Question 1 A solid circular shatt with diameter D is subjected to a bending moment of 40 kNm , and the permissible bending stress is not to exceed 50 MNm . Find the diameter of the shatt \( \mathrm{I}=\frac{\boldsymbol{a} \mathrm{D}^{4}}{64} \) for a solid circular shaft (A) 201 mm (B) 8.15 mm (C) 81.5 mm (D) 378 mm
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To find the diameter of the shaft, apply the bending stress formula: \[ \sigma = \frac{M c}{I} \] Where \( M \) is the bending moment, \( c \) is the distance from the neutral axis to the outer fiber (\( \frac{D}{2} \)), and \( I \) is the moment of inertia. Given \( I = \frac{\pi D^4}{64} \) and rewriting the bending stress equation, you can solve for \( D \): \[ \frac{M \cdot \frac{D}{2}}{\frac{\pi D^4}{64}} \leq 50 \text{ MN/m}^2 \] Substituting the values, you will eventually find out that the diameter \( D \) works out to be 201 mm, corresponding to option (A). For extra fun, consider this: if our bending scenario was a battle between iron and steel, the iron may have great tensile strength, but when it comes to bending, the steel often wins by being more ductile, allowing it to absorb and disperse energy better. Who knew your shaft had such a tale of strength! Curiously, in engineering design, the interplay of forces isn’t just technical; it’s much like a dance! Every structure or mechanical component has its own 'dance moves' adapted to balance loads without crumpling under the pressure, much like ensuring dancers don’t collide during a performance!
