Chapter 12. Photoelasticity Applications
Sample problems
12-S1
Derive the equations of Tardy’s compensation
Solution to 12-S1
The following sequence of operations is required to get the Tardy compensation. It begins with a
polarizer and analyzer that are aligned with the principal axis of the retarder. According to Figure
P12.1 we start with a vertical polarizer followed by a pair of crossed quarter wave plates. A
retarder is inserted between the quarter wave plates. Finally a rotating analyzer receives the
output of the optical arrangement. It is a circular polariscope where the direction of polarization
coincides with the fast axis of the retarder.
Figure P12.1 . Set up for the Tardy’s compensation
The solution is obtained by utilizing the instrument matrices of the different components
following the order of operations as it has been done in preceding problems in Chapter 11.

 cos 2 
 sin  cos  1  1  i e i
E( z, t )  



sin 2   2  i 1   0
 sin  cos 
0  1 1 i  i 0


 Ee 1
ei.  2 i 1
 
The first Jones vector corresponds to the vertically polarized light produced by the polarizer. The
next term, second term corresponds to the first quarter wave plate at +45o with respect to the
reference axis 0A. The third term corresponds to the retarder with the fast axis parallel to the
polarization direction. The fourth term corresponds to the second quarter wave plate. The last
term corresponds to the rotated analyzer that is turned in the –θ direction. The matrix of the
rotated analyzer is obtained b y rotating the instrument matrix (Table 11.2)
sin  cos 
cos   sin  1 0  cos  sin    cos 2 

 sin  cos   0 0  sin  cos   
sin 2  



 sin  cos 
The output of the circular polariscope can be computed by performing the different matrix
products.
1 i  0 i  e i.
i 1 1  1 , 

     0
0  i  ie  i 
   
.
ei.  1  ei 
 1  i ie  i  ie  i  ie i 
 i 1   i    i i 

 e   e  e 
The x-component of the Jones vector can be written after dividing and multiplying by 2i.
ei  e i
i(ei  ei )  2
 2 sin 
2i
Similarly the y-component,
ei  e i
(ei  ei )  2i
 2i sin 
2i
ie  i  ie i   sin  
.Utilizing the identity ei  cos   i sin  then,
 i i   

 e  e   i sin 
 sin    sin  
 i sin   cos   ei , replacing in the preceding equation 
, this vector can

i 
 i sin  cos   e 
 sin    0 
be written, 
   
cos   e 
Operating with the first vector,
 cos 2 
 sin  cos   sin    cos 2  sin   sin  cos  cos  




sin 2   cos   sin  cos  sin   sin 2  cos 
 sin  cos 
 cos2  sin   sin  cos  cos    cos  cos  sin   sin  cos  

 

2
 sin  cos  sin   sin  cos   sin cos  sin   sin  cos 
 cos  sin c   
 sin  sin e    this is the output of the Tardy compensation. The irradiance is,


I  E2 sin e2   
With the substitution sin e 2    
1  cos  
we get the final result,
2
I
1  cos  
2
The intensity will be zero when,    or k with k  0,1,2,3....
I(, ) 
Then

  (k  ) with k  0,1,2,3....

This is the final answer of the problem.
 0 
We still have the additional term   
 e 
Then
 cos 2 
 sin  cos  0
  sin  cos 

a





sin 2  
sin 2   a 

 sin  cos 
where a   e 
The final irradiance is proportional to: a 2 sin 2  but when the extinction occurs θ=kπ and then
this term is also zero.
12-S2
Analyze the effect of the mirror in the circular polariscope shown in Figure 12.43.
Solution to 12-S2
The convention of polarization orientation from the point of view of observing towards the
source as stated in section 6.4 is assumed. A point to take into consideration is the reversal of the
components of the electrical vector caused by a mirror, this effect is independent of the
convention of sign utilized to define handness. Also there is a jump of phase of π at the surface
of the mirror, but since the components of polarization experience the same change it does not
change the relative phase. The reversal caused by the mirror converts the right rotation into to
left rotation, the directions of the x axis or the y axis must be reversed. If we reverse the x-axis a
Jones matrix of the form,
 1 0
R

 0 1
Considering a circular polariscope,

1 0 1  1  i 1 1 i  i 0
E( z, t )  




 Ee 1
0 0 2  i 1  2 i 1
 
we have to introduce the mirror, then the output is

1 0 1  1  i  1 0  1 1 i  i 0 
E( z, t )  





 Ee 1 
0 0 2  i 1   0 1  2 i 1
 
This equation provides the output of the system. The bracket was added to indicate that the wave
front resulting from the indicated optical devices will be subjected to reversal. From problem
11.7 we get
 1 1 i  0  i 


     
 2 i 1 1  1
By applying the reversal,
  1 0  i   i 
 0 1 1   1 

  
If we apply the reversed beam to the rest of the polariscope
 1  i   i 
i
 i 1   1   2 0 

 
 
if we keep the crossed polaroids,
1 0  i   i 
0 0  0  0

   
Due to the reversal of the mirror the dark background polariscope has been converted in a bright
background polariscope. The beam exits the device and has experienced a retardation of π/2, due
to the fact that has traversed two quarter wave plates that due to the reversal add their
retardations. Circular polarization is pictured in the mind by thinking of the electric field as being
divided into two components which are at right angles to each other.
Figure P12.2
This is illustrated in Figure P12.2. In the way that the relative amplitude at the origin of
coordinates is given in the graphic representation the horizontal component leads by a quarter of
a wave in the vertical component. This phase difference creates the helix that the vector
describes.
Looking towards the source the vector rotates in the counterclockwise direction. The handedness
of polarized light is reversed when it is reflected by a mirror. As a result of the interaction of the
electromagnetic field with the conducting surface of the mirror, as it has been stated before, both
orthogonal components are shifted by one half of a wavelength or π. This effect does not change
the relative phase. However as a result of the change in direction, a mirror image of the wave is
created and the relationship of the phases is reversed changing the handness of the light. It is also
necessary to realize that the definition of handness is a convention and has no bearing in the
effect that we are describing. There is a physical effect that we call reflection and we are used to
it but this effect implies the change of orientation of the component vector that causes the
handness change.
12-S3
Analyze the effect of reflection on a mirror on a general instrument matrix.
Solution to 12-S3
One important aspect of the Jones matrix for an instrument from forward passage of light as
related to the backward passage, is that the backward passage matrix is the transpose of the
forward matrix. This property arises from the matrix property that allows moving from the input
to the output, or to the output towards the input.
Recalling the definition of transpose of a matrix M, MMT  1 . If one has the
relationship Mout  AM in ,
then M T out   AM in T  M T in AT ,
multiplying
by
A,
M T out A  A  M in   M T in A T A
T
AT A  AA T  I , where I is the identity matrix. Then M T out A  M T in . Applying again the
transpose operation ,
A T M out  M
in
(a)
Let us now consider a polarization vector Pin that is operated on by a matrix A yielding a matrix
Pout, Pout=APin, if the sense of the beam is reversed, taking into consideration (a). To this result
  1 0
the effect of the mirror that switches the handness must be added, calling R ev  
 then,
 0 1
Poutrev=RevATAPout.
One example of application of this equation is the case of a half wave rotator applied followed
 cos  sin  
by a mirror. The half wave rotator has the Jones matrix 
 .The total effect of the
 sin  cos 
light going through the rotator and being reflected by the mirror is,
 1 0 cos   sin   cos  sin    1 0
 0 1  sin  cos    sin  cos    0 1



 

For example if one sends a plane polarized beam through the rotator it rotates the plane of
vibration of the vector of π/2. As it comes back after reflection the rotator changes sign in the x1 1
direction. If the plane polarized beam comes at 45o the Jones vector is J 45o 
  , after going
2 1
through the rotator and the mirror,
 1 0 1 1 1  1
J out  

 
  . See Figure 6.9, δ=0 and δ=π.
2 1
 0 1 2 1
Problems to be solved
12.1
Derive the equation of the Séanarmont compensation. Utilize Figure 12.11 as a basis for the
derivation.
12.2
Analyze the effect of a set up that consists of a right circular polarizer and a mirror.
12.3
Derive an expression for the Babinet-Soleil compensator operation.
12.4
Figure P12.3 is the isochromatic patter of a beam under pure bending. The dimensions of the
cross- section of the beam are: depth 21.5 mm, width3.43 mm. The applied bending moment is
3.423 Nm. Compute the photoelastic coefficient f of the material.
Figure P12.3. Bending moment pattern of a beam subjected to pure bending.
12.5
Utilizing the isochromatic fringe pattern of figure P12.4 (Figure 12.8): Convert the data from the
English system to the ISU(International Unit System). Plot the stresses along the boundary of the
hole. Justify the maximum fringe order shown in the figure. Plot the difference of stresses along
a coordinate axis corresponding to the horizontal symmetry axis. Plot the difference of stresses
along the vertical axis of symmetry. Utilize the rules given in section 12.5 to do your plots and
Figure 12.21.
Figure P.12.4.Bar with a hole
12.6
Figure P12.5 shows a beam under bending produced by a concentrated load in the middle of the
span. The beam data are: Span L=12.7 cm. Depth h=21.1 mm, thickness t=6.35 mm.The shear
stress fringe constant is fs=2.665 N/cm2.
Figure P12.5.Beam subjected to concentrated load.
Take the line AB. As mentioned in section 5.5.1 we do not know offhand which is the maximum
principal stress and the minimum principal stress but through analysis we can find out. We will
call the maximum principal stress p and the minimum q. We know that one of these stresses is
tangent to the isoclinics and the other is perpendicular. The isoclinics are referred to the
horizontal line. Let us consider different elements in the cross-section of the beam. From the
behavior of the beam we know that in the upper boundary the horizontal stress is of the q family,
maximum compressive stress and p=0, free boundary, (see Figure P.12.6). Moving downwards,
the direction of the principal stresses change, the stress q rotates counterclockwise. In the tension
region in the principal stress p is the maximum tensile stress q=0 because of a free boundary. The
difference p-q is always positive by convention. In problem 12.4 it was shown that signs of
stresses and retardations have separate conventions. A sign convention for the positive shear
stresses must be adopted. The convention that the positive shears converge to the first diagonal is
a good choice. This is due to the fact that the value of p-q is always positive and if plotting from
the top to the bottom in a plot of the vertical section it will be always a positive quantity.
Assuming that the section AB is at a distance from the center line 0.1 of h, plot a graph of the
difference of the principal stresses vs. the depth and of the angle θ. Divide the depth into enough
points such that the curves of the corresponding quantities are well defined.
Figure P.12.6 Elements along the vertical line AB.
In addition:
1. Compute and plot the shear stresses τxy along the vertical line. Remember that as stated in
problem 12.4 the positive shear is along the diagonal that makes the angle  45 o with the
algebraic maximum stress.
2. Verify the equilibrium equation in the vertical direction.
3. Compare the actual shear distribution with the shear distribution computed with the
Strength of Materials equation.
4. Give an evaluation of this comparison and of the utilization of the Strength of Materials
equation for design purposes.
5. Plot the isostatics along the vertical axis, indicating the p and q families.
6. Compute the maximum bending stresses and compare with the bending stresses
computed with the Strength of Materials equation.
7. Give an approximated plot of the bending stresses in the vertical section. Hint: At one
point the bending stresses change sign.
12.7
Utilizing equations (12.22) determine the stresses signs along the boundary (Figure P12.7)
knowing that in the left straight boundary the principal stress p is parallel to the boundary.
Figure P12.7.Boundarystresses of a 2-D field.
12.8
Plot the difference of the stresses along the diameter of the disk. Discuss the obtained results in
view of the analysis of fringe patterns utilizing the RGB method.
Figure. P12.8.Central portion of disk under diametrical compression. Blue pattern of partially
coherent light, band-pass filtered fringe profile.
Figure P12.9. Central portion of disk under diametrical compression green pattern of partially
coherent light. Band-pass filtered cross-section of the fringes.
Figure P12.10. Central portion of disk under diametrical compression. . Red pattern of partially
coherent light. Band-pass filtered profile of the fringes.
Figure P.12.11.Central portion of disk under diametrical compression. Monochromatic pattern
of partially coherent light. Band-pass filtered profile of the fringes.
12.9
Utilizing the information contained in Figure P12.12:
1.
Determine the strains along the cantilever beam shown in the figure.
2.
Neglecting any reinforcing effect of the coating compute the bending moments
in the cantilever bam knowing that is made of aluminum alloy.
12.10
Utilizing the information contained in Figure P12.12 and assuming that the same coating has
been applied:
1.
Analyze the difference of strains distribution in the coated component shown
on the left of the figure along the symmetry axis and at the shown horizontal boundaries.
2.
What inferences can be made concerning the actual strain distribution in the
structural component.
Thickness of coating d=2.54 mm, Ke=0.15, λ=575 nm fε=757 microstrain
Figure P12.12 Example of photoelastic coating strain analysis (Courtesy of Vishay).
12.11
Figure P12.13 shows a prism in contact with a plate that is assumed to represent the half space.
The plate is large enough to simulate the half space. The prism has a width of 2 cm and a height
of 1 cm. The thickness of the specimens is 0.52 cm. A load P=81.4 N. is applied to the prism.
The upper part of the Figure P12.13 corresponds to the photoelastic fringes obtained with a
Mach-Zender interferometer. This pattern can be assumed as the superposition of two families of
fringes the isopachics and the isochromatics. The second pattern represents the isochromatics
and has been recorded to facilitate the analysis of the two systems of fringes. Observe that the
isopachics experience a change of phase of π. That is a bright fringe is transformed into a black
fringe. Hence in analyzing the data one should consider this property. Nisida and Saito provided
the equation (12.92),
Figure P12.13.Patterns of absolute and relative retardation, contact stresses
The isopachics and the isochromatic modulate each other. This equation allows the separation of
stress if one gets the isopachics and the isochromatics. For the material utilized to get the
photoelastic patterns the constants A, B and C have been measured. A=0.367 x10-11 m2/N, B=3.81x x10-11 m2/N. C=A-B/2=4.18 x10-11 m2/N. Determine the contact stresses between the plane
and the half space. Plot both contact stresses of the half space and that of the prism. Verify the
equilibrium conditions.
Hint: In a graph plot the fringe orders corresponding to the both system of fringes. Write a
program to transform the fringe orders into values of σ1 and σ2.
12.12
Figure P12.14 shows the scattering fringe pattern of a square section under torsion. The section
has 5 cm × 5 cm. The fringe value for shear is f = 296.4 KPa.
Figure P12.14.Square section under torsion scattered light pattern.
Find the stress concentration effect caused by the circular notch. Assume as reference value the
stress on the horizontal axis.