1. HYDROSTATIC FORCES PREPARED BY: AAMINA RAJPUT LECTURER CED

2. INTRODUCTION It deals with pressure at rest. Total pressure is defined as the force exerted by a static fluid on a surface either plane or curved when the fluid comes in contact with the surface. This force is always normal to the surface. Centre of pressure is defined as the point of application of the total pressure on the surface. The submerged surfaces case can be  Horizontal plane surface  Vertical plane surface  Inclined plane surface  Curved plane surface

3. HORIZONTAL PLANE SURFACE SUBMERGED IN LIQUID

4. VERTICAL PLANE SUBMERGED IN LIQUID The total pressure on the surface may be determined by dividing the entire surface into a number of small parallel strips. The force on small strip is then calculated the total pressure force on the whole area is calculated by integrating the force on small strip. Consider a strip of thickness dh and width b at a depth of h from the free surface of liquid. h’ hphp Consider a plane vertical surface of arbitrary shape immersed in a liquid, let A = total area of surface G = centre of gravity of plane surface P = centre of pressure h’ = distance of C.G of area from free surface of liquid h p = distance of centre of pressure from free surface of liquid

5. VERTICAL PLANE SUBMERGED IN LIQUID

7. It is clear that 1. Centre of pressure lies below the centre of gravity of the vertical surface 2. The distance of centre of pressure from free surface of liquid is independent of the density of the liquid.

8. PLANE SURFACEC.G. from the base AreaMoment of inertia about an axis passing through C.G. and parallel to base (I G ) Moment of inertia about base (I o ) bd R- 0.1102 R 4 -

9. PRESSURE DIAGRAMS Centre of pressure from surface, centroid of triangle = 2/3H

10. INCLINED PLANE SURFACE SUBMERGED IN LIQUID Let A = total area of inclined surface h’ = depth of C.G of inclined area from free surface h p = distance of centre of pressure from free surface of liquid Let the plane of the surface, if produced meet the free liquid surface at O. then O-O is the axis perpendicular to the plane of surface. hphp h’

11. INCLINED PLANE SURFACE SUBMERGED IN LIQUID

15. CURVED SURFACE SUBMERGED IN LIQUID Consider a curved surface, AB, submerged in a static fluid. Let dA is the area of small strip at a depth of h from water surface. But here as the direction of the forces on a small areas are not in the same direction, but varies from point to point. Hence integration of equation for curved plate is impossible. The problem can, however, be solved by resolving the force dF in two components dF x and dF y in the x- and y-directions respectively.

16. CURVED SURFACE SUBMERGED IN LIQUID

18. PROBLEM # 1 A rectangular plane surface is 2m wide and 3m deep. It lies in vertical plane in water. Determine the total pressure and position of centre pressure on the plane surface when its upper edge is horizontal and a) coincides with water surface, b) 2.5m below the free water surface.

21. PROBLEM # 2 Determine the total pressure on a circular plate of diameter 1.5m which is placed vertically in water in such a way that the centre of the plate (C.G.) is 3m below the free surface of water. Find the position of centre of pressure also.

22. PROBLEM # 3 A circular opening, 3m diameter, in a vertical side of a tank is closed by a disc of 3m diameter which can rotate about a horizontal diameter. when the head of water above the horizontal diameter is 4m (i.e. C.G.). Calculate: a)The force on the disc and b)The torque required to maintain the disc in equilibrium in the vertical position

24. PROBLEM # 4 Determine the total pressure and centre of pressure on an isosceles triangular plate of base 4m and height 4m when it is immersed vertically in an oil of sp.gr. 0.9. The base of the plate coincides with the free surface of oil.

25. PROBLEM# 5 A tank contains water upto a height of 0.5m above the base. An immiscible liquid of sp.gr. 0.8 is filled on the top of water upto 1m height. Calculate: a)Total pressure on one side of the tank b)The position of centre of pressure for one side of the tank, which is 2m wide.

27. PROBLEM# 6 A vertical sluice gate is used to cover an opening in a dam. The opening is 2m wide and 1.2m high. On the upstream side of a gate, the liquid of sp.gr. 1.45, lies upto a height of 1.5m above the top of the gate, whereas on the downstream side the water is available upto a height touching the top of the gate. Find the resultant force acting on the gate and position of centre of pressure. Find also the force acting horizontally at the top of the gate which is capable of opening it. Assume that the gate is hinged at the bottom.

30. PROBLEM # 7 (Assignment) A tank width 2m is full of water. Find: a)Total pressure on the bottom of tank b)Weight of water in the tank c)Hydrostatic paradox between the result of a) and b) Solution: a)282528 N b)70632 N c) it is observed total weight of water in the tank is much less than the total pressure at the bottom of tank. This is known as hydrostatic paradox. 0.4 m 3m 0.6m 4m

31. PROBLEM # 8 (Assignment) A closed tank, rectangular in plan with vertical sides, is 1.8m deep and contains water to a depth of 1.2 m. Air is pumped into the space above the water until the air pressure is 35kNm −2. If the length of one wall of the tank is 3 m, determine the resultant force on this wall and the height of the centre of pressure above the base. R Air = 35 x 1.8 x 3 = 189kN R H2O = area of triangle x width of tank = ½ x (rho x g x h) x 1.2 x 3 = 0.5 x (1000 x 9.81 x 1.2) x 1.2 x 3 = 21.189kN R = R Air + R H2O = (189 + 21.19) × 10 3 = 210.19 × 10 3 N. If x is the height above the base of the centre of pressure through which R acts, R × x = R Air × (1.8/2) + R H2O × (1/3 X 1.2) x = (189 × 0.9 + 21 × 0.4)/210.19 = 0.85m.

32. PROBLEM # 9 (Assignment) The angle between a pair of lock gates is 140° and each gate is 6 m high and 1.8m wide, supported on hinges 0.6 m from the top and bottom of the gate. If the depths of water on the upstream and downstream sides are 5 m and 1.5 m, respectively, estimate the reactions at the top and bottom hinges.

33. Solution F is the force exerted by one gate on the other and is assumed to act perpendicular to the axis of the lock if friction between the gates is neglected. P is the resultant of the water forces P1 and P2 acting on the upstream and downstream faces of the gate, and R is the resultant of the forces RT and RB on the top and bottom hinges. Upstream water force, P1 = ρgA 1 D 1 = 10 3 × 9.81 × (5 × 1.8) × 5/2 = 220.725 × 10 3 N, Downstream water force, P2 = ρgA 2 z 2 = 10 3 × 9.81 × (1.5 × 1.8) × 1.5/2 = 19.865 × 10 3 N, Resultant water force on one gate, P = P1 − P2 = (220.73 − 19.86) × 10 3 N = 200.87 × 10 3 N. If P acts at a distance x from the bottom of the gate, then by taking moments about O, P*x = P1 × (5/3) − P2 × (1.5/3) = (220.73 × 5/3 − 19.86 × 0.5) × 10 3 = 357.95 × 10 3 Nm. x = (357.95 × 10 3 )(200.87 × 10 3 ) = 1.782m. Assuming that F, R and P are coplanar, they will meet at a point, and, since F is assumed to be perpendicular to the axis of the lock on plan, both F and R are inclined to the gate as shown at an angle of 20° so that F = R P = F sin 20° + R sin 20° = 2R sin 20°, R = P/2sin20° = 293.65 × 10 3 If R is coplanar with P it acts at 1.78 m from the bottom of the gate. Taking moments about the bottom hinge, 4.8R T = 1.18R R T = 1.18/4.8 × 293.65 × 10 3 = 72.2 × 10 3 N = 72.2kN, R B = R − R T = 293.65 − 72.2 = 221.45kN

34. PROBLEM # 10

35. PROBLEM # 11 Compute the horizontal and vertical components of the total force acting on a curved surface AB, which is in the form of a quadrant of a circle of radius 2m as shown. Take the width of the gate as unity.

37. PROBLEM # 12 A gate having a quadrant shape of radius 2m. Find the resultant force due to water per metre length of gate. Find also the angle at which the total force will act.

39. PROBLEM # 13 A sluice gate is in the form of a circular arc of radius 6 m. Calculate the magnitude and direction of the resultant force on the gate, and the location with respect to O of a point on its line of action. Depth of water, h = 2 × 6 sin 30° = 6 m, Horizontal component of force on gate = R h per unit length = Resultant force on PQ per unit length = ρg × h × h/2 = ρgh 2 /2 = (10 3 × 9.81 × 36)/2 N/m = 176.58 kN/m, Vertical component of force on gate = R v per unit length = Weight of water displaced by segment PSQ = (Sector OPSQ − ΔOPQ)ρ g = [ (60/360) × π × 6 2 − 2 x ½ x 6 sin 30° × 6 cos 30° ] × 10 3 × 9.81N/m = 32.00kN/m, Resultant force on gate, R =sqrt (176.58 2 + 32.00 2 ) = 179.46 kN/m. If R is inclined at an angle θ to the horizontal, tan θ = R v /R h = 32.00/176.58 = 0.181 22 θ = 10.27° to the horizontal. Since the surface of the gate is cylindrical, the resultant force R must pass through O.