Question

A two hinged parabolic arch is subjected to u.d.I. 'w' over entire span. Then the horizontal thrust is

• A. $w{l}^2/6h$
• B. $w{l}^2/8h$
• C. $w{l}^2/4h$
• D. $w{l}^2/3h$

Answer 1

The correct option is B $w{l}^2/8h$



$V_a+V_b=wl$
$V_a=V_b=\frac{wl}{2}$
$M_x=\frac{wl}{2}\times x-\frac{w{x}^2}{2}$ $h=\frac{4h}{l^2}(l-x)$

$Horizontalthrust\left(H\right)=\frac{{\int }_0^{l}M.yds}{{\int }_0^l{y}^2.ds}$

$=\frac{{\int }_0^l\frac{4h}{l^2}x(l-x)\times (\frac{wlx}{2}-\frac{w{x}^2}{2})ds}{{\int }_0^l{\left(\frac{4h}{l^2}x\right(l-x\left)\right)}^{2}ds}$

$=\frac{\frac{4wh}{2l^2}{\int }_0^{l}x(l-x)(lx-x^2)ds}{\frac{16h^2}{l^4}{\int }_0^4{x}^2(l-x^2)ds}$

$=\frac{\frac{2wh}{l^2}{\int }_0^l{x}^2(l-x^2)ds}{\frac{16h^2}{l^2}{\int }_0^l{x}^2(l-x{)}^2}=\frac{2wh}{l^2}\times \frac{l^4}{16h^2}=\frac{w{l}^2}{8h}$