Question

For a linear elastic beam shown in the figure, the flexural rigidity EI, is $781250kN-m^2$. When $w=100kN/m$, the vertical reaction $R_A$ for $w=100kN/m$ is

• A. $500kN$
• B. $425kN$
• C. $250kN$
• D. $75kN$

Answer 1

The correct option is B $425kN$



Let support reaction at $B$ is $R_B$
Bending moment at a distance $x$ from $B$
$M\left(x\right)=R_{B}x-\frac{W{x}^2}{2}=(R_{B}x-50x^2)$
Strain energy,$U={\int }_0^l\frac{M^{2}dx}{2EI}={\int }_0^5\frac{(R_{B}x-50x^2{)}^{2}dx}{2EI}$
$\frac{\partial U}{\partial R_B}=-6\times {10}^{-3}m$
$\{-ve$ sign because $R_B$ is in upward direction and deflection in downward dir.$\}$
${\int }_0^5\frac{2(R_{B}x-50x^2)xdx}{2EI}=-6\times {10}^{-3}$
${\int }_0^5(R_B{x}^2-50x^3)dx=-6\times {10}^{-3}\times 781250$
${[R_B\frac{x^3}{3}-\frac{50x^4}{4}]}_0^5=-4687.5$
$\frac{125}{3}{R}_B-\frac{50\times 5^4}{4}=-4687.5$
$\Rightarrow R_B=75kN$
$\sum V=0$
$\Rightarrow R_A+R_B=100\times 5$
$R_A=500-75$
$=425kN$

Alternative approach 1:



Deflection at
$B=$ Deflection due to $UDL$
$-$ Deflection due to $R_B$
$6\times {10}^{-3}=\frac{w{L}^4}{8EI}-\frac{R_B{L}^3}{3EI}$
$6\times {10}^{-3}\times 781250=\frac{100\times 5^4}{8}-\frac{R_B{5}^3}{3}$
$\Rightarrow R_B=75kN$
$\sum V=0$
$\Rightarrow R_A+R_B=100\times 5$
$R_A=500-75$
$=425kN$

Alternative approach 2:


Deflection at $B$ due to $100kN/m$ load$=\frac{w{l}^4}{8EI}=\frac{100(5{)}^4}{8\times 781250}=10mm$
for $100kN/m$ load free end deflection on the absence of support would have been $10mm$.
Support reaction at $B$ is restraining $4mm$ deflection
$\Rightarrow \frac{R_B(5{)}^3}{3EI}=4\times {10}^{-3}m$
$\Rightarrow R_B=\frac{4\times {10}^{-3}\times 3\times 781250}{125}kN$
$=75kN$
$\Rightarrow R_A=100\times 5-75=425kN$