Question 10.7: Calculate the strain energy of the shaft as shown in Figure ...

Let us calculate the reaction torques of the given statically indeterminate truss as follows:

In Figure 10.25, we show the free-body diagram of the entire shaft and the free-body diagrams of segments AC, CD and DB. Thus, from Figure 10.25(a),

-T_{ A }+3 T+T_{ B }=0

or        T_{ A }-T_{ B }=3 T           (1)

We note from Figure 10.25(c) that

T_{ A }-2 T+T_{ B }+T

Now, angle of twist \theta_{ A }=\theta_{ B }=0 as both ends are fixed. Therefore, from Figures 10.24(b), (c) and (d), we get

\theta_{ C / A }=\theta_{ C }-\theta_{ A }=\theta_{ C }=\frac{T_{ A }(L / 4)}{G J}=\frac{T_{ A } L}{4 G J}          (2)

\theta_{ D / C }=\theta_{ D }-\theta_{ C }=\frac{\left(T_{ B }+T\right)(L / 2)}{G J}=\frac{\left(T_{ B }+T\right) L}{4 G J}             (3)

and        \theta_{ B / D }=\theta_{ B }-\theta_{ D }=-\theta_{ D }=\frac{T_{ B }(L / 4)}{G J}=\frac{T_{ B } L}{4 G J}           (4)

Adding Eqs. (2)–(4), we get

\begin{array}{r} \left\lgroup \frac{T_{ A }}{4}+\frac{T_{ B }+T}{2}+\frac{T_{ B }}{4} \right\rgroup \left\lgroup \frac{L}{G J} \right\rgroup =0 \\ T_{ A }+2 T_{ B }+2 T+T_{ B }=0 \end{array}

or        T_{ A }+3 T_{ B }=-2 T            (5)

From Eqs. (1) and (5), we obtain

-4 T_{ B }=5 T \Rightarrow T_{ B }=-\frac{5 T}{4} \Rightarrow T_{ A }=\frac{7 T}{4}

Now strain energy of the entire shaft is

U=\frac{T_{ A }^2(L / 4)}{2 G J}+\frac{\left(T_{ B }+T\right)^2(L / 2)}{2 G J}+\frac{T_{ B }^2(L / 4)}{2 G J}

Putting T_{ A }=2 T \text { and } T_{ B }=-T , we get

U=\frac{(49 / 16) T^2(L / 4)}{2 G J}+\frac{(T / 4)^2(L / 2)}{2 G J}+\frac{(5 T / 4)^2 L}{8 G J}

=\left\lgroup \frac{49}{128}+\frac{1}{64}+\frac{25}{128} \right\rgroup\left\lgroup \frac{T^2 L}{G J} \right\rgroup

or      U=\frac{19}{32} \frac{T^2 L}{G J}