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## Q. 3.SP.5

A steel shaft and an aluminum tube are connected to a fixed support and to a rigid disk as shown in the cross section. Knowing that the initial stresses are zero, determine the maximum torque $T _{0}$ that can be applied to the disk if the allowable stresses are 120 MPa in the steel shaft and 70 MPa in the aluminum tube. Use G = 77 GPa for steel and G = 27 GPa for aluminum. ## Verified Solution

Statics. Free Body of Disk. Denoting by $T _{1}$ the torque exerted by the tube on the disk and by $T _{2}$ the torque exerted by the shaft, we find

$T_{0}=T_{1}+T_{2}$                        (1)

Deformations. Since both the tube and the shaft are connected to the rigid disk, we have

$\begin{gathered}\phi_{1}=\phi_{2}: \quad \frac{T_{1} L_{1}}{J_{1} G_{1}}=\frac{T_{2} L_{2}}{J_{2} G_{2}} \\\frac{T_{1}(0.5 m )}{\left(2.003 \times 10^{-6} m ^{4}\right)(27 GPa )}=\frac{T_{2}(0.5 m )}{\left(0.614 \times 10^{-6} m ^{4}\right)(77 GPa )}\end{gathered}$

$T_{2}=0.874 T_{1}$                      (2)

Shearing Stresses. We assume that the requirement $\tau_{\text {alum }} \leq 70$ MPa is critical. For the aluminum tube, we have

$T_{1}=\frac{\tau_{\text {alum }} J_{1}}{c_{1}}=\frac{(70 MPa )\left(2.003 \times 10^{-6} m ^{4}\right)}{0.038 m }=3690 N \cdot m$

Using Eq. (2), we compute the corresponding value $T _{2}$ and then find the maximum shearing stress in the steel shaft.

$\begin{gathered}T_{2}=0.874 T_{1}=0.874(3690)=3225 N \cdot m \\\tau_{\text {steel }}=\frac{T_{2} c_{2}}{J_{2}}=\frac{(3225 N \cdot m )(0.025 m )}{0.614 \times 10^{-6} m ^{4}}=131.3 MPa\end{gathered}$

We note that the allowable steel stress of 120 MPa is exceeded; our assumption was wrong. Thus the maximum torque $T _{0}$ will be obtained by making $\tau_{\text {steel }}=120 MPa$. We first determine the torque $T _{2}$.

$T_{2}=\frac{\tau_{\text {steel }} J_{2}}{c_{2}}=\frac{(120 MPa )\left(0.614 \times 10^{-6} m ^{4}\right)}{0.025 m }=2950 N \cdot m$

From Eq. (2), we have

$2950 N \cdot m =0.874 T_{1} \quad T_{1}=3375 N \cdot m$

Using Eq. (1), we obtain the maximum permissible torque

\begin{aligned}T_{0}=T_{1}+T_{2}=3375 N \cdot m +2950 N \cdot m \\T_{0}=6.325 kN \cdot m\end{aligned}   