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J. Liu CE270 Spring 2010
CE270 Spring 2010
Stress Transformations
Often interested in maximum normal and
shear stresses at a point and orientation on
the element
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Plane Stress
Stress can be analyzed in a single plane (no
load on surface)
Typically on thin sections, such as a beamweb
x
y
z
x
y
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x
x
y
y
xy
State of planestress for elementis unique to
orientation
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How do we transform the
stress components when we
change orientation?
x
xy
Use equilibrium of
FORCES!
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First, Sign Convention
Positive (+) if pointing in (+) direction on (+)
face
Positive (+) if pointing in (-) direction on (-)face
x
y
OR normal stress
(outward)
shear stress (up on
right-hand side)
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x
xy
A x face
Asin
Acos
y
x
xy
yx
)cos)(cos(
)cos)(sin()sin)(sin()sin)(cos(0
''
A
AAAAF
x
xyy
xyxx
4/23/10 Fix! (had them backwards on
Wednesday)
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cossin22sin
sincos2cos
)cossin2(sincos
22
22' xyyxx
Repeat for other stresses
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Stress Transformations
2sin2cos22
2cos2sin2
2sin2cos22
'
''
'
xyyxyx
y
xyyx
yx
xyyxyx
x
Hibbeler Chp. 9
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Example
x
z
A
L
T=294.5 N-m
wood grainGiven:
Wooden shaft
dia. = 100 mm
Find:
stresses parallel
and perpendicular
to the grain at
point A ( to z
axis)A
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x
z
A
L
T=294.5 N-m
wood grain
JTcmax
4)05.0(
2
)05.0)(5.294(
m
mNm
Pax 6105.1
BUT! (+) on (+) face, therefore (+) shear stress
x
y
max
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x
z
A
L
T=294.5 N-m
wood grain
x
y
max
72
72
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x
z
A
L
T=294.5 N-m
wood grain
72
Perpendicular to grain (could cause splitting)
2sin2cos
22
' xyyxyx
y
MPaPaxy 88.0))72(2(sin)105.1( 6'
Compression perpendicular to grain
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x
z
A
L
T=294.5 N-m
wood grain
72
Parallel to grain (could shear along grain?)
2cos2sin
2
'' xyyx
yx
MPaPaxyx 2.1144cos)105.1( 6
''
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Parallel to grain (pulling on grain)
2sin2cos22
' xyyxyx
x
MPax 88.0'
0.88 MPa
0.88 MPa
1.2 MPa
Final
State ofStress
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Principal Stresses
Maximum and minimum normal stresses
acting at a point
Can differentiate with respect to and set=0,
etc., or can use Mohrs Circle
Equations for plane stress transformation can
be put into a graphical format
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Mohrs Circle
22
22''
2'
2
2
xyyx
yxavg
yxavgx
R
R
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Sign Convention Mohrs Circle
positive to the right
clockwise above,counterclockwise below
In the kitchen, the clock is above, and the counter is below.
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General Procedure
1. Plot C = avg
2. Plot (x
,xy
), (y
,yx
)
3. Determine R
4. Draw circle
C
( x, xy)
( y, yx)x
y
avg
Reference state
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on element = 2 on Mohrs Circle
( x, xy)
( y, yx)
x
y
( y, yx)
( x, xy)
x
y
90
180
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Principal Stresses
( x, xy)
( y, yx)
p from
reference
state
12
p
R
R
avg
avg
2
1
1
2
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Maximum Shear Stress
Ryx max''
( x, xy)
( y, yx)
s from
referencestate
xymax
s
45
9022
ps
ps
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Example
x
z
A
L
T=294.5 N-m
wood grain
Given:
Wooden shaft
dia. = 100 mm
Find:stresses parallel
and perpendicular
to the grain at
point A ( to zaxis) using Mohrs
Circle
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Reference
state
x
y
max =1.5MPa = R (0, yx)
(0, xy)
2 =144
144 -90 =54
( y, yx)
( x, xy)
MPaRx 88.054cos'
MPaRy 88.054cos'
MPaRyx 2.154sin''
C
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0.88 MPa
0.88 MPa
1.2 MPa
Final
State of
Stress
MPaRx 88.054cos'
MPaRy 88.054cos' MPaR
yx
2.154sin''