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STATICS - CASE STUDY SOLUTION

    Pressure Distribution


Solving for a Single Slice


Water Pressure Resultant Force

 

The pressure at a point on the submerged arch is perpendicular to the surface and depends on the water depth at that point. If an axis system is oriented at the center of the arch, the pressure at any height y can be determined as

     p = ρg(b - y)

where ρ is the water density, g is the acceleration of gravity and b is the total depth of the water to the floor. If the angle θ is as shown, then

     y = a sinθ

and

     p = ρg(b - a sinθ)

The pressure at any angle θ can be broken into x and y components as

     px = -p cosθ
          = ρg(a sinθ - b) cosθ

     py = -p sinθ
          = ρg(a sinθ - b) sinθ

   
    Resultant Force


Integrate Over Surface Area S
 

The resultant force is found by integrating the pressure over the arch surface, S, which means integrating θ from 0 to π and multiplying by the length of the arch.

     

For the x component:

Integrate or use integral tables to find

     Rx = 270 ρg [a/2 sin2θ - b sinθ]0π

Evaluating the integral for Rx gives

     Rx = 0 lb

The y component is

Integrating and evaluating for Ry

     Ry = 270ρg [a/2 (-cosθ sinθ + θ) b cosθ]0π

     Ry = 270ρg (aπ/2 - 2b) lb

Finally, substitute for the known density, water depth, and arch radius to give

     Ry = -436,200 lb = -436 kips

Therefore, in vector form, the resultant force is

     R = 0i - 436j + 0k kips

     
   
 
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