Question

A symmetrical 3-hinged parabolic arch of span ${}^{\prime }{l}^{\prime }$ and rise 'h' carries a point load 'W' which may be placed anywhere on the span. The distance of the section where maximum bending moment occurs from crown will be

• A. $\frac{l}{2\sqrt{3}}$
• B. $\frac{l}{4\sqrt{2}}$
• C. $\frac{l}{3\sqrt{2}}$
• D. $\frac{l}{3\sqrt{3}}$

Answer 1

The correct option is A $\frac{l}{2\sqrt{3}}$

$V_a=\frac{W(l-x)}{l},V_b=\frac{Wx}{l}$

$\Sigma M_C=0\left(Fromrightend\right)\Rightarrow H\times h=\frac{Wx}{l}\times \frac{l}{2}$

$\Rightarrow H=\frac{Wx}{2h}$

$\therefore B{M}_x=\frac{W(l-x)}{l}\times x-\frac{Wx}{2h}\times \frac{4h}{l^2}x(l-x)$

$\Rightarrow B{M}_x=\frac{Wx(l-x)}{l}-\frac{2W}{l^2}{x}^2(l-x)$

The maximum B.M. occurs under the load.

$\therefore$ $\frac{d\left(B{M}_x\right)}{dx}=0;$

$\Rightarrow \frac{W}{l}(l-2x)=\frac{2W}{l^2}(2lx-3x^2)$

For absolute maximum B.M.,

$\frac{d\left(B{M}_x\right)}{dx}=0;$

$\Rightarrow \frac{W}{l}(l-2x)=\frac{2W}{l^2}(2lx-3x^2)$

$\Rightarrow 6x^2-6lx+l^2=0$

$\Rightarrow x=\frac{6-2\sqrt{3}}{12}l(\sin x<\frac{1}{2})$

$\Rightarrow x=\frac{l}{2}-\frac{l}{2\sqrt{3}}fromA$

$\therefore \text{From crown, distance of maximum B.M. on either side}=\frac{l}{2}-x=\frac{l}{2\sqrt{3}}$