Question 4.6: Consider the simple shaft shown in Fig. 4.19, which has unif...

Because the shaft is fixed at both ends, the problem is statically indeterminate. If we let the end torques be denoted by T_{A} and T_{C}, overall equilibrium requires that T_{\omicron}  + T_{C} + T_{A}=0 (where T_{\omicron} is the known, applied torque). We need another equation to find the reactions however. Note, there-fore, that

where \Theta =0 at both  z=0 and z=L. Moreover, \Theta(L/2) is but a single value; thus,

and therefore, having two equations and two unknowns, we can solve for the two reactions FIGURE 4.19 Statically indeterminate shaft, fixed on both ends, and subjected to a single applied torque   T_{\circ} at z=L/2.   Free-body diagrams of the whole structure and the parts allow the reaction and internal torques to be isolated but not determined because we have only one nontrivial equation (the sum of the twisting moments equals zero) for the two reaction torques T_{A} and T_{C}. There is, therefore, a need for an additional equation.

As a special case, note that if J_{1}=J_{2}(i.e., the shaft has a constant cross section), then T_{C}=-T_{\omicron }/2 and T_{A}=-T_{\omicron }/2, as expected.