The ratio of the maximum deflection of a simply supported beam with a central load W to that of a cantilever of the same length with a load W at its free end is:
Try it first: pick an option, then read the explanation.
Correct answer: D. 1/16
Asked in SSC JE 27/01/2018 (Morning)
Detailed Explanation
Divide the two standard maximum deflections; W, L and EI cancel.
\[ \frac{\delta_{ss}}{\delta_{cant}} = \frac{WL^3/48EI}{WL^3/3EI} = \frac{3}{48} = \frac{1}{16} \]
Why “1/16” Is Correct
The ratio is 1/16 — the simply supported beam is 16 times stiffer for this loading.
| Beam and loading | Maximum slope | Maximum deflection | Where |
|---|---|---|---|
| Cantilever, point load W at free end | WL²/2EI | WL³/3EI | Free end |
| Cantilever, UDL w over full length | wL³/6EI | wL⁴/8EI | Free end |
| Cantilever, triangular load (w at fixed end, 0 at free end) | wL³/24EI | wL⁴/30EI | Free end |
| Cantilever, couple M at free end | ML/EI | ML²/2EI | Free end |
| Simply supported, central point load W | WL²/16EI | WL³/48EI | Slope at supports, deflection at centre |
| Simply supported, UDL w over full span | wL³/24EI | 5wL⁴/384EI | Slope at supports, deflection at centre |
| Simply supported, equal end couples M (pure bending) | ML/2EI | ML²/8EI | Slope at supports, deflection at centre |
Standard results (EI constant). In a cantilever the slope and deflection are zero at the fixed end and maximum at the free end.
Why the Other Options Are Wrong
- A. 1/8: 1/8 has no basis here.
- B. 1/10: 1/10 has no basis here.
- C. 1/12: 1/12 has no basis here.
Key Point
δ(SS, central W) : δ(cantilever, tip W) = 1 : 16.
Key Concepts for Students
- For a UDL the ratio is (5/384)/(1/8) = 5/48.
- Each half of an SS beam is a cantilever of L/2 with load W/2.
- Stiffness ratio is the inverse: 48EI/L³ vs 3EI/L³.
Practice more: this is question 45 in Slope and Deflection MCQs – Part 3. Try the whole set and check every answer instantly.