Question

If the length of a simply supported beam carrying a concentrated load at the center is doubled, the deflection at the center will become .......

• A. Two times
• B. Four times
• C. Eight times
• D. Sixteen times

Hint:

For beams under concentrated loads, the deflection at the center is directly proportional to the cube of the length. Doubling the length results in an 8-fold increase in deflection.

Answer 1

The Correct Option is C

For a simply supported beam under a concentrated load at the center, the deflection \( \delta \) at the center is given by the formula: \[ \delta = \frac{P L^3}{48 E I} \] Where:
- \( P \) is the load applied at the center,
- \( L \) is the length of the beam,
- \( E \) is the modulus of elasticity of the material,
- \( I \) is the moment of inertia of the beam's cross-section.
If the length of the beam is doubled, the new deflection \( \delta' \) becomes: \[ \delta' = \frac{P (2L)^3}{48 E I} = \frac{P \cdot 8L^3}{48 E I} = 8 \times \frac{P L^3}{48 E I} \] Therefore, the deflection at the center becomes 8 times the original deflection.