Question 3.10: A solid circular bar AB of length L is fixed at one end and ...

(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^{2} L}{2 G I_{P}}:

U_{a}=\frac{T_{a}^{2} L}{2 G I_{p}}    (a)

(b) Torque T_{b} acting at the midpoint (Fig. 3-41b). When the torque acts at the midpoint, we apply Eq. (3-55a) to segment AC of the bar:

U_{b}=\frac{T_{b}^{2}(L / 2)}{2 G I_{P}}=\frac{T_{b}^{2} L}{4 G I_{P}}      (b)

(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)] is

\begin{aligned}U_{c} &=\sum_{i=1}^{n} \frac{T_{i}^{2} L_{i}}{2 G\left(I_{P}\right)_{i}}=\frac{T_{a}^{2}(L / 2)}{2 G I_{P}}+\frac{\left(T_{a}+T_{b}\right)^{2}(L / 2)}{2 G I_{P}} \\&=\frac{T_{a}^{2} L}{2 G I_{P}}+\frac{T_{a} T_{b} L}{2 G I_{P}}+\frac{T_{b}^{2} L}{4 G I_{P}}\end{aligned}      (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.
(d) Numerical results. Substituting the given data into Eq. (a), we obtain

Recall that one joule is equal to one newton-meter (1 J = 1 N . m).
Proceeding in the same manner for Eqs. (b) and (c), we find

Note that the middle term, involving the product of the two loads, contributes significantly to the strain energy and cannot be disregarded.