For a fixed beam AB of length L, what moments MA and MB are induced when the right support sinks by Δ (EI = flexural rigidity)?
Try it first: pick an option, then read the explanation.
Correct answer: A. MA = 6EIΔ/L², MB = 6EIΔ/L²
Asked in SSC JE 30/10/2020 (Morning)
Detailed Explanation
When B sinks but neither end may rotate, the beam takes an S-shape. By antisymmetry the end moments are equal in size and the BMD is linear, crossing zero at mid-span.
\[ \text{Slope-deflection: } M_{AB} = M_{BA} = -\frac{6EI\Delta}{L^2}\quad(\theta_A = \theta_B = 0) \qquad R = \frac{12EI\Delta}{L^3} \]
Why “MA = 6EIΔ/L², MB = 6EIΔ/L²” Is Correct
Both end moments are 6EIΔ/L². (If the far end were hinged instead, the moment at the fixed end would be 3EIΔ/L².)
Why the Other Options Are Wrong
- B. MA = 3EIΔ/L², MB = 6EIΔ/L²: 3EIΔ/L² applies only at a fixed end when the other end is hinged; here both are fixed.
- C. MA = 3EIΔ/L², MB = 3EIΔ/L²: 3EIΔ/L² is the propped-cantilever value.
- D. MA = 6EIΔ/L², MB = 3EIΔ/L²: Both ends are fixed, so the moments are equal.
Key Point
Settlement Δ, both ends fixed: M = 6EIΔ/L² at each end; one end hinged: 3EIΔ/L² at the fixed end.
Key Concepts for Students
- Shear induced = 12EIΔ/L³.
- Settlement moments are added to load FEMs.
- Stiffer (larger EI) beams attract larger settlement moments.
Practice more: this is question 6 in Fixed Beam MCQs – Part 1. Try the whole set and check every answer instantly.