Transcript Document

Analysis of Stress and Strain
Review:
- Torsional shaft
- Axially loaded Bar
𝜎𝜃
p

P
𝜏𝜃
h
q
F P cos2 
  =
A
A
 
V
P cos  sin 

A
A
P
xy

nt
n
yx
 n   xy sin 2
 nt   xy cos 2
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?
Analysis of Stress and Strain
y
State of stress at one point:
Stress element:
y
z
zy
yz
yx
yx
xy
zx xz
x
x
y
xy
xy
yx
y
z
- 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.
x
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
Plane Stress
Definition: Only x and y faces are subject to stresses, and all
stresses are parallel to the x and y axes.
 z   xz   yz  0
y
1
x y 
x
1 1
1
Stresses on inclined planes
xy
x y
1 1

x
yx 
y
x
1
F
F
x
0
y
0
Transformation equations for
plane stress
 x   x cos2    y sin 2   2 xy sin  cos 
1
 x y    x   y  sin  cos    xy  cos2   sin 2  
1 1
x 
x  y
1
x y  
1 1
2

 x  y
2
 x  y
2
cos 2   xy sin 2
sin 2   xy cos 2
Transformation Equations
x 
1
y 
 x  y
2
 x  y
1
x y  
1 1
2


 x  y
2
 x  y
2
 x  y
2
cos 2   xy sin 2
cos 2   xy sin 2
 x  y   x  y
1
1
sin 2   xy cos 2
 angle between x1 and x axes, measured counterclockwise
Plane Stress – Special Cases
x
Uniaxial Stress: x
y
Pure Shear:
xy
x
xy
Biaxial Stress:
yx
y
x
x
y
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 stresses
acting on an element that is oriented at a clockwise angle of 15o with
respect to the original element.
Principal Stresses
Principal stresses: maximum and minimum normal stresses.
Principal planes: the planes on which the principal stresses act
x 
x y

2
1
x  y
2
cos 2   xy sin 2

d x1
d

x  y
2
2sin 2  2 xy cos 2  0

tan 2 p 
2 xy
x  y
 p : The angle defines the orientation of the principal planes.
Principal Stresses
tan 2𝜃𝑝 =
2𝜏𝑥𝑦
𝜎𝑥 −𝜎𝑦
⇒ cos 2𝜃𝑝 =
𝜎𝑥1
OR
tan 2𝜃𝑝 =
𝜎𝑥 −𝜎𝑦
2𝑅
𝜎𝑥 +𝜎𝑦
2
⇒ cos 2𝜃𝑝 = −
𝜎𝑥1
, sin 2𝜃𝑝 =
𝜏𝑥𝑦
𝑅
,𝑅 =
𝜎𝑥 −𝜎𝑦 2
2
+ 𝜏𝑥𝑦 2
𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦 𝜎𝑥 − 𝜎𝑦
𝜏𝑥𝑦
=
+
∙
+ 𝜏𝑥𝑦 ∙
2
2
2𝑅
𝑅
𝜎1 = 𝜎𝑥1 =
2𝜏𝑥𝑦
𝜎𝑥 −𝜎𝑦
+
𝜎𝑥 −𝜎𝑦
2𝑅
𝜎𝑥 −𝜎𝑦 2
2
+ 𝜏𝑥𝑦 2
, sin 2𝜃𝑝 = −
𝜏𝑥𝑦
𝑅
,𝑅 =
𝜎𝑥 −𝜎𝑦 2
2
𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦 −𝜎𝑥 + 𝜎𝑦
−𝜏𝑥𝑦
=
+
∙
+ 𝜏𝑥𝑦 ∙
2
2
2𝑅
𝑅
𝜎2 = 𝜎𝑥1 =
+ 𝜏𝑥𝑦 2
1   2
𝜎𝑥 +𝜎𝑦
2
−
𝜎𝑥 −𝜎𝑦 2
2
+ 𝜏𝑥𝑦 2
Shear Stress
Shear stresses on the principal planes:
x y  
 x  y
1 1
2
sin 2 p   xy cos 2 p  0
Example 2: Principal stresses in pure shear case:
y
xy
x
xy
yx
Maximum Shear Stresses
x y  
1 1
 x  y
2
sin 2   xy cos 2
d x1 y1
d
   x   y  cos 2  2 xy sin 2  0
 x  y
1
tan 2s  
 tan 2s  
2 xy
tan 2 p
tan 2𝜃𝑠 = −tanα
𝜋
2
Let 𝛼 = − 2𝜃𝑝
2𝜃𝑠 = − α
or
   y 
  2
   xy 2  1
 max   x
2
 2 
2
2𝜃𝑠 = π − α
s   p 
1
1
s   p 
2
1

4

4
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
Transformation equations:
x 
x  y
1
x y  
1 1
2

 x  y
2
 x  y
2
cos 2   xy sin 2
(1)
sin 2   xy cos 2

  x1 

(1)2 + (2)2

x  y 
2
  x  y 
2
2
   x1 y1  



2
2
2
   xy

  ave   x1 y1  R 2
2
x1
(2)
 ave 
 x  y
2
2
,
  x  y 
R  

2
2
   xy 2

Two Forms of Mohr’s Circle
x y
1 1
x
1
x
1
x y
1 1
Construction of Mohr’s Circle
Approach 1: For the given state of stresses, calculate  ave and R. The center
Of the circle is (  ave , 0) and the radius is R.
x
1
x y
1 1
Construction of Mohr’s Circle
Approach 2: Find points corresponding to  = 0 and  = 90o and then draw a line.
The intersection is the origin of the circle.
x
1
x y
1 1
Applications of Mohr’s Circle
Example 4: An element in plane stress at the surface of a large machine
is subject to stresses 𝜎𝑥 = 15000psi; 𝜎𝑦 = 5000psi; 𝜏𝑥𝑦 = 4000psi
Using Mohr’s circle, determine the following quantities: (a) the stresses
acting on an element inclined at an angle of 40o, (b) the principal stresses
and (c) the maximum shear stress.
Plane Strain
Definition: Only x and y faces are subject to strains, and all
strains are parallel to the x and y axes.
 z   xz   yz  0
Note: Plane stress and plane strain do not occur simultaneously.
Plane Strain
Transformation Equations:
x 
1
y 
x y
2
x y
2
1
x y
1 1
2



x y
2
x y
x y
2
2
cos 2 
cos 2 
sin 2 
Principal Strains:
 xy
2
 xy
sin 2
2
 xy
sin 2
2
x y  x y
1
1
cos 2
1 
2 
x y
2
x y
2
  x   y    xy 
  

 
 2   2 
2
2
  x   y    xy 
  

 
 2   2 
2
2