A simply supported beam of 9 m span carries a point load of 30 kN at 2 m from the left end of the support. It also carries a UDL of 12 kN/m over 4 m span from the right support. What is the value of shear force at 5 m from the left support?
- A4 kNCorrect
- B44 kN
- C3 kN
- D33 kN
Solution
Concept
Find the support reactions, then SF at a section = algebraic sum of vertical forces on one side.
Formula
ΣMA = 0; SFx = RA − loads left of x
Step-by-step solution
1. UDL = 12 × 4 = 48 kN, acting from x = 5 m to 9 m. Its resultant acts at x = 7 m. 2. Take ΣMA = 0: RB × 9 = 30 × 2 + 48 × 7 = 60 + 336 = 396, so RB = 44 kN. 3. RA = (30 + 48) − 44 = 34 kN. 4. SF at x = 5 m (from left) = 34 − 30 = 4 kN. 5. Check from the right: 48 − 44 = 4 kN. OK.
Explanation
First find the reactions using moment equilibrium: RA = 34 kN and RB = 44 kN. At 5 m from the left, only RA and the 30 kN load act on the left side. So SF = 34 − 30 = 4 kN.
Why the other options are wrong
B: 44 kN is RB, not the SF at 5 m. C: 3 kN is an arithmetic error. D: 33 kN does not match any correct step.
Shortcut
SF at the start of the UDL = RA − 30 = 4 kN.
Common mistake
Taking the UDL resultant from the left end (at 2 m) instead of at 7 m from the left.
Correct answer: A) 4 kN