teknik kekuatan bahan
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Transformations of Stress
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• The most general state of stress at a point may
be represented by 6 components,
),, :(Note
stressesshearing,,
stressesnormal,,
xz zx zy yz yx xy
zx yz xy
z y x
• 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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• 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,, and xy zy zx z y x
• State of plane stress also occurs on the free surface
of a structural element or machine component, i.e.,
at any point of the surface not subjected to an
external force.
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MPa
y F
MPa
x F
3.43
030sin000.2002310
0
0.75
030cos000.2002310
0
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Transformation of Plane Stress
• Consider the conditions for equilibrium of a
prismatic element with faces perpendicular to
the x, y, and x’ axes.
2cos2sin
2
2sin2cos22
2sin2cos22
xy
y x
y x
xy
y x y x
y
xy
y x y x
x
• The equations may be rewritten to yield
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sinsincossin
coscossincos0
cossinsinsinsincoscoscos0
A A
A A A F
A A A A A F
xy y
xy x y x y
xy y
xy x x x
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sinsincossincoscossincos0
cossinsinsin
sincoscoscos0
A A A A A F
A A
A A A F
xy y
xy x y x y
xy y
xy x x x
2
2
2
2
sincossin
cossincos0
cossinsin
sincoscos0
xy y
xy x y x y
xy y
xy x x x
F
F
2
2cos1
sin
cossin22sinand 2
2cos1cos
2
2
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2cos2
2sin
2cos2
2sin0
2sin2cos22
2sin2cos22
0
2
2sin
2
2cos1
2
2sin
2
2cos1
0
xy y x y x
xy y x y x
xy
y x y x
x
xy
y x y x
x
xy y xy x x
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2cos2sin2
2cos2sin2
2cos2sin2
2sin2cos2
2sin2cos22
2sin2cos22
2cos2sin2
2sin2cos22
2222
2
2
2
222
22
22
xy
y x
xy
y x
y x
xy
y x
y x
xy
y x
xy
y x y x
x
xy
y x y x
x
xy y x
y x
xy
y x y x
x
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2
2
2
2
22222
2
2
2
222
2
2
222
22
22
2sin2cos2sin2cos22
2cos2sin22cos2sin2
2sin2cos2
2sin2cos22
xy
y x
y x
y x
x
xy
y x
y x
y x
x
xy
y x
xy
y x
y x
xy
y x
xy
y x y x
x
22
222
22
where
xy y x y x
ave
y xave x
R
R
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Principal Stresses
• The previous equations are combined to
yield parametric equations for a circle,
22
222
22
where
xy y x y x
ave
y xave x
R
R
• Principal stresses occur on the principal
planes of stress with zero shearing stresses.
o
22
minmax,
90 byseparatedanglestwodefines : Note
22tan
22
y x
xy p
xy
y x y x
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Maximum Shearing Stress
Maximum shearing stress occurs for ave x
2
45 byfromoffset
and90 byseparatedanglestwodefines: Note
2
2tan
2
o
o
22
max
y xave
p
xy
y x s
xy y x
R
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Example 1
For the state of plane
stress shown, determine
(a) the principal planes,
(b) the principal stresses,(c) the maximum shearing
stress and the
corresponding normal
stress.
SOLUTION:
• Find the element orientation for the principalstresses from
y x
xy p
2
2tan
• Determine the principal stresses from
22
minmax,22 xy
y x y x
• Calculate the maximum shearing stress with
22
max 2 xy
y x
2
y x
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Example 1
SOLUTION:
• Find the element orientation for the principalstresses from
1.233,1.532
333.11050
40222tan
p
y x
xy p
6.116,6.26 p
• Determine the principal stresses from
22
22
minmax,
403020
22
xy
y x y x
MPa30
MPa70
min
max
MPa10
MPa40MPa50
x
xy x
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Example 1
MPa10
MPa40MPa50
x
xy x
2
1050
2
y x
ave
• The corresponding normal stress is
MPa20
• Calculate the maximum shearing stress with
22
22
max
4030
2
xy
y x
MPa50max
45 p s
6.71,4.18 s
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Sample Problem 2
A single horizontal force P of 150 lb
magnitude is applied to end D of lever
ABD. Determine (a) the normal and
shearing stresses on an element at
point H having sides parallel to the x
and y axes, (b) the principal planes
and principal stresses at the point H .
SOLUTION:
• Determine an equivalent force-couple
system at the center of the transverse
section passing through H .
• Evaluate the normal and shearing
stresses at H .• Determine the principal planes and
calculate the principal stresses.
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Sample Problem 2
SOLUTION:
• Determine an equivalent force-couplesystem at the center of the transverse
section passing through H .
inkip5.1in10lb150
inkip7.2in18lb150
lb150
x M
T
P
• Evaluate the normal and shearing stresses
at H .
421
4
4
1
in6.0
in6.0inkip7.2
in6.0
in6.0inkip5.1
J
Tc
I
Mc
xy
y
ksi96.7ksi84.80 y y x
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Sample Problem 2
• Determine the principal planes and
calculate the principal stresses.
119,0.612
8.184.80
96.7222tan
p
y x
xy p
5.59,5.30 p
22
22
minmax,
96.72
84.80
2
84.80
22
xy
y x y x
ksi68.4
ksi52.13
min
max
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Mohr’s Circle for Plane Stress
• With the physical significance of Mohr’s circle
for plane stress established, it may be applied
with simple geometric considerations. Criticalvalues are estimated graphically or calculated.
• For a known state of plane stress
plot the points X and Y and construct the
circle centered at C .
xy y x ,,
22
22 xy
y x y xave R
• The principal stresses are obtained at A and B.
y x
xy p
ave R
22tan
minmax,
The direction of rotation of Ox to Oa is
the same as CX to CA.
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Mohr’s Circle for Plane Stress
• With Mohr’s circle uniquely defined, the state
of stress at other axes orientations may bedepicted.
• For the state of stress at an angle with
respect to the xy axes, construct a new
diameter X’Y’ at an angle 2 with respect to XY .
• Normal and shear stresses are obtained
from the coordinates X’Y’.
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Mohr’s Circle for Plane Stress
• Mohr’s circle for centric axial loading:
0, xy y x
A
P
A
P xy y x
2
• Mohr’s circle for torsional loading:
J
Tc xy y x 0 0 xy y x
J
Tc
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Example 3
For the state of plane stress shown,
(a) construct Mohr’s circle, determine
(b) the principal planes, (c) the
principal stresses, (d) the maximumshearing stress and the corresponding
normal stress.
SOLUTION:
• Construction of Mohr’s circle
MPa504030
MPa40MPa302050
MPa202
1050
2
22
CX R
FX CF
y xave
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Example 3
• Principal planes and stresses
5020max CAOC OA MPa70max
5020max BC OC OB
MPa30max
1.532
30
402tan
p
pCP
FX
6.26 p
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Example 3
• Maximum shear stress
45 p s
6.71 s
Rmax
MPa50max
ave
MPa20
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Sample Problem 4
For the state of stress shown,
determine (a) the principal planes
and the principal stresses, (b) the
stress components exerted on the
element obtained by rotating thegiven element counterclockwise
through 30 degrees.
SOLUTION:
• Construct Mohr’s circle
MPa524820
MPa802
60100
2
2222
FX CF R
y xave
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Sample Problem 4
• Principal planes and stresses
4.672
4.220
482tan
p
pCF
XF
clockwise7.33 p
5280
max
CAOC OA
5280
max
BC OC OA
MPa132max MPa28min
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Sample Problem 4
6.52sin52
6.52cos5280
6.52cos5280
6.524.6760180
X K
CLOC OL
KC OC OK
y x
y
x
• Stress components after rotation by 30o
Points X’ and Y’ on Mohr’s circle that
correspond to stress components on therotated element are obtained by rotating
XY counterclockwise through 602
MPa341
MPa6.111
MPa4.48
y x
y
x