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UPSC Civil Service Exam Questions

Q1:

For the beam shown in figure, what are the distribution factors at joint B ?

A 0.4, 0.6

B 0.5, 0.5

C 0.6, 0.4

D 0.65, 0.35

ANS:B - 0.5, 0.5

K1 = 4EI/L1 = .5EI
K2 = 3EI/L2 = .5EI
Sum of K1 + K2.
.5+.5=1.

Now, the factors are;
.5/1 & .5/1
ie 0.5&0.5.



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