A simply supported beam of length L carries two equal and opposite (unlike) couples M at its two ends, so that the beam is in pure bending. If EI is constant, the central deflection is:
Try it first: pick an option, then read the explanation.
Correct answer: D. ML²/(8EI)
Asked in UP Jal Nigam JE 2014; SSC JE 2008
Detailed Explanation
Equal and opposite end couples give a constant bending moment M along the whole span (pure bending). The beam bends into a circular arc of radius R = EI/M.
\[ EI\frac{d^2y}{dx^2} = -M \Rightarrow EIy = -\frac{Mx^2}{2} + \frac{ML}{2}x \quad (y(0) = 0,\ y'(L/2) = 0) \]
\[ y\left(\frac{L}{2}\right) = \frac{ML^2}{8EI},\qquad \theta_{end} = \frac{ML}{2EI} \]
Why “ML²/(8EI)” Is Correct
Central deflection = ML²/(8EI) — the sagitta of a circular arc of radius EI/M over a chord L (δ = L²/8R).
Why the Other Options Are Wrong
- A. ML²/(4EI): ML²/4EI is twice the correct value.
- B. ML²/(16EI): ML²/16EI is half the correct value (it is the central deflection for a single end couple M).
- C. ML²/(64EI): ML²/64EI is far too small.
Key Point
Pure bending of a simple beam: δ = ML²/8EI, θend = ML/2EI.
Key Concepts for Students
- Circular arc: δ = L²/(8R) with 1/R = M/EI.
- Single couple M at one end: central deflection ML²/16EI.
- Pure bending has zero shear force.
Practice more: this is question 32 in Slope and Deflection MCQs – Part 3. Try the whole set and check every answer instantly.