A simply supported beam of length 4 m is subjected to a udl of intensity 10 kN/m over entire span of beam. Determine the magnitude of maximum shear force acting anywhere in the beam.
- A40/6 kN
- B10 kN
- C40 kN
- D20 kNCorrect
Solution
Concept
For a simply supported beam with full-span UDL, maximum shear force occurs at the supports and equals the support reaction wL/2.
Formula
RA = RB = wL/2; SFmax = wL/2
Step-by-step solution
1. Given: L = 4 m, w = 10 kN/m. 2. Total load W = w × L = 10 × 4 = 40 kN. 3. By symmetry, RA = RB = 40/2 = 20 kN. 4. Shear force is largest at the supports, so SFmax = 20 kN. 5. Answer is (D) 20 kN.
Explanation
The total load of 40 kN is shared equally by the two supports, so each reaction is 20 kN. Shear force is largest at the supports and becomes zero at mid-span. So the maximum shear force is 20 kN.
Why the other options are wrong
A: 40/6 kN has no basis for this loading. B: 10 kN is just the load intensity w, not a force. C: 40 kN is the total load, not the shear at a support.
Shortcut
SFmax = wL/2 = 10 × 4/2 = 20 kN.
Common mistake
Taking the total load wL (40 kN) as the maximum shear force.
Correct answer: D) 20 kN