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Analysis of Stress and Strain
2cos=
cos sin
F P
A A
V P
A A
Review:
- Axially loaded Bar - Torsional shaft
Questions: (1) Is there any general method to determine stresses on any arbitrary plane at one point if the stresses at this point along some planes are known? (2) For an arbitrary loaded member, how many planes on which stresses are known are required to determine the stresses at any plane at one point?
P P
p
q
h
sin 2
cos2
n xy
nt xy
yx
nnt
xy
Analysis of Stress and Strain
State of stress at one point:
Stress element:
- Use a cube to represent stress element. It is infinitesimal in size. - (x,y,z) axes are parallel to the edges of the element- faces of the element are designated by the directions of their outward normals.
Sign Convention:- Normal stresses: “+” tension; “-” compression. - Shear stresses: “+” the directions associated with its subscripts are plus-plus or minus-minus “-” the directions associated with its subscripts are plus-minus or minus-plus
x
y
z
x
y
z
xy
xz
yx
yz
zx
zy x
y
y
x
xy
yx
xy
yx
Plane Stress
1x
0z xz yz
1 1x y
Definition: Only x and y faces are subject to stresses, and all stresses are parallel to the x and y axes.
Stresses on inclined planes
yx
xy
y
x
0
0
x
y
F
F
1
1 1
2 2
2 2
cos sin 2 sin cos
sin cos cos sin
x x y xy
x y x y xy
1
1 1
cos 2 sin 22 2
sin 2 cos 22
x y x yx xy
x yx y xy
Transformation equations for plane stress
1x
1y
1 1x y
Transformation Equations
2cos2sin2
2sin2cos22
2sin2cos22
11
1
1
xyyx
yx
xyyxyx
y
xyyxyx
x
angle between x1 and x axes, measured counterclockwise
yxyx 11
Plane Stress – Special Cases
Uniaxial Stress:
Pure Shear:
Biaxial Stress:
x
xy
yx
xy
yx
x
x
y
y
x
Plane Stress
Example 1: A plane-stress condition exists at a point on the surface of a loaded structure, where the stresses have the magnitudes and directions shown on the stress element of the following figure. Determine the stressesacting on an element that is oriented at a clockwise angle of 15o with respect to the original element.
Principal Stresses
1
1
cos2 sin 22 2
2sin 2 2 cos2 02
2 tan 2
x y x yx xy
x x yxy
xyp
x y
d
d
Principal stresses: maximum and minimum normal stresses.Principal planes: the planes on which the principal stresses act
:p The angle defines the orientation of the principal planes.
Principal Stresses 2
22tan 2 cos2 , sin 2 ,
2 2 2xy x y xy x y
p p p xyx y
RR R
1
1
2
21
2 2 2 2
2 2
x y x y x y xyx xy
x y x yx xy
R R
OR
2
22tan 2 cos2 , sin 2 ,
2 2 2xy x y xy x y
p p p xyx y
RR R
1
1
2
22
2 2 2 2
2 2
x y x y x y xyx xy
x y x yx xy
R R
21
Shear Stress
Shear stresses on the principal planes:
1 1sin 2 cos 2 0
2x y
x y p xy p
Example 2: Principal stresses in pure shear case:
xy
yx
xy
yx
Maximum Shear Stresses
22212
2
max
xy
yx1 1
2 1
4
4
s p
s p
1 1sin 2 cos 2
2x y
x y xy
1 1 cos 2 2 sin 2 0x y
x y xy
d
d
1tan 2 tan 2
2 tan 2x y
s sxy p
Plane Stress Example 3: Find the principal stresses and maximum shear stresses and show them on a sketch of a properly oriented element.
Mohr’s Circle For Plane Stress – Equations of Mohr’s Circle
1
1 1
cos 2 sin 2 (1)2 2
sin 2 cos 2 (2)2
x y x yx xy
x yx y xy
1 1 1
2 2
2 2
2 2x y x y
x x y xy
Transformation equations:
(1)2 + (2)2
2
2
222
2 ,
2
111
xyyxyx
ave
yxavex
R
R
Two Forms of Mohr’s Circle
1x
1 1x y
1x
1 1x y
Approach 1: For the given state of stresses, calculate and R. The centerOf the circle is ( , 0) and the radius is R.
Construction of Mohr’s Circleave
ave
1x
1 1x y
Construction of Mohr’s CircleApproach 2: Find points corresponding to = 0 and = 90o and then draw a line. The intersection is the origin of the circle.
1x
1 1x y
Applications of Mohr’s Circle
Example 4: An element in plane stress at the surface of a large machine is subjected to stresses Using Mohr’s circle, determine the following quantities: (a) the stressesacting on an element inclined at an angle of 40o, (b) the principal stressesand (c) the maximum shear stress.
psi 5000 ,psi 15000 xyx
Plane Strain
Definition: Only x and y faces are subject to strains, and all strains are parallel to the x and y axes.
Note: Plane stress and plane strain do not occur simultaneously.
0z xz yz
Plane Strain
2cos2
2sin22
2sin2
2cos22
2sin2
2cos22
11
1
1
xyyxyx
xyyxyxy
xyyxyxx
22
2
22
1
222
222
xyyxyx
xyyxyx
Transformation Equations:
Principal Strains:
yxyx 11