written 5.6 years ago by
teamques10
★ 64k
|
• modified 2.2 years ago |
Subject :- Structural Analysis II
Title :- Analysis of Indeterminate truss
Difficulty :- Hard
|
written 5.6 years ago by
teamques10
★ 64k
|
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
Joint A :-
4) K-Analysis :-
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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