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Analyse Pin-joint frame by flexibility method.

Subject :- Structural Analysis II

Title :- Analysis of Indeterminate truss

Difficulty :- Hard

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1 Answer
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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

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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)

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Joint E :-

ΣFY = 0(+ve)

PECsin(53.13) + 30 = 0

PEC = -37.5KN (c)

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K1 Analysis :-

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K2 Analysis :-

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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

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