Forces on a curved surface
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1 Civil Engineering Hydraulics Forces on Curved Surfaces There is a theory which states that if ever anybody discovers exactly what the Universe is for and why it is here, it will instantly disappear and be replaced by something even more bizarre and inexplicable. There is another theory which states that this has already happened.-douglas Adams Example.9 A concrete culvert that contains water is.0 m in diameter. Determine the forces exerted on the portion labeled A B in Figure.18 if the culvert is filled halfway. Determine also the location of the forces. Culvert length (into the paper) from joint to joint is.5 m. 1
2 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 xdirection will be developed using the weight of water above the curved surface. 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.
3 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.5 m (we are looking at a single section of culvert which is noted as.5 m from joint to joint. So the area that the pressure acts over is Azy = 1.0m.5m =.5m 5 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.5m =.5m zc = 6 1.0m = 0.5m
4 So the magnitude of the force can be determined using the expression for a vertically oriented surface submerged in a fluid. Azy = 1.0m.5m =.5m zc = 1.0m = 0.5m kg m Fh = Rh = ρ gzc Azy = ( 0.5m ).5m m s ( ) Rh = 16.5N = 1.6kN 7 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. A = 1.0m.5m =.5m zy 1.0m = 0.5m kg m Fh = Rh = ρ gzc Azy = ( 0.5m ).5m m s z = zc = ( ) Rh = 16.5N = 1.6kN bh (.5m ) (1.0m ) = = m 1 1 Iz m zr = zc + c = 0.5m + = 0.667m zc Azy ( 0.5m).5m Iz = ( 8 )
5 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. π (.0m ) 1 π D 1 l =.5m = 1.96m V= 9 So the force is the weight of this volume of fluid. π (.0m ) 1 π D 1 l =.5m = 1.96m kg m Fv = Rv = ρ gv = (1.96m ) = N m s Rv = 19.17kN V= 10 5
6 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. π (.0m ) 1 πd 1 l =.5m = 1.96m kg m Fv = Rv = ρ gv = (1.96m ) = N m s Rv = 19.17kN V= x = xc = 11 D.0m = = 0.m π π When the culvert described in Example.9 was installed and still empty, it was buried halfway in mud (Figure.19). Determine the forces acting on half the submerged portion, assuming that the mud has a density equal to that of water. 1 6
7 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. 1 Example.11 Figure.0a shows a gate that is 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. 1 7
8 Problem Note: Be careful to include all of the volume above the surface DE for both the force and centroid calculations. Problem
9 Problem
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