCBs often use metal-strip laminations of aluminum or copper to conduct away the heat. In some cases, thin copper strips are bonded to a circuit board; electronic component parts such as resistors, diodes, flatpacks, and transistors are cemented to these copper strips; and their electrical lead wires are soldered to the PCB. Aluminum plates have been laminated to PCBs in order to conduct away the heat. In some cases, the components are mounted directly on the aluminum, which has clearance holes for the electrical lead wires to permit soldering to the circuit board on the back side of the aluminum plate. In other cases, the components may be mounted on a circuit board only 0.005 in. thick. The heat will then flow right through the epoxy fiberglass circuit board, with a moderate temperature rise, and be conducted away by the aluminum plate [54]. Sometimes it is necessary to thermally isolate an electronic component, such as a gyro, which must be maintained at a constant temperature to provide the required accuracy. A closely controlled heating system may be used to maintain a constant gyro temperature over a wide external temperature range. Since most structural members are made of metal, and are good heat conductors, the thermal isolation system will usually consist of a metal laminated to a plastic or a ceramic. Two very common materials that are often used in electronic box structures are aluminum and epoxy fiberglass. The natural frequency of a composite beam of these materials can be determined by considering a combination of the physical properties of both materials. Consider the case of a simply supported laminated beam that has a uniform load distribution along its length (Fig. 4.8). If the lamination of aluminum and epoxy fiberglass is side by side, as shown in the cross-section view of Fig. 4.9, then an equivalent beam of one material can be used to find the natural frequency.

FIGURE 4.8. A simply supported laminated beam.

COMPOSITE BEAMS

71

Epoxy fiberglass

h= 10

Section AA

I

FIGURE 4.9. Cross section of a lami-

I

nated beam.

The width of the epoxy fiberglass lamination can be reduced to make it equivalent in its stiffness to an aluminum beam, using the EI stiffness factor. The subscripts a and e refer to aluminum and epoxy, respectively:

E , I, E,b,h:

=E, I, -

12

E,b,hg

12

Since the height of the aluminum section, h a , is the same as the height of the epoxy section, h e , they cancel. The aluminum equivalent of the epoxy section is (4.38) where b, = width of aluminum equivalent to epoxy be = 0.50-in. (width of epoxy) E , = 2.0 X l o 6 lb/in.2 (epoxy fiberglass) E , = 10.5 X l o 6 lb/in.2 (aluminum) Thus b,

=

(0.50)

(

2.0 x l o 6 106

= 0.0952

in.

(4.39)

The equivalent width of a solid aluminum beam can now be determined by adding the increment shown above to the width of the aluminum section: be, = 0.75

+ 0.0952 = 0.8452 in.

(aluminum)

The natural frequency of this beam in the vertical direction can be determined from the equation of a uniform beam as follows: 1/2

(see Eq. 4.7)

72

BEAM STRUCTURES FOR ELECTRONIC SUBASSEMBLIES

10.5 x IO6 1b/in2 (aluminum) = b,,h3/12 = (0.8452)(l.0)3/12 = 0.0704 in.4 g = 386 in./s2 (gravity) L = 15.0 in. (length) W, = (15)(1.0)(0.75)(0.10 lb/in.3) = 1.125 lb (aluminum) We= (15~(l.0>(0.50>(0.065 lb/in.3) = 0.487 lb (epoxy) W = total beam weight = 1.612 lb

where E I

=

Thus

f"

7~

(10.5 X 106)(0.0704)(386)

2

(l.612)( 15.0)3

=-[

f , = 360 HZ

(4.40)

If the lamination of aluminum and epoxy is stacked in the vertical direction as shown in the cross-section view of Fig. 4.10, then a composite moment of inertia can be computed for the two materials. The same cross-sectional area for the aluminum and the epoxy fiberglass is used to keep the same weights. The following tabulations can be set up to simplify the calculation of the composite moment of inertia for the beam stiffness factor EI: Section

Area

Eo

AE

Y

AEY

EPOXY Aluminum

0.50 0.75

2 x lo6 10.5 X IO6

1.00 x l o 6 7.87 X IO6

0.80 0.30

0.80 x l o 6 2.36 X 106

Total

8.87 X IO6 3.16 x l o 6

CAEY

-

y=

~

CAE

-

8.87 X IO6

=

0.356 in.

3.16 X l o 6 (centroid)

Y

FIGURE 4.10. Cross section of a laminated beam.

10

Aluminum

COMPOSITE BEAMS

Section EPOXY Aluminum

C

C2

0.444 0.056

0.196 0.0031

O I 0.0066 0.0225

AEC 0.196 x lo6 0.024 X lo6 0.220 x lo6

Total

73

EO Io 0.0132 x lo6 0.2360 X lo6 0.2492 x lo6

The composite beam stiffness factor is

C A E c 2+ C E o I , ,= 0.220 X lo6 + 0.249 X lo6

EI

=

EI

= 0.469 X

lo6 in.in.2

(4.41)

Substituting into Eq. 4.7 for the natural frequency,

f

=-[ 7~

2

(0.469 X 106)(386) (1.612)(15.0)3

I”

= 286

HZ

(4.42)

Equation 4.38 can also be used with the composite cross section shown in Fig. 4.10. The result will then be a T section with only one material involved. The equivalent width of an aluminum section is

b,

=

2.0 x l o 6

I.,,(

10.5 x lo6

)

= 0.238

in.

This results in the aluminum T section shown in Fig. 4.11. A table can be set up to simplify the calculation for the moment of inertia of the cross section: Item 1 2

Area 0.095 0.750

Total

0.845

Y 0.80 0.30

--1.254

=

C

C2

0.444 0.056

0.196 0.0031

0.301

- CAY =-=-Y= EA I

AY 0.076 0.225

0.301

- 0.356 in

0.845

cAc2 + cIo

= 0.0209

Ac 0.0186 0.0023

IO 0.00127 0.02250

0.0209

0.02377

. (centroid)

+ 0.02377 = 0.04467 in.4

FIGURE 4.11. A T section used to simulate a laminated beam.

74

BEAM STRUCTURES FOR ELECTRONIC SUBASSEMBLIES

Y

\--

125--

s 10

FIGURE 4.12. Cross section of a beam with three

laminations.

I 1

,

7

Aluminum

EPOXY

0 30

Aluminum -

b

The beam stiffness factor becomes

EI

=

10.5 x 106(0.04467)

=

0.469 x l o 6 Ib.in.*

(4.43)

Comparing Eq. 4.43 with Eq. 4.41 shows the beam stiffness factors are exactly the same for both methods. If the composite lamination of aluminum and epoxy fiberglass is stacked in three layers like a sandwich, as shown in the cross-section view of Fig. 4.12, then the beam stiffness factor EZ can be calculated more easily. The same cross-sectional area for the aluminum and the epoxy fiberglass is used to keep the same weights. For the aluminurn section, (0.40)3] = 1.025 x 106 1 b . h . ' For the epoxy section,

The composite beam stiffness factor is

EI

= E,Za

EI

=

+ E,Z,

=

1.025 X l o 6

1.0383 x l o 6 Ib'in.'

+ 0.0133 X lo6 (4.44)

Use Eq. 4.7 to determine the natural frequency of the sandwich cross-section composite beam:

The composite beams shown in this section must be cemented together in order to eliminate relative motion at the interface between the aluminum and the epoxy. If bolted joints are used, relative motion will usually occur, which will reduce the stiffness and the natural frequency. It takes a large number of large bolts to prevent relative motion between two members at high frequencies and high G forces.

-

CHAPTER 5

Component Lead Wires as Bents, Frames, and Arcs

5.1

ELECTRONIC COMPONENTS MOUNTED ON CIRCUIT BOARDS

Electronic boxes are being required to occupy less space while providing more functions, so emphasis has been put on reducing the size of electronic component parts. The development of solid-state electronic parts, such as integrated circuits, sharply reduced the physical size of the parts and permitted more functions to be included in a smaller volume. Even small electronic component parts, however, must be mounted to provide the proper heat removal, accessibility, and structural integrity, depending on the environment. Because space is very limited in most aircraft, spacecraft, submarines, and even automobiles, electronic equipment must be supported by many different types of structures that can be adapted to the geometry of the system. Also, the physical size and shape of many electronic component parts themselves may permit them to be analyzed as structural members. Consider, for example, some typical electronic component parts such as integrated circuits, resistors, capacitors, and diodes mounted on printed circuit boards (PCBs) by their electrical lead wires. This is a common practice in the electronics industry because it permits low-cost production and easy maintenance. These component parts are generally soldered to the PCB by infrared, vapor phase, dip, and wave soldering, and even hand soldering. The PCBs are usually of the plug-in type, which are guided along the edges to permit easy connector engagement. Under these conditions, the edges of the PCB can usually be considered as simply supported. A typical installation might be like that shown in Fig. 5.1. During vibration in an axis perpendicular to the plane of the PCB, the acceleration forces will produce deflections in the circuit board. As the circuit board bends back and forth, bending stresses are developed in the electrical lead wires that fasten the electronic component parts to the circuit board (Fig. 5.2). Electronic component parts can be mounted on PCBs in many different ways. The ability of these components to survive a severe vibration environment will depend on many different factors such as component size, resonant 75

76

COMPONENT LEAD WIRES AS BENTS, FRAMES, AND ARCS Resistor

edge-guide

Printed-circuit board

FIGURE 5.1. Electronic component parts mounted on a printed circuit board.

-

coniporient

Electrical lead

Simply supported edge Circuit board bending

w

FIGURE 5.2. Bending in the component lead wires on a vibrating circuit board.

frequency of the circuit board, acceleration G forces, method of mounting components, type of strain relief in the electrical lead wires, location of the component, and duration of the vibration environment. Most component failures in a severe vibration environment will be due to cracked solder joints, cracked seals, or broken electrical lead wires. These failures are usually due to dynamic stresses that develop because of relative motion between the electronic component body, the electrical lead wires, and the PCB. This relative motion is generally most severe during resonant conditions that can develop in the electronic component part or in the PCB. Resonances may develop in the component part when the body of the component acts as the mass and the electrical lead wires act as the springs. These resonances are usually not too severe if the body of the component is in contact with the PCB, since this contact will sharply reduce the relative motion of the component. If resonances of this type do develop, it is an easy task to tie or cement the component part to the circuit board. If resonances develop in the circuit board, large displacements can force the electrical lead wires to bend back and forth as the circuit board vibrates up and down (Fig. 5.2). If the stress levels are high enough and if the number of fatigue cycles is great enough, then fatigue failures can be expected in the solder joints and the electrical lead wires.

BENT WITH A LATERAL LOAD-HINGED

ENDS

77

The most severe stress condition will be for a component part mounted at the center of a PCB. For a rectangular board, the most severe condition will occur when the body of the component part is parallel to the short side of the circuit board. This is due to the more rapid change of curvature for the short side of the board than for the long side, for the same displacement. Tying the electronic component part to the circuit board with lacing cord, at the center of the component body, will generally have very little effect on the relative motion between the component body and the circuit board when the circuit board is in resonance. Cementing the component body to the circuit board or tying both ends of the component part to the circuit board will greatly improve the fatigue life in the solder joints and in the electrical lead wires. Conformal coatings are quite often used on PCBs to protect the circuitry from moisture. Although most conformal coatings are quite thin, from about 0.0005 in. to about 0.010 in. thick, they will still act as a cement, so that they will also improve the fatigue life of the electronic component parts. PCB resonances can lead to high bending stresses in the lead wires on electronic component parts. Every time the circuit board bends, it forces the electrical lead wires to bend, since the wires are normally soldered to the circuit board. The greater the circuit board deflection, the greater the bending stresses in the lead wires. If the component body is cemented to the circuit board, the relative motion between the board and the component will be substantially reduced. Stresses in the component lead wires can be determined by examining the geometry of the electronic component part, the natural frequency of the circuit board, and the acceleration G forces. For example, considering the geometry, components such as resistors, capacitors, diodes, and flatpacks have small electrical lead wires compared to a large component body. Almost all of the relative deflection between the circuit board and the component will therefore occur in the electrical lead wires. If the body of the component is ignored and only the electrical lead wire deflections are examined, the resulting errors will be very small. The relative slope of the electrical lead wires, where they are soldered to the circuit board, will not change as the board vibrates up and down. The wires will always remain perpendicular to the board. Also, each electrical lead wire will always remain perpendicular to the component body where the lead joins the body, because the body is so much stiffer than the lead. Therefore, as the circuit board bends back and forth during vibration, the electrical lead wires will bend as shown in Fig. 5.2. If only the electrical lead wires are considered, they will approximate the shapes of the various frames and bents shown in the following sections. 5.2

BENT WITH A LATERAL LOAD-HINGED

ENDS

Consider a bent with hinged ends and a concentrated load acting in the lateral direction. The deflection can be broken up into two parts. The first

78

COMPONENT LEAD WIRES AS BENTS, FRAMES, AND ARCS

I"

Deflection due to bending of members AB and CD only

Deflection due to bending of member BC only

FIGURE 5.3. Bent with a lateral load and hinged ends.

FIGURE 5.4. Free-body force diagram of the vertical leg.

v

part considers bending of the vertical legs A B and CD only, with no bending in the horizontal leg BC. The second part considers bending of the horizontal leg BC only, with no bending in the vertical legs. When the horizontal leg bends, the angular displacement causes an additional displacement of the vertical leg through pure rotation (Fig. 5.3). The vertical leg A B is shown in Fig. 5.4. Considering forces in the horizontal direction,

Considering moments about points B ,

BENT WITH A LATERAL LOAD-HINGED

79

ENDS

FIGURE 5.5. Free-body force diagram of the horizontal leg.

Considering leg AB as a cantilevered beam with a concentrated end load,

a,=--H h 3

Ph3

--

3EI,

6EI,

(5.3)

The horizontal leg BC is shown in Fig. 5.5. The angular displacements are 0,

MBL 3E I ,

=-

and

MCL 02= 6E I ,

(5.4)

Substitute Eq. 5.4 into Eq. 5.5 and note that MB = M , due to symmetry:

Substituting Eq. 5.2 into Eq. 5.6,

The vertical leg will rotate through the angle 0, due to bending of the horizontal leg shown in Figs. 5.3 and 5.5:

The total deflection of the bent will then be the sum of 6, and 6,:

f3T =

6,

Ph3 Ph2L + 6,= + 6E12 12EI,

80

COMPONENT LEAD WIRES AS BENTS, FRAMES, AND ARCS

Let K

= hIl/L12.Then

Ph (5.9)

5.3 STRAIN ENERGY-BENT WITH HINGED ENDS Castigliano's strain energy theorem can be used to determine the loads, moments, and deflections in rectangular bents. This theorem states that the partial derivative of the strain energy, with respect to an applied force, will give the deflection produced by that force in the direction of the force. The total strain energy in a system can usually be accounted for by considering tension and compression, bending, torsion, and shear. Considering each one individually, they are as follows [13]:

Tension and compression:

Ut =

[g

(5.10)

Bending:

(5.11)

Torsion:

(5.12)

Shear:

q"" 2 GA

(5.13)

For example, the lateral displacement of a rectangular bent with hinged ends and a concentrated lateral load can be determined by considering only the strain energy of bending. Consider the rectangular bent shown in Fig. 5.6. The bending deflections are so much greater than all of the other deflections, such as tension and compression, torsion and shear, that only the bending need be considered unless the beams are very short (length less than about three times the depth). If stresses are within the elastic limit and deflections are small, there is very little error involved in ignoring the other deflection sources. The strain energy of bending must be considered for two vertical legs and one horizontal leg. Since the bent is symmetrical, only one-half of the system need be considered (Fig. 5.7).

STRAIN ENERGY-BENT

WITH HINGED ENDS

81

Y

Erin( I I

V

V

FIGURE 5.6. Bent with a lateral load and hinged ends.

1

FIGURE 5.7. Half of a bent with a lateral load and hinged ends.

V

The strain energy of bending was shown to be

lJb =

/te

2EI

Using Castigliano's theorem to find the deflection, we have

b2M(dM/dP)dX

-=

8P

[

2EI

1 =-

EI

["7 b

dM

dX

With the X and Y axes as shown, the bending moment at point vertical leg is

M ,= H Y

(5.14)

0 on the (5.15)

82

COMPONENT LEAD WIRES AS BENTS, FRAMES, AND ARCS

Considering all the forces in the X direction, P H=2

(5.16)

Substituting into Eq. 5.15 PY

M1 =

(5.17)

2

(5.18) With the X and Y axes as shown, the bending moment at point horizontal leg is

M 2 = HI.2 - I/x

@

on the

(5.19)

The force V can be determined by taking moments about point A in Fig. 5.6: Ph

=

VL

Ph v=-

L

(5.20)

Substituting Eqs. 5.16 and 5.20 into Eq. 5.19, (5.21) (5.22) Considering the strain energy in two vertical legs and two halves of the horizontal leg as shown in Fig. 5.7, the deflection equation 5.14 can be written as follows (note that E , = E? = E ) :

STRAIN ENERGY-BENT

WITH FIXED ENDS

83

Substituting Eqs. 5.17, 5.18, 5.21, and 5.22 into the above,

a=-+-2Ph3

12E12

2 P (h’u EI, 4

h2a2 I h2a3 2L

)

3L2

Since a =L/2

we have Ph3

Ph2L

a = - 6E12 +-1 2 E I , Let K = h I , / L 1 2 , and solve for 8: (5.23)

An examination of Eq. 5.23 shows it is exactly the same as Eq. 5.9, which was obtained by using superposition.

5.4

STRAIN ENERGY-BENT

WITH FIXED ENDS

If a rectangular bent has fixed ends, as in Fig. 5.8, it becomes statically indeterminate. Castigliano’s theorem can then be utilized by considering various physical characteristics of the structure. For example, it is obvious from Fig. 5.8 that the angular rotation at points A and D will be zero, because these ends are fixed. Also, since the structure is symmetrical, only one-half of the structure need be examined, as shown in Fig. 5.9. What may not be obvious from Fig. 5.9 is that the center of the horizontal leg BC becomes a point of inflection, at point E , so the moment ME must be zero. Also, point E will move only in the horizontal direction and not in the

84

COMPONENT LEAD WIRES AS BENTS, FRAMES, AND ARCS

FIGURE 5.8. Bent with a lateral load and fixed ends.

H

Y

h I

V

FIGURE 5.9. Half of a bent with a lateral load and fixed ends.

vertical direction. If the X and Y axes are taken as shown in Fig. 5.9. Castigliano’s theorem can be used by knowing that the vertical deflection at point E , due to the force V , must be equal to zero. This can be written as follows: (5.24) Since only bending deflections will be considered here and since there are and @ in Fig. 5.9, the bending moments to be considered at points deflection equation can be established using Eqs. 5.11 and 5.14:

0

The bending moment at point

0on the horizontal leg is M , = vx

(5.26)

STRAIN ENERGY-BENT

WITH FIXED ENDS

85

so JMI

-=x

(5.27)

av

@

The bending moment at point

on the vertical leg will be (5.28)

The horizontal force H can be determined by considering all of the forces in the X direction. This results in

P H=2

(5.29)

Substituting Eq. 5.29 into Eq. 5.28,

VL

M

, - 2

dM,

PY

(5.30)

2

_ -L

--

av

(5.31)

2

Substituting Eqs. 5.26, 5.27, 5.30, and 5.31 into Eq. 5.25,

1

VL3

EI,( 24

VL2h

VL2 611

-+-=-

pLhzI

)+g(T--

VLh I,

8

=O

Ph2 21,

Let K = hZ,/LZ, and solve for I/:

V=

3 PhK L(1+6K)

(5.32)

86

COMPONENT LEAD WIRES AS BENTS, FRAMES, AND ARCS

The moment M D can be determined from Fig. 5.9 by taking moments about point D :

CMD=O

(:i

M D + V - -Hh=O

(5.33)

Substituting Eqs. 5.32 and 5.29 into the above,

3 PhK MD'L(1+6K)(i)

Ph -0 2

With a little algebra this can be written as

(5.34) The deflection can be determined using the same method, making sure to include both vertical legs and both halves of the top horizontal leg:

The moment at point

0is &I1 = m =

3 PhK

L( 1

+ 6 K )X

(5.36)

so

aM,

3hK

aP

L(1+6K)

-=

The moment at point

@

(5.37)

X

is

VL

PY

3 PhK

PY

(5.38)

so

(5.39)

STRAIN ENERGY-BENT

WITH FIXED ENDS

87

Substituting Eqs. 5.36-5.39 into Eq. 5.35,

I(

3 PhKX

L(1+ 6 K )

EI,

+

E I l L 2 ( 1 6K)’

[ ’+

L(1+ 6 K )

0

s-

Y2

6hK

9h2K2Y

2E12 (1 + 6 K ) 2 - ( 1 + 6 K )

6=

IdX

[:IL’,

18Ph2K2

6=

3hKX

Y3

h

(T)+i], + 6 K ) + (1 + 6 K ) ’ 3( 1 + 6 K ) ,

27K2 - 9 K ( l

3 Ph2K ,L 4E11(1 + 6K)’

2EI,

Ph3 j 3 6 K 2 + 3 0 K + 4 / 24E12 3 6 K 2

+ 12K + 1

This can be written as

6=

-(

Ph

24E12 Ph

1+

+ i

( 6 K3+( 61K) ( 6 K + 1) 3

(5.40)

The previous bent with fixed ends could also have been analyzed by knowing that, for small angular displacements, the rotation angle on the vertical leg shown by 6,, will be the same as the rotation angle on the horizontal leg shown by e,, and indicated in the free-body force diagram of Fig. 5.10. A close inspection of the figure shows that the slope is positive at O B 2 , but negative at e,,; thus 6 7 2 = - OB1

(5.41)

Castigliano’s theorem can be applied here, since the partial derivative of the strain energy with respect to an applied moment will give the angular rotation produced by that moment in the direction of the moment. Since the strain energy was due only to bending, Eq. 5.11 is used again:

88

COMPONENT LEAD WIRES AS BENTS, FRAMES, AND ARCS

L

X

4

Y

FIGURE 5.10. Free-body force diagram of a bent with a lateral load.

Using Castigliano's theorem to find the angular rotation due to an applied moment,

In Fig. 5.10 the angular rotation due to moment M B acting at the top of the vertical leg will be approximately the same as the angular rotation due to moment M B acting on the horizontal leg. Then using Eqs. 5.41 and 5.42, this can be written as follows:

If the X and Y axes are taken as shown in Fig. 5.10, the bending moment at on the vertical leg is point

0

M 1 -- M

B

-fl=M

B

PY 2

--

(5.44)

so

(5.45)

STRAIN ENERGY-BENT

The bending moment at point

@

WITH FIXED ENDS

89

on the horizontal leg is

M2= -M,+Vx

(5.46)

The vertical force can be expressed in terms of M B by considering the sum of the moments about the left end, at point B , on the horizontal leg (note that from symmetry M , = M E ) : 2ME I/= -

(5.47)

L Substituting Eq. 5.47 into Eq. 5.46,

(5.48) (5.49) Substituting Eqs. 5.44, 5.45, 5.48, and 5.49 into Eq. 5.43,

MBh M E L -+-=I, 61,

Let K

= hI,/L12

Ph2

41,

and solve for M E : Ph - 4( 1 1 / 6 K )

M E

+

With a little algebra this can be written as

"(

M -_ 4

6Kl+I)

Substituting Eq. 5.47 into the above,

"(

I/=-

2L

1--

6K+1

(5.50)

90

COMPONENT LEAD WIRES AS BENTS, FRAMES, AND ARCS

With a little algebra this can be written as

V=

3 PhK L(6K

(5.51)

+ 1)

An examination of Eq. 5.51 shows it is exactly the same as Eq. 5.32.

5.5

STRAIN ENERGY-CIRCULAR

ARC WITH HINGED ENDS

Castigliano’s theorem is very convenient for determining forces, moments, and deflections in circular arcs. For example, consider the semicircular arc with hinged ends and a concentrated load (Fig. 5.11). Since the figure is symmetrical, it is possible to simplify the problem by considering only half of the system. The vertical deflection produced by the force P acting on the full system will then be the same as the vertical deflection produced by the force V acting on half of the system. The vertical force V can be determined by considering the sum of all the forces acting in the vertical direction. This leads to P v=-

(5.52)

2

The horizontal force H is not quite as easy to determine. An examination of the structure shows that the horizontal deflection at point B , due to force H , will be zero. Using Castigliano’s theorem, this can be written as follows:

a,,

IP

dU

-

JH

(5.53)

0

6 I

VI

FIGURE 5.11. Circular arc with hinged ends and a vertical load.

STRAIN ENERGY-CIRCULAR

91

ARC WITH HINGED ENDS

When the angle 0 is between 0" and 90", the bending moment at point will be M,=VR(l-cosO)

-HRsine

0

(5.54)

so (5.55) The horizontal deflection at point B can be determined from Eqs. 5.11 and 5.53: (5.56) Substituting Eqs. 5.54 and 5.55 into the equation above, 1

S""=E/O [ VR( 1 - cos e ) - HR sin 0 ] ( -R sin 6 ) R d 0 = 0 Integrating the above equation and collecting terms,

Only the factor in parentheses can be zero, so this leads to 2v H=IT

(5.57)

The vertical deflection due to force V can be determined from the following equation: (5.58) From Eq. 5.54, dM1 dV

-= R ( 1 -cos 0 ) Substituting Eqs. 5.54 and 5.59 into Eq. 5.58, 1 6= EI

0

[ VR( 1 - cos e ) - HR sin e ] [ R( 1 - cos e ) ]R d e

(5.59)

92

COMPONENT LEAD WIRES AS BENTS, FRAMES, AND ARCS

Substituting Eq. 5.57 into the above equation and integrating,

(5.60) Substituting Eq. 5.52 into Eq. 5.60, 6=-

5.6

PR

'

(5.61)

52.6EI

STRAIN ENERGY-CIRCULAR

ARC WITH FIXED ENDS

If the circular arc has fixed ends, it becomes statically indeterminate and requires more work to solve for the redundant forces. A semicircular arc with fixed ends and a concentrated load is shown in Fig. 5.12. Since the figure is symmetrical, the problem can be simplified by considering only half of the system. There are two unknowns, H and M,, so that two equations must be developed to find these forces. The force I/ is known, since an examination of all the forces in the vertical direction will show the following:

P

I/= -

(5.62)

2

The two physical properties that can be used to find the unknown forces are that the horizontal deflection at point C due to force H is zero, and that the relative slope at point C due to the moment M , is zero. The bending moment at point @ is

M , = H R ( l - c o s O ) +M,-VRsinO

forOsOsT/2

FIGURE 5.12. Circular arc with fixed ends and a vertical load.

(5.63)

STRAIN ENERGY-CIRCULAR

ARC WITH FIXED ENDS

93

Thus (5.64) The horizontal deflection at point C can be written as

( 5 -65) Substitute Eqs. 5.63 and 5.64 into Eq. 5.65:

a,,

1 z/o

v/2

=

[ HR( 1 - cos e ) + M ,

- VR

sin

e ]( R ) ( 1 - COS e ) R d e = 0

Integrating the above equation and collecting terms,

i

- 0.35618 + 0.5707-MC R EI3

R

- OSOOV

(5.66)

This is the first equation that contains the two unknowns, H and M,. The second equation can be determined from the relative slope at point C, which can be written as follows: (5.67) From Eq. 5.63, (5.68) Substituting Eqs. 5.63 and 5.68 into Eq. 5.67, 1 O = ~ j g a i i [ H R ( 1 - c o s 8 ) +MC-VRsin8](1)RdB=O

Integrating the above equation and collecting terms, 0.5707H +

(5.69)

Only the factors in parentheses in Eqs. 5.66 and 5.69 can be zero, so these two equations can be used to solve for H and Mc. These two equations are

94

COMPONENT LEAD WIRES AS BENTS, FRAMES, AND ARCS

repeated below: Mc 0.3561H + 0.5707R 0.5707H

+ 1.5707-MC R

= 0.5OOV

=

V

Solving these two equations simultaneously results in the following:

H Mc

= 0.458P =

0.152 PR

(5.70) (5.71)

The vertical deflection due to force I/ can now be determined from the following equation:

=

r2

aM, Ml-Rd9

(5.72)

-Rsin 9

(5.73)

av

From Eq. 5.63,

aMl dV

-- -

Substituting Eqs. 5.63 and 5.73 into Eq. 5.72,

1

6= EI

I

[ HR( 1 - cos 0 ) + Mc - VR sin 81 ( - R sin 8 ) R d 8

Integrating the above equation and collecting terms,

Substitute Eqs. 5.62, 5.70, and 5.71 into the above equation: PR3 PR' 6 = 0.0115 -= EI 87EI STRAIN ENERGY-CIRCULAR STRAIN RELIEF

5.7

(5.74)

ARCS FOR LEAD WIRE

Electrical lead wires are often subjected to high axial dynamic loads during vibration when the PCB natural frequency is excited. Many large surface mounted and through-hole mounted electronic components have several

STRAIN ENERGY-CIRCULAR

ARCS FOR LEAD WIRE STRAIN RELIEF

95

straight leads that extend from the bottom surface of the component body down to the PCB. When the PCB bends during its resonant condition, the lead wires will be forced to deform as shown in Fig. 5.2. This type of installation often leads to very high axial forces, high stresses, and a reduced fatigue life in the solder joints and the lead wires at the ends of a long component, where the relative displacements are the greatest. These failures can usually be avoided by adding a strain relief in the wire in the form of an offset. When the offset is properly designed, the axial load in the wire is changed to a bending load in the wire. Axial loaded wires usually have very high spring rates and early fatigue failures. Wires that are loaded in bending will typically have much lower spring rates, so they produce lower forces, resulting in an increased fatigue life. In a linear system the force P is related to the spring rate K and the deflection Y as shown below:

P=KY

(5.75)

The displacement Y is often a constant for given sets of conditions in vibration, shock, and thermal cycling environments. An examination of the above equation shows that when the displacement Y is constant, the best way to reduce the force P in the system is to reduce the spring rate K . This can usually be accomplished by adding an offset in the straight lead wire. Offsets in the form of circular arcs as shown in Fig. 5.13 work very well. The displacement and the resulting spring rate of the offset due to an axial load can be obtained with the use of Castigliano’s strain-energy methods as shown below.

FIGURE 5.13. Half of the geometry proposed for a component lead wire strain relief.

96

COMPONENT LEAD WIRES AS BENTS, FRAMES, AND ARCS

The bending moment at point M,

0is = PR( 1 - COS

e,)

(5.76) (5.77)

The bending moment at point

0 is

M 2 = PR( 1 + sin

el)

(5.78)

aM2 -=R(l+sin@,) aP

(5.79)

The deflection Y can be obtained from the strain energy of the bending in the system. ( 53 0 )

Substitute Eqs. 5.76-5.79 into Eq. 5.81: (1-cos8,)’de1+/

T I 2R 3 ( 1 + s i n e , ) * d e 2

n y=

PR

[ivi2(l -2

PR3

EI

~ 0 8, s

8, - 2sin 8,

1

+ cos2 0 , ) de, + /“12(1 + 2sin el + sin2 8,) d e , n

1

91 1 ++ -sin28, 2 4

PR3

-(O)-(-2)]=

4

4

4.71PR3 EI

The above displacement value is for only one-half of the circular arc offset. The other half of the offset is symmetrical, and it is in series, so the final displacement will be two times greater:

Y=

9.42 PR3

EI

(5.82)

STRAIN ENERGY-CIRCULAR

Sample Problem-Adding the Fatigue Life

ARCS FOR LEAD WIRE STRAIN RELIEF

97

an Offset in a Wire to Increase

A 1.2-in. long application-specific integrated circuit (ASIC) is through-hole mounted at the center of a plug-in PCB 8.0 in. long and 0.080 in. thick, as shown in Fig. 5.14. The component has several kovar wires, 0.012 in. in diameter, that extend from the bottom surface down to the PCB. The PCB has a natural frequency of 190 Hz. It must pass a 30-minute resonant dwell safety of flight test using a sinusoidal vibration input level of 3.4 G peak applied perpendicular to the plane of the PCB. Find the dynamic forces, stresses, and fatigue life in the end lead wires and solder joints. Add an offset strain relief in the wire as shown in Fig. 5.15 and recalculate the dynamic forces, stresses, and fatigue life in the end lead wires and solder joints.

kL+

0.6-

0.080 l 1 I I l l l I I l I l

t L = 8.0

1

t i c

FIGURE 5.14. Relative motion between the vibrating circuit board and the component produces strain in the lead wires.

R

i

FIGURE 5.15. Full geometry proposed for a component lead wire strain relief.

98

COMPONENT LEAD WIRES AS BENTS, FRAMES, AND ARCS

Solution Use Eqs. 2.33 and 4.1 to Find Relative Displacement Difference 6 Between the P C B and the Component as Shown in Fig. 5.14. The displacement shape of the PCB at its natural frequency will be a half sine wave. The maximum displacement Z , will be at the center. The displacement Z , at any point on the curve will be as shown below: T X

2,=&sin-

so

L

S = Z , 1 -sin-i

(5.83)

L

where G = 3.4 (peak input acceleration level) Q = V‘i!% = 13.78 (approximate transmissibility expected for PCB) f , = 190 Hz (PCB natural frequency) Z,

=

9.8Gi,Q/(f,,)’

=

(9.8)(3.4)(13.78)/(190)’

=

0.0127 in. (5.84)

x = 3.4 in. (location of edge of component) L

= 8.0

in. (length of PCB parallel to component)

(

6 = (0.0127) 1 - sin-

T(3.4) 8.0

i

= 3.51 X

in. (5.85)

The above expression represents the relative displacement between the component and the PCB, which represents the stretch in the lead wire. This conservatively assumes that the PCB deflects in a perfect sine wave, there is no bending in the component body, and the lead wire is loaded in perfect tension only. These conditions are only approximations, so they will result in values that are slightly greater than the true values. A stress concentration is not used in the calculation of the wire axial stress, because a stress concen-

tration of 2.0 will be used in the fatigue exponent b when the fatigue life of the wire is calculated in Eq. 5.90. The spring rate K of a straight wire that is loaded in pure tension can be obtained from a handbook, as shown below:

( 52 6 ) where d, A E L,

= 0.012

in. (wire diameter) in.’ (wire area) = .rr(0.012)’/4 = 1.131 X = 20 x 106 lb/in.’ (kovar wire modulus of elasticity) = 0.068 4(d,) = 0.068 4(0.012) = 0.116 in. (effective length of wire for axial load goes into component 2 diameters and into PCB 2 diameters) =

n-(d,,)’/4

+

+

K,

=

(1.131 x 10-4)(20 x 106) 0.116

QQ

ARCS FOR LEAD WIRE STRAIN RELIEF

STRAIN ENERGY-CIRCULAR

=

1.950 x lo4 lb/in.

(spring rate) (5.87)

The dynamic force in the straight wire P , can be obtained from the wire spring rate K , and the relative displacement S from Eq. 5.85 as shown below:

P,=K,S=(1.950x104)(3.51X10-4)

=6.84lb

(5 .88)

The dynamic axial stress in the wire can be obtained from the standard stress equation, the dynamic load, and the wire area shown in Eq. 5.86: 6.84

p, w- A

s

--=

1.131 x

=

60,477 lb/in

.*

(wire axial stress) (5.89)

The approximate fatigue life of the kovar wire can be obtained from Fig. 3.1 with the general fatigue equation as shown below. Note that the fatigue exponent 6.4 has a stress concentration of 2.0 included to account for any possible deep cuts or scratches due to careless handling or manufacturing problems. N,

= N2[;]’

=

(1000)

84,000

6.4

=

8189 cycles to fail

(5.90)

The approximate time for the wire to fail can be obtained from the 190-Hz PCB natural frequency as shown below: Wire fatigue life

8189 cycles to fail =

(190 cycles/s) (60 s/min)

=

0.718 minute to fail (5.91)

The qualification test requires a 30-minute dwell. The rapid wire fatigue failure shows the proposed design is not acceptable. The dynamic shear stress in the PCB wire solder joint can be obtained from the force in the wire and the average shear area of the solder joint as shown below:

ss= where P , d, d,,

p, A,

(5.92)

lb (see Eq. 5.88) in. (wire diameter) = 0.024 in. (expected diameter of the plated through-hole in the PCB) = 6.84

= 0.012

100

COMPONENT LEAD WIRES AS BENTS, FRAMES, AND ARCS

d,, t A,

= (0.012

+ 0.024)/2

0.018 in. (average solder joint diameter) = 0.080 in. (PCB thickness) = rd,,,t = r(0.018)(0.080) = 0.00452 i n 2 (average solder shear area) (5.93)

&=--

6.84

=

-

1513 lb/in.*

(solder shear stress) (5.94)

0.00452

The approximate vibration fatigue life of the solder joint can be obtained from Fig. 3.2 with the general fatigue equation as shown below:

The approximate time for the solder joint to fail can be obtained from the 190-Hz PCB natural frequency as shown below: Solder fatigue life

3.41 X 10' cycles to fail =

(190 cycles/s) (60 s/min)

= 29.9

minutes to fail (5.96)

The solder joint fatigue life is better than the lead wire fatigue life shown in Eq. 5.91, but the solder joint will not meet the 30-minute resonant dwell requirement with a safety factor. Now add an offset strain relief in the lead wire as shown in Fig. 5.15 and recalculate the dynamic forces, stresses, and fatigue life in the lead wires and solder joints. The new spring rate of the wire with the strain relief can be obtained from Eq. 5.82 as shown below: (5.97) where E = 20 x 106 lb/in.* (kovar wire modulus of elasticity) d , = 0.012 in. (wire diameter) I , = n-(d,I4/64 = .rr(O.O12)'/64 = 1.018 X i n 4 (wire moment of inertia) Y = 0.017 in. (radius of the wire offset strain relief) K=

(20 X 106)(1.018 X (9.42) (0.017)

= 440

lb/in.

(5.98)

The dynamic force in the wire can be obtained from the relative displacement shown in Eq. 5.85 and the above spring rate. P,,=K13=(440)(3.51xlO-~) =0.154lb

(5.99)

101

ARCS FOR LEAD WIRE STRAIN RELIEF

STRAIN ENERGY-CIRCULAR

Although there is an axial load in the wire, the strain relief offset will produce a bending moment and a bending stress in the wire. The offset distance will be equal to 2 radii.

M = 2rP,

=

(2)(0.017)(0.154)

= 0.00523

1b.in.

(5.100)

The bending stress equation in the wire can be obtained from a structural handbook as shown below. A stress concentration factor is not used here because it will be used in the fatigue exponent b when the fatigue life of the wire is calculated in Eq. 5.103. (5.101) where C = d,/2 = 0.012/2 = 0.006 in. (wire radius) M = 0.00523 lb-in. (see Eq. 5.100) I , = dd,I4/64 = ~ ( 0 . 0 1 2 ) ~ / 6=4 1.018 X inertia) (0.00523)(0.006) s b =

1.018 X

= 30,825

lb/in.2

in.4 (wire moment of

(wire bending stress) (5.102)

The approximate fatigue life of the kovar wire can be obtained using Fig. 3.1 with the general fatigue equation shown below. A stress concentration factor of 2.0 was included in the 6.4 fatigue exponent to account for any deep cuts or scratches that may be caused by the wire-forming dies.

Nl = N 2 [ $ I b = (lOOO)(w] 84,000

=6.115 x 10’cycles to fail (5.103)

The approximate time for the wire to fail can be obtained from the 190-Hz PCB natural frequency as shown below: Wire fatigue life

6.115 x 10’ cycles to fail =

(190 cycles/s) (60 s/min)

= 53.6

minutes to fail (5.104)

The system requires a qualification fatigue life of 30 minutes. The lead wire has a safety factor of about 1.8, which should be adequate since the analysis method was conservative. The shear stress in the solder joint can be obtained using Eq. 5.92, the average solder shear area from Eq. 5.93, and the wire force from Eq. 5.99, as shown below:

s

P, As

--=-

0.154 0.00452

= 34

lb/in

.’

(solder shear stress)

(5.105)

102

COMPONENT LEAD WIRES AS BENTS, FRAMES, AND ARCS

The approximate vibration fatigue life of the solder joint can be obtained from Fig. 3.2 and the general fatigue equation as shown in Eq. 5.95. 1012 cycles to fail (5.106) The approximate time for the solder joint to fail can be obtained from the 190-Hz PCB natural frequency as shown below: Solder fatigue life

1.33 x lo1' cycles to fail =

(190 cycles/s) (3600 s/h)

=

1.94 x 106 hours to fail (5.107)

The solder joint in this case is very safe. It is much less critical than the lead wire so the lead wire is the weaker element in this system.

-

CHAPTER 6

Printed Circuit Boards and Flat Plates

6.1

VARIOUS TYPES OF PRINTED CIRCUIT BOARDS

Electronic systems make extensive use of plug-in printed circuit boards (PCBs) because they are very easy to service. Defective circuit boards can be removed and replaced quickly and easily without bothering with wires and a soldering iron. After a defective circuit board has been removed, troubleshooting is easy and repairs can be made by a skilled technician. Many different types of PCBs are manufactured by the electronics industry. Epoxy fiberglass is the most common material used, with laminated layers of copper on one or both sides of the board to form the electrical conductors. The overall PCB thickness can vary from about 0.006 to 0.125 in. Board sizes can vary from about 2 to 16 in. Many different shapes can be found, ranging from small squares to large circular plates and triangles, depending on the shape of the electronic box used to support the circuit boards. Since electronic equipment is being packed into every available inch of space in most airplanes, missiles, and even television sets, the shape of the circuit board is often dictated by the geometry of the available space. The rectangular PCB is the most common shape used by the electronics industry, since this shape is easily adapted to the popular modular plug-in type of assembly, which utilizes an electrical connector along the bottom edge of the circuit board. PCBs with many high-power-dissipating components will run very hot unless the heat is removed. Therefore aluminum and copper, which have high thermal conductivities, are often bonded to the epoxy fiberglass circuit boards to act as heat sinks. Ribs are often added to PCBs that must operate in a severe vibration and shock environment. Ribs increase the stiffness of the circuit board, which in turn will increase the resonant frequency. This will reduce the board deflections during resonant conditions, thus reducing the stresses developed in the electronic component parts mounted on the circuit board. Ribs can be fabricated of steel, aluminum, or epoxy fiberglass. If metal ribs are used, caution must be exercised to prevent short circuits across exposed electrical 103

104

PRINTED CIRCUIT BOARDS AND FLAT PLATES

printed circuit strips. Ribs can be bolted, riveted, soldered, cemented, welded, or cast integral with heat-sink plates. PCBs may be supported by the electronic box in many different ways, depending on such factors as the environment, weight, maintainability, accessibility, and cost. For example, in a vacuum environment where there is no air to provide convection heat transfer, heat will often be conducted from the circuit board to a heat exchanger. High-pressure thermal interfaces must then be provided, using materials that have a high thermal conductivity, in order to prevent excessive temperatures from developing in the electronic components mounted on the circuit boards. Screws are ideal for providing high-pressure interfaces, but they are generally time-consuming to install, since they take many turns to insert and remove. Maintenance people prefer simply to unplug one module and plug in another. This reduces maintenance time substantially, since electronic systems can easily contain over a hundred plug-in PCBs. Sometimes a fast-acting cam-and-wedge type of clamp is used to provide a high-pressure interface, if there is enough room available in the electronic box. The manner in which the PCBs are supported in the electronic box can be an important factor in determining just how the boards will respond to vibration and shock. If the electronic box is fabricated from light sheet metal and manufacturing tolerances are loose, a loose fit is desirable for the plug-in circuit boards. This will permit easy connector engagement, with a fixed connector position, when there is a substantial mismatch in the mating box connector. This type of installation is not desirable, however, for a system that will be subjected to severe vibration and shock. A loose circuit board will often develop high acceleration loads, which will lead to high deflections and stresses in the electronic component parts mounted on the circuit boards. The edges of the PCBs should be supported if they will be subjected to severe vibration or shock environments. Board edge guides are available that will support a wide variety of PCBs. These guides are usually fabricated of beryllium copper, but they are also available in a wide range of metals and molded plastics. A board edge guide that grips the edge of a PCB firmly is very desirable. This firm grip can reduce deflections due to the edge rotation and translation, which will increase the natural frequency of the circuit board. In addition, a firm grip will tend to dissipate more energy during vibration because of friction and relative motion between the edges of the board and the edge guide. This, in turn, will reduce the transmissibility experienced by the printed circuit board during resonance. A tight board edge guide may require tight manufacturing tolerances, plus a floating chassis connector, to permit accurate connector alignment during the circuit board installation, to prevent connector damage. The transmissibility developed by a PCB during resonance will depend on many factors, such as the board material, number and type of laminations in a multilayer board, natural frequency, type of mounting, type of electronic component parts mounted on the circuit board, acceleration G levels, type of

VARIOUS TYPES OF PRINTED CIRCUIT BOARDS

105

Side guide

Circuit board laminated to

4.0 Electronic component parts

I I I l l

1 1 I I I I ~ I 1 I1 1 1 1 1 I

\Connector

FIGURE 6.1. Laminated circuit board with a stiffening frame.

conformal coating, type of connector, and shape of the board. Slight modifications in the installation geometry of the circuit board can have a sharp effect on the transmissibilities and the mode shapes developed at resonance. This can be demonstrated by changing the edge restraints of a PCB from approximately simply supported to approximately fixed. Vibration test data on one group of boards, which used Birtcher guides on two opposite sides of the board to simulate a simply supported condition, showed natural frequencies of about 260 Hz. The transmissibility was about 16 for a 5-G peak sinusoidal vibration input. This is about equal to the square root of the natural frequency. The opposite sides of the circuit boards were then clamped to simulate a fixed-edge condition. This increased the typical natural frequency to about 380 Hz and increased the typical transmissibility to about 25. This is equal to 1.28 times the square root of the typical natural frequency. The group of PCBs used in these tests had the dimensions and construction shown in Fig. 6.1. These tests were run in a rigid vibration test fixture that included a mating connector for each circuit board. Accelerometers were mounted at several points on the vibration fixture and each circuit board to monitor the input and output acceleration G forces for the system (Fig. 6.2). These same PCBs were then installed in an electronic box, and the 5-G peak sinusoidal vibration tests were repeated. The circuit boards with Birtcher guides showed a typical natural frequency of about 225 Hz with a typical transmissibility of about 14. This is equal to 0.93 times the square root of the natural frequency. The circuit boards then had their opposite sides clamped to the chassis structure to simulate a fixed-edge condition. The natural frequency increased to about 340 Hz, and the transmissibility, from the chassis to the center of the circuit board, increased to about 20. This is equal to 1.08 times the square root of the natural frequency.

106

PRINTED CIRCUIT BOARDS AND FLAT PLATES

accelerometers

Birtcher guide for a supported edge

Vibration input direction Rigid vibration test fixture

I

Vibration shaker head

FIGURE 6.2. PCB mounted in a rigid vibration fixture.

6.2

CHANGES IN CIRCUIT BOARD EDGE CONDITIONS

In order to calculate the natural frequency of a PCB, it is necessary to estimate the edge conditions of the board. Much insight can be obtained into edge conditions by using a strobe light to watch board motions during vibration. This shows very quickly whether the edges are translating or rotating. Sometimes edge conditions can change when the acceleration forces change. An edge that appears to be simply supported with a 2-G peak input may appear to be free with a 5-G peak input. The natural frequency with a 2-G input will then be different from the natural frequency with a 5-G input. Sinusoidal vibration tests were run on a group of PCBs that had the opposite sides supported by a preloaded beryllium-copper wavy spring, which acted as a board edge guide, in a channel section. The bottom edge of each board had a 100-pin connector that plugged into the sockets of a mating connector mounted in a rigid vibration fixture. The top edge of each board was restrained by three foam-rubber strips 1 in. by 0.125 in. thick and compressed about 60% when the board was installed. These rubber strips were used to restrain the unsupported top edge of the circuit board, to simulate the action of a cover using the same type of rubber strips. Strips of this type are convenient to use because they provide some support and damping without the high fabrication costs associated with close tolerances on machined parts.A sketch of this type of installation is shown in Fig. 6.3. Vibration tests were run with peak inputs of 2, 5 , 10, and 15 G using a strobe light to observe the action of the circuit board edges during resonant conditions. The 2-G tests showed that all four edges of the circuit board appeared to be simply supported. The edges were observed to be rotating without translation. The 5-G tests showed that the sides of the circuit boards acted as though they were supported, but the top edge at the rubber pads and the bottom edge at the connector showed some translation in addition to the rotation. The 10-G tests still showed that the sides of the circuit boards acted as though they were simply supported, but the top edges at the rubber strips and the bottom edges at the connectors all experienced much more

CHANGES IN CIRCUIT BOARD EDGE CONDITIONS

107

/ Side channel guide

Wavy spring

Section AA

/ l I ' l l I l ~ l ~ / ~ l l ~ l l l ~ I ~ ~ 100-pin connector

FIGURE 6.3. Typical installation of a PCB.

-*

8 0 +----

-F-

-c

0 3 8 rib ai u in i n urn

1

i 0 12 rib

NO 2-56

70 I I

! I 1

"1

I

1.1 I 1

I I 1 I I I 1 I I 1 1 I I I I 1

150-pin connector

k

fiberglass board 0 062 in

- .

FIGURE 6.4. PCB with ribs fastened by screws.

translation than rotation. The 15-G tests showed some translation at the sides in addition to the rotation, while the top and bottom edges appeared to be free because there was very little rotation, only large translation amplitudes. Resonant dwell tests at a 10-G peak resulted in many broken connector pins. Stiffening ribs were added to some of the circuit boards to increase their resonant frequencies. Some of these ribs were screwed to the faces of the circuit boards (Fig. 6.4).

108

PRINTED CIRCUIT BOARDS AND FLAT PLATES

TABLE 6.1 Input G Level

Resonant Frequency

Transmissibility

(Hz)

Q

2

215

15.0

5 10

182

11.2 8.2

161

Vibration tests were run on these boards using the same type of foamrubber strips at the top of each circuit board and the same type of wavy-spring channel guides at the sides of each board. When the input G levels were changed, there were substantial changes in the natural frequencies and transmissibilities (Table 6.1).

6.3 ESTIMATING THE TRANSMISSIBILITY OF A PRINTED CIRCUIT BOARD

Printed circuit boards can have a wide variety of sizes and shapes along with many different mounting arrangements. In the early stages of a design, these circuit boards must be analyzed to make sure they will function properly in the required environment. One critical part of the analysis relates to the dynamic loads developed in these circuit boards at their fundamental resonant frequencies. These loads are closely associated with the transmissibilities, developed by the circuit boards, but transmissibilities are very difficult to estimate without previous test data on similar boards. Obviously, test data are the best sources for information on the transmissibility characteristics for various types of circuit boards. However. there are times when test data are not available because of a new design or a geometry that has not been used before. Under these conditions it becomes necessary to estimate the response characteristics of the circuit board, at least until a model circuit board can be fabricated and tested. There are a number of factors that should be considered when the transmissibility of a PCB must be estimated without the use of test data. These factors are all related to the damping characteristics of the board, which determine the amount of energy lost during the vibrating condition. When more energy is lost, or transformed to heat, there is less energy remaining and the transmissibility is lower; hence the dynamic loads and stresses are also lower. The greatest energy losses are probably due to hysteresis and friction. Hysteresis losses are generally due to internal strains that are developed during bending deflections in the circuit board. Friction losses are generally due to relative motion between high-pressure interfaces such as mounting

ESTIMATING THE TRANSMISSIBILITY OF A PRINTED CIRCUIT BOARD

109

surfaces, stiffening ribs, and edge guides. These energy losses are greatest when the deflections are greatest and smallest when the deflections are smallest. Since higher frequencies have smaller deflections, they will also have less damping. This means higher frequencies will usually have higher transmissibilities at resonant conditions. Terms such as “low” resonant frequency and “high” resonant frequency are only relative. In general, however, the term “low” applies to resonant frequencies below about 100 Hz. The term “high” applies to resonant frequencies above about 400 Hz. Most circuit boards appear to have their fundamental resonant frequencies between about 200 and 300 Hz when used in military electronic systems. Test data show the transmissibility of a PCB can generally be related to the square root of the natural frequency of the board. For rectangular boards, the transmissibility will normally range from about 0.50 to 2.0 times the square root of the natural frequency, depending on many factors. A small circuit board with small electronic component parts and no stiffening ribs, with a resonant frequency of about 400 Hz, can have a transmissibility as high as 2.0 times the square root of the natural frequency for a low input G force below a 2-G peak. The transmissibility for this case will then be about 40. A large circuit board with large electronic component parts and several stiffening ribs, with a resonant frequency of about 100 Hz, can have a transmissibility as low as 0.50 times the square root of the natural frequency for a high input G force above a 10-G peak. The transmissibility for this case will then be about 5. Over the middle frequency ranges (200-300 Hz), the transmissibility will often be about equal to the square root of the natural frequency for a 5-G peak sinusoidal vibration input. Again, there are many other factors that will influence the transmissibility of a PCB. If there are no test data available on the particular type of circuit board being analyzed, then the approximations just outlined are suggested as a good starting point. Some of the factors that should be considered when the transmissibility of a circuit board is being estimated are as follows: 1. The Natural Frequency of the Circuit Board. A high natural frequency means low displacements and low strains, so the transmissibilities are usually higher. Conversely, a low natural frequency means high displacements and high strains, so the transmissibilities are usually lower. 2. The Input G Force for Sinusoidal Vibration. A lower input G force means low displacements and low strains, so the transmissibilities are usually higher. A high input G force means high displacements and high strains, so the transmissibilities are usually lower. 3. Ribs. Riveted and bolted ribs will generally permit some relative motion to occur at the high-pressure interface between the ribs and the circuit board, which will dissipate energy and reduce the transmissibility at resonance.

110

PRINTED CIRCUIT BOARDS AND FLAT PLATES

Welded, cast, and cemented ribs are usually stiffer than riveted and bolted ribs, so that they raise the resonant frequency more, but they do not provide as much damping. 4. Circuit Board Supports. Circuit board edge guides that grip the edges of the board firmly provide a high-pressure interface that will dissipate energy and lower the transmissibility. If mounting bolts are used to fasten the boards, high-pressure interfaces in the bolt areas will tend to dissipate energy. More mounting bolts will dissipate more energy and increase the stiffness at the same time. 5. Circuit Board Connectors. Circuit board connectors, such as the edge type or the pin-and-socket type, will provide some damping that will tend to reduce the transmissibility. A longer connector will provide more support to the circuit board edge while providing additional damping. 6. Type of Electronic Component Part. Large electronic component parts that are in intimate contact with the circuit board will tend to dissipate more energy than small electronic component parts. This is because larger components cover a bigger span on the circuit board, resulting in larger relative deflections at the circuit board interface over a larger area, so the damping is slightly greater. 7. Heat-Sink Strips. Heat-sink strips are often laminated to the circuit board to conduct the heat away from the electronic component parts. These strips will act as laminations that will dissipate energy through relative interface shear deflections. At high frequencies, where the displacements are small, the damping is small, so transmissibilities are not reduced very much. 8. Multilayer Circuit Boards. Many PCBs must use extra layers to provide all of the electrical interconnections required by the circuits. These extra layers will increase the damping slightly and thus tend to reduce the transmissibility. At high frequencies, where deflections are small, transmissibilities are not reduced very much. 9. Conformal Coatings. Conformal coatings are generally used to protect electronic components from dust accumulation. Many manufacturers of electronic equipment also use conformal coatings to protect sensitive circuits from water vapor and condensation. A thin polyurethane or epoxy film about 0.005 in. thick is usually sprayed or dipped on the circuit board. Unless the adhesion is good and unless all of the sharp points are coated, water vapor can wick in between the coating and the circuit board and condense. Once this happens, the moisture will usually stay where it is unless it is baked out at high temperatures. Small amounts of moisture will not have much effect on the electrical operating characteristics of the circuits unless the circuits have a high impedance. Conformal coatings also act as a cement, which provides good adhesion for electronic component parts to the circuit board. During vibration, relative motion between the components and the circuit board will produce strains in the conformal coating and will tend to provide some damping.

NATURAL FREQUENCY USING A TRIGONOMETRIC SERIES

11 1

z FIGURE 6.5. A flat rectangular plate supported on four sides.

6.4 NATURAL FREQUENCY USING A TRIGONOMETRIC SERIES Most PCBs can be approximated as flat rectangular plates with different edge conditions and different loading conditions. General plate equations can then be used to determine the strain energy and the kinetic energy of the vibrating plate, which leads to the natural frequency equation. One very convenient method for analyzing plates is the Rayleigh method [28]. A deflection curve is assumed that satisfies the geometric boundary conditions, which are the deflection and the slope for a particular plate. Once these boundary conditions are satisfied, the assumed deflection curve is used to obtain the strain energy and the kinetic energy of the particular plate. If there is no energy dissipated, the strain energy will be equal to the kinetic energy and the approximate natural frequency can be determined [32]. The Rayleigh method results in a natural frequency that is slightly higher than the true natural frequency for a given set of conditions, unless the exact deflection curve is used. Consider a flat, rectangular plate, with four simply supported edges and a uniformly distributed load, being vibrated in a direction perpendicular to the plane of the plate (Fig. 6.5). The deflection curve for the simply supported plate can be represented by a double trigonometric series [341: I Z mn-X nrY Z= A,,,,,sin-sina b m= n= 1.3:5. 1,3:5,

Extensive vibration test data on PCBs show that most of the damage occurs at the fundamental resonant mode where the displacements and the stresses are the greatest. The above equation can then be simplified to the

112

PRINTED CIRCUIT BOARDS AND FLAT PLATES

following expression:

Z

=Z,

ITX %-Y sin -sin a b

The assumed deflection curve must be checked to make sure it meets the geometric deflection boundary conditions. This means the edges of the plate must have a zero deflection and the center of the plate must have the maximum deflection. Then from Eq. 6.1, at X = O atX=O atX=a atX=a a at X = 2

and y = O , andY=b, andY=O, andY=b, b andY=-, 2

Z=O Z=O Z=O Z=O Z=Z,

The deflection boundary conditions are satisfied: there are no deflections at the edges and the maximum deflection is at the center. Next check the geometric boundary conditions for the slope of the plate at different points. There must be a finite slope at all four edges, but the center of the plate must have a zero slope. Considering first the slope along the X axis, the required equation can be determined from the partial derivative of Z with respect to X. Then from Eq. 6.1, az 7i 7ix 7iY 8 - - = Z,-cos-sin-ax a a b so

at X = a

b and Y = -, 2

7T

O x = -Z,-

a

The slope boundary conditions are satisfied at the edges and at the center of the plate along the X axis. The same check can be made for the slope along the Y axis. The total strain energy I/ of the vibrating plate can be represented in the following form [41:

NATURAL FREQUENCY USING A TRIGONOMETRIC

SERIES

113

where D = Eh3/12(l - p’) (plate stiffness factor) E = modulus of elasticity (lb/in.2) h = plate thickness (in.) p = Poisson’s ratio (dimensionless) The total kinetic energy T of the vibrating plate can be represented in the following form [41:

T = c / u / b Z 2dXdY 2 0 0 where p = W/abg = u h / g (mass per unit area) W = total weight of plate (lb) u = material density (lb/in.3) a = length of plate (in.) b =width of plate (in.) h = plate thickness (in.) g = acceleration of gravity (386 in./s2) a=circular frequency (rad/s) Performing the operations on Eq. 6.1 required by Eq. 6.2,

aZ

%-x

n

- = 2,-cos-sina a

7TY

ax

a22

--

n2

-

ax2

a2z 2

b 7 T x

7TY

a

b

-2 -sin-sin-

‘a2

nY

(z =ziTsin’-sin2] nX n4

b

a

az

n

nx

nY

3Y

b

a

b

- = 2,-sin-cos-

a2z

-- -

dY2

n2

-Z,,sin-sinb

nX a

~

nY b

114

PRINTED CIRCUIT BOARDS AND FLAT PLATES

From Eqs. 6.5 and 6.8,

(6.10)

From Eq. 6.4 or 6.7,

a22

T2

ax dY

= z,-cos-cos-

nx

77-Y

a

b

ab

[SY)n-x 2

7i-4

=Z~a"COS--COS

7

a

,n-Y b

(6.11)

Since these equations must be integrated, the following relations are used:

(6.12)

Following Eq. 6.2 and integrating Eq. 6.6 using Eq. 6.12,

(6.13)

Following Eq. 6.2 and integrating Eq. 6.9 using Eq. 6.12,

(6.14)

Following Eq. 6.2 and integrating Eq. 6.10 using Eq. 6.12,

NATURAL FREQUENCY USING A TRIGONOMETRIC SERIES

115

Following Eq. 6.2 and integrating Eq. 6.11 using Eq. 6.12,

(6.16)

Substituting Eqs. 6.13-6.16 into Eq. 6.2 for the strain energy of the vibrating plate,

V=

2

(6.17)

8

The kinetic energy of the vibrating plate can be determined from the deflection (Eq. 6.1) along with the kinetic energy (Eq. 6.3): (6.18) Substituting Eq. 6.12 into Eq. 6.18 for the integration, we have

( i b Z 2 dXdY

ab

=Zi-

4

(6.19)

Substituting Eq. 6.19 into Eq. 6.3 for the kinetic energy of the vibrating plate, we have (6.20) Since the strain energy of the vibrating plate must equal the kinetic energy at resonance, if there is no energy dissipated, then Eq. 6.17 must be equal to Eq. 6.20:

P

116

PRINTED CIRCUIT BOARDS AND FLAT PLATES

Solving for the natural frequency of the rectangular plate, we have

(6.21)

In this particular case the natural frequency equation shown above is exact using the Rayleigh method, because the deflection curve used for the vibrating plate is exact for the boundary conditions. 6.5

NATURAL FREQUENCY USING A POLYNOMIAL SERIES

The natural frequency equation for a uniform, flat, rectangular plate simply supported on four edges can also be derived with the use of a polynomial series. Consider the coordinate system for the rectangular plate when the X and Y axes are taken at the center of the plate (Fig. 6.6). The deflection curve for the simply supported plate can be represented by a double polynomial series as follows [30]: -x

-x

Z = h C m=O n=O

[

1 - 7

) ( &' ) [ ;) ( ;) I--

'l*

A,,,, cos R t

Since most of the damage on a PCB will occur in the fundamental resonant mode, where the displacements and the stresses are the greatest, the above

I

Supported

I

z FIGURE 6.6. A simply supported plate with the axes at the center.

NATURAL FREQUENCY USING A POLYNOMIAL SERIES

117

equation can be simplified to the following: (6.22) The assumed deflection curve must be checked to make sure it meets the geometric deflection boundary conditions. This means the edges of the plate must have a zero deflection and the center of the plate must have the maximum deflection. Then from Eq. 6.22, at X = O a n d Y = O ,

Z=Z,

at X = O a n d Y = d ,

Z=O

at X = c a n d Y = d ,

Z=O

at X = c a n d Y = O ,

Z=O

The deflection boundary conditions are satisfied, since there are no deflections at the edges and the maximum deflection is at the center. Next check the geometric boundary conditions for the slope of the plate at different points. There must be a finite slope at all our edges, but the center of the plate must have a zero slope. Considering first the slope along the X axis, the required equation can be determined from the partial derivative of Z with respect to X. Then from Eq. 6.22,

and at X=O

andY=O,

8,=0

at X=O

andY=d,

O,=O

The slope boundary conditions are satisfied at the edges and at the center of the plate along the X axis. The same check can be made for the slope along the Y axis. The strain energy equation (Eq. 6.2) and the kinetic energy equation (Eq. 6.3) can also be used with the polynomial equation (Eq. 6.22). Perform-

118

PRINTED CIRCUIT BOARDS AND FLAT PLATES

ing the operations on Eq. 6.22 as required by Eq. 6.2 for the strain energy,

( 6.23)

(6.24)

(6.25)

(6.26)

(6.27)

(6.28) From Eqs. 6.24 and 6.27,

From Eq. 6.23 or 6.26, d2Z

2x 2Y

ax aY =z,---;c 2 d' (6.30) When the integration is performed according to Eq. 6.2 for the strain energy, observe that the X and Y reference axes for this geometry are not at the edges of the plate; they are at the center of the plate. Therefore, to consider the full plate, we use Z ) dXdY = 4

jc jdf( Z ) dXdY 0 0

(6.31)

NATURAL FREQUENCY USING A POLYNOMIAL SERIES

119

From Eqs. 6.2, 6.25, and 6.31, 2

4l[[$]

16Zicd d X d Y c= r

From Eqs. 6.2, 6.28, and 6.31, 2

46/i$) From Eqs. 6.2, 6.29, and 6.31,

From Eqs. 6.2, 6.30, and 6.31,

J2Z

2

-(-I(

64Zicd 1 4 / i / d ( ~ y ) dXdY= c2d2 3 0

0

'j

(6.35)

3

The strain energy of the plate can be obtained by substituting Eqs. 6.32-6.35 into Eq. 6.2:

1 V = 64DZicdjjj-p

1

1

1

+9c2d2 15d4 +

(6.36)

The kinetic energy of the plate can be determined from Eqs. 6.3 and 6.22 as follows:

+-+,+, xc 4

y4

d

x4y4]

c4d4

(6.37)

Since the X and Y reference axes are through the center of the plate and integration must take place over the entire plate area. Equation 6.31 must be

120

PRINTED CIRCUIT BOARDS AND FLAT PLATES

used with the kinetic energy (Eq. 6.3). Then from Eq. 6.37,

Substituting Eq. 6.38 into Eq. 6.3 for the kinetic energy of the vibrating plate,

T = 0.566ZicdpR2

(6.39)

Since the strain energy must equal the kinetic energy at resonance, Eq. 6.36 must be equal to Eq. 6.39:

64DZ;cd

( 3+ 9 c 2 d 2 1

1

1

-

= 0.566Z;cdpR2

+

Solving for the natural frequency of the plate,

f

( 6.40)

=-=-

Sample Problem-Resonant

Frequency of a PCB

Determine the resonant frequency of a rectangular PCB that will be subjected to a 2-G peak sinusoidal vibration test. With a low input G level, the four edges of the circuit board will act as though they were simply supported. The board has a uniformly distributed weight of 1 lb, with dimensions as shown in Fig. 6.7.

Y

. 1

a = 83

. , 7

Supported U

?

g a

Typical electronic component part

a f n

I

Supported

-jkh=0062

%

S

E

5

I

7' O

I

FIGURE 6.7. Dimensions of a simply supported rectangular plate.

121

NATURAL FREQUENCY USING A POLYNOMIAL SERIES

Solution. Using the natural frequency equation derived with the use of a trigonometric expression,

f

-(-I

n- D

=

n

2

1 (2 +

P

$1 1

(see Eq. 6.21)

where a = 8.0 in. (board length) b = 7.0 in. (board width) W = 1.0 lb (board weight) g = 386 in./s2 (gravity) h = 0.062 in. (board thickness) E = 2.0 X l o 6 lb/in.2 (epoxy fiberglass) p = 0.12, Poisson’s ratio (dimensionless) we have

D=

Eh 12(1 - P 2 )

area

2 x 106(0.062)3 =

12[1 - (0.12)*] 1.o

W

mass

’=

-

-

gab = (386)(8.0)(7.0)

40.1 Ib.in.

= 0.463 X

lb.~’/in.~

Substituting into Eq. 6.21, 40.1

1/ 2

1

1

(61+9]

0.463~10-~]

f,, = 52.6 HZ

(6.41)

The natural frequency of the circuit board can also be determined from the equation derived with the use of a polynomial expression:

1

where D

= 40.1

lb . in.

lb . s 2 / i n 3 p = 0.463 x c = a / 2 = 8.0/2 = 4.0 in. d = b/2

= 7.0/2 = 3.5

in.

(see Eq. 6.40)

122

PRINTED CIRCUIT BOARDS AND FLAT PLATES

Substituting into Eq. 6.40, 1

+

9(4.0)2(3.5)2

15(3.5)4

f, = 56.1 HZ

(6.42)

6.6 NATURAL FREQUENCY EQUATIONS DERIVED USING THE RAYLEIGH METHOD

The Rayleigh method is very convenient for determining approximate resonant frequencies of rectangular plates with various edge conditions. All that is required is to assume a deflection curve that satisfies the geometric boundary conditions of deflection and slope for the particular plate. Consider the case of a rectangular plate that is fixed on all four edges. The deflection will be zero at all four edges and maximum at the center. The slope will also be zero at the four edges and zero at the center. If a trigonometric expression is used to describe this plate, the coordinate axes can be placed at the edges of the plate as shown in Fig. 6.8. One trigonometric expression that will satisfy these boundary conditions is

i

z=z,1-cos-

a

This deflection curve can be examined to make sure it meets the geometric

Fixed 0

2

t i

Fixed

FIGURE 6.8. A rectangular plate with four sides fixed.

1

z

I

b

NATURAL FREQUENCY EQUATIONS USING THE RAYLEIGH METHOD

123

deflection requirements: at X = O a n d Y = O ,

Z=O

at X = O a n d Y = b ,

Z=O

at X = a a n d Y = b ,

Z=O

a b at X = - and Y = -, Z 2 2

= 42,

Now examine this curve to make sure it meets the geometric slope requirements. Consider first the slope along the X axis, which can be determined by taking the partial derivative of Z with respect to X :

Then b atX=OandY=-, 2

O,=O

b at X = a and Y = -, 2

O,=O

a b at X = - and Y = -, O , = O 2 2

The same check can be made for the slope along the Y axis, which shows the curve will satisfy the geometric slope requirements. Then using Eqs. 6.2 and 6.3 for the strain energy and the kinetic energy, the fundamental natural frequency becomes

A polynomial expression can also be used to describe a rectangular plate fixed on all four sides. The coordinate axes for this plate can be shifted to the center (Fig. 6.9). One polynomial expression that will satisfy these boundary conditions is

124

PRINTED CIRCUIT BOARDS AND FLAT PLATES

Y

t--7

Fixed

u ! I

u.

>X

Fixed

--

& i Q-x

_-

c

FIGURE 6.9. Axes at the center of a plate with four fixed edges.

z

This polynomial will result in the following natural frequency:

The natural frequency of a rectangular plate with three supported edges and one free edge can be derived using a trigonometric function or a polynomial function. A sketch of the coordinate axes for each plate, along with the deflection expression and the resulting natural frequency equations, for a trigonometric function and for a polynomial function, are shown in Fig. 6.10. A comparison can be made of the deflection equation and the resulting natural frequency equation using a trigonometric function or a polynomial function for plates with combinations of free edges, supported edges, and fixed edges, as shown in Figs. 6.11 and 6.12. Trigonometric functions can also be combined with polynomial functions in the same deflection equation to describe the deflection of a uniform rectangular plate. For example, considering a plate that is simply supported on two opposite edges and fixed on two opposite edges, the coordinate axes and the deflection equation that meets the geometric boundary conditions will be as shown in Fig. 6.13:

z=z, ( l-cos-

2"x)/I c1

-

$1

NATURAL FREQUENCY EQUATIONS USING THE RAYLEIGH METHOD Trigonometric

125

Polynomial

Y

Y

a m a

a m a

X

U

Supported

X

(61

fa)

i?xsin irY z = Z,,sin 2a h

z = z , , x1-d'

y2)

FIGURE 6.10. T h r e e sides supported and o n e side free: ( a ) trigonometric and ( b ) polynomial.

Trigonometric

a a m

Polynomial

n Free

a

a

m

X

i?X

Z = Z , ,sin (I

Free

( 1 -):7

z = z,,

FIGURE 6.11. Two opposite edges free and two opposite edges supported: ( a ) trigonometric and ( b ) polynomial.

126

PRINTED CIRCUIT BOARDS AND FLAT PLATES Trigonometric

Y

I Supported

Supported

I 1 ----

I

X

(

z=z,1 - - :*2)2(

-

p)

FIGURE 6.12. Two opposite edges fixed and two opposite edges supported: ( a > trigonometric and ( b ) polynomial.

Y

Supported

FIGURE 6.13. Combining a trigonometric function and a polynomial function.

1

-x

Supported

There has been an extensive amount of work on the derivation of natural frequency equations for uniform, flat, rectangular plates with various edge conditions. Little [30], Warburton [31], and Laura and Saffel [33] made extensive use of trigonometric and polynomial series to determine the various resonant modes of different plates. If only the fundamental resonant mode of a uniform, flat, rectangular plate is desired, then the simplified types of trigonometric and polynomial expressions, just outlined, can be combined to determine the fundamentalresonant-frequency equations for many different types of plates. Many of these equations for plates with various edge conditions are shown in Figs. 6.14-6.16.

DYNAMIC STRESSES IN THE CIRCUIT BOARD

127

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Free edge

Supported edge

Fixed edge

Equation

Plate a

X X

b I

XL

xxxxxxxxxxxxxxxxxxxxxx

j b " !

xxxxxxxxxxxxxxxxxxxxxx

X

xxxxxxxxxxxxxxxxxxxxxx

xxxxxxxxxxxxxxxxxxxxxx

ZTbji X

X

xxxxxxxxxxxxxxxxxxxxxx

xxxxxxxxxxxxxxxxxxxxxx

\ " I

xxxxxxxxxxxxxxxxxxxxxx

FIGURE 6.14. Natural frequency equations for uniform plates.

6.7

DYNAMIC STRESSES IN THE CIRCUIT BOARD

Dynamic bending stresses in a uniformly loaded PCB can be determined for the resonant condition by considering a sinusoidal load variation for the dynamic load over the board surface when the edges of the board are simply supported. This follows from the relative dynamic deflections, which will be maximum at the center of the board and zero at the edges (Fig. 6.17).

128

PRINTED CIRCUIT BOARDS AND FLAT PLATES xxxxxxxxxxxxxxxxx

Free edge

Supported edge

Fixed edge Plate

Equation l

0

I

j, =

+)3 55 (i-

D p

h

I "

$ " l

xxxxxxxxxxxxxxxxxxxxxx

a

031'

b

FIGURE 6.15. Natural frequency equations for uniform plates.

The load intensity at any point can be expressed in a trigonometric form that satisfies the boundary conditions: q

?Tx

= q,, sin-sin-

a

7TY

b

The differential equation for the deflection of the circuit board then becomes [291

a4z

a4z

a4z qo

7Tx

7TY

- -sin -sin axj+2aX2aY'+iiYc-D a b

(6.43)

DYNAMIC STRESSES IN THE CIRCUITBOARD

Plate

Equation

f n

D 2 [p

= -x

D 2.45

Nib'

2.68 2 45 f n = -2 - -+-(,4 c r 2 b b "h + J x

Lzizi3 xxxxxxxxxxxxxxxxxxxxx

-+-+F)] 0.127 0.707 2.44

( u4

[, (

129

il I"'

xxxxxxxxxxxxxxxxxxxxxxx X

a

I '' \

xxxxxxxxxxxxxxxxxxxxxx

:v

xxxxxxxxxxxxxxxxxxxxxxx

:m

xxxxxxx

x x

x xx

FIGURE 6.16. Natural frequency equations for uniform plates.

FIGURE 6.17. Dynamic deflection mode of a circuit board.

130

PRINTED CIRCUIT BOARDS

A N D FLAT PLATES

The deflection of the circuit board at any point can also be represented by a trigonometric expression similar to the one shown in Eq. 6.1:

%-x

%-Y

Z = A sin ---sina

(6.44)

b

Performing the operations on Eq. 6.44 as required by Eq. 6.43,

a4z ax4 a4z --

-- -

a4z

%-Y b

a

a4

%-x

ITTT1

-

dY4

%-x

IT4

-Asin-sin-

%-Y

,Asin-sinb a

b

%-x

7i4

A sin -sin ax2ay2 a2b2 a =-

TY b

-

(6.45) (6.46) (6.47)

The deflection form factor A can be determined by substituting Eqs. 6.45-6.47 into Eq. 6.43: A=

40 1 2 n4D T + F [a a2b-

1

(6.48)

+g)

Substitute Eq. 6.48 into Eq. 6.44, and the deflection at any point on the circuit board is

%-x

q0 sin -sin Z=

ITY -

b

a

2

i l

The maximum deflection, Z,, will occur at the center of the board where X = a / 2 and Y = b / 2 . Substituting into the above equation, 40

z0= r4D

1

2

1

?+- a2b2' 2 )

(6.49)

[ a

Assuming that the circuit board acts like a single-degree-of-freedom system at its fundamental resonant mode, the maximum dynamic displacement can be approximated by Eq. 2.30 as follows:

DYNAMIC STRESSES IN THE CIRCUIT BOARD

131

Substitute this expression into Eq. 6.49 and solve for the maximum dynamic pressure intensity qo as follows: 2 4n

=

(6.50)

f,'

The plate stiffness factor D can be determined from the natural frequency equation for a uniform, simply supported rectangular plate, using Eq. 6.21: 4f,2 P 2

D=

1

(6.51)

Substitute Eq. 6.51 into Eq. 6.50, and the maximum dynamic load intensity at the center of the plate simplifies to the following expression: (6.52) The maximum dynamic bending moment in the rectangular circuit board will occur in the center. The bending moment must be greater for the section along the shorter plate dimension, b, where it has the value M y because the bending is along the Y axis. The maximum bending moment then becomes [291

My=

Sample Problem-Vibration

(6.53)

Stresses in a PCB

Determine the dynamic bending stresses in a 1.0-lb epoxy fiberglass PCB 7.0 X 8.0 X 0.062 in. thick when it is exposed to a 2-G peak sinusoidal vibration input and a transmissibility Q of 8.0 (see Fig. 6.7). Solution. Substitute into Eq. 6.53 for the dynamic bending moment, where

a = 8.0 in. (board length) b = 7.0 in. (board width) Gout= G,,Q = (2)(8.0) = 16 at center W = 1.0 lb (board weight) p = 0.12 (Poisson's ratio for G-10 epoxy fiberglass) 40 = WG,,,/ab = (1.0)(16)/(8.0)(7.0) = 0.286 Ib/in.2

132

PRINTED CIRCUIT BOARDS AND FLAT PLATES

Substituting into Eq. 6.53 for the bending moment, we have

M y=

= 0.496

in.lb/in.

(6.54)

The dynamic bending stress at the center of the plate can then be determined from the standard plate equation. Since PCBs normally have holes in them for the component lead wires, a stress-concentration factor should be used: (6.55) where M y = 0.496 in: lb/in. (see Eq. 6.54) h = 0.062 in. (circuit board thickness) K , = 3.0 (theoretical stress-concentration factor for a small hole in the circuit board) Substituting into the above equation,

s, =

6(3.0)(0.496) =

(0.062)

2320 lb/in.’

(6.56)

An examination of the fatigue S-N curve for G-10 epoxy fiberglass shows that the circuit board will never fail with a stress so low. 6.8

RIBS ON PRINTED CIRCUIT BOARDS

If the stiffness of a PCB is increased without a significant weight increase, the natural frequency will also increase and the deflection at the center of the board will decrease rapidly. A decrease in the circuit board deflection means a decrease in the electrical lead wire stresses and the circuit board stresses during vibration. One simple method for increasing the stiffness of a circuit board is to add ribs. If the ribs are made of thin steel, copper, or brass, they can be cemented or soldered to the copper cladding on the circuit board. This forms a very stiff section because these materials have a high modulus of elasticity. These ribs can be undercut between supports to allow the printed circuits to run under the ribs without causing electrical shorts. Consider the PCB shown in Fig. 6.7. If two steel ribs are soldered to the board in a symmetrical pattern along the length, the circuit board will appear as shown in Fig. 6.18.

133

RIBS ON PRINTED CIRCUIT BOARDS

Board supported on 4 sides

Y h=O062 epoxy

/

"if

n

a

and cemented to board

b=70

-

t

= 0040 rib (steel)

1

b/2 = : 5 0 = d -L

U, -

b/4 = 175

1

X

'

Connector

FIGURE 6.18. Stiffening ribs soldered to a circuit board.

If a 2-G peak sinusoidal vibration input is used to excite the circuit board, the four edges of the board will act as though they were simply supported. If a substantially higher G force is used, such as 10 G, the connector edge may act more like a free edge. Considering four simply supported edges, the natural frequency for the circuit board with ribs can be determined by making a slight modification in Eq. 6.21 to take account of the different stiffnesses along the X and Y axes. The natural frequency equation then takes the following form [291:

+3 ] l ' *

(6.57)

Here D, is the bending stiffness of the composite board along the X axis, and D , is the bending stiffness along the Y axis. Since the circuit board ribs are parallel to the X axis, they will increase the bending stiffness along the X axis but not along the Y axis. The torsional stiffness of the plate-and-rib combination is represented by DxY, If the ribs are in a symmetrical pattern on one side of the board, then the moment of inertia of one T section can be used to compute the bending stiffness along the X axis. A table (see Tables 6.2 and 6.3) is used to make it easier to compute the moment of inertia of the composite T section shown in Fig. 6.19.

z f

0062J

L-----35

-_I

Epoxy board

@

FIGURE 6.19. Dimensions for a rib section soldered to a circuit board.

134

PRINTED CIRCUIT BOARDS AND FLAT PLATES

TABLE 6.2

Item

Area

E X 10'

Z

A E x 10'

A E Z X 10'

I,, x 10-3 3.5(0.062)'

1

0.2170

2.0

0.031

0.434

0.0134

2

0.0152

29.0

0.252

0.441

0.1110

0.875

0.1244

= 0.070

12 0.040(0.380)3

Total

= 0.183

12

TABLE 6.3

2

EI,, x 103

C

2

0.140 5.310

0.111 0.110

Total

5.450

Item

AEC'

x 10'

~~~~~~~~~

1

0.0123

5.34 5.34

0.0121

10.68

The centroid of the T section is at

z=XCAAEEZ

0.1244 -

X

30'

0.875 X l o 6

= 0.142

in.

The bending stiffness of the T section is

C E I = EI, +AEc'

= 5.450

x 10'

+ 10.68 x l o 3 (6.58)

C E I = 16.13 x l o 3 lb.in.2

The bending stiffness of the circuit board along the X axis, with the ribs, is

D,=T where EI d

EI

(6 3 9 )

16.13 x l o 3 1 b . i n 2 (see Eq. 6.58) = b/2 = 3.5 in. (see Fig. 6.18) =

Substituting into the above equation, D,

16.13 X 10' =

3.5

= 4610

1b.h.

(6.60)

RIBS ON PRINTED CIRCUIT BOARDS

135

The bending stiffness along the Y axis will be approximately the same as the epoxy board:

Eh

D,= where E h

(6.61)

12(1 - p2)

=2X

l o 6 1b/in2 (G-10 epoxy fiberglass) = 0.062 in. (circuit board thickness) p = 0.12 (Poisson’s ratio)

Substituting into the above equation, (2 X 106)(0.062)3

D,=

12[ 1 - (0.12)?]

= 40.2

1b.in.

(6.62)

The torsional stiffness can be determined by considering a unit board width along with one rib (Fig. 6.19). The subscripts e and r refer to the epoxy board and to the rib, respectively:

D,,

= G, J ,

Gr +2d Jr

(6.63)

where G, = 0.90 X l o 6 1b/ine2 (shear modulus, epoxy fiberglass) J, = $h3 (unit torsional stiffness, epoxy fiberglass) in3 J, = 3(0.062>3= 79.3 X GI = 12 X l o 6 lb/in.2 (shear modulus, steel rib) J , = i L t 3 (torsional stiffness, steel rib) JI = $(0.38>(0.040>3= 8.10 x l o p 6 in.4 d = b/2 = 3.5 in. (rib spacing; see Fig. 6.18) Substituting into the above equation,

D,,=

(0.90 X 106)(79.3 X

D,,=

85.3 1b.in.

+ (12

X

106)(8.10 X 2( 3 S ) (6.64)

With the addition of the two steel ribs, the circuit board weight will increase to about 1.06 lb. The mass per unit area will then be mass P = --

area

p = 0.490 X

W

-

1.06 (386)(8.0)(7.0)

lb

in.^

gab

(6.65)

136

PRINTED CIRCUIT BOARDS AND FLAT PLATES

The natural frequency of the circuit board with the two ribs can be determined by substituting Eqs. 6.60, 6.62, 6.64, and 6.65 into Eq. 6.57: 4610

1

f,

=

1/2

4( 85.3)

251 HZ

(6.66)

It would appear from Eq. 6.66 that the natural frequency of the circuit board with two ribs will be about 251 Hz. However, since the center section of the circuit board, between the ribs, is only 0.062 in. thick, as shown in Fig. 6.18, it may be possible for this section to develop a resonance below 251 Hz. This can be checked by considering the center section of the circuit board, between the ribs, as a simply supported rectangular plate. Equation 6.21 can then be used to determine the natural frequency as follows:

-[-I 7T

f

"

=

2

D P

1

1

(2+ 2 )

(seeEq.6.21)

where a = 8.0 in. (board length) b = 3.5 in. (board width) D = 40.2 lb .in. p = 0.463 x lb . s 2 / i d Substituting into the above equation,

'= f,

7T

=

40.1 0.463 X l o p 4

142 HZ

1

1

)1'2[m m] +

(6.67)

Since the natural frequency of the unstiffened center section is only 142 Hz, a third rib may have to be added to raise the circuit board fundamental resonant frequency to 251 Hz. When the fundamental resonant frequency of the circuit board is increased to 251 Hz, the transmissibility of the circuit board will also increase. Considering the geometry and the mounting of the circuit board, if a low-input vibration-acceleration force with a 2-G peak is used, the transmissibility at resonance will probably be slightly greater than the square root of the natural frequency, as explained in Section 6.3. Vibration test data on many circuit boards indicate that the transmissibility is related to the natural

RIBS FASTENED TO CIRCUIT BOARDS WITH SCREWS

137

frequency as follows: Q = 1.2(fn)”’ Q = 1.2(251)”’

=

19.0

The dynamic deflection at the center of the circuit board with the ribs can be determined from Eq. 2.30 by approximating the circuit board as a single-degree-of-freedom system in its fundamental resonant mode:

6,=

9.8GinQ

f,’

where Gin= 2.0 (peak input acceleration) Q = 19.0 (transmissibility at resonance) f, = 251 Hz (natural frequency of board with ribs) Thus 6, =

(9.8) (2.0) (19.0) (251)*

= 0.00591

in.

(6.68)

Electronic component parts mounted on the circuit board will experience lower stresses in their electrical lead wires when the circuit board deflections are reduced. These stresses are directly proportional to the circuit board deflections at the fundamental resonant mode. This means that the deflections can be used in a direct ratio to determine the new stresses developed in the circuit board components when the ribs are added. 6.9 RIBS FASTENED TO CIRCUIT BOARDS WITH SCREWS

Extensive vibration tests on bolted assemblies show that bolted joints will experience a substantial amount of relative motion during major structural resonances. This relative motion may add damping to the system, which reduces the transmissibility at resonance. It also reduces the stiffness of the structure, and this tends to reduce the natural frequency. The ability of two bolted interfaces to remain rigid is a function of the stiffness of the joint, the number of screws, the screw size, the screw spacing, the screw torque, the interface conditions, the vibration force, and the vibration frequency. In order to correlate test data with a theoretical analysis, a bolted efficiency factor is very convenient. This efficiency factor will range from 0%, for a joint that has no physical connection, to loo%, for a joint that is welded. Evaluating the types of bolted joints normally found in electronic

138

PRINTED CIRCUIT BOARDS AND FLAT PLATES

Z 7 k 0040 original thickness

' 2

-~+0010

equivalent thickness

Bolted steel'

__

._._ ~.

I

b

y

r Bolted ~ interface

Y

4

L = 0 38 -T

h=0062

subassemblies such as brackets, covers, ribs, and stiffeners, which must transfer an alternating load, the typical efficiency of a bolted joint used for airborne electronic structures appears to be about 25%. It may be as high as 50% for small sheet-metal panels with large, closely spaced screws, and as low as 10% for large sheet-metal panels held down with quarter-turn quickdisconnect fasteners. The natural frequency of a PCB, with stiffening ribs bolted to the board, can be determined by using a bolted efficiency factor. For example, if the ribs on the circuit board shown in Fig. 6.19 were bolted to the board, instead of soldered, the resulting T section through one rib would be similar to the section shown in Fig. 6.19, except for the efficiency of the bolted interface. The 25% efficiency factor for this bolted joint will be analyzed by reducing the effective thickness of the rib to 25% of its true thickness (Fig. 6.20). The moment of inertia of the composite T section, with the bolted rib, will be determined with the use of Tables 6.4 and 6.5. Note that the equivalent thickness of the rib has been reduced from 0.040 to 0.010 in. to account for the 25% bolted efficiency factor. The centroid of the T section is at '=

CAEZ 0.O411X1O6 CAE 0.544 X 106

=

0.0756 in

TABLE 6.4 ~~

Item

Area

E X lo6

Z

1

0.2170

2.0

0.031

0.434

0.0134

2

0.0038

29.0

0.252

0.110

0.0277

0.544

0.0411

Total

A E X l o 6 A E Z X lo6

I" x lo-' 3.5(0.062)3 12

(0.010)(0.380)3 12

= 0.070

= 0.046

RIBS FASTENED TO CIRCUIT BOARDS WITH SCREWS

139

TABLE 6.5

EI, x 103

Item

C

C2

A E C x~ io3

0.0446 0.1764

0.00199 0.0311

0.864 3.420

~~

1 2

0.140 1.335

Total

1.475

4.284

The bending stiffness of the T section is

LEI

= El,

XEI

= 5.76

+ A E c 2 = 1.475 X l o 3

+ 4.28 X l o 3

x lo3 lb.in.2

(6.69)

The bending stiffness of the circuit board along the X axis, with the bolted rib, is

where EI d

= 5.76 X

lo3 1 b . h 2 (see Eq. 6.69) in. (see Fig. 6.18)

= b/2 = 3.5

Substituting into the above equation,

D,

5.76 x 103 =

3.5

=

1645 1b.in.

(6.70)

The bending stiffness along the Y axis will be approximately the same as that of the epoxy board, which is shown by Eq. 6.62 to be DY=40.2lb.in.

(6.71)

The torsional stiffness of the bolted T section can be determined by considering a unit board width, along with one rib, as shown in Fig. 6.20. The subscripts e and r refer to the epoxy board and to the rib, respectively: Gr Jr D,yy=G,J,+ - (see Eq. 6.63) 2d where G, = 0.90 X l o 6 lb/in.’ (shear modulus, epoxy fiberglass) J, = +A3 = +(0.06213= 79.3 X in3 G, = 12 X lo6 1b/in2 (shear modulus, steel rib) Jr = iLt3 3 = f(0.38)(0.010)3 = 0.126 X in.4 d = b/2 = 3.5 in. (rib spacing; Fig. 6.18)

140

PRINTED CIRCUIT BOARDS AND FLAT PLATES

Substituting into the above equation,

D,,

=

+

(0.90 X 106)(79.3 X

(12 X 106)(0.126 X

2( 3 S)

DX,=71.6lb.in.

(6.72)

If the circuit board weight does not change, the mass per unit area will be the same as that shown in Eq. 6.65. The natural frequency of the circuit board with the bolted ribs can then be determined by substituting Eqs. 6.70-6.72 and 6.65 into Eq. 6.57:

f=Z[ ”

f,

2 0.490 =

[-+1645

1 X

lo-‘

(8.0)4

“il

1/2

4( 71.6) (8.0)2(7.0)2

160 HZ

+

(7.0)‘

(6.73)

Since the center section of the circuit board is not stiffened between the ribs, as shown in Fig. 6.18, it may be possible for this section to develop a resonance below 160 Hz. The natural frequency of this section was previously checked in the approximation of a simply supported rectangular plate. These results are shown in Eq. 6.67, which shows a natural frequency of 142 Hz. If a natural frequency of 160 Hz is required, then a third rib may have to be bolted to the circuit board. The deflection at the center of the circuit board, with the bolted ribs, can again be determined from Eq. 2.30 by approximating the board as a singledegree-of-freedom system at its resonance. The transmissibility developed with the bolted ribs should be slightly less than with the soldered ribs because of the additional damping at the rib interfaces and the bolt interfaces. Vibration test data on several groups of circuit boards using a 2-G peak sinusoidal vibration input indicate the transmissibility was approximately related to the natural frequency as follows:

Q

=

( 160)1’2 = 12.7

Substituting into Eq. 2.30. the single-amplitude displacement with the bolted ribs becomes 6, = 6,

9.8G,,Q - 9.8(2.0)(12.7) f ;1 (160)‘

= 0.00972

in.

(6.74)

Circuit board displacements are important because they can be used to estimate the fatigue life, or the change in the fatigue life, of the electronic components mounted on the circuit board.

PROPER USE OF RIBS TO STIFFEN PLATES AND CIRCUIT BOARDS

141

Y

t

I

Circuit

board

h-rrmrmrrrml FIGURE 6.21. A circuit board with ribs in two directions.

6.10 PRINTED CIRCUIT BOARDS WITH RIBS IN TWO DIRECTIONS

If two sets of ribs are fastened to one side of a circuit board in a symmetrical pattern, a T section will have to be analyzed to determine the bending stiffness along the Y axis as well as the X axis (Fig. 6.21). Again, be sure to check each plate section between the ribs to make sure each section has a natural frequency greater than the full-plate natural frequency including the ribs.

6.11 PROPER USE OF RIBS TO STIFFEN PLATES AND CIRCUIT BOARDS

Ribs will not increase the stiffness of a plate or a circuit board if they are not properly used. In order for a rib to be effective, it must carry a load. Since all loads must eventually be transferred to the supports, the rib is most effective when it carries a load directly to the supports. For example, consider a flat square plate simply supported on two opposite edges and free on two opposite edges. If ribs are added to the plate so that they join the two free edges, the ribs will not carry the load directly to the supports and these ribs will not be very effective in stiffening the plate. If the ribs are added to the plate so that they join the two supported edges (Fig. 6.22), then these ribs will carry the load directly into the supports and they will be very effective in stiffening the plate. If ribs cannot be carried directly into the supports, then a secondary member should be provided to carry the load to the supports to make the

142

PRINTED CIRCUIT BOARDS AND FLAT PLATES Poor

Good Free

r-----l

Free

Free

FIGURE 6.22. Adding ribs to stiffen a flat plate effectively.

Poor

Good

Free

Free

FIGURE 6.23. Adding an additional rib to stiffen a flat plate.

ribs more effective. For example, consider a flat square plate simply supported on three sides and free on a fourth side. If ribs must be added to the plate so they end at the free edge, then an additional rib should be placed across the free edge to act as a secondary member, which will carry the load to the supports (Fig. 6.23).

6.12 QUICK WAY TO ESTIMATE THE REQUIRED RIB SPACING FOR CIRCUIT BOARDS

A quick method for estimating the required spacing for stiffening ribs on flat plates was shown by Bruhn [13]. H e uses stiffening rib spacing criteria based on the thickness of the plate. The stiffness (or modulus of elasticity) of the rib material should be at least equal to or greater than the stiffness of the plate material. The height and thickness of the ribs and the attachment to the plate must add enough local stiffness to the assembly to substantially reduce the displacements when the plate is subjected to a transverse load. The effective plate width B that the rib can reinforce is shown in Fig. 6.24 as 30t, where t is the plate thickness. The required rib height for a metal rib will vary from about 0.25 in. for a small PCB about 4 in. square, to a metal rib height of about 0.50 in. for a large PCB about 10 in. square.

143

QUICK WAY TO ESTIMATE THE REQUIRED RIB SPACING FOR CIRCUIT BOARDS

Rib,

\n

,Bonded

I

I

if

I

I-

B----l

B = 30t

t

FIGURE 6.24. Effective width of a flat uniform plate with a stiffening rib.

An examination of the sample problem rib pattern in Figs. 6.18 and 6.19 shows the rib is centered on a PCB section that is 3.5 in. wide. When the Bruhn criteria are used for the spacing, the allowable PCB plate width should be limited to the value shown below:

B

= 30t =

(30)(0.062) = 1.86 in.

(6.75)

The obvious conclusion from this quick approximation is that the rib spacing in Figs. 6.18 and 6.19 is inadequate. More stiffening ribs will be required to reach the 251-Hz natural frequency shown in Eq. 6.66. The same conclusion was reached after a great deal of work in the paragraph following Eq. 6.67, and in the paragraph following Eq. 6.73. If a third rib is added to the PCB as shown in Fig. 6.25, the effective plate width associated with each rib will be 2.33 in. This is still greater than the maximum allowable plate width of 1.86 in. shown in Eq. 6.75. This means that even with three ribs on the PCB, the natural frequency will still be

T 7.0 I

1 -

FIGURE 6.25. Suggested stiffening rib placement to achieve effective circuit board frequency.

144

PRINTED CIRCUIT BOARDS AND FLAT PLATES

slightly under the desired 251-Hz level. If the thickness of the PCB is increased to about 0.0776 in., the effective allowable plate width will be as shown below:

B

= 30t =

( 3 0 ) ( 0 . 0 7 7 6 ) = 2.33 in.

(6.76)

The desired plate width of 2.33 in. now matches the plate width shown in Fig. 6.25 for three ribs.

6.13 NATURAL FREQUENCIES FOR DIFFERENT PCB SHAPES WITH DIFFERENT SUPPORTS

Any time the natural frequency of a PCB is excited, problems can develop. Electrical lead wires, solder joints, and connector pins can crack, closely spaced PCBs can impact against each other and cause short circuits, and screws can become loose. Most of the PCBs made today are rectangular in shape. Some of these PCBs are plug-in types and other are bolted to some type of support. Natural frequencies for uniform rectangular plates and PCBs with various supports are readily found in technical literature. However, there has been a large increase in the use of electronics in unusual new applications, where volume and size are limited, so new shapes with new support methods are often required. There is very little data available for finding natural frequencies for odd-shaped plates and PCBs with different edge and point supports. Computers are available with various FEM software programs that can calculate the natural frequencies and mode shapes for plates with different shapes and supports. When access to a computer is available, when the operator is familiar with the FEM software, and when there is time available, odd-shaped plates and PCBs are easy to evaluate. When time is short, or when there are no computers available, the analysis methods shown here can be very convenient for obtaining the natural frequencies of some odd-shaped PCBs. The analysis methods shown here can be used for a sanity check of FEM analyses of odd-shaped plates. When the computer analyst is tired, especially right after lunch or late in the afternoon, it is very easy to make data entry errors. One bad data entry error on a computer can ruin weeks and months of hard work. In addition, it is often very difficult to locate and correct the bad data entry point. Natural frequencies for a variety of PCB shapes with different supports were calculated using a computer. The results for the first three natural frequencies are shown in Figs. 6.26-6.29 for different shapes. The first number is for the fundamental or the lowest natural frequency. The next two numbers are for the higher harmonics. Most of the vibration and shock damage will usually occur at the lowest natural frequency where the dynamic displacements are the greatest. The higher harmonics usually have smaller

NATURAL FREQUENCIES FOR DIFFERENT PCB SHAPES

145

Rectangles

[XI

Case 1

(1) 450.60 (2) 647.31 (3) 887.04

Hz

Case 5

Case 2

Case 3

243.10 Hz 452.94 731.52

72.21 Hz 123.16 208.36

1

Case 4

1

111.27 Hz 136.32 252.29

I

1

Case 6

iU (1) 72.12 (2) 200.80 (3) 223.21

71.78 174.72 210.68

Case 9

Case 10

174.72 210.68 243.48

217.88 270.59 278.93 I

a3 (1) 270.59 (2) 278.93 (3) 281.00

42.8 1 164.27 275.81

Case 13

Case 14

, , 107.72 225.38 343.32

Caai5

(1) 217.30 (2) 255.83 (3) 423.79

142.20 212.87 305.91

160.49 166.69 176.11

164.74 217.30 342.75

I

[ a i 7

160.49 176.11 178.21

FIGURE 6.26. Natural frequencies for flat uniform rectangular plates with different types of supports.

displacements so they produce less damage. These frequencies were tabulated for easy reference. The frequencies are for 0.10-in. thick epoxy fiberglass PCBs with a 24-in.’ area and a density of 0.208 Ib/ine3. This resulted in slightly different dimensions for the different shapes. The density was based on the total weight of a fully assembled PCB. This included all of the electronic component parts as well as the weight of the bare epoxy circuit board. The density was based on the total weight divided by the volume of

146

PRINTED CIRCUIT

BOARDS AND FLAT PLATES Hexagons

Case 1

( 1 ) 379.62 Hz ( 2 ) 718.78 (3) 8 0 6 . 3 1

Case 5

00 G Case 2

Case 3

Case 4

212.42 Hz 513.24 534.05

72.52 H z 89.17 235.97

117.56 Hz 134.47 272.42

Case 6

Case 7

Case 8

00 (1) 61.36 (2) 166.94 ( 3 ) 184.45

97.55 142.27 233.46

142.27 169.85 36 1.43

147.86 184.45 216.28

Case 9

Case 10

Case 11

Case 12

170.27 210.54 350.92

135.45 340.55 36 1.43

@ ( 1 ) 184.45 (2) 216.45 (3) 260.60

0 55.11

208.86 229.82

Case 1 3

( 1 ) 340.55 (2)361.43 ( 3 )3 9 3 . 3 1

189.24 265.18 265.98

@ 184.45 194.95 214.54

184.45 214.54 233.23

FIGURE 6.27. Natural frequencies for flat uniform hexagonal plates with different types of supports.

the epoxy board by itself. The PCB shapes are rectangles, hexagons, triangles, and circles with 16 different support arrangements for each shape shown in Figs. 6.26-6.29. The natural frequencies of a PCB that matches the dimensions, material, and support method of one of the tabulated PCBs can be obtained directly from the tabulated natural frequencies shown for each geometry. The approximate natural frequencies for PCBs with similar shapes and similar supports, but with different sizes and different materials, can be obtained with some simple conversion calculations using the equation shown below.

NATURAL FREQUENCIES FOR DIFFERENT PCB SHAPES

147

Triangles

I

Case 3

I

Case 1

Case 2

Case 4

(1) 443.23 Hz (2) 858.95 (3) 971.22

270.02 Hz 657.43 759.54

44.01 Hz 87.93 99.66

75.25 Hz 139.12 150.74

Case 5

Case 6

Case 7

Case 8

(1) 88.10 (2) 89.97 (3) 132.27

88.70 118.79 241.04

88.74 120.78 259.15

88.13 89.97 134.44

10

Case 11

Case 12

( 1 ) 39.31 ( 2 ) 98.49 (3) 126.02

77.21 290.66 335.03

263.63 306.07 456.64

226.94 237.77 261.01

Case 13

Case 14

Case 1 5

Case 16

(1) 95.48 (2)127.65 13) 227.91

227.54 245.94 326.47

AAAA Case 9

Case

A AIA1A I

I

115.57 122.67 199.19

122.50 141.72 210.29

--1 1

1

~~

Free edge

Supported edge

Fixed edge

Point support ~

FIGURE 6.28. Natural frequencies for flat uniform triangular plates with different types of supports.

148

PRINTED CIRCUIT BOARDS AND FLAT PLATES

Circles Case 1

Case 2

@ (1) 384.96Hz (2)775.09 (3)775.40 Case

5

(1) 59.55

(2)183.34 ( 3 ) 196.05 Case

9

4

Case

@ 177.02Hz 505.87 506.06 Case

6

119.73 166.89 249.89 Case

10

81.15Hz 94.16 258.04 Case

124.97Hz 148.03 306.04

7

166.89 182.12 441.24 Case

8

Case

156.94 196.05 255.55

11

Case

12

~

(1) 196.05 (2) 255.56 (3)255.91 Case

67.04 219.59 234.61

13

172.81 441.23 442.49

162.43 234.61 342.68 Case

15

Case

@olf '

(1) 441.23 (2)442.49 (3)458.35

Free edge

210.37 290.44 335.79

16

!$-

196.05 216.08 252.96

196.05 252.96 253.17

Fixed edge

Point s u p p o r t

............//////////// -I

Supported edge

FIGURE 6.29. Natural frequencies for flat uniform circular plates with different types of supports.

NATURAL FREQUENCIES FOR DIFFERENT PCB SHAPES

149

The subscript 1 refers to the tabulated values and geometry shown for the four different PCB shapes. The subscript 2 refers to the properties of a new specimen plate or PCB.

(6.77)

where f = natural frequency (Hz) a = width (in.) b = height (in.) E = modulus of elasticity (lb/in.,> h = thickness (in.) y = density (lb/in.3) p = Poisson’s ratio (dimensionless) Sample Problem-Natural Three Point Supports

Frequency of a Triangular PCB with

Find the natural frequency of a polyimide glass triangular PCB that is 8.5 in. wide, 7.5 in. high, and 0.090 in. thick, with a total weight of 0.70 Ib. It is fastened to a support with three small screws located at the center of the three sides, near the edges, similar to Fig. 6.28 Case 5. The following information is required to obtain a solution: a, = 7.44 in. wide

a, = 8.5 in.

b , = 6.45 in. high E , = 2.0 X l o 6 lb/in.’ modulus h , = 0.10 in. thick p1 = 0.12 dimensionless

in. E , = 3.0 X l o 6 lb/in.’ h , = 0.090 in. p2= 0.12 dimensionless

y , = 0.208 1b/ins3

f,= 88.10 Hz

b,

= 7.5

0.70 lb

yz

(8.5)(7.5)(0.090)(0.50) in.

0.244 1 b / h 3

f, = ? (3.0 X 106)(0.090)2(0.208) (2.0 X 106)(0.10)2(0.244)

= 67.5

Hz (6.78)

I CHAPTER 7

Octave Rule, Snubbing, and Damping to Increase the PCB Fatigue Life

7.1 DYNAMIC COUPLING BETWEEN THE PCBs AND THEIR SUPPORT STRUCTURES

Vibration and shock failures in PCBs can often be avoided by understanding how energy is transferred directly to the PCBs from their support structures. The support structures are used to provide easy access to the PCBs for troubleshooting, repairs, or replacement. Since the PCBs are attached to the support structure, the dynamic response of the support structure becomes the input to the PCBs. In any type of dynamic vibration or shock environment, the natural frequencies of the support structure and the PCBs can become excited. If the natural frequencies of the PCBs are close to the natural frequency of the support structure, their transmissibility Q values can couple, which means that their Q values will multiply. This will force the PCB response acceleration G levels to be magnified substantially, resulting in rapid PCB failures. For example, when the chassis has an input acceleration level of 10 G and a transmissibility Q of 10, the chassis will experience 10 X 10 or 100 G. If the PCB also has a transmissibility Q of 10 at the same frequency, the PCB will respond at a level of 100 X 10 or 1000 G. Acceleration levels this high can result in very rapid fatigue failures. The natural frequencies of the chassis and the various PCBs must be well separated to prevent severe coupling. An octave apart represents a good separation. Octave means to double. The natural frequencies of the chassis and the PCBs should be separated by a ratio of more than 2 : 1. The properties, locations, and arrangements of the various PCBs and the properties of the support structure will affect the responses and the fatigue life of the PCBs. There are also feedback effects, where the dynamic feedback from the PCBs to the support structure can produce acceleration G levels high enough to damage the support structure, in poorly designed electronic systems. Electronic systems with various PCBs enclosed within some type of chassis, form a complex dynamic structure. This usually requires the use of a 150

DYNAMIC COUPLING BETWEEN PCBs AND SUPPORT STRUCTURES

-,

151

PCB

Chassis mass 1

FIGURE 7.1. Chassis and PCB modeled as a two-degree-of-freedom lumped mass dynamic

system.

high-speed computer with the proper software to perform a dynamic frequency response analysis and to obtain the coupling properties between the chassis and the PCBs. This can be a very expensive and time-consuming analysis. This study has already been done using a computer to perform a parametric analysis of many two-degree-of-freedom systems with springs, masses, and dampers. The chassis is represented as mass 1, since it receives the energy first, so it is the first degree of freedom. The PCB is represented as mass 2, since it receives its energy from the chassis, so it is the second degree of freedom. This is shown in Fig. 7.1. The spring rates were changed to change the natural frequencies, the masses were changed to change the weight ratios between the chassis and the PCBs, and the dampers were changed to change transmissibility Q values of the chassis and the PCBs. The energy transferred from the chassis to the PCB for different dynamic combinations was used to find the acceleration G levels acting on the PCB. The information was arranged in the form of curves with several different PCB and chassis weight ratios, shown in Fig. 7.2-7.5, to make it easier to find the dynamic acceleration G response of different types of PCBs mounted in different types of chassis. Vibration test data show that most of the damage in electronic assemblies occurs at the fundamental natural frequency of the PCBs. These data also show that the response of the chassis is important since this represents the input to the PCB. The manner in which the chassis influences the PCB response determines the coupled acceleration G level acting on the PCB. The coupled response of the PCB is shown as Q,. The uncoupled response of the chassis acting by itself is shown as ql. The uncoupled response of the PCB acting by itself is shown as q2. The ratio R is used to show the Q2/q1 values that were obtained from the computer parametric studies. The fundamental natural frequency for the chassis is shown as f l and the fundamental natural frequency of the PCB is shown as f i . An examination of Fig. 7.2 shows that when the natural frequency of the PCB is about the same as the natural frequency of the chassis, the frequency ratio f2/fi is about 1.0. The

152

OCTAVE RULE TO INCREASE THE PCB FATIGUE LIFE

-

Dangerous area

-

T

E=4.0

/k

\\

\

7 %=0.05

l.-==u.3u

i

1

0 0

0.5

2

1

3

Ratio = f2/fl

FIGURE 7.2. Dynamic coupling effects between the chassis and the P C B for various ratios of transmissibility when the weight of the chassis is 20 times the P C B weight.

0

1

2

3

Ratio = f2/fi

FIGURE 7.3. Dynamic coupling effects between the chassis and the P C B for various ratios of transmissibility when the weight of the chassis is four times the P C B weight.

DYNAMIC COUPLING BETWEEN PCBs AND SUPPORT STRUCTURES

153

FIGURE 7.4. Dynamic coupling effects between the chassis and the PCB for various ratios of transmissibility when the weight of the chassis is two times the PCB weight.

FIGURE 7.5. Dynamic coupling effects between the chassis and the PCB for various ratios of transmissibility when the weight of the PCB is two times the chassis weight.

154

OCTAVE RULE TO INCREASE THE PCB FATIGUE LIFE

weight ratio of the PCB to the chassis is shown as W,/W,. When the weight of one of the PCBs is less than one-tenth the weight of the chassis, the weight ratio will be less than 0.10. Under these conditions the dynamic coupling between the PCB and the chassis, shown as R = Q 2 / q 1 ,can reach a value that is close to about 5.0. This means that the acceleration G levels acting on the PCB will be about 5 times greater than the expected acceleration G levels acting on the chassis. These conditions can produce high forces and stresses on the PCB, which can result in very rapid PCB fatigue failures. This condition can be avoided by following the octave rule. 7.2

EFFECTS OF LOOSE EDGE GUIDES ON PLUG-IN TYPE PCBS

The computer parametric studies for the octave rule were based on plug-in types PCBs that have three or four restrained edges that are simply supported (hinged) or clamped (fixed). Vibration test data show that a typical plug-in type of electrical connector on one edge of the PCB will act like a support. If there are any two unsupported PCB edges that are free to translate instead of being restrained in some way, the increased displacement amplitude will increase the period of vibration. This condition can occur, for example, when loose PCB edge guides are used. This tends to decrease the effective natural frequency of the PCB. This condition often brings the PCB natural frequency closer to the chassis natural frequency, which increases the dynamic coupling between the chassis and the PCB. This is a dangerous condition that can sharply increase the acceleration G levels produced in the PCB, resulting in more rapid PCB fatigue failures. The purpose of the octave rule is to reduce the dynamic coupling effects between the chassis and the PCBs, to improve the fatigue life of the PCBs in severe vibration and shock environments. Loose PCB edge guides will often result in increased coupling between the PCB and the chassis, which is not desirable. Loose PCB edge guides should be avoided to improve the reliability of the PCBs in dynamic conditions. DESCRIPTION OF DYNAMIC COMPUTER STUDY FOR THE OCTAVE RULE 7.3

Dynamic response data generated by the computer for the chassis and the PCB were translated into a graphic form with curves, to cover a wide range of parameters in order to simplify analysis methods that are normally very complex. The purpose of this technique was to permit the use of hand calculations for finding quick. but approximate, dynamic acceleration G levels acting on the PCB. These analysis methods have also been used as a sanity check for finite element modeling (FEM) calculations performed on the computer. When there are large differences in the results of the computer analysis and the hand calculations, something is usually wrong, so

THE REVERSE OCTAVE RULE MUST HAVE LIGHTWEIGHT PCBs

155

further investigations should be made. It is very easy to make data entry errors on a large computer program. Data input errors are very easy to make, but very hard to find in a large program. One bad data entry can ruin months of hard work. This is called GIGO, for garbage in garbage out, by computer users. 7.4

THE FORWARD OCTAVE RULE ALWAYS WORKS

Natural frequencies for the chassis and the PCBs should be planned in advance to take advantage of the octave rule. Once the design has been completed and orders have been placed, it is usually too late to try to utilize the octave rule. In the analysis phase the natural frequency for the chassis, with the weights of the PCBs included, must be obtained. Then the natural frequency of each PCB, assumed to be acting by itself outside the chassis, must be obtained. The resulting natural frequencies of the chassis and the various PCBs should be separated by more than one octave to avoid severe dynamic coupling. For example, when the chassis considered as mass 1 has a natural frequency of 100 Hz, then each PCB considered as mass 2 should have a natural frequency greater than 2 X 100 or greater than 200 Hz. This is called the forward octave rule. The forward term is used to describe the direction of the load path. The chassis, mass 1, gets the dynamic load first and transfers the load in the forward direction to the PCBs, mass 2. An examination of Figs. 7.2-7.5 shows that the dynamic coupling ratio R between the PCBs and the chassis is always sharply reduced when the forward octave rule is used, where the frequency ratio f2/fi is greater than 2.0. This means that the acceleration G levels transferred from the chassis to the PCBs are also sharply reduced. This reduction increases the fatigue life of the PCBs in severe dynamic environments.

7.5 THE REVERSE OCTAVE RULE MUST HAVE LIGHTWEIGHT PCBs

The reverse octave rule is used to describe the load path in the opposite direction from the PCB (mass 2) back to the chassis (mass 1). For the reverse octave rule to work the chassis should have a natural frequency that is more than two times greater than the natural frequency of the PCB. When the chassis has a natural frequency of 200 Hz, then the PCB should have a natural frequency less than 100 Hz. An examination of Fig. 7.2 reveals that a frequency ratio f2/fl of 0.5 or less represents the reverse octave rule. This shows the ratio R is approximately 1.0. This means that the acceleration level acting on the PCB will be about the same as the acceleration level acting on the chassis. This is good for vibration and shock, because the acceleration G levels transferred to the PCB from the chassis are low. Remember, these conditions are for a very small PCB weight, where the weight ratio W,/W, is

156

OCTAVE RULE TO INCREASE THE PCB FATIGUE LIFE

typically less than about 0.10. This covers the vast majority of the electronic equipment produced. Care must be used when the PCB weight ratio is increased to a value of 0.25, as shown in Fig. 7.3. Then the reverse octave rule, with a frequency ratio of f 2 / f l of 0.5, shows an R value of about 4.5. This means that the acceleration level acting on the PCB will be 4.5 times greater than the acceleration level acting on the chassis. The reverse octave rule in this case is a very dangerous condition because it can lead to high PCB acceleration G levels with a reduced PCB fatigue life. Some companies in the automobile industry tend to use a single PCB clamped to a sheet metal tray, where the PCB weight may be half the weight of the tray, This results in a weight ratio W,/W, of 0.50. An examination of Fig. 7.4 shows that if the reverse octave rule is used with the frequency ratio of 0.50, the coupling ratio R is about 4.0. The PCB acceleration G level under these conditions can be about 4 times greater than the G levels developed in the supporting tray. This condition may result in a reduced PCB fatigue life. An examination of Fig. 7.4 for the forward octave rule, with a frequency ratio of 2 or more, shows a coupling R value of about 1.2. This means that acceleration G levels acting on the PCB will be about 1.2 times greater than the acceleration G levels acting on the chassis. This shows that the forward octave rule will always work well, even for conditions that do not work well for the reverse octave rule. Sample Problem-Vibration

Problems with Relays Mounted on PCBs

Sinusoidal vibration tests were run on a small electronic chassis that contained several relays, with an acceleration rating of 75 G peak, mounted at the center of a plug-in PCB. During a 5 G peak qualification test, using a sinusoidal vibration sweep from 20 to 2000 Hz, the relays chattered, resulting in a test failure. An examination of the PCB by itself showed it had a weight of 2.5 lb, a natural frequency of 81 Hz, and a transmissibility Q of 9.0. An examination of the chassis by itself showed it had a weight of 10.0 lb, a natural frequency of 120 Hz, and a transmissibility Q of 3.0. The questions here are why did the relays chatter, and what can be done to the system so it will pass the test? Solution The Curves in Fig. 7.3 Can Be Used to Solve This Problem. Several different relations must be obtained to solve this problem as shown below.

fi/fl W,/W,

=

=

81/120 = 0.675 (uncoupled natural frequencies of PCB to chassis)

(7.1)

2.5/10

(7.2)

=

0.25 (weight ratio of PCB to chassis)

q 2 / q 1= 9/3 = 3.0 (uncoupled transmissibility ratio of PCB to chassis)

(7.3)

PROPOSED CORRECTIVE ACTION FOR RELAYS

G,

= Ginql

G 2 = G,Q, R

= Q2/q1

157

(uncoupled acceleration response of chassis)

(7.4)

(coupled acceleration response of PCB)

(7.5)

(ratio of the coupled transmissibility of PCB Q2 to uncopuled transmissibility of chassis q11

(7.6)

(coupled transmissiblility of the PCB) (rewriting Eq. 7.6) (7.7)

Q2 = Rq,

Substitute Eqs. 7.4 and 7.7 into Eq. 7.5 to obtain the coupled response of the PCB: G,

=

(Gi,q,)( Rq,)

= RGin(q , ) ,

(coupled response of the PCB) (7.8)

An examination of Fig. 7.3 shows there is no curve with a ratio q2/ql = 3, so a curve with this ratio must be generated by following the profiles between uncoupled transmissibility ratios 2 and 4. When this new curve is added with a dashed line, using the frequency ratio of 0.675 from Eq. 7.1, it results in the coupled transmissibility ratio R from the chassis to the PCB as shown below:

Substitute Eq. 7.9 into Eq. 7.8 along with an input Gin level of 5.0 and a chassis uncoupled transmissibility q , of 3.0 to obtain the coupled acceleration G, response of the PCB as shown below:

G,

=

(3.6)(5.0)(3.0),

=

162-G peak response of PCB

(7.10)

The 162-G peak response of the PCB far exceeds the 75-G peak rating for the relays, so the relays are not compatible with the PCB design and the environment. Some type of corrective action must be taken to allow the system to pass the qualification test. 7.6 PROPOSED CORRECTIVE ACTION FOR RELAYS

1. Mount the relays away from the center of the PCB, where acceleration levels are the highest. Mount the relays at the edges of the PCB where the acceleration levels are the lowest. This will require new PCB layouts with new PCB fabrication and assembly costs and delayed delivery schedules. 2. Make a search of the various relay manufacturers to see if they may have another relay off the shelf, with the same form, fit, and function, that can meet the requirements. 3. Examine the PCB layout to see if it is possible to add stiffening ribs to increase the natural frequency to a value that is two times the chassis natural

158

OCTAVE RULE TO INCREASE THE PCB FATIGUE LIFE

frequency. This would result in a new PCB natural frequency of 2 X 120 = 240 Hz, which would then follow the forward octave rule. Increasing the PCB natural frequency will also increase the PCB Q value. 4. The reverse octave rule with a PCB natural frequency of 120/2 = 60 Hz will not work here because the coupling ratio R for the PCB would be about 3.3. An examination of Eq. 7.10 shows the PCB acceleration response would be about 148 G peak, which is still well above the 75-G peak relay rating. 5. Adding snubbers to the PCB will reduce the PCB dynamic displacement and stress, but that is not the problem here. The problem is high acceleration levels. Snubbers can cause impacting shocks when the snubbers strike each other, which can produce high shock acceleration pulses. This can make the relays chatter and fail. 6. Damping is often recommended. Increased damping will decrease the transmissibility Q value during vibration. The best damping is constrained layer damping. This consists of alternate layers of thin high-energy dissipating viscoelastic strips separated by thin strips of aluminum foil. Damping works well on structures with low natural frequencies, below about 50 Hz. The dynamic displacements are then large enough to produce large relative shear displacements that convert kinetic energy into heat. When the natural frequencies are above about 100 Hz, the relative displacements are not large enough to provide significant damping. The amount of room required by the constrained layer dampers can usually be better utilized by employing stiffening ribs to increase the PCB natural frequency. The best option is corrective action item 3 above, adding stiffening ribs to the PCB to increase the natural frequency to 240 Hz. This may be difficult to do on a PCB that is fully populated so it may require a new design. This change will follow the forward octave rule. The new information required is shown below. =

f2/f,

g,

ql/q,

R

=

=

=

240/120

=

2.0 (frequency ratio of PCB/chassis)

(7.11)

= 15.5 (approximate uncoupled transmissibility of PCB)

(7.12)

15.5/3 = 5.16 (uncoupled transmissibility ratio of PCB/chassis)

(7.13)

1.4 (value of Q2/g1 estimated from Fig. 7.3)

(7.14)

Substitute Eq. 7.14 into Eq. 7.8 to find the coupled acceleration response of the PCB: G,

=

( 1.4) (5 .O) ( 3 ) 2 = 63 .O-G peak PCB response

(7.15)

USING SNUBBERS TO REDUCE PCB DISPLACEMENTS AND STRESSES

159

The peak response acceleration level of 63.0 G for the modified PCB is now well below the relay rating of 75 G peak, so the new PCB design should be acceptable. 7.7 USING SNUBBERS TO REDUCE PCB DISPLACEMENTS AND STRESSES

Snubbers can often be used to make PCBs work smarter in severe vibration and shock environments. Sometimes electronic equipment designers pay no attention to the octave rule so their PCBs have a high failure rate when they are exposed to vibration. Once the design is in full production and equipment is failing in the field, it can be very difficult to find a cost-effective fix. This is where snubbers can often save the day [431. High procurement costs for military electronic equipment have forced governments to go to the best available commercial electronic equipment manufacturers for many of the new military programs. Sometimes the commercial equipment lacks the ruggedness necessary for military programs, so the ruggedness must be increased, but at a reasonable cost. Again snubbers can often save the day. Snubbers are very effective for reducing problems in PCBs associated with large dynamic displacements. This condition can cause high lead wire and solder joint stresses, cracking of components, short circuits due to impacting between adjacent PCBs, and broken pins on electrical connectors. Snubbers, sometimes called bumpers, are small devices that can be attached to adjacent PCBs so they strike each other in dynamic conditions. This reduces the bending displacements in severe vibration and shock conditions. Reducing the deflection reduces the dynamic forces and stresses. This increases the fatigue life of the assembly. The best locations for snubbers and bumpers are close to the centers of the various PCBs, where the maximum displacements are expected. Soft rubber snubbers work best on PCBs with natural frequencies below about 50 Hz. Half-sphere shaped snubbers with diameters of about 0.20 in. can be used on small PCBs. Large PCBs work well using similar half-sphere shapes with diameters up to about 0.50 in. Soft rubber snubbers should be installed so they interfere slightly with other snubbers on adjacent PCBs for improved performance, as shown in Fig. 7.6. PCBs with natural frequencies above about 100 Hz must use more rigid materials for snubbers to reduce the PCB relative motion. Higher natural frequencies result in smaller dynamic displacements, so soft rubbery materials will not work. High-level vibration and shock qualification test programs using plug-in PCBs with epoxy fiberglass snubbers have been very successful. The snubbers in these programs were short posts with a cross-section diameter of 0.25 in., epoxy bonded to the center of each PCB. Hard snubber materials require a small clearance between adjacent snubbers to allow the PCBs to be inserted and removed with no interference. Smaller clearances

160

OCTAVE RULE TO INCREASE THE PCB FATIGUE LIFE

FIGURE 7.6. Soft rubber snubbers recommended for low natural frequency circuit boards.

work better. The best snubbers are ones that just touch each other with no space between them. This acts like a center support, which substantially increases the natural frequency. Small gaps are very desirable for reducing displacements, but they can be expensive because they require very close manufacturing tolerance control. The clearance between adjacent snubbers should be less than half of the dynamic PCB displacement expected in some particular dynamic environment, as shown in Fig. 7.7. Most snubber applications will be related to modifying existing hardware. The best location is at the center of the PCB. It will be very difficult to find any space at the center of any existing hardware. Therefore, a search must be made of the various PCBs to find free spaces for attaching the snubbers. One method that works well is to make transparent copies of all the PCB layouts. These are then arranged in their proper assembled order. Looking through the transparent assembled layouts will often reveal desirable places for locating adjacent snubbers.

Small gap

--I&

Rigid

FIGURE 7.7. Rigid epoxy glass rod snubbers with a small clearance gap, recommended for high natural frequency circuit boards.

USING SNUBBERS TO REDUCE PCB DISPLACEMENTS AND STRESSES

Sample Problem-Adding

161

Snubbers to Improve PCB Reliability

Several different military and commercial electronic systems have been experiencing component failures on plug-in type PCBs during sinusoidal vibration tests. The PCBs that have been experiencing problems are listed below. 1. PCBs with a natural frequency of 125 Hz during a 5-G peak sine vibration test. 2. PCBs with a natural frequency of 160 Hz during a 5-G peak sine vibration test. 3. PCBs with a natural frequency of 260 Hz during a 6-G peak sine vibration test.

An examination of the problem PCBs shows there is room available near the center for mounting small snubbers. The proposed plan is to use 0.25-in. diameter epoxy glass dowel rods as snubbers, bonded to the PCBs, leaving a nominal clearance gap of 0.012 in. between adjacent snubbers. Is this an acceptable course of action?

Solution Find the Expected Dynamic Displacements and Compare with 0.012-in. Gap. PCB dynamic displacements can be obtained from the general deflection equation: 2, =

9.8GinQ

(see Eq. 2.33)

(fn)'

Table 7.1 lists the results when values are inserted in above equation. The recommended gap between the snubbers is 0.012 in. for all three problem PCBs. The first PCB needs a gap of 0.017 in. or less so a smaller gap of 0.012 in. is good. The second PCB needs a gap of 0.012 in. or less so it just meets the requirements. The third PCB needs a gap of 0.007 in. or less, so a gap of 0.012 in. does not meet the requirements. Such a small gap will be very difficult and very expensive to achieve due to the large number of tolerances involved in the materials, the manufacturing processes, and the

TABLE 7.1 PCB

Gin

fn

1 2 3

5.0

125 Hz 160 Hz 260 Hz

5.0 6.0

Q

dm = 11.2 dm 12.6 =

= 16.1

ZO

Gap = Z0/2

Acceptable

0.035 in. 0.024 in. 0.014 in.

0.017 in. 0.012 in. 0.007 in.

Yes Yes no

162

OCTAVE RULE T O INCREASE T H E PCB FATIGUE LIFE

assembly. However, it can be done with the use of matched assemblies. This involves matching each individual PCB with its own set of snubbers in its own chassis. There can be no switching of similar PCBs to another chassis because the accumulation of tolerances might result in a much large gap that can reduce the PCB fatigue life.

7.8 CONTROLLING THE PCB TRANSMISSIBILITY WITH DAMPING

Damping is usually defined as the conversion of kinetic energy into heat. Material damping relates to the energy lost due to internal friction or hysteresis in the molecular structure of the material. Structural damping relates to the energy lost due to friction in rubbing, scraping, slapping, and impacting at various interfaces and joints. The total system damping is the sum of the material damping and the structural damping. All real systems produce some damping when they are vibrated. When a system with light damping is disturbed, it will continue to oscillate for a long time after the disturbing force has been removed. A system with heavy damping may oscillate once or twice after the disturbing force has been removed, similar to a good automobile shock absorber. The damping in all systems will always bring the system back to a state of rest some time after the disturbing force has been removed. Damping removes some of the energy within the system so there is less energy available to distort and damage the system. This means there is less energy available to deform structures such as PCBs in vibration and shock environments. A reduction in the kinetic energy reduces the forces and stresses in PCBs, which increases the fatigue life, so damping increases the PCB reliability. Increasing the damping in a PCB increases the amount of energy that is lost, which decreases the transmissibility, and further increases the fatigue life and the reliability of the PCB. 7.9

PROPERTIES OF MATERIAL DAMPING

Internal energy is lost, or converted into heat, any time a structure is deformed. Applying a tensile load to a bar does not instantly deform the bar. There is a small time lag as the bar elongates to its new stabilized length. This entire process is reversed when the applied load is removed. However, the bar never returns to its exact original length since some of the strain energy has been converted into heat, so less energy is available to restore the original bar length. Applying an equal compressive load on the bar reverses the direction of the events just described. The positive and negative strains can be plotted on a graph to show the typical hysteresis loop formed by materials that are deformed. The area in the enclosed hysteresis loop becomes a measure of the energy lost with each stress cycle. A larger area

CONSTRAINED LAYER DAMPING WITH VISCOELASTIC MATERIALS

163

enclosed within the loop shows there is more damping. Again, more damping reduces the PCB transmissibility, which increases the fatigue life and the reliability. 7.10 CONSTRAINED LAYER DAMPING WITH VISCOELASTIC MATERIALS

Most of the damage produced in a vibrating PCB occurs at the natural frequency, where the transmissibility is the greatest, so the displacements, forces, and stresses are also the greatest. When the damping is increased, the PCB reliability is increased. More damping in a PCB is desirable, if it does not result in an increase in the size, weight, and cost. One good method for increasing the damping is with the use of viscoelastic materials with constrained layers to increase the damping. Viscoelastic materials are very similar to rubber materials. They are quite elastic so they can deform, compress, and stretch through large displacements without failing. These materials can also dissipate large amounts of energy when they are deformed. Even more energy can be dissipated with the use of materials such as asbestos mixed in with the viscoelastic materials. Pure natural rubber has similar problems-it does not have much strength or much damping. The addition of carbon black improves these properties. Low temperatures and very high frequencies make viscoelastic materials act like hard materials. Their ability to perform well under these conditions is reduced. High temperatures are also a problem, because these materials lose much of their damping properties under these conditions. Their modulus of elasticity is difficult to measure because it changes with the speed of the applied load. The modulus for a rapidly applied load can be three times greater than for a slowly applied load. Viscoelastic materials can be effective in reducing the dynamic displacements, forces, and stresses in vibrating beam and plate structures that are not required to operate over a wide range of high to low temperatures. These materials are available in various forms of paints and tapes that can be applied in single and multiple layers. Multiple layers of damping materials are much more effective than single-layer applications. More damping layers are more effective but they also add more weight and may make repairs more difficult. One of the most effective methods for increased damping is constrained layer damping. This is the application of many layers of damping material that are applied in the form of a high narrow strip. Alternate layers of viscoelastic materials are separated by layers of aluminum foil about 0.005 in. thick. The typical height is about 0.50 in. This can often be added across the center of a plug-in type PCB, in the form of a strip similar to a stiffening rib, as shown in Fig. 7.8. The center of the PCB is the best place for a damping strip because the displacements are the greatest at that position. This results in large relative shear deflections in the alternate damping layers, which improves the damping.

164

OCTAVE RULE TO INCREASE THE PCB FATIGUE LIFE Constrained layer viscoelastic

Adhesive bonding

PCB

Alumi

fo ‘Connector

Section BB enlarged

FIGURE 7.8. Constrained layer viscoelastic damper for possible use on low-frequency

boards.

7.1 1 WHY STIFFENING RIBS ON PCBs ARE OFTEN BETTER THAN DAMPING

Vibration testing experience with damping techniques has shown that damping can be quite effective for PCBs with low natural frequencies, below about 50 Hz. However, the tests show that stiffening ribs usually work even better. When the natural frequencies are well above about 100 Hz damping is not a very effective means for reducing the transmissibility. High operating temperatures also tend to reduce the damping properties and effectiveness of viscoelastic materials. Constrained layer damping strips take away valuable space that could be used for mounting additional electronics. When this space is already lost, experience has shown that it is often much better to use the space for bonding stiffening ribs to the PCB, instead of constrained layer damping strips. Stiffening ribs will increase the natural frequency, which will more rapidly reduce the dynamic displacements, because displacements are inversely related to the square of the natural frequency.

7.12

PROBLEMS WITH PCB VISCOELASTIC DAMPERS

A very large supplier of sophisticated electronic systems received a large contract for a military electronic system that had to operate with a high reliability in severe thermal, vibration, and shock environments. There was not enough sway space available for an isolation system, so the mechanical design engineers selected to use viscoelastic damping techniques to reduce the plug-in PCB response accelerations and displacements. The design ap-

PROBLEMSWITH PCB VISCOELASTIC DAMPERS

165

proach used two, one-sided surface mounted PCBs assembled back to back, with a constrained layer viscoelastic material bonding both halves of the PCBs together into one double PCB plug-in module. Vibration and shock tests were run on prototype models using extensive instrumentation. The test data looked very good, so the design was released to the production group. The first few production models were shipped to the customer for a preliminary inspection and evaluation. Production unit number 8 was selected for the qualification test program. Many military qualification test programs require running electrically operating vibration tests at low temperature, room temperature, and high temperature. The first two vibration tests were satisfactory. However, the high-temperature vibration test resulted in many PCB failures. Component lead wires were breaking, solder joints were cracking, and components were flying off the PCBs. The high-temperature vibration tests were a complete disaster. The mechanical engineers were stunned. They could not understand what went wrong. The previous vibration tests on prototype models had been very successful. An investigating failure analysis team was formed and additional vibration tests were run on the viscoelastic damped PCBs. The tests were instrumented and run at room temperature. The resulting test data looked good. The vibration tests were then run again at high temperatures. The test results were dramatically different, with substantially lower PCB natural frequencies producing sharp increases in the dynamic displacements, with components flying off the PCBs. This was caused by the dramatic reduction in the stiffness and the damping properties of the viscoelastic materials at the high temperatures. The system had to be completely redesigned, using stiffening ribs on the PCBs, to obtain a high natural frequency to keep the dynamic displacements and stresses low to achieve the required fatigue life. The production group had to redesign their tools for the new production program. Needless to say that a lot of money was lost on this program, and product deliveries were substantially delayed, because the mechanical engineering group overlooked the adverse effects of high temperatures on the damping properties of the viscoelastic materials.

DCHAPTER 8

Preventing Sinusoidal Vibration Failures in Electronic Equipment

8.1

INTRODUCTION

Sinusoidal vibrations are often found in machines and devices that rotate or oscillate, such as electric motors. wheels, engines, turbines, gears, washing machines, blenders, oil drilling equipment, springs, and even bridges. Sinusoidal vibrations are also extremely useful as a diagnostic tool for evaluating the dynamic characteristics of various types of structures, and for examining resonant conditions with the use of a strobe light. Vibration failures are often difficult to trace, since there are so many different factors to be considered. Not only must the environment be considered, but also the mechanical and electrical design, manufacturing tolerances, resonant frequency of the outer housing structure, resonant frequency of internal elements, possible dynamic coupling effects, sensitivity of electronic components, cables and harnesses, effects of conformal coatings or lack of coatings, humidity effects, and dozens of other factors. Sometimes there does not appear to be any connection between parts that are failing and exposure to vibration, as shown in the following case history. A large auto manufacturer was having problems with dead batteries on a particular automobile model, when the auto was shipped across the country by rail. The batteries were charged when they left the factory and were discharged when they arrived at their destination. The most obvious sources for the problem were the batteries themselves and the electrical connections related to the battery. When these suspected sources were cleared of blame, the investigation was expanded to other areas, but always within the engine compartment where the battery was located. It appeared logical that the battery failures must be related to the battery location. After many hours of investigation and many meetings, one clever individual observed that the horn ring on this particular car model had a very soft touch. The horn would sound when a small force was applied to the horn ring. H e had a hard time convincing the company officials to send an observer along with the next rail shipment of cars across the country. His theory was correct. Along one long stretch of track through a desolate area, the speed of the train and the 166

ESTIMATING THE VIBRATION FATIGUE LIFE

167

condition of the tracks set up a disturbing force that excited the resonant frequency of the horn ring on its springs. Over an extended period of many hours, the constant sounding of the horn drained the battery. The springs on the car horn ring were made stiffer, which raised the resonant frequency of the horn ring above the disturbing force, and the problem was solved. ESTIMATING THE VIBRATION FATIGUE LIFE

8.2

The approximate fatigue life of a vibrating system can often be estimated from the fatigue characteristics of the various members that carry the major structural loads. These fatigue properties are usually obtained from controlled stress cycle tests performed on many parts that have been manufactured from the same material, with close dimensional tolerances. These parts are stress-tested to failure; then the data points are plotted on log-log paper, with stress on the vertical axis and the number of cycles to failure on the horizontal axis. A straight line that represents the best average fatigue properties is drawn through the scattered data points as shown in Fig. 8.1. The equation for the sloped portion of the curve can be expressed as follows [l]:

N,S,b = N,S,b where N = number of stress cycles to produce a fatigue failure S = stress level at which these failures occurred b = fatigue exponent related to the slope of the line

(8.1)

The exponent b is related to the fatigue life of the structure, and it is useful in predicting the expected fatigue life of other structures fabricated of the same material and exposed to similar environments. In order to reflect the real structural properties of these materials, a stress-concentration factor K must be included in the evaluation of the exponent b. A typical stress-concentration factor for an electronic box structure or for a PCB within the box is about 2.0. Since the stress concentration has little importance with a small number of fatigue cycles, and more importance with more fatigue cycles, the slope of the S-N curve increases with the stress concentration as shown by the dashed line in Fig. 8.1.

I Cycles to failure.

I

FIGURE 8.1. Typical plot of the data obtained

A’

from fatigue tests of metal structures.

168

PREVENTING SINUSOIDAL VIBRATION FAILURES

The value of the exponent b can be determined for aluminum by using typical physical properties for aluminum alloys, where the endurance stress is typically one-third of the ultimate tensile stress. With a stress concentration of 2, the endurance is one-sixth of the ultimate tensile stress. Rewrite Eq. 8.1 as follows:

where Nl = lo8 cycles N2 = l o 3 cycles S , = S, = +St, (one-sixth the ultimate tensile stress) S, = S,, = ultimate tensile stress Substitute into Eq. 8.2 and simplify:

Take the logarithm of both sides and solve for the exponent b:

b=

log,, 105

--

log,, 6

5

0.778

=

6.4

This fatigue exponent b can be used to determine the approximate fatigue life for a number of different conditions. For linear systems, the number of fatigue cycles will be directly related to the time, so the time T can be used to replace the number of cycles ( N ) .Also, the stress S will be directly related to the acceleration G and to the displacement Z . The dynamic load can also be included if it is desired. These relations can be written as follows:

T,G:

=

N,ZF

= N2Z!

(8.4b)

N,G!

=N , G ~

(8.4~)

Sample Problem-Qualification

T,G~

(8.4a)

Test for an Electronic System

An electronic control system is to be used on subway trains that are expected to operate 6 hours per day, 4 days per week, for 52 weeks per year, with a desired life of 20 years. Field test data show that the subway track system produces harmonic vibrations over a frequency range from 10 to 300 Hz, with typical accelerations of 1.1 G peak. A sinusoidal vibration qualification test is

ELECTRONIC COMPONENT LEAD WIRE STRAIN RELlEF

169

desired to prove that the electronic system will be capable of operating in this environment for the 20-year period. The proposed qualification test plan specifies a sinusoidal test where the forcing frequency sweeps back and forth over the observed bandwidth using a 4.5-G peak acceleration input level. Determine the length of time that should be used for these sweep tests to qualify the electronic system. Solution. Equation 8.4a can be used to determine the period of time for the qualification test. The following information is required:

T2 = (6 h)(4 day)(52 wk)(20 yr) = 24,960 h (life desired) G, = 1.1 G (peak field test data) GI = 4.5 G (peak proposed qualification test level) b = 6.4 (vibration fatigue exponent for electronic structures) Substitute into Eq. 8.4a and solve for the desired vibration qualification test time period:

T , = (24,960)

(

:::r4

-

= 3.03

hours

(8.5)

The 3.03 hours represents the testing time with no safety factor. Since there are many variables involved in the determination of the field test data and in the design, manufacture, and assembly of this electronic equipment, some safety factor must be included. Therefore, recommendations should be made to vibrate the electronic equipment for 3.03 hours along each of three mutually perpendicular axes for a total of 9.09 hours of vibration for the qualification test. The above recommendation is equivalent to a safety factor of 3.0, which will require a slightly heavier structural assembly to meet these levels. The weight penalty is not very large here, because the fatigue life changes very rapidly with a slight reduction of the stress level. For example, when the stress level changes by a factor of 2, the fatigue life changes by a factor of 2 raised to the 6.4 power, which is 84.4.

8.3 ELECTRONIC COMPONENT LEAD WIRE STRAIN RELIEF Electronic systems that are used in any vibration or shock environments will expose the electronic components to some level of vibration and shock. These levels may be amplified or attenuated, depending on the characteristics of the dynamic system. When PCBs are used in these electronic systems, the resonant frequencies of the PCBs are often excited, which forces the PCBs to bend back and forth. This motion forces the electrical lead wires on the components to also bend back and forth, as shown in Fig. 8.2.

170

PREVENTING SINUSOIDAL VIBRATION FAILURES

Strained lead wires

+L

\

Relative

PCB Bending’

1

B -_____i_/

FIGURE 8.2. Relative motion produced between component and circuit board when the board resonance is excited.

Different lead wire geometries will result in different stress levels in the lead wires and solder joints of these components. Even when the dynamic displacements of a group of PCBs are the same, the stresses developed in the lead wires and solder joints of similar components can still be sharply different, depending on the geometry of the wire. When the dynamic displacement of a particular PCB has been determined for a particular environment and a particular PCB resonant frequency, the displacement amplitude can be treated as a fixed condition. Under these circumstances, it can be shown that the best way that the forces and stresses can be reduced in the lead wires is to reduce the spring rate of the wires. This appears to be contrary to the normal approach generally taken when a structure fails in a vibration or shock environment. Stronger is normally considered to be better, so most of the time the structure will be made stiffer (stronger) when a failure occurs. However, in the case of lead wires, a stiffer wire will fail faster. This can be demonstrated by examining the force P developed in a structure as a function of the spring rate K and the displacement Y as follows:

P=KY

(8.6)

When the displacement is held constant, the best way that the force can be reduced is to reduce the spring rate. This is the same condition that exists in the component lead wires for a specific dynamic displacement generated by the resonant frequency and acceleration level. In the typical application, where the electrical lead wires are forced to bend back and forth during a resonant condition, the lead wire stiffness is related to the following parameters:

K wire .

EI = -

L3

(wire flexing stiffness parameters)

The above relation shows that increasing the length L of the wire will reduce the stiffness K very rapidly, since it is a cubic function. For example, when the wire length is doubled, the stiffness, the force, and the stress in the wire are reduced by a factor of two cubed, or eight. The above relation also shows

DESIGNING PCBs FOR SINUSOIDAL VIBRATION ENVIRONMENTS

171

that decreasing the moment of inertia I of the electrical lead wire will rapidly reduce the wire stresses, since the moment of inertia varies as the cube of the wire height. When the wire height, or thickness, is reduced by a factor of two, the spring rate, stiffness, force, and stress in the wire are decreased by a factor of two cubed, or eight. This can be accomplished by coining the lead wire, that is, squeezing it into a flat thin metal strip. The cross-sectional area of the wire stays the same, but its thickness is reduced, so the spring rate, stiffness, force, and stress in the wire are also reduced. The above relation also applies to surface mounted electronic components as well as the poke-through type of mounting, where electrical lead wires are forced to bend during resonant conditions. It also applies to leadless ceramic chip carrier (LCCC) devices, which are surface mounted with or without electrical lead wires. In applications that do not use lead wires, the above relations then apply to the shear stiffness of the solder joints that bond the component to the PCB. Here again the stiffness of the solder joints must be reduced to reduce the forces and stresses in them. An examination of the shear stiffness will show which parameters will affect the solder stresses: AG Kshear =

(solder shear stiffness) L

(8.8)

where A = cross-sectional area of the solder joints (he2) L = height of solder joint (in.) G = shear modulus of the solder (lb/in.’> The shear stiffness can be decreased by reducing the cross-sectional area of the solder, reducing the shear modulus of the solder, or increasing the height of the solder joint. Some experiments are being carried out to reduce the solder shear modulus, but so far this has been unsuccessful. Some success has been achieved by reducing the solder cross-sectional area. The greatest success in improving the fatigue life of the solder has been in increasing the height of the solder joint. The typical solder joint height for LCCC devices is about 0.003-0.005 in. This has successfully been raised to 0.010 in. with the use of small solder mask pads, 0.010 in. thick, under the LCCC devices. 8.4 DESIGNING PCBs FOR SINUSOIDAL VIBRATION ENVlRONMENTS

In typical electronic systems, the electronic components are mounted on plug-in PCBs. This makes it easy to repair an electronic system if an electrical failure occurs and the failure can be traced to a faulty PCB. The faulty PCB is simply replaced with a good one.

172

PREVENTING SINUSOIDAL VIBRATION FAILURES

f

PC pcB

2nd

degree

Chassis

I

1st degree Rigid fixture

3

1 Vibrat

3n

lnpu!

I

1

V i brat ion

input

FIGURE 8.3. A simple electronic chassis assembly with PCBs, simulated as a

lumped-mass system. When the plug-in PCBs are mounted within a chassis housing, and the system is required to operate in a sinusoidal vibration environment, then the chassis will act as the first degree of freedom, because the vibration energy will excite the chassis structure first. Since the PCBs are normally attached to the chassis structure (side walls or base of the chassis), the PCBs receive their dynamic excitation from the chassis. This makes the PCBs act like the second degree of freedom. A typical system is shown in Fig. 8.3. Resonant conditions can occur in the chassis and also in the PCBs during excitation by sinusoidal vibration. Since the chassis and the PCBs are coupled together, any resonant condition in any one of the masses will force some response in the other masses. When the resonant frequency of the chassis is close to the resonant frequency of any of the PCBs, and when the chassis and the PCBs have high transmissibilities Q, then the chassis response can augment the PCB response. This will produce very high acceleration levels in the PCBs, which can result in rapid fatigue failures. In order to prevent this from occurring, the chassis resonance and the PCB resonances must be well separated. This can be accomplished by using the octave rule. An octave means a factor of two in frequency. When the resonances of the chassis and the PCBs are separated by one octave or more, then the coupling effects between those members are sharply reduced. It makes very little difference whether the chassis or the PCBs have the higher resonant frequency, as long as their resonant frequencies are well separated. Vibration testing experience and associated finite element studies have shown that the fatigue life of many different types of electronic components can be related to the dynamic displacements developed by the PCBs during vibration [38]. These studies have shown that these components can achieve a fatigue life of about 10 million stress reversals in a sinusoidal vibration environment, when the peak single-amplitude displacement of the PCB is limited to the value shown below and in Fig. 8.4 [371:

Z=

0.00022 B Chra

(maximum desired PCB displacement)

(8.9)

DESIGNING PCBs FOR SINUSOIDAL VIBRATION ENVIRONMENTS Bending lead wires

I

FIGURE 8.4. Axial-leaded component wires will bend when the PCB resonance is excited.

-2 displacement where B

173

of PCB edge parallel to component (in.) L = length of electronic component (in.) h = height or thickness of PCB (in.) C = constant for different types of components (see Eq. 9.50) r = relative position factor for components (see Eq. 9.50) = length

The peak single-amplitude displacement expected at the center of a PCB can be estimated by assuming the PCB acts like a single-degree-of-freedom system at its resonant condition, as shown in Eq. 2.30: 9.8G 9.8Gi,Q z=-

f'

f,'

The displacement value obtained from the above relation will be a little less than the correct value obtained from the actual dimensions of the PCB plate geometry. When the PCB is approximated as a single degree of freedom, and a single mass is used to represent the PCB, the single mass will show the displacement of the centroid of the PCB and not the actual displacement of the PCB itself. Vibration test data have shown that the transmissibility Q of a PCB can typically be related to its resonant frequency. Although the test data show that high input acceleration levels (above 8 G) will reduce the Q values, and low input acceleration levels (below 2 G ) will increase the Q values, a good approximation of the Q value can still be obtained with the following relation (also see Eq. 14.21):

Q=&-

(8.10)

The desired PCB resonant frequency that will provide a fatigue life of about 10 million stress cycles for the component lead wires and their