Question

Consider the beam subjected to different loads as shown in the figure


The ratio of fixed end moments (magnitude) i.e. $\frac{M_A}{M_B}$ is_____.

• A. 1.17
• B. 6.33
• C. 3.00
• D. 0.3

Answer 1

The correct option is D 0.3


$M_{FAB}=-\frac{30\times (3{)}^2\times 1}{4^2}=-16.875kN-m$
$M_{FBA}=+\frac{30\times 1^2\times 3}{4^2}=5.625kN-m$


$M_{FAB}=\frac{60(2\times 3-1)\times 1}{4^2}=18.75kN-m$
$M_{FBA}=\frac{60\times (2\times 1-3)\times 3}{4^2}=-11.25kN-m$
$M_A=-16.875+18.75=+1.875kN-m$
$M_B=5.625-11.25=-5.625kN-m$


$\therefore \frac{M_A}{M_B}=\frac{1.875}{5.625}=0.333$