Stress Transformation

Chapter Objectives
 Navigate between rectilinear co-ordinate systems for

stress components
 Determine principal stresses and maximum in-plane
shear stress
Copyright © 2011 Pearson Education South Asia Pte Ltd

GENERAL EQUATIONS OF PLANE-STRESS
TRANSFORMATION
• The state of plane stress at a point is uniquely represented by
three components acting on an element that has a specific
orientation at the point.
• Sign Convention:
– Positive normal stress acts outward
from all faces
– Positive shear stress acts upwards
on the right-hand face of the element

TRANSFORMATION (cont)
• Sign convention (continued)
• Both the x-y and x’-y’ system follow the right-hand rule
• The orientation of an inclined plane (on which the normal and

shear stress components are to be determined) will be defined
using the angle θ. The angle θ is measured from the positive x to
the positive x’-axis. It is positive if it follows the curl of the right-
hand fingers.

• Normal and shear stress components:
– Consider the free-body diagram of the segment

+ΣFx’ = 0; σx’ ∆A – (τxy ∆A sin θ) cos θ – (σy ∆A sin θ) sin θ

– ( τxy ∆A cos θ) sin θ – (σx ∆A cos θ) cos θ = 0
σx’ = σx cos2 θ + σy sin2 θ + τxy (2 sin θ cos θ)
+ΣFy’ = 0; τx’y’ ∆A + (τxy ∆A sin θ) sin θ – (σy ∆A sin θ) cos θ

– ( τxy ∆A cos θ) cos θ + (σx ∆A cos θ) sin θ = 0
τx’y’ = (σy – σx) sin θ cos θ + τxy (cos2 θ – sin2 θ)
σx + σy σx – σy
σx’ = + cos 2θ + τxy sin 2 θ
2 2
σx + σy
τx’y’ = – sin 2θ + τxy cos 2 θ
2
σx + σy σx – σy
σy’ = – cos 2θ – τxy sin 2 θ
2 2

VARIABLE SOLUTIONS
Please click the appropriate icon for your computer to access the
variable solutions

IN-PLANE PRINCIPAL STRESS
• The principal stresses represent the maximum and
minimum normal stress at the point.
• When the state of stress is represented by the principal
stresses, no shear stress will act on the element.
d x '  x  y
 2 sin 2   2 xy cos 2
d 2
Solving this equation leads to θ = θp

 xy
tan 2 p 
 x   y / 2
2
 x  y   x  y 
 1, 2       xy2
2  2 

IN-PLANE PRINCIPAL STRESS (cont)
 xy
Solving this equation leads to θ = θp; i.e tan 2 p 
 x   y / 2
2
 x  y   x  y 
 1, 2       xy2
2  2 
MAXIMUM IN-PLANE PRINCIPAL STRESS
• The state of stress can also be represented in terms of the
maximum in-plane shear stress. In this case, an average
stress will also act on the element.
d x ' y '  x  y
 2 cos     xy 2 sin 2   0
d 2
   y / 2
Solving this equation leads to θ = θs; i.e tan 2 s
x
• 
 xy
2
  x  y 
 max in -plane      xy2
 2 
• And there is a normal stress on the

plane of maximum in-plane shear stress
 x  y
 avg 
2
Summary
 xy
• Principal plane orientations: tan 2 p 
 x   y / 2
• Principal stresses:
2
 x  y  x  y 
 1, 2       xy2
2  2 
• Plane orientation for maximum shear stress
tan 2 s 
 x   y / 2
 xy
• Maximum Shear stress
2
  x  y 
 max in -plane      xy2
 2 
Example
SOLUTION:
Find the element orientation for the
principal stresses from
2 xy
tan 2 p 
 x  y
Determine the principal stresses from

Plane stress element. 2
x  y  x  y  2
 max,min       xy
2  2 
For the state of plane stress
shown, determine (a) the
Calculate the maximum shearing stress
principal planes, (b) the
with  x  y 
2
principal stresses, (c) the  max   2
   xy
maximum shearing stress and  2 
the corresponding normal stress. x  y
 
2
Example
SOLUTION:
Find the element orientation for the
principal stresses from
2 xy 2 40 
tan 2 p    1.333
 x  y 50   10 
2 p  53.1, 233.1
Plane stress element.
 p  26.6, 116 .6
 x  50 MPa  xy  40 MPa
 x  10 MPa Determine the principal stresses from
2
x  y  x  y  2
 max,min       xy
2  2 
 20  302  402
 max  70 MPa
 min  30 MPa
Plane stress element oriented in
7 - 13
principal directions.
Example
Calculate the maximum shearing stress
with 2
 x  y  2
 max      xy
 2 
 302  402
Plane stress element.
 max  50 MPa
 x  50 MPa  xy  40 MPa  s   p  45
 x  10 MPa  s  18.4, 71.6
The corresponding normal stress is

 x   y 50  10
    ave  
2 2
   20 MPa
Plane stress element showing

7 - 14
maximum shear orientation.
MOHR’S CIRCLE OF PLANE STRESS
• A geometrical representation of equations 9.1 and 9.2;
i.e.
  x  y    x  y 
 x '       cos 2   xy sin 2
 2   2 
  x  y 
 x ' y '    sin 2   xy sin 2
 2 
• Sign Convention:
σ is positive to the right, and τ is positive downward.

EXAMPLE 7
The state of plane stress at a point is shown on the element
in Fig. 9–20a. Represent this state of stress on an element
oriented 30°counterclockwise from the position shown.

EXAMPLE 7 (cont)
Solutions
Construction of the Circle
We first construct of the circle,
 x  8,  y  12 and  xy  6
The center of the circle C is on the axis at

 8  12
 avg   2 MPa
2
From point C and the A(-8, -6) are plotted, we have
R  10 2  6 2  11 .66

EXAMPLE 7 (cont)
Solutions
Stresses on 30° Element
From the geometry of the circle,

6
  tan1
 30.96   60  30.96  29.04
10
 x '  2  11.66 cos 29.04  8.20 MPa (Ans)
 x ' y '  11.66 cos 29.04  5.66 MPa (Ans)
The stress components acting on the adjacent face DE of the element,

which is 60° clockwise from the positive x axis, Fig. 9–20c, are represented
by the coordinates of point Q on the circle.
 x '  2  11 .66 cos 29.04  12.22 MPa (Ans)

 x'y'  11 .66 sin 29.04  5.66 MPa (check) (Ans)

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Chapter Objectives

 Navigate between rectilinear co-ordinate systems for

Copyright © 2011 Pearson Education South Asia Pte Ltd

Copyright © 2011 Pearson Education South Asia Pte Ltd

• The orientation of an inclined plane (on which the normal and

Copyright © 2011 Pearson Education South Asia Pte Ltd

Copyright © 2011 Pearson Education South Asia Pte Ltd

+ΣFx’ = 0; σx’ ∆A – (τxy ∆A sin θ) cos θ – (σy ∆A sin θ) sin θ

+ΣFy’ = 0; τx’y’ ∆A + (τxy ∆A sin θ) sin θ – (σy ∆A sin θ) cos θ

Copyright © 2011 Pearson Education South Asia Pte Ltd

Copyright © 2011 Pearson Education South Asia Pte Ltd

Solving this equation leads to θ = θp

Copyright © 2011 Pearson Education South Asia Pte Ltd

• And there is a normal stress on the

Determine the principal stresses from

The corresponding normal stress is

Plane stress element showing

Copyright © 2011 Pearson Education South Asia Pte Ltd

Copyright © 2011 Pearson Education South Asia Pte Ltd

We first construct of the circle,

The center of the circle C is on the axis at

From point C and the A(-8, -6) are plotted, we have

Copyright © 2011 Pearson Education South Asia Pte Ltd

From the geometry of the circle,

The stress components acting on the adjacent face DE of the element,

 x '  2  11 .66 cos 29.04  12.22 MPa (Ans)

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