CLASS NOTES - concrete canoeconcretecanoe.org/Webmaster/Resume/CourseNotes/NOTES6.pdfCLASS NOTES Version 5.0 by John A. Gilbert, Ph.D. Professor of Mechanical Engineering University

MAE 477/577

- EXPERIMENTAL TECHNIQUES

IN SOLID MECHANICS -

CLASS NOTES Version 5.0

by

John A. Gilbert, Ph.D.

Professor of Mechanical Engineering

University of Alabama in Huntsville

FALL 2013

i

Table of Contents

Table of Contents i

Introducing MAE 477/577 iv

MAE 477/577 Outline v

MAE 477 ABET Syllabus vi

MAE 577 SACS Syllabus viii

Chapter 1. Stress

1.1 Introduction 1.1

1.2 Types of Forces 1.1

1.3 Traction Vector, Body Forces/Unit Mass, Surface and Body Couples 1.1

1.4 Resolution of the Traction Vector – Stress at a Point 1.3

1.5 Equilibrium Equations – Conservation of Linear Momentum 1.5

1.6 Stress Symmetry – Conservation of Angular Momentum 1.7

1.7 Transformation Equations – Mohr’s Circle 1.7

1.8 General State of Stress 1.16

1.9 Stresses in Thin Walled Pressure Vessels 1.19

1.10 Homework Problems 1.22

Chapter 2. Strain


2.2 Strain-Displacement Equations 2.1

2.3 Strain Equations of Transformation 2.2

2.4 Compatibility Equations 2.3

2.5 Constitutive Equations 2.3


Chapter 3. Linear Elasticity

3.1 Overview 3.1

3.2 Plane Stress 3.1

3.3 Plane Strain 3.2


ii

Chapter 4. Light and Electromagnetic Wave Propagation

4.1 Introduction to Light 4.1

4.2 Electromagnetic Spectrum 4.3

4.3 Light Propagation, Phase, and Retardation 4.4

4.4 Polarized Light 4.5

4.5 Optical Interference 4.6

4.6 Complex Notation 4.6

4.7 Intensity 4.7

4.8 Superposition of Wavefronts 4.8

4.9 Reflection and Refraction 4.11

4.10 Double Refraction – Birefringence 4.12


Chapter 5. Photoelasticity


5.2 Plane Polariscope 5.2

5.3 Circular Polariscope 5.6

5.4 Calibration Methods for Determining the Material Fringe Value 5.10

5.5 Compensation Methods for Determining Partial Fringe Order Numbers 5.12

5.6 Calculation of the σ1 Direction 5.14

5.7 Birefringent Coatings 5.15

5.8 Three-Dimensional Photoelasticity 5.17


5.10 Classroom Demonstration in Photoelasticity 5.24

Chapter 6. Photography


6.2 Single Lens Reflex Cameras 6.1

6.3 Depth of Field 6.2

6.4 Photographic Processing 6.3

6.5 Polaroid Land Film Cameras 6.5

6.6 Digital Cameras 6.9

Chapter 7. Brittle Coatings


7.2 Testing Procedures 7.2

7.3 Calibration 7.3

7.4 Measurements 7.8

7.5 Coating Selection 7.9

7.6 Application 7.11

7.7 Crack Patterns 7.11

7.8 Relaxation Techniques 7.11

iii


Chapter 8. Moiré Methods


8.2 Analysis 8.2

8.3 Optical Filtering 8.4

8.4 Stress Analysis 8.5

8.5 Shadow Moiré 8.6


Chapter 9. Strain Gages


9.2 Electrical Resistance Strain Gages 9.2

9.3 Length Considerations 9.4

9.4 Transverse Sensitivity Corrections 9.5

9.5 Temperature Considerations 9.5

9.6 Rosettes 9.6

9.7 Gage Selection and Series 9.8

9.8 Dummy Gages and Transducer Design 9.9

9.9 P-3500 Strain Indicator 9.11

9.10 Two-Wire and Three-Wire Strain Gage Circuits 9.13


Appendix I – Review of Statics and Mechanics of Materials

A.1.1 2-D Forces in Rectangular Coordinates A1.1

A.1.2 3-D Rectangular Coordinates A1.1

A1.3 Addition of Concurrent Forces in Space/3-D Equilibrium of Particles A1.2

A1.4 Moments and Couples A1.3

A1.5 Equilibrium of Rigid Bodies A1.4

A1.6 Center of Gravity and Centroid A1.4

A1.7 Moments of Inertia A1.6

A1.8 Shear and Moment Diagrams A1.8

A1.9 Deflection of Beams A1.8

A1.10 Synopsis of Mechanics of Materials A1.10

A1.11 Homework Problems A1.13

Appendix II – Basic Formulas for Mechanics of Materials A2.1

iv

Introducing MAE 477/577

EXPERIMENTAL TECHNIQUES IN SOLID MECHANICS - 3 hrs.

Required Text: Experimental Techniques in Solid Mechanics, Version 4.1 by Gilbert

Reference Text: Experimental Stress Analysis, 4th. Edition by Dally and Riley

Prerequisite: MAE/CE 370; Junior Standing

Overview:

When dealing with the majority of real engineering systems, it is not always sufficient, or

advisable, to rely on analytical results alone. The experimental determination of stress, strain,

and displacement is important in both design and testing applications. MAE 477/577,

Experimental Techniques in Solid Mechanics, presents techniques which are valuable

complements to the design and analysis process; and, in some difficult or complex situations,

provide the only practical approach to a real solution.

The bulk of the course constitutes a detailed treatment of the more conventional methods

currently used for experimental stress analysis (photoelasticity, brittle coatings, moiré methods,

strain gages, etc.); however, more recent developments in the field are also introduced (hybrid

methods, speckle metrology, holographic interferometry, moiré interferometry, fiber optics,

radial metrology, and STARS). In-class laboratory exercises are included so that students gain

some practical experience. The lecture and laboratory exercises are designed to provide enough

exposure for participants to secure an entry level position in the field of experimental mechanics.

Attendance:

Students are required to be present from the beginning to the end of each semester, attend all

classes, and take all examinations according to their assigned schedule. In case of absence,

students are expected to satisfy the instructor that the absence was for good reason. For

excessive cutting of classes (3 or more class periods), or for dropping the course without

following the official procedure, students may fail the course.

Homework:

Homework assignments must be done on only one side of 8 1/2" x 11" paper. Each problem

shall begin on a separate page and each page must contain the following (in the upper right hand

corner): Your name, the date, and page __ of __. All final answers must be boxed and converted

(SI to US or US to SI). Homework is due at the beginning of the class on the date prescribed.

Work must be legible and should be done in pencil. Problems shall be restated prior to solution

and free-body diagrams (FBDs) shall be drawn for problems requiring such. Loose sheets shall

be stapled together in the upper left hand corner.

Note: For further information on the course and registration procedure, call Professor Gilbert by

telephone at (256) 824-6029, or, contact him by e-mail at [email protected].

mailto:[email protected]

v

MAE 477/577 Experimental Techniques in Solid Mechanics

Fall 2013

Time/Place: Friday 8:00 a.m. - 10:40 a.m.; TH S117

Instructor: Dr. John A. Gilbert

Office: OB 301E

Telephone: (256) 824-6029 (direct); (256) 824-5117/5118 (Beth/Cindy)

Fax/E-mail: (256) 824-6758; [email protected]

Office Hours: To be announced.

Grading: 10% Attendance (100% for being in class; 0% for not being there); 30%

Homework and Labs; 30% Midterm (10/11/13); 30% Final (11/29/13).

Required Text: Gilbert, J.A., Experimental Techniques in Solid Mechanics, Volume 5.0,

University of Alabama in Huntsville, 2013.

Reference Text: Dally, J.W., Riley, W.F., Experimental Stress Analysis, 4th Edition,

College House Enterprises, 2005, ISBN 0-9762413-0-7.

Course Outline

1. Overview of Solid Mechanics - Stress, Strain, and Displacement; Equilibrium,

Transformation, Strain-Displacement, Compatibility, and Constitutive Equations.

2. Stress Analysis - Method of Attack; Stress Concentration, Failure, Design, and Examples.

3. Light and Electromagnetic Wave Propagation – Amplitude, Phase, Polarization,

Coherence, Interference, Reflection, Refraction, Birefringence, and Stress Optic Law.

4. Photoelasticity – Plane and Circular Polariscopes; Calibration and Compensation

Methods; Reflection Polariscope, Birefringent Coatings, and 3-D Photoelasticity.

5. Photography – Cameras, Lenses, and Photographic Development; Digital Cameras,

Image Acquisition and Processing.

6. Brittle Coatings – Theory, Calibration, Application, and Measurement.

7. Moiré Methods – Geometrical Considerations and In-Plane Displacement Measurement;

Diffraction, Optical Filtering, and Stress Analysis; Out-of-Plane Displacement

Measurement and Shadow Moiré.

8. Electrical Resistance Strain Gages - Parametrical Studies, Transverse Sensitivity,

Rosettes, Circuitry, Installation, and Transducer Design.

9. Advanced Topics – Moiré Interferometry, Speckle Metrology, Holographic

Interferometry, Fiber Optics, Digital Image Processing, Hybrid Methods, Panoramic

Imaging Systems, Radial Metrology, and STARS.

vi

MAE 477 Experimental Techniques in Solid Mechanics

Responsible Department:

Mechanical and Aerospace Engineering

Catalog Description:

477 Experimental Techniques in Solid Mechanics 3 hrs.

Experimental methods to determine stress, strain, displacement, velocity, and

acceleration in various media. Theory and laboratory applications of electrical

resistance strain gages, brittle coatings, and photoelasticity. Application of

transducers and experimental analysis of engineering systems. (Same as CE 477)

Prerequisites:

MAE/CE 370 and junior standing.

Textbook:

Required Text: Gilbert, J.A., Experimental Techniques in Solid

Mechanics, Volume 5.0, University of Alabama in

Huntsville, 2013.

Reference Text: Dally, J.W., Riley, W.F., Experimental Stress Analysis,

4th Edition, College House Enterprises, 2005, ISBN 0-

9762413-0-7.

Course Objectives:

1. To present techniques that are valuable complements to the design and

analysis process; and in some difficult or complex situations, provide the

only practical approach to a real solution.

2. The bulk of the course constitutes a detailed treatment of the more

conventional methods currently used for experimental stress analysis

(photoelasticity, brittle coatings, moiré methods, strain gages, etc.);

however, more advanced topics are introduced (hybrid methods, speckle

metrology, holographic interferometry, moiré interferometry, fiber optics,

radial metrology, and STARS).

3. A portion of the course is devoted to laboratory work.

Topics Covered:

1. Overview of solid mechanics

2. Stress analysis

3. Light and electromagnetic wave propagation

4. Photoelasticity

5. Photography

6. Brittle coatings

7. Moiré methods

8. Electrical resistance strain gages

9. Advanced topics

vii

Class Schedule:

Once per week; class is 2 hours 40 minutes.

Contribution of Course to Meeting the Professional Component:

Basic Mathematics & Science: 0 credits.

Engineering Science: 3 credits.

Engineering Design: 0 credits.

Relationship of Course to Program Outcomes:

In this course each undergraduate student will have to:

a) apply their knowledge of mathematics, science, and engineering;

b) apply a knowledge of calculus-based physics;

c) apply advanced mathematics through differential equations;

d) apply a knowledge of linear algebra;

e) design a system, component or process to meet desired needs;

f) work professionally in mechanical systems;

g) use the techniques, skills, and modern engineering tools necessary for

engineering practice;

h) conduct experiments and to analyze and interpret data;

i) identify, formulate and solve engineering problems;

k) communicate effectively.

Person Preparing this Description:

John A. Gilbert, Ph.D., Course Coordinator, Professor

MAE Program

28 November 2012

viii

MAE 577 Experimental Techniques in Solid Mechanics

Responsible Department:

Mechanical and Aerospace Engineering

Catalog Description:

577 Experimental Techniques in Solid Mechanics 3 hrs.

Experimental methods to determine stress, strain, displacement, velocity, and

acceleration in various media. Theory and laboratory applications of electrical

resistance strain gages, brittle coatings, and photoelasticity. Application of

transducers and experimental analysis of engineering systems. (Same as CE 577)

Prerequisites:

MAE 370 and junior standing

Textbook:

Required Text: Gilbert, J.A., Experimental Techniques in Solid

Mechanics, Volume 5.0, University of Alabama in

Huntsville, 2013.

Reference Text: Dally, J.W., Riley, W.F., Experimental Stress Analysis,

4th Edition, College House Enterprises, 2005, ISBN 0-

9762413-0-7.

Course Educational Objectives:

1. To present techniques that are valuable complements to the design and

analysis process; and in some difficult or complex situations, provide the

only practical approach to a real solution.

2. The bulk of the course constitutes a detailed treatment of the more

conventional methods currently used for experimental stress analysis

(photoelasticity, brittle coatings, moiré methods, strain gages, etc.);

however, more advanced topics are introduced (hybrid methods, speckle

metrology, holographic interferometry, moiré interferometry, fiber optics,

radial metrology, and STARS).

3. A portion of the course is devoted to laboratory work.

Topics Covered:

1. Overview of solid mechanics

2. Stress analysis

3. Light and electromagnetic wave propagation

4. Photoelasticity

5. Photography

6. Brittle coatings

7. Moiré methods

8. Electrical resistance strain gages

9. Advanced topics

ix

Class Schedule:

Once per week; class is 2 hours 40 minutes.

Relationship of Course to Program Outcomes:

In this course each graduate student will have to:

a) apply mathematics to solve engineering problems;

b) successfully perform advanced analysis in a technical area;

c) communicate effectively.

Person Preparing this Description:

John A. Gilbert, Course Coordinator, Professor

MAE Program

28 November 2012

1.1

CHAPTER 1 - STRESS

1.1 Introduction

When dealing with the majority of real engineering systems, it is not always sufficient, or

advisable, to rely on analytical results alone. The experimental determination of stress, strain,

and displacement is important in both design and testing applications. Experimental Techniques

in Solid Mechanics presents techniques which are valuable complements to the design and

analysis process; and, in some difficult or complex situations, provide the only practical

approach to a real solution.

The bulk of the course constitutes a detailed treatment of the more conventional methods

currently used for experimental stress analysis (photoelasticity, brittle coatings, moiré methods,

strain gages, etc.); however, more recent developments in the field are also introduced (hybrid

methods, speckle metrology, holographic interferometry, moiré interferometry, fiber optics,

radial metrology, and STARS). In-class laboratory exercises are included so that students gain

some practical experience. The lecture and laboratory exercises are designed to provide enough

exposure for participants to secure an entry level position in the field of experimental mechanics.

1.2 Types of Forces

The forces which act to produce stress in a material are classified according to their manner of

application. Surface forces act on the surface of the stressed body and are usually generated

when two bodies come into contact. Body forces, on the other hand, act on each element of the

body and are generated by force fields such as gravity. In a great many cases, the body forces

are small as compared to the surface forces and can be neglected in the analysis.

1.3 Traction Vector, Body Force/Unit Mass, Surface and Body Couples

When a solid body is subjected to external loads, the material from which the body is made

resists and transmits these loads. The strength of the body is measured in terms of stress.

Stress represents force intensity and is a mathematical quantity that must be computed by

dividing an applied force by the area over which the force acts. For a given force; the smaller the

area, the higher the stress. In general, the stress is different at every point in a loaded body.

The stress distribution at a given point is relatively complex and may be represented by a second

order tensor. Each direction in space (typically defined by a unit vector perpendicular to a plane

passed through the point in question) is associated with a stress vector which can be resolved into

components normal (normal stress) and parallel (shear stress) to the plane.

1.2

(1.3-1)

(1.3-2)

(1.3-3)

Figure 1, for example, shows an arbitrary internal or external surface with a small area ΔA

enclosing a point P. As mentioned above, the orientation of the plane is defined with respect to a

reference axes system by a unit vector, n, drawn perpendicular to the surface. In general, the

distributed forces acting on ΔA can be reduced to a force-couple system at point P where ΔFn

and ΔMn are the resultant force and couple, respectively. In most cases, the resultant force, ΔFn,

is not coincident with n and the couple vector lies at an arbitrary angle with respect to the line of

action of the resultant force.

Figure 1. Forces and moments act on a small area surrounding a point.

The average stress (force intensity) acting over the area is:

The traction (or resultant stress) vector, tn, at point P is found by taking the limit of the quantity

in Equation (1.3-1) as ΔA goes to zero:

Since the magnitude and direction of the terms on opposite sides of Equation (1.3-2) are equal,

the traction vector is aligned with ΔFn. Simply put, tn depends upon the point in question and the

plane defined by n.

The surface couple at point P is given by:

Figure 2, on the other hand, shows a three-dimensional body with point Q surrounded by a

volume element ΔV having a corresponding mass, Δm. ΔFB and ΔMB are the resultant body

force and body couples acting on the volume.

1.3

(1.3-5)

(1.3-4)

Figure 2. Forces and moments act on a small volume corresponding to a given mass.

Following arguments similar to those taken in obtaining Equations (1.3-2) and (1.3-3), the

resultant body force per unit mass acting at point Q is:

whereas, the resultant body couple per unit mass is:

While formulating the governing equations, surface and body couples are almost always

neglected. As a consequence, higher order terms disappear thereby simplifying the equations.

1.4 Resolution of the Traction Vector - Stress at a Point

Figure 3 shows the traction vector resolved into components normal and tangent to the plane

under consideration. The normal component, σ, is called the normal stress, whereas the

tangential component, τ, is called the shear stress.

Figure 3. The traction vector can be resolved into normal and tangential components.

1.4

(1.4-1)

(1.4-2)

A more meaningful description is obtained when one axis of a Cartesian system is aligned with

the normal to the plane under consideration. Figure 4, for example, shows the traction vector

acting at point P on a plane having a normal n acting along the Z axis. In this case, the tangential

component of the traction vector can be broken into two in-plane components. A double

subscript notation is used to define the stress components; the first subscript denotes the normal

to the surface under consideration while the second subscript denotes the direction in which the

stress component acts. For the configuration shown in Figure 4, the shear and normal stress

components are denoted by τzx, τzy and σzz, respectively.

Figure 4. The traction vector is resolved into three Cartesian components.

Two additional sets of components are obtained for planes having their normal along the X and

Y axes. The corresponding stress components are σxx,τxy,τxz and τyx,σyy,τyz , respectively. The

stress distributions on these three mutually perpendicular planes can be summarized in a stress

tensor as follows:

In general, the magnitude and direction of the traction vector at a point depend upon the plane

considered. The rows in the stress tensor, on the other hand, correspond to the scalar

components of the traction vectors which act on planes having their normal vector acting along

the X,Y,Z axes, respectively.

Even though an infinite number of traction vectors may be defined by passing different planes

through a single point, each can be characterized in terms of the 9 components of stress defined

on three mutually perpendicular planes. Mathematically, the traction vector, tn, on a plane

defined by its normal, n, may be determined from the stress tensor, Τ, using:

.

1.5

The infinitesimal region surrounding a point can be modeled by a rectangular parallelepiped. If

the sides are aligned with a Cartesian axes system, the 9 components of stress can be graphically

represented.

Figures 5, 6, and 7, for example, show the stress components on planes with their outer normals

parallel to the X,Y,Z axes, respectively. In this case, each parallelepiped is infinitesimal and the

stress distribution on opposing faces is equal and opposite, thus maintaining equilibrium.

By convention, stress components are drawn along the positive coordinate direction when the

outer normal to the plane on which they act lies along a positive coordinate direction; otherwise,

the component is drawn along the negative coordinate direction.

Each component is labeled using the double subscript notation described above. Normal stresses

acting away from the element correspond to tensile stresses while normal stresses acting toward

the element correspond to compressive stresses.

1.5 Equilibrium Equations - Conservation of Linear Momentum

The overall distribution of forces acting on the body must be such that the stress distribution

satisfies equilibrium. Figure 8, for example, shows a small element of volume taken from a

stressed body.

Figure 8. A finite element taken from a stressed body.

Figure 5. Stresses on planes

with their normal along X.


with their normal along Y.


with their normal along Z.

1.6

(1.5-1)

(1.5-2)

(1.5-4)

(1.5-3)

(1.5-5)

(1.5-6)

As opposed to the parallelepipeds depicted in Figures 5-7, the one is Figure 8 has finite

dimensions. To simplify the analysis, only the stress components which act in the X direction

have been included on the figure along with Fx, the body force per unit mass.

Assuming that the body is in equilibrium:

.

The stresses can be converted to forces by multiplying each stress by the area of the face over

which it acts and the body force per unit volume can be converted to force by multiplying it by

the volume. When this is done and Equation (1.5-1) is applied:

Dividing by the volume,

When a similar argument is applied to analyze the stresses along Y:

and for those along Z:

Summarizing,

The expressions in Equation (1.5-6) are referred to as the equilibrium equations.

1.7

(1.6-1)

(1.7-1)

1.6 Stress Symmetry - Conservation of Angular Momentum

Neglecting surface and body couples and taking moments about an XYZ axes located at the

centroid of an element subjected to stresses:

The expressions in Equation (1.6-1) show that the stress tensor, Τ, is symmetrical. Hence, there

are 6 independent components of stress.

1.7 Transformation Equations - Mohr's Circle

In Section 1.4, it was shown that for every point in a loaded member, the components of shear

and normal stress change as different planes are passed through the point. Even though an

infinite number of planes can be passed through the point, it is sufficient to determine only the

stresses on three mutually perpendicular planes to completely characterize the state of stress on

an arbitrary plane. There are, however, certain critical planes, the analysis of which help to

determine the structural integrity. This section considers the transformation equations required

to pinpoint the orientations of these critical planes and the corresponding stresses which act on

them.

This exercise is important in experimental mechanics where the first step is usually to identify

critical points on the loaded structure. For an interior point, the measurements must be made on

three mutually perpendicular planes. In general, the stresses obtained on these planes are not the

maximum stresses at the point under consideration and a coordinate rotation is required.

When the stress tensor is formulated for a Cartesian axes system and the axes system is rotated, a

new stress tensor, Τ', is obtained. This is illustrated in Figures 9 and 10.

The stress tensor is modified because the shear and normal stress components on the rotated

planes are different from those on the original planes. However, the stresses on the rotated

planes may be obtained from those on the original planes by applying the law of transformation

for a tensor of the second rank. Mathematically, the transformation takes the form,

where τij' and τkl are the components of the rotated and original tensors, respectively; λik and λjl

correspond to the terms contained in the matrix of direction cosines and in its transpose,

respectively.

1.8

(1.7-2)

(1.7-3)

Fortunately, most experimental measurements are taken on a free surface where the components

of traction on the plane defined by the outer normal vanish. This condition is often referred to as

a plane, 2-D, or, biaxial stress state. In this case, for a point contained in the XY plane, the stress

tensor becomes:

.

From Equation (1.7-2) it is apparent that for plane stress, there are 3 independent stress

components. Thus, from an experimental mechanics standpoint, the complete determination of

the stress distribution at the point requires that 3 independent measurements be made.

The stress transformation can be studied analytically using Equation (1.7-1) or graphically by

manipulating these expressions and using Mohr's circle. Consider, for example, the stress

element shown in Figure 11.

Equation (1.7-1) can be applied to determine the stresses on the rotated element shown in Figure

12 as follows:

Figure 10. The stress distribution referred to

a rotated X'Y'Z' axes system. Figure 9. The stress distribution referred to

an XYZ axes system.

1.9

(1.7-4)

(1.7-5)

(1.7-6)

Applying the expressions in Equation (1.7-3) to the configuration shown in Figure 12, noting

that the X’ and Y’ axes correspond to θ = 0o and 90

o, respectively,

The expressions in Equation (1.7-4) are referred to as the transformation equations for a point in

plane stress.

An alternate approach is to manipulate each of the expressions included in Equation (1.7-3) as

follows:

and

Adding Equations (1.7-5) and (1.7-6):

Figure 12. The stress distribution on a plane

stress element referred to a rotated X'Y' axes.

Figure 11. The stress distribution on a plane

stress element referred to an XY axes system.

1.10

(1.7-7)

(1.7-8)

(1.7-9)

(1.7-10)

Equation (1.7-7) is the equation of a circle drawn in σ-τ space. This stress circle is commonly

referred to as Mohr's circle.

In X-Y space, the expression

corresponds to a circle centered at x = a and y = b with radius c.

Comparing the terms in Equation (1.7-7) to those in Equation (1.7-8), the Mohr's circle in σ-τ

space is centered at:

with radius

As illustrated in Figure 13, the circle is drawn by plotting points X with coordinates σxx and τxy

and Y with coordinates σyy and -τxy. Normal stresses are assumed positive when they act away

from the element; positive shear stresses produce a counterclockwise rotation of the element.

Figure 13. A typical Mohr's circle for graphical transformation of stress.

1.11

(1.7-11)

(1.7-12)

(1.7-13)

(1.7-14)

Each radial line in the circle represents the orientation of the unit vector normal to one of the

infinite number of planes which can be passed through the point. The intersection of the line

with the circle defines the shear and normal stresses acting on that plane. The orientation of the

planes are usually measured with respect to the X axis and are determined by computing angles

on the circle and dividing these values in half. The resulting angle is measured in the opposite

sense (clockwise or counterclockwise) on the element.

Points A and B are of special interest, since they correspond to the principal planes on which the

principal stresses act. Note that the shear stress is zero at these points. Other points of interest

lie at D and E where the maximum in-plane shearing stress occurs. As discussed later in Section

1.8, this value is the absolute maximum shearing stress when the principal normal stresses are of

opposite sign (i.e., the Mohr’s circle encompasses the origin). Note that the normal stress for

both points corresponds to the location of the center of the circle.

The values of normal and shear stress at these critical locations can be computed graphically, as

demonstrated in the example problem outlined at the end of this section, or calculated on the

basis of Equation (1.7-3) as described below.

The orientation of the principal planes with respect to the X axis, θp, are obtained by

differentiating the first expression in Equation (1.7-3) with respect to θ and setting the result

equal to zero. This operation produces,

The magnitudes of the principal stresses are found by substituting the values found from

Equation (1.7-11) back into the first expression in Equation (1.7-3). This operation produces:

A similar argument can be applied to the second expression in Equation (1.7-3). That is, the

planes of maximum in-plane shear stress are oriented at θs with respect to the X axis, where

Substituting the values in Equation (1.7-13) into the expressions in Equation (1.7-3) produces:

It should be noted, as illustrated in the following example, that the principal planes are oriented

at an angle of 45o with respect to the planes of maximum shear.

1.12

(1)

(2)

(3)

Example: For the stress state shown in Figure 14, determine the:

(a) orientation of the principal planes.

(b) magnitude of the principal stresses.

(c) orientation of the planes of maximum in-plane shearing stress.

(d) magnitude of the maximum in-plane shearing stress.

(e) magnitude of the normal stress acting on the maximum in-plane shear planes.

Figure 14. A point in a state of plane stress.

Solution: The problem can be solved analytically. From Equation (1.7-11):

.

Thus, (a) the principal planes are oriented at:

.

From the first expression in Equation (1.7-12):

1.13

(4)

(5)

(6)

(7)

(8)

(9)

.

Thus, (b) the principal stresses are:

.

It is not apparent, however which of the stresses in Equation (4) correspond to the orientations

given in Equation (2). To determine the correspondence, it is necessary to substitute the values

in Equation (2) into the first expression in Equation (1.7-3). This operation shows that σmax lies

at θp = - 13.3o while σmin corresponds to θp = 76.7

o. It should be noted that, from the second

expression in Equation (1.7-12), the shear stresses on the principal planes are zero.

From Equation (1.7-13):

.

Thus, (c) the orientations of the maximum in-plane shearing stresses are:

.

From the first expression in Equation (1.7-14), (d) the maximum in-plane shearing stress is:

.

Again, it is not apparent which value of θs corresponds to τmax. This can only be determined by

substituting the values in Equation (6) into the second expression of Equation (1.7-3). This

process reveals that τmax corresponds to θs = 31.7o; τmin = - τmax lies at θs = 121.7

o.

From the second expression in Equation (1.7-14), (e) the corresponding normal stress is:

The graphical solution is obtained by drawing the Mohr's circle shown in Figure 15. This is

accomplished by plotting the two points corresponding to the planes labeled as 1 and 2 on

Figure 14 in σ-τ space. The points are connected and a circle is drawn centered at C with radius

r. The location of the center of the circle is computed from Equation (1.7-9) as:

1.14

(11)

(12)

(10)

Figure 15. A Mohr’s circle can be drawn for the point in question.

The radius is computed from Equation (1.7-10) as:

The maximum and minimum values of stress are determined by adding and subtracting the

magnitude of the radius to the value of σ at point C, respectively. This operation produces:

.

The orientations of the principal planes are obtained by determining the orientation of the point

labeled 3 on Figure 15 with respect to the point labeled 1. Referring to the figure:

Note that on the circle, 2α is drawn counterclockwise from point 1 to point 3. Also note that

point 3 corresponds to σmin. Therefore, the normal to the plane corresponding to σmin lies at α =

13.3o measured in the clockwise direction from the normal to the plane labeled as 1 in Figure 14.

These results are summarized in Figure 16 which shows the rotated element corresponding to

the principal planes.

1.15

(13)

(14)

Referring to Figure 15, the value of τ at the point labeled 5 is equal to the radius while the

corresponding value of σ is equal to the normal stress of point C. At the point labeled 5:

.

The orientation of the maximum in-plane shearing stresses may be determined by finding the

angle labeled 2β on Figure 15 as:

.

Note that 2β is drawn from point 1 to point 5 in a clockwise direction. Thus, on the element, the

plane corresponding to the maximum in-plane shearing stress lies at β = 31.7o measured

counterclockwise from the plane labeled as 1 on Figure 14. Since the shear is positive, it is

drawn such that it creates a counterclockwise rotation on the element. These results are

summarized in Figure 17.

It is apparent by comparing Figures 16 and 17 that the planes of maximum in-plane shearing

stress are oriented at 45o with respect to those corresponding to the principal stresses. It should

also be noted from Figures 14-17 that the sum of the normal stresses acting on two

perpendicular planes is constant. The sum corresponds to the trace of the stress tensor (σxx + σyy

+ σzz) which is one of three quantities which remain invariant during coordinate rotation (in this

case, about the Z axis).

Figure 17. The maximum and minimum in-

plane shear stresses and their orientations.

Figure 16. The principal stresses and their

orientations.

1.16

(1.8-1)

1.8 General State of Stress

As discussed in Sections 1.4 and 1.6, the stress tensor consists of 9 components of stress, 6 of

which are independent. The 9 values correspond to 3 normal and 6 shear stress components

acting on three mutually perpendicular planes.

As illustrated in Figure 18, the coordinate axes can be rotated, in much the same manner as in

plane stress, to determine principal planes on which the shear stress is zero. In mathematical

terms, this is called an eigenvalue problem. One obtains a characteristic equation which has

three roots called eigenvalues; each eigenvalue is equal in magnitude to one of the principal

stresses. Eigenvectors are determined for each eigenvalue, and these vectors define the

orientation of the normal to each of the principal planes on which the principal stresses act.

Figure 19 shows one possible configuration for graphically depicting the stress distribution at a

point. In this case, the three eigenvalues are all positive. Coordinate rotations around the three

eigenvectors are pictorially represented by the three circles, and points on their circumferences

define the stress distribution on different planes on a rotated element. A more general coordinate

transformation leads to a point inside of the area between the inner circles and the outer circle.

In all cases,

Figure 19. The Mohr’s circle for an arbitrary

point at which the three eigenvalues are all

positive.

Figure 18. A rotated element showing the

three principal stresses.

1.17

Figure 22. Orientation of the maximum

shear stress.


shear stress.

Two cases are of interest when evaluating Equation (1.8-1) for the case of plane stress (when one

of the eigenvalues is zero).

Figure 20 shows that when the principal stresses obtained from the bi-axial transformation (the

2-D Mohr's circle shown as the solid line in the figure) are of opposite sign, the in-plane

maximum shear stress is the absolute maximum shear stress at the point under study. As

illustrated in Figures 21 and 22, the corresponding planes are oriented at 45 degrees with respect

to the principal directions.

Figure 20. A Mohr's circle for a point at which the principal stresses are of opposite sign.

1.18


shear stress.


shear stress.

Figure 23, on the other hand, illustrates that when the in-plane principal stresses are of the same

sign (see the solid circle), the maximum shear stress (shown at D') must be calculated based on

the knowledge that the third eigenvalue is zero; i.e. equal to one-half the stress at A. The

orientations of the maximum shear stress planes are shown in Figures 24 and 25.

Figure 23. A Mohr's circle for a point at which the principal stresses are of the same sign.

1.19

(1.9-1)

1.9 Stresses in Thin Walled Pressure Vessels

A thin walled pressure vessel provides an important application of the analysis of plane stress.

The following discussion will be confined to cylindrical and spherical vessels.

The cylindrical vessel shown in Figure 26 has inside radius r and wall thickness t. The vessel

contains a fluid under pressure, p.

Figure 26. A cylindrical pressure vessel.

The hoop stress, σ1, and the longitudinal stress, σ2, are given by:

The Mohr's circle corresponding to the cylindrical pressure vessel is shown in Figure 27.

Figure 27. Mohr’s circle for a cylindrical pressure vessel.

1.20

(1.9-2)

(1.9-3)

(1.9-4)

(1.9-5)

Note,

and

Figure 28, on the other hand shows a spherical pressure vessel of inside radius r and wall

thickness t, containing a fluid of gage pressure p.

Figure 28. A spherical pressure vessel.

In this case,

The Mohr's circle for the spherical pressure vessel is shown in Figure 29.

Note,

1.21

Figure 29. Mohr’s circle for a spherical pressure vessel.

1.22

1.10 Homework Problems

1. Draw a rectangular parallelepiped whose sides are aligned with a Cartesian axis system

passing through the point in question. Assign stresses on each of the six faces consistent

with convention.

Answer: Combine the stress states shown in Chapter 1, Figures 5 through 7.

2. Consider a small element of volume in a continuum. Assign the stresses and the body

force/unit volume which act in the Y direction. Use this figure as a basis and derive the

corresponding equilibrium equation (1.5-4) by summing forces along Y.

Answer: See Section 1.5.

3. Consider a small element of volume in a continuum. Assign the stresses and the body

force/unit volume which act in the Z direction. Use this figure as a basis and derive the

corresponding equilibrium equation (1.5-5) by summing forces along Z.

Answer: See Section 1.5.

4. The following stress distribution has been determined for a machine component:

Is equilibrium satisfied in the absence of body forces?

Answer: Expand expressions in Equation (1.5-6); yes.

5. The following stress distribution has been determined for a machine component:

Is equilibrium satisfied in the absence of body forces?

Answer: Expand expressions in Equation (1.5-6); yes.

6. If the state of stress at any point in a body is given by the equations:

1.23

What equations must the body-force intensities Fx, Fy, and Fz satisfy?

Answer: Apply expressions in Equation (1.5-6); Fx = - ( a + 2 p z ), Fy = - ( m + 2 e y ),

Fz = - ( l + 2 n x + 3 i z2 ).

7. If the state of stress at any point in a body is given by the equations:

What equations must the body-force intensities Fx, Fy, and Fz satisfy?

Answer: Apply expressions in Equation (1.5-6); Fx = - 4y (3y2+x), Fy = x, Fz = 0.

8. A two-dimensional state of stress (σzz = τzx = τzy = 0) exists at a point on the surface of a

loaded member. The remaining Cartesian components of stress are:

Using Mohr’s circle:

(a) Determine the principal stresses and sketch the element on which they act.

(b) What is the maximum shear stress at the point in question?

Answer: (a) σ1 = 95.14 MPa, σ2 = - 85.14 MPa, θp = - 9.72o (b) τmaximum = 90.14 MPa.



σxx = 90 MPa σyy = 40 MPa τxy = 60 MPa .




Answer: (a) σ1 = 130 MPa, σ2 = 0 MPa, θp = 33.7o (b) τmaximum = 65 MPa.




1.24

25o

8'

30"



Answer: (a) σ1 = 107 MPa, σ2 = 43 MPa, θp = - 19.3o (b) τmaximum = 53.5 MPa.






Answer: (a) σ1 = 11 ksi, σ2 = 1 ksi, θp = - 63.4o (b) τmaximum = 5.5 ksi.

12. A pressure tank is supported by two cradles as shown; one of the cradles is designed so that

it does not exert any longitudinal force on the tank. The cylindrical body of the tank is

fabricated from a 3/8" steel plate by butt welding along a helix which forms an angle of 25o

with respect to a transverse plane (the vertical direction). The end caps are spherical and

have a uniform wall thickness of 5/16". The internal gage pressure is 180 psi.

Figure 30. A pressure vessel with helical welds.

(a) Determine the normal stress and the maximum shearing stress in the spherical

caps.

(b) Determine the normal and shear stresses perpendicular and parallel to the

weld, respectively.

Answer: (a) σ = 4230 psi, τmaximum = 2115 psi, (b) σweld = 4137 psi, τweld = 1345 psi.

13. A pressure vessel of 10-in. inside diameter and 0.25-in. wall thickness is fabricated from a

4-ft section of spirally welded pipe. The vessel is pressurized to 300 psi and a centric axial

1.25

35o

10 kip

4 ft

A

B

compressive force of 10-kip is applied to the upper end through a rigid end plate.

Figure 31. A compressed pressure vessel with helical welds.

(a) Draw a stress element for a point located along the weld with horizontal and

vertical faces.

(b) Determine σ and τ in directions, respectively, normal and tangent to the weld.

(c) What is the maximum shear stress in the wall?

Answer: (a) σweld = 3.15 ksi, (b) τweld = - 1.993 ksi, (c) τmax = 3 ksi.

14. Square plates, each of 5/8” thickness, may be welded together in either of the two ways

shown to form the cylindrical portion of a compressed air tank. Knowing that the

allowable stress normal to the weld is 9 ksi, determine the largest allowable gage pressure

for each case.

Figure 32. Two different pressure vessels constructed from square plates.

Answer: Case No. 1: 62.9 psi, Case No. 2: 83.9 ksi.

2.1

X

Y

Z

O

d_

u_

v_

w_

CHAPTER 2 - STRAIN

2.1 Introduction

In Chapter 1, stress relationships were based on equilibrium. No restrictions were made

regarding the deformation of the body or the physical properties of the material from which it

was made. Strain is a purely geometric quantity and no restrictions need be placed on the

material of the body; however, restrictions must be placed on the allowable deformations in

order to linearize the strain-displacement relations. The type of material and/or material

constraints are introduced in the constitutive equations which relate stress to strain.

2.2 Strain-Displacement Equations

Displacement is a vector quantity which characterizes the movement of an arbitrary point. As

illustrated in Figure 1, each point has a displacement vector, d, which can be resolved along

Cartesian axes X,Y,Z such that the corresponding scalar components of displacement are u,v,w,

respectively.

Figure 1. The movement of a point is described by a displacement vector, d, which has scalar

components u,v,w.

The motion of the points in the body consists of two parts. Rigid body motion includes

translations and rotations during which there is no relative motion between neighboring points.

The movement of points relative to one another is called deformation.

Strain is a geometric quantity which describes the deformation which occurs in the body. There

are two types of strain. Shear strain is a measure of the angular change which takes place

between two line segments which were originally perpendicular. Since shear strain is expressed

in non-dimensional terms, the angle is measured in radians. Normal strain, on the other hand, is

defined as the change in length of a line segment divided by the original length of the segment.

2.2

(2.2-1)

(2.3-1)

(2.3-2)

(2.3-3)

(2.3-4)

The displacement components shown in Figure 1 are related to the strain components by the

strain-displacement equations which are given as follows:

2.3 Strain Equations of Transformation

The general transformation for a second order symmetric tensor illustrated by Equation (1.7-1)

can also be applied to transform strain provided that a slight modification is made when defining

the shear strain. To this end, the following definition is introduced:

In terms of Cartesian components, Equation (2.3-1) is written:

Then,

where εij' and εkl are the components of the rotated and original tensors, respectively; λik and λjl

correspond to the components in the matrix of direction cosines and its transpose, respectively.

From a rudimentary standpoint, the same relations developed in Section 1.7 for stress apply to

strain when the following substitutions are made:

Mohr's circle, for example, is plotted with ε as the abscissa and γ/2 as the ordinate.

2.3

(2.4-1)

2.4 Compatibility Equations

Equation (2.2-1) demonstrates that the strain field can be obtained from a given displacement

field by differentiating the displacement. However, if the strain field is measured, it becomes

necessary to integrate to obtain the displacement field. In this case, a unique displacement field

is obtained if and only if the body is simply connected (two points on the boundary can be

connected by never leaving the body or traversing it) and the strains satisfy compatibility

equations derived from the strain-displacement relations. The compatibility equations are:

2.5 Constitutive Equations

In general, stress is related to strain through 81 material constants. The linearity of stress versus

strain reduces this number to 36 while strain energy considerations further reduce the number to

21. If the body is linearly elastic, homogeneous and isotropic, the number reduces to 2.

A linearly elastic body is one which is made of a material which has a linear stress versus strain

curve. The body is homogeneous if it is made of the same material throughout. Isotropic refers

to the fact that a material property at a given point is independent of the direction in which it is

measured. For such a body, the equations which relate the stress to the strain are written in terms

of two independent material constants and are referred to as Hooke's laws. These constitutive

equations may be expressed for strain as a function of stress as:

2.4

(2.5-1)

(2.5-2)

or expressed for stress in terms of strain as:

2.5


1. The central portion of the tensile specimen shown in Figure 2 represents a uniaxial stress

condition; P is the applied load and A is the cross-sectional area. For a point located on the

surface and in the central portion of the specimen:

Figure 2.

(a) Draw a two-dimensional stress element having horizontal and vertical faces. Show the

stress distribution on this element. Are these the principal stress planes?

(b) Construct the Mohr’s circle for the point and draw a rotated element showing the

maximum in-plane shear stress. Be sure to label all stresses and specify the angle of

orientation with respect to, say, the horizontal direction.

(c) How does the maximum in-plane shear stress compare with the absolute maximum

shear stress at the point? (d) Develop the constitutive equation(s) for this uniaxial stress condition from the three

dimensional Hooke’s Law (see Eqn. 2.8-2).

(e) Develop the mathematical relationships for Young's Modulus and Poisson's Ratio and

describe how these quantities could be calculated by making physical measurements.

(f) Specify the six components of strain if the specimen is made of steel (E = 30 x 106 psi, ν

= 0.3) and loaded to yield (σy = 45 x 103 psi).

(g) What would be the angular orientation of the fracture surface with respect to a

horizontal axis if the material were ductile and failed along the planes of maximum shear?

What would happen if the material were brittle?

Answer: (a) σxx = 0, σyy = P/A, τxy = 0, yes, (b) element oriented at 45o with τ = σ = P/2A, (c)

equal, (d) σyy = Eεyy ..., (e) E = σyy/εyy, υ = -εxx/εyy = - εzz/εyy, measure strain and

compute stress, (f) εxx = εzz = - 450 με, εyy = 1500 με, γxy = γxz = γyz = 0 με, (g) ductile

@ 45o, brittle @ 0

o.

2.6

A

B

_T

_T’

C

ττ

Axis of shaft

Axis of shaft

2. Figures 3 and 4 show that when a plane is passed through a circular shaft perpendicular to its

longitudinal axis, a state of pure shear exists.

Figure 3. Figure 4.

(a) Specify the formula used to determine the shear stress and define all terms. Draw a

diagram showing how the stress varies over the circular cross section. (b) Express the magnitude of the maximum shear stress at the surface in terms of the applied

torque and the radius.

(c) Assuming that the shaft is oriented so that its longitudinal axis is horizontal, draw a two-

dimensional stress element having horizontal and vertical faces, and show the stress

distribution on this element. Are these the principal stress planes? (d) Construct the Mohr’s circle for the point and draw a rotated element showing the

principal stresses. Be sure to label all stresses and specify the angle of orientation with

respect to, say, the horizontal direction.

(e) What would be the angular orientation of the fracture surface with respect to a

horizontal axis if the material were ductile and failed along the planes of maximum shear?

What would happen if the material were brittle?

Answer: (a) τ = Tρ/J, (b) τ = 2T/πc3, (c) σxx = σyy = 0, τxy = 2T/πc

3, no, (d) element oriented at

45o with σ = 2T/πc

3, (e) ductile @ 90

o, brittle @ 45

o.

3.1

Y

X Z

Y

O

CHAPTER 3 – LINEAR ELASTICITY

3.1 Overview

At each point in a loaded body, there are 15 unknowns: 3 displacements (u,v,w), 6 strains

(εxx,εyy,εzz,γxy,γxz,γyz) and 6 stresses (σxx,σyy,σzz,τxy,τxz,τyz). A complete solution to the general

elasticity problem must satisfy 15 equations (3 equilibrium, 6 compatibility and 6 constitutive)

plus boundary conditions formulated in terms of displacement and/or traction.

In general, it is possible to obtain a closed-form solution only for problems having relatively

simple loading and geometry. In more complex situations, it is necessary to resort to an

analytical formulation using a technique such as finite element analysis. In this method, the

body is mathematically modeled by partitioning it into a finite number of elements connected

together at points called nodes. Displacements and/or tractions are specified at the boundary and

a mathematical wave front is passed over the mesh. Stresses are predicted based on either a

flexibility or stiffness approach.

An alternative solution to the problem is to experimentally measure the displacements, strains or

stresses. When displacements are measured, the strain-displacement equations are used to

determine the strains. Stresses are obtained using the constitutive equations, after which the

transformation equations may be applied to obtain the principal stresses and the maximum shear

stress. When strains are measured, the compatibility equations must be satisfied to obtain a

unique displacement field.

3.2 Plane Stress

The geometry of the body and the nature of the loading allow two important types of problems to

be defined. Figure 1 illustrates an example of plane stress in which the geometry is essentially a

flat plate with a thickness much small than the other dimensions. The loads applied to the plate

act in the plane of the plate and are uniform over the thickness.

Figure 1. An example of plane stress.

3.2

(3.2-1)

(3.2-3)

(3.3-1)

(3.2-2)

Mathematically, for a plate contained in the XY plane,

.

Although the normal stress perpendicular to the surface vanishes, the normal strain, εzz, does not.

The relationship between the latter and the other two in-plane strain components is found by

setting σzz = 0 in the expression given for it in Equation (2.5-2). This results in:

An alternate form of Hooke’s law for plane stress is found by substituting Equation (3.2-2) into

the remaining expressions contained in Equation (2.5-2) as

3.3 Plane Strain

Figure 2, on the other hand, illustrates the condition of plane strain. In this case, the body is

considered a prismatic cylinder with a length much larger than the other dimensions. The loads

applied to the cylinder are distributed uniformly with respect to the large dimension and act

perpendicular to it.

Mathematically, for a point in the XY plane,

.

3.3

Y

X

Z

O

(1)

(2)

(3)

Example: A flat plate (located in the XY plane of a Cartesian coordinate system, with

normal along Z) was designed and built as part of a compressor system using steel

(Elastic Modulus = 30 x 106 psi; Poisson's Ratio = 0.3). The allowable shear

stress of the material is 12.5 ksi but the manufacturer wants a factor of safety of

2.0 to be incorporated into the design.

Preliminary tests conducted by the manufacturer indicated that the most critical location on the

structure was located at x = 1.0, y = 2.0, z = 0.0. This point was on a traction free surface where

the in-plane displacements were found to be expressed as follows:

(a) Determine the magnitudes and directions of the principal stresses at the critical location.

Sketch your results on a rotated element and label the stress distribution. Specify the angle

of orientation of the element with respect to the x direction.

(b) Does the part meet the required specifications?

Solution: The point is in plane stress and the in-plane strain components may be obtained

using the strain-displacement equations given by Equation (2.2-1) as

For plane stress, σzz = 0, and

Figure 2. An example of plane strain.

3.4

(4)

(5)

(6)

The in-plane stresses are computed using Equation (2.5-2) [or Equation (3.2-3)] as

Figure 3 shows the corresponding stress element; and, Figure 4 shows the Mohr’s circle for the

stress distribution at the critical point.

The circle is drawn by plotting the two points corresponding to the planes labeled as 1 and 2 on

Figure 3 in σ-τ space. The points are connected and a circle is drawn centered at C with radius r.

The location of the center of the circle is computed from Equation (1.7-9) as:

The radius is computed from Equation (1.7-10) as:

The maximum and minimum values of stress are determined by adding and subtracting the

magnitude of the radius to the value of σ at point C, respectively. This operation produces:

Figure 3. Stress element. Figure 4. Mohr’s circle.

3.5

(7)

(8)

(9)

(10)

σ

σ .

The orientations of the principal planes are obtained by determining the orientation of the point

labeled 3 on Figure 4 with respect to the point labeled 1. Referring to the figure:

Note that on the circle, 2α is drawn clockwise from point 1 to point 3. Also note that point 3

corresponds to σmax. Therefore, the normal to the plane corresponding to σmax lies at σ = 10.9o

measured in the counterclockwise direction from the normal to the plane labeled as 1 in Figure

3. These results are summarized in Figure 5 which shows the rotated element corresponding to

the principal planes.

The in-plane shear stress is equal in magnitude to the radius. However, since the circle does not

encompass the origin, the absolute maximum shear stress is equal to one-half of the maximum

value,

The allowable stress, on the other hand, is found by dividing the maximum shear stress that the

material can take by the factor of safety,

Figure 5. Principal planes.

3.6

εxx, εyy, εzz, γxy, γxz, γyz

σxx, σyy, σzz, τxy, τxz, τyz

u, v, w

σ1, σ2, σ3

1

2

3

4

5 6

7

8

Since the maximum stress is greater than the allowable stress, the part does not meet the design

specifications.


1. Figure 6 shows a flowchart for the solution to the general elasticity problem. Add an

appropriate description to the items numbered 1 through 8.

Answer: 1. displacements; 2. strain, 3. stress, 4. principal stress, 5. compatibility equations, 6.

strain-displacement equations, 7. constitutive equations, and 8. transformation

equations.

2. Assume that a state of plane stress exists such that σzz = τyz = τzx = 0.

(c) Reduce the equilibrium equations given in Equation (1.5-6) for plane stress.

(d) Reduce the constitutive equations given in Equation (2.5-1) for plane stress.

(e) Knowing σzz = 0, develop an expression for εzz = f (εxx, εyy) for plane stress and show by

substituting this expression into Equation (2.5-2) that the constitutive equations become:

Answer: (a) ∂σxx/∂x + ∂τyx/∂y + Fx = 0 ..., (b) εxx = 1/E [σxx - ν σyy] ..., (c) εzz = - ν/(1-ν) [εxx +

εyy] ...

Figure 6. The general elasticity problem.

3.7

3. A critical point on the surface of a part fabricated using steel (E = 30 x 106 psi; ν = 0.3) is

under plane stress conditions with σzz = τyz = τzx = 0. Knowing that εxx = 505 x 10-6

in/in [or,

505 με], εyy = 495 x 10-6

in/in [or, 495 με] and γxy = 100 x 10-6

in/in [or, 100 με], determine

the

(a) strain εzz.

(b) principal stresses and their orientation with respect to the XYZ axes system.

(c) maximum shear stress at the point.

Answer: (a) εzz = - 428.6 με, (b) 22.59 ksi, 20.27 ksi, 0 ksi, φx1 = 42o (c) τmax = 11.3 ksi.

4.1

E

Hz

(4.1-1)

(4.1-2)

CHAPTER 4 – LIGHT AND ELECTROMAGNETIC WAVE

PROPAGATION

4.1 Introduction to Light

Electromagnetic radiation is predicted by Maxwell's theory to be a transverse wave motion

which propagates in free space with a velocity of approximately c = 3 x 108 m/s (186,000 mi/s).

The wave consists of oscillating electric and magnetic fields which are described by electric and

magnetic vectors E and H, respectively. These vectors are in phase, perpendicular to each other,

and at right angles to the direction of propagation, z.

A simple representation of the electric and magnetic vectors associated with an electro-magnetic

wave at a given instant of time is illustrated in Figure 1. As described below, the wave is

assumed to have a sinusoidal form and may be characterized by either the electric or magnetic

vector. In the arguments that follow, light propagation is described in terms of the electric

vector, E.

Figure 1. An electromagnetic wave.

A mathematical representation for the magnitude of the electric vector as a function of position,

z, and time, t, can be determined by considering the one-dimensional wave equation

and setting E = u. The De’Lambert solution to this partial differential equation is

where z is the position along the axis of propagation and t is time.

In Equation (4.1-2), the quantity f (z - ct) represents a wave moving at velocity c in the positive z

direction while g (z + ct) represents a wave moving at velocity c in the negative z direction.

4.2

(4.1-3)

a

E

z

E = a cos 2π/λ (z - ct)

ct λ

t = 0

t > 0

(4.1-4)

(4.1-5)

(4.1-6)

Many of the optical effects of interest in experimental mechanics and optical metrology can be

described by considering a harmonic vibration propagating along positive z such that

where a is the instantaneous amplitude, λ is the wavelength, and 2π/λ is the wave number.

Equation (4.1-3) neglects the attenuation associated with the expanding spherical wavefront and

is rigorously valid only for a plane wave.

Figure 2 shows a graphical representation of the magnitude of the light vector as a function of

position along the positive z axis, at two different times. The wavelength, λ, is defined as the

distance between successive peaks. The time required for the wave to propagate through a

distance equal to the wavelength is called the period, T.

Figure 2. A wave, originally at time t = 0, propagates through space.

Since the wave is propagating with velocity, c,

The angular frequency of the wave is defined as

where the frequency,

represents the number of oscillations which take place per second.

4.3

(4.1-7)

The above definitions make it possible to express E(z,t) in several ways. For example,

4.2 Electromagnetic Spectrum

Although the electromagnetic spectrum has no upper or lower limits, the radiations commonly

observed have been classified in the table shown in Figure 3.

Figure 3. The electromagnetic spectrum.

The visible range of the spectrum is centered about a wavelength of 550 nm and extends from

approximately 400 to 700 nm. Figure 4 lists the different colors observed by the human eye as a

function of wavelength.

Figure 4. The visible spectrum.

4.4

(4.3-1)

E1 = a cos 2π/λ (z - ct + δ1)

E2 = a cos 2π/λ (z - ct + δ2)

E = a cos 2π/λ (z - ct)ct

δδ2

δ1

a

E

z

(4.3-2)

The characteristics of an electromagnetic wave such as color depend upon the frequency; since,

the frequency is independent of the medium through which propagation occurs. The wavelength,

λ, on the other hand, depends upon the medium; since the velocity of propagation changes from

medium to medium (see Section 4.9).

When the light vectors corresponding to E(z,t) are all of one frequency, the light is referred to as

monochromatic. Light consisting of several frequencies is recorded by the eye as white light.

4.3 Light Propagation, Phase, and Retardation

In Equation (4.1-3) [the first form of E(z,t) shown in Equation (4.1-7)], the term in the argument

of the cosine function represents the angular phase, α, while the term in the parenthesis

corresponds to the linear phase, δ. It is apparent that these quantities are related through the

wave number as follows:

Figure 5, for example, shows two waves having the same wavelength, λ, and equal amplitudes.

In this case, the waves are plotted as a function of position, along the axis of propagation.

Figure 5. Two waves out of phase.

Assuming that each wave has different amplitude given by a1 and a2, respectively,

4.5

(4.3-3)

Ellipical helixLight vector

z

x

y

a

b

Circular helix

Light vector

z

x

y

a

where δ1 (α1) and δ2 (α2) are the initial linear (angular) phases of E1 and E2, respectively. The

linear phase difference, δ, is given by

Since one wave trails the other, the linear phase difference is often referred to as the retardation.

4.4 Polarized Light

Most light sources consist of a large number of randomly oriented atomic or molecular emitters.

The rays emitted in any direction from such sources will have electric fields that have no

preferred orientation. In this case, the light beam is said to be unpolarized.

If, however, a light beam is made up of rays with electric fields that show a preferred direction of

vibration, the beam is said to be polarized. Figure 6, for example, shows the most general case

of elliptical polarized light where the tip of the electric vector describes an elliptical helix as it

propagates along z.

Figure 6. Elliptically polarized light. Figure 7. Circularly polarized light.

Figure 7 shows a more restrictive case where the tip of the light vector describes a circular helix

as it propagates along z. This condition, referred to as circularly polarized light, is studied in

Section 4.8. It is important because it will be encountered in Chapter 5 when studying the

circular polariscope in conjunction with the method of photoelasticity.

The most restrictive case, illustrated in Figure 8, occurs when E(z,t) vibrates in a single known

direction. This condition, referred to as linearly polarized light, is perhaps the most important

case as far as experimental mechanics is concerned. The figures used to characterize light in the

beginning of this chapter, for example, were drawn by making the assumption that the light

vector was linearly polarized. This state of polarization can be produced by using sheet-

polarizers, a Nicol prism, or, by reflection at the Brewster angle.

4.6

Plane of polarization

z

x Tip sweeps out a sine curve

(4.6-1)

Figure 8. Linearly polarized light.

4.5 Optical Interference

The light emitted by a conventional light source, such as a tungsten-filament light bulb, consists

of numerous short pulses originating from a large number of different atoms. Each pulse

consists of a finite number of oscillations known as a wave train. Each wave train is thought to

be a few meters long with duration of approximately 10-8

s. Since the light emissions occur in

individual atoms which do not act together in a cooperative manner, the wave trains differ from

each other in the plane of vibration, frequency, amplitude, and phase. Radiation produced in this

manner is referred to as incoherent light and the wavefronts simply add as scalar quantities.

For other light sources, such as a laser, the atoms act cooperatively in emitting light and produce

coherent light. In this case, the wave trains are monochromatic, in phase, linearly polarized, and

extremely intense. When the light coming from these sources is split and later recombined, the

wavefronts superimpose as vectors to produce optical interference.

Although coherent light must be monochromatic (of a single frequency) to interfere, light

coming from two different monochromatic sources is not, in general, coherent. In most cases,

interference will not take place unless the wavefronts in question originate from the same source.

4.6 Complex Notation

A convenient way to represent both the amplitude and phase of the light wave described by

Equation (4.1-3) for calculations involving a number of optical elements is through the use of

complex or exponential notation. To this end, the Euler identity is

where i2 = -1. The first term on the right hand side is called the real part of the complex function

whereas the term associated with “i” is the imaginary part.

Using this notation, the sinusoidal wave characterized by Equation (4.1-3) can be represented by

4.7

(4.6-2)

(4.6-3)

(4.6-4)

(4.7-1)

(4.7-2)

(4.7-3)

When the operations performed are linear, the symbol Re is dropped and calculations are

performed using the complex function

where α and δ are the angular and linear phase, respectively.

If the waves have an initial phase, such as those described by Equation (4.3-2), they can be

expressed as

4.7 Intensity

In many practical applications, the light vector is related to a measurable quantity called the

intensity where

Assuming that k = 1, and using the form of E(z,t) given in Equation (4.6-3),

where E(z,t)* is the complex conjugate of E(z,t) and the operation is a dot product.

E(z,t)* is obtained from E(z,t) by changing the sign of the imaginary part of the latter which

amounts to changing the sign in the exponent of the exponential function. When referring to a

single wavefront, the dot product operation becomes a simple multiplication and

Equation (4.7-3) shows that the intensity of light is proportional to the square of the

instantaneous amplitude of the light vector.

4.8

(4.8-1)

(4.8-2)

(4.8-3)

(4.8-4)

(4.8-5)

4.8 Superposition of Wavefronts

As mentioned in Section 4.5, incoherent wavefronts superimpose as if they were scalars.

Assuming that two waves are represented by vectors E1 and E2, the resulting intensity for

incoherent waves is simply

If the wavefronts are coherent, the light vectors add as vectors. In this case, E1 and E2 can be

combined into a resultant ER and the intensity is

Consider, for example the superposition of two coherent (monochromatic; same frequency),

rectilinear harmonic vibrations (plane polarized wavefronts) that are of unequal amplitude and

out of phase. In this case, the expressions in Equation (4.6-4) can be used to characterize the

wavefronts; and, Equation (4.8-2) represents the intensity.

Since the two wavefronts are co-linear, the dot product becomes a simple multiplication, and

Equation (4.8-2) can be expanded as follows:

From complex variables, the last term in brackets on the right hand side of Equation (4.8-3) is

proportional to the cosine of the argument in the exponential function and the intensity becomes

where the term in the brackets of Equation (4.8-4) represents the difference in angular phase of

the wavefronts.

Two cases are of special interest. In the case in where the angular phase difference is a multiple

of π, Equation (4.8-4) becomes,

If a1 = a2, then I = 0. In this case, there is no measurable intensity; implying that light plus light

results in darkness. Destructive interference is said to take place.

4.9

(4.8-6)

Ex = ax cos 2π/λ (z - ct + δx)Ex

Ey

δ

ax

ay

Ey = ay cos 2π/λ (z - ct + δy)

(4.8-7)

(4.8-8)

If, on the other hand, the angular phase difference is either zero or a multiple of 2π, then

Equation (4.8-4) becomes

If a1 = a2, then I = 4a2, thereby resulting in constructive interference.

Thus, the intensity observed when two rectilinear coherent wavefronts are superimposed depends

upon the amplitudes of the wavefronts and the relative difference in phase between them. In the

case considered above, the electric vector is restricted to a single plane and the superposition

results in a plane polarized, or, linearly polarized light. Interference effects have important

applications in optical metrology; for example, in photoelasticity, moiré, holography, and

speckle metrology.

Two other important forms of polarized light, previously discussed in Section 4.4, arise as a

result of the superposition of two linearly polarized light waves having the same frequency but

mutually perpendicular planes of vibration. This condition is depicted in Figure 9.

Figure 9. Superposition of two wavefronts having perpendicular planes of vibration.

In this case, at a fixed position along the axis of propagation, the magnitude of the light vector

can be expressed as

where α is the angular phase and ω is the angular velocity at which the electric vector rotates. To

achieve a better understanding of this case, the vector can be resolved into components along the

x- and y-axes, and the wavefronts shown in Figure 9 represented by

4.10

(4.8-9)

(4.8-10)

(4.8-11)

(4.8-12)

Ellipical helix

Light vector E

z

x

y

where αx and αy are the phase angles associated with waves in the xz and yz planes, and ax and ay

are the amplitudes of waves in the xz and yz planes.

The magnitude of the resulting light vector is given by vector addition as

Considerable insight into the nature of the light resulting from the superposition of two mutually

perpendicular waves is provided by studying the trace of the tip of the resulting electric vector on

a plane perpendicular to the axis of propagation at points along the axis of propagation. An

expression for this trace can be obtained by eliminating time from Equation (4.8-8). Pursuing

this argument in terms of angular phase results in the relation

Or, since

Equation (4.8-10) becomes

Equation (4.8-12) is the equation of an ellipse, and light exhibiting this behavior is known as

elliptically polarized light.

As illustrated in Figure 10, tips of the electric vectors at different positions along the z axis form

an elliptical helix. As mentioned previously, the electric vector rotates with an angular velocity,

ω. During an interval of time t, the helix will translate a distance z = ct in the positive direction.

As a result, the electric vector at position z will rotate in a counter-clockwise direction as the

translating helix is observed in the positive z direction.

Figure 10. Elliptically polarized light.

4.11

(4.8-13)

(4.9-1)

(4.9-2)

A special case of elliptically polarized light occurs when the amplitudes of the two waves are

equal [ax = ay = a] and α = ± π/2 [or, δ = ± λ/4]. In this case, Equations (4.8-10) and (4.8-12)

reduce to the equation of a circle given by

As mentioned previously in Section 4.4, light exhibiting this behavior is known as circularly

polarized light, and the tips of the light vectors form a circular helix along the z axis. For α = π/2

[δ = λ/4], the helix is a left circular helix, and the light vector rotates counter-clockwise with time

when viewed from a distant position along the z axis. For α = - π/2 [δ = - λ/4], the helix is a right

circular helix, and the electric vector rotates clockwise.

4.9 Reflection and Refraction

The discussions so far have been limited to light propagating in free space. Most optical effects

of interest, however, occur as a result of the interaction between a beam of light and some

physical material.

As mentioned in the beginning of the chapter, light propagates in free space with a velocity c

equal to approximately 3 x 108 m/s (186,000 mi/s). In any other medium, the velocity is less

than the velocity in free space.

The ratio of the velocity in free space to the velocity in a medium, v, is called the index of

refraction, n, of the medium. The latter is defined as

The index of refraction for most gases is only slightly greater than unity (for air, n = 1.0003).

Values for liquids range from 1.3 to 1.5 (for water, n = 1.33) and for solids range from 1.4 to 1.8

(for glass, n = 1.5). The index of refraction for a material is not constant but varies slightly with

wavelength of the light being transmitted. This dependence of index of refraction on wavelength

is referred to as dispersion.

Figure 11 illustrates that when a beam of light strikes a surface between two transparent

materials with different indices of refraction, it is divided into a reflected ray and a refracted ray.

The reflected and refracted rays lie in the plane formed by the incident ray and the normal to the

surface, known as the plane of incidence. The angle of incidence, α, the angle of reflection β, and

the angle of refraction γ are related. For reflection, light bounces off the surface with

Refraction is governed by Snell’s law and can be quantified in terms of the velocity of

propagation and/or index of refraction of the media as follows:

4.12

(4.9-3)

(4.10-1)

(4.10-2)

(4.10-3)

Reflected rayIncident ray

Wavefront

Material 1

Material 2

Refracted ray

α β

γ

Figure 10. Reflection and refraction.

4.10 Double Refraction - Birefringence

Light traveling through an isotropic medium propagates with the same velocity in all directions;

however, when light enters an anisotropic or crystalline medium along an optical axis it is

divided or refracted into two plane polarized components which are orthogonal and are

propagating at different velocities. This phenomenon is known as double refraction or

birefringence, and occurs in photoelastic models subjected to stress.

'The angular phase difference that occurs between these two wavefronts when light passes

through a plate of thickness d is

where λ is the wavelength and n is the index of refraction.

The corresponding linear phase difference, δ, is given by Equation (4.3-1); and,

The fringe order, R, to be defined later in conjunction with the method of photoelasticity,

provides a measure of the relative retardation as follows:

4.13

(4.10-4)

(4.10-5)

(4.10-6)

The difference in indices of refraction is proportional to the difference in principal stress; in

particular,

where C is the stress optic coefficient.

A more useful description of this phenomenon is given by the relation

where c is the velocity of propagation outside the model (usually air, n = l), d is the thickness of

the model, R is the fringe order, F = fσ/c is the model fringe value, and fσ = λ/c is the material

fringe value.

Equation (4.10-5) is used to determine the stress state in a loaded photoelastic model from the

analysis of the isochromatic fringe pattern. As illustrated in Chapter 5, the most commonly used

form of this equation is

where d is found by direct measurement, R is determined from the photoelastic fringe pattern,

and fσ is determined by performing a calibration test on a sample of the photoelastic material in

question.

4.14

E1 = a cos 2π/λ (z - ct + δ1)

E2 = a cos 2π/λ (z - ct + δ2)

E = a cos 2π/λ (z - ct)ct

δδ2

δ1

a

E

z


1. Figure 12 shows two coherent (monochromatic; same frequency), rectilinear harmonic

vibrations (plane polarized wavefronts) which have unequal amplitude and are out of phase.

(a) Express the magnitudes of these light vectors in complex notation using the linear phase

descriptions labeled in the figure.

(b) Using complex notation, derive the intensity distribution resulting from their

superposition.

Hint: See Equation (4.6-4); apply Equation (4.8-2), and follow the arguments used to derive

Equation (4.8-4).

2. Analyze the intensity distribution derived in Problem 1 and specify the linear phase and any

other conditions that must occur to produce destructive and constructive interference.

Quantify the magnitude of the intensity for each of these conditions.

Hint: See the arguments made in Section 4.8. The results can be checked by applying Equation

(4.3-1).

Figure 12.

5.1

CHAPTER 5 - PHOTOELASTICITY

5.1 Introduction

Photoelasticity has long been a widely used experimental technique for determining stresses and

strains. The technique is applicable to any state of stress, but historically it has been for the most

part limited to the determination of static two-dimensional stress distributions. Later

developments led to its application to three dimensional stress problems, two dimensional

dynamic problems including both cyclic and transient, and some plasticity and viscoelastic

problems.

The technique is limited to a class of materials which are birefringent; however, the technique

can be applied to other materials and actual prototypes by cementing films or thin sheets of

photo-elastic material directly to the surface of the specimen. These techniques are covered in

this chapter.

The photoelastic technique is based upon the stress-optic law (see Section 4.10) that applies to

many transparent materials when they are stressed. Depending on the material, the phenomena

may be due to stress or strain, or both. If the material is linear and elastic the effect can be

referred to either stress or strain. If the material is viscoelastic both the strain-optic and the

stress-optic relationships must be known.

The materials in question exhibit a temporary birefringence when they are under a temporary

load. The principal optical axes coincide when the directions of the principal stresses and strains

are linearly related to the refractive indices along their two respective principal axes. The result

is that if a plane-polarized light wave enters a birefringent material under load, it will be resolved

into two plane-polarized waves that propagate through the stressed material at different

velocities. When the two waves emerge from the material they are out of phase, where the phase

difference is proportional to the difference between the principal stresses and the thickness of the

material. A detailed analysis is included herein.

Photoelastic materials that exhibit birefringence are commonly analyzed with respect to their

stress distributions in a device called a polariscope. The simplest type of polariscope is a plane

polariscope which makes use of two polarizers whose axes are crossed with respect to each

other. Light is passed through the first polarizer and then through the loaded specimen where it

is resolved into two plane-polarized waves which are analyzed by the second polarizer. As

discussed in Section 5.2, the resulting pattern is an isochromatic fringe pattern superimposed

upon an isoclinic fringe pattern. Isoclinic fringes are the loci of points where the principal stress

directions coincide with the axis of the polarizer and analyzer; isochromatic fringes are the loci

of constant principal stress difference.

A widely used polariscope is the circular polariscope. It has the advantage that the isoclinic

fringe pattern is eliminated leaving only the isochromatic fringes and is discussed in Section 5.3.

5.2

Principal stressdirection σ1

Principal stressdirection σ2

Analyzer

Axis ofpolarization

Axis ofpolarization

Model

Polarizer

Lightsource

z

θ

(5.2-1)

5.2 Plane Polariscope

As mentioned above, many transparent noncrystalline materials which are optically isotropic

when free of stress become optically anisotropic and display characteristics similar to crystals

when they are stressed. These characteristics persist while loads on the material are maintained

but disappear when the loads are removed. This behavior is known as temporary double

refraction or birefringence.

The state of stress in birefringent materials can be analyzed by using a polariscope. Figure 1, for

example, shows a plane polariscope which consists of a light source (white or monochromatic)

and two linear polarizers called the polarizer and the analyzer. The configuration shown in the

figure is referred to as a dark field polariscope because the crossed polarizers result in a dark

background.

Figure 1. A dark field plane polariscope.

Light coming from the source passes through the first linear polarizer (called the polarizer)

shown in the figure with its plane of vibration in the vertical direction.

In accordance with the arguments presented in Chapter 4, the amplitude of the light leaving the

polarizer can be expressed represented by its electric vector, E1. The magnitude of the latter can

be expressed as a rectilinear harmonic vibration of the form

It should be noted that the ensuing argument could just as easily have been pursued using the

cosine function as illustrated in Experimental Stress Analysis by Dally and Riley, 4th

Edition,

College House Enterprises, 2005, ISBN 0-9762431-0-7.

A doubly refracting plate, labeled as the model in Figure 1, is located between the polarizer and

the analyzer with one of its principal axes making an angle θ with respect to the horizontal.

5.3

(5.2-2)

(5.2-3)

(5.2-4)

σ1

σ2

E1

E3

E2 θ

Figure 2 illustrates that when E1 enters the model, it is resolved into two vector components, E2

and E3 with magnitudes

Figure 2. Linearly polarized light coming from the polarizer is resolved along the principal

stress directions in the model.

The orientation of these components remains constant as these plane waves pass through the

plate; since, the model is assumed to be in a state of plane stress. But, because the model is

birefringent and each of the waves propagate with a different velocity as they travel through the

thickness, d, the components emerge with an angular phase difference, α, given by

where λ is the wavelength.

Assuming that E4 and E5 correspond to the wavefronts which emerge as a result of E2 and E3,

respectively,

The wavefronts continue to propagate until they reach the analyzer; another linear polarizer

oriented with its plane of vibration in the horizontal direction.

As illustrated in Figure 3, the analyzer allows only the horizontal components of the vectors E4

and E5 to pass.

5.4

E5

E4 θ

θ

(5.2-5)

(5.2-6)

(5.2-7)

(5.2-8)

Figure 3. The two components of light emerging from the model are resolved by the analyzer.

The resulting wavefront is characterized by a resultant vector, say E6, with magnitude

Equation (5.2-5) can be simplified somewhat by using the following trigonometric identities:

Applying the identities in Equation (5.2-6) to the relation in Equation (5.2-5),

In addition to the identities stipulated in Equation (5.2-6), the following relations hold true:

Applying the latter to the expression contained in Equation (5.2-7),

5.5

(5.2-9)

(5.2-10)

(5.2-11)

(5.2-12)

(5.2-13)

By recognizing that

Equation (5.2-9) can be written as

where

The term in Equation (5.2-12) is the instantaneous amplitude of the emerging beam; and, a

measure of the intensity, observed by looking through the analyzer, is given by the square of its

amplitude:

There are two conditions of extinction. The first occurs when sin2 2θ = 0. This happens when 2θ

= nπ, where n is an integer; or, when θ = nπ/2. For n = 0 and n = 1, θ = 0 and θ = π/2,

respectively.

For the configuration studied in Figure 1, θ represents the orientation of the principal stress

measured counterclockwise from the horizontal. Thus, the critical values found for θ coincide

with the directions of the polarizer and analyzer. The implication is that all points which have

their principal stress direction aligned with the polarizing elements appear dark. The loci of all

such points create families of fringes called isoclinics.

The second condition of extinction occurs when sin2 α/2 = 0. This occurs when α/2 = nπ, where

n is an integer; or, when α = 2nπ. The integer multiples of 2π correspond to linear phase

differences which are integer multiples of the wavelength, λ. Thus, extinction also occurs when

linear phase changes by an integral number of wavelengths. An isochromatic fringe pattern

5.6

(5.2-14)

σ2

AnalyzerModel

Polarizer

Lightsource

z

2nd λ/4 plate

F S

FS

A

P

σ1

β

1st λ/4 plate

π/4

results in which the fringe order number, R, is related to the principal stress difference by the

equation developed in Section 4.10. In particular,

In Equation (5.2-14), R is the fringe order number (measured at a point in the isochromatic fringe

pattern), d is the thickness of the model (found by direct measurement), and fσ is the material

fringe value (determined by performing a calibration test on a sample of the photoelastic material

used to construct the model). Integer order fringes correspond to the color of the background.

In conclusion, two families of fringes result when a loaded model is studied with a plane

polariscope: isoclinic fringes represent the loci of points which have their principal stress

directions aligned with the polarizer and analyzer; isochromatic fringes represent the loci of

points which have a constant principal stress difference. Since the difference between the two

principal stresses is twice the in-plane shear stress, the isochromatic pattern also represents the

loci of constant shear stress.

5.3 Circular Polariscope

The isoclinics are not always needed in photoelastic analysis and often obscure the isochromatics

to the point where a stress analysis becomes impossible. Figure 4 shows a device called a

circular polariscope which is used to eliminate the isoclinics.

Figure 4. A dark field circular polariscope.

The dark field polariscope depicted in the figure represents one of many possible configurations

for a circular polariscope. In this case, light comes from the source and passes through the first

5.7

(5.3 -1)

F

E1

E3

E2450

S

(5.3-2)

(5.3-3)

polarizer with its plane of vibration in the vertical direction. The amplitude of the light vector

leaving the polarizer can be represented by

As illustrated in Figure 4, the light hits a quarter wave plate having its fast and slow axes

oriented at 450 with respect to the axis of the polarizer.

The plate is made of a birefringent material and is produced so that every point has the same

principal stress directions. Thus, the incoming wave is split into two equal components which

are plane polarized at 900. The components are represented by E2 and E3 in Figure 5.

Figure 5. Linearly polarized light coming from the polarizer is resolved along the fast and slow

axes of the first quarter wave plate.

The magnitudes of these vectors are

where b = a/ √2.

The thickness of the quarter wave plate is adjusted so that after the waves travel through the plate

and emerge, the one along the fast axis leads the other by λ/4; a linear phase shift corresponding

to an angular phase shift of π/2. Since E2 is assumed to lie along the fast axis, the waves coming

out of the plate are represented by

where the out-coming waves E4 and E5 correspond to the incoming waves E2 and E3,

respectively.

5.8

E5

E4

450

ββ

x

y

F S

(5.3-4)

(5.3-5)

Although the above argument was made by adding the angular phase into the argument of the

expression for E4, this could have been done by including -π/2 into the argument of E5.

The two waves continue to propagate along their respective directions until they encounter the

model. Referring to the argument made in Section 4.8, the model is illuminated with circularly

polarized light; that is, the components in Equation (5.3-3) are plane polarized in perpendicular

directions, in phase, of equal amplitude, and have a relative phase difference of π/2.

As illustrated in Figure 6, it is assumed that, for the point in question, one of the principal axes

makes an angle of β with respect to the fast axis of the first quarter wave plate.

Figure 6. The two light vectors exiting the first quarter wave plate are resolved along the

principal directions in the model.

Light is resolved into vector components along directions x and y; and, if these components are

denoted by E7 and E6, respectively,

The orientation of these components remains the same as these plane waves pass through the

plate; since, the model is assumed to be in a state of plane stress. But, because the model is

birefringent and each of the waves propagate with a different velocity as they travel through the

thickness, d, the components emerge with an angular phase difference, α, given by Equation

(5.2-3). Assuming that E8 and E9 correspond to the wavefronts which emerge as a result of E6

and E7, respectively,

5.9

E9

E8

450

ββ

x

y

S F

(5.3-6)

(5.3-7)

(5.3-8)

The resulting vectors are now aligned along the principal stress directions and propagate until

they reach the second quarter wave plate. As illustrated in Figure 7, the second quarter wave

plate is oriented having its fast and slow axes crossed with respect to the first quarter wave plate.

Figure 7. The two components of light emerging from the model are resolved along the fast and

slow axes of the second quarter wave plate.

When the vectors E8 and E9 reach the plate, they are resolved into vector components E10 and E11

along the slow and fast axis, respectively, where

These beams emerge from the quarter wave plate with an additional phase difference equal to π/2

which can be added into the argument of E11. Assuming that vectors E12 and E13 correspond to

the out-coming wavefronts associated with the incoming wavefronts E10 and E11, respectively,

The analyzer allows only the horizontal components of these vectors to pass. Since both vectors

make an angle of 450 with respect to the analyzer, the resulting vector, E14, is

5.10

(5.3-9)

(5.3-10)

(5.4-1)

where

The term in Equation (5.3-9) is the instantaneous amplitude of the emerging beam; and, a

measure of the intensity (observed by looking through the analyzer) is given by the square of its

amplitude:

There is only one condition of extinction corresponding to sin2 α/2 = 0. This occurs when α/2 =

nπ, where n is an integer; or, when α = 2nπ. The integer multiples of 2π correspond to linear

phase differences which are multiples of the wavelength, λ. Thus, extinction occurs when the

linear phase changes by an integral number of wavelengths. The isochromatic fringe pattern

which results is governed by Equation (5.2-14). As mentioned previously, integer order fringes

correspond to the color of the background.

In conclusion, isochromatic fringes are observed when a loaded model is studied with a

circular polariscope. The isoclinic pattern is eliminated, since the model is illuminated with

circularly polarized light.

5.4 Calibration Methods for Determining the Material Fringe Value

It is evident from Equation (5.2-14) that the material fringe value, fσ, must be known before an

isochromatic fringe pattern can be interpreted. Since this quantity may vary for a given material

because of its age, supplier, batch of resin, temperature, etc., a test is usually conducted to

determine fσ even if a value is provided by the manufacturer. The associated calibration

procedure is conducted by constructing a model and testing it in a known stress state which is

easily applied.

A tensile test is one of the most popular methods for the calibration of the material fringe value.

In this case, the principal stresses are given by

where the load, P, is applied along the “1" direction; “w” and “d” are the width and thickness of

the cross section, respectively. Substituting the quantities in Equation (5.4-1) into Equation (5.2-

14) yields

5.11

(5.4-2)

P P

21 4

3

0

P P

3 4

4.5

4 3

Reference point

(5.4-3)

(5.4-4)

where m is the slope of the curve obtained by plotting the load versus fringe order number.

Typical tensile specimens are shown in Figures 8 and 9. A reference point must be selected for

the tapered specimen, since the stress varies over the shank.

Figure 8. A standard tensile specimen. Figure 9. A tapered tensile specimen.

Another popular calibration specimen is a beam in pure bending. In this case, the principal

stresses at the free surface on the tensile side of the beam are

where M is the moment, c is the distance between the neutral axis and the free surface (equal to

half the height, w, for a rectangular cross section), and I is the centroidal moment of inertia about

the axis around which the moment is applied (equal to 1/12 dw3, for a rectangular cross section

of thickness, d). Substituting the quantities in Equation (5.4-3) into Equation (5.2-14) gives

where Rbdry is the fringe order number at the boundary on the tensile side of the specimen.

Figure 10 shows a typical calibration specimen recorded with a dark field circular polariscope.

The integer order fringes are black and correspond to the background. The principal stresses are

parallel and perpendicular to the longitudinal axis of the beam.

5.12

P/2

P/2 P/2

P/2

a

Neutral axis

0 1 2 3 4 5

Tensile side

(5.5-1)

Figure 10. The central portion of the beam is in pure bending.

At each point in the central section of the span, the stress in the vertical direction is zero while

the stress in the horizontal direction increases linearly with thickness. The neutral axis is located

at the center of the beam. For the case shown, the material above the neutral axis is in uniaxial

compression while the material below the axis is in uniaxial tension. For the beam shown, the

applied moment is equal to Pa/2; and, Rbdry is approximately 5.0.

Note that at the corners, there are two perpendicular free surfaces. Since both principal stresses

are zero there, the fringe order number is also zero. These isotropic points (where σ1 = σ2 = 0)

are very good places to begin labeling the fringe order numbers.

5.5 Compensation Methods for Determining Partial Fringe Order Numbers

In photoelasticity, it is necessary to accurately determine the fringe orders in the isochromatic

fringe pattern. There are three basic techniques used to obtain partial fringe order numbers.

The color sequence method is accurate to approximately 1/10th

of a fringe. It involves observing

the fringe orders of the isochromatic fringes surrounding the point in question, as well as the

color at that point. The color is compared with those listed on the chart shown in Table 1.

As an example, consider a point which lies between fringes having R = 2 and R= 3 where the

color is yellow. From the above chart, the retardation is 57 x 10-6

in. The partial fringe order at

the point in question, RP, is computed based on the retardation corresponding to R = 1.0 (22.7 x

10-6

in.) as follows:

5.13

(5.5-2)

RETARDATION (10-6

in) COLOR OBSERVED R

0 Black 0

12 Yellow

18 Red

22.7 1st Fringe 1

25 Blue-Green

35 Yellow

40 Red

45.4 2nd Fringe 2

50 Green

57 Yellow

63 Red

68.1 3rd. Fringe 3

73 Green

Table 1. Color compensation chart.

Once a fringe is located (usually 0 or 1st) the 2

nd then 3

rd fringe order can be found by tracking

the color sequence yellow-red-green toward areas of higher strain. The sequence green-red-

yellow corresponds to decreasing strain, leading to progressively lower fringe orders.

The second method is called the Tardy compensation method. The procedure is accurate to

approximately 1/100th

of a fringe and is executed by first using a dark field plane polariscope to

position the polarizer and analyzer in the directions of principal stress. This is accomplished by

rotating the crossed wave plates until an isoclinic fringe passes through the point in question.

The quarter wave plates are then inserted at 450 relative to the polarizer and analyzer directions

to form a circular polariscope. Once this has been accomplished, the analyzer is rotated until one

of the adjacent isochromatic fringes moves to the point. The rotation angle is measured and the

fringe order computed based on

where Radj is the fringe order number of the fringe which moved to the point; the plus sign is

used if this is the lower fringe order number while the negative sign is used if it is the higher

fringe order number. In Equation (5.5-2), the rotation angle, δ, is measured in radians.

5.14

P λ/4 λ/4 AModel

Compensator

As illustrated in Figure 11, digital compensation may also be performed by placing a small

variable wave plate between the model and one of the quarter wave plates. The Babinet-Soleil

compensator shown in the figure, for example, consists of two adjustable wedges. The unit is

placed along one of the principal stress directions with the help of a plane polariscope. Then the

quarter wave plates are inserted to form a circular polariscope and the compensator is adjusted

until its relative retardation is equal and opposite to that of the point in question. The partial

fringe order number is based on a digital read out and determined from a calibration chart/curve

provided with the compensator by the manufacturer. Units typically have a linear response and

can determine fringe orders to within 1/100th

of a fringe.

Figure 11. A Babinet-Soeil compensator.

5.6 Calculation of the σ1 Direction

In photoelasticity, the isoclinics pinpoint the directions of the principal stresses but they do not

specify which one of them corresponds to the algebraically largest eigenvalue, σ1. However, a

beam in pure bending can be used to establish a rule that can later be applied to any isochromatic

pattern observed in the polariscope for which the rule was formulated.

The procedure is to align the polarizer with the maximum principal stresses found on the tensile

side of the beam. After the quarter waves plates are inserted to form a circular polariscope, the

analyzer is rotated in a known direction, say counterclockwise. It is observed whether the higher

order isochromatics move toward the lower orders on the tensile side of the beam, or, visa versa.

If, for example, the higher order isochromatics move toward the lower orders (fringes on the

boundary move toward the neutral axis) the following rule would be established:

The rule is tested at points on an arbitrary specimen. To do this, a plane polariscope is used to

position the polarizer and analyzer in the principal stress directions by passing an isoclinic

The polarizer is in the direction of σ1 if for a counterclockwise

rotation of the analyzer, higher order isochromatics move

toward lower orders.

5.15

2nd λ/4Analyzer

Observer

1st λ/4Polarizer

Light source

Coating

Part

Loads

Thickness, d

(5.7-1)

through the point. Then, the quarter wave plates are inserted to form a circular polariscope. The

rotation is executed and the movement of the isochromatics noted. If the rule holds true at the

point, the polarizer is along σ1. If it does not hold true, then the analyzer is oriented along σ1

(i.e., the polarizer is not).

5.7 Birefringent Coatings

In general, the method of photoelasticity relies on a birefringent model which must be fabricated

and placed in a state of plane stress. However, tests can be conducted on a prototype made from

another engineering material, such as steel or aluminum, by using a reflective adhesive to bond a

thin layer of photoelastic plastic to the surface of the specimen. During this procedure, the

assumption is made that the photoelastic coating does not reinforce the structure to the point

where the overall structural response is changed.

As illustrated in Figure 12, the photoelastic coating is studied with a reflection polariscope.

Note that, since the light reflects off the cement, it passes through the coating twice.

Figure 12. A reflection polariscope.

The stresses in the specimen can be related to those in the coating by making the assumption that

the strains in the coating are the same as those in the specimen. This assumption can be

expressed mathematically as

where the superscripts “c” and “s” represent the coating and specimen, respectively.

For convenience, Equation (5.7-1) is expressed in terms of the principal directions. Applying

Hooke’s laws separately to the specimen and the coating [Equation (2.5-1) with σxx = σ1, σyy =

σ2, and σzz = σ3 = 0] gives

5.16

(5.7-2)

(5.7-3)

(5.7-4)

(5.7-5)

(5.7-6)

where E is the elastic modulus and ν is the Poisson’s ratio. When the expressions contained in

Equation (5.7-2) are used to satisfy Equation (5.7-1) and the resulting equations solved

simultaneously for the stresses in the coating,

Subtracting the second expression in Equation (5.7-3) from the first yields

The term on the left hand side of Equation (5.7-4) is governed by a modified from of Equation

(5.2-14), by taking into account that light passes through the thickness twice, as

Substituting Equation (5.7-5) into Equation (5.7-4) and solving for the stress difference in the

specimen gives

where R is the fringe order observed in the coating, d is the thickness, and fσ is the material

fringe value obtained by calibrating the coating on a simple cantilever beam. Figure 13, for

example, shows a typical calibration specimen. In this case, the stress on the upper surface of the

beam is uniaxial with σ1s = Mc/I and σ2

s = 0. The latter are principal stresses directed parallel

and perpendicular to the longitudinal axis of the beam, respectively.

As illustrated in the figure, a reference point is selected at a convenient location along the span.

The moment is computed at this location based on statics and the material fringe value is

obtained from the isochromatic fringe order at the point in question by applying Equation (5.7-

6).

5.17

3 2 1 0

Increasing stress/strain

R

Cantilever beam

W

Reference point

Color

Figure 13. A cantilever beam is used for calibration.

Problems may occur when using this technique when the coating is applied and used under

different environmental conditions. Edge effects may also be present at free boundaries, since it

takes a finite distance for the strains to transfer from the specimen to the coating. It is also

important to keep the thickness constant to obtain a stress plot; otherwise, Equation (5.7-6) is

only valid on a point-wise basis.

5.8 Three-Dimensional Photoelasticity

Most of the studies done in photoelasticity involve a model in plane stress. In a more general

case, the method of stress freezing can be applied. The procedure consists of thermally locking

in stresses, slicing the model into pieces sufficiently thin for them to be assumed to be in a state

of plane stress, and analyzing the slices using conventional methods.

The method works based on the di-phase behavior exhibited by most photoelastic materials. The

material is heated beyond its transition temperature to break its secondary bonds. After loads are

applied, the material is cooled slowly to reform the bonds and lock in the stresses. The model is

then sliced and analyzed.

A few factors need to be taken into account. First, since the material fringe value is a function of

temperature, a calibration specimen must be placed in the oven along with the specimen.

Second, since the slices are usually thin and not too birefringent, compensation methods that

work well at low fringe order are usually required. The Babinet-Soleil compensator is an

excellent tool for this task.

5.18


1. During the 1930's, Max Frocht developed the stress concentration factor chart included below

in Figure 14 by conducting a series of tests on photoelastic specimens having different

geometries.

Figure 14. Stress concentration factor charts for bars containing a hole and fillets.

The isochromatic pattern for one of these tests is illustrated in the photographs shown in

Figure 15. A blow-up of the region adjacent to the hole is shown in the photo to the left.

Figure 15. Isochromatic fringes form around the hole.

5.19

P

P

Dr. Frocht specified that the bar had a width of 0.789 in. and a thickness of 0.136 in. The

hole was 0.258 in. in diameter and the material fringe value was 86 psi/fringe/in. For the

case shown, the applied load was 146.6 lb. What should the fringe order number be

adjacent to the hole if his chart is accurate? Does this appear to be the case?

Answer: R ~ 7.2; yes.

2. Figure 16 shows the isochromatic pattern created when a disk is subjected to a diametrical

compressive load, P = 275 lb. The disk has a diameter, D = 1.25 in., and a thickness, t =

0.2 in.

Figure 16. Isochromatic fringes for a disk under diametrical compression.

From symmetry, it is known that the principal stress directions at the center point of the

disk are aligned with the horizontal and vertical directions; assumed as x and y,

respectively. The theory of elasticity predicts that at the center point

where P is the applied load, D is the diameter, and t is the thickness.

(a) Sketch the configuration for the polariscope used to conduct this investigation.

(b) Calculate the material fringe value. Be sure to include units in your answer.

Answer: (a) light field circular polariscope, (b) fσ = 86.2 psi/fringe/in.

5.20

C

A

B R = 4.0

D

3. Figure 17 shows the stress pattern in a beam with fillets subjected to pure bending. The

depths of the beam are 1.222 in. and 0.622 in., respectively; and, the radii of the fillets are

0.0675 in.

A stress concentration factor chart for this case is included below as Figure 18.

(a) What are the fringe orders at the points labeled as A, B and C?

(b) Assuming that the fringe order at the boundary of the shank is R = 4.0, use the

chart to predict the fringe order at point D.

Figure 17. Isochromatic fringes in filleted bar. Figure 18. Stress concentration factor chart.

Answer: (a) RA = 0, RB = 1.5, RC = 0; (b) RD = 7.6.

4. Figure 19 shows the isochromatic pattern for a portion of a beam with shallow grooves

subjected to a bending moment equal to 26.5 in-lb. A stress concentration factor chart is

shown in Figure 20. The pattern was recorded using a light field circular polariscope. The

overall depth of the beam is 0.655 in.; the depth of the minimum section located between

the grooves is 0.432 in.; the radii of the grooves are 0.222 in.; and the thickness of the

specimen is 0.246 in.

(a) Determine the material fringe value of the specimen based on the pattern

observed in the region farthest away from the grooves.

(b) Compute the stress concentration factor at the root of the notch based on the

nominal bending stress in the critical section containing the discontinuity.

(c) Determine the stress concentration factor from the chart and then compare the

result with that obtained from part (b). Present the answer in terms of

percentage error.

(d) What might the error computed in part (c) be due to?

5.21

0

4

2

12

Figure 19. Bar with shallow grooves. Figure 20. Stress concentration factor chart.

Answer: (a) fσ = 90 psi/fringe/in.; b) K = 1.27; c) Kchart = 1.22, error = 4%; d) chart

formulated based on deep as opposed to shallow grooves.

5. Figure 21 shows the bending calibration specimen used to compute the material fringe

value for a material used to study the isochromatic fringe pattern for a ring subjected to

diametrical compression (see Figure 22). The four loads applied to the calibration

specimen are each 96 lb.

Figure 21. Isochromatic pattern in a beam in bending.

The height (vertical dimension) of the calibration beam is 0.763" and the thickness is

0.180". The distance between the upper loads is 3.0" while the distance between the lower

loads is 4.0". Tardy compensation showed that the fringe order number at the top/bottom

of the beam was 5.5.

The isochromatic pattern in the ring was created by applying loads equal to 202.5 lb along

the vertical diameter. The ring was cut from the same sheet of photoelastic material as the

calibration specimen; it has an inner diameter equal to 0.766" and an outer diameter of

1.532". The fringe order number at the critical point located on the inner radius and along

the horizontal diameter is 16.75.

5.22

202.5 lb

202.5 lb

(a) Determine the material fringe value of the calibration specimen.

(b) Determine the stress concentration factor at the critical location, assuming that

the area along the horizontal diameter is subjected to pure compression.

(c) What is the maximum in-plane shear stress at the critical location.

Answer: (a) fσ = 90 psi/fringe/in.; b) K = 5.7; c) τmax = 4185 psi.

6. Figure 23 shows a 6 in. long aluminum calibration specimen being used to compute the

material fringe value of a photoelastic coating. A 50-lb load, labeled as W, is applied to the

beam. The beam is 1.0 in. wide and 0.25 in. thick. It has an elastic modulus of 10 x 106 psi

and a Poisson’s ratio of 0.3.

The coating is 0.10 in. thick. It has an elastic modulus of 450 ksi and a Poisson’s ratio of

0.42. The color at the reference point is yellow; a color compensation chart is included as

Figure 24.

Compute the material fringe value based on the stress at the reference point using the

5.23

3" 3"

relation

Be sure to include units in your answer.

Figure 23. A cantilever beam is coated with a birefringent coating.

Figure 24. Color compensation chart.

Answer: fσ = 77 psi/fringe/in.

5.24

5.10 Classroom Demonstration in Photoelasticity

1. Photoelasticity requires a basic knowledge of plane and circular polariscopes. In

conjunction with a classroom demonstration of these units:

a) Sketch the configuration for a dark field plane polariscope. Explain how the device

was constructed in class. After observing the isochromatics and isoclinics in a beam

subjected to pure bending, discuss the meaning of each system of fringes and

comment on any observations made during the demonstration.

b) Sketch four possible configurations for a circular polariscope. Explain how these

light and dark field configurations were constructed in class.

c) What is the main advantage of using a circular polariscope over a plane polariscope?

2. The isoclinics characterize the principal directions; however, it is impossible to determine

which direction corresponds to the algebraically largest component of the stress unless a

rule is established with a well-known stress state.

a) Establish a rule for the polariscope demonstrated in class using a beam in bending.

b) Discuss the procedure for applying the rule to an arbitrary specimen.

3. To evaluate the stress distribution at a point, it is necessary to know the material fringe

value, fσ; and, accurately determine the fringe order number. Calibration and

compensation methods are used to obtain fσ and interpolate between fringes, respectively.

a) Outline three basic methods that can be used for compensation.

b) After having placed a beam with a circular hole in uniaxial tension, calibrate the

specimen using the fringe order observed in the uniformly stressed portion of the

specimen removed from the hole. Specify the apply load and the fringe order

number.

c) Assume that the beam is loaded along the vertical direction and consider a point

adjacent to the hole along the horizontal centerline. Determine the sign of the

boundary stresses adjacent to the hole using the rule established in part 2 of the

demonstration.

6.1

Viewer

Film

Diaphragm

Lenses

Shutter

CHAPTER 6 - PHOTOGRAPHY

6.1 Introduction

The use of a camera is an effective means of recording experimental data for storage or later

analysis. Conventional photography relies on recording the image of an object by focusing it

onto the emulsion side of a silver-based photographic film and/or paper. Any silver salt particle

in the emulsion which is exposed to light undergoes a physical change.

During development in a darkroom, the physical change becomes a chemical change leading to

the formation of metallic silver atoms. This transition takes place in a solution called the

developer. The action of the developer is stopped after a specified period using a second

solution called a “stop bath.” A third solution, called the “fixer,” dissolves any unexposed silver

salts remaining in the emulsion.

An alternate approach is to capture the image using a digital camera. In this case, the image is

captured on a focal plane array and stored in a computer as a three-dimensional array consisting

of the pixel locations (x,y) of points in the image and their intensities (z).

6.2 Single Lens Reflex Cameras

Figure 1 illustrates that the basic components of a camera are the lens, the diaphragm, the

shutter, and the film plane. The lens (usually compound) is used to focus the image of the object

onto a photographic film. The specifications of the lens are given in terms of focal length and

maximum f-number (f-numbers are defined as the focal length divided by the aperture). The

diaphragm adjusts the lens aperture while the shutter controls the amount of time that the film is

exposed to the admitted light.

Figure 1. Basic components of a camera.

6.2

Film speed dialFocusing screen

Rewind knob

Pentaprism

Shutter speeed ringFocusing ringgAperture ring

Filter thread

Reflex mirrorSelf timerShutter releaseFilm advance leverFrame counter

The amount of light allowed to react with the film can be controlled by varying the shutter speed

(in seconds or fractions of a second) or the lens aperture (the larger the aperture, the more light

admitted). When the f-number setting is increased, the aperture decreases. On most cameras,

the shutter speeds and apertures are adjusted so that a change in one f-stop setting corresponds to

one step change in shutter speed.

Figure 2, for example, shows a single lens reflex camera in more detail. With the SLR the

image is viewed through the lens (TTL). Depressing the shutter release button first causes a

hinged mirror to swing upwards clear of the light path before the shutter opens, and the film

records the image exactly as it appeared in the viewfinder. SLR cameras using 35 mm film are

most popular for general, good-quality photography. Large-format SLR cameras are used mostly

by professionals. Many have optional prism-finder attachments, with built-in exposure meters.

Figure 2. A single lens reflex camera.

As illustrated in Figure 3, the controls found on most lens barrels are the aperture control ring

and the focus control ring. Other information to be found on lenses is a depth of field scale, in

both feet and meters, and an infra-red focusing mark. Apertures are aligned against a fixed

reference mark and the focusing ring should be turned until the viewfinder shows a-sharp image.

As mentioned above, the lower the f-number, the larger the aperture.

6.3 Depth of Field

The depth of field is the zone extending in front of and behind the plane of sharpest focus. In

this zone, blur is negligible, and everything may be considered as being “in focus.”

6.3

Focus reference markDepth of field scaleFocusing distancesFocus control ring

Aperture control ring

f2.8 f4 f5.8 f8 f11 f16

Figure 3. A typical compound lens.

The depth of field depends on the focal

length, lens to subject distance, and

aperture. The shorter the focal length, the

greater the depth of field. The closer the

subject, the smaller the depth of field.

And, the smaller the aperture, the greater

the depth of field.

In most engineering applications, focusing

is performed with an open aperture to

obtain the best focus and then the lens is

stopped down to take the picture with a

larger depth of field.

Figure 4 illustrates that the greater a lens's

focal length, the less its angle of view. As

the focal length increases the minimum

focusing distance also increases.

6.4 Photographic Processing

As explained in data sheet included as Figure 6, there are three basic steps in processing black

and white photographic film. First, the exposed film is placed in a developer solution which

reacts with and sensitizes the emulsion (this reaction varies with developer temperature,

concentration, and with the amount of time the film remains in the bath). Second, the exposed

film is placed in a "stop bath" which stops the action of the developer. Third, the film is placed

Figure 4. Lens’ characteristics.

6.4

in a "fixer" solution, which removes the backing and excess emulsion from the negative. The

negative is now ready for printing.

Figure 5. Data sheet for a 35mm photographic film.

6.5

Figure 6 shows a typical configuration for the darkroom facility used to process the film. Once

this is done, an enlarger is used to expose a piece of photographic paper by passing light through

the negative. The same development process used to process the film is used to process the

paper. Figure 7 includes the data sheet for the latter.

Figure 6. A typical darkroom.

6.5 Polaroid Land Film Cameras

Some cameras are designed to allow a user to develop the film immediately after it has been

exposed. A Polaroid Land camera relies on a packet filled with developer to do this. It takes

approximately 30 seconds to develop the image. Most of the film packets for land cameras do

not provide the user with a negative making it difficult to make copies of the print. For details,

see Figures 8 and 9.

6.6

Figure 7. Data sheet for a photographic paper.

6.7

Figure 8. Data sheet for a Polaroid Land Camera.

6.8

Figure 9. Data sheet for a Polaroid Land Camera (continued).

6.9

6.6 Digital Cameras

An alternate approach to conventional photography is to capture the image using a digital

camera, such as the ones shown in Figure 6.10. In this case, the image is captured on a focal

plane array and may be stored in a computer as a three-dimensional array consisting of the pixel

locations (x,y) of points in the image and their intensities (z).

Figure 10. Digital cameras provide an alternative to conventional photographic processing.

Digital photography has many advantages over traditional film photography. Digital photos are

convenient, allow users to see the results instantly, don't require the costs of film and developing,

and are suitable for software editing and uploading to the Internet.

The basic attribute of a digital camera that determines image quality is its megapixel rating. This

number refers to the amount of information that the camera sensor can capture in a single

photograph. Cameras with high megapixel ratings take larger pictures with more detail. Those

photos will also look better when printed, especially in bigger sizes. At five megapixels, image

quality gets in the neighborhood of film.

It should be noted that the focal plane of most cameras contains rectangular pixels which have a

length to height ratio of 4:3. In this case, care must be taken while making measurements, since

the different scale factors must be taken into consideration. For this reason, some manufacturers

provide cameras having arrays with square pixels.

There are also many additional features available on digital cameras, including image

stabilization, on-board image editing, color correction functions, auto-bracketing and burst

modes. Many of these can be handled by image editing software, and so they can be

unnecessary (and often inferior) when built into a camera.

http://electronics.howstuffworks.com/camera-stuff-pictures.htm

7.1

CHAPTER 7 – BRITTLE COATINGS

7.1 Introduction

Brittle coatings are lacquers which are sprayed onto a test part. When dried, they become brittle

and crack well below the yield stress of the material that they are deposited on. As illustrated in

Figure 1, cracks occur in tension, initially perpendicular to the maximum principal stress. The

cracks form isostatics; lines tangent to the principal stress directions.

Figure 1. Cracks form along isostatics in a coating applied to a tubular structure.

The brittle coating technique is widely employed in experimental stress analysis because it is

inexpensive and easy to use; and, quickly provides an engineer with vital information about the

distribution and severity of stresses in a test part. The coating provides a visual overall picture of

the strain distribution and can be applied to almost any part or structure, irrespective of the size,

shape, or material to highlight areas of stress concentration. The technique locates the critically

stressed areas, and gives the principal strain directions, as well as information on the strain

gradients; all of which are needed for the proper selection, location, and orientation of strain

gages for accurate measurement of peak stresses. Perhaps more importantly, the technique is

inexpensive and simple to use, and requires no instrumentation. In most cases, the coating has

negligible reinforcing effect on the test part. Most coatings dry to vibrant colors, offering good

contrast for observation of the crack patterns.

7.2

7.2 Testing Procedures

In most practical applications, the brittle coating is sprayed onto the surface of the part or

structure to be tested, and allowed to air-dry. After drying, the part is subjected to test loads

usually applied in incremental fashion. As the incremental loading proceeds, the lacquer first

cracks in the region where the tensile strain is greatest; the cracks are always oriented along

directions perpendicular to the maximum tensile strain.

With each additional increment of load, the cracks multiply and grow outward from the area of

peak stress, terminating at points where the strain on the surface of the test part falls just below

the lacquer’s threshold strain. The loci of the crack ends is typically outlined with a grease

pencil after each load increment, and identified with the corresponding load magnitude.

Figure 2 illustrates this testing procedure and shows a line called an isoentatic being drawn on a

loaded ring. The isoentatic, P, encloses an area where the strain equals or exceeds the threshold

strain. As the load is increased incrementally, several isoentatics (Pl, P2, P3 ... ) can be traced.

Figure 2. Isoentatics are drawn as the specimen is loaded incrementally.

For field testing under service loads (as in the case of aircraft or spacecraft, for example), it is

common practice to apply the brittle coating over all areas of interest and, after the lacquer has

dried thoroughly, subject the structure to the prescribed working conditions. The coating is then

carefully examined to locate the regions where the strains induced by service loads were high

enough to produce cracks in the lacquer. Strain gages can subsequently be installed in these

critical areas for accurate strain measurements. Figure 3, for example, shows the crack pattern

developed on a nuclear heat exchanger as a result of hydrostatic pressure.

7.3

Figure 3. Cracks on a heat exchanger.

7.3 Calibration

Quantitative measurement of the strains in the test part requires that the coating be calibrated to

obtain the strain level at which cracking occurs. This is accomplished by spraying a number of

calibration beams with the lacquer at the same time the test part is sprayed. The part and the

calibration beams are then allowed to dry together so that the two coatings experience identical

curing conditions. Coincident with loading the test part, the beams are placed in a cantilever

calibration fixture and subjected to a predetermined deflection at the free end. This imposes a

known linear strain gradient along the length of the beam, from which the threshold strain level

is read on the calibrator measuring scale. Figure 4 shows a typical calibration test fixture.

As the deflection is applied, cracks begin to form perpendicular to the longitudinal axis of the

beam at the fixed end where the stress is the highest. The threshold strain is determined by

pinpointing the location of the last full crack and calculating the strain at this point.

Figure 5, for example, shows a cantilever beam of length, L, fixed into a wall at the left end and

subjected to a concentrated load, P, applied at the right end.

7.4

L

AB

Y

X

P_

L

A

B

P_

_RAx

_RAy

_MA

(7.3-1)

Figure 4. A cantilever calibration fixture.

Figure 5. An end-loaded cantilever beam. Figure 6. FBD of the cantilever beam.

Figure 6, on the other hand, shows the free body diagram corresponding to the beam illustrated

in Figure 5. Since all of the forces are contained in a single plane, the reactions at the fixed

support consist of two force reactions and one moment reaction.

Referring to Figure 6, and taking into account overall equilibrium,

Expressions for the shear force, V, and bending moment, M, can be obtained for all points along

the span by passing a section through a point, O, located at an arbitrary distance, x, from the

fixed support.

7.5

x

A O

_V

P_M

PL

(7.3-2)

V

M

A B

BA

X

X

P

-PL

(7.3-3)

Figure 7 shows this procedure and illustrates the standard convention for assigning the shear and

moment distribution on the cut section.

Figure 7. A cut at position “x”.

Applying the equilibrium equations to the section:

The expressions in Equation (7.3-2) can be used to plot the shear and moment diagrams shown in

Figure 8. The shear force is constant over the span and equals the applied load. The bending

moment is maximum at the fixed support and progressively decreases with increasing x.

Figure 8. Shear and moment diagrams.

The diagrams are consistent with the fact that the shear force at each section is equal to the

derivative of the bending moment (the slope of the moment diagram). Conversely, the difference

in the bending moment between two sections is equal to the area under the shear curve. That is,

for small deflections,

The shear force creates a shear stress at every section that can be approximated by dividing the

7.6

(7.3-4)

(7.3-5)

(7.3-6)

force by the cross-sectional area. However, this approximation only gives an average value for

the stress and does not adequately define the actual distribution through the thickness. A better

description is given by,

where V is the shear force, Q is the first moment of the area measured about the neutral axis of

the portion of the cross section located either above or below the point under consideration, I is

the centroidal moment of inertia of the entire cross section, and t is the width of a cut made

through the point in question perpendicular to the applied load.

When Equation (7.3-4) is applied to a beam of rectangular cross section, it is found that the shear

stress distribution is parabolic through the thickness. The shear stress is zero at the upper and

lower surfaces of the beam and maximizes at the neutral axis.

Shear stresses always come in pairs, and it should be noted that the vertical shear force creates a

horizontal shear flow. That is, horizontal shear stresses develop between the longitudinal fibers

in the beam. This effect is zero on the free surface and maximum at the neutral axis.

The bending moment creates normal stresses parallel to the longitudinal axis of the beam. The

magnitude of these stresses is given by the elastic flexure formula,

where M is the moment, y is the distance measured from the neutral axis to the point under

consideration, and I is the centroidal moment of inertia corresponding to the axis about which the

moment is applied.

Equation (7.3-5) shows that the normal stress is linearly distributed through the thickness. It

becomes a maximum at the free surfaces and reduces to zero at the neutral axis.

Equation (7.3-5) can be applied to determine the stress on the upper surface of the cantilever

beam shown in Figure 5. Assume that the beam has width, b, and thickness, t. Since the neutral

axis is located at the geometric center of the beam, y = t/2. The centroidal moment of inertia

measured around the bending axis (z) is I = 1/12 bt3; the moment is given as a function of x by

Equation (7.3-2). Substituting these values into Equation (7.3-5),

Equation (7.3-6) shows that the normal stress at the top of the section is tensile. Since the fibers

on this surface do no experience any shear stress, the points located there are in a state of

uniaxial stress. In this simple case, the constitutive relations (Hooke’s Laws) reduce to

7.7

(7.3-7)

(7.3-8)

(7.3-9)

(7.3-10)

(7.3-11)

(7.3-12)

(7.3-13)

(7.3-14)

where E is the elastic modulus and εxx is the normal strain measured along the longitudinal axis

of the beam.

Substituting Equation (7.3-7) into (7.3-6) and solving for the strain,

Equation (7.3-8) implies that the strain along the beam varies linearly with x.

The load can also be obtained by knowing the deflection under the load. The governing

expression can be found by applying the equation,

where M(x) is the expression for the applied moment given by Equation (7.3-2). Thus,

Integrating once,

Integrating again,

Applying the boundary conditions [@ x = 0, dy/dx = 0] and [@ x = 0, y = 0] to the expressions

in Equations (7.3-11) and (7.3-12), respectively, reveals that C1 = C2 = 0. Therefore,

Under the load at x = L,

Denoting the absolute value of this deflection by δend, and solving for P,

7.8

(7.3-15)

(7.4-1)

(7.4-2)

(7.4-3)

7.4 Measurements

Although considered primarily a qualitative tool, the method can provide an approximate

measure of surface strain magnitudes. If care is taken in the preparation, application, and testing

procedure, a coating that is properly calibrated can yield quantitative results accurate to ± 100 με.

Under optimum conditions, accuracy on the order of ± 50 με can be achieved.

The nominal sensitivity, or threshold strain, εt*, of most brittle coatings is around 500 uε. The

stress in the specimen is calculated based on its elastic modulus, Es, from the simple relation

Equation (7.4-1) assumes that the stress at the target point is uniaxial; whereas, in general, the

part is in a state of biaxial, or plane, stress. This, in itself, may give rise to substantial error. But

other errors exist due to a mismatch in the Poisson’s ratio between the specimen and coating

materials.

In general, the stress in the specimen is not equal to that in the coating. This can be

demonstrated by making the assumption that the strains in the coating are the same as those in

the specimen. This assumption can be expressed mathematically as

where the superscripts “c” and “s” represent the coating and specimen, respectively.

By assuming plane stress (σzz = 0) and applying the Hooke’s laws, previously given in Equation

(2.5-1), separately to the specimen and the coating:

where E is the elastic modulus and ν is the Poisson’s ratio. When the expressions contained in

Equation (7.4-3) are used to satisfy Equation (7.4-2) and the resulting equations solved

simultaneously for the stresses in the coating,

7.9

(7.4-4)

(7.4-5)

(7.4-6)

(7.4-7)

During calibration, the stress on the upper surface of the cantilever beam is in uniaxial tension

and, when the x axis is directed along the span, characterized by

Since the strain in the specimen is equal to the strain in the coating,

Substituting these stresses into Equation (7.4-4) reveals

Equation (7.4-7) shows that, in general, a biaxial stress state is produced in the coating even

when the stress state is uniaxial in the specimen. The only condition under which Equation (7.4-

7) predicts a uniaxial stress state occurs when νc = ν

s. But this is far from the case for most

engineering materials that have a Posson’s ratio of approximately 0.3; since, the Possion’s ratio

of brittle coatings are typically 0.42.

7.5 Coating Selection

As mentioned previously, the threshold strain of most brittle coatings is around 500 uε.

However, the actual threshold for a particular test depends upon the application and the test

environment. Because its cracking behavior is affected by both temperature and humidity, many

different coatings are manufactured to provide optimum performance over a wide range of

environmental conditions.

The proper coating for a given test condition is obtained from a selection chart, such as the one

shown in Figure 9 (for Tens-Lac manufactured by Photolastic Inc.).

If the anticipated test temperature were 700 F (21

0 C) and the relative humidity 50%, for

example, the appropriate coating indicated by the selection chart would be TL-500-75. At the

same temperature, but with 30% relative humidity, either TL-500-75 or TL-500-70 would be

7.10

suitable. The lacquer with the higher number (TL-500-75) would have greater sensitivity (that

is, a lower threshold), while the TL-500-70 would begin cracking at higher strains.

Figure 9. Coating selection chart.

The lacquer normally begins to crack when a nominal strain level of 500 με is reached.

However, an increase in temperature or humidity will raise the threshold of the brittle coating

and cause a reduction in sensitivity. At thresholds greater than 1200 uε, the cracks will not

always remain open after removing the load, and it may be necessary to observe the crack

patterns on the test part while under load. The threshold of most brittle lacquers can be lowered

by decreasing the temperature. However, there is a practical lower threshold limit of about 300

με.

7.11

7.6 Application

A brittle coating can be applied to almost any test material. Application of the lacquer must be

preceded, however, by proper preparation of the surface to render it completely free of dirt, oils,

rust, loose paint, etc. After cleaning, an undercoating is sprayed on the part and allowed to air

dry (15 to 30 minutes). The undercoating is an aluminized covering agent which provides a

uniformly reflective surface for promoting crack visibility in the overlying lacquer.

When the undercoating has dried, the brittle lacquer is sprayed over it in a succession of thin, wet

layers, allowing about two minutes between coats. For most test situations a uniform thickness

of. about 0.15 mm (0.006 in.) is optimum. The thickness is monitored by comparing the color to

a test strip provided by the manufacturer. Thicker coatings appear as darker colors whereas

thinner coatings appear lighter. After spraying the final layer, the lacquer must then be dried at

3-60 C (5-10

0 F) above the expected test temperature. Best performance of the lacquer is

achieved after a drying period of 18-24 hours.

7.7 Crack Patterns

Cracks only occur in areas where the threshold strain is reached. Since the surface is in plane

stress, σzz = 0. In the case where the in-plane principal stresses are of opposite sign, cracks will

occur perpendicular to the maximum tensile stress provided, of course, that the threshold is

reached. When both principal stresses are positive, two families of cracks may result. Cracks

will initially form perpendicular to the largest stress and then, if the strain is high enough,

perpendicular to the lower tensile stress. For the special case where both principal stresses are

positive and equal, the cracks will appear random but will always intersect at 900.

7.8 Relaxation Techniques

A brittle coating cracks only when placed in tension. To test for compressive loads, the

specimen is first loaded and then coated. As the load is relaxed, the coating cracks.

7.12

3 in.

6 in.

0.25 in.

P

2 in.


1. Figure 10 shows the isostatic fringe pattern observed in a brittle coating that was applied to

the top surface of a 9” long end-loaded cantilever beam (Elastic Modulus = 30 x 106 psi;

Poisson's Ratio = 0.3). When a load of P = 100 lb was applied, it was observed that the

crack pattern extended 3" from the fixed end.

(a) Determine the reactions at the fixed support.

(b) Sketch the shear and moment diagrams for the beam.

(c) Determine the threshold strain for the coating.

\

Figure 10. Isostatic pattern in a brittle coating applied to an end-loaded cantilever.

Answer: (a) Rx = 0, Ry = 100 lb ↑, M = 900 in.-lb clw, b) shear is constant; moment is linear, c)

ε t* = 480 με.

8.1

CHAPTER 8 – MOIRÉ METHODS

8.1 Introduction

Moiré patterns result from interference caused by superposing two geometrical patterns. These

interactions are very common in everyday life and may be observed if one looks closely for

them. Figure 1, for example, shows a moiré formed by two wire mesh bowls used in food

preparation. The fringes correspond to the superposition of two amplitude gratings, comprised

of opaque bars and clear spaces. The pattern is indicative of both the spacing between the bowls

and their differences in shapes.

Figure 1. A moiré pattern is formed by two wire bowls.

The arrays used to produce moiré fringes may be a series of straight parallel lines, a series of

radial lines emanating from a point, a series of concentric circles, or a pattern of dots. In stress-

analysis, arrays typically consist of straight parallel lines, ideally formed with opaque bars

having clear spaces of equal width.

The principles underlying the geometrical moiré methods described in this chapter are relatively

straightforward. However, these methods suffer from low sensitivity and are mainly used to

measure displacement, as opposed to strain.

Recent advancements in grating replication techniques, such as interferometry and micro-

lithography, have led to the development of extremely sensitive methods such as moiré

interferometry and diffractive-optic interferometry. But these techniques require studying the

8.2

(8.2-1)

(8.2-2)

(8.2-3)

complex superposition of coherent wavefronts diffracted by high frequency gratings deposited or

projected onto test surfaces.

8.2 Analysis

In conventional moiré analysis, the procedure used to measure in-plane displacement of a test

specimen is to fasten or deposit a master grating onto its surface. The distance between the lines

in the master when it is undeformed is referred to as the pitch, p. The frequency, f, of the grating

is the reciprocal of the pitch. Hence, the relation between pitch and frequency (density) is

Displacement is always measured perpendicular to the lines in the master grating. When the

frequency of the grating is low [0.4 lines per mm (10 lines per inch) or less], image analysis

techniques are normally employed to measure the changes in spacing at each point before and

after loading. Displacements and strains are extracted from these data. For example, the strain

in the direction perpendicular to the lines in the master grating is simply

where Δp is the change in pitch.

For gratings with higher frequency and smaller pitches [2 to 40 lines per millimeter (50 to 1000

lines per inch)], the specimen grating is typically superimposed with a master grating to form a

moiré fringe pattern. The master grating can be another grating placed in front of the specimen

or a stored copy (photographically or digitally) of the undeformed specimen grating.

Assuming that the master grating is placed in the xy plane with its lines parallel to the y

direction, the displacement component along the x direction is given by

where nx is the fringe order number.

Figure 2, for example, shows the interference produced by a uniform elongation of the

specimen. In this example, it is assumed that p is the pitch of the master grating while p1 is the

pitch of the specimen grating.

A dark fringe appears at the points where an opaque strips of the master grating fall over the

transparent strips of the specimen grating. When the opaque strips of the gratings coincide, there

is a maximum of light intensity and a light fringe appears. Light fringes represent integer orders.

8.3

Center of alight fringe

Center of adark fringe

Center of a light fringe

Center of adark fringe

Specimengrating

Mastergrating

p1

p

Master grating

0

-2-1

12

3

Figure 2. A moiré pattern is formed due to specimen elongation.

Figure 3, on the other hand, shows a more general case where the specimen deforms differently

at each point. In this case, light fringes appear along the shorter diagonals of the intersecting

grating lines. Even though the superposition is more complex than that illustrated in Figure 2,

Equation (8.2-3) still holds.

Figure 3. A moiré pattern for the case of general deformation.

The fringe orders are typically established with respect to a reference point. The black dot on

Figure 3, for example, can be selected as a starting point and the fringe passing through it

labeled as zero (implying that the specimen grating did not move relative to the master grating at

this point. As one moves along fringe order “0" progressively to the right, all intersections

coincide to points where the lines in the master did not move relative to those on the specimen.

At the intersections of the gratings along fringe number one, all points in the specimen move one

pitch to the right relative to the master, etc.

In general, two in-plane displacement maps are required to completely determine the overall

8.4

(8.2-4)

p

Pointsource

43210-1-2-3-4

Transform lensCollimating lens

Gratingλ

θ1

θ2

strain state; and, a cross grating is usually deposited on the specimen.

Assuming that the gratings have equal pitch, the lines parallel to the x axis produce an additional

moiré fringe pattern which corresponds to the displacement component in the y direction, v,

governed by

where ny is the fringe order number.

8.3 Optical Filtering

When a cross grating is used the two moiré patterns are superimposed but they can be separated

by sampling the diffraction spectrum in an optical filtering arrangement. To understand this,

Figure 4 shows that when light emanates from a point source, spherical wavefronts are

produced.

Figure 4. Light is diffracted by a grating.

Plane waves are created when a lens is placed at a focal distance from the source; and, the light

beam is referred to as collimated. When the collimated beam is used to illuminate a grating, the

light is bent due to diffraction. Each clear space in the grating produces a separate source of

light and when the light is collected by another lens, the wavefronts from the different sources

combine to produce interfere in the lens’ transform plane. The dots represent points where

constructive interference occurs. The dots are oriented perpendicular to the lines in the grating

and spaced at a distance, d, apart. The spacing is given by

8.5

(8.3-1)

Transform plane

Moiréand

gratings

Moiréless

gratings

Input Output

4

3

2

1

1

2

3

4

+10

-1

where λ is the wavelength, p is the pitch, and f is the focal distance of the transform lens.

Equation (8.3-1) shows that the smaller the pitch, the larger the spacing between the diffraction

orders.

Figure 5 shows an optical filtering arrangement that can be used to separate the moiré pattern

from the gratings that form it. Coherent light from a laser is expanded and collimated by placing

a lens at a focal distance away from a spatial filter. The light is used to illuminate a photographic

film containing the master and specimen gratings, as well as the moiré pattern that they produce.

Figure 5. An optical filtering arrangement.

The gratings coincide at the center of a light moiré fringe and produce diffraction that contributes

to the dot pattern observed in the transform plane. A filter is positioned in this plane that allows

only the +1 diffraction order to pass through the system. When the imaging lens creates the

filtered image, moiré fringes are observed.

A similar procedure is followed when a crossed grating is used to study the displacement of the

specimen. In this case, the diffraction pattern is relatively complex consisting of a rectangular

array of dots. However, the moiré patterns are separated by allowing the corresponding +1

diffraction order to pass through the system.

8.4 Stress Analysis

Once the in-plane displacements have been determined, the strains can be computed based on a

reduced form of the general strain-displacement relations given in Equation (2.2-1). For small

strains measured on a traction free surface in plane stress, these relations become

8.6

(8.4-1)

(8.4-2)

(8.5-1)

The terms in Equation (8.4-1) can be evaluated by plotting u versus x, u versus y, v versus x, and

v versus y over the full field and then computing the slopes of the curves at the point(s) in

question, Once the strain components are established, the stresses can be computed by applying

the constitutive equations for plane stress. Principal values are obtained using the transformation

equations.

If the strains are large, higher order terms must be taken into account. In the Lagrangian

description, for example,

The derivatives of the in-plane displacement components are determined from the slopes of the

plots produced by analyzing the moiré patterns produced by a cross grating. The derivatives of

the out-of-plane displacement can be determined by plotting the information obtained from the

shadow moiré technique described below.

8.5 Shadow Moiré

As illustrated in Figure 6, the out-of-plane displacement, w, can be measured by projecting a

grating onto the surface of a specimen. In formulating the argument, the assumption is made that

the light source and camera are located at equal distances from the surface.

The displacement is typically measured relative to a flat reference surface. In Figure 6, for

example, a moiré pattern is generated by superimposing the image of the grating projected onto

the reference surface (AB) with that projected onto the specimen (A’B’). The displacement

normal to the surface is given by

8.7

(8.5-2)

Lightsource

Camera

z

OS

L

D

x’x

dz

α’ β’

βα

A’

A

a b c d e

B’

B

x

α’ β’

(8.5-3)

(8.5-4)

where p is the pitch and n is the fringe order number. Although the angle of illumination, α, and

the angle of observation, β, vary from point to point, the sum of their tangents is constant, and

Equation (8.5-1) holds over the full field.

Figure 6. Set up for the shadow moiré method.

The perspective effect of the coordinates can be corrected for a large specimen by using the

relations

where x’,y’ and x,y are the apparent and real coordinates, respectively.

For the special case where the angle of illumination is set equal to 450, and the surface is viewed

normal to the reference surface at β = 0, Equation (8.5-1) reduces to

The shadow moiré method can be used to obtain the thickness change in a specimen. If a

specimen of thickness, t, is in a state of plane stress, the strain in the direction normal to the

surface of a specimen is given by

8.8

P

P

1.5 in.

1.5 in.


1. Figure 7 shows the moiré pattern created by printing a grating of equispaced circles with a

pitch of 1/10,000" on a 1.5" wide, 1/4" thick, flat uniaxial tensile specimen. The moiré

pattern shown in the figure was observed when a circular master of the original size was

placed in front of the deformed specimen. A load, P = 4,500 lb, was applied to create the

fringe pattern.

(a) What is the Young's modulus of the specimen?

(b) What is the Poisson's ratio of the specimen?

(c) What kind of material do you think this is?

Figure 7. A moiré pattern created by concentric circles.

Answer: (a) E = 10 x 106 psi, (b) ν = 0.3, (c) aluminum.

2. A moiré analysis was performed on the tension specimen shown in Figure 8. Gratings

having a spatial frequency of 1000 lines/in. were used in the analysis. The moiré fringe

patterns are shown superimposed on the specimen with the orientation of the grating used

to generate the pattern depicted to the right of each figure. Knowing,

t = 0.5" w = 3" d = 6.0" P = 1,875 lb.

(a) What is the Young's modulus of the specimen?

(b) What is the Poisson's ratio of the specimen?

Answer: (a) E = 500 x 103 psi, (b) ν = 0.4.

8.9

Figure 8. Moiré patterns created by linear gratings.

9.1

(9.1-1)

(9.1-2)

(9.1-3)

(9.1-4)

CHAPTER 9 – STRAIN GAGES

9.1 Introduction

As explained in Chapter 2, there are six components of strain which may be expressed in terms

of the displacements via Equation (2.2-1) as

When measurements are made on a free surface contained in the x,y plane, the plane stress

assumption,

holds, and the relations in Equation (9.1-1) reduce to

Since there are three independent non-zero relations in Equation (9.1-3), it is sufficient to make

three independent strain measurements at the point in question to completely define the strain

state there.

One approach is to measure the normal strain along three different directions and use the strain

transformation relations

9.2

BackingEncapsulation

Copper-coated tabs

(9.2-1)

to extract the required information. Once this is done, the constitutive equations [see Equation

(3.2-3)] can be applied to obtain the stress; transformation equations can be used to obtain the

principal values and/or maximum shear stress.

There are many different types of strain gages and they are nominally classified into five basic

groups: mechanical, electrical, optical, acoustical, and pneumatic. Most of these devices are

designed to obtain normal strain in one direction by measuring the change in length which occurs

over a relatively large gage length.

The most important characteristics of a strain gage are its length, sensitivity, accuracy, and

range. Since most units measure the average strain over the length, it is advantageous to make

the latter as small as possible. The sensitivity is the smallest reading that can be measured on the

scale associated with the gage. But, sensitivity does not necessarily mean accuracy. The latter

has to do with the smallest quantity that can be extracted when measurements are repeated.

Range, on the other hand, is the extent to which measurements can be made without having to

replace and/or realign the gage.

It is also advantageous to have an output which is linearly proportional to strain, so that

calibration curves and/or look-up tables are not required.

9.2 Electrical Resistance Strain Gages

The electrical resistance strain gage is perhaps the most widely employed strain measurement

device. As illustrated in Figure 1, a typical strain gage consists of a metal foil alloy etched into

a grid pattern. The gage is bonded to the test beam so that the foil experiences the same normal

strain (ΔL/L) as the specimen.

Figure 1. An electrical resistance strain gage.

The resistance of the gage, R, is measured in Ohms (Ω) and may be expressed as

9.3

(9.2-2)

(9.2-3)

V

R1 R2

R3R4

G

where ρ is the resistivity [measured in Ω-m (Ω-in.)], L is the length [m (in.)], and A is the cross-

sectional area [m2 (in.

2)].

When the strain parallel to the grid is positive, the length of the foil increases while its cross-

sectional area decreases. Assuming that the resistivity remains constant, Equation (9.2-1) shows

that the resistance of the gage will increase. When a constant current is passed through the gage,

this change in resistance, ΔR, produces a change in voltage (i.e., Ohm’s law; V = IR). The

voltage change is converted to strain by using a strain indicator.

The manufacturer of the gage provides a gage factor, Sg, so that the change in resistance can be

correlated with the strain. Sg is a dimensionless quantity typically having a magnitude of 2.0.

The gage factor equation is given by,

where R is the resistance and ε is the strain. Rearranging terms,

Typically, R = 120 Ω and Sg = 2.0; strain ranges from εmax = 10,000 με to εmin = 5 με.

Substitution of these values into Equation (9.2-2) shows that the resistance changes to be

measured range from 2.4 to 0.012 Ω. The most practical means of accurately measuring these

relatively small resistance changes is by using a Wheatstone bridge. The bridge circuit, shown in

Figure 2, has an input voltage source, V, four resistors, R, and a galvanometer, G.

Figure 2. A Wheatstone bridge circuit.

9.4

(9.2-4)

(9.2-5)

(9.2-6)

(9.2-7)

In most cases, R1 is replaced by an active strain gage; R2 is a variable resistor used to balance or

zero the bridge; and, R3 and R4 remain constant.

The galvanometer is balanced and displays a zero reading when the resistance ratio is satisfied:

As the gage in R1 is strained, the bridge becomes unbalanced and

However, the bridge can be rebalanced by adjusting R2 through a ΔRm such that

If R = R1 = R2 = R3 = R4, the circuit is balanced when ΔR1 = ΔRm. In this case,

ΔRm is measured by using a commercially available instrument called a strain indicator. The

input must include specifying the gage factor and the resistance of the gage. Detailed operating

instructions for a commercially available unit are included in Section 9.9.

9.3 Length Considerations

The gage length of a strain gage is the active or strain-sensitive length of the grid. The endloops

and solder tabs are considered insensitive to strain because of their relatively large cross-

sectional area and low electrical resistance.

As mentioned previously, a strain gage measures the average strain over its length;

consequently, it is desirable to make this dimension as small as possible. Figure 3 illustrates that

the average of any nonuniform strain distribution is always less than the maximum.

Consequently, a strain gage that is noticeably larger than the maximum strain region will indicate

a strain magnitude which is too low.

9.5

Peak strain

Indicated strain

x

(9.4-1)

Figure 3. Strain is averaged over the length of the gage.

9.4 Transverse Sensitivity Corrections

In order to produce a small gage length and achieve the necessary resistance for the Wheatstone

Bridge circuit, the manufacturer is forced to etch the foil into loops. The ends of each loop (see

Figure 1) result in a sensitivity to transverse strain.

This effect, referred to as transverse or cross sensitivity, can be eliminated by introducing a

correction factor found based on the transverse sensitivity factor, Kt, and the ratio of transverse

to axial strain. Kt is provided by the manufacturer, and can be used along with charts (also

provided by the manufacturer) to obtain the true strain readings.

An alternate approach is to read strain from two gages oriented perpendicular to one another.

The true strain measurements are obtained from the measured readings by applying

where Kt is the transverse sensitivity factor, and νo is the Poisson’s ratio of the beam used to

calibrate Kt. Most strain gage manufacturers use a calibration beam having νo = 0.285.

9.5 Temperature Compensation

Of all modem strain gage alloys, constantan is the oldest, and still the most widely used. This

alloy has the best overall combination of properties needed for many strain gage applications. It

9.6

has, for example, an adequately high strain sensitivity, or gage factor, which is relatively

insensitive to strain level and temperature. Its resistivity is high enough to achieve suitable

resistance values in even very small grids, and its temperature coefficient of resistance is not

excessive. In addition, constantan is characterized by good fatigue life and relatively high

elongation capability. Very importantly, constantan can be processed for self-temperature

compensation to match a wide range of test material expansion coefficients.

Self-temperature-compensated strain gages are designed to produce minimum thermal output

(temperature-induced apparent strain) over the temperature range from about -450 to +200

0 C (-

500 to +400

0 F). When selecting either constantan (A-alloy) or modified Karma (K-alloy) strain

gages, the self-temperature-compensation (S-T-C) number must be specified. The S-T-C number

is the approximate thermal expansion coefficient in ppm/0C (ppm/

0F) of the structural material

on which the strain gage will display minimum thermal output. The strategy of the designer is to

balance apparent strains caused by differences in the coefficients of expansion of the gage and

backing materials with the fluctuations in resistivity with temperature to obtain linear response

over an extended temperature range.

9.6 Rosettes

At every point in a solid body there are three mutually perpendicular planes that are called the

principal planes. When a Cartesian axes system is assigned, the unit vectors aligned normal to

these planes are called eigenvectors. By definition, there are no shear stresses on the principal

planes; consequently, the normal stresses are aligned with the eigenvectors. The magnitudes of

the normal stresses are referred to as eigenvalues. The three principal stresses (eigenvalues)

include the maximum and minimum values of the normal stress at the point in question. As

discussed in Chapter 1, the eigenvectors and eigenvalues are found by using transformation

equations which depend on rotations of the coordinate axes.

When a point is located on a free surface (plane stress), one of the eigenvectors is normal to the

surface. The corresponding eigenvalue is equal to zero. The other two eigenvectors (principal

stress directions) and eigenvalues (principal stresses) are found by considering a rotation of

coordinates around this eigenvector. This can be done analytically by using the transformation

equations or graphically by using Mohr’s circle.

Since the in-plane principal stresses are always perpendicular to one another, it is sufficient to

determine the angle from a reference axis to only one of them. Since the magnitudes of the

stresses are also unknown, three independent equations are required to fully establish the stress

state. In the case of a linearly elastic, homogeneous, and isotropic material, the directions of

principal stress and strain coincide, and three independent strain measurements can be made to

determine the two in-plane principal stresses and their directions. In most cases, this is done by

measuring the normal strains along three different directions.

The three axes along which strains are measured can be arbitrarily oriented about the point of

interest. For computational convenience, however, it is preferable to space the measurement

9.7

(9.6-1)

(9.6-2)

(9.6-3)

axes apart by submultiples of π, such as π/3 (60o) or π/4 (45

o). Integral arrays of strain gages

intended for simultaneous multiple strain measurements about a point are known as rosettes.

Figure 4 shows two of the most common commercially available strain rosettes. The rosette at

the left is called a "delta" or equiangular rosette while the rosette at the right is a 45o rectangular

rosette. The configurations are analyzed by the manufacturer who provides the governing

equations required to extract critical information regarding the strain and the stress.

Figure 4. The delta (left) and rectangular (right) rosette.

To perform this analysis, an axis is typically aligned with one of the gages. The general strain

transformation

is applied to each of the gages with θ measured counterclockwise from the x axis. The three

equations are manipulated to obtain εxx, εyy, and γxy in terms of the three normal strain readings

taken from the gages (ε1, ε2, ε3). Then, a final transformation is made to obtain the principal

values.

For the delta rosette, the principal strains (εp and εq) and the orientation of the principal axes (θp

and θq) are given in terms of the three measured strains as,

and

9.8

(9.6-5)

(9.6-4)

(9.6-6)

(9.6-7)

The rectangular rosette, on the other hand, is characterized by,

and

The algebraically maximum (εp) and minimum (εq) principal strains correspond to the plus and

minus alternatives, respectively. As discussed in Chapter 3, even though there is no stress

normal to the surface (i.e., σzz = 0), the surface deforms in this direction and the corresponding

strain (εzz) is

where υ is the Poisson’s ratio.

The information in Equation (9.6-6) can be introduced into the generalized Hooke’s Laws to

obtain the simplified expressions for the biaxial case at hand. For plane stress, the principal

stresses are given by

where E is the elastic modulus.

Since the principal directions of stress and strain coincide for the material in question, Equations

such as (9.6-3) and (9.6-5) can be used to obtain the principal stress directions.

Several other configurations are commercially available. Some of these have two pair of gages

with the gages in each pair oriented perpendicular to one another. The configuration provides an

extra reading for redundancy. The main advantage is that the strains in each gage pair can be

corrected for transverse sensitivity using Equation (9.4-8).

9.7 Gage Selection and Series

Most strain gages are characterized by a gage designation series which is provided by the

manufacturer. Figure 5, for example shows how Micro-Measurements, Inc. designates their

gages. It is evident that there are but five parameters to select, not counting options. These are:

the gage series, the S-T-C number, the gage length and pattern, and the resistance.

9.9

(9.8-1)

Figure 5. Gage designation.

Of the preceding parameters, the gage length and pattern are normally the first and second

selections to be made, based on the space available for gage mounting and the nature of the stress

field in terms of biaxiality and expected strain gradient. The next step is to select the gage series,

thus determining the foil and backing combination, and any other features common to the series.

After selecting the gage series (and option, if any), reference is made to a catalog to record the

gage designation of the desired gage size and pattern in the recommended series. The resistance

and S-T-C number are then specified.

9.8 Dummy Gages and Transducer Design

More than one strain gage can be inserted into the Wheatstone bridge circuit previously shown in

Figure 2. In general, the reading from the bridge is proportional to the net strain given by

where the constant K is typically set to one by specifying the resistance and gage factor on the

instrument used to measure the strain.

Equation (9.7-1) shows that the strains in gages placed in opposite arms add while the strains in

gages placed in adjacent arms subtract. This property can be used to compensate for

temperature, as well as, for transducer design.

Figure 6, for example, shows a cantilever beam which is part of a transducer used to measure the

load, P. The beam is equipped with a single strain gage attached to its upper surface.

For argument sake, assume that the strain reading consists of two parts: one corresponding to the

actual strain produced by the load, εP, the other an apparent strain due to temperature, εT. If the

9.10

(9.8-3)

(9.8-2)

L

P

1

(9.8-4)

gage is placed in Arm No. 1 of the bridge and the bridge adjusted so that K = 1.0, Equation (9.8-

1) predicts

Figure 6. A load transducer.

Since the net strain includes the apparent strain due to temperature, the system is not

compensated for environmental effects.

One approach which can be taken to compensate for environmental effects is to place a dummy

gage in Arm No. 2 of the bridge. The dummy gage is placed on a similar material that is not

loaded but subjected to a similar environment so that it experiences εT only. In this case,

Equation (9.8-1) predicts

The dummy gage is not the only strategy that can be applied. Figure 7, on the other hand,

depicts a beam equipped with two gages, one on the upper surface and one on the lower surface.

If the beam has a rectangular cross section, the mechanical strain in the gage on the lower surface

is of opposite sign from that on the upper surface; both gages experience the same apparent strain

due to temperature. By placing the second gage into Arm No. 2 with the original gage still in

Arm No. 1, Equation (9.8-1) predicts

Equation (9.8-3) shows that not only is the transducer temperature compensated, it is twice as

sensitive to load.

9.11

L

P

1

2

(9.8-5)

Figure 7. A temperature compensated unit.

The exercise demonstrates how environmental effects can be compensated using the fact that

similar readings in adjacent arms of a Wheatstone bridge subtract from one another. The

transducer could have been designed with four active gages, two on the upper surface and two on

the lower surface of the cantilever. If they are placed in Arms No. 1 and 3, and 2 and 4,

respectively, Equation (9.8-1) predicts

Other transducers can be designed to measure bending, torsion, pressure, etc. using the unique

properties of the Wheatstone bridge and possibly incorporating dummy gages.

9.9 P-3500 Strain Indicator

The P-3500, shown in Figure 8, is a portable, battery-powered precision instrument which can

be used to monitor strain in a surface mounted electrical resistance strain gage. As illustrated in

Figures 9, the unit features an LED output. The push-button controls provide an easy-to-follow,

logical sequence of setup and operational steps.

Connect Strain Gage

Resistive strain gages are normally connected at the binding posts located on the right of the

front panel. These binding posts are color-coded in accordance with conventional practice, and

are clearly labeled. Input connections for full-, half-, and quarter-bridge configurations are

shown on the inside cover of the instrument.

To use this unit, connect the strain gage leadwires to the binding posts. Select the desired bridge

configu1ration using the BRIDGE push button. Set the MULT push button to X1 position.

9.12

Figure 8. P-3500 strain indicator. Figure 9. P-3500 push-button control panel.

Initially adjust using AMP ZERO - Orange

Depress the AMP ZERO push button. Allow the instrument to warm up for at least two minutes.

Rotate the AMP ZERO finger-tip control for a reading of ±0000. [To save time, the instrument

may be left in the AMP ZERO position while the gage(s) is being connected.]

Set the GAGE FACTOR - Orange

Set the GAGE FACTOR switch to the desired gage factor range and depress the GAGE

FACTOR push button. The gage factor will be displayed on the readout. Rotate the GAGE

FACTOR potentiometer for the exact desired gage factor, and lock the control in place. The

knob utilizes a lever which must be rotated clockwise to lock it in place. The knob can be

unlocked simply by rotating the lever back to the counterclockwise stop.

Select RUN - Green

Select the X1 or X10 MULT position as required and depress the RUN push button. In this

position all internal circuitry is configured to make an actual strain measurement. Set the

BALANCE range switch and rotate the BALANCE potentiometer to obtain a reading of ±0000

with no load on the test structure, and lock the control in place. The knob utilizes a lever which

must be rotated clockwise to lock it in place. The knob can be unlocked simply by rotating the

lever back to the counterclockwise stop.

Note: A reading of ±0000 with flashing colons indicates an off-scale condition that is usually

caused by improper input wiring or a defective strain gage installation.

The test structure may now be loaded and the reading recorded.

9.13

9.10 Two-Wire and Three-Wire Strain Gage Circuits

All commercial static strain indicators employ some form of the Wheatstone bridge circuit to

detect the resistance change in the gage with strain. When a single active gage is connected to

the Wheatstone bridge with only two wires, as illustrated in Figure 10, both of the wires must be

in the same leg of the bridge circuit with the gage. One of the effects of this arrangement is that

temperature induced resistance changes in the lead wires are manifested as apparent strain by the

strain indicator. The errors due to lead wire resistance changes in single-gage installations with

two-wire circuits can be minimized by minimizing the total lead wire resistance; that is, by using

short lead wires of the largest practicable cross-section.

Figure 10. Two-wire circuit for connecting a single active strain gage (quarter-bridge circuit).

When two matched gages are connected as adjacent legs of the bridge circuit (with the same

length lead wires, maintained at the same temperature), the temperature effects cancel since they

are the same in each leg, and like resistance changes in adjacent legs of the bridge circuit are self

-nullifying. The two-wire circuit for connecting strain gages is shown in Figure 11.

Figure 11. Two-wire circuit for connecting two strain gages (half-bridge circuit).

Lead wire effects can be virtually eliminated in single active gage installations by use of the

"three- wire" circuit. In this case, a third lead (representing the center point connection of the

9.14

bridge circuit), is brought out to one of the gage terminals. Resistance changes in the bridge

center point lead do not affect bridge balance. For this method of lead wire compensation to be

effective, the two lead wires in the adjacent bridge arms should be the same length, and should

be maintained at the same temperature. The three-wire circuit is the standard method of

connection for a single active temperature-compensated strain gage in a quarter-bridge

arrangement. This approach is illustrated in Figure 12.

Figure 12. Three-wire circuit for connecting a single active gage (quarter-bridge circuit).

Contact resistance at mechanical connections within the Wheatstone bridge circuit can lead to

errors in the measurement of strain. Connections should be snugly made. Following bridge

balance, a "wiggle" test should be made on wires leading to mechanical connections. No change

in balance should occur if good connections have been made.

If necessary, contact surfaces may be cleaned of oils with a low residue solvent such as isopropyl

alcohol. If long periods of disuse have caused contact surfaces to tarnish, clean them by scraping

lightly with a knife blade.

9.15


1. Figure 13 shows how a strain gage is attached to measure the strain in the longitudinal

direction on the cylindrical surface of a pressure vessel of 23.5 inch outside diameter and

0.3 inch wall thickness. The strain gage reads 120 x 10-6

in/in and the material properties

are E = 30 x 106 psi and υ = 0.25.

(a) Determine the three principal strains on the cylindrical surface of the vessel.

(b) Compute the principal stresses in the wall.

(c) What is the gage pressure inside the vessel?

Figure 13. A strain gage mounted on a cylindrical pressure vessel.

Note: The biaxial stress distribution in a pressure vessel is given in terms of the gage pressure, p,

the inner radius, ri, and the thickness, t, by

Answer: (a) εxx = 120 με, εyy = 420 με, εzz = -180 με, (b) σxx = 7200 psi, σyy = 14400 psi, (c) p =

377.3 psi.

2. Figure 14 shows how a strain gage is attached to measure the strain in the longitudinal

direction on the cylindrical surface of a pressure vessel of 30 inch outside diameter and 0.5

inch wall thickness. The strain gage reads 630 x 10-6

in/in and the material properties are E

= 10 x 106 psi and υ = 0.3.

(a) Determine the three principal strains on the cylindrical surface of the vessel.

(b) Compute the principal stresses in the wall.

(c) What is the gage pressure inside the vessel?

Figure 14. A strain gage mounted on a cylindrical pressure vessel.

Note: The biaxial stress distribution in a pressure vessel is given in terms of the gage pressure, p,

the inner radius, ri, and the thickness, t, by

9.16

Answer: (a) εxx = 148 με, εyy = 630 με, εzz = -334 με, (b) σxx = 3706 psi, σyy = 7412 psi, (c) p =

255.6 psi.

3. Normal strains are measured along three directions in the

rosette shown in Figure 15. Express the principal

strains (ε1 and ε2) and the angle of orientation of the

principal direction, φ, measured with respect to the x

direction in terms of the readings taken from the gages

(εA, εB, and εC). Be sure to show all work.

Answer: ε1,2 = (εA + εC)/2 ± ½ {(εA - εC)2 + [(4εB - εA -

3εC)/(3)½ ]

2}

½, φ = ½ tan

-1 {(4εB - εA - 3εC)/[(3)

½ (εA - εC)]}.

Figure 15. A strain gage rosette.

4. Normal strains are measured along three directions in the

rosette shown in Figure 16. Express the principal

strains (ε1 and ε2) and the angle of orientation of the

principal direction, φ, measured with respect to the x

direction in terms of the readings taken from the gages

(εA, εB, and εC). Be sure to show all work.

Answer: ε1,2 = (εA + εC)/2 ± 1/√2 [(εA - εB)2 + (εB - εC)}

2]1/2

; φ

= ½ tan-1

[2εB - εA - εC)] / (εA - εC).

Figure 16. A strain gage rosette.

5. Four electrical-resistance strain gages have been mounted on a torque wrench and wired

into a Wheatstone bridge as shown in Figure 17. The designer of the system claims that

the output signal from the bridge is proportional to the torque applied by the wrench

irrespective of the point of application of the load P so long as it remains to the right

of the gages. Analyze the system and prove or disprove this claim.

9.17

Figure 17. A torque wrench.

Hint: The applied moment at given point on the wrench is given by multiplying the

magnitude of the applied load, P, by the distance between the point of application of the

load and the point in question.

Answer: The claim is valid.

A1.1

(A1.1-1)

APPENDIX I – REVIEW OF STATICS

AND MECHANICS OF MATERIALS

A1.1 2-D Forces in Rectangular Coordinates

A force may be represented by a directed line segment. Since it is a vector quantity, the length

represents the magnitude. The direction must be specified relative to a given axes system.

As illustrated in Figure 1, one of the most effective methods for working with concurrent,

coplanar forces is to resolve the forces into components along orthogonal coordinate axes.

Figure 1. A force may be resolved into components along rectangular coordinates.

In this case,

When using rectangular coordinates, a scalar component is assumed to be positive when its

corresponding vector is in a positive coordinate direction; otherwise, it is negative.

A1.2 3-D Rectangular Coordinates

When adding three or more forces, there is no practical trigonometric solution from the polygon

which describes the resultant. The best approach for finding the resultant is to resolve each force

into rectangular components. The components are then composed into their resultant.

A1.2

(A1.2-1)

(A1.2-2)

(A1.3-1)

Figure 2 shows the case in which the line of action of a force is defined by two points. The

force may be resolved into rectangular components using a vector approach.

Figure 2. A force may be directed along a line segment.

Referring to Figure 2,

where

A1.3 Addition of Concurrent Forces in Space/3-D Equilibrium of Particles

When two or more forces act on the same particle, the resultant can be determined from their

components; since,

xx FR yy FR zz FR

222

zyx RRRR

R

R x

x cos R

R y

y cos .cosR

R z

z

Equilibrium exists when a particle remains at rest or moves with a constant velocity. In this

case,

A1.3

(A1.3-2)

(A1.4 -1)

(A1.4-2)

.000 zyx FFF

The method of attack is to resolve all forces into their components, sum the components and then

apply the appropriate equations.

A1.4 Moments and Couples

Referring to Figure 3, the moment of a force F acting at point A about point O may be expressed

as either

dFM )

or

FxrM O

where d is the perpendicular distance from the reference point O to the line of action of the force

F; and, r is a position vector drawn from O to any point along the line of action of F.

These equations can also be applied to compute the moment produced by the couple shown in

Figure 4.

Varigon's theorem states that the moment of a force about any axis is equal to the moments of its

components about that axis. Conversely, the moment of several concurrent forces about any axis

is equal to the sum of the moment produced by their resultant about that axis.

It should be noted that the maximum moment for a given force, or the least force necessary to

produce a given moment, occurs when the force is perpendicular to the moment arm.

Figure 3. The moment of a force may be

computed as a scalar according to convention

or expressed in terms of a vector cross

product.

Figure 4. The two diametrically opposed

forces form a couple that causes a rigid body

to rotate about an axis perpendicular to the

plane formed by the forces.

A1.4

(A1.5-1)

(A1.5-2)

(A1.5-3)

(A1.5-4)

(A1.5-5)

A1.5 Equilibrium of Rigid Bodies

In general, distributed forces and applied couples cause a rigid body to translate and rotate. For

the body to remain at rest or to move with a constant velocity,

.00 MF

The expressions in Equation (A1.5-1) are vector equations which correspond to six scalar

equations; namely,

0

0

0

z

y

x

F

F

F

0

.0

0

z

y

x

M

M

M

When dealing with a 2-D force and moment system, the equilibrium equations become:

0xF 0yF .0Az MM

where A is an arbitrary point in space, usually taken in the X-Y plane. The other three

equilibrium equations are automatically satisfied. Alternate equations may be applied to

determine these unknowns provided that they result in independent equations. One could use,

for example,

0xF 0AM 0BM

or

0AM 0BM .0 CM

To eliminate unknowns and thus simplify the analytical work, one could select the moment

center so that the line of action of one or more unknowns passes through the point and/or align

an axis with one of the unknowns and take the summation of forces in the perpendicular

direction.

A1.6 Center of Gravity and Centroid

For the purposes of computing the reactions at the points of constraint, the distributed weight of

a body can be considered to be a concentrated force acting through a single point called the

center of gravity. For the flat plate shown in Figure 5,

A1.5

(A1.6-1)

(A1.6-2)

(A1.6-3)

n

i

i

n

i

ii

n

i

ii

w

wx

W

wx

x

1

11.

1

11

n

i

i

n

i

ii

n

i

ii

w

wy

W

wy

y

For a homogeneous plate of constant thickness, the center of gravity is called the centroid. In

this case,

n

i

i

n

i

ii

n

i

ii

A

Ax

A

Ax

x

1

11.

1

11

n

i

i

n

i

ii

n

i

ii

A

Ay

A

Ay

y

As the number of plates increases to infinity, the plates have differential areas and the finite sum

becomes an integral. In this case,

dA

dAx

A

dAxx

elel.

dA

dAy

A

dAyy

elel

The expressions xeldA and yeldA are called the first moment of the area about the Y and X axis,

respectively. If the centroid is located on a coordinate axis, the first moment of the area with

respect to that axis is zero. The converse also holds true.

The center of gravity or centroid is determined using either the method of integration or the

method of composites. When the method of composites is used, a rectangular or polar coordinate

axes system is established. The coordinates of the centroid of each individual portion of the area

are assigned as positive or negative depending upon their position with respect to the coordinate

axes system. Areas and/or lengths are considered as positive or negative depending upon

whether the shape must be added or subtracted, respectively, to form the overall shape desired.

Figure 5. Finding the centroid of a flat plate located in the XY plane.

A1.6

(A1.7-1)

(A1.7-2)

A1.7 Moments of Inertia

Inertia has significance when used in conjunction with other quantities (equations of motion in

dynamics, flexure formula for beam stress in mechanics of materials, etc.). Basically, the term

inertia is used to describe the tendency of matter to show resistance to change. Referring to the

X and Y axes shown in Figure 6, the area moments of inertia are,

AdyI x

2

.2 AdxI y

The units associated with the area moment of inertia are in4, cm

4, etc. Since ρ

2 > 0 (y

2 and x

2 in

the above equation), the moment of inertia is always positive.

The polar moment of inertia is used to characterize torsional behavior in cylindrical shafts, the

rotation of slabs, etc. It is computed for an axis passing through point O perpendicular to the x,y

plane (i.e., around a z axis). By definition,

yxO IIAdrJ 2

where r is the distance from the element of area to the point O.

Imagine that the area shown in Figure 6 is squeezed either into a strip oriented parallel to one of

the coordinate axes as shown in Figures 7 and 8, or, into a thin ring centered at O as shown in

Figure 9.

Figure 6. The moment of inertia is defined with respect to a rectangular coordinate system.

Figure 7. The area is

assumed to be a strip

parallel to the X axis.

Figure 9. The area is assumed to

be a thin ring centered at the

origin.

Figure 8. The area is

assumed to be a strip

parallel to the Y axis.

A1.7

(A1.7-3)

(A1.7-4)

(A1.7-5)

(A1.7-6)

For the area to have the same Ix, Iy and Jo, the respective strips must lie at distances kx, ky and ko

respectively, such that,

AkI xx

2 AkI yy

2 .2 AkJ OO

The distances kx, ky and ko are called the radii of gyration. Note that kx and ky are measured

perpendicular to the x and y axes, respectively, not along them.

As illustrated by Figure 10, it is often convenient to transfer the moment of inertia between a

centroidal axis and an arbitrary parallel axis. This transformation can be accomplished using a

simple formula known as the parallel axis theorem:

2dAII

where I is the area moment of inertia about the arbitrary axis, is the moment of inertia about a

parallel axis passing through the centroid of the area, and d is the perpendicular distance between

the two parallel axes.

Utilizing the expressions for the radii of gyration, the parallel axis theorem may also be

expressed as follows,

.222 dkk

This approach can also be taken to transfer the polar moment of inertia and the polar radius of

gyration, respectively, as follows,

.2222 dkkdAJJ OOOO

If the area under consideration is made up of several parts, its moment of inertia about an axis is

found by adding the moments of inertia of the composite areas about that axis. In general, it is

Figure 10. The moment of inertia may be transferred between two parallel axes.

A1.8

(A1.8-1)

(A1.9-1)

necessary to transfer each moment of inertia to the appropriate axis by the parallel axis theorem

before adding.

Note: The radius of gyration is not the sum of the component areas' radius of gyration. It is first

necessary to perform the superposition of the moments of inertia and then use the

definition the k2 = Itotal/Atotal.

A1.8 Shear and Moment Diagrams

A beam is a structural member that is designed to support loads at various points along its span.

In most cases, loads are perpendicular to the beam. A portion of mechanics of materials deals

with selecting a cross section to resist these shearing forces and bending moments. This often

requires drawing shear and moment diagrams for the beam.

The method of attack is to determine the reactions at points of constraint, then cut the beam at

various sections along the span, indicating on which portion of the beam internal forces are

acting. A convention must be established; the shear V and the bending moment M at a given

point of a beam are assumed to be positive when the internal forces and couples acting on each

element of the beam are directed as shown in Figure 11.

The following relationships are often helpful:

Vxd

Mdw

xd

Vd

where w is the distributed load per unit length.

A1.9 Deflection of Beams

Recall that a prismatic beam in pure bending bends into a circular arc with curvature

.)(1

IE

xM

Figure 11. The sign convention used for constructing shear and bending moment diagrams.

A1.9

(A1.9-2)

(A1.9-3)

(A1.9-4)

(A1.9-5)

For pure bending, M(x) is constant over the span. However, when the beam is subjected to a

transverse loading, Equation (A1.9-1) remains valid but both the bending moment and curvature

of the neutral surface vary from section to section.

The elastic curve of the beam is the relation between the deflection y measured at a given point

Q on the axis of the beam and the distance x of that point from some fixed origin. The elastic

curve is illustrated for a beam in Figure 12.

It can be shown from calculus that

2

21

dx

yd

and Eqn. (A1.9-1) becomes

.)(

2

2

IE

xM

dx

yd

For a prismatic beam in which E and I are constant, this second order linear differential equation

can be integrated to obtain the parametrical representation of the elastic curve as follows

x

CdxxMxIEdx

dyIE

01)()(

where dy/dx is the slope of the curve and

.)()(0 0

2120 0

1

x xx x

CxCdxxMdxCdxCdxxMyIE

Figure 12. The elastic curve represents the deflection of the beam.

A1.10

The constants are determined from the boundary conditions imposed on the deflection and slope

by considering the beam and the position of its supports. The three types of supports shown in

Figure 13 need be considered for complete analysis of statically determinate beams.

Concentrated loads, reactions at supports, or discontinuities in a distributed load make it

necessary to divide the beam into several portions, and to represent the moment by a different

function M(x) in each portion of the beam. Additional equations are generated at the interface

between different sections, since the slope and deflection must be continuous at these locations

(the shear and bending moment may be discontinuous).

Recall that in statically indeterminate problems, the reactions may be obtained by considering the

deformations of the structure involved. In these cases, the equilibrium equations may be

combined with the expressions for slope and deflection to evaluate both the constants in the

parametrical equations and the reactions at points of constraint.

A1.10 Synopsis of Mechanics of Materials

When loads are applied to a deformable body they produce stresses. The stresses represent the

force intensity and are computed by dividing the force by the area over which it acts. A normal

stress is produced when the force is perpendicular to the surface under consideration. A tensile

stress results when the force is directed along the outer normal to the surface; a compressive

stress results when the force is directed toward the element. Shear stress results when the force

is tangent to the surface.

The stresses produce changes in shape (deformations) characterized by a quantity called strain.

Normal stresses produce normal strains defined as the change in length of a line segment

divided by the original length of the segment. Shear stresses produce shear strains defined as

the change in angle between two line segments that were originally perpendicular to one another.

Figure 13. The three basic supports required to determine the equations of the elastic curve.

A1.11

(A1.10-1)

(A1.10-2)

(A1.10-3)

(A1.10-4)

There are three basic loading conditions that can be superimposed to produce a general state of

deformation. Axial loading produces either tensile or compressive stresses assumed to be

uniformly distributed across a plane cut normal to the longitudinal axis of the member.

Combinations of normal and shear stresses occur on oblique planes. The stresses are computed

by dividing the load (P) by the area (A) considered; thus,

.A

P

A

P

Bending produces a uniaxial stress condition in which normal stresses occur parallel to the

longitudinal axis of the member. For a prismatic member possessing a plane of symmetry,

subjected at its ends to equal and opposite couples acting in a plane of symmetry, the stress

distribution is linear through the thickness; compressive stresses occur on one side of the neutral

axis and tensile stress occur on the other side. The stress is computed using the flexure formula

I

cM

where M is the moment, c is the distance measured from the neutral axis to the point under

consideration, and I is the centroidal moment of inertia measured around the axis about which

the moment is applied.

Torsion produces shear stresses. In a prismatic member of circular cross section subjected to

couples (torques) of magnitude T, the shear stress acts in the direction of the applied torque. It is

zero at the center of the shaft and maximizes at the outer surface. The stress is computed using

the elastic torsion formula,

J

T

where ρ is the distance measured from the center of the shaft to the point under consideration and

J is the polar moment of inertia.

Shear stresses also result in prismatic members subjected to transverse loads. In this case, the

stress is given by

tI

QV

m

max

where V is the shear force, Q is the first moment of the area measured about the neutral axis of

the portion of the cross section located either above or below the point under consideration, I is

the centroidal moment of inertia of the entire cross section, and t is the width of a cut made

through the point perpendicular to the applied load.

A1.12

The stress is related to the strain through constitutive equations (Hooke’s Laws) that depend

upon material properties. In general, these are very complex relations especially when dealing

with composite materials; however, only two independent material constants [typically Young’s

modulus (E) and Poisson’s ratio (υ)] are required when dealing with a linearly elastic (stress-

strain curve is linear), homogeneous (mechanical properties are independent of the point

considered), and isotropic (material properties are independent of direction at a given point)

material. Other constants, such as the shear modulus [G = E/2(1 + υ)], are sometimes used in

these equations. Material constants are evaluated under simple loading conditions (tension test,

bending test, torsion tests) on specimens having simple geometry (typically rods, bars, and

beams).

It is often necessary to find the weakest link in a structure and this can be done by studying the

stress distribution at a critical location. This is not as easy as it may sound, since stress is a

tensor and must be studied using transformation equations. That is, the stress distribution

changes from point to point in a loaded member. However, there is usually a critical point or

region in the structure where the stress is the highest. The stress distribution at this critical

location depends upon the plane considered. There is a set of planes on which the normal stress

becomes maximum; shear stress is zero on these planes. The planes of maximum shear typically

have a non-zero normal stress component. Analytical equations (transformation equations) or

graphical techniques (Mohr’s circle) can be employed to pinpoint the maximum stresses and the

planes on which they act. These quantities are used in conjunction with failure criteria to assess

structural integrity.

In some cases, a member may fail due to its unstable geometry as opposed to the imposed stress

state. Column buckling is one of the areas where such considerations come into play. Design

codes have been developed to prevent these cataclysmic failures.

A1.13

60 mm

15 mm

40 mm

15 mm

6"

2"

8"

2"

1 kip/ft

A

CB

D

15'

5'

3 kips

6' 4'

5 kips

300 kN/m

A C

0.20 m 0.15 m

B

A1.11 Homework Problems

1. As illustrated in Figure 14, a prismatic beam is subjected to a uniformly distributed load over

portion AB of its span. The dimensions of the cross section are shown in Figure 15.

Figure 14. Figure 15.

(a) Draw a complete free body diagram showing an equivalent system constructed by

replacing the distributed load, and determine the reactions at the roller (B) and pin (C).

(b) Establish the analytical relationships for the shear force and bending moment distribution

for all points along the span of the beam illustrated in the figure.

(c) Draw the shear and bending-moment diagrams for the beam illustrated in the figure.

(d) Compute the centroid and the centroidal moment of inertia of the cross section.

(e) Describe the exact location (along the span and on the cross section) where the maximum

normal stress due to bending occurs and compute this quantity.

(f) Describe the exact location (along the span and on the cross section) where the maximum

transverse shearing stress in the beam occurs and compute this quantity.

Answer: (a) RB = 100 kN ↑, RCx = 0, RCy = 40 kN ; (b) AB: V = -300x kN and M = -

150x2 kN-m; BC: V = 40 kN and M = (40x-14) kN-m; (c) see part b; (d) centroid

is 30 mm above bottom and IC = 787 x 10-9

m4; (e) σmax = 343 MPa at top with x

= 0.2 m; (f) τmax = 77.2 MPa at centroid with x = 0.2 m.

2. A prismatic beam is loaded as shown in Figure 16. The dimensions of the cross section are

shown in Figure 17.


A1.14

80 lb/ft

1.5 ft.

A B

9 ft.

1600 lb

11.5 in.

1.5 in.


replacing the distributed load, and determine the reactions at the pin (B) and roller (D).

(b) Establish the analytical relationships for the shear force and moment distribution for all

points along the span of the beam illustrated in the figure.


(d) Compute the centroid and the centroidal moment of inertia of the cross section.

(e) Describe the exact location (along the span and on the cross section) where the maximum

normal stress due to bending occurs and compute this quantity.

(f) Describe the exact location (along the span and on the cross section) where the maximum

transverse shearing stress in the beam occurs and compute this quantity.

Answer: (a) RBx = 0, RBy = 13.7 k ↑, RD = 4.3 k ↑; (b) AB: V = -5 k and M = -5x ft-k; BC: V =

13.7-x k and M = -81+13.7x-x2/2 ft-k; CD: V = 10.7-x k and M = -48+10.7x-x

2/2 ft-k;

(c) see part b; (d) centroid is 5.29 in. above bottom and IC = 151 in4; (e) σmax = - 10.5

ksi at bottom with x = 5 ft; (f) τmax = .81 ksi at centroid with x = 5 ft.

3. A prismatic beam is loaded as shown in Figure 18. The dimensions of the cross section are

shown in Figure 19.



replacing the distributed load, and determine the reactions at the pin (A) and roller (B).

(b) Establish the analytical relationships for the shear force and moment distribution for all

points along the span of the beam illustrated in the figure.


(d) Determine the maximum transverse shearing stress in the beam assuming that it has the

cross section shown.

(e) Determine the maximum normal stress due to bending.

Answer: ( a) RAx = 0, RAy = 1670 lb ↑, RB = 650 lb ↑; (b) in the first segment between A and the

1600 lb load, for example, V = 1670 lb (↓) and M = 1670 x ftlb (cclw), etc.; (c) plot

the relationships found in part (c) along the span; (d) τmax = 145.2 psi; (e) σmax = 959

psi.

A1.15

w0

L

A B

Y

X

w

LA

B

Y

X

4. As shown in Figure 20, a cantilever beam that is fixed into a wall at B is subjected to

uniformly distributed load. The weight of the beam can be neglected. The origin of the

coordinate system is at point A.

Figure 20.

(a) Draw a complete free body diagram of an equivalent beam and determine the reactions

at the fixed support (in terms of w and L).

(b) Draw the shear and bending moment diagrams for the beam illustrated in the figure.

(c) Determine the equation of the elastic curve (in terms of x, L, E, I, w).

(d) Determine the slope at the free end (in terms of L, E, I, w).

Answer: (a) RBx = 0, RBy = wL ↑, MB = wL2/2 clw; (b) V is linear; M parabolic; (c) y = -w/24EI

[x4-4L

3x+3L

4]; (d) θA = wL

3/6EI.

5. As illustrated in Figure 21, a cantilever beam that is fixed into a wall at B is subjected to

uniformly distributed load as shown. The weight of the beam can be neglected. The origin

of the coordinate system is at point A.

Figure 21.

(a) Draw a complete free body diagram of the beam and determine the reactions at the fixed

support (in terms of w0 and L).


(c) Determine the equation of the elastic curve (in terms of x, L, E, I, w0).

(d) Determine the deflection at the free end (in terms of L, E, I, w0).

Answer: (a) RBx = 0, RBy = w0L/2 ↑, MB = w0L2/6 clw; (b) V is parabolic; M cubic; (c) y = -

w0/120EIL [x5-5L

4x+4L

5]; (d) yA = -w0L

4/30EI.

A1.16

L

A B

Y

X

P

6. As illustrated in Figure 22, a cantilever beam that is fixed into a wall at B is subjected to a

concentrated load at its free end, A. The weight of the beam can be neglected. The origin of

the coordinate system is at point A.

Figure 22.

(a) Draw a complete free body diagram of the beam and determine the reactions at the fixed

support (in terms of P and L).


(c) Determine the equation of the elastic curve (in terms of x, L, E, I, P).

(d) Determine the deflection at the free end (in terms of L, E, I, P).

Answer: (a) RBx = 0, RBy = P ↓, MB = PL cclw ; (b) V = P and M = Px; (c) y = P/6EI [x3-

3L2x+2L

3] ; (d) yA = PL

3/3EI.

A2.1

APPENDIX II – BASIC FORMULAS FOR

MECHANICS OF MATERIALS

Axial Loading:

A

P

A

Pave

dt

P

A

Pb

loadallowable

loadultimateSFsafetyofFactor

..

L

EA

LP

strainaxial

strainlateralv

Thermal Deformation:

LTT )( TL

T

T

Generalized Hooke's Laws:

EE

v

E

v

E

v

EE

v

E

v

E

v

E

zyx

z

zyxy

zyx

x

G

xy

xy

G

yz

yz

G

zx

zx

where

)1(2 v

EG

Stress Concentration Factor (Axial Loading):

ave

K

max

Torsion of Circular Shafts:

L

J

TG

GJ

LT

solid shaft: 4

2

1cJ hollow shaft: )(

2

1

2

1

2

1 4

1

4

2

4

1

4

2 ccccJ

A2.2

Transmission Shafts:

TfP 2 slbinslbfthpHzsrpm /6600/550160

1

60

11 1

Stress Concentration Factor (Stepped Shaft in Torsion):

J

cTKmax

Pure Bending (Homogeneous Beams):

IE

MI

1

curvaturecanticlasti

yx

I

yMx

S

Mm

c

IS

Pure Bending (Composite Beams):

1

2

E

En

I

yMnx

IE

M

1

1

Stress Concentration Factor (Notched Beam in Bending):

I

cMKm

Eccentric Axial Loading:

y

y

z

z

xI

zM

I

yM

A

P tantan

y

z

I

I

Shear and Moment Considerations:

I

cM

m

max

tI

QV

m

max w

xd

Vd V

dx

dM

Transverse Loading:

I

QVq

tI

QVave

A2.3

Transformation Equations:

2sin2cos22

2sin2cos22

xy

yyxxyyxx

yy

xy

yyxxyyxx

xx

2cos2sin2

xy

yyxx

xy

2

2

minmax,22

xy

yyxxyyxx

and 0

xy with yyxx

xy

p

22tan

2

2

max2

xy

yyxx

and

2

yyxx

ave

with

xy

yyxx

s

22tan

Mohr's Circle:

02

c

yyxx

c

2

2

2xy

yyxxR

minmaxmax2

1

Pressure Vessels:

Cylindrical: t

rp1

t

rp

22 Spherical:

t

rp

221

Design of Prismatic Beams:

all

MS

max

min

Deflection of Beams:

IE

xM

dx

yd )(2

2

Buckling of Columns:

2

2

e

crL

IEP

2

2

)/( rL

E

e

cr

2rAI

A2.4

Design of Steel Columns:

2rAI

y

c

EC

71.4

:/r ce CLfor ycr

e

y

)(

658.0 67.1

crall

:/rL e cCfor ecr 877.0 67.1

crall

2

2

)/( rL

Ewhere

e

e

AP allall

Conversions:

g = 32.2 ft/s2 = 9.81 m/s

2; 1 slug = 14.59 kg; 1 ft = .3048 m; 1 lb = 4.45 N; 1 psi = 6.895 kPa

Documents

CLASS NOTES - concrete canoeconcretecanoe.org/Webmaster/Resume/CourseNotes/NOTES6.pdfCLASS NOTES Version 5.0 by John A. Gilbert, Ph.D. Professor of Mechanical Engineering University