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Determine the degree of static and kinematic indeterminacy of the following structure

Subject*: Structural Analysis II

Topic: Flexibility Method

Difficulty: Medium / High

1 Answer
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58views

enter image description here

$D_{S}$

$D_{s}=D_{se}+D_{s};$

enter image description here

=$(5-3)-(2-7)$

=$2-1$

=1

$D_{k}$

$D_{k}=4$

$D_{si}=0$[For beam D_{si}=0]

$D_{s}=D_{se}+D_{sd}$

=$1+0=1$

$D_{s}=1$

enter image description here

$D_{s}$

$D_{s}=D_{se}+D_{si}$

$D_{se}=r-3-(m^\prime-1)-(m^\prime-1)$

=$6-3-(2-1)-(2-1)$

=$3-1-1=2-1=1$

$D_{si}=0$[For beam D_{si}=0]

$D_{s}=1+0=1$

$D_{s}=1$

$D_{k}$

$D_{k}=6$

enter image description here

$D_{s}$

$D_{s}=D_{se}+D_{si}$

$D_{se}=r-3-(m^\prime-1)$

$5-3-(2-1)$

$2-1$

=1

$D_{si}=0$

$D_{s}=1+0=1$

$D_{s}=1$

$D_{k}$

$D_{k}=5$

enter image description here

$D_{s}$

$D_{s}=D_{se}+D_{si}$

$D_{se}=r-3$

=$7-3$

=4

$D_{si}=0$

$D_{s}=4+0$

$D_{s}=4$

$D_{k}$

$D_{k}=1$

enter image description here

$D_{s}$

$D_{s}=D_{se}+D_{si}$

$D_{se}=r-3-(m^\prime-1)$

=$6-3-(2-1)$

=$3-1$

=2 $D_{si}=0$

$D_{s}=2+0$

=2

$D_{s}=2$

$D_{k}$

$D_{k}=5$

enter image description here

$D_{s}$

$D_{s}=D_{se}+D_{si}$

$D_{se}=r-3-(m^\prime-1)$

=$6-3-(2-1)$

=$3-1$

=2

$D_{si}=0$

$D_{s}=2+0$

$D_{s}=2$

$D_{k}$

enter image description here

$D_{s}$

$D_{s}=D_{se}+D_{si}$

$D_{se}=r-3$

=$5-3=2$

$D_{si}=3\times c$[c=closed loop]

=$3\times3$

=9

$D_{s}=9+2=11$

$D_{s}=11$

$D_{k}$

$D_{k}=13$

enter image description here

$D_{se}=r-3=6-3=3$

$D_{si}=3 \times c =3\times 4=12 $

$D_{s}=3+12=5$

$D_{s}=15$

$D_{k}$

$D_{k}=15$

enter image description here

$D_{s}$

$D_{s}=D_{se}+D_{si}$

$D_{se}=r-3=6-3=3$

$D_{si}=3\times c$

=$3\times 2=6$

$D_{s}=3+6=9$

$D_{s}=9$

$D_{k}$

$D_{k}=9$

enter image description here

$D_{s}=D_{se}+D_{si}$

$D_{se}=r-3=8-3=5$

$D_{si}=3\times c=3\times 2=6$

$D_{s}=5+6=11$

$D_{s}=11$

$D_{k}$

$D_{k}=9$

enter image description here

$D_{s}$

$D_{s}=D_{se}+D_{si}$

$D_{se}=r-3=3-0$

$D_{si}=3\times c=3 \times 1 =3$

$D_{s}=0+3$

$D_{s}=3$

$D_{k}$

$D_{k}=9$

enter image description here

$D_{s}$

$D_{s}=D_{se}+D_{si}$

$D_{se}=r-3-(m^\prime-1)$

=$6-3-(3-1)$

=$3-2$

=1

$D_{si}=3\times c$

=$3\times 6=18$

=$3-2$

=1

$D_{si}=3\times c$

=$3\times 6=18$

$D_{s}=1+18=19$

$D_{s}=19$

$D_{k}$

$D_{k}=20$

enter image description here

$D_{s}$

$D_{s}=D_{se}+D_{si}$

$D_{se}=r-3=4-3=1$

$D_{si}=m-2j-3$ (formula)

=$10-2(6)-3$

=$10-12-3$

=$10-9$

=1

$D_{s}=2$

$D_{s}=1+1=2$

$D_{k}=2{j}-r$

=$2\times 6-4$

=$12-4$

=8

$D_{k}=8$

enter image description here

$D_{s}$

$D_{s}=D_{se}+D_{si}$

$D_{se}=r-3=3-3=0$

$D_{si}=m-2j-3$

=$11-2(6)-3$

=$11-12-3$

=$11-9$

=2

$D_{s}=2$

$D_{k}$

$D_{k}=2j-r$

=$2\times 6-3$

=$12-3$

$D_{k}=9$

enter image description here

$D_{s}=D_{se}+D_{si}$

$D_{se}=r-3=5-3=2$

$D_{si}=M-2j-3$

=$8-(2\times 5)-3$

=$8-10-3$

=$8-7$

=1

$D_{s}=2+1=3$

$D_{s}=3$

$D_{k}$

$D_{k}=2j-r$

=$2\times 5-5$

=$10-5$

=5

$D_{k}=5$

enter image description here

$D_{s}$

$D_{se}=r-3=4-3-1$

$D_{si}=m-2j-3$

=$5-(2\times 4)-3$

=$5-8-3$

=$5-5=0$

$D_{s}=1+0$

$D_{s}=1$

$D_{k}$

$D_{k}=2j-r$

=$2\times 4-4$

=$8-4$

$D_{k}=4$

enter image description here

$D_{se}=r-3=4-3=1$

$D_{si}=m-2j-3$

=$15-2\times 8-3$

=$15-16-3$

=$15-13$

=2

$D_{s}=1+2=3$

$D_{s}=3$

$D_{k}=2j-r$

=$2\times 8-4$

=$16-4$

=12

$D_{k}=12$

enter image description here

$D_{s}$

$D_{s}=D_{se}+D_{si}$

$D_{se}=r-3$

=$7-3=4$

$D_{si}=3\times c =3\times 0=0$

$D_{s}=4+0=4$

$D_{s}=4$

$D_{k}$

$D_{k}=2$

enter image description here

$D_{s}$

$D_{s}=D_{se}+D_{si}$

$D_{se}=r-3=4-3=1$

$D_{si}3\times c=0$

$D_{s}=1+0$

$D_{s}=1$

$D_{k}$

$D_{k}=4$

enter image description here

$D_{s}$

$D_{s}=D_{se}+D_{si}$

$D_{se}=r-3=3-3=0$

$D_{si}=3\times c=3\times 0=0$

$D_{s}=0+0=0$

$D_{s}=0$

$D_{k}$

$D_{k}=3$

enter image description here

$D_{s}$

$D_{s}=D_{se}+D_{si}$

$D_{se}=r-3=9-3=6$

$D_{si}=3\times c=0$

$D_{s}=6$

$D_{k}$

$D_{k}=4$

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