written 6.6 years ago by | • modified 3.2 years ago |
Subject*: Structural Analysis II
Topic: Flexibility Method
Difficulty: Medium / High
written 6.6 years ago by | • modified 3.2 years ago |
Subject*: Structural Analysis II
Topic: Flexibility Method
Difficulty: Medium / High
written 6.6 years ago by | • modified 6.6 years ago |
DS
Ds=Dse+Ds;
=(5−3)−(2−7)
=2−1
=1
Dk
Dk=4
Dsi=0[For beam D_{si}=0]
Ds=Dse+Dsd
=1+0=1
Ds=1
Ds
Ds=Dse+Dsi
Dse=r−3−(m′−1)−(m′−1)
=6−3−(2−1)−(2−1)
=3−1−1=2−1=1
Dsi=0[For beam D_{si}=0]
Ds=1+0=1
Ds=1
Dk
Dk=6
Ds
Ds=Dse+Dsi
Dse=r−3−(m′−1)
5−3−(2−1)
2−1
=1
Dsi=0
Ds=1+0=1
Ds=1
Dk
Dk=5
Ds
Ds=Dse+Dsi
Dse=r−3
=7−3
=4
Dsi=0
Ds=4+0
Ds=4
Dk
Dk=1
Ds
Ds=Dse+Dsi
Dse=r−3−(m′−1)
=6−3−(2−1)
=3−1
=2 Dsi=0
Ds=2+0
=2
Ds=2
Dk
Dk=5
Ds
Ds=Dse+Dsi
Dse=r−3−(m′−1)
=6−3−(2−1)
=3−1
=2
Dsi=0
Ds=2+0
Ds=2
Dk
Ds
Ds=Dse+Dsi
Dse=r−3
=5−3=2
Dsi=3×c[c=closed loop]
=3×3
=9
Ds=9+2=11
Ds=11
Dk
Dk=13
Dse=r−3=6−3=3
Dsi=3×c=3×4=12
Ds=3+12=5
Ds=15
Dk
Dk=15
Ds
Ds=Dse+Dsi
Dse=r−3=6−3=3
Dsi=3×c
=3×2=6
Ds=3+6=9
Ds=9
Dk
Dk=9
Ds=Dse+Dsi
Dse=r−3=8−3=5
Dsi=3×c=3×2=6
Ds=5+6=11
Ds=11
Dk
Dk=9
Ds
Ds=Dse+Dsi
Dse=r−3=3−0
Dsi=3×c=3×1=3
Ds=0+3
Ds=3
Dk
Dk=9
Ds
Ds=Dse+Dsi
Dse=r−3−(m′−1)
=6−3−(3−1)
=3−2
=1
Dsi=3×c
=3×6=18
=3−2
=1
Dsi=3×c
=3×6=18
Ds=1+18=19
Ds=19
Dk
Dk=20
Ds
Ds=Dse+Dsi
Dse=r−3=4−3=1
Dsi=m−2j−3 (formula)
=10−2(6)−3
=10−12−3
=10−9
=1
Ds=2
Ds=1+1=2
Dk=2j−r
=2×6−4
=12−4
=8
Dk=8
Ds
Ds=Dse+Dsi
Dse=r−3=3−3=0
Dsi=m−2j−3
=11−2(6)−3
=11−12−3
=11−9
=2
Ds=2
Dk
Dk=2j−r
=2×6−3
=12−3
Dk=9
Ds=Dse+Dsi
Dse=r−3=5−3=2
Dsi=M−2j−3
=8−(2×5)−3
=8−10−3
=8−7
=1
Ds=2+1=3
Ds=3
Dk
Dk=2j−r
=2×5−5
=10−5
=5
Dk=5
Ds
Dse=r−3=4−3−1
Dsi=m−2j−3
=5−(2×4)−3
=5−8−3
=5−5=0
Ds=1+0
Ds=1
Dk
Dk=2j−r
=2×4−4
=8−4
Dk=4
Dse=r−3=4−3=1
Dsi=m−2j−3
=15−2×8−3
=15−16−3
=15−13
=2
Ds=1+2=3
Ds=3
Dk=2j−r
=2×8−4
=16−4
=12
Dk=12
Ds
Ds=Dse+Dsi
Dse=r−3
=7−3=4
Dsi=3×c=3×0=0
Ds=4+0=4
Ds=4
Dk
Dk=2
Ds
Ds=Dse+Dsi
Dse=r−3=4−3=1
Dsi3×c=0
Ds=1+0
Ds=1
Dk
Dk=4
Ds
Ds=Dse+Dsi
Dse=r−3=3−3=0
Dsi=3×c=3×0=0
Ds=0+0=0
Ds=0
Dk
Dk=3
Ds
Ds=Dse+Dsi
Dse=r−3=9−3=6
Dsi=3×c=0
Ds=6
Dk
Dk=4