Question

The flexibility matrix for beam system shown below for the assigned co -ordinate system is $\left[\begin{array}{cc}\frac{1000}{3EI}& \frac{2500}{3EI}\\ \frac{2500}{3EI}& \frac{3000}{3EI}\end{array}\right]$

The reaction at support B is:

• A. 10.71 kN
• B. 7.64 kN
• C. 5.77 kN
• D. 1.28 kN

Answer 1

The correct option is C 5.77 kN


${\Delta }_B=\frac{24\times 5^3}{3EI}+\frac{24\times 5^2}{2EI}\times 5=\frac{2500}{EI}$

${\Delta }_C=\frac{24\times 5^3}{3EI}+\frac{24\times 5^2}{2EI}\times 15=\frac{5500}{EI}$

$\left[\Delta \right]+\left[f\right]\left[R\right]=0\left[\begin{array}{c}{\Delta }_B\\ {\Delta }_C\end{array}\right]+\left[f\right]\left[\begin{array}{c}{R}_B\\ R_C\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]\left[\begin{array}{c}\frac{2500}{EI}\\ \frac{5500}{EI}\end{array}\right]+\left[\begin{array}{cc}\frac{1000}{EI}& \frac{2500}{EI}\\ \frac{2500}{3EI}& \frac{8000}{3EI}\end{array}\right]\left[\begin{array}{c}{R}_B\\ R_C\end{array}\right]=0$

$\frac{1000}{3EI}{R}_B+\frac{2500}{3EI}{R}_C=-\frac{2500}{EI}\frac{2500}{3EI}{R}_B+\frac{8000}{3EI}{R}_C=-\frac{5500}{EI}{R}_B=-10.71kN{R}_C=1.28kN$