1. Chapter 4 FLUID STATICS HYDROSTATIC FORCES AND BUOYANCY

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 – hydrostatics; Gas – aerostatics.

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.
Applications: Floating or submerged bodies,
water dams and gates, liquid storage tanks,
etc.

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

5. HYDROSTATIC FORCES
What about the areas where the pressure is non-uniformly distributed ??
FIND THE AVERAGE PRESSURE AND WHERE IT ACTS

6. 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.

7. The Hoover Dam is an example of an arch-gravity dam.
The Gathright Dam in Virginia is a rock-fill embankment dam.

8. The Grand Coulee Dam is an example of a solid gravity dam.
Gordan Dam, Tasmania is an arch dam
The Koshi Barrage dam

9. 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.
The centroid gives a definition of the mean position of an area (volume). It is closely related to the center of
mass a body.

10. 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 z axis.

11. 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,
dFR = PdA =∫ (Po + ρgy sin Ө)dA = PoA + ρg sin Ө ∫ydA
if Po, ρ, g and Ө are constant
**∫ydA is the first moment of area with respect the z axis can be written as, yc A = ∫ ydA
where yc is the y coordinate of the centriod of area A measured from the z axis which passes through O
The 1st moments of areas are the average displacement of an area about an axis of rotation. They are closely related to the centroid.
FR = (Po + ρgyc sin Ө) A = (Po + ρghc) A
= PcA=PavgA
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.

12. CENTROID AND CENTROIDAL OF INERTIA MOMENTS(2nd MOMENT OF AREAS) FOR SOME COMMON GEOMETRIES

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

14. 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.

15. The vertical location of the line of action is
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 = ∫A yP dA =∫A y(Po + ρgy sin Ө) dA
= Po ∫A y dA + ρgsin Ө ∫A y2 dA
= PoycA + ρgsin Ө Izz,o

16. The 2nd moments of areas are the average (displacement)2 of an area about an axis of rotation. Has units of m4
Izz,o = ∫y2dA is actually the second moment of area (also called the area moment of inertia)about the z axis passing through point O.

17. 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 Izz,o and Izz,c
Therefore,
If Po = 0(atmospheric pressure)

18. SUBMERGED RECTANGULAR PLATE : HOW TO DETERMINE THE C.O.P.

19. If Ө=90

20. Example 11-1; Calculate FR, C. O
Example 11-1; Calculate FR, C.O.P and discuss whether the driver can open the door or not.
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.

21. EXAMPLE
A 5-m-high, 5-m-wide rectangular plate blocks the end of a 4-m-deep freshwater channel, as shown below. The plate is hinged about a horizontal axis along its upper edge through a point A and is restrained from opening by a fixed ridge at point B. Determine the force exerted on the plate by the ridge.

22. Then the resultant hydrostatic force on each wall becomesFR
Fridge
s = 1 m
h = 4 m A B
Then the resultant hydrostatic force on each wall becomes
The line of action of the force passes through the pressure center, which is 2h/3 from the free surface,
Taking the moment about point A and setting it equal to zero gives

23. Solving for Fridge and substituting, the reaction force is determined to be
Discussion The difference between FR and Fridge is the force acting on the hinge at point A.

24. EXAMPLE
The 200-kg, 5-m-wide rectangular gate shown below is hinged at B and leans against the floor at A making an angle of 45° with the horizontal. The gate is to be opened from its lower edge by applying a normal force at its center. Determine the minimum forces F required to open the water gate.

25. FR F
45 B
0.5 m
3 m A
The average pressure on a surface is the pressure at the centroid (midpoint) of the surface, and multiplying it by the plate area gives the resultant hydrostatic on the surface,
The distance of the pressure center from the free surface of water along the plane of the gate is

26. The distance of the pressure center from the hinge at point B is
Taking the moment about point B and setting it equal to zero gives
Solving for F and substituting, the required force to overcome the pressure is
In addition to this, there is the weight of the gate itself, which must be added. In the 45o direction,
Thus, the total force required in the 45o direction is the sum of these two values,

27. EXAMPLE
Below shows an inclined plane wall 4 meter long, partially immersed in oil. The wall is subjected to a resultant hydrostatic force exerted by oil pressure.
Calculate the resultant hydrostatic force on the wall due to the oil pressure.
Determine the location of the centre of pressure (COP). Show the resultant hydrostatic force on the wall.
Answer:
FR = 46.8 kN
Yp = 1.32 m

28. Hydrostatic Forces on Curved Surfaces
Liquid Above Surface
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.

29. 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.

30. 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=rgV.
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.

31. Hydrostatic Forces on Curved Surfaces
Magnitude of force FR=(FH2+FV2)1/2
Angle of force is a = tan-1(FV/FH)

32. EXAMPLE Given: Surface AB goes 1 m into the paper
Find Fv, FH, W, Fx, Fy, F, Line of action for FH & Fv
Given: Surface AB goes 1 m into the paper
Fx= FH = (5 x 9810) (2 x 1) = 98.1 kN
Pres. at the cenroid AC side area
Fy= W + Fv
Fv= (9810 x 4) x (2 x 1) = 78.5 kN
W= γVABC= 9810 (1/4 x  r2) 1 = 30.8 kN
Fy= = kN
The hydrostatic force acting on AB
is equal and opposite to the force
F shown

33. The centroid of the Quarter-circular area
Location of the resultant force

34. BUOYANCY
An objects lighter and weigh less in fluid compare to in an air.
Fluid exerts an upward force on a body immersed in it.
This force that tends to lift the body is called the buoyant force or the force of buoyancy.

35. Difference between gage pressure at the top and bottom surface is
Buoyant force is caused by the increase of pressure with depth in fluid.
Figure 10-13
Difference between gage pressure at the top and bottom surface is
FB = Fbottom – Ftop
= ρfg(s+h)A – ρfgsA = ρfghA = ρfgV,
V = hA, volume of the plate.
Buoyant force

36. We conclude that the buoyant force acting on the plate is equal to the weight of the liquid displaced by the plate.

37. Archimedes principal : The buoyant force acting on a body immersed in a fluid is equal to the weight of the fluid displaced by the body, and it acts upward through the centroid of the displaced volume.
FB=ρfgV
Density of fluid

38. The tendency of fluid to exert a supporting force on a body placed in the fluid.
The force = weight of the fluid displaced by the body. Its act upward through centroid of the displaced volume.
FB=rfgV
Density of fluid

39. Bouyancy of Floating Bodies
FB = W (The weight of the entire body must be equal to the buoyant force)
ρf gVsub = ρave,bodygVtotal
Vsub / Vtotal = ρave,body / ρf (The submerged volume fraction is equal to the ratio of the average density of the body to the density of the fluid)
Completely submerged when, ρave,body >= ρf
Submerged portion (yellow)
LIQUID
BOUYANT FORCE =
WEIGHT OF DISPLACED FLUID
Displaced fluid whose volume = yellow volume

40. An object will
sink if the weight of the object is greater than the buoyant force
move upwards if the buoyant force is greater than its weight
remain stationary in a liquid or float on the surface of the liquid if the buoyant force is equal to its weight.

41. A ship floats on the surface of sea water if the upward buoyancy force that it receives is great enough to overcome the downward force of its weight. A steel ship is hollow inside so that the overall density of the ship is less than that of sea water. The level a ship sinks in the sea depends on the buoyant force that acts on it.
The deeper a ship sinks in the sea, the larger the volume of sea water that is displaced. Hence the buoyant force increases. The density of sea water varies with location. This will vary the level that the ship sinks in the sea. The denser the sea water, the larger the buoyant force. The ship will submerge deeper in fresh water than in sea water. The density of sea water varies with the region it is in and the season it sails. Therefore, most ships are marked with the Plimsoll line(waterline). The Plimsoll line is painted on ships to show the depth a ship should travel at.

42. To float a submarine: a) water is forced out from the ballast tank with compressed air b) this decreases the density and weight of the submarine c) the buoyant force is larger than the weight of the submarine, causing the submarine to float. If the density of the submarine is the same as the density of the sea water, the depth of the submarine is maintained at one position. A submarine moves forward with the help of a propeller.
Cross section of a vessel with a single ballast tank at the bottom.
A submarine has a large ballast tank used to control the position and depth of the submarine from the surface of the sea. To sink a submarine: a) the ballast tank is filled with water b) this increases the density and weight of the submarine c) the submarine sinks when its weight is larger than the buoyant force.

43. BUOYANCY Buoyance force by air is so small,
0.1m3, ρair = 1.2 kg/m3 , 1.2N. If mass = 80 kg, weigth = 788 N. Ignore the buoyancy, error is 0.15%, so small.
Rise of warm air – natural convection currents.
The rise of hot air or Helium balloons, and air movements in the atmosphere.

44. Example

45. Example - Wood block (1) has dimensions of 10‐mm x 50 mm x 50 mm
Specific gravity of 0.3
- Metal object (2) has volume of 6600 mm3
– Find the tension in the cable
and mass of object 2.

47. Example
A rectangular pontoon has a width B of 6 m, a length l of 12 m, and a draught D of 1.5 m in fresh water (density 1000 kg/m3). Calculate :
a) The weight of the pontoon
b) Its draught in sea water (density 1025 kg/m3)
c) The load (in kiloNewtons) that can be supported by the pontoon in fresh water if the maximum draught permissible is 2 m.
Solution
When the pontoon is floating in an unloaded condition,
Up trust on immersed volume = weight of pontoon
Since the up trust is equal to weight of the fluid displaced,
Weight of pontoon = weight of fluid displaced,
So,

48. In fresh water, and D=1.5 m
Therefore,
Weight of pontoon,
b) In the sea water, ;therefore
Draught in sea water,
c) For maximum draught of 2 m in fresh water,
Total uptrust = weight of water displaced
Load which can be supported = Upthrust – weight of pontoon

49. Example
A raft is make by tying 5 empty steel barrels to a bamboo platform, as shown in next slide. The bamboo weigh 100 kg and each barrel is 0.5 m in diameter, 1 m in length and has a mass of 15 kg. What is the maximum load in kilogram that the raft can carry before it sinks in fresh water?
Data:
Number of barrels = 5
d = 0.5 m
l = 1.0 m
Mass of each barrel = 15 kg
Mass of bamboo = 100 kg

50. w Fu
SOLUTION
Total weight of barrels,
5 x 15 x 9.81 = N
Weight of bamboo,
100 x 9.81 = 981 N
Total weight, N

51. The volume of water displaced by each barrel,
The volume of water displaced by 5 barrels,
5 x = 0.98 m 3
Therefore, upthrust,
Fu = 0.98 x 9.81 x 1000 N
= N
The nett upward force,
Fu – w = – = 7897 N
Thus the mass that the raft can carry,
7897/9.81 = 805 kg

52. Example
A river barge, whose cross section is approximately rectangular, carries a load of grain. The barge is 5.6 m wide and 18 m long. When unload its draft (depth of submergence) is 1 m, and with the load of grain the draft is 1.4 m. Determine: (a) the unloaded weight of the barge, and (b) the weight of the grain.
Example
The homogeneous wooden block A is 0.7 m by 0.7 m by 1.3 m and weighs 2.4 kN. The concrete block B (Specific Weight=23.6 kN/m3) is suspended from A by means of the slender cable causing A to float in the position indicated. Determine the volume of B.