Hydrostatic Forces on Curved, Submerged Surfaces

Pressure is always acting perpendicular to the solid surface since there is no shear motion in static condition.

P

Px

Pz

x

Z

x

x

P=Pcos( ) i+Psin( )j

Integrate over the entire surface

dF=dF

(Pcos( ) i+Psin( )j)dA

dF cos( )

sin( )

z

x

z z

i dF j PdA

PdA PdA

dF PdA PdA

dAx=dAcos()

dAz=dAsin()

Projected Forces

x

x z

dF cos( ) , sin( )

Integrate over the entire surface:

F , F

x Z Z

x x x z z z

PdA PdA dF PdA PdA

dF PdA ghdA dF PdA ghdA

x

ZIntegrated overall elements

h

dAz

where is the fluid volume enclosed

between the curved surface and the

free surface. The force is equal to the

weight of the total colume of fluid

directly above the curved surface.

The li

z zF g hdA g

ne action passes through the

center of gravity of the volume

of liquid being displaced.

Buoyancy

Force acting down FD= gV1

from

Force acting up FU = gV2

from

Buoyancy = FU-FD

=g(V2-V1)=gVV: volume occupied by the object

Horizontal Forcesx

ZIntegrated overall elements

h

dAz

xF x x xdF PdA ghdA

h

dAx

Equivalent system: A plane surface perpendicular to the free surface

Projectedarea Ax 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.

xF x x xdF PdA ghdA

Examples

Determine the magnitude of the resultant force acting on the hemispherical surface.

2 m R=0.5 m

z

xx

x

2 2x

F

A is the projection of the sphere along the x direction

and it has a shape of a circle A = (0.5) 0.785( )

(1000)(9.8)(2)(0.785) 15386( )

( )

(1000)

x x C x

x

z z z

PdA g ydA gh A

m

F N

F PdA g zdA g g

31 4(9.8) 2564( )2 3 R N

minusequals

15386 2564

The line of action of the force

must go through the center of the

hemisphere, O (why?)

x zF F i F k i k

Line of Action

Horizontal direction: line of action goes through z’

z

zC

4

ˆ ˆ2

4' (2)(2)

2.03125( )

xxC

C

RIz z

Az R

m

Vertical direction: the line of action is 3R/8away from the center of the hemisphere

Projection in x-direction

location of the centroid for a hemisphereis 3R/8=0.1875(m) away from the equator plane

The resultant moment of both forces with respect to the center of the hemisphere should be zero:Fx(2.03125-2)-Fz(0.1875)=15386(0.03125)-2564(0.1875)=0

z’