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Civil Engineering Hydraulics

Forces on Curved Surfaces

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Forces on a curved surface

Forces on Curved Surfaces 2

Example 2.9 A concrete culvert that contains water is 2.0 m in diameter. Determine the forces exerted on the portion labeled A–B in Figure 2.18 if the culvert is filled halfway. Determine also the location of the forces. Culvert length (into the paper) from joint to joint is 2.5 m.

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Forces on a curved surface

Forces on Curved Surfaces 3

We typically break up the force resulting from the action of the fluid on the curved surface into components normal to the fluid surface and tangential to the fluid surface. In this case, our axis has z oriented vertically and x oriented horizontally. The force in the z-direction will be developed using pressure and the force in the x-direction will be developed using the weight of water above the curved surface.

Forces on a curved surface

Forces on Curved Surfaces 4

Consider the y-direction as into the page/slide. There would be a projected area in the y-z plane which would be the same as we considered previously. This means that to determine the Force labeled Rh below, we can use the same expressions that we used previously.

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Forces on a curved surface

Forces on Curved Surfaces 5

The magnitude of the force will be determined by the force of the pressure on the projection of the curved surface into the y-z plane. In the example this would be a rectangle with a depth of 1.0 m and a width of 2.5 m (we are looking at a single section of culvert which is noted as 2.5 m from joint to joint. So the area that the pressure acts over is

Azy =1.0m × 2.5m = 2.5m2

Forces on a curved surface

Forces on Curved Surfaces 6

We also need the distance to the centroid of the projected area from the surface of the fluid. In this example, the fluid is at the top of the rectangle so the distance to the centroid will be ½ the height of the projected area.

Azy =1.0m × 2.5m = 2.5m2

zc =1.0m2

= 0.5m

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Forces on a curved surface

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So the magnitude of the force can be determined using the expression for a vertically oriented surface submerged in a fluid.

Azy =1.0m × 2.5m = 2.5m2

zc =1.0m2

= 0.5m

Fh = Rh = ρgzcAzy = 1000 kgm3

⎛⎝⎜

⎞⎠⎟ 9.81m

s2⎛⎝⎜

⎞⎠⎟ 0.5m( ) 2.5m2( )

Rh =12262.5N =12.26kN

Forces on a curved surface

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And the location of the force can be determined using the expression for a vertically oriented surface submerged in a fluid. In this case the projected surface is a rectangle with the top at the liquid surface. Azy =1.0m × 2.5m = 2.5m2

z = zc =1.0m2

= 0.5m

Fh = Rh = ρgzcAzy = 1000 kgm3

⎛⎝⎜

⎞⎠⎟ 9.81m

s2⎛⎝⎜

⎞⎠⎟ 0.5m( ) 2.5m2( )

Rh =12262.5N =12.26kN

Iz =bh3

12=2.5m( ) 1.0m( )3

12= 2.08×10−1m4

zr = zc +IzczcAzy

= 0.5m + 2.08×10−1m4

0.5m( ) 2.5m2( ) = 0.667m

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Forces on a curved surface

Forces on Curved Surfaces 9

The magnitude of the vertical force on the curved surface is the weight of fluid above the curved surface. Be careful to include all the fluid above the curved surface.

V = πD2

414

⎛⎝⎜

⎞⎠⎟ l =

π 2.0m( )24

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⎛⎝⎜

⎞⎠⎟ 2.5m =1.96m3

Forces on a curved surface

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So the force is the weight of this volume of fluid.

V = πD2

414

⎛⎝⎜

⎞⎠⎟ l =

π 2.0m( )24

14

⎛⎝⎜

⎞⎠⎟ 2.5m =1.96m3

Fv = Rv = ρgV = 1000 kgm3

⎛⎝⎜

⎞⎠⎟ 9.81m

s2⎛⎝⎜

⎞⎠⎟ 1.96m

3( ) =19261.89NRv =19.17kN

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Forces on a curved surface

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The line of action of the force is through the centroid of the volume of fluid. In this case it will be through the xc of the quarter circle. Take care to know where the centroid is taken in reference to.

V = πD2

414

⎛⎝⎜

⎞⎠⎟ l =

π 2.0m( )24

14

⎛⎝⎜

⎞⎠⎟ 2.5m =1.96m3

Fv = Rv = ρgV = 1000 kgm3

⎛⎝⎜

⎞⎠⎟ 9.81m

s2⎛⎝⎜

⎞⎠⎟ 1.96m

3( ) =19261.89NRv =19.17kN

x = xc =4 D2

⎛⎝⎜

⎞⎠⎟

3π=4 2.0m

2⎛⎝⎜

⎞⎠⎟

3π= 0.42m

Forces on a curved surface

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When the culvert described in Example 2.9 was installed and still empty, it was buried halfway in mud (Figure 2.19). Determine the forces acting on half the submerged portion, assuming that the mud has a density equal to that of water.

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Forces on a curved surface

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In this case the fluid is acting on the outside or the curved surface. We can calculate the forces and lines of action in exactly the same way. The only difference is that the directions of the forces are reversed.

Example 2.11

Forces on Curved Surfaces 14

Figure 2.20a shows a gate that is 4 ft wide (into the page) and has a curved cross section. When the liquid level gets too high, the moments due to liquid forces act to open the gate and allow some liquid to escape. For the dimensions shown, determine whether the liquid is deep enough to cause the gate to open. Take the liquid to be castor oil.

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Problem 6-1

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Note: Be careful to include all of the volume above the surface DE for both the force and centroid calculations.

Problem 6-2

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Problem 6-3

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