Giáo trình Strength of Materials - Chapter 4: Stress transformation

CHAPTER 4: STRESS TRANSFORMATION 
4.1 Introduction 
4.2 Plane Stress State 
4.3 Transformation of plane stress 
4.4 Morh’s circle for plane stress 
4.5 Hooke’s Laws 
4.6 Transformation of plane strain 
4.7 Morh’s circle for plane strain 
4.1 INTRODUCTION 
Stress state at a point is the set of all stresses acting on all faces passing through 
this point. 
• The most general state of stress at a point may be 
represented by 6 components, 
),, :(Note
stresses shearing,,
stresses normal,,
xzzxzyyzyxxy
zxyzxy
zyx




• 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. 
4.1 INTRODUCTION 
SIGN CONVENTION: 
Normal stress: 
 Tension is positive 
 Compression is Negative 
Shear stress: two subscripts 
 + First subscript denotes the face on which 
the stress acts 
 + Second gives the direction on the stress 
vector 
Positive face (+): normal axis follows the 
positive direction of the original axis 
Negative face (-): normal axis follows the 
negative direction of the original axis 
4.1 INTRODUCTION 
SIGN CONVENTION: 
Positive direction (+): stress vector follows 
positive direction of the axis 
Negative direction (-): stress vector follows 
negative direction of the axis 
positive direction - positive face = positive stress 
negative direction-negative face = positive stress 
positive direction-negative face = negative stress 
negative direction-negative face = negative stress 
4.2 PLANE STRESS STATE 
• 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 
.0,, and xy  zyzxzyx 
4.2 PLANE STRESS STATE 
• 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. 
4.2 PLANE STRESS STATE 
• State of plane stress occurs in a thin plate subjected to forces acting in the 
midplane of the plate. 
4.3 TRANSFORMATION OF PLANE STRESS 
   
   
   
    



sinsincossin
coscossincos0
cossinsinsin
sincoscoscos0
AA
AAAF
AA
AAAF
xyy
xyxyxy
xyy
xyxxx






• Consider the conditions for 
equilibrium of a prismatic element 
with faces perpendicular to the x, 
y, and x’ axes. 
4.3 TRANSFORMATION OF PLANE STRESS 
cos 2 sin 2
2 2
cos 2 sin 2
2 2
sin 2 cos 2
2
x y x y
x xy
x y x y
y xy
x y
x y xy
   
   
   
   
 
   


 
 
  
 
  

  
• The equations may be rewritten to yield 
θ is positive if the rotation is counter 
clockwise from x to x’ 
4.3 TRANSFORMATION OF PLANE STRESS 
• The previous equations are combined to 
yield parametric equations for a circle, 
 
2
2
222
22
where
xy
yxyx
ave
yxavex
R
R









 



 
• Principal stresses occur on the principal 
planes of stress with zero shearing stresses. 
o
2
2
minmax,
90by separated angles twodefines :Note
2
2tan
22
yx
xy
p
xy
yxyx













 



Principal Stresses 
4.3 TRANSFORMATION OF PLANE STRESS 
Maximum Shearing Stresses 
2
45by fromoffset 
and 90by separated angles twodefines :Note
2
2tan
2
o
o
2
2
max
yx
ave
p
xy
yx
s
xy
yx
R


















 

4.3 TRANSFORMATION OF PLANE STRESS 
4.3 TRANSFORMATION OF PLANE STRESS 
For the state of plane stress shown, 
determine (a) the principal panes, 
(b) the principal stresses, (c) the 
maximum shearing stress and the 
corresponding normal stress. 
SOLUTION: 
• Find the element orientation for the principal 
stresses from 
yx
xy
p





2
2tan
• Determine the principal stresses from 
2
2
minmax,
22
xy
yxyx


 




 



• Calculate the maximum shearing stress with 
2
2
max
2
xy
yx


 




 

2
yx 



EXAMPLE 4.01 
4.3 TRANSFORMATION OF PLANE STRESS 
EXAMPLE 4.01 
SOLUTION: 
• Find the element orientation for the principal 
stresses from 
 
 







1.233,1.532
333.1
1050
4022
2tan
p
yx
xy
p




 6.116,6.26p
• Determine the principal stresses from 
   22
2
2
minmax,
403020
22






 


 xy
yxyx



MPa30
MPa70
min
max




MPa10
MPa40MPa50


x
xyx


4.3 TRANSFORMATION OF PLANE STRESS 
EXAMPLE 4.01 
MPa10
MPa40MPa50


x
xyx


2
1050
2




yx
ave


• The corresponding normal stress is 
MPa20
• Calculate the maximum shearing stress with 
   22
2
2
max
4030
2






 
 xy
yx



MPa50max 
45 ps 
 6.71,4.18s
4.4 MORH’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. 
• The principal stresses are obtained at A and B. 
yx
xy
p
ave R







2
2tan
minmax,
 The direction of rotation of Ox to Oa is 
the same as CX to CA. 
• For a known state of plane stress 
plot the points X and Y and construct the 
circle centered at C. 
xyyx  ,,
2
2
22
xy
yxyx
ave R 

 




 



4.4 MORH’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’. 
4.4 MORH’S CIRCLE FOR PLANE STRESS 
• Mohr’s circle for centric axial loading: 
0,  xyyx
A
P

A
P
xyyx
2
 
• Mohr’s circle for torsional loading: 
J
Tc
xyyx   0 0 xyyx
J
Tc

4.4 MORH’S CIRCLE FOR PLANE STRESS 
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 
   
    MPa504030
MPa40MPa302050
MPa20
2
1050
2
22 






CXR
FXCF
yx
ave


EXAMPLE 4.02 
4.4 MORH’S CIRCLE FOR PLANE STRESS 
EXAMPLE 4.02 
• Principal planes and stresses 
5020max  CAOCOA
MPa70max 
5020max  BCOCOB
MPa30max 


1.532
30
40
2tan
p
p
CP
FX


 6.26p
4.4 MORH’S CIRCLE FOR PLANE STRESS 
EXAMPLE 4.02 
• Maximum shear stress 
 45ps 
 6.71s
Rmax
MPa 50max 
ave 
MPa 20 
4.4 MORH’S CIRCLE FOR PLANE STRESS 
EXAMPLE 4.03 
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 
given element counterclockwise 
through 30 degrees. 
SOLUTION: 
• Construct Mohr’s circle 
        MPa524820
MPa80
2
60100
2
2222 





FXCFR
yx
ave


4.4 MORH’S CIRCLE FOR PLANE STRESS 
EXAMPLE 4.03 
• Principal planes and stresses 


4.672
4.2
20
48
2tan
p
p
CF
XF


clockwise7.33 p
5280
max

 CAOCOA
5280
max

 BCOCOA
MPa132max  MPa28min 
4.4 MORH’S CIRCLE FOR PLANE STRESS 
EXAMPLE 4.03 







6.52sin52
6.52cos5280
6.52cos5280
6.524.6760180
XK
CLOCOL
KCOCOK
yx
y
x




• 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  602
MPa3.41
MPa6.111
MPa4.48






yx
y
x



4.4 MORH’S CIRCLE FOR PLANE STRESS 
4.4 MORH’S CIRCLE FOR PLANE STRESS 
4.4 MORH’S CIRCLE FOR PLANE 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 
principal stresses on the principal planes 
(shearing stress is zero) 
minmaxmax
2
1
 
• Radius of the largest circle yields the 
maximum shearing stress. 
Application of Morh’s circle to the Three-Dimensional Analysis of Stress 
4.4 MORH’S CIRCLE FOR PLANE STRESS 
Application of Morh’s circle to the Three-Dimensional 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). 
c) planes of maximum shearing stress 
are at 45o to the principal planes. 
b) the maximum shearing stress for the 
element is equal to the maximum “in-
plane” shearing stress 
a) the corresponding principal stresses 
are the maximum and minimum 
normal stresses for the element 
• If the points A and B (representing the 
principal planes) are on opposite sides of 
the origin, then 
4.4 MORH’S CIRCLE FOR PLANE STRESS 
Application of Morh’s circle to the Three-Dimensional Analysis of Stress 
• If A and B are on the same side of the 
origin (i.e., have the same sign), then 
c) planes of maximum shearing stress are 
at 45 degrees to the plane of stress 
b) maximum shearing stress for the 
element is equal to half of the 
maximum stress 
a) the circle defining max, min, and 
max for the element is not the circle 
corresponding to transformations within 
the plane of stress 
4.5 HOOKE’S LAW 
• For an element subjected to multi-axial loading, 
the normal strain components resulting from the 
stress components may be determined from the 
principle of superposition. This requires: 
 1) strain is linearly related to stress 
2) deformations are small 
EEE
EEE
EEE
zyx
z
zyx
y
zyx
x









• With these restrictions: 
GENERALISED HOOKE’S LAW 
4.5 HOOKE’S LAW 
DILATATION: BULK MODULUS 
• Relative to the unstressed state, the change in volume is 
      
 
 e)unit volumper in volume (change dilatation 
21
111111






zyx
zyx
zyxzyx
E
e




• For element subjected to uniform hydrostatic pressure, 
 
 
modulusbulk 
213
213








E
k
k
p
E
pe
• Subjected to uniform pressure, dilatation must be 
negative, therefore 
2
10 
4.5 HOOKE’S LAW 
SHEARING STRAIN 
• A cubic element subjected to a shear stress will 
deform into a rhomboid. The corresponding shear 
strain is quantified in terms of the change in angle 
between the sides, 
 xyxy f  
• A plot of shear stress vs. shear strain is similar the 
previous plots of normal stress vs. normal strain 
except that the strength values are approximately 
half. For small strains, 
zxzxyzyzxyxy GGG  
where G is the modulus of rigidity or shear modulus. 
4.5 HOOKE’S LAW 
Relation Among E, , and G 
• An axially loaded slender bar will 
elongate in the axial direction and 
contract in the transverse directions. 
  1
2G
E
• Components of normal and shear strain are 
related, 
• If the cubic element is oriented as in the 
bottom figure, it will deform into a 
rhombus. Axial load also results in a shear 
strain. 
• An initially cubic element oriented as in 
top figure will deform into a rectangular 
parallelepiped. The axial load produces a 
normal strain. 
4.6 TRANSFORMATION FOR PLANE STRAIN 
• Plane strain - deformations of the material 
take place in parallel planes and are the 
same in each of those planes. 
• Example: Consider a long bar subjected 
to uniformly distributed transverse loads. 
State of plane stress exists in any 
transverse section not located too close to 
the ends of the bar. 
• Plane strain occurs in a plate subjected 
along its edges to a uniformly distributed 
load and restrained from expanding or 
contracting laterally by smooth, rigid and 
fixed supports 
 0
 :strain of components
x  zyzxzxyy 
4.6 TRANSFORMATION FOR PLANE STRAIN 
• State of strain at the point Q results in 
different strain components with respect 
to the xy and x’y’ reference frames. 
 
   
 yxOBxy
xyyxOB
xyyx






2
45
cossinsincos
2
1
22














2cos
2
2sin
22
2sin
2
2cos
22
2sin
2
2cos
22
xyyxyx
xyyxyx
y
xyyxyx
x
















• Applying the trigonometric relations 
used for the transformation of stress, 
4.7 MORH’S CIRCLE FOR PLANE STRAIN 
• The equations for the transformation of 
plane strain are of the same form as the 
equations for the transformation of plane 
stress - Mohr’s circle techniques apply. 
• Abscissa for the center C and radius R , 
22
222 











 



xyyxyx
ave R


• Principal axes of strain and principal strains, 
RR aveave
yx
xy
p







minmax
2tan
4.7 MORH’S CIRCLE FOR PLANE STRAIN 
Three-Dimensional Analysis of Strain 
• Previously demonstrated that three principal 
axes exist such that the perpendicular 
element faces are free of shearing stresses. 
• By Hooke’s Law, it follows that the 
shearing strains are zero as well and that 
the principal planes of stress are also the 
principal planes of strain. 
• Rotation about the principal axes may be 
represented by Mohr’s circles. 
4.7 MORH’S CIRCLE FOR PLANE STRAIN 
Three-Dimensional Analysis of Strain 
• For the case of plane strain where the x and y 
axes are in the plane of strain, 
- the z axis is also a principal axis 
- the corresponding principal normal strain 
is represented by the point Z = 0 or the 
origin. 
• If the points A and B lie on opposite sides 
of the origin, the maximum shearing strain 
is the maximum in-plane shearing strain, D 
and E. 
• If the points A and B lie on the same side of 
the origin, the maximum shearing strain is 
out of the plane of strain and is represented 
by the points D’ and E’. 
4.7 MORH’S CIRCLE FOR PLANE STRAIN 
Three-Dimensional Analysis of Strain 
• Consider the case of plane stress, 
0 zbyax 
• If B is located between A and C on the 
Mohr-circle diagram, the maximum 
shearing strain is equal to the diameter CA. 
• Strain perpendicular to the plane of stress 
is not zero. 
• Corresponding normal strains, 
   babac
ba
b
ba
a
E
EE
EE















1
4.7 MORH’S CIRCLE FOR PLANE STRAIN 
Three-Dimensional Analysis of Strain 
• Strain gages indicate normal strain through 
changes in resistance. 
 yxOBxy   2
• With a 45o rosette, x and y are measured 
directly. xy is obtained indirectly with, 
333
2
3
2
3
222
2
2
2
2
111
2
1
2
1
cossinsincos
cossinsincos
cossinsincos



xyyx
xyyx
xyyx



• Normal and shearing strains may be 
obtained from normal strains in any three 
directions,