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Determine the supports moments for the continous girder shown in fig. if the support B sinks by 2.50mm For all members I=3.50107mm4,E=200KN/mm2.

Subject: Structure Analysis 2

Topic: Slope Diflection method

Difficulty: Hard

enter image description here

1 Answer
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FEM's (Fixed End Moments)

AB=Mab=wl212=403212=30KN.m

AB=Mba=wl212=403212=30.m

BC=Mbc=wl8=10028=25KN.m

BC=Mcb=wl8=10028=25KN.m

CD=Mcd=wl212=503212=37.5KN.m

AB=Mdc=wl212=503212=37.5.m

2) Slop deflection equation:

Span AB:

Mab=Mab+2EIl(2Qa+Qb3γl)

=30+2(7000)3(Qb32.53000)

=41.667+140003Qb

Mba=Mba+2EIl(2Qb+Qc+3γl)

=30+2(7000)2(2Qb+032.53000)

=18.333+280003Qb

Span BC:

Mbc=Mbc+2EIl[2Qb+Qc+3γl]

=25+2(7000)2[2Qb+Qc+32.52000

=1.25+14000Qb+7000Qc

Mcd=Mcd+2EIl[2Qc+Qb+3γl]

=25+2(7000)2[2Qc+Qb+32.53000

=51.25+7000Qb+14000Qc

Span CD:

Mcd=Mcd+2EIl[2Qc+Qd]

=37.50+2(7000)3[2Qc]

=37.50+280003Qc

Mdc=Mdc+2EIl[2Qd+Qc]

=37.50+2(7000)3[Qc]

=37.50+140003Qc

3) Apply Equilibrium Condition for joints:

Joint B:

Mba+Mbc=0

(18.3333+280003Qb)+(1.25+14000Qb+7000Qc)=0

700003Qb+7000Qc=19.5833

10QB+3Qc=0.008392848429-----------(i)

Joint C:

Mcb+Mcd=0

(51.25+7000Qb+14000Qc)37.50+280003Qc)=0

7000Qb+70003Qc=13.75

3Qb+10Qc=0.0058928571-----------(i)

Solving (i) & (ii) we get

Qb=0.000728

Qc=0.000370879

Substituting value of Qb & Qc in moment equation (i) to (iii)

Mab=41.6667+140003(0.000728)=45.11KN.m

Mba=11.54KN.m

Mbc=11.5KN.m

Mcb=40.96

Mcd=40.96

Mdc=35.77

enter image description here

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