Transcript Hydrostatic Forces on Curved, Submerged Surfaces

Hydrostatic Forces on Curved, Submerged Surfaces
x
Pz
P
q
Pressure is always acting
perpendicular to the solid surface
since there is no shear motion in
static condition.
Px
P=Pcos(q )i+Psin(q )j
Z
Integrate over the entire surface
dF=dFx i  dFz j  PdA
q
q dA =dAcos(q)
x
 (Pcos(q )i+Psin(q )j)dA
dFx  PdA cos(q )  PdAx
dFz  PdA sin(q )  PdAz
dAz=dAsin(q)
Projected Forces
dFx  PdA cos(q )  PdAx , dFZ  PdA sin(q )  PdAZ
Integrate over the entire surface:
Fx   dFx   PdAx    ghdAx , Fz   dFz   PdAz    ghdAz
x
where  is the fluid volume enclosed
h
dAz
Z
Fz   g  hdAz   g
between the curved surface and the
free surface. The force is equal to the
weight of the total colume of fluid
Integrated over
all elements
directly above the curved surface.
The line action passes through the
center of gravity of the volume
of liquid being displaced.
Buoyancy
Force acting down FD= gV1
from
Buoyancy = FU-FD
=g(V2-V1)=gV
V: volume occupied by the object
Force acting up FU = gV2
from
Horizontal Forces
x
Fx   dFx   PdAx    ghdAx
h
Projected
area Ax
dAz
Z
Integrated over
all elements
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.
h
dAx
Equivalent system:
A plane surface perpendicular
to the free surface
Fx   dFx   PdAx    ghdAx
Examples
Determine the magnitude of the resultant force acting on the hemispherical surface.
x
2m
R=0.5 m
Fx   PdAx   g  ydAx   ghC Ax
A x is the projection of the sphere along the x direction
and it has a shape of a circle A x = (0.5) 2  0.785( m 2 )
Fx  (1000)(9.8)(2)(0.785)  15386( N )
Fz   PdAz   g  zdAz   g    g (   )
 
 (1000)(9.8) 1

4  R 3  2564( N )
2
3
z
F  Fx i  Fz k  15386i  2564k
The line of action of the force
equals
minus
must go through the center of the
hemisphere, O (why?)
Line of Action
Horizontal direction: line of action goes through z’
R
I xxˆ ˆ
4
z '  zC 
 (2) 
AzC
 R 2 (2)
4
z’
zC
Projection
in x-direction
 2.03125( m)
Vertical direction: the line of action is 3R/8
away from the center of the hemisphere
z
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
location of the centroid for a hemisphere
is 3R/8=0.1875(m) away from the equator plane