4 stress transformation 2015 bach khoa

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,
 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.

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

 x ,  y ,  xy and  z   zx   zy  0.


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
• 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 


4.3 TRANSFORMATION OF PLANE STRESS
θ is positive if the rotation is counter
clockwise from x to x’

• 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


4.3 TRANSFORMATION OF PLANE STRESS
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

4.3 TRANSFORMATION OF PLANE STRESS
Maximum Shearing Stresses
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


4.3 TRANSFORMATION OF PLANE STRESS


4.3 TRANSFORMATION OF PLANE STRESS
EXAMPLE 4.01

SOLUTION:

• Find the element orientation for the principal
stresses from
2 xy
tan 2 p 
 x  y
• Determine the principal stresses from
 max, min 

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.

x  y
2

2

 x  y 
2
   xy
 
2



• Calculate the maximum shearing stress with
2

 x  y 
2
   xy
 max  
2



x  y

 
2


4.3 TRANSFORMATION OF PLANE STRESS
EXAMPLE 4.01
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 

30 2  40 2

 max  70 MPa
 min  30 MPa

2

 x  y 
2
   xy
 

2




4.3 TRANSFORMATION OF PLANE STRESS
EXAMPLE 4.01
• 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

   20 MPa

2


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.
• 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.


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:

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

4.4 MORH’S CIRCLE FOR PLANE STRESS
EXAMPLE 4.02

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


4.4 MORH’S CIRCLE FOR PLANE STRESS
EXAMPLE 4.02

• Principal planes and stresses
 max  OA  OC  CA  20  50

 max  70 MPa

 max  OB  OC  BC  20  50
 max  30 MPa
FX 40

CP 30
2 p  53.1

tan 2 p 

 p  26.6


4.4 MORH’S CIRCLE FOR PLANE STRESS
EXAMPLE 4.02

• Maximum shear stress
 s   p  45

 max  R

    ave

 s  71.6

 max  50 MPa

   20 MPa


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 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


4.4 MORH’S CIRCLE FOR PLANE STRESS
EXAMPLE 4.03

• 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

 max  OA  OC  BC
 80  52

 min  28 MPa


4.4 MORH’S CIRCLE FOR PLANE STRESS
EXAMPLE 4.03

• 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


4.4 MORH’S CIRCLE FOR PLANE STRESS