GATE 2022 AE – Question 59
A 0.5 m long thin-walled circular shaft of radius 2 cm is to be designed for an axial load of 7.4 kN and a torque of 148 Nm applied at its tip, as shown in the figure. The allowable stress under uniaxial tension is 100 MPa. Using maximum principal stress criterion, the minimum thickness, t, of the shaft so that it does not fail is _____ mm (rounded off to the nearest integer).
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Correct answer: 1
Explanation
The shaft is thin-walled, with mean radius $r=2$ cm and thickness $t$.
**Stresses.**
- Axial stress from the load $F=7.4$ kN: $\sigma=\dfrac{F}{2\pi r t}$.
- Shear stress from the torque $T=148$ N m: $\tau=\dfrac{T}{2\pi r^2t}$.
Numerically (with $t$ in metres):
$$\sigma=\frac{7400}{2\pi(0.02)\,t}=\frac{58\,892}{t}\ \text{Pa},\qquad\tau=\frac{148}{2\pi(0.02)^2t}=\frac{58\,892}{t}\ \text{Pa}.$$
(They happen to be equal.)
**Maximum principal stress criterion:**
$$\sigma_1=\frac\sigma2+\sqrt{\left(\frac\sigma2\right)^2+\tau^2}\leq100\text{ MPa}.$$
With $\sigma=\tau=s$: $\sigma_1=s\left(\tfrac12+\sqrt{\tfrac14+1}\right)=1.618\,s$.
$$1.618\,s=100\times10^6\;\Rightarrow\;s=61.8\text{ MPa}=\frac{58\,892}{t}\;\Rightarrow\;t=0.953\times10^{-3}\ \text{m}.$$
The minimum thickness is 0.953 mm, which rounds to **1 mm**.