Question

A fixed beam ABC with internal pin at point 'B' is shown in the figure. Due to system of forces acting over the Beam, the observed rotation of span AB at point 'B' was found to be 0.00765 radians in clockwise direction. If $EI=3800kN-m^2$, then moment developed at end 'A' is

• A. 48.64 KN-m
• B. -61.16 KN-m
• C. 58.12 KN-m
• D. 69.69 KN-m

Answer 1

The correct option is B -61.16 KN-m


$GivenEI=3800kN-m^2$

$\begin{array}{cc}\text{Member}& \text{FEM(kN-m)}\\ AB& -11.25\\ BA& 11.25\\ BC& -8\\ CB& 16\end{array}$

Slope deflection equation :

$M_{AB}=-11.25+\frac{2E\left(2I\right)}{3}[{\theta }_{BA}-\delta ]....\left(1\right)$

$M_{BA}=11.25+\frac{4EI}{3}[2{\theta }_{BA}-\delta ]....\left(2\right)$

$M_{BC}=-8+\frac{2EI}{3}[2{\theta }_{BC}+\delta ]....\left(3\right)$

$M_{CB}=16+\frac{2EI}{3}[{\theta }_{BC}+\delta ]...\left(4\right)$

Under equilibrium,
$M_{BA}=0$

$11.25+77.52-5066.67\delta$ = 0

$\Rightarrow \delta =0.0175m$

Hence, from (1), we get

$M_{AB}=-11.25+\frac{4\times 3800}{3}[0.00765-0.0175]$

$M_{AB}=-61.16kN-m$

'-ve' sign indicates moment in counter clockwise direction.