Pressure at a Point: Pascal’s Law Pressure is the normal force per unit area at a given point acting on a given plane within a fluid mass of interest. Blaise Pascal (1623-1662) How does the pressure at a point vary with orientation of the plane passing through the point? Pressure Forces F.B.D. Gravity Force Wedged Shaped Fluid Mass p is average pressure in the x, y, and z direction. Ps is the average pressure on the surface q is the plane inclination is the length is each coordinate direction, x, y, z ds is the length of the plane g is the specific weight V = (1/2dydz)*dx

Pressure at a Point: Pascal’s Law For simplicity in our Free Body Diagram, the x-pressure forces cancel and do not need to be shown. Thus to arrive at our solution we balance only the the y and z forces: Rigid body motion in the y-direction Pressure Force in the y-direction on the y-face Pressure Force on the plane in the y-direction Pressure Force in the z-direction on the z-face Pressure Force in the plane in the z-direction Rigid body motion in the z-direction Weight of the Wedge Now, we can simplify each equation in each direction, noting that dy and dz can be rewritten in terms of ds:

Pressure at a Point: Pascal’s Law Substituting and rewriting the equations of motion, we obtain: Math Now, noting that we are really interested at point only, we let dy and dz go to zero: Pascal’s Law: the pressure at a point in a fluid at rest, or in motion, is independent of the direction as long as there are no shearing stresses present.

Pressure at a Point: Pascal’s Law p1dxds p2dxds psdxds ps = p1 = p2 Note: In dynamic system subject to shear, the normal stress representing the pressure in the fluid is not necessarily the same in all directions. In such a case the pressure is taken as the average of the three directions.

Pressure Field Equations How does the pressure vary in a fluid or from point to point when no shear stresses are present? Consider a Small Fluid Element Surface Forces p is pressure is specific weight Taylor Series Body Forces V = dydzdx For simplicity the x-direction surface forces are not shown

Pressure Field Equations Looking at the resultant surface forces in the y-direction: Similarly, looking at the resultant surface forces in the x and z-direction, we obtain: Expressing these results in vector form:

HYDROSTATIC FORCES F=PA

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

WHAT IS A RESULTANT FORCE AND CENTRE OF PRESSURE ? . در اين بخش راجع به مقدار نيروي برآيند و خط اثر آن (مركز فشار) بحث مي كنيم. .

HOW TO DETERMINE THE RESULTANT FORCE  

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.

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

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.

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

Moment of the resultance force = Moment of the distributed pressure force about the x axis. Ixx,o = ∫y2dA is actually the second moment of area about the x axis passing through point O.

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 Therefore, If Po = 0

CENTROID AND CENTROIDAL MOMENTS OF INERTIA FOR SOME COMMON GEOMETRIES

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

If Ө=90

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.

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.

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.

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.

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

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

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

BUOYANCY An objects lighter in fluid compare to in an air. Fluid exerts an upward force on a body immersed in it.

BUOYANCY Buoyant force is caused by the increase of pressure with depth in fluid. Figure 10-13 Difference between pressure at the top and bottom surface is ρg(s+h)A - ρghA = ρgsA = ρgV, V = hA, volume of the plate.

Buoyancy 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.

Buoyancy 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

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

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.

Buoyancy Calculation Example 11-3 and 4 Q 11-33 and 11-34