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MECHANICS OF
MATERIALSFerdinand P. Beer
E. Russell Johnston, Jr.
John T. DeWolf
CHAPTER
7
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Introduction
The most general state of stress at a point may
be represented by 6 components,
),,:(Note
stressesshearing,,
stressesnormal,,
xzzxzyyzyxxy
zxyzxy
zyx
Same state of stress is represented by adifferent set of components if axes are rotated.
The first part of the chapter is concerned with
how the components of stress are transformed
under a rotation of the coordinate axes. The
second part of the chapter is devoted to a
similar analysis of the transformation of the
components of strain.
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Introduction
Plane Stress- state of stress in which two faces of
the cubic element are free of stress. For theillustrated example, the state of stress is defined by
.0,, andxy zyzxzyx
State of plane stress occurs in a thin plate subjected
to forces acting in the mid-plane of the plate.
MECHANICS OF MATERIA S
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Transformation of Plane Stress
sinsincossin
coscossincos0
cossinsinsin
sincoscoscos0
AA
AAAF
AA
AAAF
xyy
xyxyxy
xyy
xyxxx
Consider the conditions for equilibrium of a
prismatic element with faces perpendicular to
thex,y, and xaxes.
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Transformation of Plane Stress
MECHANICS OF MATERIALS
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Transformation of Plane Stress
MECHANICS OF MATERIALS
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Transformation of Plane Stress
MECHANICS OF MATERIALS
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Transformation of Plane Stress
MECHANICS OF MATERIALS
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Transformation of Plane Stress
MECHANICS OF MATERIALS
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Example 7.01
For the state of plane stress shown,
determine (a) the principal planes,
(b) the principal stresses, (c) the
maximum shearing stress and thecorresponding normal stress.
SOLUTION:
Find the element orientation for the principalstresses from
yx
xyp
2
2tan
Determine the principal stresses from
22
minmax,22 xy
yxyx
Calculate the maximum shearing stress with
2
2
max2
xyyx
2
yx
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Example 7.01
SOLUTION:
Find the element orientation for the principalstresses from
1.233,1.532
333.11050
40222tan
p
yx
xyp
6.116,6.26p
Determine the principal stresses from
22
22
minmax,
403020
22
xy
yxyx
MPa30
MPa70
min
max
MPa10
MPa40MPa50
x
xyx
MECHANICS OF MATERIALS
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Example 7.01
MPa10
MPa40MPa50
x
xyx
2
1050
2
yx
ave
The corresponding normal stress is
MPa20
Calculate the maximum shearing stress with
22
22
max
4030
2
xy
yx
MPa50max
45 ps
6.71,4.18s
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Mohrs Circle for Plane Stress
Mohr's circlea useful graphical technique for finding principal stresses and
strains in materials. Mohr's circle also tells you the principal angles
(orientations) of the principal stresses without your having to plug an angleinto stress transformation equations.
Starting with a stress or strain element in the XY plane, construct a grid with a
normal stress on the horizontal axis and a shear stress on the vertical. (Positive
shear stress plots at the bottom.) Then just follow these steps:
Plot the vertical face coordinates V(xx, xy). Plot the horizontal coordinates H(yy, xy).
You use the opposite sign of the shear stress from Step 1 because the
shear stresses on the horizontal faces are creating a couple that
balances (or acts in the opposite direction of) the shear stresses on the
vertical faces. Draw a diameter line connecting Points V (from Step 1) and H (from Step
2).
Sketch the circle around the diameter from Step 3.
The circle should pass through Points V and H as shown here.
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Mohrs Circle for Plane Stress
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Example 7.02
For the state of plane stress shown,
(a) construct Mohrs circle, determine
(b) the principal planes, (c) the
principal stresses, (d) the maximumshearing stress and the corresponding
normal stress.
SOLUTION:
Construction of Mohrs circle
MPa504030
MPa40MPa302050
MPa202
1050
2
22
CXR
FXCF
yxave
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Example 7.02
Principal planes and stresses
5020max CAOCOAMPa70max
5020min BCOCOB
MPa30min
1.532
30
402tan
p
pCP
FX
6.26p
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Example 7.02
Maximum shear stress
45ps
6.71s
Rmax
MPa50max
ave
MPa20
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MECHANICS OF MATERIALS Beer Johnston DeWolfPractice Problems
7 18
Practice Problems:
7.1-7.16 and 7.31-7.38
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