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Analyze the frame shown in fig for the forces in values members. All the members may be assumed to have the same $\frac{L}{AE}$ value.

Subject :- Structural Analysis II

Title :- Analysis of Indeterminate truss

Difficulty :- Hard

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1 Answer
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1) Degree of redundancy :-

$D_{se}$ (Degree of external redundancy) = R-3 = 3-3 = 0

$D_{si}$ (Degree of internal redundancy) = M-[2j-3] = 6-[2(4)-3] = 6-[8-3] = 6-5 = 1

Total = 0+1 =1

2) Let AD be the redundant :-

Remove member AD

3) P-Analysis :-

Σ$M_c$ = 0 (+ve) = 54 - $V_B$4 = 0

$V_B$ = 5KN

Σ$F_y$ = 0 (+ve) = $V_C$-5+5 = 0

$V_C$ = 0

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Joint A :-

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4) K-Analysis :-

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Table :-

Member P K PK $K^{2}$ F=P+$R_K$
AB 0 -0.71 0 0.504 0.62
AC 0 -0.71 0 0.504 0.62
AD 0 1 0 1 -0.88
BD -5 -0.71 3.55 0.504 -4.37
BC 0 1 0 1 -0.88
CD 0 -0.71 0 0.504 0.62
Σ$P_K$ = 3.55 Σ$K^2$ = 4.016

R = -[$\frac{ΣP_k}{ΣK^2}$] = -$\frac{(3.55)}{4.016}$ = -0.88

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