§2.5 Total hydrostatic pressure of liquid acting on a horizontal surface

Determining the magnitude, direction and action point of the total hydrostatic pressure of liquid acting on a horizontal surface is a mechanical problem which many engineering techniques have to solve.The methods used to calculate the total hydrostatic pressure on a horizontal surface include analytic method and graphic method.

2.5.1 Analytic method

As shown in Fig.2-16, MN is a proj ection of an inclined plane at an angleθwith the horizontal plane.The area of the compressed face in the horizontal plane is A. Atmospheric pressure is exerted on the surface of water.The intersection between the extended plane of MN and free surface of liquid is designated as Ox axis.Oy axis is downward and perpendicular to Ox axis.The coordinate plane, in which the horizontal surface locates, rotates 90°around Oy axis in order to exhibit the geometric relationship of plane xOy.h is the depth of water at an arbitrary point.y is the distance from this point to Ox axis.C represents the centroid of the compressed face.

1．Magnitude and direction of force

An infinitesimal area dA in a horizontal strip was considered as the obj ect of study at the depth h of water.The pressure distribution is uniform in this infinitesimal surface.Thus, the pressure exerted on the infinitesimal surface is:

d F=p d A=ρg h d A=ρg y sinθd A

The direction of the pressure is orthogonal to dA and same as that of interior normal.

Because MN is a surface, the direction of the pressure acting on any infinitesimal area is parallel to each other.Their algebraic sum can be calculated by integrating the equation above.The total pressure acting on the surface is:

whereis the static moment of the compressed face with respect to Ox axis.Its magnitude equals the area A of the compressed face multiplied by the coordinate yC of centroid.

Withbeing substituted into the equation above, and hC=yC sinθ, we can write where F is the total hydrostatic pressure acting on the surface, hC is the submerged depth of the centroid of the compressed face, and pC is the pressure at the centroid of the compressed face.

Equation(2-19)indicates the magnitude of the total hydrostatic pressure acting on the surface of arbitrary shape in arbitrary direction is equal to the hydrostatic pressure at the centroid of the compressed face multiplied by the area.In other words, the average pressure acting on any compressed face is equal to the pressure exerted on its centroid. Note that the magnitude of the total hydrostatic pressure is independent of the inclined angleθand depends only on the specific weight of the fluid, the area and the depth of the centroid below the surface.The direction of total hydrostatic pressure is the same as that of interior normal of the compressed face.

2．Acting point of total hydrostatic pressure—center of pressure

The point through which the total hydrostatic pressure acts is called center of pressure denoted as D, as shown in Fig.2-16.The acting point can be located according to the law of resultant moment, i.e.the moment of the resultant force on an arbitrary axis is equal to the algebraic sum of the moment of each component force on the axis. On the x axis, the following equation can be established as:

whereis the moment of inertia of the compressed face A with respect to Ox axis, i.e.After this equation being substituted into the equation above, we can write

F· yD=ρg sinθ· Ix

After F=ρg sinθ· yC· A being substituted into the equation, we can write

According to the law of parallel motion, we can write

ICx is the moment of inertia of the compressed face with respect to the centroidal axis which passes through its centroid and parallel to Ox axis.Thus, the distance from the acting point of the total hydrostatic pressure on the surface to Ox axis can be calculated as:

This equation indicates the location of the center of pressure is independent of the inclined angle θ of the compressed face.Because , the center of pressure is always below the centroid (yD≥yC)and move closer to the centroid with the increase of the immersion depth.Only when the compressed face is horizontal, the center of pressure and centroid overlap.

In practical engineering, compressed face is generally axis-symmetric plane, where the axis is parallel to Oy axis.The acting point of the total pressure F must locate on the symmetrical axis, i.e.xD=xC.Thus, the center of pressure can be determined as long as the position of the center of pressure in the direction of y is calculated.

Example 2-5 A vertical rectangular door was placed in the water (see Fig.2-17). The distance from the gate top to the water surface is h1=1 m.The height of the gate is h2=2 m.The width of the gate is b=1.5 m.Calculate the total hydrostatic pressure and acting point on the gate.

Solution:According to equation(2-19), the total hydrostatic pressure is:

After the data being substituted, F can be calculated as:

According to equation (2-19), the location of the center of pressure can be determined by

The moment of inertia with respect to the rectangular centroidal axis is . After the data being substituted into the equation above, can be calculated as:

The location of the center of pressure is 2.17 m below the water surface, and the direction is horizontal to the right.

Example 2-6 There is a vertical semi-circle plane.As shown in Fig.2-18, the front is facing water.The diameter of water attaining surface exactly lies in the surface of liquid.Calculate the magnitude and acting point of the total hydrostatic pressure (hC= ).

Solution:

Substitute and into equation(2-20), Then, we can write

The total hydrostatic pressure is and the center of pressure is located below water surface.

2.5.2 Graphic method

Graphic method is more suitable for the calculation of total hydrostatic pressure and acting point on the specification plane (for example, rectangle).Pressure distribution graph need be plotted in order to calculate the total hydrostatic pressure.

1．Hydrostatic pressure distribution map

The law of hydrostatic pressure distribution can be described by geometrical figure. The length of the line indicates the magnitude of the pressure at a point.The arrow at the line ends represents the direction of the pressure at a point, i.e.the direction of the interior normal of the compressed face. The figure is a combination of lines perpendicular to the action surface, which is called hydrostatic pressure distribution map.It is necessary to note the hydrostatic pressure in the hydrostatic pressure distribution map is relative pressure.Because all around the building is in the atmosphere, the atmospheric pressure in each direction counteracts each other.It is known the hydrostatic pressure is directly proportional to the depth, i.e.the relationship between the hydrostatic pressure and depth is linear correlation.If the compressed face is a plane, the pressure distribution map must be described by straight lines and two points can determine a straight line.If the compressed face is a curved surface, the distance between envelope and the curved surface represents the magnitude of pressure, which is directly proportional to the depth of the action point on the curved surface.If the curved surface is circular arc, each action line of pressure must pass through the center of the circle.Fig.2-19 shows different pressure distribution maps for different scenarios.

2．Graphic method

There is a vertical rectangular gate with the height of h and the width of b.The top edge is on an even height with the water surface.The plane below the water is ABCD, as shown in Fig.2-20.

According to equation(2-19),

Where the magnitude of ρg h 2 is equal to the area of the hydrostatic pressure distribution graphΩ, and the unit ofΩis N/m.

Thus, the equation above can be rewritten as:

Equation(2-21)indicates the magnitude of the total hydrostatic pressure is equal to the volume of the distribution graph of pressure acting on the plane.

where h is the depth of the centroid of the pressure distribution graph below the water surface.The action line of the total hydrostatic pressure passes through the centroid of the pressure distribution graph and points to the action surface.

Example 2-7 Use graphic method to calculate the total hydrostatic pressure and the location of the center of pressure in the example 2-5.

Solutions:Plot the pressure distribution graph of rectangular gate:the bottom is the area of the compressed face and the height is the pressure at each point as shown in Fig.2-21b.

According to the principle of graphic method, the magnitude of the total pressure is the volume of pressure distribution graph.Then:

Action line passes through the center of the pressure distribution graph, i.e.the centroid of the trapezoid.The equation for the centroid of the trapezoid is:

where a, b and h is the width of the upper and the bottom and the height of the trapezoid, respectively (a=ρg h 1, b=ρg(h 1+h 2), h=h 2).

Example 2-8 A rectangular plane is tilted in the water(Fig.2-22a).The top of the rectangular plane is 1 m below the water surface (h=1m)and the bottom is 3 m below the water surface (H=3m).The width of the rectangle is b=5m.Use analytic method and graphic method to calculate the total hydrostatic pressure acting on the plane and locate the action point of the total hydrostatic pressure.

Solution:(1)Analytic method

The length of the rectangular plane along the y axis is:

and

According to equation(2-19), F=ρg hC· A=9800N/m3×2m×20m2=392000N=392kN

and

where , and

and hD=yD sin 30°=2.17 m

The total hydrostatic pressure is 392 kN.The center of the pressure locates 2.17 m below water surface and the direction of the pressure is the direction of interior normal of the compressed face.

(2)Graphic method

Because this is a regular rectangular plane, graphic method can be used to calculate the total hydrostatic pressure.The pressure distribution graph was plotted and divided into two parts including a rectangle and triangle, as shown in Fig.2-22b.

It is known.The area of the trapezoid can be calculated as:

Then, F=Ω b=8m2×ρg×5m=40m3×9800N/m3=392000N=392kN

Because the equation for the centroid point coordinate of the trapezoid is complicated, the trapezoid can be divided in to a simple triangle and rectangle in order to calculate its area and the center of pressure.

For the triangle,

For the rectangle, F 2=ρg h b l=20m3×ρg

According to the law of resultant moment, F·yD=F1 yD1+F2 yD2

After the data being substituted into this equation, yD can be calculated as:

and hD=2.17 m

The results are the same as that calculated by means of the analytic method.