Question

The value of ${\theta }_B$ (magnitude only ) for the beam shown in the figure is then the value of K is_____$\left(\frac{k}{EI}\right)$

• A. 30

Answer 1

The correct option is A 30

Using slope deflection

$M_{BA}=M_{FBA}+\frac{2EI}{L}(2{\theta }_B+{\theta }_A)$

$M_{BA}=0+\frac{2EI}{6}(2{\theta }_B+0)$

$M_{BA}=\frac{2EI{\theta }_B}{3}$

$M_{BC}=M_{FBC}+\frac{2EI}{L}(2{\theta }_B+{\theta }_C)$

$M_{BC}=0+\frac{2EI}{6}(2{\theta }_B+0)$

$\Rightarrow M_{BC}=\frac{2}{3}EI{\theta }_B$

$M_{BD}=20\times 2=40kNm$

Now,
$M_{BA}+M_{BC}+M_{BD}=0$

$\Rightarrow \frac{2EI{\theta }_B}{3}+\frac{EI{\theta }_B}{3}+40=0$

$\Rightarrow \frac{4EI{\theta }_B}{3}=-40$

$\Rightarrow {\theta }_B=-\frac{30}{EI}$

$\therefore K=30$
(Magnitude only)