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TAM 326—Experimental Stress AnalysisJames W. Phillips

Copyright© 1998 Board of TrusteesUniversity of Illinois at Urbana-Champaign

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6. Photoelasticity

Photoelasticity is a nondestructive, whole-field,graphic stress-analysis technique based on an opto-mechanical property called birefringence, possessedby many transparent polymers.

Combined with other optical elements andilluminated with an ordinary light source, a loadedphotoelastic specimen (or photoelastic coatingapplied to an ordinary specimen) exhibits fringepatterns that are related to the difference betweenthe principal stresses in a plane normal to the light-propagation direction.

The method is used primarily for analyzing two-dimensional plane problems, which is the emphasisin these notes. A method called stress freezingallows the method to be extended to three-dimensional problems. Photoelastic coatings areused to analyze surface stresses in bodies ofcomplex geometry.

Advantages and disadvantages

Advantages.—Photoelasticity, as used for two-dimensional plane problems,

• provides reliable full-field values of thedifference between the principal normalstresses in the plane of the model

• provides uniquely the value of the non-vanishing principal normal stress along theperimeter(s) of the model, where stresses aregenerally the largest

• furnishes full-field values of the principal-stress directions (sometimes called stresstrajectories)

• is adaptable to both static and dynamicinvestigations

• requires only a modest investment inequipment and materials for ordinary work

• is fairly simple to use

Disadvantages.—On the other hand, photo-elasticity

Phillips TAM 326—Photoelasticity 6–2

• requires that a model of the actual part bemade (unless photoelastic coatings are used)

• requires rather tedious calculations in orderto separate the values of principal stresses ata general interior point

• can require expensive equipment for preciseanalysis of large components

• is very tedious and time-consuming forthree-dimensional work

Procedure

The procedure for preparing two-dimensionalmodels from pre-machined templates will bedescribed. Alternatively, specimens may bemachined “from scratch,” in which case a computer-controlled milling machine is recommended.

1. Selecting the material. Many polymers exhibitsufficient birefringence to be used as photo-elastic specimen material. However, suchcommon polymers as polymethylmethacrylate(PMMA) and polycarbonate may be either toobrittle or too intolerant of localized straining.Homalite®-100 has long been a popular general-purpose material,1 available in various thick-nesses in large sheets of optical quality.PSM-1® is a more recently introduced material2

that has excellent qualities, both for machiningand for fringe sensitivity. Another goodmaterial is epoxy, which may be cast betweenplates of glass, but this procedure is seldomfollowed for two-dimensional work.

2. Making a template. If more than 2 or 3 piecesof the same shape are to be made, it is advisableto machine a template out of metal first. Thistemplate may then be used to fabricate multiplephotoelastic specimens having the same shapeas that of the template. The template should beundercut by about 0.050 in. through about halfthe template thickness from one side to avoidcontact with the router bit (explained below).

3. Machining the specimen. If the specimen ismachined “from scratch,” care must be taken to

1Manufactured by Homalite Corporation, Wilmington, Del.2Marketed by Measurements Group, Raleigh, N.C.


take very light cuts with a sharp milling cutter inorder to avoid heating the specimen undulyalong its finished edges. A coolant, such asethyl alcohol, kerosene, or water, should be usedto minimize heating.

If a template is used, then a bandsaw with asharp, narrow bandsaw blade is used to roughout the shape of the specimen. A generousallowance of about 1/8 in. should be marked onthe specimen all around the template edge, sincethe blade will heat the material and nick theedge. Then a router with a high-speed carbiderouter bit, preferably with fine multiple flutes,should be used to fabricate the edge of themodel (Fig. 1). A succession of two centeringpins—the first having a diameter larger than thatof the router bit (as shown in the figure), and thesecond one the same size—should be used sothat excess material can first be removedquickly, and then in a very controlled manner,leaving the specimen with the same dimensionsas those of the template.

High-speedcarbide bit

Table

Guide pin

Locknut

Specimen

Double-sticktape

Template

Fig. 1. Use of a template, router, and guide pinto rout the edge of a specimen.

The piece should always be forced into thecutting edge of the bit, that is, from front to backif the piece is on the right side of the bit (as inthe figure). The final router passes should besmooth and very light so as to avoid heating ofthe specimen edges.

4. Drilling the specimen. If the specimen hasholes, such as those used for load-application


points using pins, then these holes should bedrilled carefully with a sharp bit with plenty ofcoolant, such as ethyl alcohol, kerosene, orwater; otherwise unwanted fringes will developaround the edge of the hole. As illustrated inFig. 2, the specimen should be backed with asacrificial piece of similar material in order toavoid chipping on the back side of the specimenas the drill breaks through.

High-speeddrill bit

Table

Sacrificialback piece Specimen

Double-stick tape

Template

Fig. 2. Drilling a hole in the specimen.

A series of 2 or 3 passes of the drill bit throughthe specimen, with coolant added each time, willminimize heat-induced fringes.

5. Viewing the loaded specimen. After thespecimen is removed from the template andcleaned, it is ready for loading. A polariscope(to be described later) is needed for viewing thefringes induced by the stresses. The elements ofthe polariscope must be arranged so as to allowlight to propagate normal to the plane of thespecimen. If a loading frame is needed to placea load on the specimen, then this frame must beplaced between the first element(s) and the lastelement(s) of the polariscope. Monochromaticlight should be used for the sharpest fringes;however, the light source does not need to becoherent, and the light may or may not becollimated as it passes through the specimen.

6. Recording the fringe patterns. An ordinarystill camera or a videocamera may used torecord the fringe patterns.

7. Calibrating the material. The sensitivity of aphotoelastic material is characterized by itsfringe constant fσ , which relates the value Nassociated with a given fringe to the thickness hof the specimen in the light-propagation


direction and the difference between theprincipal stresses σ σ1 2− in the plane normal tothe light-propagation direction:

σ σ σ1 2− =

Nf

h.

By means of an experiment using a model ofsimple geometry subjected to known loading,the value of fσ is determined. The disk indiametral compression is a common calibrationspecimen.

8. Interpreting the fringe patterns. Two types ofpattern can be obtained: isochromatics andisoclinics. These patterns are related to theprincipal-stress differences and to the principal-stress directions, respectively. Details are givenlater in the notes below.

Wave theory of light

The theory of photoelasticity is based on thewave nature of light. Light is regarded as asinusoidal electromagnetic wave having transverseamplitude a and longitudinal wavelength λ, propa-gating in the z direction with velocity v (Fig. 3).

av

z

λ

Fig. 3. Light wave.

A wave propagating in the +z direction may berepresented in trigonometric notation as

a a z vtcos cos ( ) ,Φ = −FH

IK

2πλ

where the quantity Φ = −2πλ

( )z vt is called the

phase of the wave. Terms related to the wave-length and speed are the ordinary frequency f (inHz), the angular frequency ω (in rad/s), and thewave number k, as follows:

kv

fv= = =2

2π

λω π

λ λ, , . (1)


Thus the phase has the alternative representations

Φ = − = − = −22

πλ

ω π( ) .z vt kz t kz ft

The speed of light v in a vacuum is approxi-mately 299.79 Mm/s, independent of its wavelengthor amplitude. From the last of the expressions inEqns. (1), it will be seen that the frequency of agiven light wave must depend on its wavelength:

Color

Speed, v(Mm/s)

Wavelength,λ (nm)

Frequency,f (THz)

Deep violet 299.79 400 750

Green 299.79 550 550

Deep red 299.79 700 430

Notice that the visible spectrum covers a nearly2-to-1 ratio of wavelengths, the blue–violet wave-lengths being much shorter than the orange–redones.

An equivalent way to express the trigonometricform of a wave is in complex notation as

Re Re Re .( )

( )ae ae aeii z vt

i kz tΦn s o t=RS|T|

UV|W|

=− −

2πλ ω

Since the complex notation is much easier tomanipulate when phases and amplitudes undergochanges, we shall use it in these notes. The operatorRe{ } will be omitted for convenience, with theunderstanding that, if at any time a quantity is to beevaluated explicitly, the real part of the expressionwill be taken. A typical expression for a light wavewill therefore be simply

A ae ae aeii z vt

i kz t= = =− −Φ

2πλ ω( )

( ) . (2)

Refraction

When light passes through any medium, itsvelocity decreases to a value

vv

n11

= , (3)

where n1 denotes the index of refraction of themedium (Fig. 4). However, the frequency f of the


wave is unaffected. Therefore the wavelength λ1must also decrease proportionally:

λ λ1

1

1 1= = =v

f

v

n f n.

v vv1

n1n n

z

λ λ1

Air AirMedium

Fig. 4. Refraction.

Note that the time t1 required for light topropagate through a thickness h of medium 1 havingindex of refraction n1 is

th

vn

h

v11

1= = .

v vv1

n1n n

z

v vv2

n2n n

z

δh

Fig. 5. Double refraction.

If similar light waves pass through the samethickness h of two media having indices of


refraction n1 and n2 (Fig. 5), and if n n2 1> (asillustrated in the figure), then the difference intransit times t t2 1− will be

t th

v

h

v

h

vn n2 1

2 12 1− = − = −( ) .

Therefore, the phase difference δ between the twowaves after they emerge from the media will be

δ = − = −v t t h n n( ) ( ) .2 1 2 1 (3)

We will need this relation later when we considerbirefringence of materials.

Polarization

A given light wave has an amplitude vector thatis always perpendicular to its propagation direction.However, for ordinary light, the orientation of theamplitude vector in the plane perpendicular to thepropagation direction is totally random.

Plane-polarized light

If the amplitude vectors of all light raysemanating from a source are restricted to a singleplane, as in Fig. 6, the light is said to be planepolarized.

a

v

z

λ

Plane ofpolarization

z

Fig. 6. Plane-polarized light.

An observer viewing the light wave head-onwould see the wave with its amplitude vectorrestricted to a single plane, which is called the planeof polarization. This plane is not necessarily


vertical, as shown in the figure, but vertical polari-zation is quite common. Polaroid® sunglasses, forexample, employ vertically polarizing media in bothlenses to block the horizontally polarized light thatis reflected from such horizontal surfaces ashighways and lakes.

Addition of two plane-polarized waves in phase

If two waves are propagating in the samedirection, vector algebra may be applied to the waveamplitudes to determine the resultant waveamplitude. Consider first the addition of two plane-polarized waves that are in phase, but that havedifferent planes of polarization (Fig. 7).

Plane ofpolarizationz

z

Resultant

λ

Fig. 7. Addition of two plane-polarized light wavesthat are in phase.

The vector addition of these two waves pro-duces a new plane-polarized wave having the samefrequency, wavelength, and phase as the componentwaves. Note that the two planes of polarizationneed not be orthogonal in order for this result tohold.

Elliptically polarized light

A more interesting case arises when two plane-polarized waves of arbitrary amplitude and differentphase are combined (Fig. 8).


zEllipse

λ

δ

Fig. 8. Elliptically polarized light.

In the figure, the horizontally polarized wave isahead of the vertically polarized wave by a distanceδ, if we regard the positive senses for the horizon-tally and vertically polarized waves as being to theright and upward, respectively. At the instantshown, at the leading edge of the wave, thehorizontal component is negative and the verticalcomponent vanishes; therefore the resultant is in thenegative horizontal direction. A tiny instant later,the horizontal component becomes slightly morenegative and the vertical component rapidlybecomes negative; therefore the resultant is in thefourth quadrant as viewed backwards along the +zaxis. With increasing time, an elliptical path istraced by the amplitude vector of the resultant wave,as shown.

Thus the result of adding two plane-polarizedwaves that are neither in phase nor in the sameplane is a special kind of rotating wave, called anelliptically polarized wave, having the samefrequency as the component waves, but which is notrestricted to a single plane.

Circularly polarized light

A very important special case of ellipticallypolarized light is circularly polarized light, whichcan be (and usually is) created by combiningorthogonal plane-polarized waves of equal ampli-tude that are out of phase by exactly one-quarter of awavelength, i.e. δ λ= / 4 . See Fig. 9.


z

Circle

λ

δ =λ 4

ω

Fig. 9. Circularly polarized light.

For this special combination of plane waves, theresultant wave is a rotating wave having constantamplitude and constant angular frequency ω.

Optical elements

The method of photoelasticity requires the useof two types of optical element—the polarizer andthe wave plate.

Polarizer

A polarizer (Fig. 10) is an element that convertsrandomly polarized light into plane-polarized light.It was the introduction of large polarizing sheets byPolaroid Corporation in 1934 that led to the rapidadvance of photoelasticity as a stress-analysis tool.Prior to that, small naturally occurring crystals wereused for this purpose.

z

P

Incoming light

Transmittedlight

Passed

Rejected

Polarizer

Direction ofpolarization

Fig. 10. Polarizer.


In the figure, a single light wave with arbitrarilyoriented wave amplitude approaches the polarizerfrom the left. As this wave encounters the polarizer,it is resolved into two vector components—oneparallel to the polarizing direction of the polarizer,and one perpendicular to it. The parallel componentis passed, but the perpendicular one is rejected.Light emanating from the polarizer is thereforeplane-polarized in the direction of polarization ofthe polarizer.

Viewed in ordinary (unpolarized) light, a polar-izer always looks dark because half the light strikingit is rejected.

Wave plate

A wave plate (Fig. 11) resolves incident lightinto two components, but instead of rejecting one ofthese components, it retards it relative to the othercomponent.

FastSlow

z

α

δ

sf

Waveplate

Incoming light

Transmitted light

Fig. 11. Wave plate.

In the figure, the “fast” axis of the wave platemakes an angle α with respect to an arbitrarilychosen reference direction. The component f of theincident light with amplitude vector in this orien-tation is retarded somewhat as it passes through thewave plate. However, the orthogonal component sis retarded even more, resulting in a phase lag δbetween this “slow” component and the “fast” one.The term double refraction is often used to describethis behavior.

Wave plates may be either permanent ortemporary. A permanent wave plate has a fixedfast-axis orientation α and a fixed relative retarda-


tion δ. Such wave plates have their use in photo-elasticity, as will be seen subsequently. A tempo-rary wave plate has the ability to produce doublerefraction in response to mechanical stimulus.Photoelastic specimens are temporary wave plates.

Quarter-wave plate

A quarter-wave plate (Fig. 12) is a permanentwave plate that induces a phase shift δ equal toλ / 4, where λ is the wavelength of the light beingused. The quarter-wave plate is an essentialelement in a circular polariscope (discussed later).

P

Fast

Slow

Quarter-waveplate

Incoming light

Transmitted light

z

fs

α = π4

δ = λ4

Fig. 12. Quarter-wave plate.

In Fig. 12, the incident light is assumed to haveits amplitude vector in a vertical plane, and the fastaxis of the quarter-wave plate has been oriented at45º with respect to the horizontal axis. Thesespecial conditions have no bearing on the definitionof a quarter-wave plate, but are often employed witha quarter-wave plate to produce circularly polarizedlight (Fig. 9)—observe that the orthogonal f and scomponents emanating under these special con-ditions have equal amplitude and are separated byexactly one-quarter wavelength.

The phase shift δ induced by any permanentwave plate is usually independent of the wave-length λ. Therefore, if a wave plate has been


designed to be a quarter-wave plate for green light(say λ = 550 nm), it will induce a relative phaseshift δ equal to (550 nm)/4 or 138 nm. This samewave plate, if used for red light (say λ = 650 nm),will still induce an absolute phase shift of 138 nm,which is now only 138/650, or about 0.212, of thewavelength λ of red light. Quarter-wave plates aretherefore said to be imperfect, not because of anymanufacturing defect, but because the retardationexpressed as a fraction of the wavelength iswavelength dependent.

Birefringence

Photoelastic materials are birefringent, that is,they act as temporary wave plates, refracting lightdifferently for different light-amplitude orientations,depending upon the state of stress in the material.

In the unloaded state, the material exhibits anindex of refraction n0 that is independent oforientation. Therefore, light of all orientationspropagating along all axes through the materialpropagate with the same speed, namely v n/ 0.

In the loaded state, however, the orientation of agiven light amplitude vector with respect to theprincipal stress axes, and the magnitudes of theprincipal stresses, determine the index of refractionfor that light wave.

Birefringentmaterial

σ1σ2

α

Fast

Slow

xz

y

Fig. 13. Principal-stress element.


Effectively, a birefringent material acts as atemporary wave plate (Fig. 13). The index ofrefraction n1 for light having its amplitude vector inthe direction of the maximum principal normalstress σ1 is given by

n n c c1 0 1 1 2 2 3− = + +σ σ σ( ) , (4a)

where c1 and c2 are called the stress-optic coeffi-cients, and, if birefringence is to occur, c c1 2≠ .

In a similar way, the index of refraction n2 forlight having its amplitude vector in the direction ofthe minimum principal normal stress σ2 is given by

n n c c2 0 1 2 2 3 1− = + +σ σ σ( ) , (4b)

and for light having its amplitude vector in the out-of-plane direction,

n n c c3 0 1 3 2 1 2− = + +σ σ σ( ) . (4c)

Equations (4) are called Maxwell’s equations.

Here, the light waves of interest are those propa-gating in the z direction, or σ3 direction. Thesewaves necessarily have their amplitude vectors inthe xy plane, and we will therefore be interestedmainly in Eqns. (4a) and (4b). It is always possibleto resolve a given light amplitude vector intocomponents aligned with the σ1 and σ2 axes. Letus suppose that light waves with amplitude vector inthe σ2 direction propagate more slowly through thematerial than those with amplitude vector in the σ1direction. Then n n2 1> . Now consider the emerg-ing phase difference δ between orthogonal compo-nents M1 and M2 of a light wave that entered thematerial from the back in phase and that werealigned in the principal stress directions (Fig. 14).From Eqns. (4a,b),

n n c c

c2 1 2 1 1 2

1 2

− = − −= −

( )( )

( ) ,

σ σσ σ

(5)

where c is called the relative stress-optic coefficient,which is a material property. It is important to seein Eqn. (5) that the refraction-index differencen n2 1− is independent of σ3. The results to followtherefore hold for arbitrary value of σ3; it is notnecessary to assume that the material is in a state ofplane stress.


Birefrin

gent

material

σ1

σ2α

Fast

Slow x

y

z

δ

Μ1Μ2

h

Lightdirection

Fig. 14. Birefringent effect.

From Eqn. (5), and from Eqn. (3) derived ear-lier, we see that the phase difference δ between M1and M2 is given by

δσ σ

= −= −

h n n

hc

( )

( ) ,2 1

1 2(6)

where h is the thickness of the material in the light-propagation direction (Fig. 14). This equation is offundamental importance in the theory of photo-elasticity. It is often written in terms of the numberN of complete cycles of relative retardation, or,equivalently, in terms of the angular phase differ-ence ∆, as follows:

Nhc

hf

= = = − = −∆2

1 2 1 2

πδλ

σ σλ

σ σ

σ

( ), (7)

where fσ is called the fringe “constant” of thematerial. The relative stress-optic coefficient c isfound to be rather insensitive to the wavelength ofthe light. Therefore, the fringe “constant”

fcσλ= (8)

is anything but constant if light composed of manycolors is used.

In summary, a birefringent material does twothings:


• It resolves the incoming light into 2 com-ponents—one parallel to σ1 and the otherparallel to σ2, and

• It retards one of the components, M2, withrespect to the other, M1, by an amount δ thatis proportional to the principal stress differ-ence σ σ1 2− .

Polariscopes

A polariscope is an optical setup that allows thebirefringence in specimens to be analyzed. Itconsists of a light source, a polarizer, an optionalquarter-wave plate, a specimen, another optionalquarter-wave plate, and a second polarizer called theanalyzer. Two types of polariscope are commonlyemployed—the plane polariscope and the circularpolariscope.

The plane polariscope

The plane polariscope (Fig. 15) consists of alight source, a polarizer, the specimen, and ananalyzer that is always crossed with respect to thepolarizer.

Polarizer

Light

source

ObserverAnalyzer

Model

Fig. 15. The plane polariscope.

The direction of polarization of the polarizer isassumed to be vertical in the figure. We shall seelater that it may be useful to rotate the polarizer andanalyzer together in order to determine principal-stress directions, but in the derivation of thepolariscope equations, there is no loss of generalityby assuming that the light leaving the polarizer(Fig. 16) is vertically polarized.


P

Fig. 16. Light leaving polarizer.

Let P designate the light leaving the polarizer.The orientation of P is vertical; its amplitude is a,and its phase Φ is that of a traveling wave, namely,

Φ = − = −2πλ

ω( ) .z ct kz t

Therefore, its complex representation is

P aei= Φ . (9)

Now consider the light leaving the specimen(Fig. 17).

P

M1

M2

α

Fig. 17. Light leaving model (plane polariscope).

This light has been resolved into a fast com-ponent M1 having amplitude

M P1 = sinα (10a)

and a retarded component M2 having amplitudePcosα . However, this retarded component is notin phase with the fast component; it lags the fastcomponent by an angular phase shift

∆ = =kδ π δλ

2 .

Therefore its correct complex representation is

M Pei2 = ∆ cos .α (10b)

The correct sign of the angular phase shift can bechecked by examining two waves propagating in the+z direction (Fig. 18). The blue wave is lagging thered wave by the amount δ as both waves propagate


to the right. If, at time t = 0, the red wave has therepresentation cos[ ]kz , then the blue wave has therepresentation cos[ ( )]k z+ δ .

δ a cos[kz−ωt]a cos[k(z+δ)−ωt]

z

Re

Im

kδ

Increasing t

Fig. 18. Phase shift at time t = 0.

The Argand diagram in the figure illustratesperhaps more clearly the relations between z, δ, andtime t. With increasing time, the vectors repre-senting the amplitude and phase of the slow and fastwaves rotate clockwise. The projections of theamplitudes on the real axis give the actual ampli-tudes at any instant t and position z. The red waveis clearly ahead of the blue wave if the retardation∆ = kδ is taken in the positive sense as indicated.

Finally, we consider the light leaving theanalyzer (Fig. 19). In a plane polariscope, the ana-lyzer is always crossed with respect to the polarizer.Since the polarizer was assumed to be verticallyoriented, the analyzer is horizontally oriented.

M1

M2

α

A

Fig. 19. Light leaving analyzer (plane polariscope).


The analyzer rejects all vertically polarizedcomponents of light and passes the horizontallypolarized components, so that

A M M

P Pe

P e

i

i

= −

= −

= −

1 2

1

cos sin

( sin )cos ( cos )sin

sin cos .

α α

α α α α

α α

∆

∆d iWe note that

sin cos sinα α α= 12 2

and that

1

22

2 2 2

2

− = −

= −FHIK

−e e e e

e i

i i i i

i

∆ ∆ ∆ ∆

∆ ∆

/ / /

/ sin .

d i

Therefore, by making use of the original repre-sentation of P, Eqn. (9), the expression for Asimplifies to

A ie ae

ae

i i

i

= −

= − + +

∆ Φ

∆ Φ

∆

∆

/

( / / )

sin sin

sin sin .

2

2 2

22

22

α

απ

In this expression, as in many like it to follow, it isimportant to realize that a factor like ei( / / )− + +π 2 2∆ Φ

has a magnitude equal to unity. Therefore, the mag-nitude of A is simply

| | sin sin .A a= 22

α ∆

The intensity of light I leaving the analyzer is givenby the square of the light amplitude. Thus, for theplane polariscope, the intensity of the light seen bythe observer is

I A a= =| | sin sin .2 2 2 222

α ∆(11)

Isoclinics and isochromatics

The human eye is very sensitive to minima inlight intensity. From Eqn. (11), it is seen that eitherone of two conditions will prevent light that passesthrough a given point in the specimen from reachingthe observer, when a plane polariscope is used.

The first condition is that


sin ,

, , , , .

2 2 0

20 1 2

α

α π=

= = ± ±

i.e.

m m L(12)

Since α is the angle that the maximum principalnormal stress makes with the polarizing direction ofthe analyzer, this result indicates that all regions ofthe specimen where the principal-stress directionsare aligned with those of the polarizer and analyzerwill be dark. The locus of such points is called anisoclinic because the orientation, or inclination, ofthe maximum principal normal stress direction isthe same for all points on this locus. By rotatingboth the analyzer and polarizer together (so thatthey stay mutually crossed), isoclinics of variousprincipal-stress orientations can be mapped through-out the plane. Several examples are given in thetext (Dally and Riley 1991).

The second condition is that

sin ,

, , , , .

2

20

20 1 2

∆

∆

=

= = =

i.e.

N n nπ

L

(13)

The locus of points for which this condition is metis called an isochromatic, because (except for n = 0)it is both stress and wavelength dependent. Recallfrom Eqn. (7) that

Nhc

hf

= = = − = −∆2

1 2 1 2

πδλ

σ σλ

σ σ

σ

( ).

Therefore, points along an isochromatic in a planepolariscope satisfy the condition

σ σ λ σ1 2− = = =N

ch

Nf

hN n, . (14)

The number n is called the order of the iso-chromatic. If monochromatic light is used, then thevalue of λ is unique, and very crisp isochromatics ofvery high order can often be photographed. (Exam-ples will be given later.) However, if white light isused, then (except for n = 0), the locus of points forwhich the intensity vanishes is a function of wave-length. For example, the locus of points for whichred light is extinguished is generally not a locus forwhich green or blue light is extinguished, andtherefore some combination of blue and green willbe transmitted wherever red is not. The result is a


very colorful pattern, to be demonstrated bynumerous examples in class using a fluorescentlight source.3

The circular polariscope—dark field

The circular polariscope (Fig. 20) consists of alight source, a polarizer, a quarter-wave plateoriented at 45º with respect to the polarizer, thespecimen, a second quarter-wave plate, and ananalyzer.

Polarizer

Light

source

ObserverAnalyzer

Modelλ/4 plate

λ/4 plate

F

S s

f

Fig. 20. Circular polariscope (dark field).

The two quarter-wave plates are generallycrossed (as shown in the figure) to minimize errordue to imperfect quarter-wave plates. The analyzeris either crossed with respect to the polarizer (asshown in the figure), or parallel to the polarizer.

Again, the direction of polarization of thepolarizer is assumed to be vertical (Fig. 16), and wetake as a representation of the polarized lightleaving the polarizer the expression (Eqn. (9))

P aei= Φ .

This light enters the first quarter-wave plate andleaves with the two components

3However, this interference of colors results in a “washing out”of the higher-order isochromatics and is particularly trouble-some if a panchromatic black-and-white film is used to recordthe isochromatic patterns. A way to minimize the loss of datadue to washing out is to photograph the isochromatic patternswith color film and to scan the color negative digitally,separating out the red, green, and blue components. The com-ponent with the sharpest fringes can then be used for analysis.


FP

SP

e

iP

i

=

=

=

2

2

2

2

,

,

/π (15)

as illustrated in Fig. 21. Comparison of this figurewith Fig. 9 shows that this arrangement producescircularly polarized light that is rotating counter-clockwise.

P

π4

FS

Fig. 21. Light leaving first quarter-wave plate(circular polariscope).

Light components leaving the specimen are

shown in Fig. 22. The angle φ π α= −4

that F

makes with M1 is introduced for convenience.

π4

FS

M1

M2

φ

α

Fig. 22. Light leaving model (circular polariscope).

Taking into account the relative retardation ∆that the specimen introduces with respect to theprincipal stress planes, we have


M F S

Pi

Pe

M F S e

Pi e

iP

i e

iP

e

i

i

i

i

i

1

2

2

2

2

2

2

= −

= −

=

= +

= +

= −

=

−

−

cos sin

(cos sin )

,

( sin cos )

(sin cos )

(cos sin )

.( )

φ φ

φ φ

φ φ

φ φ

φ φ

φ

φ

∆

∆

∆

∆

(16)

Now consider the light leaving the secondquarter-wave plate (Fig. 23). As mentioned earlier,the second quarter-wave plate is usually crossedwith respect the first quarter-wave plate. Therefore,it has the effect of “derotating” the light that was“rotated” by the first quarter-wave plate.

π4

sf

M1

M2

φ

α

Fig. 23. Light leaving the second quarter-wave plate(circular polariscope).

Components s and f have the expressions

s M M e

Pe ie i

f M M

Pe ie

Pe i e i

i

i i

i i

i i

= +

= +

= − +

= − +

= +

−

−

−

( cos sin )

(cos sin ) ,

sin cos

( sin cos )

( sin cos ) .

/1 2

2

1 2

2

2

2

φ φ

φ φ

φ φ

φ φ

φ φ

π

φ

φ

φ

∆

∆

∆

(17)

Finally, if the analyzer is crossed with respect tothe polarizer, as shown in Fig. 24, then its output Ais given by the expression


π4

s

f

A

Fig. 24. Light leaving analyzer (circular polariscope).

A s f

Pe i ie i e

Pe i i e

Pe ie e

Pe ie i

Pe e

i i i

i i

i i i

i i

i i

= −

= + − −

= − −

= −

= −FHIK

=

−

−

− −

−

−

1

2

2

21

21

22

2

2

2 2

2 2

( )

cos sin sin cos

cos sin

sin

sin .

/

/

φ

φ

φ φ

φ

φ

φ φ φ φ

φ φ

∆ ∆

∆

∆

∆

∆

∆

∆

d i

a fd i

d i

Since P aei= Φ as before, the magnitude of A is just

| | sin ,A a= ∆2

and the intensity I of the light leaving the analyzer is

I A a= =| | sin .2 2 2

2∆

(18)

This result is similar to that for the plane polari-scope, with the notable absence of the factorsin2 2α . Therefore, a circular polariscope producesisochromatics but not isoclinics. The lack ofisoclinics is often desirable, since the dark isoclinicsin a plane polariscope often obscure large areas ofthe model. For this reason, most of the photoelasticpatterns published in the literature are obtained witha circular polariscope.

An examination of Eqn. (18) shows that, whenthere is no model, or when the model is unstressed,i.e. ∆ = 0 everywhere, the entire field is dark. Thecircular polariscope configuration just studied istherefore called the dark field configuration. Theisochromatic pattern is analyzed in exactly the samemanner as that for the plane polariscope, i.e.

σ σ λ σ1 2− = = =N

ch

Nf

hN n, . (19)


The circular polariscope—light field

An important simple variation on the arrange-ment of elements in a circular polariscope is one inwhich the quarter-wave plates remain crossed andthe analyzer is aligned parallel to the polarizer(Fig. 25).

Polarizer

Light

source

ObserverAnalyzer

Modelλ/4 plate

λ/4 plate

F

S s

f

Fig. 25. Circular polariscope (light field).

In this case, the output A of the analyzerbecomes

A s f= +1

2( ) ,

where s and f are the slow and fast componentsleaving the second quarter-wave plate (Eqns. (17)).One can show that the intensity of light leaving theanalyzer is now given by

I A a= =| | cos .2 2 2

2

∆(20)

Therefore, in the absence of a model, or for a modelthat is unloaded, the intensity is at its maximum.This configuration is called the light field configura-tion. In many cases the light-field arrangement ispreferred over the dark-field one because theboundary of the model is seen more clearly. Theonly difference in isochromatic interpretation is that

the intensity now vanishes wherever cos22

0∆ = , i.e.

N n n= = + =∆2

0 1 212π

, , , , .L (21)


The dark isochromatics in a light-field circularpolariscope therefore correspond to the orders 1

2 ,

112 , 2 1

2 , etc., instead of integer values, i.e.

σ σ λ σ1 2

12− = = = +N

ch

Nf

hN n, . (22)

The light isochromatics in a light-field circularpolariscope correspond to the integer values of N.Since fσ is wavelength dependent, the only iso-chromatic that is unaffected by the color content ofthe light source is the zero order one, which is light.

Calibration

The value of the fringe constant fσ can be deter-mined experimentally by inducing a known stressdifference σ σ1 2− in a model that is made of thesame material as the specimen of interest, byobserving the corresponding value of N, and bysolving Eqn. (7) for fσ :

f hNσ

σ σ= −1 2 . (23)

Observe that strongly birefringent materials willhave low values offσ , since the stresses required toproduce a given value of N will be small.

A common calibration specimen is the circulardisk of diameter D and thickness h loaded indiametral compression (Fig. 26).

P

D

x

y

σ1

σ2

Fig. 26. Disk in compression.


The horizontal and vertical normal stressesalong the x axis are principal stresses because theshear stress τxy vanishes due to symmetry about the

x axis. Also, σx is positive, while σy is negative.

We therefore take σ σ1 = x and σ σ2 = y so as to

render σ σ1 2 0− ≥ . From theory of elasticity, thesolutions for the normal stresses along thehorizontal diameter are (after Dally and Riley 1991)

σπ

ζζ

σπ

ζ ζ

ζ

1

2

2

2

2

2 13

2

2 2

2 1

1

6 1 1

1

= −+

FHG

IKJ

= −− +

+

P

hD

P

hD

,

,d id i

d i

(24)

where ζ = =x R x D/ /2 . These stresses are plottedin Fig. 27.

−8

−6

−4

−2

0

2

4

6

8

10

−1.0 −0.5 0 0.5 1.0Horizontal position, ζ = x/R = 2x/D

Nor

mal

ized

str

ess,

σ/(P

/πhD

)

σ1

σ2

σ1 − σ2

Fig. 27. Stress distribution along horizontal diameter.

Along the horizontal diameter, the maximumdifference σ σ1 2− occurs at the center, that is, atζ = 0. At this point,

σ σπ1 28− = P

hD. (25)

Combining this result with the basic photoelasticrelation (Eqn. (7)) gives

Nf

h

P

hDσ σ σ

π= − =1 2

8,

or

fD

P

Nσ π= 8

. (26)


Notice that the specimen thickness h does notappear in this equation. The reason is that therelative retardation is proportional to h, but for agiven force P, the stresses are inversely proportionalto h. The net effect is a result for fσ that isindependent of h. A photograph of a light-fieldisochromatic pattern for a diametrally loaded diskmade of PSM-1® is shown in Fig. 28.

Fig. 28. Light-field isochromatics in adiametrally loaded circular disk.

For this specimen, the diameter D was63.50 mm (2.500 in.), and the load was 1.33 kN(298 lb). The value of N at the center of the disk, asseen in Fig. 28, is approximately 7.0. Therefore thefringe constant for this material is approximately

fD

P

Nσ π π= = = ⋅8 8

2 500

298

7 043

( . ) ..

psi in.

fringe

A more accurate way of determining fσ using thisspecimen is to record several readings of increasingload P as the fringe value N at the center takes oninteger or half-integer values. The saddle shape ofthe central fringe allows rather precise determina-tion of N for this purpose. Then P is plotted as afunction of N, and the best straight-line fit of thedata is used to determine the ratio P N/ to be usedin Eqn. (26).

The disk in diametral compression is a favoritespecimen for calibration because it is simple tofabricate (at least with a template); it is easy to load;


it is not likely to fracture; it produces a fringepattern that, in the region of interest, is insensitiveto edge imperfections; and it is simple to analyze. Italso tests the limit of fringe density that can berecorded photographically, by producing very largefringe orders in the vicinity of the contact regions.

The image in Fig. 28 was obtained with amercury-vapor light source, which is rich in greenlight, but which also contains other colors as well.A Tiffen #58 Green filter was placed on the 35 mmcamera to reduce the transmission of the othercolors. Kodak Gold ASA 200 color film was usedto record the image at f/8 with an exposure of 0.7 s.This arrangement results in distinct fringes up toabout N = 15, which is adequate for most work inphotoelasticity.

A somewhat enhanced image is shown inFig. 29. This image was produced from the samenegative as that used to produce Fig. 28. However,only the green component of the red-green-blue(RGB) digital scan was retained, and this compo-nent was then enhanced by increasing the contrast.

Fig. 29. Enhanced image using only the green componentof the light used in Fig. 28.

Fringe orders up to about N = 20 can bediscerned in this figure. The low-order values of Nare marked along the horizontal diameter. Note thatthe center of the specimen is the location of a truesaddle point in the function σ σ1 2− : to the left and


right of this point, the function decreases, and aboveand below this point, the function increases. Suchsaddle points are common in photoelastic patterns,as will be seen in other examples to be given below.

The stress distribution in a disk in diametralcompression is unique in that the fringe number N isequal to zero everywhere along the unloadedboundary.

Examples

Several photoelastic patterns will be presentedfor common structural shapes. These include theuniform beam in 4-point bending, as well as beamshaving cutouts; a compact-tension specimen; and astraight–curved “U” specimen.

A light-field circular polariscope with therecording and processing methods described inconnection with Fig. 29 will be used.

Beam in bending

Consider first a uniform rectangular beam sub-jected to 4-point bending (Fig. 30).4 Let x and ydenote positions along the horizontal and verticalaxes having their origin at the center of thespecimen.

Fig. 30. Beam in 4-point bending.

Within the central portion of the beam, thebending moment is constant and, according toelementary beam theory, the axial stress σx is givenby −My I/ , varying from −Mc I/ at the top surfaceto Mc I/ at the bottom surface. Also within thisregion, the transverse shear V vanishes, and there-fore τxy = 0 in this region. In addition, σy = 0.

4Specimen material, PSM-1. Load, 150 lb. Beam length, 5 in.Beam height, 1.000 in. Beam thickness, 0.213 in. Distancebetween lower supports, 4.00 in. Distance between uppersupports, 2.00 in.


Since τxy = 0, the normal stresses σx and σy are

principal stresses. It follows that

σ σ σ σ1 2 0− = − = − − =x yMy

I

M

Iy| | ,

with the understanding that M is positive.

We have to be careful about the signs of theprincipal stresses because we want σ σ1 2 0− ≥ .With this convention, the fringe number N willalways be positive or zero. If one argues thatnegative values of N should be allowed (whether ornot σ σ1 2 0− ≥ ), one runs into inconsistencies.Consider, for example, the centrally located fringesin Fig. 30. There is a zero-order (light) fringe at thevery center, along the neutral axis of bending. It issurrounded by a continuous dark fringe of orderN = 1

2 . If we assigned a value of N equal to − 12 on

the top portion (where the axial stress is negative),and a value equal to + 1

2 on the bottom portion

(where the axial stress is positive), we would not beable to reconcile the value of N at the ends of theloop. It cannot vary along the loop because bydefinition the isochromatic is a locus of constant N.

Let us calculate the stress at the top and bottomof the beam. We know that fσ is about 43 psi·in./fringe for this material, and by counting fringes inFig. 30, we determine that N = 9 at both the top andbottom of the beam. To calculate σ σ1 2− , we alsoneed to know the thickness h. For the specimen inFig. 30, h = 5.41 mm (0.213 in.). Therefore, at theextreme fibers,

σ σ σ σx

Nf

h= − = = =1 2

9 43

0 21318

( )

( . ). .ksi

At the top, σx is negative, so it is equal to –1.8 ksithere; whereas, at the bottom, σx is positive, so it isequal to +1.8 ksi. This simple example serves toillustrate the point that photoelastic analysisrequires some knowledge about the stress state inorder to interpret the fringes correctly.

We can check these results. For the given beam,M Pa= 1

2 , I hH= 112

3, c H= 12 , and therefore


σxMc

I

Pa H

hH

Pa

hH= = =

= =

12

12

112

3 2

2

3

3 150 1 00

0 213 1 00021

c hc h

( )( . )

( . )( . ). .ksi

The predicted stresses are slightly higher than thoseobserved, suggesting perhaps that friction at thesupports may be reducing the effective value of M.

Beam with a keyhole notch

The stress distribution is altered drastically if afillet, crack, or notch is machined into the specimen.The stress-concentration effect of a “keyhole” notchis shown in Fig. 31.5

Fig. 31. Beam with keyhole notch.

The resolution of this image does not permit anexact determination of the stress-intensity factor forthe notch, but it is clear that the stress at the root ofthe notch is much larger in tension than thecompressive stress at the top of the beam, where thefringe number N happens to have the value 9 as itdid in the previous uniform beam at a higher load.A reduced load and/or an improved imaging methodwould be needed to resolve the fringes at the root ofthe notch.

This specimen provides a good lesson in fringecounting. At all exterior corners, including theupper and lower corners on each side of the notch,N = 0. Also, N = 0 in the tiny triangular “island”just above the notch, and N = 0 in the “eyes”between the notch and each upper support.Remnants of the uniform bending field are seenbetween the upper supports.

There are 4 saddle points to the left of thecenterline, and a matching set of 4 to the right. Onthe left, starting from the left, they have the approxi-mate values N = 3 1

2 , 3, 4 12 , and 2.

5Load, 100 lb. Other parameters are the same as for theuniform beam.


Beam with a sharp notch

An even sharper discontinuity is illustrated inFig. 32.6 Here, the notch was produced with amilling cutter having a 45º tip. The tip has a smallradius; if the tip had been absolutely sharp, thespecimen would probably have broken before thispicture could have been taken.

Fig. 32. Beam with sharp notch.

The stress field away from the sharp notchresembles very closely the stress field for thekeyhole notch. The 4 saddle points seen earlier areevident in this photograph as well. Details of thestress field around the notch would require a smallerload on the beam and an improved method ofrecording the fringe patterns.

Compact-tension specimen

Fatigueprecrack

Thickness=

Notch forCOD gage

d

h

a

P

P

B

w

Fig. 33. Compact-tension specimen.

6Load, 100 lb. Other parameters are the same as for theuniform beam.


Perhaps the most common specimen used todetermine experimentally the fracture toughnessK CI of materials is the compact-tension specimen(Fig. 33).

In the region of the crack tip, the stress field isdominated by the KI singularity (Fig. 34).

Crack

x

y

σy

σx

τxy

rθ

σ

σ

Fig. 34. Stress element near the crack tipin mode I loading.

Under plane strain or plane stress conditions, ifthe stresses remain elastic, the stress distributioncan be shown to be

σστ

πθ

θ θ

θ θ

θ θ

x

y

xy

K

r

RS|

T|

UV|

W|=

−

+

R

S|||

T|||

U

V|||

W|||

I

2 2

12

3

2

12

3

2

2

3

2

cos

sin sin

sin sin

sin cos

, (27)

where KI is the stress-intensity factor, which is afunction of the specimen geometry and the remotelyapplied load. For this distribution, the in-planeprincipal-stress difference σ σ1 2− can be calcu-lated; the result is

σ σσ σ

τ

πθ

1 2

222

2

2

− =−F

HGIKJ +

=

x yxy

K

rI sin .

(28)


Combining this theoretical formula with the basicequation of photoelasticity gives

Nf

h

K

rσ σ σ

πθ= − =1 2 2

I sin

or

N rh

f

K

r( , ) sin .θ

πθ

σ= I

2(29)

Alternatively, this equation may be solved para-metrically for r in terms of θ :

rhK

Nf=

FHG

IKJ

1

2

22

πθ

σ

I sin .

This solution allows us to construct the theoreticalphotoelastic fringes, as shown in Fig. 35. Thesefringes look like nested ellipses, but the shapes arenot really ellipses—they become almost pointed atthe crack tip.

Crack

P

P

x

y

N = 1

N = 2

N = 3

(etc.)

N = 0

N = 1

Fig. 35. Theoretical fringes for a KI -dominant field.

We can now compare the theoretical fringepattern with an actual fringe pattern from a photo-elastic model (Fig. 36).7 The features of thesingular field are shown near the crack tip for valuesof N greater than about 4. Outside this region,however, the stress distribution is dominated by the

7Material, PSM-1. Load, 60 lb. Overall specimen size,2.50 in. square, 0.213 in. thick; crack length a, 0.875 in., widthw (Fig. 35), 2.00 in.


bending field and is also affected by the freeboundaries of the specimen.

Fig. 36. Photoelastic model of a compact-tension specimen.

To compute the value of KI for the specimen inFig. 36, we first scale the photograph and determinethat for the matching fringe loops of order N = 5, forexample, the radius r at θ = ± °90 is about 0.23 in.Therefore, from Eqn. (29),

Kr f N

hI

ksi in

= =

=

2 2 0 23 43 5

0 213

1 2

π πσ ( . )( )( )

.

. .

Meanwhile, according to Tada, Paris, andIrwin’s Stress Analysis of Cracks Handbook (1973),the stress-intensity factor for this geometry andloading is given by

KP

h wf

a

wI = =( ) , ,λ λ

where in this case (again from the photograph) a =0.86 in. and w = 2.00 in., so that λ = 0.43. For thisλ, f ( )λ is about 7.8. Therefore, KI should have thevalue

KI ksi in= =60

0 213 2 007 8 155

. .( . ) . .

The photoelastic analysis is seen to underpredictthe value of KI by about 25%, probably because thecrack tip has a finite wedge angle and is rather


blunt. Dally and Riley (1991: Chapter 14) discussvarious ways to improve KI calculations fromphotoelastic data. One method, attributed toSchroedl, Smith, and others, results in a differenc-ing equation (Eqn. (14.14) of Dally and Riley’s text)that provides an improved estimate of about1.4 ksi in if (in addition to the N = 5 data previ-ously used) the radius r for the N = 7 fringe is takento be about 0.13 in. The apparent underprediction istherefore reduced to about 8%, which would bemarginally acceptable for most fracture-mechanicswork.

Simulated photoelastic fringes

Considerable insight into the analysis of photo-elastic fringes can be gained by constructing simu-lated fringes for known stress distributions.

Analytical

These constructions can be in the form of linesof constant N determined analytically, as in Fig. 35,where the solution r( )θ to Eqn. (29) is foundexplicitly, with N as a parameter.

Numerical—bitmap

Alternatively, a bitmap image can be con-structed by calculating the intensity I numerically ateach point in a large array of regularly spacedpoints, and by observing the resulting pattern. Anexample is shown in Fig. 37 for the dark-field iso-chromatics in a KI stress field, based on Eqn. (29).

Fig. 37. Simulated KI field (green light).


An advantage of this method is that it can beused even if the theoretical stress field is so com-plicated that it is impossible to determine the locusof constant N analytically.

The program used to generate the image inFig. 37 is given below. The program was developedfor arbitrary intensity and wavelength values of thecolors red, blue, and green. A succession ofseparate bitmap files are created for steadily increas-ing load. The file format is BMP, which is thesimplest of the standard image file types.

#include <windows.h>#include <stdio.h>#include <math.h>

// Windows-based C program for creating a color BMP file// of the photoelastic isochromatics for a KI field

// Matthew J Phillips and James W Phillips// December 1997

// Pixel order is RGB for Mac, BGR for PCtypedef struct {// unsigned char red, green, blue;

unsigned char blue, green, red;} pixel;

// These image_* functions maintain the pixels for us:typedef struct {

int width,height,rowbytes;

char *imageptr;} image;

image *image_new(int width, int height)

{image *pimNew;int rowbytes;

rowbytes = sizeof(pixel)*width;// Make rowbytes a multiple of 4rowbytes = (rowbytes+3)&~3;

pimNew = (image*)malloc(sizeof(image));pimNew->width = width;pimNew->height = height;pimNew->rowbytes = rowbytes;pimNew->imageptr = malloc(rowbytes*height);

// Technically this isn't needed because we're going to// fill in the whole bitmap eventually, but just to be// nice ...

memset(pimNew->imageptr,0xFF,rowbytes*height);

return (pimNew);}

void image_setpixel(image *pim, int x, int y, pixel *color)

{pixel *pixptr;

pixptr = (pixel*)(pim->imageptr+pim->rowbytes*y+sizeof(pixel)*x);

*pixptr = *color;}

void image_writetofile(image *pim, FILE *fp)

{int y;

for(y = 0; y < pim->height; y++) {fwrite(pim->imageptr+pim->rowbytes*y,

1,pim->rowbytes,fp);}

}

void image_dispose(image *pim)

{free(pim->imageptr);


free(pim);}

// This function writes a Windows BMP file using the// specified image as its source.

void write_BMP_file(image *pim, char *filename)

{BITMAPINFOHEADER bmi;BITMAPFILEHEADER fh;FILE *fp;

fp = fopen(filename,"wb");

fh.bfType = 'MB';fh.bfSize = sizeof(BITMAPFILEHEADER)

+sizeof(BITMAPINFOHEADER)+pim->rowbytes*pim->height;

fh.bfReserved1 = 0;fh.bfReserved2 = 0;fh.bfOffBits = sizeof(BITMAPFILEHEADER)

+sizeof(BITMAPINFOHEADER);

fwrite(&fh,1,sizeof(fh),fp);

bmi.biSize = sizeof(BITMAPINFOHEADER); bmi.biWidth = pim->width; bmi.biHeight = pim->height; bmi.biPlanes = 1; bmi.biBitCount = 24; bmi.biCompression = BI_RGB; bmi.biSizeImage = 0; bmi.biXPelsPerMeter = 2835; bmi.biYPelsPerMeter = 2835; bmi.biClrUsed = 0; bmi.biClrImportant = 0;

fwrite(&bmi,1,sizeof(bmi),fp);

image_writetofile(pim,fp);

fclose(fp);}

typedef struct {double red, green, blue;

} dblpixel;

int main()

{image *pim;int px,py,n_pic;pixel c;double x,y,k,base_n;dblpixel intensity, wavelength, strength;char filenamebuf[32];

float red, green, blue;

#define n_pics 10#define k_max 1000.0#define PI 3.14159265359#define IMGWIDTH 256#define IMGHEIGHT256

printf("\r\nWavelengths (red, green, blue) (nm): ");scanf ("%f %f %f", &red, &green, &blue);wavelength.red = red;wavelength.green = green;wavelength.blue = blue;printf("%f %f %f", wavelength.red,

wavelength.green, wavelength.blue);

printf("\r\nStrengths (red, green, blue) (0-255): ");scanf ("%f %f %f", &red, &green, &blue);strength.red = red;strength.green = green;strength.blue = blue;printf("%f %f %f", strength.red,

strength.green, strength.blue);

for(n_pic = 1; n_pic <= n_pics; n_pic++) {

pim = image_new(IMGWIDTH,IMGHEIGHT);

k = k_max*n_pic/n_pics;

printf("Generating image number %02d (k = %g).\r\n",n_pic,k);

for(py = 0; py < IMGHEIGHT; py++) {// y = ((double)py/(IMGHEIGHT-1))*2.0-1.0;

y = ((double)py/IMGHEIGHT)*2.0-1.0;for(px = 0; px < IMGWIDTH; px++) {

x = ((double)px/(IMGWIDTH-1))*2.0-1.0;


// base_n is N(lambda) without// the 1/lambda factorif (x == 0.0 && y == 0.0) base_n = 0.0;else

base_n = k*fabs(y)/pow(x*x+y*y,3./4.);

intensity.red = strength.red*pow(sin(PI*base_n/wavelength.red),2);

intensity.green = strength.green*pow(sin(PI*base_n/wavelength.green),2;

intensity.blue = strength.blue*pow(sin(PI*base_n/wavelength.blue),2);

c.red = (unsigned char)intensity.red;c.green = (unsigned char)intensity.green;c.blue = (unsigned char)intensity.blue;

image_setpixel(pim,px,py,&c);}

}

// Overprint white crack

c.red = 255;c.green = 255;c.blue = 255;

y = 0.0;py = (int)(IMGHEIGHT/2*(y+1.0));for(px = 0; px < IMGWIDTH/2; px++) {

image_setpixel(pim,px,py,&c);}

sprintf(filenamebuf,"image_%02d.bmp",n_pic);write_BMP_file(pim, filenamebuf);image_dispose(pim);

}return (0);

}

Note how the quantity |sin |/θ r is calculated inthe code:

|sin | | | | | | |./ /

θr

y

r r

y

r

y

x y= = =

+3 2 2 2 3 4d i

This method is used to avoid trigonometric functioncalls and the problem of determining the correctinverse tangent of the angle θ corresponding to agiven point ( , )x y .

Once the BMP files are created, they may beedited by a graphics program, such as AdobeIllustrator®, and saved in an alternative compressedformat, such as Graphics Interchange Format (GIF).The fringe numbers that appear in Fig. 37 wereadded in this manner. An animated GIF can then beconstructed using a program such as Microsoft GIFAnimator®.

Numerical—Finite-element method

Finite-element codes, such as ABAQUS, havecontour-plotting capabilities that allow varioustypes of contours to be constructed so as to visualizethe output.

If an option is available to plot the in-planeprincipal-stress difference, or, equivalently, twicethe in-plane maximum shear stress, then this optioncan be used to generate simulated isochromatics.

http://www.tam.uiuc.edu/Courses/TAM326/1998.1/Lectures/animated_KI_green.gif


A variable that is sometimes plotted to simulatephotoelastic fringe patterns is the Tresca “equivalentstress” σTresca given by

σ τ

σ σ σ σ σ σTresca =

= − − −

2

1 2 2 3 3 1

max

max | |, | |, | | .b gIn regions where the out-of-plane normal stress is anintermediate principal stress, the in-plane principalstress difference will in fact determine the value ofσTresca, and therefore a contour of constant σTresca

corresponds to an isochromatic of some order N.However, in regions where the out-of-plane normalstress is not intermediate, that is, it is either theminimum or maximum principal stress, the σTresca

contour will not coincide with an isochromatic.

As an example, consider the stresses in a curvedbeam due to pure bending (Fig. 38).8 Only thecurves for R h/ = 1 are shown, but the observationswe are about to make are independent of the valueof R h/ .

−0.6−0.4−0.200.20.40.6−1.5

−1.0

−0.5

0

0.5

1.0

1.5

2.0

2.5

Normalized position, (y+e)/h = (R−r)/h

Nor

mal

ized

str

ess,

σ/(M

c/I)

R

M

h

|σθ − σr|

σθ

σθ

σr

σz

Tresca

Fig. 38. Comparison of Tresca equivalent stress within-plane principal-stress difference in a curved beam.

The hoop stress σθ varies from a positive valueat the inside radius to a negative value at the outsideradius. The radial stress σr vanishes at theseextremes and reaches a maximum positive valuenear the center. If we let σ1 and σ2 denote the in-plane principal stresses, then

σ σ σ σθ1 2− = −| | ,r

8For a more complete discussion of the stresses in a curvedbeam, refer to the Appendix.


as shown with the dashed curve. It is this differencethat would give rise to an isochromatic pattern.Note that the zero-order fringe is shifted consid-erably towards the center of curvature from theneutral axis, that is, from the point where the hoopstress σθ vanishes.

Meanwhile, the third principal stress σ3, whichis equal to the out-of-plane normal stress σz, hasthe value zero if we assume a state of plane stress.Wherever σ3 is intermediate between σ1 and σ2,the Tresca equivalent stress and the differenceσ σ1 2− are the same thing—this is the situation forall of the section where σθ is compressive. How-ever, near the center of the section, σθ becomes theintermediate principal stress and the Tresca equiva-lent stress is numerically equal to the radial stressσr since σz vanishes. Finally, near the insideradius, the radial stress σr becomes the interme-diate principal stress, the hoop stress σθ dominates,and the Tresca equivalent stress becomes numeri-cally equal to the hoop stress because, again, σzvanishes.

In this simple example, it is seen that the Trescaequivalent stress and the in-plane principal stressdifference can be, but are often not, the same thing.One can expect that in more complicated shapesanalyzed by the finite-element method, the matchingof these two quantities would need to be checkedcarefully.

Photoelastic photography

Next we consider some practical aspects ofobserving and recording photoelastic fringepatterns.

Collimated vs. diffuse light

One choice to be made is whether the light thattraverses the specimen is collimated. In the tradi-tional optical arrangement (Fig. 39), the light sourceproduces diverging light from a tiny aperture, and alarge field lens placed at is focal length from thissource produces collimated light that is passedthrough all the polariscope elements and is collectedby a second large field lens.


Lens P Q Q A LensM

Pinhole Pinhole

Source Camera

Collimated light

Fig. 39. Collimated light source.

A pinhole placed at the focal point of the secondfield lens eliminates stray light and allows a realimage of the specimen to form on a piece of paperor photographic film. Smaller pinholes producesharper images of reduced intensity.

The light source can be a specially designedmercury-vapor or incandescent lamp in a housinghaving various-sized pinhole openings, or it can bea laser beam passed through an expanding lens.

This arrangement has the advantage that theedges of the specimen are in sharp focus, since thecollimated light propagates parallel to these edges.Also, no focusing is required. Disadvantagesinclude the cost of the large field lenses, and the factthat all the optical elements in the region ofcollimated light are in focus; therefore all imperfec-tions (such as scratches and dust) appear in sharpfocus in the recorded image. Also, rather large filmsizes (such as 5x7-in.) are required for good imagereproduction.

An alternative arrangement that allows the useof an ordinary 35 mm film camera or a CCD videocamera employs a diffuse light source (Fig. 40). Aground glass or other translucent material is placedbetween the light source (which need not be apinhole type) and the other optical elements. Everypoint on this diffusing surface produces light thatpropagates in all directions.

Ground glass

P Q Q AM

Source Camera

Diffused light

Fig. 40. Diffuse light source.


One of the advantages of this method is that anordinary camera can be used—and with 1-hourphoto finishing being commonplace today, this factalone makes the arrangement attractive. (Alterna-tively, a video camera can be used to record theimage “live” or allow a frame-grabber to acquire asequence of fringe patterns.) Another advantage isthat this arrangement is generally more compactthan the collimated one.

The principal disadvantage is that the light is nolonger passing parallel to the specimen edges andtherefore the edges do not appear sharp. Thisparallax problem can be reduced by moving thecamera farther away from the specimen, but then theimage size is reduced. A long focal-length lens willhelp, but then focusing becomes critical.

White vs. monochromatic light

A quite separate issue is whether the light usedis white (that is, rich in many colors) or is mono-cromatic. There are advantages and disadvantagesto either type.

Fig. 41. Approximate white-light light-field isochromaticsfor a disk in diametral compression.

Consider the fringe pattern in a circular diskloaded in diametral compression between photo-


elastic strips of the same material, and photo-graphed using white light (Fig. 41).9

The image is striking but is difficult to analyzebecause the colors interfere. Now compare thisfigure with that in Fig. 42, which is for the samespecimen at the same load with the same lightsource, but with a Tiffen® #58 dark green filterplaced over the camera lens.

Fig. 42. Filtered green-light light-field isochromaticsfor a disk in diametral compression.

Observe that a light fringe of order N = 7 is atthe center of the disk when approximately mono-chromatic green light is used. Now refer back toFig. 41 and observe that the center of the disk has agreen fringe. Apparently at this point, the red andblue components have low intensity, allowing greento dominate.

To see why isochromatics are so colorful whenviewed in white light, consider the absolute retarda-tion given by Eqn. (6):

δ σ σ= −hc( ) ,1 2

9Material, PSM-1. Load, 298 lb. Diameter, 2.489 in. Lightsource, mercury vapor. Blue and red components artificiallyenhanced to simulate a white-light source.


where c is the relative stress-optic coefficient. Thevalue of c is independent of wavelength, and there-fore δ is a function only of the stress differenceσ σ1 2− . However, when the retardation δ isexpressed as a fraction of the light wavelength λ, soas to determine the number of cycles N of relativeretardation

N = δλ

,

the value of N is seen to vary inversely as thewavelength. It is the value of N, and not δ, thatdetermines the intensity of light at a given point inthe specimen. For example, in a dark field (circularpolariscope), the intensity varies as

I a a N

a

= =

=

2 2 2 2

2 2

2sin sin

sin .

∆ π

π δλ

If we decompose white light into the primary colorsred, green, blue (RGB) and examine the dependenceof I on δ for the representative wavelengths, say

λλ

λ

red

green

blue

nm ,

nm ,

nm ,

==

=

650

550

450

then we obtain the results shown in Fig. 43.

0

0.2

0.4

0.6

0.8

1.0

1.2

0 500 1000 1500 2000 2500Absolute retardation, δ = ch(σ1−σ2) (nm)

No

rma

lize

d in

ten

sity

, I/a2 B G R

Fig. 43. Dark-field intensity variationsfor representative primary-color wavelengths.

With the important exception of δ = 0, weobserve that the value of δ that causes the intensityof a given color to vanish generally does not cause


the intensity of the other two colors to vanish, andtherefore some combination of the other two colorswill be visible. We conclude that in a dark-fieldpolariscope (plane or circular), only the zero-orderisochromatic is black. All other isochromatics havelocations that depend on the wavelength.

In a light-field polariscope (which must becircular), the situation is similar, except now theintensity varies as the cosine squared instead of thesine squared:

I a a N

a

= =

=

2 2 2 2

2 2

2cos cos

cos .

∆ π

π δλ

The corresponding results are shown in Fig. 44.

0

0.2

0.4

0.6

0.8

1.0

1.2

0 500 1000 1500 2000 2500Absolute retardation, δ = ch(σ1−σ2) (nm)

BGR

No

rma

lize

d in

ten

sity

, I/a2

Fig. 44. Light-field intensity variationsfor representative primary-color wavelengths.

In this figure we see that there really is nolocation where the intensity vanishes for all colors.Instead, for δ = 0, the intensity is a maximum for allcolors and therefore the zero-order isochromaticappears light. The 12 -order fringe has the narrowest

range of δ for which the intensities are minimal andtherefore is the “blackest” of the isochromatics.

Note that the order of colors in an isochromaticpattern is well defined. As δ increases, the relativeretardation for blue light outpaces that of green andred because blue light has the shortest wavelength.Therefore, regardless of the field (dark or light),blue fringes appear ahead of green and red fringes.Sometimes an individual primary color is seendistinctly, but this condition requires that the other


two components vanish, and such a condition israre. As a consequence, the fringe pattern consistsof continuously varying colors, not bands of primarycolors. Blue and green, for instance, combine toform a blue–green called cyan, green and redcombine to form yellow, and red and blue combineto form magenta. All three combine to form white.The first-order “light” fringe in a light-field pattern(Fig. 44) is really a band of colors consisting ofblack–blue–cyan–green–yellow–red–magenta, fol-lowed by a strong blue and a strong green. Thesefeatures can be seen in the diametrally loaded disk(Fig. 41) near the outside edge. Observe, forinstance, the strong yellow band, where blue isreaching a relative retardation of 11

2 cycles and is

therefore vanishing, leaving green and red to com-bine and form yellow.

Further insight is gained by decomposing anunfiltered mercury-vapor light-source isochromaticpattern into its component RGB colors (Fig. 45).The decomposition is achieved electronically fromthe scanned negative.

Fig. 45. Central band of a disk in diametral compressionin its original form (Hg) and decomposed

into its RGB components.

At the center of the disk, the green fringe oforder 7 is seen to correspond to the red fringe oforder 6.5 and to the blue fringe of order 9.5. Now,if the characteristic wavelength of green light istaken to be 550 nm, the absolute retardation δ at thecenter of the disk must be

δ λ= =

=

Ngreen green nm

nm

7 0 550

3850

. ( )

.


If red light is observed to produce a fringe order of6.5 at the same point, then the characteristic wave-length of this light must be

λ δred

red

nm

6.5nm= = =

N

3850590 .

Similarly, if blue light is observed to produce afringe order of 9.5 at the same point, then thecharacteristic wavelength of this light must be

λ δblue

blue

nm

9.5nm= = =

N

3850400 .

It should be emphasized that a mercury light sourcehas many lines in its spectrum, and the effectsobserved here are weighted averages of thestrengths of these lines, the spectral sensitivity ofthe color film, and the spectral sensitivities of thered, green, and blue sensors in the film scanner.Nevertheless, it is remarkable that the wavelengthseparation between blue and green is observed to bemuch larger than that between green and red.

The beautiful rainbow colors of the white-lightisochromatics are fascinating but they complicatethe data analysis, particularly for high-order fringes,which tend to “wash out” because the almostrandom combinations of the primary colors virtuallyguarantee that light of some wavelength will not beextinguished. This problem is particularly acute ifthe recording system responds in a panchromaticfashion (that is, to all colors of light), but recordsthe image in monochromatic form (as, for example,a black-and-white CCD camera or black-and-whitefilm). For the sharpest fringe recording, mono-chromatic light is preferred.

There are advantages to the white-light source,however. In a plane polariscope, the isochromaticsare brightly colored while the isoclinics are black;therefore the two patterns are easily distinguished inthe laboratory or in a color photograph, whereasthey would not be easily distinguished if mono-chromatic light were used.

Also, in a dark-field (plane or circular) polari-scope, only the zero-order isochromatic fringe isblack and therefore this important fringe order canbe distinguished from all the higher-order ones; thesame cannot be said if the light source is mono-chromatic. To see this effect, compare the white-light simulated isochromatics for a KI field


(Fig. 46) with those using a monochromatic lightsource (Fig. 37).

Fig. 46. Simulated KI field (white light).

Similarly, in a light-field (circular) polariscope,the 12 -order isochromatic fringe is the “blackest” of

the dark fringes—observe the color intensities inFig. 44 for retardations in the range of 250 nm. Thefringe pattern in Fig. 41 illustrates this point well.

Compensation

The term compensation refers to any techniqueused to determine fractional fringe orders.

Interpolation and extrapolation

x

N

5

5

4

6

7

867

Interpolation

Extrapolation

Boundary

Fig. 47. Interpolation and extrapolation.

http://www.tam.uiuc.edu/Courses/TAM326/1998.1/Lectures/animated_KI_white.gif


The simplest compensation techniques are inter-polation and extrapolation of data based on whole-order (and/or half-order) fringe locations (Fig. 47).

Interpolation is the more reliable of thesemethods but it is necessarily restricted to interiorpoints in the specimen.

Extrapolation is the less reliable method but isrequired if fractional values of N are to be deter-mined along the boundary of the specimen. Thismethod is less reliable than interpolation for tworeasons:

• It requires a guess of the proper analyticalapproximation outside the domain in which suchan approximation can be deduced on the basis ofknown points.

• It is subject to errors in fringe values near thesurface. Such errors may be due to surfaceimperfections or to residual heat- or moisture-induced birefringence that has nothing to dowith the applied stresses. Residual edge stressesappear in an unloaded specimen; they cause acharacteristic sudden turning of obliquely inter-secting isochromatics at the edge, and a localshift of near-edge isochromatics that are parallelto the edge.

Since the maximum stresses (which are usually thestresses of primary interest) often occur at theboundary, it behooves the specimen fabricator toprepare the surface with as much care as possible,leaving it nick-free and devoid of heat-inducedresidual fringes.

Tardy compensation

A technique that takes the guesswork out ofdetermining fractional fringe orders is Tardy com-pensation. It works equally well for interior andboundary points, although for boundary points it isadversely affected by heat- or moisture-inducedresidual stresses. The method requires 3 steps forevery point analyzed:

1. The polarizing directions of the polarizer andanalyzer are aligned with the principal-stressdirections at the point. This step is accom-plished by using a plane polariscope and byrotating the polarizer and analyzer together untilan isoclinic passes through the point, therebyleaving


α = 0 .

2. The polariscope is converted from plane tocircular by rotating the quarter-wave plates totheir 45º orientations with respect to thepolarizer and analyzer. The polarizer andanalyzer are held fixed during the conversion,leaving the polariscope in the dark-field circularconfiguration, still with α = 0, but with no iso-clinics.

3. The analyzer alone is rotated through someangle γ, as shown in Fig. 48, until a neighboring(dark) isochromatic of integer order n passesthrough the point in question.

π4

sf

A

ψγ

Fig. 48. Light leaving the analyzer(Tardy compensation).

The fractional fringe order is then equal to theratio γ π/ , as will be shown subsequently.Therefore the value of N at the point is

N n= ± γπ

. (30)

The sign of the fractional correction depends onthe orientation of the quarter-wave plates. ForCCW rotating light, the + sign applies, and forCW rotating light, the – sign applies.

To derive Eqn. (30), we start with the expres-sions for the slow and fast wave amplitudes s and f,respectively, that leave the second quarter-waveplate in a circular polariscope. Since α = 0, the

angle φ has the special value φ π α π= − =4 4

; there-

fore, from Eqns. (17),


sP

e i e

Pe i e

fP

e ie

Pe ie

i i

i i

i i

i i

= −FH

IK

= −

= − +FH

IK

= − +

−

−

−

−

2 4 4

2

2 4 4

21

4

4

4

4

π

π

π

π

π π

π π

/

/

/

/

cos sin

,

sin cos

.

∆

∆

∆

∆

d i

d iFrom Fig. 48, the amplitude A of light leaving theanalyzer is given by

A s f= −cos sin ,ψ ψ

where the angle ψ π γ= −4

is introduced for con-

venience:

A s f

Pe i e ie

Pe i e i

Pe i i e i

Pe e e

Pe e i

Pe e

i i i

i i

i i

i i i

i i

i i

= −

= − − − +

= + − +

= − − +

= −

= ⋅ − − −FH

IK

FHG

IKJ

=

−

−

−

− − +

− − + +

cos sin

cos sin

cos sin (cos sin )

(cos sin ) (cos sin )

sin

sin

/

/

/

/ ( / ) ( )

/ ( / )/

/

ψ ψ

ψ ψ

ψ ψ ψ ψ

ψ ψ ψ ψ

π ψ ψ

π

π

π

π

π π ψ ψ

π π ψ ψ

π

21

2

2

2

22

1

2 2

4

4

4

4

4 2

4 2 2

2

∆ ∆

∆

∆

∆

∆

∆

∆

d i d i

− −FH

IKψ ∆

2.

Since P aei= Φ and π ψ γ4

− = , the magnitude of A

is simply

| | sin ,A a= −FH

IKγ ∆

2

and therefore the intensity I of light leaving theanalyzer is

I A a= = −FH

IK| | sin .2 2 2

2γ ∆

(31)

This light intensity vanishes wherever γ − ∆2

is

equal to −nπ , where n is an integer. Solving for therelative angular retardation ∆ and dividing it by 2πgives


N n= = +∆2π

γπ

. (32a)

The sign of the γ π/ term is correct for counter-clockwise rotating circularly polarized light passingthrough the specimen (Fig. 21). If the slow and fastaxes of the quarter-wave plates are reversed, thenthe circularly polarized light passing through thespecimen will rotate clockwise, and it can be shownthat

N n= = −∆2π

γπ

. (32b)

In practice, the sign of the correction is determinedby first interpolating visually to obtain an estimateof the fractional fringe order (with the analyzer andpolarizer crossed), then by rotating the analyzeruntil the closest whole-order fringe moves to thepoint.

Typically, the value of γ can be resolved towithin 2º, so the fractional part γ π/ can be resolvedto within 2º/180º or about 0.01. If the value of N isconsiderably smaller than 1, the relative error in Ncan be appreciable, but if N is of order 1 or more,then the relative error in N is less than 1%, which isquite acceptable.

Separation

The term separation refers to any technique fordetermining one of the principal stresses uniquely.It must be remembered that when we use the basicrelation

σ σ σ1 2− =

Nf

h,

we determine only the difference between the prin-cipal stresses—not these stresses separately.

Boundary condition

Perhaps the most important separation methodconsists of applying the boundary condition at a freeedge of the specimen (Fig. 49). If t and n denote thetangential and normal directions at a free boundary,then τnt = 0 because there is no shear tractionapplied at the boundary, and therefore the normalstresses σn and σt are principal stresses.


t

n

σt

Free boundary

Fig. 49. Stress state at a free boundary.

Furthermore, σn = 0 because there is no normaltraction applied to the surface, either. The principalstresses are now separated because one of them, σn,is equal to zero and the other one, σt , must accountfor the difference σ σ1 2− :

σ

σ σ σ σ

n

tNf

h

=

− = =

0

1 2

,

.(33)

Note that the sign of the tangential normal stress isnot resolved; it may be positive, negative, or zero.

As an example of the application of this separa-tion technique, consider the U-shaped specimen inbending (Fig. 50).10 All of the visible boundary is afree boundary and therefore the normal stresses maybe separated anywhere along this boundary.

Fig. 50. Light-field isochromatics in a U-shaped specimensubjected to an opening load.

10Material, PSM-1. Load, approximately 3 lb. Thickness,0.213 in.


In particular, at the inside curvature, the maxi-mum tangential stress is known to be positive.Since by extrapolation N is approximately 4.2 at thispoint, we have

σ

σ σ

n

t t

=

= = =

0

4 5 43

0 213910

,

. ( )

..psi

At the outside straight edge, N is approximately 3.8,and the tangential stress is known to be negative.Therefore,

σ

σ σ

n

t t

=

= − = =

0

3 8 43

0 213770

,

. ( )

.,psi

that is, σt = −770 psi .

The fact that the principal stresses are easilyseparated along a free boundary led to the early useof photoelasticity to determine stress-concentrationfactors for cutouts, fillets, holes, etc., in a widevariety of geometrical shapes subjected to differentloads. The technique is applicable to both static anddynamic photoelastic investigations.

Caution. The technique applies only to a freeboundary. It does not hold at contact points, wherethe normal stress σn becomes large; and it does nothold for interior points. One might be tempted, forexample, to extend the principle to all points alongthe plane of symmetry of the model in Fig. 50,arguing that τxy vanishes along the plane of sym-

metry, and that σx is equal to zero at the left- andright-hand ends; therefore σx must be equal to zeroeverywhere along the symmetry plane. But σx doesnot vanish along the symmetry plane if at least oneof the surfaces is curved, as can be seen by anequilibrium argument that is used, in fact, to derivean expression for the radial stress component in acurved beam in bending.11

Shear-difference method

To separate the principal stresses in the interiorof a specimen, more work is required. A methodthat is particularly useful because it does not requireany additional experimental apparatus is the shear-

11See the Appendix.


difference method, which is based on stressequilibrium, as will be shown subsequently.

Before we examine this method, let us firstobserve what can be determined easily at any pointin the specimen using a combination of isochro-matics and isoclinics. The isochromatics providethe principal-stress difference

σ σ σ1 2− = Nf

h

and therefore the radius R of Mohr’s circle

RNf

h= − =σ σ σ1 2

2 2, (34)

as shown in Fig. 51.

σσ1σ2

τ

2αR

Nfσh

C

τxy (σx, τxy)

Fig. 51. Calculation of τxy from isochromatics andisoclinics at an arbitrary point.

Meanwhile, a complete family of isoclinics willprovide values of the principal-stress orientation αthroughout the specimen. (Examples of isoclinicsare given in Fig. 5212 for the disk in diametralcompression.) Thus, at any point in the specimen,the angle 2α in Mohr’s circle can be found.

Observe from Fig. 51 that the shear stress τxy

has the value

τ α ασxy R

Nf

h= =sin sin2

22 (35)

and that sufficient information is available from theisochromatics and isoclinics to determine this valueanywhere in the specimen.

12Material, PSM-1. Load, 128 lb. Diameter, approximately2.5 in. Light source, mercury vapor, unfiltered.


Fig. 52(a). The 0º/90º isoclinics for a diskin diametral compression.

Fig. 52(b). The 10º/100º isoclinics for a diskin diametral compression.

We now consider how to use values of τxy along

certain lines in the specimen to separate the princi-pal stresses. Recall from stress equilibrium(Fig. 53) that, in the absence of body forces,


∆x

∆yσx σx + ∆x

∂σx

∂x

τyx + ∆y∂τyx

∂y

τyxx

y

Fig. 53. Differential stress element.

∂σ∂

+∂τ∂

=x xy

x y0

everywhere in the body. Along a line y = const(Fig. 54), dy = 0, and therefore

d d d

d

d

σxx x

xy x

xy

xx

yy

yx

y

yx

= ∂σ∂

+ ∂σ∂

= −∂τ∂

+ ∂σ∂

⋅

= −∂τ∂

0

.

(36)

∆yσx(x0)

∆xy=y+

y=y−

x

y

x0

y = const

Fig. 54. Integration path for the shear-difference method.

In the shear-difference method, we write thecentral-difference approximation for ∂τ ∂xy y/ as

∂τ∂

≅−+ −

xy xy xy

y y

τ τ∆

,


where τxy x+ ( ) and τxy x− ( ) denote the values of shear

stress along the lines y y= + and y y= − , respec-tively, and then integrate Eqn. (36) from x x= 0 toan arbitrary point x along the line y = const:

σ στ τ

x xxy xy

x

xx x

yx( ) ( ) .− = −

−′

+ −

z00 ∆

d (37)

This is the shear-difference formula for finding thevalue of σx anywhere along the line y = const.With σx known at a point, all the normal stressescan be found at that point because the center C ofMohr’s circle is located at

C Rx= −σ αcos ,2

and with C known, the entire Mohr’s circle can beconstructed.

The initial value σx x( )0 in general must befound by using a Mohr’s circle construction afterthe principal stresses have been separated at x x= 0

using the traction-free boundary condition there.

The shear-difference formula is often writtendirectly in terms of isochromatic and isoclinic data.

Recalling that τ αxy R= sin 2 , where RNf

h= σ

2, we

find that

σ σ

α α

σx x

x

x

x xf

h y

N N x

( ) ( )

sin sin .

= − ×

− ′+ + − −z0 2

2 20

∆

d i d

(38)

To avoid error propagation, it is a good idea to plotsmoothed curves through the data for N+ +sin 2αand N− −sin 2α , then to use a high-order integrationscheme, such as Simpson’s rule, to carry out theintegration of the difference between these terms.

Photoelastic coatings

We close the discussion of photoelasticity withan important application that permits surface analy-sis of actual components that have irregularsurfaces.

A photoelastic coating can be molded to thesurface contour of a complicated part and bonded toit (Fig. 55). Light is reflected at the coating–component interface and therefore propagates twice


through the coating thickness h, giving an effectivepath length of 2h in the coating.

P Q

QA

Lightsource

Observer

Photoelastic coating

Component

h

Fig. 55. Photoelastic coating.

The component is now the primary load-carrying member, not the photoelastic material. Thein-plane coating strains are assumed to be equal tothe in-plane surface strains in the component, andthe analysis of the photoelastic patterns is based onthe principal strain difference, which is related tothe principal stress difference in the componentthrough the elastic constants of the componentmaterial. In this respect, photoelastic coatings areregarded in much the same way as brittle coatings.

As discussed in Dally and Riley (1991), com-promises must be made. For example, the coatingmust be thick enough to generate a reasonablenumber of fringes in response to the componentstrains, yet not so thick that the average strains inthe coating deviate significantly from the interfacestrains and the coating begins to reinforce thecomponent.

For components that undergo very small strainswhen loaded, very sensitive coatings must be used,and a careful analysis of the colors of the isochro-matics may be needed to determine fractional fringeorders.

References

Dally, J. W., and W. F. Riley. 1991. ExperimentalStress Analysis, 3rd ed. New York: McGraw-Hill.

Tada, H., P. C. Paris, and G. R. Irwin. 1985. StressAnalysis of Cracks Handbook, 2nd ed. St.Louis, Mo.: Paris Productions, Inc.