Question

For the continuous beam ABC as shown in figure, the reaction developed at rolled support is


261 KN 36 kN 3m 2m 4m

• A. $36.47kN$
• B. $34.39kN$
• C. $39.67kN$
• D. $32.14kN$

Answer 1

The correct option is C $39.67kN$



$\text{Let us consider redundants as}{R}_B$ & $R_C$
Here, compatibility is
$Verticaldeflection\left({\Delta }_B\right)=0$
$Verticaldeflection\left({\Delta }_C\right)=\frac{R_C}{k}$

At 'B'



${\Delta }_B=\frac{36\times 4^3}{3EI}+\frac{36\times 4^2}{2EI}\times 2+\frac{(26-R_C-R_B)\times 6^3}{3EI}+\frac{\left\{\right(26-R_C)\times 3\}\times 6^2}{2EI}=0$

$\Rightarrow 4608+3456+11232-432(R_C+R_B)+8424-324R_C=0$

$\Rightarrow 432R_B+756R_C=27720..\left(1\right)$

At 'C'



${\Delta }_C=\frac{36\times 4^3}{3EI}+\frac{36\times 4^2}{2EI}\times 5-\frac{R_B\times 6^2}{3EI}-\frac{R_B\times 6^2}{2EI}\times 3+\frac{(26-R_C)\times 9^3}{3EI}=\frac{9R_C}{EI}$

$\Rightarrow 8526-126R_B-243R_C=9R_C$

$\Rightarrow 126R_B+252R_C=8526...\left(2\right)$
From (1) & (2)
$R_B=39.67kN$