Find the support reactions acting on the given simply supported beam of length 6 m, which is subjected to a uniformly distributed load of intensity 20 kN/m on the left half span as shown in the diagram.
- A15 kN (left), 45 kN (right)
- B45 kN (left), 15 kN (right)Correct
- C-20 kN (left), 80 kN (right)
- D30 kN (left), 30 kN (right)
Solution
Concept
Use the equilibrium equations ΣM = 0 and ΣV = 0 to find the reactions of a simply supported beam.
Formula
W = w × a; RB = W × x̄ / L; RA = W − RB
Step-by-step solution
1. UDL = 20 kN/m on the left half, length a = 3 m. 2. Total load W = 20 × 3 = 60 kN, acting at 1.5 m from the left support A. 3. Take moments about A: RB × 6 = 60 × 1.5, so RB = 15 kN. 4. ΣV = 0: RA = 60 − 15 = 45 kN. 5. Left reaction = 45 kN and right reaction = 15 kN. Answer (B).
Explanation
The UDL acts only on the left half, so its resultant is closer to the left support. The left support therefore carries more load. RA = 45 kN and RB = 15 kN.
Why the other options are wrong
A: The values are swapped. The support nearer to the load carries more. C: A negative reaction is not possible for this downward load. D: Equal reactions happen only for a symmetric load, which this is not.
Shortcut
For a UDL on half span (total W): near support = 3W/4, far support = W/4. So 3×60/4 = 45 and 60/4 = 15.
Common mistake
Taking the resultant at mid-span (3 m) instead of at the centre of the loaded length (1.5 m), or swapping the left and right values.
Correct answer: B) 45 kN (left), 15 kN (right)