Presentation on theme: "Hydrostatic Forces on Curved, Submerged Surfaces"— Presentation transcript:
1 Hydrostatic Forces on Curved, Submerged Surfaces xPressure is always acting perpendicular to the solid surface since there is no shear motion in static condition.PzPqPxZqqdAx=dAcos(q)dAz=dAsin(q)
2 Projected ForcesxhdAzZIntegrated overall elements
3 Buoyancy Force acting down FD= rgV1 from Buoyancy = FU-FD =rg(V2-V1)=rgVV: volume occupied by the objectForce acting up FU = rgV2from
4 Horizontal Forces x Z h Projected area Ax dAz Finding Fx is to determine the force acting on a plane submerged surface oriented perpendicular to the surface. Ax is the projection of the curved surface on the yz plane. Similar conclusion can be made to the force in the y direction Fy.ZIntegrated overall elementshdAxEquivalent system:A plane surface perpendicular to the free surface
5 ExamplesDetermine the magnitude of the resultant force acting on the hemispherical surface.x2 mR=0.5 mzequalsminus
6 Line of Action zC z’ Projection in x-direction Horizontal direction: line of action goes through z’zCz’Projectionin x-directionVertical direction: the line of action is 3R/8away from the center of the hemispherezThe resultant moment of both forces with respect to the center of the hemisphere should be zero:Fx( )-Fz(0.1875)=15386( )-2564(0.1875)=0location of the centroid for a hemisphereis 3R/8=0.1875(m) away from the equator plane
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