Because most civilizations use submerged surfaces such as dams to control flooding and manage their water resources, engineers must understand pressure changes and design their equipment to perform under extreme situations.
When you dive into the deep end of a swimming pool, you can feel the pressure pushing all around, and the deeper you swim (or sink, if you're not a strong swimmer), the stronger the pressure you feel. At large depths (such as the bottom of the ocean), these fluid pressures can be downright deadly, which is why professional deep-sea divers must use specialized equipment to survive.
In this chapter, I show you some of the basic calculations that you can perform on a submerged surface. I explain the types of forces that are created by fluid pressures and how to calculate their quantities. I also show you how to apply fluid pressures to your free-body diagrams (F.B.D.s). Finally, I explain how to calculate partial fluid pressures on gates and openings.
For the purposes of this text, I deal only with incompressible fluids (or fluids that don't change volume) such as water. The study of incompressible fluids and their pressures can be broken into two categories: dynamic fluids and static fluids.
Dynamic fluids:Dynamic fluids include all fluids in motion or subject to a pressure. The flow of water within the water lines inside your home and the flow of gasoline from the pump to your car are examples of a dynamic fluid.
Static fluids: Static fluids include fluids at rest or fluids subjected to non-pressurized flow. Examples of static fluids include a lake or reservoir and the milk in your morning cereal bowl.
In this text, I deal exclusively with static fluids. The forces from static fluids are classified into two categories: forces from hydrostatic pressure and forces from fluid self weight.
Hydrostatic pressure: Hydrostatic pressure is the pressure associated with the depth of the fluid below the fluid surface.
Fluid self weight: Self weight is gravitational effects acting on the mass particles of the fluid.
In Chapter 9, I define a relationship relating the specific weight (γ) of a material to its density (ρ): γ = ρg, where g is the gravitational acceleration constant.
The material properties of most fluids are dependent on the temperature of the fluid, but for the purposes of this text, I take them as the following:
SI units:ρ = 1,000 kilograms per cubic meter and γ = 9,810 Newton per cubic meter
U.S. customary units: γ = 62.4 pounds per cubic foot. (The U.S. customary units for ρ are really ugly, so I just use specific weight in U.S. customary units.)
Hydrostatic pressure is the pressure of a fluid associated with its depth and is a function of the type of fluid, the gravitational constant, and the depth of the fluid below the fluid surface. Figure 23-1 shows a typical hydrostatic pressure distribution for a fluid.
Hydrostatic pressure is a linear distribution that starts at a value of zero at the fluid's surface and increases linearly with vertical depth. The relationship between pressure p and depth z is given by p = ρgz = γz, where ρ is the density of the fluid, g is the gravitational constant, γ is the specific weight of the fluid, and z is the depth below the surface of the fluid.
So at an arbitrary depth of zA, the pressure is γza. At a depth of 10 feet, the hydrostatic pressure is p10 = (62.4 pcf) · (10 feet) = 624 psf.
The resultant (or total force) of this entire distribution occurs at a distance of (depth/3) above the bottom. This distance is also the centroid (geometric center — see Chapter 11) of a triangular distribution. For more information on resultants of distributions, flip to Chapter 10.
In pressure calculations, you have two types of pressure readings: absolute pressure and gauge pressure.
Absolute pressure is the pressure exerted on an object, including a body of water, by the air and atmosphere above it. However, over small areas, this pressure is usually fairly constant and is taken as an approximate value of 14.68 pounds per square inch or 101.325 kilopascals.
In submerged fluid calculations, you usually make another type of pressure calculation known as gauge pressure, which is a measure of the pressure above atmospheric pressure. The gauge pressure at the surface of a fluid is taken to be zero (because the depth is zero at the surface) and increases with depth.
In Chapter 10, I explain that all linear distributed loads have an intensity that is measured in units of force per length. Notice, however, the hydrostatic pressure calculation earlier in the chapter produced a pressure that was measured in pounds per square foot (which is a force per area unit), which are not the correct units for a linear distributed load.
To work around this units issue, you need to include a width dimension. Submerged surface problems (especially dams and wall structures) are often very long or have an unspecified total length, so you typically assume the forces you're calculating are acting along a unit width, which is equal to either 1 foot or 1 meter (depending on your system of units). Multiplying the previous pounds per square foot units by a unit width of 1 foot produces units of pounds per linear foot, which satisfies the units for a distributed load.
The self weight from a fluid is the weight of a region of fluid acting directly above the object of interest and is the second source of load from fluids. Imagine collecting all the water above an object and storing it in a bucket. If you place that filled bucket on a scale, clearly it has weight which you must apply to the object of interest. Consider the dam shown in Figure 23-2 subjected to a water depth of 10 feet and a horizontal width on the face of the dam of 6 feet.
If you want to determine the weight of water acting on the dam, you include the volume of water multiplied by the specific weight of the fluid:
W = (Specific weight of fluid) · (Volume of fluid).
If you don't know the width of the dam, you'll need to convert the volume calculation to an area calculation multiplied by a unit width, as shown in the following equation:
W = (Specific weight of fluid) · (Two-dimensional area) · (unit width)
If you plug in the numbers for Figure 23-2, you get
Ww = (62.4 pcf) · (0.5 · (10 ft) · (6 ft)) · (1 ft) = 1,872 lb per unit width
The resultant force of this self weight occurs at the center of mass (also the centroid) of the area of the fluid, or in this case at 2/3(6 feet) = 4 feet from the front of the dam.
Note
When you calculate the resultant of the water's self weight, you must also find the centroid of the water's area to know where the resultant is acting.
As with any statics problem, the first step is always to draw the correct free-body diagram, and submerged surface problems are no different. Check out the dam in Figure 23-3a.
To determine the loads acting on the dam's foundation, you must draw an F.B.D. — such as the one in Figure 23-3b — showing all forces acting on the dam, and the assumed support reactions (restraints), which are the foundation forces, consisting of a moment, shear, and normal force. The external loads appearing on this structure include the water forces from the linear hydrostatic pressure distribution, the self weight of the water above the dam, and the self weight of the concrete dam itself.
When you're drawing a free-body diagram for a submerged surface, you must remember that both hydrostatic fluid pressure and self weight from fluids can occur simultaneously on the same structure. However, when the face of the structure is vertical, there is no fluid self weight acting on it. After all, there's no fluid area vertically above a vertical face.
Consider the sloped concrete dam (γCONC = 150 pcf) with dimensions shown in Figure 23-3. The dam holds back water (γH20 = 62.4 pcf) 20 feet deep. A design engineer needs to determine the forces on the base of the dam in order to design the proper foundation, so in the coming sections, I illustrate the steps required to compute these design values.
The hydrostatic pressure for the problem in Figure 23-3 is a linearly varying distribution with a pressure of zero at the water surface and increases to 20(γH20) at the base of the dam.
Tip
This hydrostatic pressure acts along the entire height of the dam. However, to help keep the representation of this distribution clear, I like to draw it slightly off the structure horizontally and align the pressures relative to a vertical reference line. When I make this move, I can clearly see the distribution as a triangular distribution, and then I know how to compute the distribution's resultant. This trick is actually legal because from a statics point of view, all of the forces of the distribution are remaining on their original lines of action.
The first calculation you need to perform determines the hydrostatic fluid pressure distribution. The pressure at the water surface, where z = 0, is p0 = (62.4 pcf) · (0 ft) · (1 ft) = 0 plf.
In Figure 23-3, you know that the depth of the water is 20 feet, so you can calculate the pressure intensity at the bottom of the distribution as p20 = (62.4 pcf) · (20 ft) · (1 ft) = 1,248 plf.
You can then compute the resultant force PRES of the linear pressure distribution (see the preceding section) by determining the area of the linear hydrostatic pressure diagram:
In this case, the pressure distribution is triangular (or linearly distributed), with a value of zero pounds per linear foot at the water surface and 1,248 pounds per linear foot at the bottom. The resultant hydrostatic pressure for a unit width for this example is thus:
This load is acting at the centroid of the pressure distribution, which is located above the bottom at a distance of:
Next, you calculate the weight of the water volume acting directly above the concrete dam. This weight is applied as a single concentrated point load at the center of mass of the water volume (see Chapter 11 for more information about centroids and centers of mass).
To determine the weight of the water above the dam in Figure 23-3, you can compute the area of the water in two dimensions and multiply by the specific weight of the fluid and then multiply by the unit width, using the following formula:
This calculation provides the weight of water per unit width for the water acting directly above the concrete dam.
You also must include the self weight of the dam on the free-body diagram. For convenience of calculation, you can break the dam into two regions, a rectangular region with weight WC1 and a triangular region with weight WC2. The calculations (which follow) are very similar to the calculation for the self weight of water (which you can find in Chapter 9) except that the specific weight for normal-weight concrete is approximately 150 pounds per cubic foot. For the rectangular region of the dam:
This weight is acting at a distance of 5 feet from the back of the dam. For the triangular region:
This weight is acting at a distance of 10 feet from the front toe of the dam.
The reactions for the base of the dam can be modeled as a fixed support and have a horizontal force and a normal force (acting vertically) to prevent translation and a moment to prevent rotation.
The horizontal component of the reaction in Figure 23-3 is a shear force V acting parallel to the base along the interface of the dam and the foundation. Though you draw this force as a single concentrated load, it's actually spread along the entire length of the dam.
The vertical component of this reaction is a normal force N acting perpendicular to the base of the dam. At this time, the point of application (point where the force is acting in space) of this normal force is unknown. But for reasons I explain in Chapter 24, I assume it to be acting at the bottom corner at the back of the dam, or Point O in Figure 23-3b.
Finally, because all of the forces are eccentric to(or not acting at) the back corner of the dam, an equivalent moment M must be present at that point in order to maintain equilibrium and to prevent the dam from overturning.
After you determine the F.B.D. of the dam in Figure 23-3 and apply the forces and support reactions (including the forces from the water and the self weight of the concrete dam), the final step is to actually apply the equations of equilibrium.
Shear V along the base of the dam is computed to ensure horizontal translational equilibrium:
The positive sign on this calculation illustrates that the assumed direction of the base shear of the dam was correct. Next, the normal force N on the dam is computed to verify vertical translational equilibrium:
Finally, you can compute the overturning moment (or the moment that must be resisted to prevent the dam from toppling) M, by summing moments about the tipping point, or Point O, in Figure 23-3. You find the tipping point by examining the dam to determine which point the dam will rotate about should it start to fall over.
The negative sign indicates that the moment direction assumed is backwards on the free-body diagram. This discrepancy actually has significant meaning with respect to overturning of the dam. As the negative sign implies, in order to actually make the dam overturn about Point O, an additional clockwise moment would have to be applied to overcome the weight of the water and self weight of the concrete dam. This finding means that the dam is in a stable (or equilibrium) condition, because in reality that additional moment doesn't actually exist. Hence, the dam can't overturn. Good news for the folks downstream, no doubt!
Dams and hydroelectric installations also employ systems of gates and valves to use forces from flowing water to spin turbines and provide electric power. These gates and openings allow water to flow through to mechanical machinery inside of the structure.
Unless the gate or opening is located at the water surface, the pressure distribution on the opening is no longer triangular, although it does remain linear. Instead, the pressure becomes a trapezoidal shape, with a unique pressure at the top of the opening and another at the bottom of the opening. Consider the pivot gate (a mechanical gate that is hinged or pivoted on one end) shown in Figure 23-4a.
The F.B.D. for the pivot gate contains the reaction forces Bx and By at the pivot and the normal contact force of the gate on the ground Cy. The external forces acting on the gate include the weight of the water volume, which has been broken into a rectangular and a triangular portion (WW1 and WW2) respectively, and the trapezoidal hydrostatic distribution acting directly on the gate or opening.
Note
This trapezoidal region has a pressure of γzB per unit width at the top elevation of the gate and a pressure of γzC per unit width at the bottom elevation. The following calculations give you the pressures for the trapezoidal region in Figure 23-4b:
Tip
To find the resultant, you can calculate the area of this trapezoidal region directly and then perform a separate calculation to determine the centroid of the single force resultant. However, a simpler method is to break this trapezoidal region into a simple rectangular and triangular region and find the resultant and location of each region individually (keeping the math calculations a lot simpler!).
In the example of Figure 23-4, PRES1 represents the resultant of the rectangular subregion and PRES2 represents the resultant of the triangular subregion. You can calculate PRES1 per unit width from the uniform distribution, which has an intensity of γzB and height of (zC – zB).
Note
Because this distribution is rectangular, you also know that its resultant location is at a height of 2 meters (or one half of the total) from the base of the distribution.
Similarly, you can calculate PRES2 per unit width from the triangular region, which has a maximum intensity of γ(zC – zB), and a height of (zC – zB).
You know that the centroid of the triangular region will be at a distance of one third of the height of the opening above the base of the opening.
The self weight of the water acting on the gate extends from Point C to Point B vertically up to the water surface in Figure 23-4a. In this example, the volume of water is a trapezoidal area with a unit width. Just as with the trapezoidal hydrostatic pressure I discuss earlier in this section, you can also easily break this trapezoidal volume into a rectangular region and triangular region.
You calculate the weight of the water in the rectangular region per unit width, WW1, from WW1 = (9.81 kN/m) · (3 m) · (5 m) · (1 m) = 147.2 kN and is located at a distance of 1.5 meters horizontally from Point B (or half of the 3-meter horizontal dimension of the pivot gate). The weight of the water in the triangular region, WW2, per unit width is calculated as
which acts at a horizontal distance of 1 meter (to the right) from Point C.
At this point you have all the forces computed that are acting on the gate. You can now apply the equilibrium equations as I do in the preceding section to compute the vertical reaction at Point C, or the forces on the pin at Point B, depending on what information you want to determine.