gtu fm civil hydrostatics ppt topic: static forces on surface

1. PREPARED BY,
ENROLLMENT NO :-
150490106033
150490106034
150490106035
150490106036
150490106037
TOPIC - STATICS FORCES ON SURFACE
SUBJECT :- FLUID MECHANICS
SNPIT&RC,UMRAKH

2. FLUID STATICS
Deals with problems associated with fluid at
rest.
No relative motion between adjacent layers.
Thus, no shear stress (tangential stress) to
deform the fluid.
Liquid – hydrostatic; Gas – aerostatic.

3. FLUID STATICS
The only stress in fluid statics is normal
stress (perpendicular to surface)
 Normal stress is due to pressure (Pressure:
gravity field-weight of fluid)
 Variation of pressure is due only to the weight of
the fluid → fluid statics is only relevant in
presence of gravity fields.

4. HYDROSTATIC FORCES
F=PA
Only when the
pressure distribution
is uniform over the
entire area of interest.

5. HYDROSTATIC FORCES ON PLANE
SURFACES
Problem
 To find the resultant force FR (magnitude) and its point of
application (center of pressure) for non-uniformly distributed
pressure.
**Atmospheric pressure Patm can be neglected when it acts on
both sides of the surface.

6. WHAT IS A RESULTANT FORCE AND CENTRE OF
PRESSURE ?
Engineering Mechanics
Resultant force:
A force that results from the
combination of two or
more forces.
COP:
A point where the entire
forces (Resultant force)
were concentrated at a
single point.

7. HOW TO DETERMINE THE RESULTANT FORCE
Consider a flat plate completely
submerged in a liquid.
Plane of the top surface intersects
with horizontal surface with an
angle Ө, the line of intersection
will be the x axis.

8. HOW TO DETERMINE THE RESULTANT FORCE, FR
Absolute pressure at any point of the fluid,
P = Po + ρgh, h = y sin Ө
= Po + ρgy sin Ө,
FR = ∫PdA =∫ (Po + ρgy sin Ө)dA = PoA + ρg sin Ө ∫ydA
**∫ydA is the first moment of area is related to the y
coordinate of the centroid (or centre) of the surface by ,
yc= 1/A ∫ydA.
FR = (Po + ρgyc sin Ө) A = (Po + ρghc) A
= PcA
The magnitude of the resultant force, FR acting on a plane surface of a completely
submerged plate in a homogeneous (constant density) fluid is = the product of the
pressure Pc at the centroid of the surface and the area A of the surface.

9. CENTROID AND CENTROIDAL MOMENTS OF
INERTIA FOR SOME COMMON GEOMETRIES

10. SUBMERGED RECTANGULAR
PLATE : HOW TO DETERMINE
THE RESULTANCE FORCE, FR.

11. HOW TO DETERMINE THE LOCATION OF THE COP?
Line of action of resultant force
FR=PCA does not pass
through the centroid of the
surface. In general, it lies
below the centroid because
pressure increase with
depth.

12. HOW TO DETERMINE THE LOCATION OF THE RESULTANT
FORCE (CENTER OF PRESSURE) ?
The vertical location of the line of action is determined by
equating the moment of the resultant force, FR to the
moment of the distributed pressure force about the x-
axis.
Then,
ypFR = ∫yPdA =∫ y(Po + ρgy sin Ө)dA
= Po ∫ ydA + ρgsin Ө ∫y2dA
= PoycA + ρgsin Ө Ixx,o

13. Moment of the resultance force = Moment of the
distributed pressure force about the x axis.
xx,oco
A
2
A
o
A
o
A
Rp
ρgsinθIAyP
dAyρgsinθydAP
ρgysinθ)dAy(P
yPdAFy







Ixx,o = ∫y2dA is actually the second moment of area
about the x axis passing through point O.

14. Normally, the second moment of area is given about the axes
passing through the centroid of the area. Therefore, we need
parallel axis theorem to relate the Ixx,o and Ixx,c
AyII ccxxoxx
2
,, 
 AgPy
I
yy
oc
cxx
cp
)sin/(
,


Therefore,
If Po = 0
Ay
I
yy
c
cxx
cp
,


15. SUBMERGED RECTANGULAR PLATE : HOW TO
DETERMINE THE C.O.P.
 AgPy
I
yy
oc
cxx
cp
)sin/(
,


3
,
12
;;
2
ab
IabA
b
sy cxxc 
abgP
b
s
abb
sy
o
p






)sin/(
2
12
2
3


16. IF Ө=90
ab
b
s
abb
syp






2
12
2
3

17. Example 11-1; Calculate FR, C.O.P and discuss whether the driver can open
the door or not.
APF CR
]m1*m2.1A[; 


101.3kNAPF cR
2
cc
84.4kN/m
b/2)ρg(sρghP
my
ab
b
s
abb
sy
p
p
61.8
2
12
2
3







The moment acting on the mid point of the submerged
door is 101.3kN x 0.5m = 50.6kNm, which is 50 times
of the moment the driver can possibly generate. CAN’T
OPEN.

18. HYDROSTATIC FORCES ON CURVED SURFACES
 Complicated: FR on a curved surface requires
integration of the pressure forces that change
direction along the surface.
 Easiest approach: Determine horizontal and
vertical components FH and FV separately.

19. HYDROSTATIC FORCES ON CURVED SURFACES
1. Vertical surface of the liquid block,
BC = projection of the curved surface
on a vertical plane (vertical
projection).
2. Horizontal surface of the liquid
block, AB = projection of the curved
surface on a horizontal plane
(horizontal projection).
3. Newton’s 3rd Law – Action and
reaction. The resultant force acting on
the curve liquid surface = the force on
curved solid surface.

20. • Assume that the direction to the right and up is
positive,
Horizontal force component on curved surface:
FH – FX = 0.
The horizontal component of the hydrostatic
force (FH) acting on a curved surface is (both
magnitude and line of action) equal to the
hydrostatic force (FX) acting on the vertical
projection of the curved surface.
• Vertical force component on curved surface:
FV - FY – W = 0, where W is the weight of the liquid
in the enclosed block W=gV.
The vertical component of the hydrostatic force
(FV) acting on a curved surface is (both magnitude
and line of action) equal to the hydrostatic force
(FY) acting on the horizontal projection of the
curved surface, plus or minus (depend on
direction) the weight of the fluid block.

21. HYDROSTATIC FORCES ON CURVED SURFACES
Magnitude of force FR=(FH
2+FV
2)1/2
Angle of force is a = tan-1(FV/FH)

22. A GRAVITY CONTROLLED CYLINDRICAL GATE
Example 11-2 A long solid cylinder, R=0.8, hinged at point A.
When water level at 5m, gate open.
Determine:
1. The hydrostatic force acting on the cylinder and its line of action when the gates open
2. The weight of cylinder per m length of the cylinder

23.  Friction at hinge is negligible
 The other side of the gate is
exposed to the atmosphere,
therefore, the Patm is
cancelled out.
A) Determine the net horizontal and vertical force, FH and FV respectively.
FH = FX = PCA = ρghcA; hc = 4.2 + 0.8/2
FY = ρghcA; hc = 5m ; W=ρg(R2 – (πR2/4))
FY - FV - W = 0;
Solve for resultant force and angle.
B) Think about the moment acting at point A
due to the cylinder weight and also resultant
force

24. THANK YOU