Question

The moment reaction developed at end 'C' for the beam given below is:

• A. $2.5kN-m$
• B. $38kN-m$
• C. $35kN-m$
• D. $40kN-m$

Answer 1

The correct option is A $2.5kN-m$

Let us apply unit load method:

Portion	Origin	Range of $x$	$EI$	$M$	$m$
$AB$	$A$	$0-2$	$EI$	$-20$	$0$
$BD$	$B$	$0-3$	$EI$	$-20-6.67x$	$\frac{1}{6}x$
$CD$	$C$	$0-3$	$EI$	$6.67x$	$1-\frac{x}{6}$

${\theta }_C=\underset{0}{\overset{3}{\int }}\frac{(\frac{-20x}{6}-\frac{6.67x^2}{6})}{EI}+\underset{0}{\overset{3}{\int }}\frac{(6.67x-\frac{6.67x^2}{6})}{EI}$

$\Rightarrow {\theta }_C=-\frac{25.005}{EI}+\frac{20.01}{EI}=-\frac{4.995}{EI}$

$\text{Let the rotation due to unit load be}{\varphi }_C$

${\varphi }_C=\underset{0}{\overset{3}{\int }}\frac{{\left(\frac{x}{6}\right)}^{2}dx}{EI}+\underset{0}{\overset{3}{\int }}\frac{{(1-\frac{x}{6})}^{2}dx}{EI}=\frac{2}{EI}$

Now applying compatibility

${\theta }_C+M_C{\varphi }_C=0$

$\Rightarrow -\frac{4.995}{EI}+M_C\times \frac{2}{EI}=0$

$\Rightarrow M_C=2.5kN-m$

Alternatively:

$\text{[Note: As far end is fixed, so carry over factor is}\frac{1}{2}]$