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 the 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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For thin-walled shafts subjected to both axial and torsional loading, use the maximum principal stress criterion to combine the effects of both stresses.
The maximum principal stress criterion is used to determine the minimum shaft thickness. The total stress is composed of both the axial and torsional components.
1. The axial stress is given by:
\[
\sigma_{axial} = \frac{P}{A} = \frac{7.4 \, \text{kN}}{\pi \left( 0.02^2 \right)} = \frac{7400}{\pi \times 0.0004} \approx 5.91 \times 10^6 \, \text{Pa} = 5.91 \, \text{MPa}.
\]
2. The torsional stress is:
\[
\sigma_{torsion} = \frac{T}{J} \cdot r = \frac{148 \, \text{Nm}}{\frac{\pi t^3}{3}} \times 0.02.
\]
3. Using the maximum principal stress criterion, the total stress:
\[
\sigma_{total} = \sqrt{\sigma_{axial}^2 + \sigma_{torsion}^2}.
\]
Equating this total stress to the allowable stress of 100 MPa:
\[
100 = \sqrt{(5.91)^2 + \left( \frac{148}{\frac{\pi t^3}{3}} \times 0.02 \right)^2}.
\]
Solving for \( t \), we find the minimum thickness to be:
\[
\boxed{1 \, \text{mm}}.
\]
