TGPSC HMWSSB Manager 2015 Paper 2 · Question 93
Question 93 of the TSPSC Manager (Engineering), HMWSSB, 2015: Paper 2 (Civil Engineering) (1 November 2015). Try it first, then open the answer and explanation.
A simply supported beam of span L, carries a UVL with intensity varying from w/unit downward at left support to w/unit upward at right support. The reactions at left and right supports are respectively:
- wL/4 upward and wL/4 downward
- wL/4 downward and wL/4 upward
- wL/6 upward and wL/6 downward
- wL/8 upward and wL/8 downward
Show answer
Answer: (3) wL/6 upward and wL/6 downward
The load is a downward triangle wL/4 acting at L/6 from A and an upward triangle wL/4 acting at 5L/6 from A, so the net load is zero; moments about A give R_B × L = (wL/4)(L/6) − (wL/4)(5L/6) = −wL²/6, so R_B = wL/6 downward and R_A = wL/6 upward.