written 6.6 years ago by | • modified 3.2 years ago |
Subject :- Structural Analysis II
Title :- Analysis of Indeterminate truss
Difficulty :- Hard
written 6.6 years ago by | • modified 3.2 years ago |
Subject :- Structural Analysis II
Title :- Analysis of Indeterminate truss
Difficulty :- Hard
written 6.6 years ago by | • modified 6.6 years ago |
1) Degree of Static indeterminacy :-
Dse = r-3 = 3-3 = 0
Dsi = M - (2j - 3) = 11 - (2(6) - 3) = 11 - 9 = 2
Ds = 0+2 =2
2) Treating diagonal member BF and DF are redundentes let us remove them
let R1 = Force BF
let R2 = Force DF
3) P-Analysis:-
ΣMA = 0(+ve)
=154 + 253 + 15*3
-ve*6 = 0
VE = 30KN
ΣFX = 0(+ve)
VA - 25 -15 + 30 =0
VA = 10KN
Joint A :- Apply C.D.E to joint A
tan@=43
@=53.13o
ΣFY = 0(+ve)
10 + PACsin(53.13) = 0
PAC = -12.5 (c)
ΣFX = 0(+ve)
PAF + (-12.5cos(53.13)) - 15 = 0
PAF = 22.5KN (T)
Joint E :-
ΣFY = 0(+ve)
PECsin(53.13) + 30 = 0
PEC = -37.5KN (c)
K1 Analysis :-
K2 Analysis :-
Table :-
Member | P | K1|K2|L|PK1L|PK2L|K21L|K22L|R1|R2 | ||||||||
---|---|---|---|---|---|---|---|---|---|---|
AB | 0 | -0.8 | 0 | 4 | 0 | 0 | 2.56 | 0 | 7.17 | 15.97 |
BC | -15 | -0.6 | 0 | 3 | 27 | 0 | 1.08 | 0 | 7.17 | 15.97 |
CF | 15 | -0.8 | -0.8 | 4 | -48 | -48 | 256 | 2.56 | 7.17 | 15.97 |
FA | 22.5 | -0.6 | 0 | 3 | -40.5 | 0 | 1.08 | 0 | 7.17 | 15.97 |
AC | -12.5 | 1 | 0 | 5 | -62.5 | 0 | 5 | 0 | 7.17 | 15.97 |
CD | 0 | 0 | -0.6 | 3 | 0 | 0 | 0 | 18.08 | 7.17 | 15.97 |
DE | 0 | 0 | -0.8 | 4 | 0 | 0 | 0 | 2.56 | 7.17 | 15.97 |
EF | 22.5 | 0 | -0.6 | 3 | 0 | -40.5 | 0 | 1.08 | 7017 | 15.97 |
CE | -37.5 | 0 | 1 | 5 | 0 | -18.75 | 0 | 5 | 7.17 | 15.97 |
DF | 0 | 0 | 1 | 5 | 0 | 0 | 0 | 5 | 7.17 | 15.97 |
FB | 0 | 1 | 0 | 5 | 0 | 0 | 5 | 0 | 7.17 | 15.97 |
-124 | -276 | 17.28 | 17.28 |
R1=ΣPkLAEK21LAE = 7.17
R2=ΣPkLAEK22LAE = 15.97
F = P + R1K1+R2K2