Principle and Maximum
Shearing Stresses (7.1-7.3)
MAE 314 – Solid Mechanics
Yun Jing
Principle and Maximum Shearing Stresses
1
Transformation of Stress



Recall the general state of stress at a point can be written in terms of 6
components: σx, σy, σz, τxy= τyx, τxz= τzx, τyz= τzy
This general “stress state” is independent of the coordinate system
used.
The components of the stress state in the different directions do
depend on the coordinate system.
Principle and Maximum Shearing Stresses
2
Transformation of Stress

Consider a state of plane stress: σz=τxz=τyz=0

Where does this occur?
Thin plate
Outer surface
Principle and Maximum Shearing Stresses
3
Transformation of Stress

What do we want to calculate?




Principle stresses (σ maximum and σ minimum)
Principle planes of stresses (orientation at which they occur)
Slice cube at an angle θ to the x axis (new coordinates x’, y’).
Define forces in terms of angle and stresses.
Principle and Maximum Shearing Stresses
4
Transformation of Stress

Sum forces in x’ direction.
 x 'A   x A cos cos   y A sin  sin 
  xy A sin   cos   xy A cos sin 
 x'   x cos2    y sin 2   2 xy sin  cos

Sum forces in y’ direction.
 x' y'A   x A cos sin    y Asin  cos  xy Asin  sin   xy A cos cos
 x' y '   y   x cos sin   xy cos2   sin 2  
Principle and Maximum Shearing Stresses
5
Transformation of Stress
 x'   x cos2    y sin 2   2 xy sin  cos
 x' 
 x  y

2
 x  y
2
cos2    xy sin 2 
To get σy’, evaluate σx’ at θ + 90o.
 y' 
 x  y
2

 x  y
2
cos2    xy sin 2 
Trig identities
1  cos2 
2 sin  cos  sin 2  cos2  
2
1  cos2 
cos2   sin 2   cos2  sin 2  
2
 x' y '   y   x cos sin   xy cos2   sin 2  
 x' y'  
 x  y
2
sin 2    xy cos2 
Principle and Maximum Shearing Stresses
6
Transformation of Stress
 x' 
 x  y
2

 x  y
2
cos2    xy sin 2 
 y' 
 x  y
2

 x  y
2
cos2    xy sin 2 
 x' y'  
 x  y
2
sin 2    xy cos2 
Now, let’s perform some algebra:
  y 

   y 
   y 
 x '  x
   x2' y '   x
 cos2 2    xy2 sin 2 2   2 x
 cos2 sin 2  
2 

 2 
 2 
2
2
  x  y 
   y 

 sin 2 2    xy2 cos2 2   2 x
 cos2 sin 2 
 2 
 2 
2
Constants (we can
find these stresses).
  y 

   y 
2
  x '  x
   x2' y '   x
   xy
2 

 2 
2
2
Variables
Principle and Maximum Shearing Stresses
7
Principle and Max Shearing Stress

Define
 ave

1
  x   y 
2
  x  y 
   xy2
R 2  
 2 
2
Plug into previous equation
  y 

   y 
2
  x '  x
   x2' y '   x
   xy
2 

 2 
2
2
 x'   ave 2   x2' y '  R 2

Which is the equation of a circle with center at (σave,0) and radius R.
Principle and Maximum Shearing Stresses
8
Principle and Max Shearing Stress

We learned before that the principle stresses (maximum and
minimum σ) occur when τx’y’ = 0.
 max   ave  R
 x  y
 x' y'  
sin  2 P    xy cos  2 P   0
 min   ave  R
2
 max  R
 x  y
 xy cos  2 P  
sin  2 P 
 min   R
2
2 xy
 x  y
tan  2 P  
tan  2 s   
 x  y
2 xy
 max,min 
 x  y
2
  x  y 
   xy2
 
 2 
2
  x  y 
 max,min   

2
2
   xy2

Principle and Maximum Shearing Stresses
9
Example Problem
For the state of plane stress shown below, determine (a) the principal
planes, (b) the principal stresses, (c ) the maximum shearing stress
and the corresponding normal stress.
Mohr's Circle
10
Mohr’s Circle (7.4-7.6)
MAE 314 – Solid Mechanics
Yun Jing
Mohr's Circle
11
Constructing Mohr’s Circle
1
  x   y 
2

Given: σx, σy, τxy at a particular orientation
 ave

Draw axes
Plot Point O, center, at (σave,0)
Plot Point X (σx,τxy),
this corresponds to θ = 0°
Plot Point Y (σy,-τxy),
this corresponds to θ = 90°
Draw a line through XY
(passes through O):
This is the diameter of the circle
Draw circle
  x  y 
2
   xy2
R  
 2 





Mohr's Circle
2
 Y

O
2P
X
12
Sign Convention
Mohr's Circle
13
Uses of Mohr’s Circle

Find stresses on any inclined plane.

Rotate by 2θ1 to point X’.



This points gives the stress state (σx1, τxy1).
The point Y’ gives (σy1, -τxy1).
Find principle stresses.





Q1 is the point with the highest
normal stress.
Q2 is the point with the smallest
Q2
normal stress.
(min, 0)
S1 is the highest shear
stress = R.
S2 is the lowest shear
(y1, -xy1) Y’
stress = -R.
Mohr's Circle
2Q2  2Q1 180
2S 2  2Q1  90
S1 (ave, max)
X’ (x1, xy1)
21 O
2Q1
2P
2Q2
S2
(ave, min)
Q1

(max, 0)
X
14
Example Problem
For the state of plane stress shown below, determine (a) the principal
planes, (b) the principal stresses, (c ) the maximum shearing stress
and the corresponding normal stress.
Mohr's Circle
15
Special Cases of Mohr’s Circle

Uniaxial tension




σmax = σx
σmin = 0
τmax = σx / 2
Hydrostatic pressure




R


2
R0
σmax = σx
σmin = σx
τmax = 0
Pure torsion

x
R
σmax = σx
σmin = -σx
τmax = σx
Mohr's Circle
x
2
16
Example Problem
For the given state of stress, determine the normal and shearing
stresses after the element shown has been rotated (a) 25o clockwise
and (b) 10o counterclockwise.
Mohr's Circle
17
Example Problem
For the element shown, determine the range of values of xy for which the
maximum tensile stress is equal to or less than 60 Mpa.
Mohr's Circle
18
3D Stress States



Mohr’s circle can be drawn for rotation about any of the three principle axes.
These axes are in the direction of the three principle stresses.
The maximum shear stress is found from the largest diameter circle.
x-c plane
y-c plane
x-y plane
Mohr's Circle
 max 
1
 max   min
2
19
Applied to Plane Stress



Recall σz = 0 (z-axis is one of the principle axes)
The third σ value is always at the origin.
Where does τmax occur?

2 possibilities
max is in the x-y plane
Mohr's Circle
max is out of the x-y plane
20
Example Problem
For the state of stress shown, determine two values of σy for which the
maximum shearing stress is 75 MPa.
Mohr's Circle
21
Thin-Walled Pressure Vessels
(7.9)
MAE 314 – Solid Mechanics
Yun Jing
Thin-Walled Pressure Vessels
22
Thin-Walled Pressure Vessels
Cylindrical vessel with capped ends

Spherical vessel
Assumptions




Constant gage pressure, p = internal pressure – external pressure
Thickness much less than radius (t << r, t / r < 0.1)
Internal radius = r
Point of calculation far away from ends (St. Venant’s principle)
Thin-Walled Pressure Vessels
23
Cylindrical Pressure Vessel
Circumferential (Hoop) Stress: σ1
Sum forces in the vertical direction.
21 tx   p(2rx)  0
pr
1 
t
Longitudinal stress: σ2
Sum forces in the horizontal direction:
 2 2rt   p(r 2 )  0
2 
pr
2t
Thin-Walled Pressure Vessels
24
Cylindrical Pressure Vessel

There is also a radial component of stress because the gage pressure
must be balanced by a stress perpendicular to the surface.



σr = p
However σr << σ1 and σ2 , so we assume that σr = 0 and consider this a
case of plane stress.
Mohr’s circle for a cylindrical pressure vessel:

Maximum shear stress (in-plane)
 max

2
pr


2
4t
Maximum shear stress (out-of-plane)
 max   2 
pr
2t
Thin-Walled Pressure Vessels
25
Spherical Pressure Vessel
Sum forces in the horizontal direction:
 2 t 2r   p(r 2 )  0
In-plane Mohr’s circle is just a point
pr
1   2 
2t
 max 
Thin-Walled Pressure Vessels
1
2

pr
4t
26
Example Problem
A basketball has a 9.5-in. outer diameter and a
0.125-in. wall thickness. Determine the normal
stress in the wall when the basketball is inflated to
a 9-psi gage pressure.

Thin-Walled Pressure Vessels
27
Example Problem
A cylindrical storage tank contains liquefied
propane under a pressure of 1.5MPa at a
temperature of 38C. Knowing that the tank has an
outer diameter of 320mm and a wall thickness of
3mm, determine the maximum normal stress and
the maximum shearing stress in the tank.

Thin-Walled Pressure Vessels
28
Example Problem
The cylindrical portion of the
compressed air tank shown is
fabricated of 8-mm-thick plate
welded along a helix forming an
angle β = 30o with the horizontal.
Knowing that the allowable stress
normal to the weld is 75 MPa,
determine the largest gage
pressure that can be used in the
tank.
Thin-Walled Pressure Vessels
29