 Ch 2. Fluid Statics Multimedia Engineering Fluids PressureVariation PressureMeasurement HydrostaticForce(Plane) HydrostaticForce(Curved) Buoyancy
 Chapter 1. Basics 2. Fluid Statics 3. Kinematics 4. Laws (Integral) 5. Laws (Diff.) 6. Modeling/Similitude 7. Inviscid 8. Viscous 9. External Flow 10. Open-Channel Appendix Basic Math Units Basic Equations Water/Air Tables Sections Search eBooks Dynamics Fluids Math Mechanics Statics Thermodynamics Author(s): Chean Chin Ngo Kurt Gramoll ©Kurt Gramoll FLUID MECHANICS - THEORY

In this section, the discussion is focused on how to determine hydrostatic forces exerted on a plane surface submerged in a static fluid. Determining this hydrostatic force is particularly crucial in the design of engineering structures, such as dams, storage tanks and hydraulic systems. The concept of using a pressure prism to obtain the hydrostatic force will also be introduced.

Hydrostatic Force on a Plane Surface Hydrostatic Force on an
Inclined Arbitrary Plane Surface

Consider an inclined plane submerged in a static fluid as shown in the figure. The resultant force FR is acting perpendicular to the plane since no shear force is present when the fluid is at rest. FR has a line of action that passes through the point (xcp , ycp), which is called the center of pressure. Note that the pressure acting perpendicular to the plane surface is also referred to as the normal stress.

Now take a small differential element dA at a depth of h. The differential force dF acting on dA is given by

dF = ρgh dA

The magnitude of the resultant force can be obtained by integrating the differential force over the whole area The integral represents the first moment of the area about the x axis, which is equal to where yc is the y coordinate of the centroid of the plane surface.

From trigonometry,

hc = yc sinθ

where hc is the vertical distance from the fluid surface to the centroid of the plane surface. The resultant force is simplified to

 FR = ρghcA Parallel Axis Theorem

The center of pressure, xcp and ycp can be obtained by summing moments about the y and x axis, respectively. First, by equating the sum of moments of all pressure forces about the x axis to the moment of the resultant force: where is the second moment of the area or the area moment of inertia (Ix) about the x axis. According to the parallel axis theorem, the moment of inertia can also be written as

Ix = Ix' + Ayc2

where Ix' is the second moment of the area with respect to the centroidal axis, which is parallel to the x axis.

Hence, the center of pressure coordinate ycp is given by Similarly, xcp can be obtained by equating the sum of moments of all pressure forces about the y axis to the moment of the resultant force where is the product of inertia (Ixy) of the area about the x and y axes. Once again according to the parallel axis theorem, it can also be written as

Ixy = Ixy' + Axcyc

where Ixy' is the product of inertia of the area with respect to the centroidal axes.

Hence, the coordinate xcp is given by The values for the second moment of inertia and product of inertia with respect to the centroidal axes for various common geometries are given in the appendix. From the formulations of xcp and ycp, it is noted that the center of pressure is always lower in the liquid than the centroid of the plane area.

Pressure Prism Hydrostatic Pressure Force Exerted
on a Vertical Plane Surface Pressure Prism

An alternate approach of determining the hydrostatic force is by means of a pressure prism. Consider a vertical plane submerged in a static fluid, as shown in the figure. The pressure increases linearly with the depth. One can then easily construct a corresponding three-dimensional diagram of the pressure distribution, and such a volume is called a pressure prism. The resultant force is the total volume of the pressure prism, that is

FR = Volume = 1/2 (ρgh) (bh)

The resultant force passes through the centroid of the pressure prism. For this particular example, the centroid of a triangular element is located at a distance of h/3 from its base and lies in the vertical symmetry axis.

As illustrated, this method is particularly convenient when the shape of the pressure prism is a common geometry, in which the volume and centroid can be readily obtained.

Practice Homework and Test problems now available in the 'Eng Fluids' mobile app
Includes over 250 problems with complete detailed solutions.
Available now at the Google Play Store and Apple App Store.