Question 3.9: The bar ACB shown in Figs. 3-37a and b is fixed at both ends...

Equation of equilibrium. The load T_{0} produces reactions T_{A} and T_{B} at the fixed ends of the bar, as shown in Figs. 3-37a and b. Thus, from the equilib-rium of the bar we obtain

T_{A}+T_{B}=T_{0}   (f)

Because there are two unknowns in this equation (and no other useful equations of equilibrium), the bar is statically indeterminate.
Equation of compatibility. We now separate the bar from its support at end B and obtain a bar that is fixed at end A and free at end B (Figs. 3-37c and d). When the load T_{0} acts alone (Fig. 3-37c), it produces an angle of twist at end B that we denote as \phi_{1}. Similarly, when the reactive torque T_{B} acts alone, it produces an angle \phi_{2} (Fig. 3-37d). The angle of twist at end B in the original bar, equal to the sum of \phi_{1} and \phi_{2}, is zero. Therefore, the equation of compatibility is

\phi_{1}+\phi_{2}=0    (g)

Note that \phi_{1} and \phi_{2} are assumed to be positive in the direction shown in the figure.
Torque-displacement equations. The angles of twist \phi_{1} and \phi_{2} can be expressed in terms of the torques T_{0} and T_{B} by referring to Figs. 3-37c and d and using the equation \phi = TL/GI_P. The equations are as follows:

\phi_{1}=\frac{T_{0} L_{A}}{G I_{p A}} \quad \phi_{2}=-\frac{T_{B} L_{A}}{G I_{p A}}-\frac{T_{B} L_{B}}{G I_{p B}}  (h,i)

The minus signs appear in Eq. (i) because T_{B} produces a rotation that is opposite in direction to the positive direction of \phi_{2} (Fig. 3-37d).
We now substitute the angles of twist [Eqs. (h) and (i)] into the com-patibility equation [Eq. (g)] and obtain

or  \frac{T_{B} L_{A}}{I_{p A}}+\frac{T_{B} L_{B}}{I_{p B}}=\frac{T_{0} L_{A}}{I_{p A}}    (j)

Solution of equations. The preceding equation can be solved for the torque T_{B} , which then can be substituted into the equation of equilibrium [Eq. (f)] to obtain the torque T_{A}. The results are

T_{A}=T_{0}\left(\frac{L_{B} I_{p A}}{L_{B} I_{p A}+L_{A} I_{p B}}\right) \quad T_{B}=T_{0}\left(\frac{L_{A} I_{p B}}{L_{B} I_{p A}+L_{A} I_{p B}}\right)   (3-49a,b)

Thus, the reactive torques at the ends of the bar have been found, and the statically indeterminate part of the analysis is completed.
As a special case, note that if the bar is prismatic (I_{P A}=I_{P B}=I_{P}) the pre-ceding results simplify to

T_{A}=\frac{T_{0} L_{B}}{L} \quad T_{B}=\frac{T_{0} L_{A}}{L}    (3-50a,b)

where L is the total length of the bar. These equations are analogous to those for the reactions of an axially loaded bar with fixed ends [see Eqs. (2-13a) R_{A}=\frac{P b}{L} and (2-13b) R_{B}=\frac{P a}{L} ].
Maximum shear stresses. The maximum shear stresses in each part of bar are obtained directly from the torsion formula:

Substituting from Eqs. (3-49a) and (3-49b) gives

\tau_{A C}=\frac{T_{0} L_{B} d_{A}}{2\left(L_{B} I_{P A}+L_{A} I_{P B}\right)} \quad \tau_{C B}=\frac{T_{0} L_{A} d_{B}}{2\left(L_{B} I_{P A}+L_{A} I_{P B}\right)}      (3-51a,b)

By comparing the product L_{B} \mathrm{~d}_{A} with the product L_{A} \mathrm{~d}_{B}, we can immediately determine which segment of the bar has the larger stress.
Angle of rotation. The angle of rotation \phi_{C} at section C is equal to the angle of twist of either segment of the bar, since both segments rotate through the same angle at section C. Therefore, we obtain

\phi_{C}=\frac{T_{A} L_{A}}{G I_{P A}}=\frac{T_{B} L_{B}}{G I_{P B}}=\frac{T_{0} L_{A} L_{B}}{G\left(L_{B} I_{P A}+L_{A} I_{P B}\right)}      (3-52)

In the special case of a prismatic bar (I_{P A}=I_{P B}=I_{P}), the angle of rotation at the section where the load is applied is

\phi_{C}=\frac{T_{0} L_{A} L_{B}}{G L I_{P}}    (3-53)

This example illustrates not only the analysis of a statically indetermi-nate bar but also the techniques for finding stresses and angles of rotation. In addition, note that the results obtained in this example are valid for a bar consisting of either solid or tubular segments.