Subject:- Structural Analysis II

Title:- Frames

Difficulty:- Hard

Step 1:-

Degree of btatic indeterminacy($D_{SE}$)

($D_{SE}$) = r-3 = 5-3 = 2

Step 2:-

Select redundant as

$V_C = R_1$

$H_C = R_2$

Step 3:- Draw M and $M_1 and M_2$ diagram remove the redundant part only.

To draw $M_1 and M_2$ remove all the external load and apply 1 KN on removed load.

Step 4:-

^$R_1 = \int_{0}^{l} \frac{M.M_1}{EI}$dx $\frac{1}{EI} \int_{0}^{4.47}(0)(0.45)dx + \int_{0}^{4}(-5x^2)(x+2)dx$ [^$R_1 = \frac{533}{EI}$] ^$R_2 = \int_{0}^{l} \frac{M.M_1}{EI}$dx = $\frac{1}{EI} { \int_{0}^{4.47} (0)(0.89x)dx + \int_{0}^{4}(-5x^2)(4)dx }$ [^$R_2 = \frac{-426.67}{EI}$] *Step 5:-* ***Flexibility Coefficient:-*** $f_{11} = \int_{0}^{l} \frac{(m_1)^2}{EI} dx$ = $\int_{0}^{4.47} \frac{(0.45x)^2}{EI}dx + \int_{0}^{4} \frac{(0.89)^2}{EI}dx$ $f_{11} = \frac{75.36}{EI}$ $f_{22} = \int_{0}^{l} \frac{(m_2)^2}{EI} dx$ = $\int_{0}^{4.47} \frac{(0.89x)^2}{EI}dx + \int_{0}^{4} \frac{(4^2}{EI}dx$ $f_{22} = \frac{85.53}{EI}$ $f_{12} = f_{21} = \int_{0}^{l} \frac{M_1.M_2}{EI}dx$ $f_{12} = f_{21} = \frac{75.92}{EI}$ *Step 6:-* **Apply flexibility equation** ^I = {^R} + [F]{R} 0 = -533.33 + 75.36$V_C + 75.92 H_C$ 0 = -426.67 + 75.92 $V_C + 87.57 H_C$ 75.36 $V_C + 75.92 H_C$ = 533.33 ----(1) 75.92 $V_C + 87.53 H_C$ = 426.67 -----(2) *Solving (1) and (2) we get, $R_1 = V_C = 17.29 KN$ $R_2 = H_C = -10.12KN$ *Step 7:-* Draw BMD ![enter image description here][5] Free Bonding moment (FBM) =$\frac{we^2}{8} = \frac{104^2}{8}$ = 20KN.m ***BM Calculation:-** $B.M_C = 0$ $B.M_B$ = 17.92*2 - 10.12*4 = -5.9 KN.m $B.M_A = 17.296 - 10.124 - 104*\frac{4}{2}$ = -16.74 kN.m