The three-moment equation was originally developed by:
Try it first: pick an option, then read the explanation.
Correct answer: A. Clapeyron
Asked in SSC JE 28/10/2020 (Evening)
Detailed Explanation
For a continuous beam, the three-moment equation links the bending moments at three consecutive supports using the slope-continuity condition over the middle support.
For constant EI and no settlement: \[ M_A L_1 + 2M_B(L_1 + L_2) + M_C L_2 = -\frac{6A_1\bar{x}_1}{L_1} - \frac{6A_2\bar{x}_2}{L_2} \]
Why “Clapeyron” Is Correct
The theorem of three moments is due to Clapeyron (1857).
| Name | Known for |
|---|---|
| Clapeyron | Three-moment equation (continuous beams) |
| Euler | Buckling load of columns |
| Mohr | Mohr's circle; moment-area theorems |
| Hardy Cross | Moment distribution method |
Why the Other Options Are Wrong
- B. Euler: Euler is known for the critical buckling load of columns.
- C. Mohr: Mohr gave Mohr's circle and the moment-area theorems.
- D. Hardy Cross: Hardy Cross developed the moment distribution method (1930).
Key Point
Three-moment theorem → Clapeyron.
Key Concepts for Students
- The equation is written once for each interior support.
- End simple supports have zero moment; fixed ends are handled with an imaginary zero-length span.
- 6Ax̄/L terms come from the free BMD of each span.
Practice more: this is question 24 in Slope and Deflection MCQs – Part 2. Try the whole set and check every answer instantly.