A fixed-fixed beam ‘AB’ of length 3 m is subjected to a point load of 45 kN at a distance 2 m from left support ‘A’. What are the vertical reaction forces at both the supports ‘A’ and ‘B’ ?
- ARA = 15 kN and RB = 30 kN
- BRA = 15 kN and RB = 11·67 kN
- CRA = 33·33 kN and RB = 30 kN
- DRA = 11·67 kN and RB = 33·33 kNCorrect
Solution
Concept
Reactions of a fixed beam with an eccentric point load
Formula
RA = W b² (3a + b) / L³; RB = W a² (a + 3b) / L³
Step-by-step solution
1. a = 2 m, b = 1 m, L = 3 m, W = 45 kN. 2. RA = 45 × 1² × (6 + 1) / 27 = 11.67 kN. 3. RB = 45 × 2² × (2 + 3) / 27 = 33.33 kN. (Check: 11.67 + 33.33 = 45 kN.)
Explanation
In a fixed beam the end moments change the reactions from the simply supported values. The support nearer to the load carries more than its simply supported share, which is 30 kN.
Why the other options are wrong
Options (a), (b) and (c) do not add up to 45 kN, which violates vertical equilibrium.
Shortcut
Reaction at A uses b² (3a + b); at B uses a² (a + 3b). They must sum to W.
Common mistake
Using simply supported reactions (15 and 30 kN).
Correct answer: D) R_A = 11·67 kN and R_B = 33·33 kN