Question

A naturally curved steel beam AB having Young's modulus 208 GPa, area moment of inertia \( I = 26.7 \, \text{cm}^4 \) and radius \( R = 2 \, \text{m} \) is subjected to a vertical load \( P = 1000 \, \text{N} \) at B. The end A at \( \theta = 90^\circ \) is rigidly fixed. The bending strain energy of the beam (in Nm, rounded off to two decimal places) is \(\underline{\hspace{2cm}}\).
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Hint:

Use the bending strain energy formula \( U = \frac{P^2 L^2}{2EI} \) to calculate the strain energy in beams subjected to loads.

Answer 1

Correct Answer: 56.5

The strain energy in bending is given by the formula:
\[ U = \frac{P^2 L^2}{2EI} \] where:
- \( P = 1000 \, \text{N} \),
- \( L = R = 2 \, \text{m} \),
- \( E = 208 \times 10^9 \, \text{Pa} \),
- \( I = 26.7 \, \text{cm}^4 = 26.7 \times 10^{-8} \, \text{m}^4 \).
Substituting these values into the equation:
\[ U = \frac{1000^2 \times 2^2}{2 \times 208 \times 10^9 \times 26.7 \times 10^{-8}} \approx 56.58 \, \text{Nm}. \] Thus, the bending strain energy is \( \boxed{56.58} \, \text{Nm}. \)