Chapter 9, Problem 56P

To determine

Height of the pointed arch.

Answer to Problem 56P

Solution: $12m$

Explanation of Solution

Considering the centers of gravities of the left half of the circular positions is same as that of right halves from below diagram.

 

Given:

  

Fis the load of the arch

  $F_{H(round)}$ is the horizontal force exerted from the base of the round arch.

  $F_{H(point)}$ is the horizontal force exerted from the base of the point arch.

  $L=8\text{m}$ is the width of the space

Calculations:

The horizontal force exerted to the point arch is one third of the round arch.

From the given statement

  $F{}_{H(point)}=\frac{1}{3}{F}_{H(round)}$

A round arch is simply a semicircle (top half of a circle, diameter 8m) resting on 2 supports. The arch itself can be thought of as 2 halves.

From diagram, the line of action of F (pointing left and acting through the top of L) is a distance $\frac{8}{2}\text{m}$ (radius of arch) from the base of L. So, it produces a moment (torque) about the base of L equal to $F\times 8/2.$

If the height of the pointed arch above the base is h.

The horizontal force exerted to the point arch is one third of the round arch.

From the given statement,

  $F_{H(point)}=\frac{1}{3}{F}_{H(round)}$

Moment about this force is = $\frac{1}{3}{F}_{H(round)}\times h$

These two must be equal as they balance the moment

  $\begin{array}{l}\frac{1}{3}{F}_{H(round)}\times h=F_{H(round)}\times 8/2\\ h=8\times 3/2\\ h=12m\end{array}$

Conclusion:Height of the pointed arch must be $12m$ .