chapter 7 transformations of stress and strain
DESCRIPTION
Chapter 7 Transformations of Stress and Strain. 7.1 Introduction. 3 normal stresses. -- x , y , and z. General State of Stress. 3 shearing stresses -- xy , yz , and zx. Plane Stress condition. 2-D State of Stress. Plane Strain condition. - PowerPoint PPT PresentationTRANSCRIPT
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Chapter 7
Transformations of Stress and Strain
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7.1 Introduction
General State of Stress
3 normal stresses
3 shearing stresses -- xy, yz, and zx
-- x, y, and z
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2-D State of Stress
Plane Stress condition
Plane Strain condition
A. Plane Stress State:
B. Plane Stress State:
z = 0, yz = xz = yz = xz = 0z 0, xy 0
z = 0, yz = xz = yz = xz = 0z 0, xy 0
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Examples of Plane-Stress Condition:
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Thin-walled Vessels
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Max. x & y
Max. xy
(Principal stresses)
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7.2 Transformation of Plane Stress
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0
0
' ': ( cos )cos ( cos )sin
( sin )sin ( sin )cos
x xyx x
y xy
F A A A
A A
0
0
' ' ': ( cos )sin ( cos )cos
( sin )cos ( sin )sin
x xyy x y
y xy
F A A A
A A
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2 2 2' cos sin sin cosx y xyx
2 2' ' ( )sin cos (cos sin )x y xyx y
2 22 2 2sin sin cos , cos cos sin
After rearrangement:
(7.1)
(7.2)
2 21 2 1 22 2
cos coscos , sin
Knowing
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2 22 2
' cos sinx y x yxyx
2 22
' ' sin cosx yxyx y
2 22 2
' cos sinx y x yxyy
Eqs. (7.1) and (7.2) can be simplified as:
(7.5)
(7.6)
'y Can be obtained by replacing with ( + 90o) in Eq. (7.5)
(7.7)
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1. max and min occur at = 0
2. max and min are 90o apart. max and min are 90o apart.
3. max and min occur half way between max and min
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7.3 Principal Stresses: Maximum Shearing Stress
Since max and min occur at x’y’ = 0, one can set Eq. (7.6) = 0
2 2 02
' ' sin cosx yxyx y
22tan xy
x y
1 22 2
22
4/
( ) /cos
( ) /
x y
x y xy
(7.6)
It follows,
Hence, 1 22 22
4/sin
( ) /
xy
x y xy
(a)
(b)
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Substituting Eqs. (a) and (b) into Eq. (7.5) results in max and min :
2 2
2 2max, min ( )x y x yxy
2 x y
ave
2
2( )x y
xyR
This is a formula of a circle with the center at:
and the radius of the circle as:
(7.14)
(7.10)
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Mohr’s Circle
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The max can be obtained from the Mohr’s circle:
Since max is the radius of the Mohr’s circle,
2
2max ( )x yxyR
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Since max occurs at 2 = 90o CCW from max,
Hence, in the physical plane max is = 45o CCW from max.
In the Mohr’s circle, all angles have been doubled.
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7.4 Mohr’s Circle for Plane Stress
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Sign conventions for shear stresses:
CW shear stress = and is plotted above the -axis,
CCW shear stress = ⊝ and is plotted below the -axis
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7.5 General State of Stress – 3-D cases
Definition of Direction Cosines:
cos , cos , cosx x y y z z
with2 2 2 1x y z
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0
0
: ( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
n n x x x xy x y xz x z
yx y z y y y yz y z
zx z x zy z y z z z
F A A A A
A A A
A A A
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Dividing through by A and solving for n, we have
2 2 2 2 2 2n x x y y z z xy x y yz y z zx z x
2 2 2 n a a b b c c
(7.20)
We can select the coordinate axes such that the RHS of Eq. *7.20) contains only the squares of the ’s.
(7.21)
Since shear stress ij = o, a, b, and c are the three principal stresses.
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7.6 Application of Mohr’s Circle to the 3-D Analysis of Stress
A > B > C
1 12 2max max min A C = radius of the Mohr’s circle
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7.9 Stresses in Thin-Walled Pressure Vessels
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1
prhoop stress
t
0 2 2 0: ( ) ( )z lF t x p r x
(7.30)
Hoop Stress 1
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Longitudinal Stress 2
220 2 0: ( ) ( )xF rt p r
2 2prt
Solving for 2 (7.31)
Hence 1 22
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Using the Mohr’s circle to solve for max
2
12 4max( )in plane
prt
2 2max( )out of plane
prt
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1 2
1 2 2
pr
t
max 1
1
2 4
pr
t
max 1
1( ) (1 )
2 2 t
r
prp
t