Question

Consider the following beam


The magnitude of final end moment at B is _______ kNm.

• A. 56.67

Answer 1

The correct option is A 56.67
Fixed end moments,

${\overline{M}}_{AB}=\frac{M}{4}=\frac{200}{4}=50kNm$

${\overline{M}}_{BA}=\frac{M}{4}=\frac{200}{4}=50kNm$

${\overline{M}}_{BC}=\frac{-w{L}^2}{12}=\frac{-50\times 4^2}{12}=-66.67kNm$

${\overline{M}}_{CB}=66.67kNm$

Consider of distribution factor

Joint	Member	k	$\Sigma k$	Distribution factor
B	BA	$\frac{4EI}{6}$	$\frac{5EI}{3}$	$\frac{2}{5}$
	BC	$\frac{4EI}{4}$	$\frac{3}{5}$

MDM table


$\therefore \text{Moment at B}=56.67kNm$