ENG5312 – Mechanics of Solids II
11
Strain Transformation
Plane Strain

The general state of strain at a point in a body is composed of three
components of normal strain (  x ,  y and  z ), and three components of
shear strain (  xy ,  xz and  yz ).

As for stress the given strains may be transformed to components along


alternate co-ordinatedirections.




One common use of this technique is to transform strains measured along
particular directions on a strain gauge rosette into other directions.

Consider only strain in the x, y plane (i.e.  x ,  y and  xy ):
 



Note: Plane stress does not cause plane strain. If an element is
subjected to normal stresses  x and  y , normal strains  x ,  y and  z are
produced. Normal strain  z results from the Poisson effect.



 

ENG5312 – Mechanics of Solids II
12
General Equations of Plane-Strain Transformation

Sign Convention
o  x and  y are positive if they cause elongation along the x and y
axes, respectively.

o  xy is positive if the interior angle AOB becomes smaller than 90 o.






The goal will be to obtain strain transformation to the x', y' plane at an
angle  to the x, y plane.


ENG5312 – Mechanics of Solids II

13
To determine  x' we need to find the change in length of dx' due to  x ,  y
and  xy .


 


Normal strain  x will cause a change in length of dx' : x dx cos



Normal strain  y will cause a change in length of dx' : y dy sin 





Shear strain  xy will cause a change in length of dx' :  xy dy cos



ENG5312 – Mechanics of Solids II


Adding the three components to give the total change in length ( x' ) of
dx' :
x' x dx cos  y dy sin    xy dy cos

 The normal strain in the x' direction is defined as  x'  x' / dx' , and since
dx  dx'cos and dy  dx'sin  :



14


(14)
Similarly:
y' x dx sin   y dy cos   xy dy sin 


 x'   x cos 2    y sin 2    xy sin  cos 
The normal strain in the y' direction is  y'  y' / dy', and since
dx  dy'sin  and dy  dy'cos :




 y'   x sin 2    y cos 2    xy sin  cos 

(15)
  x'y' we need to determine the amount of rotation line
To evaluate
segments dx' and dy' undergo due to  x ,  y and  xy .





 

For small angles:   y' dx' , and using dx  dx'cos and dy  dx'sin  :
y'
 x dx'sin  cos    y dx'sin  cos    xy dx'sin 2 


dx'
dx'
2
 x   y sin  cos   
sin

xy


(16)
ENG5312 – Mechanics of Solids II
15
The angle   x' dy', where dx  dy'sin  and dy  dy'cos :



dy'

dy'


  x   y sin  cos   xy cos2 


2
x'  x dy'sin  cos   y dy'sin  cos   xy dy'cos 
(17)
The total change in angle will give  x'y' and since  and  are measured
in opposite directions:
 x'y'      2 x   y sin  cos    xy cos2   sin 2  
(18)



 Substitution of the trigonometric identities: sin 2  2sin  cos ;
cos 2   1 cos2 / 2 ; sin 2  cos2  1 into Eqs. (14), (15) and (18)
 gives:
 x' 


 x   y  x  y 
 
 2
2
 x' y'

x y
 
2
 2
 y' 

 
 2
(20)

xy
sin 2
cos2 
2

(21)
Eqs.(19) to (21) are the general equations of plane-strain transformation.
Note the similarity to the equations for plane-stress transformation, Eqs.
(1) to (3). Note:  x ,  y ,  x' and  y' correspond to  x ,  y ,  x' and  x' while
 xy and  x'y' correspond to  xy /2 and  x'y' /2 .
  

(19)


 sin 2  xy cos 2
2

 x   y  x   y 
2

xy
sin 2
cos2 
2


  




ENG5312 – Mechanics of Solids II
16
Principal Strains

As for plane stress, an element may be oriented such that the plane strain
is represented by two normal strains only (i.e. principal strains), and no
shear strains.

Due to the similarity between the plane-stress and plane-strain
transformation equations, the orientation of the principal axes and the
principal strains are:
tan 2 p 

1,2 
x  y
2
 xy
x  y
 x   y 2  xy 2
 
   
 2   2 
(22)
(23)
Maximum In-Plane
Shear Strain


Similarly, the maximum in-plane shear strain is;
   
tan 2 s   x y 
  xy 
max
 in
plane

2
 x   y 2  xy 2
 
   
 2   2 
 avg 


x  y
2
(24)
(25)
(26)
Again the axes for the maximum in-plane shear strain are directed 45o to
the axes for the principal strains.

ENG5312 – Mechanics of Solids II
17
Mohr’s Circle for Plane Strain

Due to the similarity between the plane-stress and plane-strain
transformation equations, Mohr’s circle may be used to solve graphically
for plane-strain transformation.

Following the procedure used for developing the equations used in Mohr’s
circle for plane stress gives the following equations for Mohr’s circle for
plane strain:
 x   avg 
2
 xy 2
    R 2
 2 
 avg 

x  y
2
 x   y 2  xy 2
R  
   
 2   2 


(27)
(27a)
(27b)
Eq. (27) is the equation of a circle with radius R that is offset from the
origin in the  direction by  avg on the , /2 axes.





ENG5312 – Mechanics of Solids II
18

Point A is the known state of strain  and  xy /2 .

Point C is the center, offset by  avg   x   y /2 .


 x   y 2  xy 2
 Radius R  
   
 2 
  2 


Rotation of an element by 180o will give  x'   x and  x'y'   xy . To obtain
this result on Mohr’s circle requires a rotation of 360o, therefore, a rotation
of an element by  requires an angle of 2 on Mohr’s circle.

 at points
Principal normal strains are located
 B (  1, i.e. maximum) and D
(  2 , i.e. minimum), i.e. points of no shear strain. These strains occur at
 from   0 .
2 p1 and 2 p2 as measured
angles 

Maximum in-plane shear strain (  R) occurs at E and F, with the
associated normal strain  avg.






ENG5312 – Mechanics of Solids II



19
State of strain at an orientation of  counter-clockwise from the known
state is given by point P (at 2 counterclockwise from   0 ).  x' and
 x'y' /2 can be read from Mohr’s circle or found from trigonometry.
 at 900 to points B and D, as expected since
Note: points E and F occur
 in-plane shear stress are
 oriented
 45o to the
the axes of maximum
principal axes.
Absolute Maximum Shear Strain (3D)

If a body is subjected to a general 3D state of stress, an element at a point
within the body may be oriented such that it is subjected to normal
stresses only (i.e. the principal stresses  max ,  int and  min ). These
normal stresses would cause the principal strains (with no shear strain)
since there is no shear stress on the principal planes.


 to the
 x' , y' and z' directions,
Assuming  max ,  int and  min correspond
respectively:



 

ENG5312 – Mechanics of Solids II
20

This state of strain can be illustrated on Mohr’s circle:

The absolute maximum shear strain will occur on the largest Mohr’s circle
(here the x',z' plane) and it has magnitude:
max
 abs
  max   min

(28)

With an associated normal strain:


 avg 
 max   min
2
(29)
ENG5312 – Mechanics of Solids II
21
Absolute Maximum Shear Strain (Plane Strain)

Consider an element exposed to plane strain ( x', y' plane). Define  max ,
 int and  min (  0) in the x' , y' and z' directions, respectively. Assuming
 max and  int have the same sign (positive here):
 




 



The absolute maximum shear strain is located out of plane (i.e. not on the
x', y' plane) and it has magnitude:
max
 abs
  max


(30)
ENG5312 – Mechanics of Solids II

If the in-plane principal strains have opposite signs:

Then:
max
max
 abs
  max   min   in
plane

22
(31)
i.e. when the principal strains have opposite signs the absolute maximum
shear strain occurs in-plane (here x', y' ) and is equivalent to the maximum

in-plane shear strain.

ENG5312 – Mechanics of Solids II
23
Strain Gauge Rosettes

A strain gauge can be used to measure the normal strain in a specimen
loaded in tension (i.e. measurement of the change in resistance in a thin
wire as it is stretched).

The normal strains on the surface of a body are measured with a strain
gauge rosette, and this information is used to specify the state of strain at
a point.

Using the strain transformation equation (Eq. (14) for  x' ) for each gauge:
 a   x cos2  a   y sin 2  a   xy sin  a cos  a
 b   x cos2  b   y sin 2 
 b   xy sin  b cos  b
(32)
 c   x cos 2  c   y sin 2  c   xy sin  c cos  c

Solving Eq. (31) simultaneously will give the state of strain (  x ,  y and
 xy ). Then the plane-strain transformation equations, Eqs. (19) to (21), can

be used to determine the strains along other co-ordinate directions.
 

ENG5312 – Mechanics of Solids II
24

Often strain gauges rosettes are arranged in 45o or 60o configurations.

For the 45o rosette,  a  0 o ,  b  45 o , and  c  90 o , and Eq. (32) gives:



x  a
y  c
 xy  2
 b   a   c 
(33)
For the 60o rosette,  a  0 o ,  b  60 o , and  c 120 o , and Eq. (32) gives:



x  a
1
3
 y  
2 b  2 c   a 
 xy 

2
 b   c 
3
(34)