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Transcript of Chapter 6 - COMPLEX STRESS.ppt
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CHAPTER 6
COMPLEX
STRESS
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INTRODUCTION
The most general state of stress at apoint may be represented by 6
components,
,,:(Note
stressesshearing,,
stressesnormal,,
xzxzyyzyxxy
zxyzxy
zyx
Same state of stress is represented by adifferent set of components if axes arerotated.
9.1 Plane Stress Transformati
In practice, approximations andsimplificationsare done to reduce the stress components
to a single plane.
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Plane Stress
Plane Stress state of stress in which two facesof the cubic element are free of stress. For theillustrated example, the state of stress isdefined by
.0,, andxy
zyzxzyx
State of plane stress occurs in a thin platesubjected to forces acting in the midplane of theplate.
State of plane stress also occurs on the freesurface of a structural element or machinecomponent, i.e., at any point of the surface notsubjected to an external force.
9.1 Plane Stress Transformati
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Plane Stress Transformation
The material is then said to be subjected toplane stress. For general state of plane stress at a pt, werepresent it via normal-stress components, x,
yand shear-stress
component xy.
Transforming stress components from one orientationto the other is similar in concept to how we transform
force components from one system of axes to theother.
Note that for stress-component transformation, weneed to account for- the magnitude and direction of each stress
component, and- the orientation of the area upon which each
component acts. 9.1 Plane Stress Transformati
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Procedure of Analysis
(Method of Equilibrium)
If state of stress at a pt is known for a given orientation of an element ofmaterial, then state of stress for another orientation can be determined.
9.1 Plane Stress Transformati
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Transformation of Plane Stress
consider the stress element has rotated through anangle as shown.
As linear equations, the transformedstresses are given by
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Sign Convention
Positive normal stresses, xand y, actsoutward from all faces
Positive shear stress xyacts upward on theright-hand face of the element.
The orientation of the inclined plane isdetermined using the angle .
Establish a positive xand yaxes usingthe right-hand rule.
Angle is positive if it movescounterclockwise from the +xaxis to the
+xaxis.
General Equation
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Transformation of Plane Stress
2cos2sin2
2sin2cos22
2sin2cos22
xy
yx
yx
xy
yxyx
y
xy
yxyx
x
The equations may be rewritten to yield
General Equation
1
2
3
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Procedure of Analysis
General Equation
To apply equations -1 and -2, just substitute the knowndata for x, y, xy, andaccording to established signconvention.
Ifxandxy are calculated as positive quantities, thenthese stresses act in the positive direction of the xandyaxes.
Tip: For your convenience, equations -1 to -3 can beprogrammed on your pocket calculator.
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Example 6.1
The state of plane stress at a point is represented by
the element shown in figure. Determine the state ofstress at the point on another element oriented 30clockwise from the position shown.
25 MPa
80 MPa
50 MPa
147
AD
CB
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119.2 General Equation
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Principal Stresses
Principal stresses,1/2 - the maximum and minimum normal
stresses. Principal planes - the planes on which the principalstresses act.
the principal stresses will occur on mutually perpendicular
planes. the shear stresses are zero on the principal planes.
There are two way to determine the principal stresses
(a) Calculation method
(b) Mohrs Circle method
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Principal Stresses The previous equations are
combined to yield parametricequations for a circle,
22
222
22
where
xyyxyx
ave
yxavex
R
R
Principal stressesoccur on theprincipalplanes of stresswith zero shearingstresses.
9.3 Principle Stresses and Max. In-Plane Shear
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MaximumShearing Stress
Maximum shearing stressoccurs for avex
Principle Stresses and Max. In-Plane ShearStress
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22
22tan
22
21
212
2
m a x
yx
ave
xy
yx
s
xy
yxR
Mohr's Circle for Plane Stress
a graphical representation of the stress transformationequations
all stresses on Mohr's circle are in-plane stresses
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Steps in constructing Mohr's circle:1. Draw axes. ( positive to the right and
xy positive down)
2. Plot first point on circle. (x , xy )3. Plot second point on circle. (y , - xy)4. Draw a line between the two points.
This line is the diameter of the circle,
and it passes through the center ofthe circle.5. Draw a circle through the two points.
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Stress Transformations
Principal Stresses and Maximum Shear Stresses
153
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Example 6.2
The state of plane stress at apoint on a body is shown onthe element in the figure given.Represent this stress in termsof the :(a) principal stresses.(b) maximum in-plane shear
stress and associatedaverage normal stress
9.3 Principle Stresses and Max. In-Plane Shear
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199.3 Principle Stresses and Max. In-Plane Shear
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209.3 Principle Stresses and Max. In-Plane Shear
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219.3 Principle Stresses and Max. In-Plane Shear
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229.3 Principle Stresses and Max. In-Plane Shear
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239.3 Principle Stresses and Max. In-Plane Shear
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Example 6.3
9.3 Principle Stresses and Max. In-Plane Shear
For the state of plane stressshown, determine (a) theprincipal planes, (b) the
principal stresses, (c) themaximum shearing stress andthe corresponding normalstress.
SOLUTION:
Find the element orientation for theprincipal stresses from
yx
xyp
22tan
Calculate the maximum shearing
stress with2
2
max2
xyyx
2
yx
Determine the principal stressesfrom
22
minmax,22
xyyxyx
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SOLUTION:
Find the element orientation for theprincipal stresses from
1.233,1.532
333.11050
40222tan
p
yx
xyp
6.116,6.26p
Determine the principal stressesfrom
22
22
minmax,
403020
22
xy
yxyx
MPa30
MPa70
min
max
MPa10
MPa40MPa50
x
xyx
9.3 Principle Stresses and Max. In-Plane Shear
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MPa10
MPa40MPa50
x
xyx
2
1050
2
yxave
The corresponding normal
stress is
MPa20
Calculate the maximum shearingstress with
22
22
max
4030
2
xyyx
MPa50max
45 ps
6.71,4.18s
9.3 Principle Stresses and Max. In-Plane Shear
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Exercise 1
The state of plane stress at a point is represented bythe element shown in figure. Represent this state ofstress on an element oriented 30 counterclockwise fromthe position shown.
6 MPa
8 MPa
12 MPa
157
EXAMPLE FOR L SHAPED
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EXAMPLE FOR L SHAPED
QUESTION:
A single horizontal force Pof 600N magnitude is applied to end D of lever
ABD. Determine (a) the normal and shearing stresses on an element at pointHhaving sides parallel to the x and y axes, (b) the principal planes and
principal stresses at the point H.
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SOLUTION:
1. Determine an equivalent force-couple system at the center ofthe transverse section passingthrough H.
2. Evaluate the normal and shearingstresses at H.
3.Determine the principal planes and
calculate the principal stresses.
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x
P 600N
T 600N 0.45m 270 N.m
M 600N 0.25m 150 N.m
Evaluate the normal and shearing stresses at H.
y 414
xy 412
150N.m 0.015mMy
I 0.015m
56.6 MPa
270 N.m 0.015mTc
J 0.015m
50.9 MPa
a)
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x y xy0 56.6 MPa 50.9 MPa
Note that, the shearing force P Does not cause any
shearing stress at point Hb) Determine the principal planes and calculate the principal
stresses. xy
p
x y
p
p
2 2 50.9tan 2 1.8
0 56.6
2 61.0 ,119
30.5 , 59.5
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2
x y x y 2
max, min xy
2
2
max
min
2 2
0 56.6 0 56.6 50.92 2
86.5 MPa
29.9 MPa
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Example 2 (L shape) Refer to word file