Find the total twist at a distance z in each of the members in Figs. 4.16 and 4.17.

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Accepted Answer

[latex] \Theta (z)-\Theta (0)=\int_{0}^{z}{d heta }, d heta =\frac{T}{JG}dz.[/latex]

The first shaft is homogeneous and acted upon by a constant torque; it does not have a constant cross-sectional area however. The area changes from [latex]A_{1}[/latex] to [latex]A_{2}[/latex] at a length of L/2 from the wall. Therefore, J=J(z) and the angle of twist becomes

[latex]\Theta (L)-\Theta (0)=\int_{0}^{L}{\frac{T}{J(z)G} }dz=\frac{T}{G}\int_{0}^{L/2}{\frac{1}{J_{1}} }dz +\frac{T}{G}\int_{L/2}^{L}{\frac{1}{J_{2}} }dz.[/latex]

FIGURE 4.16 Two idealized circular cylinders of length L are acted upon by a single, constant end torque T. The cylinder on the left has a nonconstant cross section, whereas the one on the right is nhomogeneous in composition. FIGURE 4.17 A LEHI circular cylinder subjected to multiple applied torques. A free-body diagram of the whole allows the reaction support [latex]T_{w}[/latex] to be determined; free-body diagrams of judiciously selected parts allows internal torques to be determined as a function of z. Remember that judicious cuts are typically those taken between abrupt changes in applied loads. Because the integral was broken into a sum of integrals for the discontinuity in cross-sectional area, each new integral contains terms that are constant along the range of integration and can be moved outside the integral and evaluated. Given that the twist at the fixed end is zero [latex][i.e., \Theta (0)=0],[/latex] we have

[latex]\Theta (L)-\Theta (0)=\frac{T}{J_{1}G}\int_{0}^{L/2}{dz}+\frac{T}{J_{2}G}\int_{L/2}^{L}{dz} ightarrow \Theta (L)=\frac{TL}{2J_{1}G}+\frac{TL}{2J_{2}G}.[/latex]

The second shaft has a constant cross-sectional area and is acted on by a constant torque; it is not homogeneous however. The material properties change at a distance of L/2 from the wall. Therefore, G=G(z) and the twist becomes

[latex] \Theta (L)-\Theta (0)=\int_{0}^{L}{\frac{T}{JG(z)} }dz=\frac{T}{J}\int_{0}^{L/2}{\frac{1}{G_{1}} }dz+\frac{T}{J}\int_{L/2}^{L}{\frac{1}{G_{2}} }dz,[/latex]

or

[latex] \Theta (L)=\frac{TL}{2J}\left(\frac{1}{G_{1}}+\frac{1}{G_{2}} ight).[/latex]

The third shaft is homogeneous and has a constant cross-sectional area; it is not under a constant loading however. The applied load changes at a distance of L/2 from the wall; thus, T=T(z). Before we solve for the twist at the end of the shaft, we must determine the internal torques at each z. From equilibrium of the whole (Fig. 4.17b), we see that the reaction torque at the wall [latex]T_{w}[/latex] must balance the combined effects of the 2T and the T that are applied at z=L/2 and z=L, respectively. Equilibrium of parts (note: when we have discrete changes in loads, geometry, or properties, judicious cuts are those between the abrupt changes) reveals further that the left half has an internal torque 3T and the right half only T. Hence, the end twist becomes

[latex]\Theta (L)-\Theta (0)=\frac{3T}{JG}\int_{0}^{L/2}{dz}+\frac{T}{JG}\int_{L/2}^{L}{dz} ightarrow \Theta (L)=\frac{3TL}{2JG}+\frac{T}{JG}\left(\frac{L}{2} ight)=2\frac{TL}{JG}.[/latex]

Question 4.5: Find the total twist at a distance z in each of the members ...

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