Subject : Structural Analysis 1

Topic : Deflection of beam

Difficulty : High

1. Support reaction calculation:

$\sum M_A=0(\circlearrowright+ve)\\ 50\times4-V_B\times5.5=0\\ \boxed{V_B=36.36\ kN}(\uparrow)\\ \\ \sum F_Y=0(\uparrow +ve)\\ V_A-50+36.36=0\\ \boxed{V_A=13.64}(\uparrow)$

2. Draw BMD$[\circlearrowright|\circlearrowleft+ve]$

$B.M_A=0\\ B.M_C=13.64\times2=27.28\ kNm\\ B.M_D=36.36\times1.5=54..54\ kNm\\ B.M_B=0$

3. Draw $\frac{M}{EI}$ diagram

4. Draw conjugate beam and consider $\frac{M}{EI}$ diagram as loading on the beam

$\sum M_A=0(\circlearrowright+ve)\\ \left( \frac{1}{2}\times2\times\frac{27.28}{EI} \right) \left( \frac{2}{3}\times2\right)+\left( \frac{1}{2}\times2\times\frac{13.6}{EI}\right) \left( 2+\frac{2}{3}\times2\right)+\left( 2\times\frac{13.64}{EI}\right) \left( 2+\frac{2}{2}\right)+\left(\frac{1}{2}\times1.5\times\frac{54.54}{EI} \right)\left( 4+\frac{1}{3}\times1.5\right)-V_B'\times 5.5=0\\ 5.5V_B'=\frac{1}{EI}\left[ 36.37+45.33+81.84+184.07\right]=\frac{1}{EI}[347.61]\\ V_B'=\frac{347.61}{5.5EI}=\frac{63.20}{EI}\\ \boxed{V_B'=\frac{63.20}{EI}}$

$\sum F_Y=0(\uparrow+ve)\\ V_A'-\left( \frac{1}{2}\times2\times\frac{27.28}{EI}\right)-\left( \frac{1}{2}\times2\times\frac{13.6}{EI}\right)-\left( 2\times\frac{13.64}{EI}\right)-\left(\frac{1}{2}\times1.5\times\frac{54.54}{EI}\right)+\frac{63.20}{EI}\\ V_A'=\frac{1}{EI}\left[ 27.28+13.6+27.28+40.90-63.20\right]\\ \boxed{V_A'=\frac{45.86}{EI}}$

5. $Q_A$ in real beam = S.F @ A in conjugate beam.

$S.F_{AL}=0\\ \boxed{S.F_{AR}=\frac{45.86}{EI}}(\circlearrowright)$

6. $y_D$ in real beam = B.M @ D in conjugate beam

$B.M_D=\frac{63.20}{EI}\times1.5-\left[\frac{1}{2}\times1.5\times\frac{54.54}{EI} \right]\left(\frac{1}{3}\times1.5\right)= \frac{1}{EI}[94.8-20.45]\\ \boxed{B.M_D=\frac{74.35}{EI}}(\downarrow)$