written 6.5 years ago by | • modified 3.2 years ago |
Subject: Structure Analysis 2
Topic: Slope Diflection method
Difficulty: Hard
written 6.5 years ago by | • modified 3.2 years ago |
Subject: Structure Analysis 2
Topic: Slope Diflection method
Difficulty: Hard
written 6.5 years ago by |
FEM's (Fixed End Moments)
AB=Mab=−wl212=−40∗3212=−30KN.m
AB=Mba=wl212=40∗3212=30.m
BC=Mbc=−wl8=−100∗28=−25KN.m
BC=Mcb=wl8=100∗28=25KN.m
CD=Mcd=−wl212=−50∗3212=−37.5KN.m
AB=Mdc=wl212=50∗3212=37.5.m
2) Slop deflection equation:
Span AB:
Mab=Mab+2EIl(2Qa+Qb−3γl)
=−30+2(7000)3(Qb−3∗2.53000)
=−41.667+140003Qb
Mba=Mba+2EIl(2Qb+Qc+3γl)
=30+2(7000)2(2Qb+0−3∗2.53000)
=18.333+280003Qb
Span BC:
Mbc=Mbc+2EIl[2Qb+Qc+3γl]
=−25+2(7000)2[2Qb+Qc+3∗2.52000
=1.25+14000Qb+7000Qc
Mcd=Mcd+2EIl[2Qc+Qb+3γl]
=25+2(7000)2[2Qc+Qb+3∗2.53000
=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