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Forces due to Static Fluids

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Presentation on theme: "Forces due to Static Fluids"— Presentation transcript:

1 Forces due to Static Fluids
Pressure =Force/Area (definition) Force = Pressure*Area Example: If a cylinder has an internal diameter of 50 mm and operates at a pressure of 20 bar, calculate the force on the ends of the cylinder.

2 Example: Cylinder Calculate the force on the bottom of the container.
Oil, S.G. = 0.90 1.5 m Water, S.G. =1.0

3 Example: Cone Shaped Container
Calculate the force on the bottom of the container. 2.4 m Oil, S.G. = 0.90 1.5 m Water, S.G. =1.0 3 m

4 Force-Pressure: Rectangular Walls
Patm Vertical wall DP = g*h d

5 Derivation of Resultant Force (FR) on a rectangular (vertical) wall
FR = g*(d/2)*A = Pavg*A Pavg = g*(d/2) LP = d/3 Patm DP = g*h d FR Center of Pressure (CP) d/3

6 Procedure for computing the resultant force on a rectangular (vertical) wall
Calculate the magnitude of FR FR = g*(d/2)*A = Pavg*A; Pavg = g*(d/2) Locate center of pressure at a vertical distance d/3 from the bottom of the wall (comes from sums of forces and moments, SF = 0; SM = 0) Show the resultant force acting at the center of pressure perpendicular to the wall

7 Example Fluid is gasoline (S.G. = 0.68) and the total depth is 30 m. The wall is 12 m long. Calculate the magnitude of the resultant force on the wall and the location of the center of pressure.

8 Example A dam, 30.5 m long, which retains 8 m of water and is inclined at an angle of 60°. Calculate the magnitude of the resultant force on the dam and the location of the center of pressure. 60° d = 8 m

9 Submerged Plane Areas Definitions FR: Resultant Force
CP: Center of Pressure CT: Centroid of the area - q: Angle of inclination hc: Depth of fluid from free surface Lc: Distance from free surface to centroid of area Lp: Distance from free surface to center of pressure B, H: Dimensions of area

10 Procedure for computing FR on a submerged plane area
Identify the point S, where inclined plane intersects free surface Locate the centroid of the interested area Determine hC Determine LC, hc = Lcsinq Calculate A, area of interest, A = BH Calculate FR, FR = ghcA Calculate IC, Moment of Inertia, Ic = BH3/12

11 Procedure for computing FR on a submerged plane area
Determine LP; LP = LC + IC/LCA, comes from SF and SM Show FR, LP, LC Compute and show hp, hp = Lpsinq

12 Example Lubricating oil with a S.G. of The rectangular gate with the following dimensions B = 2m and H = 0.5 m is placed in the inclined wall of the tank (q = 60°). The centroid of the gate is at a depth of 1.5 m from the surface. Calculate FR and LP.

13 Piezometric Head hA hA = PA/g hAC = hA + hC hAC hC

14 Example Consider previous example, except the tank is sealed at the top and the pressure is 10 k.Pa gauge above the oil.

15 Forces on Curved Surfaces
h1 h2

16 Submerged Curved Surface
Resultant force can be broken up into horizontal and vertical components or vectors Horizontal component: FH = g*s*w*(h + s/2), Where, h = Depth to the top of rectangle (beginning of curve surface) s = projected rectangle height w = projected rectangle length or width Center of pressure hp = hC + IC/(hCA) hC = h + s/2 hp = hC + s2/(12hC)

17 Submerged Curved Surface
Vertical Component FV = g*Volume = g*A*w Where, A = entire area of fluid w = projected rectangle length or width Location of centroid Resultant Force

18 Procedure for computing the Force on Submerged Curved Surface
Isolate the volume of fluid above the surface Compute the weight of the isolated volume Compute the vertical component of force and its centroid Draw a projection of the curved surface onto a vertical plane and determine its height, s.

19 Procedure for computing the Force on Submerged Curved Surface
Compute the depth to the centroid of the projected area, hc Compute the horizontal component Compute the depth to the line of action (center of pressure, hp) Compute the resultant force Compute the angle of inclination Show or draw vector on diagram

20 Example Compute the horizontal and vertical components, and resultant force on the curve surface h1 = 3.00 m h2 = 4.50 m w = 2.50 m g = 9.81 kN/m3 h1 h2


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