Ó 2000, W. E. Haisler Chapter 12: Torsion of Circular Bars 48
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) The internal torques in each of the bars are labeled , , and (from left to right). Make cuts in each bar and isolate the free-bodies. Now use equilibrium to relate the internal torques.
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.
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.
d) Now combine the two equilibrium equations and the one boundary condition equation.
or in matrix notation,
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,
The internal torque diagram can now be drawn:
Important note on torque diagrams: Notice that the internal torque diagram has a discontinuity at the location of an 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,
Stress in bar 1:
Stress in bar 2:
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).
a) First construct the distributed applied torque diagram ( vs. x). Note that this is NOT the internal torque!
b) Now construct the internal torque diagram by using integration of
At x=13",
For ",
At x=5" (from equation above),
For ",
Now construct the internal torque diagram for the structure.
Note: is also: .
Important Note on torque diagrams. Compare the and diagrams. Note that when is a constant, varies linearly. Also, slope of vs. x curve is equal to the value of at any point x. This is true since we have the relationship: . Note that the bar has a torque of 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:
for ",
\ f(5") = -0.0029(5) = -0.0145 rad -= - 0.83 deg
for ",
Evaluate the above at x=13":
\ f(13") = -0.0145 + 0.304 = 0.289 rad = 16.6 deg
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),
At x = 5 in (in bar BC),
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 is same).
At x = 9 in, (get Mt from torque diagram)
At x = 13 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 ), and repeat calculations for angles of twist and stress in section BC. Section AB calculations would be unaffected since was not changed.
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.
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.