Transcript shear stress

APPLICATIONS/
MOHR’S CIRCLE
1
Special Forms of the Stress Tensor
• Uniaxial stress is given by:

0

0
Fig. 1. Examples of special
state of stress: (a) Uniaxial; (b)
biaxial; (c) hydrostatic; (d)
pure shear
0
0
0
0

0

0 
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• Biaxial Stress:
 1 0 0
0  0 
2


0 0 0 
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• Hydrostatic Pressure:
0 0
 p
 0  p 0


0  p 
 0
• It is a occurs in liquids; it is a
special case of triaxial stress,
when the three principal stresses
are equal.
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• Pure Stress:
0



0

0
0
0
0

0

• It is a special case of biaxial
stress, as will be seen by
performing a 45o rotation.
• Students are required to show
that the above statement is
correct.
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• Applying the transformation law, we have:
old
new
l11 =
2;
2
From Eq. 4-14
l11
l21
l12
l22
2;
2
2
l21 = 2
l12 =
l22 = 2 ;
2
 11 = 2l11l1212 = 12
 22 = 2l21l2212 = -12
Hence, we have
0  
 0


=
 0 
0   


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• Fig. 2. A state of pure shear stress and strain in an isotropic
material.
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Important Stresses in Plasticity
• It is often useful to decompose ij into two components:
 ij   ij'   ij ij''
Deviatoric, or Deviator,
or Shear stress Tensor
Spherical, or Hydrostatic,
or Dilatational stress Tensor
 12
 13
( 11   m )



 12 ( 11   m )
 23


 12
 12
( 11   m )

[Responsible for Failure]
(4-33)
0 
 m 0
0
m 0 


0
 m 
0
[Does Not cause Deformation]
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• It can be found that m has to be equal to:
m 
 11   22   33
3
I1

3
(4-34)
• The deviatoric (or shear) components of stress are
responsible for failure (under shear) or distortion.
• The hydrostatic (or spherical) stress produces volume
change and does not cause any plastic deformation.
This explains why shrimp can live in great ocean depths
without problems.
• The above stress components are often used in plasticity.
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• Consider a shrimp under two conditions:
(a) depth of 100 m, with hydrostatic pressure of 1.0 MPa
acting on it, and
(b) squeezed between our fingers, with an applied stress
(uniaxial) of about 0.5 MPa.
– While the Hydrostatic pressure will not bother the
shrimp, it will certainly crush under the applied tensile
stress.
– The difference between the two cases is the deviatoric
stress, which is 0 and 0.25 MPa for cases (a) and (b)
respectively.
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2-D TRANSFORMATION
• Recall the three equations obtained for the 2-D
transformation of stress:
 x   y  x  y
 x' 

cos 2   xy sin 2
2
2
 y' 
 x  y
 x' y ' 
2

 x  y
2
 x  y
2
cos 2   xy sin 2
sin 2   xy cos 2
(4-35a)
(4-35b)
(4-35c)
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• ***As in all Transformation cases, Eqs. 4-35 give the
variation of  and  with direction in the material***.
• The direction being specified by the angle  relative to
the originally chosen x-y coordinate system.
• Taking the derivative d/d of Eq. 4-35(a) and equating
the result to zero gives the coordinate axes rotations for
the maximum and minimum values of .
tan 2 n 
2 xy
 x  y
(4-36)
• Two angles n separated by 90o satisfies Eq. 3-36
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• The corresponding maximum and minimum normal
stresses from Eq. 4-35, called the principal normal
stresses can be obtained by substituting 4-36 into 4-35,
and is given as:
 1,  2 
x  y
2

   x   y 2

2





xy

 

2



(4-37)
• Note that the shear stress at the n orientation is found to
be zero.
• Conversely, if the shear stress is zero, then the normal
stresses are the principal normal stresses
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• Similarly, Eq. 4-35c and d/d = 0 gives the coordinate
axes rotation for the maximum shear stress.
 x  y
tan 2 s  
2 xy
(4-38)
• The corresponding shear stress from Eq. 4-35c is
 max 
   x   y 2

2





xy

 

2



(4-39)
• This is the maximum shear stress in the x-y plane and
is called the principal shear stress.
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• max can be expressed in terms of the principal normal
stresses 1 and 2 by substituting Eq. 4-39 into 4-37 to
obtain
1   2
(4-40)
 max 
2
• The absolute value is necessary for max due to the two
roots in Eq. 4-39
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Example
• At a point of interest on the free surface of an
engineering component, the stresses with respect to a
convenient coordinate system in the plane of the surface
are x = 95, y = 25, xy = 20 MPa. Determine the
principal stresses and the orientations of the principal
planes.
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Solution (A)
• Substitution of the given values into Eq. 4-36 gives the
angle to the coordinate axes of the principal normal
2 xy
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stresses.
tan 2 n 

 x  y
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n = 14.9o (CCW)
Substituting into Eq. 4-37 gives the principal normal
stresses
x  y
 x  y 2
2 
 1,  2 
  (
)   xy 
2

2

1, 2 = 60  40.3 = 100.3, 19.7 MPa
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Solution (B)
• Alternatively, a more rigorous procedure is to use  = n
= 14.9 in Eq. 4-35a, which gives
 = x’ = 1 = 100.3 MPa
• Use of  = n + 90o = 104.9o in Eq. 4-35a then gives the
normal stress in the other orthogonal direction.
 = y’ = 2 = 19.7 MPa
The zero value of  at  = n can also be verified by using
Eq. 4-35c.
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Mohr’s Circle
• A convenient graphical representation of the
transformation equations for plane stress.
• On  versus  coordinates, these equations can be shown
to represent a circle, called Mohr’s circle.
• The Development is as follows:
– Isolate all terms of Eq. 4-35a containing  on one side.
– Then square both sides of 4-35a and 4-35c and sum.
– Invoke simple trigonometric identities to eliminate , and
we obtain
x  y

  
2


 x  y
2
    
2


2
2

2
   xy

(4-41)
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• This equation is of the form:
  a 2    b 2  r 2
(4-42)
which is the equation of a circle on a plot of  versus 
with center at the coordinates (a, b) and radius r, where
2


x  y





x
y
2 
   xy
a
, b  0, r   


2
2



(4-43)
Comparison with Eq. 4-37 and 4-39, revealed that the
maximum and minimum normal stresses can be written
as:
x  y
(4-44)
 1,  2  a  r 
  max
2
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• Confusion concerning the signs in Mohr’s circle can be
resolved as follws:
– For shear, the portion that causes clockwise rotation is
considered positive, and the portion that causes counterclockwise rotation is considered negative
– For normal stresses, tension is positive, and compression
is negative.
• Graduate students are required to read up the 3-D
Mohr’s circle.
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Problem
• Repeat the previous problem using Mohr’s circle
method. Recall that the original state of stress is
x = 95, y = 25, and xy = 20 MPa
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Solution
• The circle is obtained by plotting two points that lie at
opposite ends of a diameter
 ,    x ,   xy   95,  20 MPa
 ,    x , xy   25, 20 MPa
– A negative sign is applied to xy for the point associated
with x because the shear arrows on the same planes as x
tend to cause a counter-clockwise rotation.
– Similarly, a positive sign is used for xy when associated
with y, due to the clockwise rotation.
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• Recall Eq. 4-43
x  y
a
2
   x   y 2

2 
   xy
, b  0, r   


2



• The center of the circle is (a, b), which lies on the 
axis:
 x   y 
 , ctr.  
, 0   60, 0  MPa

2

• The hypotenuse of the triangle/ radius of the circle is
also the principal shear stress, which is given as:
r   max 
   x   y 2

2

  352  202  40.3 MPa



xy
  2 



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• The angle with the  - axis is
tan 2 n
20

35
2 n  29.74o (CCW )
• A counter-clockwise rotation of the diameter of the
circle by this 2n gives the horizontal diameter that
corresponds to the principal normal stress.
• Their values are obtained from Eq. 4-44:
x  y
 1,  2  a  r 
  max
2
1, 2 = 100.3 and 19.7 MPa
25
, MPa
60
(60, 40.3)
(25, 20)
(100.3, 0)
(19.7, 0)
2n
, MPa
(95, -20)
(60, -40.3)
Mohr’s circle corresponding to the state of stress
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