Subject:- Structural Analysis II

Title:- Flexibility Method

Difficulty:- Hard

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

Treating $V_C$ as redunded r$x^n$. Let us remove it and obtain statically determine as below :

Σ$M_A$ = 0

$5*3*\frac{3}{2} - V_B*3 - 8$ =0

$V_B$ = 4.899KN

Σ$F_Y$ = 0

$V_A -(5*3) +4.833$ =0

$V_A$ = 10.167KN

Now draw the diagram with no external load. and apply 1KN source which we have remove and find reaction. [$M_1$ analysis]

Σ$M_A$ = 0

-$V_B$3 - 17 = 0

$V_B$ = -2.33KN

Σ$F_Y$ 0

$V_A$ -2.33 +1 =0

$V_A$ = 1.33KN

2) Table:- Finding displacement of released structure

3) ^$R_1 = \int_{0}^{l} \frac{M_1.M_2}{EI}dx$ = $\int_{0}^{3} (10.167x - 2.5x^2)(1.33x)dx + \int_{0}^{2} (10.167(3+x) - 53(x+5) + 4.833x)(1.33(3+x) - 2 - 3.3x)dx + \int_{0}^{2} 0(1x)dx$= 54.40 + 47.92 +0 ^$R_1 = \frac{102.32}{EI}$4)$f_{11} = \int_{0}^{l} \frac{M_1.M}{EI}dx = \int_{0}^{l} \frac{(M_1)^2}{EI}dx$=$\int_{0}^{3} \frac{(1.33)^2}{EI} dx + \int_{0}^{2} \frac{(1.33(x+3)-2.33x^2 )^2}{EI}dx + \int_{0}^{2} \frac{(1x)^2}{EI}dx$=15.92 + 18.55 + 2.67$f_{11} = \frac{37.14}{EI}$*5) Applying Compatibility Condition:-* [^I] = ^R + [F][R] [^$I_1$] =^$R_1$_ [$F_{11}$][$R_1$] Due to roller support at D the net displacement at c in the direction of redundant is 0: i.e. ^$I_1$=0 0 =$\frac{102.32}{EI} + [\frac{37.14}{EI}][R_C]R_C$= -2.75KN$R_C$= 2.75KN [$V_C$= 2.75 KN] ![enter image description here][3] Σ$M_A$= 0 5*3*$\frac{3}{2} - V_B3 - 8 - 2.757 = 0$[$V_B$= 11.25KN] Σ$F_Y$= 0$V_A$- (5*3) + 11.25 -2.75 =0 [$V_A$=6.5KN] ![enter image description here][4] *B.M Cal:-* B$M_A$= 0 B$M_B = 6.53 - 53*\frac{3}{2}$=-3 KN.m B$M_B$= 3KN.m B.$M_C$=0