7 stress transformations lecture thursday
DESCRIPTION
Mohr's circle and stress transformationTRANSCRIPT
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7/18/2019 7 Stress Transformations Lecture Thursday
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7 - 1
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Mohrs Circle for Plane Stress
7 - 2
With the physical significance of Mohrs
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
plot the pointsXand Yand construct the
circle centered at C.
The principal stresses are obtained atAandB.
The direction of rotation of Oxto Oais
the same as CXto CA.
Shear angle always (angle +45O)
xyyx ,,
xy
yxyxave R
+
=
+=
yx
xyp
ave R
=
=
tan
minma!,
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Mohrs Circle for Plane Stress
7 - 9
With Mohrs circle uni"uely defined, the state
of stress at other a!es orientations may bedepicted.
For the state of stress at an angle with
respect to thexya!es, construct a new
diameterXYat an angle with respect toXY.
#ormal and shear stresses are obtained
from the coordinatesXY.
Observe direction of rotationObserve double of the element angle
Shear angle always (angle +45O
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Mohrs Circle for Plane Stress
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Mohrs circle for centric a!ial loading$
Mohrs circle for torsional loading$
%, === xyyxA
P
A
Pxyyx
===
J
!xyyx === % %=== xyyx
J
!
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Example 1
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For the state of plane stress shown,
&a' construct Mohrs circle, determine
&b' the principal planes, &c' the
principal stresses, &d' the ma!imumshearing stress and the corresponding
normal stress.
()*+T)#$
Construction of Mohrs circle( ) ( )
( ) ( ) M-a%/%0%
M-a/%M-a0%%%
M-a%
1%%
=+==
===
=+
=+
=
CXR
"XC"
yxave
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Example 1
7 - 12
-rincipal planes and stresses
%%ma! +=+== CAOCOAM-a2%ma! =
%%ma! === BCOCOB
M-a0%ma! =
=
==
1.0
0%
/%tan
p
pCP
"X
= 3.3p
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Example 1
7 - 13
Ma!imum shear stress (hear angle always &angle 4/)
+= /ps
= 3.21s
R=ma!
M-a%ma! =
ave =
M-a%=
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Sample Problem 7.2
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For the state of stress shown,
determine &a' the principal planes
and the principal stresses, &b' the
stress components e!erted on the
element obtained by rotating thegiven element counterclockwise
through 0% degrees.
()*+T)#$
Construct Mohrs circle
( ) ( ) ( ) ( ) M-a/5%
M-a5%
3%1%%
=+=+=
=+
=+
=
"XC"R
yxave
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Sample Problem 7.2
7 - 15
-rincipal planes and stresses
=
===
/.32
/.%
/5tan
p
pC"
X"
clockwise2.00 =p
5%
ma!
+=+== CAOCOA
5%
ma!
=== BCOCOA
M-a10ma! += M-a5min +=
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Sample Problem 7.2
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(tress components after rotation by 0%o
-ointsXand Yon Mohrs circle that
correspond to stress components on therotated element are obtained by
rotatingXYcounterclockwise through
==
+=+=====
==
3.sin
3.cos5%
3.cos5%
3./.323%15%
X#
C$OCO$
#COCO#
yx
y
x
= 3%
M-a0./1
M-a3.111
M-a/./5
=
+=+=
yx
y
x
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Yield Criteria for Ductile Materials nder Plane Stress
7 - 17
Failure of a machine component
sub6ected to unia!ial stress is directly
predicted from an e"uivalent tensile test
Failure of a machine component
sub6ected to plane stress cannot be
directly predicted from the unia!ial state
of stress in a tensile test specimen
t is convenient to determine the
principal stresses and to base the failure
criteria on the corresponding bia!ial
stress state
Failure criteria are based on the
mechanism of failure. 7llows
comparison of the failure conditions for
a unia!ial stress test and bia!ial
component loading
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Yield Criteria for Ductile Materials nder Plane Stress
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Ma!imum shearing stress criteria$
(tructural component is safe as long as thema!imum shearing stress is less than the
ma!imum shearing stress in a tensile test
specimen at yield, i.e.,
For aand %with the same sign,
For aand %with opposite signs,
ma!Y
Y
=