Question

What will rotation of joint B for the frame shown below?

• A. $\frac{27.45}{EI}$
• B. $\frac{32.5}{EI}$
• C. $\frac{43.33}{EI}$
• D. $\frac{61.25}{EI}$

Answer 1

The correct option is B $\frac{32.5}{EI}$

Fixed end moments:
$M_{FBC}=-\frac{20\times 6^2}{12}=-60kN-m$
$M_{FCB}=+60kN-m$
$M_{FBA}=+\frac{50\times 4}{8}=+25kN-m$
$M_{FAB}=-\frac{50\times 4}{8}=-25kN-m$
Slope deflection equations,
$M_{BC}=-60+\frac{2\times \left(2EI\right)}{6}(2{\theta }_B+{\theta }_C)$
$M_{CB}=+60+\frac{2\times \left(2EI\right)}{6}({\theta }_B+2{\theta }_C)$
$M_{BA}=+25+\frac{2EI}{4}\left(3{\theta }_B\right)$
Equilibrium equations:
$M_{CB}=0$
${\theta }_B+2{\theta }_C=-\frac{90}{EI}....\left(1\right)$
And, $M_{BC}+M_{BA}=0$
$\frac{7}{3}{\theta }_B+\frac{2}{3}{\theta }_C=\frac{35}{EI}....\left(2\right)$
From equation (1) and equation (2) we get
${\theta }_B=\frac{32.5}{EI}$