CHAPTER
7
MECHANICS OF
MATERIALS
Transformations of
Stress and Strain
Dr. Atta ur Rehman Shah
(atta.shah@hitecuni.edu.pk)
Website: https://sites.google.com/view/atta85
Courtesy: © 2002 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Transformations of Stress and Strain
Introduction
Transformation of Plane Stress
Principal Stresses
Maximum Shearing Stress
Example 7.01
Sample Problem 7.1
Mohr’s Circle for Plane Stress
Example 7.02
Sample Problem 7.2
General State of Stress
Application of Mohr’s Circle to the Three- Dimensional Analysis of Stress
Yield Criteria for Ductile Materials Under Plane Stress
Fracture Criteria for Brittle Materials Under Plane Stress
Stresses in Thin-Walled Pressure Vessels
7-2
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Introduction
• The most general state of stress at a point may
be represented by 6 components,
 x , y , z
normal stresses
 xy ,  yz ,  zx shearing stresses
(Note :  xy   yx ,  yz   zy ,  zx   xz )
• Same state of stress is represented by a
different set of components if axes are rotated.
• The first part of the chapter is concerned with
how the components of stress are transformed
under a rotation of the coordinate axes. The
second part of the chapter is devoted to a
similar analysis of the transformation of the
components of strain.
7-3
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Introduction
• Plane Stress - state of stress in which two faces of
the cubic element are free of stress. For the
illustrated example, the state of stress is defined by
 x ,  y ,  xy and  z   zx   zy  0.
• State of plane stress occurs in a thin plate subjected
to forces acting in the midplane of the plate.
• State of plane stress also occurs on the free surface
of a structural element or machine component, i.e.,
at any point of the surface not subjected to an
external force.
7-4
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Transformation of Plane Stress
• Consider the conditions for equilibrium of a
prismatic element with faces perpendicular to
the x, y, and x’ axes.
 Fx  0   xA   x A cos  cos   xy A cos sin 
  y A sin  sin    xy A sin   cos
 Fy  0   xyA   x A cos sin    xy A cos  cos
  y A sin   cos   xy A sin  sin 
• The equations may be rewritten to yield
 x 
 y 
 x  y
2
 x  y
 xy  
2


 x  y
2
 x  y
2
 x  y
2
cos 2   xy sin 2
cos 2   xy sin 2
sin 2   xy cos 2
7-5
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Principal Stresses
• The previous equations are combined to
yield parametric equations for a circle,
 x   ave 2   x2y  R 2
where
 ave 
2
 x  y
 x  y 
2
   xy
R  
2


2
• Principal stresses occur on the principal
planes of stress with zero shearing stresses.
 max, min 
tan 2 p 
 x  y
2
2
 x  y 
2
   xy
 
2


2 xy
 x  y
Note : defines two angles separated by 90o
7-6
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Maximum Shearing Stress
Maximum shearing stress occurs for
 x   ave
2
 x  y 
2
   xy
 max  R  
2


 x  y
tan 2 s  
2 xy
Note : defines two angles separated by 90o and
offset from  p by 45o
    ave 
 x  y
2
7-7
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Example 7.01
SOLUTION:
• Find the element orientation for the principal
stresses from
2 xy
tan 2 p 
 x  y
• Determine the principal stresses from
 max, min 
x  y
2
 x  y 
2
   xy
 
2


2
For the state of plane stress shown,
determine (a) the principal planes, • Calculate the maximum shearing stress with
(b) the principal stresses, (c) the
2





x
y
2
maximum shearing stress and the
   xy
 max  
2
corresponding normal stress.


x  y

 
2
7-8
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Example 7.01
SOLUTION:
• Find the element orientation for the principal
stresses from
tan 2 p 
2 xy
 x  y

2 40 
 1.333
50   10 
2 p  53.1, 233 .1
 x  50 MPa
 x  10 MPa
 xy  40 MPa
 p  26.6, 116 .6
• Determine the principal stresses from
 max, min 
x  y
2
 20 
2
 x  y 
2
   xy
 
2


30 2  40 2
 max  70 MPa
 min  30 MPa
7-9
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Example 7.01
• Calculate the maximum shearing stress with
2
 x  y 
2
   xy
 max  
2



30 2  40 2
 max  50 MPa
 x  50 MPa
 x  10 MPa
 xy  40 MPa
 s   p  45
 s  18.4, 71.6
• The corresponding normal stress is
 x   y 50  10
    ave 

2
2
   20 MPa
7 - 10
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 7.1
SOLUTION:
• Determine an equivalent force-couple
system at the center of the transverse
section passing through H.
• Evaluate the normal and shearing stresses
at H.
• Determine the principal planes and
calculate the principal stresses.
A single horizontal force P of 150 lb
magnitude is applied to end D of lever
ABD. Determine (a) the normal and
shearing stresses on an element at point
H having sides parallel to the x and y
axes, (b) the principal planes and
principal stresses at the point H.
7 - 11
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 7.1
SOLUTION:
• Determine an equivalent force-couple
system at the center of the transverse
section passing through H.
P  150 lb
T  150 lb 18 in   2.7 kip  in
M x  150 lb 10 in   1.5 kip  in
• Evaluate the normal and shearing stresses
at H.
y 
1.5 kip  in 0.6 in 
Mc

1  0.6 in 4
I
4
 xy  
2.7 kip  in 0.6 in 
Tc

1  0.6 in 4
J
2
 x  0  y  8.84 ksi  xy  7.96 ksi
7 - 12
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 7.1
• Determine the principal planes and
calculate the principal stresses.
tan 2 p 
2 xy
 x  y

27.96 
 1.8
0  8.84
2 p  61.0,119 
 p  30.5, 59.5
 max, min 
x  y
2
2
 x  y 
2
   xy
 
2


2
0  8.84
 0  8.84 
2

 
  7.96 
2
 2 
 max  13.52 ksi
 min  4.68 ksi
7 - 13
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Mohr’s Circle for Plane Stress
• With the physical significance of Mohr’s circle
for plane stress established, it may be applied
with simple geometric considerations. Critical
values are estimated graphically or calculated.
• For a known state of plane stress  x , y , xy
plot the points X and Y and construct the
circle centered at C.
 ave 
 x  y
2
2
 x  y 
2
   xy
R  
2


• The principal stresses are obtained at A and B.
 max, min   ave  R
tan 2 p 
2 xy
 x  y
The direction of rotation of Ox to Oa is
the same as CX to CA.
7 - 14
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Mohr’s Circle for Plane Stress
• With Mohr’s circle uniquely defined, the state
of stress at other axes orientations may be
depicted.
• For the state of stress at an angle  with
respect to the xy axes, construct a new
diameter X’Y’ at an angle 2 with respect to
XY.
• Normal and shear stresses are obtained
from the coordinates X’Y’.
7 - 15
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Mohr’s Circle for Plane Stress
• Mohr’s circle for centric axial loading:
x 
P
,  y   xy  0
A
 x   y   xy 
P
2A
• Mohr’s circle for torsional loading:
 x   y  0  xy 
Tc
J
x y 
Tc
 xy  0
J
7 - 16
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Example 7.02
For the state of plane stress shown,
(a) construct Mohr’s circle, determine
(b) the principal planes, (c) the
principal stresses, (d) the maximum
shearing stress and the corresponding
normal stress.
SOLUTION:
• Construction of Mohr’s circle
 ave 
x  y

50    10   20 MPa
2
2
CF  50  20  30 MPa FX  40 MPa
R  CX 
30 2  40 2  50 MPa
7 - 17
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Example 7.02
• Principal planes and stresses
 max  OA  OC  CA  20  50
 max  70 MPa
 min  OB  OC  BC  20  50
 min  30 MPa
FX 40

CF 30
2 p  53.1
tan 2 p 
 p  26.6 , 116.6
7 - 18
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Example 7.02
• Maximum shear stress
 s   p  45
 max  R
    ave
 s  71.6
 max  50 MPa
   20 MPa
7 - 19
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 7.2
For the state of stress shown,
determine (a) the principal planes
and the principal stresses, (b) the
stress components exerted on the
element obtained by rotating the SOLUTION:
given element counterclockwise • Construct Mohr’s circle
through 30 degrees.
 x   y 100  60
 ave 

 80 MPa
2
R
2
CF 2  FX 2  20 2  482  52 MPa
7 - 20
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 7.2
• Principal planes and stresses
XF 48

 2.4
CF 20
2 p  67.4
tan 2 p 
 p  33.7 clockwise
 max  OA  OC  CA
 80  52
 max  132 MPa
 min  OA  OC  BC
 80  52
 min  28 MPa
7 - 21
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 7.2
• Stress components after rotation by 30o
Points X’ and Y’ on Mohr’s circle that
correspond to stress components on the
rotated element are obtained by rotating
XY counterclockwise through 2  60
  180   60  67.4  52.6
 x  OK  OC  KC  80  52 cos 52.6
 y  OL  OC  CL  80  52 cos 52.6
 xy  KX   52 sin 52.6
 x  48.4 MPa
 y  111 .6 MPa
 xy  41.3 MPa
7 - 22
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
General State of Stress
• Consider the general 3D state of stress at a point and
the transformation of stress from element rotation
• State of stress at Q defined by:  x , y , z , xy , yz , zx
• Consider tetrahedron with face perpendicular to the
line QN with direction cosines: x ,  y , z
• The requirement  Fn  0 leads to,
 n A   x Ax x   xy Ax  y   xz Ax z   yx A y x   y A y  y
  yz A y z   zx Az x   zy Az  y   z Az z  0
 n   x 2x   y 2y   z 2z  2 xy x  y  2 yz  y z  2 zxz x
• Form of equation guarantees that an element
orientation can be found such that
 n   a2a   bb2   cc2
These are the principal axes and principal planes
and the normal stresses are the principal stresses.
7 - 23
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Application of Mohr’s Circle to the ThreeDimensional Analysis of Stress
• Transformation of stress for an element
rotated around a principal axis may be
represented by Mohr’s circle.
• The three circles represent the
normal and shearing stresses for
rotation around each principal axis.
• Points A, B, and C represent the
• Radius of the largest circle yields the
principal stresses on the principal planes
maximum shearing stress.
1
(shearing stress is zero)
 max   max   min
2
7 - 24
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Application of Mohr’s Circle to the ThreeDimensional Analysis of Stress
• In the case of plane stress, the axis
perpendicular to the plane of stress is a
principal axis (shearing stress equal zero).
• If the points A and B (representing the
principal planes) are on opposite sides of
the origin, then
a) the corresponding principal stresses
are the maximum and minimum
normal stresses for the element
b) the maximum shearing stress for the
element is equal to the maximum “inplane” shearing stress
c) planes of maximum shearing stress
are at 45o to the principal planes.
7 - 25
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Application of Mohr’s Circle to the ThreeDimensional Analysis of Stress
• If A and B are on the same side of the
origin (i.e., have the same sign), then
a) the circle defining max, min, and
max for the element is not the circle
corresponding to transformations within
the plane of stress
b) maximum shearing stress for the
element is equal to half of the
maximum stress
c) planes of maximum shearing stress are
at 45 degrees to the plane of stress
7 - 26
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Concept Application 7.3
SOLUTION:
• Construct Mohr’s circle for the
transformation of stress in the xy plane
For the state of plane stress shown,
determine (a) the three principal
planes and principal stresses and (b)
the maximum shearing stress.
 ave 
R
 x  y
2

6  3.5
 4.75 ksi
2
CF 2  FX 2

1.252  32
 3.25 ksi
7 - 27
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Since the faces of the element
perpendicular to the z axis are free of
stress, they define one of the principal
planes,
and
the
corresponding
principal stress is σz = 0.
• Principle planes
FX
3

CF 1.25
2 p  67.4  p  33.7 Clockwise
tan 2 p 
The principal stresses in the plane of
stress are
 a  OA  OC  CA  4.75  3.25  8.00 ksi
 b  OB  OC  BC  4.75  3.25  1.50 ksi
7 - 28
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
• Maximum Shearing Stress
Now draw the circles of diameter OB
and OA that correspond to rotations
of the element about the a and b axes.
 max  12  a  12 8.00 ksi   4.00 ksi
(This is out-of-plane shear stress.)
7 - 29
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Yield Criteria for Ductile Materials Under Plane Stress
• Failure of a machine component
subjected to uniaxial stress is directly
predicted from an equivalent tensile test
• Failure of a machine component
subjected to plane stress cannot be
directly predicted from the uniaxial state
of stress in a tensile test specimen
• It is convenient to determine the
principal stresses and to base the failure
criteria on the corresponding biaxial
stress state
• Failure criteria are based on the
mechanism of failure. Allows
comparison of the failure conditions for
a uniaxial stress test and biaxial
component loading
7 - 30
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Yield Criteria for Ductile Materials Under Plane Stress
Maximum shearing stress criteria (Tresca
criterion):
Structural component is safe as long as the
maximum shearing stress is less than the
maximum shearing stress in a tensile test
specimen at yield, i.e.,
 max   Y 
Y
2
For a and b with the same sign,
 max 
a
2
or
b

 Y
2
2
For a and b with opposite signs,
 max 
a b
2

 Y
2
7 - 31
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Yield Criteria for Ductile Materials Under Plane Stress
Maximum distortion energy criteria (von Mises
criterion):
Structural component is safe as long as the
distortion energy per unit volume is less than
that occurring in a tensile test specimen at yield.
ud  uY



1
1
 a2   a b   b2 
 Y2   Y  0  02
6G
6G
 a2   a b   b2   Y2

7 - 32
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Fracture Criteria for Brittle Materials Under Plane Stress
Brittle materials fail suddenly through rupture
or fracture in a tensile test. The failure
condition is characterized by the ultimate
strength U.
Maximum normal stress criteria
(Coulomb’s criterion):
Structural component is safe as long as the
maximum normal stress is less than the
ultimate strength of a tensile test specimen.
 a  U
 b  U
This criterion is based on the assumption that the ultimate strength of the
material is the same in tension and in compression, which is a shortcoming
of this theory.
7 - 33
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Fracture Criteria for Brittle Materials Under Plane Stress
Mohr’s Criterion:
σb
σa
σa
σb
The Mohr theory of failure is used to predict
the fracture of a material having different
properties in tension and compression when
results of various types of tests are available
for that material.
If both principal stresses are positive, the
state of stress is safe as long as
 a   UT and  b   UT
If both principal stresses are negative, the
state of stress is safe as long as
 a   UC and  b   UC
7 - 34
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Fracture Criteria for Brittle Materials Under Plane Stress
For σa and σb having opposite signs, (torsion test), a circle centered at O is
drawn representing the state of stress corresponding to the failure of the
torsion-test specimen
According to Mohr’s criterion, a state of
stress is safe if it is represented by a circle
located entirely within the area bounded by
the envelope of the circles corresponding to
the available data.
7 - 35
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 7.4
The state of plane stress shown occurs at a critical point of a steel machine
component. As a result of several tensile tests, the tensile yield strength is σY =
250 MPa for the grade of steel used. Determine the factor of safety with
respect to yield using (a) the maximum-shearing stress criterion, (b) the
maximum-distortion-energy criterion.
7 - 36
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 7.4
SOLUTION:
• Firstly, calculate principal stresses by using Mohr’s circle or by formulas
 a ,b 
 x  y
2
 20 
  x  y 
   xy2
 
 2 
2
602  252
 a  85 MPa ,  b  45 MPa
a) Maximum-Shearing-Stress Criterion
Sincea and b have opposite signs,
 a b 
Y
F .S
250
85  45 
F .S
F .S  1.92
7 - 37
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
b) Maximum-Distortion-Energy Criterion.
2
 
 a2   a b   b2   Y 
 F .S 
852  85 45  452   250 
 F .S 
F .S  2.19
2
REFLECT and THINK
Point H denotes coordinates σa = 85 MPa
and σb = -45 MPa.
OT
 1.92
OH
OM
(b) F .S 
 2.19
OH
(a) F .S 
7 - 38
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Stresses in Thin-Walled Pressure Vessels
• Cylindrical vessel with principal stresses
1 = hoop stress
2 = longitudinal stress
• Hoop stress:
 Fz  0  12t x   p2r x 
1 
pr
t
• Longitudinal stress:
 
2
 Fx  0   2 2 rt   p  r
pr
2 
2t
 1  2 2
7 - 39
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Stresses in Thin-Walled Pressure Vessels
• Points A and B correspond to hoop stress, 1,
and longitudinal stress, 2
• Maximum in-plane shearing stress:
1
2
 max( in plane )   2 
pr
4t
• Maximum out-of-plane shearing stress
corresponds to a 45o rotation of the plane
stress element around a longitudinal axis
 max   2 
pr
2t
7 - 40
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Stresses in Thin-Walled Pressure Vessels
• Spherical pressure vessel:
1   2 
pr
2t
• Mohr’s circle for in-plane
transformations reduces to a point
   1   2  constant
 max(in -plane)  0
• Maximum out-of-plane shearing
stress
 max  12 1 
pr
4t
7 - 41
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 7.5
A compressed-air tank is supported by two cradles as shown. One of the
cradles is designed so that it does not exert any longitudinal force on the
tank. The cylindrical body of the tank has a 30-in. outer diameter and is made
of a 3/8-in. steel plate by butt welding along a helix that forms an angle of
25º with a transverse plane. The end caps are spherical and have a uniform
wall thickness of 5/16 in. For an internal gage pressure of 180 psi, determine
(a) the normal stress and the maximum shearing stress in the spherical caps,
(b) the stresses in directions perpendicular and parallel to the helical weld.
7 - 42
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 7.5
SOLUTION:
a.
Spherical Cap
p  180 psi, t 
5
in.  0.3125in., r  15  0.3125  14.688 in.
16
The state of stress within any point in the spherical section is
pr 180 psi 14.688 in.

2t
20.3125in.
  4230 psi
1   2 
We know that in-plane shearing stresses are
zero in spherical bodies. The maximum
shearing stress is equal to the diameter (CD́)
of the outer circle.
1
 max  4230  2115 psi
2
7 - 43
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 7.5
b. Cylindrical Body of the Tank
p  180 psi, t 
3
in.  0.375 in., r  15  0.375  14.625 in.
8
We determine the hoop stress σ1 and the longitudinal
stress σ2.
pr 180 psi 14.625 in.

 7020 psi
t
0.375 in.
1
 2   1  3510 psi
2
1 
Both the hoop stress and the
longitudinal stress are principal stresses,
we draw Mohr’s circle as shown
1
 1   2   5265 psi
2
1
R   1   2   1755 psi
2
 ave 
7 - 44
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 7.5
An element having a face parallel to the weld is obtained by rotating the
face perpendicular to the axis Ob counterclockwise through 25º.
Therefore, on Mohr’s circle, point X́
corresponds to the stress components on the
weld by rotating radius CB counterclockwise
through 2θ = 50º.
 w   ave  R cos50  5265 1755cos50  4140 psi
 w  R sin 50  1755sin 50  1344 psi
7 - 45