The total strain energy stored in a beam of length L, moment of inertia I, subjected to a bending moment M is:
Try it first: pick an option, then read the explanation.
Correct answer: C. ∫₀ᴸ (M²/2EI) δx
Asked in SSC JE 25/01/2018 (Evening)
Detailed Explanation
A short element dx bends through an angle dθ = M dx/EI under the moment M. The work stored is half the moment times the rotation.
\[ dU = \tfrac12 M\,d\theta = \tfrac12 M\frac{M\,dx}{EI} = \frac{M^2}{2EI}dx \qquad U = \int_0^L \frac{M^2}{2EI}\,dx \]
Why “∫₀ᴸ (M²/2EI) δx” Is Correct
The total energy is the integral over the whole length, ∫₀ᴸ (M²/2EI) dx. Option (b) is only the energy in one element, and option (a) misses the factor 1/2.
Why the Other Options Are Wrong
- A. (M²/EI) δx: Missing the factor 1/2 and the integral.
- B. (M²/2EI) δx: This is the energy in an elemental length δx only, not the total.
- D. none of these: Option (c) is correct.
Key Point
Bending strain energy: U = ∫ M²dx/(2EI); for constant M, U = M²L/(2EI).
Key Concepts for Students
- Axial: U = P²L/(2AE); torsion: U = T²L/(2GJ).
- U = ½ × load × deflection for a single load.
- Castigliano: δ = ∂U/∂P.
Practice more: this is question 26 in Slope and Deflection MCQs – Part 2. Try the whole set and check every answer instantly.