Question:-
A straight bar is fixed at edges A and B. Its elastic modulus is E and cross-section is A. There is a load P = 120 N acting at C. Determine the reactions at the ends.

Option (A)
60 N at A, 60 N at B
Option (B)
30 N at A, 90 N at B
Option(C)
40 N at A, 80 N at B
Option(D)
80 N at A, 40 N at B
Correct Option:
(D)
Question Solution:
RA = 120 x (BC/AB) = 80 N/mm2
RB = 120 x AC/AB = 40 N/mm2
F.B.D.

R1 – R2 = 120 ..........(1)
and (dl)1 + (dl)2 = 0
{(R1 x l) / (A x E)} + {(R2 x 2l) / (A x E)} = 0
\ R1 = -2R2 ........(2)
Form eq (1) and (2)
R2 = - 40
R2 = 40 N (opposite direction to our assumption) and R1 = 80 N
question-answer-faq-9370