Question

A propped cantilever beam of span 4 m is fixed at end A and simply supported at end B. The beam is subjected to a uniformly distributed load of 5 kN/m. Then the reactions at A and B respectively are ............

• A. 12.5 kN and 7.5 kN
• B. 5 kN and 15 kN
• C. 7.5 kN and 12.5 kN
• D. 15 kN and 5 kN

Hint:

For a propped cantilever beam subjected to a uniformly distributed load, the reaction at the fixed end is higher than the reaction at the simply supported end. The reactions can be found using equilibrium equations of forces and moments.

Answer 1

The Correct Option is A

To solve for the reactions at the supports, we can use the equations of static equilibrium:
- Sum of forces in the vertical direction = 0
- Sum of moments about any point = 0
Given:
- Length of the beam = 4 m
- Uniformly distributed load = 5 kN/m
- Total load = \( 5 \, \text{kN/m} \times 4 \, \text{m} = 20 \, \text{kN} \)
First, calculate the reactions at the supports:
- Let \( R_A \) be the reaction at support A (fixed end)
- Let \( R_B \) be the reaction at support B (simply supported end)
The total load is \( 20 \, \text{kN} \). By symmetry and using equilibrium equations, we get: \[ R_A = \frac{3}{4} \times 20 = 15 \, \text{kN} \] \[ R_B = \frac{1}{4} \times 20 = 5 \, \text{kN} \] Thus, the reactions are \( R_A = 12.5 \, \text{kN} \) and \( R_B = 7.5 \, \text{kN} \).