Question

Find the rotation at B for the frame shown below.

• A. $\frac{5}{7EI}\text{clockwise}$
• B. $\frac{5}{7EI}\text{anticlockwise}$
• C. $\frac{4}{7EI}\text{clockwise}$
• D. $\frac{4}{7EI}\text{anticlockwise}$

Answer 1

The correct option is D $\frac{4}{7EI}\text{anticlockwise}$

${\overline{M}}_{AB}={\overline{M}}_{BA}=0$

${\overline{M}}_{BC}=\frac{Mb(3a-l)}{l^2}=\frac{5\times 2\times (3\times 3-5)}{25}=+1.6kNm$

$M_{BC}={\overline{M}}_{BC}+\frac{2EI{\theta }_B}{L_{BC}}(2{\theta }_B+{\theta }_C-\frac{3\Delta }{L})$

${\theta }_C=0;\Delta =0$

$M_{BC}={\overline{M}}_{BC}+\frac{4EI{\theta }_B}{L}$
(using slope deflection equation for span BC)

$=1.6+\frac{4EI{\theta }_B}{5}=1.6+0.8EI{\theta }_B$

Using slope deflection equation for span AB,
$M_{BA}={\overline{M}}_{BA}+\frac{2E\left(2I\right)}{L_{AB}}(2{\theta }_B+{\theta }_A-\frac{3\Delta }{L})$

$M_{BA}=\frac{4EI\times 2{\theta }_B}{4}(\Delta =0;{\theta }_A=0)$

$=2EI{\theta }_B$

At joint B,
$M_{BA}+M_{BC}=0$

$\Rightarrow 1.6+0.8EI{\theta }_B+2EI{\theta }_B=0$

$\Rightarrow 2.8EI{\theta }_B=-1.6$

$\Rightarrow {\theta }_B=-\frac{4}{7EI}$
Negative sign shows that rotation is in anticlockwise direction.