ENG5312 – Mechanics of Solids II
Stress Transformation – A Review
Plane Stress

The general state of plane stress at a point has six independent normal
and shear stress components (  x , y , z ,  xy ,  xz ,  yz )


A state of plane stress can be defined if the general state of stress at a
point can be reduced to two normal stresses and one shear stress.
1
ENG5312 – Mechanics of Solids II

2
The state of plane stress can be expressed by stress components along
x' y' axes, which are not coincident with the x, y axes.


General Equations of Plane Stress

Sign convention
o Positive normal and shear stresses act in the positive co-ordinate
direction on a positive face, and in the negative co-ordinate
direction on a negative face.
o Angles are measured positive counter-clockwise.
ENG5312 – Mechanics of Solids II

Using an FBD of a sectioned element along the inclined plane.

Assuming static equilibrium and summing forces in the x' and y'
directions gives:
Fx'  0   x' A  ( xy A sin  )cos   ( xy A cos  )sin 
( x A cos  )cos   ( y A sin  )sin  

Fy'  0   x'y' A  ( xy A cos  )cos   ( xy A sin  )sin 
( x A cos  )sin   ( y A sin  )cos 


Or
 x'   x cos 2    y sin 2    xy (2 sin  cos  )
 x'y'  ( y   x )sin  cos    x y (cos2   sin 2  )

3
ENG5312 – Mechanics of Solids II
4
Using the trigonometric identities: sin 2   (1 cos2 ) / 2;
cos2   (1 cos2 ) / 2; and sin 2  2sin  cos :
      y 
 x'  x y   x
cos2   xy sin 2
2
2 



   y 
 x'y'   x
sin 2   xy cos2


 y' 

 x   y  x   y 


2
2
 

2
cos2   xy sin 2

(1)
(2)
(3)
These
 are the general equations of plane-stress transformation. They
permit the calculation of the plane stresses acting on an element oriented
at an angle  to the given stress components.
 Stresses
Principal

Given a state of plane stress, what are the maximum and minimum values
of normal stress acting at the point, and in which direction are these
stresses oriented?

Differentiate Eq. (1) for  x' wrt  , and set the result equal to zero and
solve:


tan 2 p 
 xy
 x   y  2
(4)
i.e. there are two values of  p ,  p1 and  p2 which are oriented at 90o to
each other.


Substituting expressions for sin 2 p and cos2 p into Eq. (1) for  x' gives:
 
 1,2 

(5)


 x   y 2
2
 
   xy
2

 2 
x y
ENG5312 – Mechanics of Solids II
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i.e. the maximum (  1) and minimum (  2 ) in-plane normal stress acting at
a point.
 1 and  2 are the principal stresses, which act on the principal planes
 and  measured
 clockwise positive from the co-ordinate
defined by
p
p

1
2
system for the given stresses (i.e. x, y ).



Note: No shear stress acts on the principal planes.



Maximum In-Plane Shear Stress

Given a state of stress, what is the maximum in-plane shear stress acting
at the point, and in which direction does the stress act?

Differentiate Eq. (2) for  x'y' wrt  , set the result equal to zero, and solve:



tan 2 s 
  x   y  2
 xy
(6)
Solving for  s1 and  s2 shows that the planes of maximum in-plane shear
stress are directed 45o to the principal planes.


The maximum in-plane shear stress is:




max
in plane
 x   y 2
2
 
   xy
 2 
(7)
Substitution  s1 and  s2 into Eq. (1) for  x' gives:




 avg 
x y
2
(8)
i.e. a normal stress acts on the plane of maximum in-plane shear stress.

Note: The existence of 
avg is quite apparent on Mohr’s Circle.

ENG5312 – Mechanics of Solids II
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Mohr’s Circle – Plane Stress

Mohr’s circle provides a graphical solution to the general equations for
plane-stress transformation.

Rewrite Eqs. (1) and (2) for  x' and  x'y' :

 x' 
x y

2
   y 
  x
cos2   xy sin 2
 2 
   y 
 x'y'   x
sin 2   xy cos2



Square and add these equations to eliminate  :


2
2

 x   y 
2
 x' 
   x'y'
2 

 x   y 2
2
 
   xy
 2 
Or
 x'   avg 
2

2
  x'y'
 R2
(9)
Where



 avg 
x y
(9a)
2
 x   y 2
2
R  
   xy
 2 
(9b)
Equation (9) is the equation of a circle with radius R offset in the 
direction by  avg on the ,  axes.




ENG5312 – Mechanics of Solids II
7

Point A is the known state of stress  x and  xy .

Point C is the offset at  avg 
x y
2


 x   y 2
2
 Radius R  
   xy .
 2 


Rotation of an element by 90o will give  x'   y and  x'y'   xy . To obtain
this result on Mohr’s circle requires a rotation of 180o, therefore, must
 rotate by angle 2 on Mohr’s circle. Maximum (principal) stress occurs at
B, minimum normal stress occurs at D, i.e. point with no shear stress.

Maximum in-plane shear stress occurs at points E and F, i.e.  avg exists
on the
plane.

A state of stress at an orientation of  counter-clockwise from the known
state is given by the point P ( 2 on Mohr’s circle). x' and  x'y' can be
read from the circle or calculated from trigonometry.






ENG5312 – Mechanics of Solids II
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Absolute Maximum Shear Stress (3D)


If a body is subjected to a general 3D state of stress there is a unique
orientation of a plane that will give the principal stresses
(  min   int   max ), and there will be no shear stress components acting
on the principal planes.

Assuming:  max in the x' ,  min in the z' and  int in the y' directions,
respectively.


 


Drawing Mohr’s circle for each plane.

ENG5312 – Mechanics of Solids II

9
The absolute maximum shear stress is located on the largest circle:
max
 abs

 max   min
(10)
2
And it has an associated normal stress:


 avg 
 max   min
(11)
2
The absolute maximum shear stress occurs on a plane oriented 45o (i.e.
90o on Mohr’s circle) to a principal plane (here the x',z' plane).

 Stress)
Absolute Maximum Shear Stress (Plane

Consider an element exposed to plane stress ( x', y' plane). Define  max
in the x' ,  int in the y' , and  min ( 0) in the z' directions, respectively.
Assuming  max and  int have the same sign:
 








Here
max
 abs
  x'z' max 

 max
2
(12)
ENG5312 – Mechanics of Solids II
10

max
Note: The maximum in-plane shear stress  in
plane   max   int / 2 is not
equal to the maximum shear stress in the body.

Consider the case where  max and  min have opposite signs (  max in the
y' directions, respectively).
x' ,  int in the z' , and  min in the 


 




Here
max
max
 abs
  in
plane 


 max   min
2
(13)
max
These values for  abs
are important for ductile materials as they fail in
shear.

