Subject: Structural Analysis II

Topic: Flexibility Method

Difficulty: Medium / High

1) Free Bending moment: [considering each span simply supported]

• For Span AB B mat $E=\frac{wab}{l}$ =$\frac{80(2)(4)}{6}=\underline{106.6 kN.m}$
• For Span BC B mat $f=\frac{wl^{2}}{8}$=$\frac{20(5)^{2}}{8}=\underline{62.5 kN.m}$
• For Span CD B mat $G=\frac{wab}{l}$=$\frac{60(2)(3)}{5}=\underline{72kN.m}$

2) Free Bending moment diagram

:

3) Three moment Theorem

• For A-B-C $m_{A}(l_{1})+2m_{B}(l_{1}+l_{2})+m_{c}(l_{2})+\frac{6A_{1}x_{1}}{l_{1}}+\frac{6A_{2}x_{2}}{l_{2}}=0$ $0+2MB(6+5)+M_{c}(5)+\frac{6 \times (320.01)(2.67)}{6}+\frac{6(208.33)(2.5)}{5}=0$ $22M_{B}+5M_{c}=-1479.42 \tag{1}$

  -For A-B-C $M_{B}(l_{1})+2m_{c}(l_{1}+l_{2})+M_{D}(l_{2})+\frac{6A_{1}x_{1}}{l_{1}}+\frac{6A_{2}x_{2}}{l_{2}}=0$ $M_{B(5)}+2M_{c}(5+5)+0+\frac{6(208.33)(2.5)}{5}+\frac{6(180)(2.67)}{5}=0$ $5M_{B}+20M_{c}=-1201.71 \tag{2}$

  Solving 1 & 2 we get

$M_{B}=-5682 kN.m$

$M_{c}=-45.88kN.m$

4) Find Reaction at Support

Reaction at $A=\underline{43.86kN}$

Reaction at $B=\underline{36.14+52.188=88.33kN}$

Reaction at $C=\underline{47812+45.17=92.98KN}$

Recation at $D=\underline{14.83kN}$

5) Draw B.M.D

Type-II:- Continous beam with over hanging ends

In continuous beam with overhanging on oneside or on both sides,the overhang portion of beam are treated as cantilever and three moments theorem is applied on the rest of the portions to determine the support moments