Question

16
A continuous beam of uniform flexural rigidity is
fixed at A and supported over B and C such that
the fixed end moments at BA = 54.72 KNm and
at BC = -29.34 KNm . The distribution factors at
BA = 0.52 and at BC = 0.48. Calculate final
moment at B.
(1 Point)​

Answer 1

Answer:

The correct option is A Mab=2EIL(θB)−19.2,Mba=4EIL(θB)+28.8



Since at end A is fixed, θA=0,δ=0

Given

MFAB=−19.2 kN−m, MFBA=28.8kN−m

MAB=MFAB+2EIL(2θA+θB−3δL)

MAB=−19.2+2EIL×θB (∴θA=0,δ=0)

MBA=MFBA+2EIL(2θB+θA−3δL)

MBA=28.8+2EIL(2θB) (∴θA=0,δ=0)

MBA=28.8+4E

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