Question

For the beam given below, the location of point of contraflexure from end B would be:

• A. 0.38 L
• B. 0.58 L
• C. 0.65 L
• D. 0.73 L

Answer 1

The correct option is D 0.73 L

Let, the reaction at B be $R_B$
${\Delta }_B=0$
Downward deflection due P = upward deflection $R_B$
$\frac{P{\left(\frac{L}{2}\right)}^3}{3EI}+\frac{P{\left(\frac{L}{2}\right)}^2}{2EI}\times \left(\frac{L}{2}\right)=\frac{R_B\times L^3}{3EI}$
$\frac{P{L}^3}{EI}(\frac{1}{24}+\frac{1}{16})=\frac{R_B{L}^3}{3EI}$
$\frac{5P}{48}=\frac{R_B}{3}\Rightarrow R_B=\frac{5P}{16}$
Let, the distance of point of contraflexure be at a distance x from B
$R_B\times x-P(x-\frac{L}{2})=0$
$\frac{5Px}{16}-Px=-\frac{PL}{2}$
$\Rightarrow x=0.727L\approx 0.73L$