Question

A slender elastic rod of length $1 \, m$, circular cross-section $d = 50 \, mm$, is subjected to equal and opposite end moments $M$. The midpoint lateral deflection is $10 \, mm$. Find maximum longitudinal strain in the rod, expressed as $p \times 10^{-3}$.

Hint:

Maximum strain in bending is proportional to curvature $\tfrac{1}{R}$ and distance from neutral axis.

Answer 1

Correct Answer: 1.9

Step 1: Relation of bending strain.
Maximum longitudinal strain: \[ \varepsilon_{max} = \frac{y}{R} \] where $y = c = \tfrac{d}{2} = 25 \, mm$.
Step 2: Radius of curvature from deflection.
For pure bending: \[ \delta = \frac{L^2}{8R} \] with $\delta = 10 \, mm$, $L = 1000 \, mm$. \[ R = \frac{L^2}{8\delta} = \frac{1000^2}{8 \times 10} = \frac{10^6}{80} = 12500 \, mm \]
Step 3: Strain.
\[ \varepsilon_{max} = \frac{25}{12500} = 0.002 = 2 \times 10^{-3} \] So $p = 2.0$. Final Answer: \[ \boxed{p = 2.0} \]