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Question

The members shown in the diagram below are to a moment 50 kNm at joint E, the moments carried by members EA and EB are x and y respectively.
Then, 2x+y =_____kNm.

  1. 66.3

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Solution

The correct option is A 66.3
MemberStiffnessDistributionMoment(kNm)=D.F.×50EA4E(2I)3=8EI3=2.67EI2.672.67+2.25+0.8=0.46623.3EB3E(1.5I)2=2.25EI2.252.67+2.25+0.8=0.39419.7EC4EI5=0.8EI0.82.67+2.25+0.8=0.147ED0
x=23.3y=19.72x+y=2×23.3+19.7=66.3kNm

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