 2000, W. E. Haisler
28
Chapter 12: Torsion of Circular Bars
Example 2: Consider the following aluminum bar in
torsion. The diameters of sections AB, BC and CD are 0.5
in, 0.75 in and 0.5 in, respectively. The shear modulus for
aluminum is 4 million psi.
A
B
C
2
1
40 in-lb
20 in
30 in
D
3
75 in-lb
20 in
a) The internal torques in each of the bars are labeled M t1,
M t 2 , and M t 3 (from left to right). Make cuts in each bar
 2000, W. E. Haisler
29
Chapter 12: Torsion of Circular Bars
and isolate the free-bodies. Now use equilibrium to
relate the internal torques.
A
Mt
40 in-lb
1
B
free-body 1
Mt
75 in-lb
2
C
free-body 2
free  body #1:  M  0  M t 2  40  M t1
free  body #2 :  M  0  M t 3  75  M t 2
M
t
3
D
 2000, W. E. Haisler
Chapter 12: Torsion of Circular Bars
30
Note that this structure is STATICALLY INDETERMINATE
since we CANNOT find all internal torques by equilibrium
alone.
b) Determine J and angle of twist of each bar.
J1  J 3   D 
32
4
 (.5)4
32
 0.0061in4
 (.75)
4

D
J2 

 0.031in
4
4
32
32
M t1L1
M t1(20)
1 

 0.00082M t1
6
J1G1 0.0061(4 x10 )
M t 2 L2
M t 2 (30)
2 

 0.000242M t 2
6
J 2G2
0.031(4 x10 )
 2000, W. E. Haisler
Chapter 12: Torsion of Circular Bars
31
M t 3 L3
M t 3 (20)
3 

 0.00082 M t 3
6
J 3G3 0.0061(4 x10 )
Note that each  above is a relative twist; i.e., the
angle of twist of one end of a bar relative to its other
end.
c) Apply the boundary condition. Since the bar is fixed
between two rigid walls, the total twist must be zero.
total angle of twist  0  1  2  3
0  0.00082 M t1  0.000242M t 2  0.00082M t 3
 2000, W. E. Haisler
Chapter 12: Torsion of Circular Bars
32
d) Now combine the two equilibrium equations and the one
boundary condition equation.
 M t1  M t 2  40
 M t 2  M t 3  75
0.00082 M t1  0.000242M t 2  0.00082 M t 3  0
or in matrix notation,
1
0   M t1   40in  lb 
 1

 

 0

1
1
M t 2   75in  lb 




0
0.00082 0.000242 0.00082   M t 3  

 2000, W. E. Haisler
Chapter 12: Torsion of Circular Bars
33
Solving for the unknown torques, one obtains (using Maple)
> with (linalg):
> A := array([[-1,1,0],[0,-1,1],[0.00082,0.000242,0.00082]]);
[ -1
1
0
[
A := [ 0
-1
1
[
[.00082 .000242
]
]
]
]
.00082]
> b := array([40,-75,0]);
b := [40, -75, 0]
> linsolve(A,b);
[10.10626992, 50.10626992, -24.89373008]
Thus,
 2000, W. E. Haisler
34
Chapter 12: Torsion of Circular Bars
M t1  10.11in  lb, M t1  50.11in  lb, M t 3  24.89in  lb
The internal torque diagram can now be drawn:
A
B
1
C
40 in-lb
75 in-lb
D
3
2
M t (in  lb)
50.11
10.11
Internal
Torque
Diagram
70
x(in)
20
24.89
Important note on torque diagrams: Notice that the internal
torque diagram M t has a discontinuity at the location of an
 2000, W. E. Haisler
Chapter 12: Torsion of Circular Bars
35
applied external torque equal to the magnitude of the applied
torque. For example at x=20”, there is a torque of 40 in-lb
applied (at x=50”, a torque of 75 in-lb is applied).
e) We can now solve for the individual angles of twist or
the shear stress. For example,
Rotation of point A  1  0.00082 M t1  0.00082(10.11)
 0.00829rad  0.47deg
Rotation of point B  1  2  0.00829rad  0.000242 M t 2
 0.00829  0.000242(50.11)
 0.0204rad  1.17deg
 2000, W. E. Haisler
Chapter 12: Torsion of Circular Bars
Stress in bar 1:
M t1r1 10.11in  lb(0.5in / 2)
 x 1 

 414 psi
4
J1
0.0061in
Stress in bar 2:
M t 2r2 50.11(0.75/ 2)
 x 2 

 606 psi
J2
0.031
36
 2000, W. E. Haisler
37
Chapter 12: Torsion of Circular Bars
Example 3: Bar with distributed torque of 60 in-lb/in
applied from A to B and a concentrated torque of 400 in-lb
at C. Material is steel with a shear modulus of 11.5 million
psi. Bars are cylindrical with diameters of 0.4 in (A-B) and
0.25 in (B-C).
60 in-lb/in
1
A
x
5 in
400 in-lb
B
C
8 in
a) First construct the distributed applied torque diagram ( mt
vs. x). Note that this is NOT the internal torque!
 2000, W. E. Haisler
38
Chapter 12: Torsion of Circular Bars
mt
5
13 in.
0
x
Applied
Torque
Diagram
-60 in-lb/in
b) Now construct the internal torque diagram by using
 Mt
 mt
integration of
x
M t (13in)  400 in  lb
At x=13",
For 5  x  13",
x
x
M t ( x)  M t (13)  13
mt dx  400  13
(60)dx  400  60( x  13)
At x=5" (from equation above),
M t (5)  80 in  lb
 2000, W. E. Haisler
39
Chapter 12: Torsion of Circular Bars
For 0  x  5",
M t ( x)  M t (5)  5x mt dx  80  5x (0)dx  80 in  lb
Now construct the internal torque diagram for the
structure.
60 in-lb/in
1
A
B
Mt
C
400+60(x-13)
5 in.
-80 in-lb
400 in-lb
400 in-lb
slope is
60 in-lb/in
x
Internal
Torque
Diagram
13 in.
lb ( x  5) .
Note: M t is also: M t ( x)  80 in  lb / in  4808in
in
 2000, W. E. Haisler
Chapter 12: Torsion of Circular Bars
40
Important Note on torque diagrams. Compare the mt and
M t diagrams. Note that when mt is a constant, M t varies
linearly. Also, slope of M t vs. x curve is equal to the value
of mt at any point x. This is true since we have the
 Mt
 mt . Note that the bar has a torque of
relationship:
x
400 in-lb applied at the end -- Hence the internal torque
starts at 400 in-lb and then decreases linearly due to the 60
in-lb/in applied distributed torque.
c) Now determine the angle of twist using integration of the
internal torque.
First obtain the polar moment of inertia for section:
 2000, W. E. Haisler
Chapter 12: Torsion of Circular Bars
41
 (.4)
4

D
J1 

 0.00251 in
4
4
32
32
4  (.25) 4
4

D
J2 

 0.000383 in
32
32
for 0  x  5",
 ( x)   (0)  
x
0
x
Mt
(80)
dx  0  
dx


0.0029
x
0 (0.00251)(11x106 )
J1G1
 (5") = -0.0029(5) = -0.0145 rad -= - 0.83 deg
 2000, W. E. Haisler
Chapter 12: Torsion of Circular Bars
for 5  x  13",
 ( x)   (5)  
x
5
42
x [400  60( x  13)]
Mt
dx  .0145  
dx
5
J 2G2
J 2G2
2 x
1
 .0145 
(380 x  30 x )
6
5
.000383(11x10 )
 0.0145  1 [380 x  30 x 2  1,150]
4,213
Evaluate the above at x=13":
 (13") = -0.0145 + 0.304 = 0.289 rad = 16.6 deg
 2000, W. E. Haisler
Chapter 12: Torsion of Circular Bars
43
d) Now determine stresses at various x points. Be sure to
use Mt, r and J for desired x value (get Mt from the
internal torque diagram).
At x = 5 in (in bar AB),
M t ( x  5)r 80in  lb(0.4"/ 2)
 x ( x  5") 

 6,375 psi
4
J1
0.00251in
At x = 5 in (in bar BC),
M t ( x  5)r 80in  lb(0.25"/ 2)
 x ( x  5") 

 26,110 psi
4
J2
0.000383 in
Note: the negative sign on shear stress does NOT mean compression.
Note that there is a significant difference in shear stress at point
B where the bars change diameter (even thought M t is same).
 2000, W. E. Haisler
Chapter 12: Torsion of Circular Bars
44
At x = 9 in, (get Mt from torque diagram)
M t ( x  9")r 160 in  lb(0.25"/ 2)
 x ( x  9") 

 52,219 psi
4
J2
0.000383 in
At x = 13 in,
M t ( x  13)r 400 in  lb(0.25"/ 2)
 x ( x  13") 

 130,550 psi
4
J2
0.000383 in
Note: depending upon the type of steel, the material may
fail before reaching a shear stress of 130 ksi.
If the stress in section BC is too high, must increase the
diameter of section BC (new J 2 ), and repeat calculations for
angles of twist and stress in section BC. Section AB
calculations would be unaffected since J1 was not changed.
 2000, W. E. Haisler
45
Chapter 12: Torsion of Circular Bars
Example 2 with MDSolids: Consider the following
aluminum bar in torsion. The diameters of sections AB, BC
and CD are 0.5 in, 0.75 in and 0.5 in, respectively. The
shear modulus for aluminum is 4 million psi.
A
B
C
2
1
40 in-lb
20 in
30 in
D
3
75 in-lb
20 in
MDSolids is a multipurpose program that solves truss
problems, does axial, torsion and bending problems, Mohr’s
circle, moments of inertia, pressure vessels and other nifty
things. Below are 3 screen captures of the above problem
solves with MDSolids.
 2000, W. E. Haisler
Chapter 12: Torsion of Circular Bars
46
 2000, W. E. Haisler
Chapter 12: Torsion of Circular Bars
47
 2000, W. E. Haisler
Chapter 12: Torsion of Circular Bars
48