Question 3.10: A solid circular bar AB of length L is fixed at one end and ......
A solid circular bar AB of length L is fixed at one end and free at the other (Fig. 3-41). Three different loading conditions are to be considered: (a) torque T_a acting at the free end; (b) torque T_b acting at the midpoint of the bar; and (c) torques T_a and T_b acting simultaneously.
For each case of loading, obtain a formula for the strain energy stored in the bar. Then evaluate the strain energy for the following data: T_a =100 N⋅m , T_b= 150 N⋅m , L = 1.6 m, G =80 GPa, and I_p=79.52 × 10³ mm^4 .
Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
1. Conceptualize, Categorize:
Part (a): Torque T_a acting at the free end (Fig. 3-41a).
In this case, the strain energy is obtained directly from Eq. (3-55a):
U=\frac{T^2L}{2GI_{p}} (3-55a)
U_{a}=\frac{T^2_aL}{2GI_{p}} \hookleftarrow (a)
Part (b): Torque T_b acting at the midpoint (Fig. 3-41b).
When the torque acts at the midpoint, apply Eq. (3-55a) to segment AC of the bar:
U_b=\frac{T^2_b(L/2)}{2GI_{p}}=\frac{T^2_bL}{4GI_{p}} \hookleftarrow (b)
Part (c): Torques T_a and T_b acting simultaneously (Fig. 3-41c).
When both loads act on the bar, the torque in segment CB is T_a and the torque in segment AC is T_a+T_b. Thus, the strain energy [from Eq. (3-57)]
U=\sum\limits_{i=1}^{n}\frac{T^2_iL_i}{2G_i(I_{p})_i} (3-57)
is
U_c=\sum\limits_{i=1}^{n}\frac{T^2_iL_i}{2G(I_{p})_i}=\frac{T^2_a(L/2)}{2GI_{p}}+\frac{(T_a+T_b)^2(L/2)}{2GI_{p}}=\frac{T^2_aL}{2GI_{p}} +\frac{T_aT_bL}{2GI_{p}}+\frac{T^2_bL}{4GI_{p}} \hookleftarrow (c)
A comparison of Eqs. (a), (b), and (c) shows that the strain energy produced by the two loads acting simultaneously is not equal to the sum of the strain energies produced by the loads acting separately. As pointed out in Section 2.7, the reason is that strain energy is a quadratic function of the loads, not a linear function.
3. Analyze:
Part (d): Numerical results.
Substitute the given data into Eq. (a) to obtain
U_a=\frac{T^2_aL}{2GI_{p}} =\frac{(100 \ N\cdot m)^2(1.6 \ m)}{2(80\ GPa)(79.52 × 10³ \ mm^4)}=1.26\ J \hookleftarrow
Recall that one joule is equal to one newton-meter (1 J =1 N·m).
Proceed in the same manner for Eqs. (b) and (c) to find
U_b=1.41\ J
U_c=1.26\ J +1.89\ J +1.41\ J=4.56\ J \hookleftarrow
4. Finalize: Note that the middle term, involving the product of the two loads, contributes significantly to the strain energy and cannot be disregarded.