Question

$\text{For the figure given below,}{\theta }_B\text{(Disregarding sign ) is}$ ____?

• A. $\frac{7P{L}^2}{48EI}$
• B. $\frac{5P{L}^2}{48EI}$
• C. $\frac{5P{L}^2}{96EI}$
• D. $\frac{7P{L}^2}{96EI}$

Answer 1

The correct option is C $\frac{5P{L}^2}{96EI}$

$M_{BA}=\frac{3PL}{16}+\frac{3EI}{L}{\theta }_B$

$M_{BC}=\frac{\left(PL\right)}{8}+\frac{3EI}{L}{\theta }_B$

$M_{BA}+M_{BC}=0$

$\Rightarrow \frac{5PL}{16}+\frac{6EI{\theta }_B}{L}=0$

$\Rightarrow {\theta }_B=-\frac{5}{96}\frac{P{L}^2}{EI}=\frac{5}{96}\frac{P{L}^2}{EI}\left(anticlockwise\right)$