Question

If the support B of continous beam ABC (as shown in figure) settles downward by an amount ${}^{\prime }{\delta }^{\prime }$ , then the reaction developed at point B is:

• A. $\frac{24EI\delta }{l^3}$
• B. $\frac{16EI\delta }{l^3}$
• C. $\frac{6EI\delta }{l^3}$
• D. $\frac{12EI\delta }{l^2}$
  

Answer 1

The correct option is A $\frac{24EI\delta }{l^3}$

Let, the reaction on developed at ${}^{\prime }{B}^{\prime }$ is $R_B$ (downwards)


We know that, mid-span deflection of fixed beam due to concentrated load,

${\delta }_B=\frac{R_B\times (2l{)}^3}{192EI}$

This must be equal to settlement of support $B$ i.e. $\left(\delta \right)$

Hence, $\frac{R_B\times (2l{)}^3}{192EI}=\delta$

$\therefore R_B=\frac{24EI\delta }{l^3}$