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Analyse pin-joint frame by flexibility Method AE=constant
written 6.2 years ago by gravatar for teamques10 teamques10 ★ 68k •   modified 2.8 years ago

Subject :- Structural Analysis II

Title :- Analysis of Indeterminate truss

Difficulty :- Hard

enter image description here
structural analysis
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written 6.2 years ago by gravatar for teamques10 teamques10 ★ 68k

Degree of static indeterminacy :-

$D_{SE}$ = r-3 = 4-3 = 1

$D_{Si}$ = M-(2j-3) = 9(2(6)-3) = 0

$D_S$ = 1+0 = 1

Selecting redundant force(R)

Let R = $H_E$(->)

Convert hinged int roller

P-Analysis:

enter image description here

Σ$M_A$ =0(+ve)

203 + 407 - $V_E$*7 =0

Σ$F_Y$ =0(+VE)

$V_A$ - 0 + 48.57 =0

$V_A$ = -8.54KN

Joint A:-

enter image description here

Σ$F_Y$ =0(+ve)

-8.57 + $P_{AC}$Sin(36.86) =0

$P_{AC}$ = 14.28KN (T)

Σ$F_Y$ =0(+ve)

$P_{AF}$ - 20 +(14.28cos38.86) =0

$P_{AF}$ = 8.57KN (T)

Joint E:-

enter image description here

Σ$F_Y$=0(+ve)

48.57 - 40 + $P_{EC}$sin45 =0

$P_{EC}$ = -12.11KN

$P_{EC}$ = 12.11 KN(C)

K-Analysis:

enter image description here

Table:-

Member P K L PKL $K^2$L R RK $P_f$=P+PK
AB 0 0 3 0 0 8.57 0 0
BC -20 0 4 0 0 8.57 0 -20
CD 0 0 3 0 0 8.57 0 0
DE -40 0 3 0 0 8.57 0 40
EF 8.57 1 3 25.71 3 8.57 -8.57 0
FA 8.57 1 4 34.28 4 8.57 -8.57 0
AC 14.28 0 5 0 0 8.57 0 14.28
CE -12.11 0 4.24 0 0 8.57 0 -12.11
CF 0 0 3 0 0 8.57 0 0
Σ59.99 7

R = -[$\frac{\frac{ΣPKL}{AE}}{\frac{K^2L}{AE}}$]

= -[$\frac{60}{7}$] = -8.57

R = i.e = -8.57KN

He= 8.57KN<-

Actual $H_A$ = 20-8.57

$H_A$ = 11.43KN
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