Bending moment (M) and Torque (T) are applied on a solid circular shaft. If maximum bending stress equals to maximum shear stress developed, then M is equal to ..........
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For shafts under both bending and torsion, equating the maximum bending stress to the maximum shear stress allows you to relate the applied bending moment and torque.
For a solid circular shaft subjected to both bending moment (M) and torque (T), the maximum shear stress due to the torque is given by the formula:
\[
\tau_{\text{max}} = \frac{T \cdot c}{J}
\]
Where:
- \( T \) is the applied torque,
- \( c \) is the radius of the shaft,
- \( J \) is the polar moment of inertia for a solid circular shaft, which is \( J = \frac{\pi d^4}{32} \).
The maximum bending stress is given by:
\[
\sigma_{\text{max}} = \frac{M \cdot c}{I}
\]
Where:
- \( M \) is the applied bending moment,
- \( c \) is the radius of the shaft,
- \( I \) is the second moment of area for a solid circular shaft, which is \( I = \frac{\pi d^4}{64} \).
Now, according to the problem, the maximum bending stress equals the maximum shear stress. Therefore, we can set the two equations equal to each other:
\[
\frac{M \cdot c}{I} = \frac{T \cdot c}{J}
\]
Substituting the values of \( I \) and \( J \):
\[
\frac{M}{\frac{\pi d^4}{64}} = \frac{T}{\frac{\pi d^4}{32}}
\]
Simplifying:
\[
M = \frac{T}{2}
\]
Thus, the bending moment \( M \) is equal to \( \frac{T}{2} \).
