Question

For the continuous beam as shown in figure, the slope at end 'A' (in radians, upto four significant decimals) is ____
$(TakeEl=4142kN-m^2)$

• A. 0.0196

Answer 1

The correct option is A 0.0196

$\because$ The beam is symmetrical with antisymmetrical loading, we can write :

$K_{BA}=\frac{3EI}{6};K_{BC}=\frac{6E\left(2I\right)}{6}$

$D{F}_{BA}=0.2;D{F}_{BC}=0.8$

$M_{BA}=67.5+\frac{3EI}{6}{\theta }_B=54$

$\Rightarrow {\theta }_B=\frac{-27}{EI}$

$M_{AB}=\frac{-5\times 24\times 6^2}{96}+\frac{2EI}{6}[2{\theta }_A+{\theta }_B]=0$

$\Rightarrow {\theta }_A=0.0196rad$