Question 3.9: The bar ACB shown in Figs. 3-37a and b is fixed at both ends......
The bar ACB shown in Figs. 3-37a and b is fixed at both ends and loaded by a torque T_{0} at point C. Segments AC and CB of the bar have diameters d _{A} and d _{B}, lengths L_{A} and L _{B}, and polar moments of inertia I_{pA} and I_{pB}, respectively. The material of the bar is the same throughout both segments. Obtain formulas for
(a) the reactive torques T_{A} and T_{B} at the ends,
(b) the maximum shear stresses τ _{AC} and τ _{CB} in each segment of the bar, and
(c) the angle of rotation \phi _{C} at the cross section where the load T_{0} is applied.
Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
1,2. Conceptualize, Categorize:
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. Equilibrium of the bar gives
T_{A}+ T_{B}=T_0 (a)
Because there are two unknowns in this equation (and no other useful equations of equilibrium), the bar is statically indeterminate.
Equation of compatibility: 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 denoted 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, which is equal to the sum of \phi _{1} and \phi _{2}, is zero. Therefore, the equation of compatibility is
\phi _{1}+\phi _{2}=0 (b)
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
\phi_{1}=\frac{T_0L_A}{GI_{pA}} \phi_{2}=-\frac{T_BL_A}{GI_{pA}}-\frac{T_BL_B}{GI_{pB}} (c,d)
The minus signs appear in Eq. (d) because T_{B} produces a rotation that is opposite in direction to the positive direction of \phi _{2} (Fig. 3-37d).
Now substitute the angles of twist [Eqs. (c) and (d)] into
the compatibility equation [Eq. (b)] and obtain
or
\frac{T_BL_A}{I_{pA}}+\frac{T_BL_B}{I_{pB}}=\frac{T_0L_A}{I_{pA}} (e)
3. Analyze:
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. (a)] to obtain the torque T_{A}. The results are
\hookleftarrow (3-49a,b)
T_{B}=T_{0}\left\lgroup\frac{L_AI_{pB}}{L_BI_{pA}+L_AI_{pB}}\right\rgroupThus, 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_{pA}=I_{pB}=I_{p}) , the preceding results simplify to
T_{A}=\frac{T_0L_B}{L} T_{B}=\frac{T_0L_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. (2-12a and b)].
R_{A}=\frac{Pb}{L} \ \ \ \ R_{B}=\frac{Pa}{L} (2-12a,b)
The TMD for this special case is shown in Fig. 3-37e.
Maximum shear stresses: The maximum shear stresses in each part of the bar are obtained directly from the torsion formula:
\tau_{AC}=\frac{T_Ad_A}{2I_{pA}} \tau_{CB}=\frac{T_Bd_B}{2I_{pB}}
Substituting from Eqs. (3-49a and b) gives
\tau_{AC}=\frac{T_0L_Bd_A}{2(L_BI_{pA}+L_AI_{pB})}\tau_{CB}=\frac{T_0L_Ad_B}{2(L_BI_{pA}+L_AI_{pB})} \hookleftarrow (3-51a,b)
By comparing the product L_B d_A with the product L_Ad_B, you 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, the angle of rotation is
\phi _{C}=\frac{T_AL_A}{GI_{pA}}=\frac{T_BL_B}{GI_{pB}} =\frac{T_0L_AL_B}{G(L_BI_{pA}+L_AI_{pB})} \hookleftarrow (3-52)
In the special case of a prismatic bar (I_{pA}=I_{pB}=I_{p}) , the angle of rotation at the section where the load is applied is
\phi _{C}=\frac{T_0L_AL_B}{GLI_{p}} (3-53)
The TDD for the case of a prismatic bar is shown in Fig. 3-37f.
4. Finalize: This example illustrates not only the analysis of a statically indeterminate 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.