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Question

Using slope deflection method, the end rotations at A and B for a horizontal member 4m long with flexural regidity EI are found as EI/100 and - EI/2000 respectively. The beam carries a 80 kN vertical downward load of 1m from A. The support moments at A and B are (assuming clockwise moments and rotations as positive)
taking EI=5×104 N mm2 are respectively.

A
-45 kN.m, +15 kN.m
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B
-7.5 kN.m, +15 kN.m
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C
+30 kN.m, -15 kN.m
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D
+15 kN.m, -15 kN.m
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Solution

The correct option is A -45 kN.m, +15 kN.m
MAB=MFAB+2EIL(2θA+θB3δL)

MFAB=80×1×324245 kN m

MFAB=80×30×1242=15 kN m

MAB=45+2EI4[2θAθA2](θB=θA2)=45+EI2(32θA)

=45+EI2(32×EI1000)

=45+34(EI1000)EI

=45+34(5×104)21000=43.12545

MBA=15+2EI4(2θB+θA)

=15+2EI4(2θA2+θA)=15 kN m

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