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Chapter 3

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.

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 T0 that can be applied to the disk if the allowable

Step-by-Step

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}

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 T0 that can be applied to the disk if the allowable
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 T0 that can be applied to the disk if the allowable
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 T0 that can be applied to the disk if the allowable