A unit rotation is to be caused at end A of a beam AB of length 5 m whose far end B is hinged. The moment that has to be applied at A is (EI = flexural rigidity):
Try it first: pick an option, then read the explanation.
Asked in SSC JE 05/06/2024 (9 am–11 am)
Detailed Explanation
The moment needed to rotate one end of a member through one radian is its rotational (flexural) stiffness. It depends on how the far end is held: 4EI/L when the far end is fixed, 3EI/L when the far end is hinged (pinned).
Far end hinged: \[ M = \frac{3EI}{L}\,\theta = \frac{3EI}{5}\times 1 = 0.6\,EI \]
Why “0.6 EI” Is Correct
Because B is hinged, it cannot carry a moment, so the member is "softer" than one fixed at B. Its stiffness is 3EI/L, and with L = 5 m the moment for unit rotation is 0.6 EI.
| Far-end condition | Stiffness (moment for θ = 1) | Carry-over factor | Value for L = 5 m |
|---|---|---|---|
| Fixed | 4EI/L | 1/2 | 0.8 EI |
| Hinged / pinned | 3EI/L | 0 | 0.6 EI |
| Guided (sliding), symmetric case | 2EI/L | — | 0.4 EI |
Why the Other Options Are Wrong
- B. 0.9 EI: 0.9 EI does not correspond to any standard stiffness for L = 5 m.
- C. 0.4 EI: 0.4 EI = 2EI/L, the stiffness used for a symmetric member at the axis of symmetry.
- D. 0.3 EI: 0.3 EI would be 1.5EI/L, which is not a stiffness of a prismatic member.
Key Point
Stiffness: 4EI/L (far end fixed), 3EI/L (far end hinged).
Key Concepts for Students
- Stiffness and carry-over factor are the basis of the moment distribution method.
- Carry-over factor is 1/2 to a fixed far end and 0 to a hinged far end.
- Distribution factor at a joint = member stiffness / sum of stiffnesses.
Practice more: this is question 2 in Bending and Shear Stress MCQs – Part 1. Try the whole set and check every answer instantly.