Question

A symmetrical parabolic arch of span L and rise 'h' is hinged at both the supports. The arch carries a uniformly distributed load of w/unit length along the entire span.
Which one of the following is correctly matched?

• A. $\text{Horizontal thrust:}\frac{w{L}^2}{8h}$
• B. $\text{Vertical reaction at supports: wL}$
• C. $\text{Bending moment at crown:}\frac{w{L}^2}{8h}$
• D. $\text{radial shear at springing:}(\frac{wL}{2}-\frac{w{L}^2}{8h})$

Answer 1

The correct option is A $\text{Horizontal thrust:}\frac{w{L}^2}{8h}$


Equation of parabolic arch

$y=\frac{4h}{l^2}(lx-x^2)$

$H=\frac{\int \frac{Mydx}{EI}}{\int \frac{y^{2}dx}{EI}}$

$M_x=(\frac{wl}{2}x-\frac{w{x}^2}{2})$

$H=\frac{\underset{0}{\overset{l}{\int }}\frac{w}{2}(lx-x^2)\frac{4h}{l^2}(lx-x^2)dx}{\underset{0}{\overset{l}{\int }}{\left[\frac{4h}{l^2}(lx-x^2)\right]}^{2}dx}$

$=\frac{\frac{w}{2}\left(\frac{4h}{l^2}\right)\underset{0}{\overset{l}{\int }}{(lx-x^2)}^{2}dx}{{\left(\frac{4h}{l^2}\right)}^2\underset{0}{\overset{l}{\int }}{(lx-x^2)}^{2}dx}$

$H=\frac{w{l}^2}{8h}$

Moment will be zero in the parabolic arch.
Since loading is symmetric both vertical reactions are equal so $V_A=V_B=\frac{wl}{2}$