Question

For the rigid jointed frame ABC as shown in the figure, the moment developed $M_{BA}$ is, $(TakeEI=1432kN-m^2)$.

• A. $-13.056kN-m$
• B. $4.139kN-m$
• C. $2.462kN-m$
• D. $15.142kN-m$

Answer 1

The correct option is A $-13.056kN-m$

Member	FEM(kN-m)
AB	$-10$
BA	$10$
BC	$10+\frac{10}{2}=15$

$M_{BA}=10+\frac{2EI}{4}\left[2{\theta }_B\right]$
$M_{BC}=15+\frac{3EI}{5}\left[{\theta }_B\right]$
Under euilibrium,
$M_{BA}+M_{BC}+12=0$

$25+\frac{2\times 1432}{4}\times 2{\theta }_B+\frac{3\times 1432}{5}{\theta }_B+12=0$

$\Rightarrow {\theta }_B=-0.0161$

$M_{BA}=10+\frac{2\times 1432}{4}\times 2\times (-0.0161)=-13.0552kN-m$