Question 6.3.5: Figure1 shows a cross section through a 100-ft-long concrete...

Goal Find the magnitude of the total force F represented by hydrostatic gage pressure acting on the seawall and its location (center of pressure).
Given Information about the geometry of the seawall the specific gravity of seawater.
Assume The water is at rest, the system is in equilibrium, and we can treat the system as planar.
Draw Based on the information given in the problem and our assumptions, we draw the water acting on the seawall (Figure 2). We define y positive downward. We chose to work with gage pressure; therefore we don’t draw atmospheric pressure on the back side of the seawall.
Formulate Equations and Solve Because hydrostatic pressure must always act perpendicular to a surface, the direction of the pressure varies along the height of the wall. As shown in Figure 2, the pressure acts horizontally at the top of the wall and vertically at the bottom. To account for this we need to integrate the horizontal components and the vertical components separately and combine them to calculate the total force \left\|F\right\| =\sqrt{F^{2}_{x} + F^{2}_{y} } :
At any point on the seawall the differential force dF is the pressure at that point multiplied by the differential area dA:

dF = p(x, y)dA

Breaking dF into its rectangular components gives

dF_{x} = p(x, y) dA cosθ                   (1)
dF_{y} = p(x, y) dA sinθ                   (2)

We represent the hydrostatic gage pressure in terms of cylindrical coordinate system so that ultimately we can integrate (1) and (2) with respect to θ. Based on (6.22A) (written in terms of y)

P_{gage} = \rho_{liq} gh                        (6.22A)
p(x, y) = ρgy = ρgr sinθ                           (3)

We define a differential element r dθ along the length of the seawall (L) as

dA = L(rdθ )                             (4)

We substitute (3) and (4) into (1) and (2) to get

dF_{x} =ρ  gr sinθ (Lrdθ ) cosθ                             (5)
dF_{y}= ρ gr sin θ (Lrdθ ) sinθ                             (6)

and integrate (5) and (6) from 0 to π /2 to find the total force in the horizontal and vertical directions:

F_{x}=\int_{0}^{\pi/2}{\rho gLr^{2}\cos \theta \sin \theta d\theta } =- \rho gLr^{2} \left[\frac{\cos ^{2}\theta }{2} \right] ^{\pi/2}_{0} =\frac{\rho gLr^{2}}{2}
F_{y}=\int_{0}^{\pi/2}{\rho gLr^{2} \sin^{2} \theta d\theta }=\rho gLr^{2}\left[\frac{\theta }{2} – \frac{\sin 2\theta }{4} \right] ^{\pi/2}_{0} =\frac{\rho gLr^{2}\pi }{4}

The total force is found from:

\left\|F\right\|=\sqrt{F^{2}_{x} + F^{2}_{y} }=\sqrt{\left\lgroup\frac{\rho gLr^{2}}{2} \right\rgroup ^{2}+ \left\lgroup\frac{\rho gLr^{2}\pi}{4} \right\rgroup ^{2} }=\frac{\rho gLr^{2}}{2}\sqrt{1+ \frac{\pi ^{2} }{4} }

Substituting numerical values gives

\left\|F\right\|=\frac{\left\lgroup 64.3 \frac{lb }{ft^{3}} \right\rgroup \left(100  ft\right) \left(25  ft\right) ^{2} }{2}\sqrt{1+ \frac{\pi ^{2} }{4}}= 3740  kip

The total force F acts perpendicular to the quarter-circle surface at angle \theta _{CP} , as shown in Figure 3. The center of pressure (CP) will then be at

X _{CP}=r \cos \theta _{c}    ,   Y _{CP}=r \sin \theta _{c}

and \theta _{CP} is defined in terms of F_{y} andF_{x} as

\theta _{CP}=\tan ^{- 1} \frac{F_{y}}{F_{x}}=\tan ^{- 1}\frac{\frac{\rho gLr^{2}\pi}{4} }{\frac{\rho gLr^{2}}{2} }= \tan ^{- 1}\frac{\pi }{2}=57.5^{\circ }              (7)

Therefore

X _{CP}=r \cos \theta _{CP}=25  ft cos(57.5° )= 13.4 ft
Y _{CP}=r \sin \theta _{CP}=25 ft sin(57.5°) = 21.1 ft

Note: We avoided integrating to find the center of pressure, because we were able to identify the relationship in (7) between the force components and their resultant, which must be perpendicular to the quarter-circle surface.
Check We could check our result by using another approach to solve this problem. This approach, which is based on calculating forces acting on projected areas, is explored in Exercise 6.3.20.