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Analysis the Continuous beam loaded and supported as shown in fig 1)Using clapeyron's theorem of three moments method Note that support C settles by 8mm during loading Take EI=1600kN/M2

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Subject: Structural Analysis II

Topic: Flexibility Method

Difficulty: Medium / High

1 Answer
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1)Free Bending moment

For AB, BM at E=wl28=24×328=27kN.m

For BC enter image description here    ML=7.5×2=15kN

MR=7.5×2

=15kN.M

For CD BM at G=wabl=15×2×35=30kN.m

2) Free Bending moment diagram

enter image description here

3)Three moment theorem

For A1AB

MA1(l1)+2MA(l1+l2)+MB(l2)+6A1x1l1+6A2x2l2=0

0+2MA(0+3)+MB(3)+6×0×006×54×1.53=0

6MA+3MB=162_

For A-B-C

MA(l1)+2MB(l1+l2)+Mc(l2)+5A1x1l1+6A2x2l2+61E1x1l1+62E2x2l2=0

A2x2={12×15×2×(23×2)}{12×15×2×(2+13×2)}

=2040

=20

1=BA

=00

=0

2=BC

$ …

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