Analyze the pin-joint frame by flexibility method as shown in fig. All the members may have assumed to the same AE=Constant.

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written 7.3 years ago by

teamques10 ★ 70k

1) Degree of static indeterminacy :-

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

$D_{si}$ = M-[2j-3] = 6-[2(4)-3] = 6-5 = 1

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

2) Let AC be the redundant :-

Remove Member AD

3) P-Analysis :-

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Σ$M_D =0 =20*4-V_c$*4 =0

$V_c$ = 20KN

Σ$F_y =0 =V_D$-30+20 =0

$V_D$ = 10KN

Joint D:-

Σ$F_y =0$

10-30+$P_{DB}$sin45 =0

$P_{DB}$ = 28.28KN

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4) K-Analysis:-

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5) Table:-

Member	P	K	L	PKL	$k^2$L	R	PK	$P_f$=P+PK
AB	-20	-0.7	4	56	1.96	-18.58	13	-7
BC	-20	-0.7	4	56	1.96	-18.58	13	-7
CD	0	-0.7	4	0	1.96	-18.58	13	13
DA	-30	-0.7	4	84	1.96	-18.58	13	17
AC	0	1	5.65	0	5.66	-18.58	-18.58	-18.58
DB	28.28	1	5.65	160.06	5.66	-18.58	-18.56	9.7
				ΣPKL =356.06	Σ$K^2$L = 19.16	-18.58

R= -[$\frac{ΣPKL}{ΣK^2L}$] = -[$\frac{356.06}{19.16}$] = -18.58KN(c)

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