Analyse Pin-joint frame by flexibility method.

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written 6.2 years ago by

teamques10 ★ 68k

• modified 6.2 years ago

1) Degree of Static indeterminacy :-

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

$D_{si}$ = M - (2j - 3) = 11 - (2(6) - 3) = 11 - 9 = 2

$D_{s}$ = 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:-

Σ$M_A$ = 0(+ve)

=154 + 253 + 15*3

-ve*6 = 0

$V_E$ = 30KN

Σ$F_X$ = 0(+ve)

$V_A$ - 25 -15 + 30 =0

$V_A$ = 10KN

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Joint A :- Apply C.D.E to joint A

tan@=$\frac{4}{3}$

@=$53.13^o$

Σ$F_Y$ = 0(+ve)

10 + $P_{AC}$sin(53.13) = 0

$P_{AC}$ = -12.5 (c)

Σ$F_X$ = 0(+ve)

$P_{AF}$ + (-12.5cos(53.13)) - 15 = 0

$P_{AF}$ = 22.5KN (T)

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Joint E :-

Σ$F_Y$ = 0(+ve)

$P_{EC}$sin(53.13) + 30 = 0

$P_{EC}$ = -37.5KN (c)

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$K_1$ Analysis :-

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$K_2$ Analysis :-

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Table :-

$R_1 = \frac{\frac{ΣP_kL}{AE}}{\frac{K_1^2L}{AE}}$ = 7.17

$R_2 = \frac{\frac{ΣP_kL}{AE}}{\frac{K_2^2L}{AE}}$ = 15.97

F = P + $R_1K_1 + R_2K_2$

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