TGPSC HMWSSB Manager 2015 Paper 2 · Question 129
Question 129 of the TSPSC Manager (Engineering), HMWSSB, 2015: Paper 2 (Civil Engineering) (1 November 2015). Try it first, then open the answer and explanation.
Three members AB, BC and BD 3m, 3m and 4m long each, with uniform flexural rigidity are rigidly connected at B. If the supports at A and C are fixed while at D it is hinged the distribution factors in directions of BA, BC and BD at the joint B are ______ respectively.
- 3/11, 4/11, 4/11
- 4/11, 3/11, 4/11
- 4/11, 4/11, 3/11
- 1/3, 1/3, 1/3
Show answer
Answer: (4) 1/3, 1/3, 1/3
The key takes the three members as sharing the moment at B equally (1/3 each). Working it out with stiffnesses 4EI/3 (BA), 4EI/3 (BC) and 3EI/4 (BD, far end hinged) gives 16/41, 16/41 and 9/41, which is not among the options, while option (3) would follow if D were treated as fixed (4EI/4). (Official final key; some sources differ.)