🟦 12.1 What is fluid statics?
Fluid statics is a topic that is covered in two different courses: Statics and Fluid Mechanics. In a static analysis, we only consider fluids at rest. We do not study fluids in motion (a waterfall, system of pipes, etc.) in Statics. You will (likely) study fluids in motion in other courses.
In Statics, we wish to investigate the structure (the container or the tank) that holds the fluid. For instance, we may want to determine whether or not a drinking glass is strong enough to contain the fluid pressure generated by the contents. Before we can tackle that sort of problem, we first need to calculate the load the fluid imposes on the submerged surface/s.
❏ Fluids and their quest for a low energy state
We aren't studying the fluid itself -- we are studying the effect of the fluid on the solid bodies that surround it.
Our analysis procedure typically has four steps:
Understand the context: cut a FBD that isolates the portion of the tank geometry you're interested in from the fluid itself (cut the fluid away)
Compute the pressure: construct a pressure distribution on the surface of interest
Equivalent systems: convert the area load into a statically equivalent line load or point load (resultant force)
E.o.E.: apply the E.o.E. to the FBD to solve for unknowns (e.g. forces A and B)
❏ The four basic steps of a fluid statics problem
🟦 12.2 Mass density vs. weight density
❏ Conventional assumptions
The theory in this lesson applies to all fluids, but for simplicity, we will principally study water. There are two different ways to communicate density:
We typically use 𝜌 (pronounced “rho”) for the mass density (mass per volume) of a fluid.
We typically use 𝛄 (pronounced “gamma”) for the weight density (weight per volume) of a fluid. This is also called the unit weight.
The density of water does change as a function of salinity (salt water vs. fresh water) and temperature. With that in mind, it is conventional in fluid statics computations to make the following assumptions regarding the weight of water.
SI units:
mass density: 𝜌 = 1000 kg / m³
weight density: 𝛄 = 𝜌g = 9.81 kN / m³
U.S. customary units:
mass density: not commonly used
weight density: 𝛄 = 62.4 # / ft³
❏ An alternate approach
As the example illustrates, any numeric analysis for problems involving the self-weight of water will generate decimals.
There is an optional math trick you can use to speed up your work on fluid problems. Instead of using the more precise 𝛄 = 9.81 kN / m³ , round up to 10 kN / m³ . Once you get to the final answer, you multiply it by the adjustment factor of (9.81/10). This is permissible because the density parameter has a linear effect on the analysis. In U.S. Customary Units, you could use 𝛄 = 100 # / ft³, and then adjust your final answer by multiplying by (62.4/100). If you choose to use this approach, be sure to effectively communicate your approach to your audience with commentary.
❏ Weight densities to memorize
❏ Example
🟦 12.3 Fluid statics in the wild
It's hard to find good photos of submerged surfaces ... all that water gets in the way! Here are two structures that are safely supporting loads imposed on them by the water.
❏ The Hoover Dam, Boulder City, Nevada, USA
🟦 12.4 Fluid pressure
We can think of a fluid as a continuum of differential-scale fluid particles. Imagine cutting a tiny cubic FBD of a particle of fluid by cutting out all six faces (+x, -x, +y, -y, +z, and -z).
If we then take the limit as dx, dy, and dz all approach zero, we zoom in to a single particle (or point) of fluid.
Each fluid particle exerts a push on the neighboring particles. Per Newton's Third Law, the neighboring particles push right back.
All of this pushing is given the term fluid pressure.
Fluid pressure (force per area) is a subcategory of area load.
Of course, the fluid particle that is in contact with the fluid's container (or tank) delivers a push to that surface.
❏ A continuum of fluid particles
Conceptually, when we are talking about the fluid, we say that fluid pressure is directionless. That means that the pressure is equal in all directions (up, down, north, south, east, west, as well as any inclined geometry you can imagine).
In Statics, we wish to study the fluid's container -- or any submerged surface (perhaps a concrete dam or the hull of a submarine). We integrate the fluid pressure over the surface area in order to generate a normal force that lies perpendicular to the surface.
Inspect these examples of resultant forces (the force that results from the fluid pressure) on the interior surface of different shaped containers.
Did you notice the blue triangle at the top of the image? That symbol is commonly used to communicate the location of the top surface of the fluid.
❏ The orientation of resultant forces on various surfaces
🟦 12.5 Pascal's Law
We use Pascal's Law to compute the static fluid pressure at a specified depth. Pascal's Law only applies to incompressible fluids (like water). It isn't applicable to compressible fluids (like gases).
Pascal's Law states that fluid pressure is a function of two parameters: (1) the weight density (𝛄), and (2) the head (or height) above the point, measuring straight up to the surface of the fluid.
p = fluid pressure at a depth (or at a point) = 𝛄h
where h = "head" or height of the column of fluid above the particle
and 𝛄 is the weight density (or unit weight) of the fluid
❏ Definition of "head"
You can think of fluid pressure as the weight of the column of fluid that lies above a unit horizontal plane that containing the point (e.g. a flat square that measures 1x1 meters or 1x1 feet).
Here is Pascal's Law, paraphrased for fluid statics:
the magnitude of fluid pressure increases linearly with depth and acts perpendicular to the geometry of each submerged surface.
❏ How to compute pressure
❏ Pascal's Law illustrated
In the example below, Pascal's Law is used to compute the fluid pressure at points A, B, C, and D in four tanks of water.
❏ Pascal's Law in practice
Some texts incorrectly use the term hydrostatic pressure for both water and non-water fluids. Since “hydro” means water, “hydrostatic” means that the fluid is water. It's better to use the generic term static fluid pressure to refer to fluids in general, and hydrostatic pressure when you are specifically referring to water.
🟦 12.6 Pressure prisms: horizontal and vertical surfaces
In a FBD, we have often isolated a body (a member, a pin, a portion of a cable) by cutting away the context to free the body -- while being certain to replace everything we cut away with its effect on the body.
In fluid statics, we will cut away the fluid particles in contact with a solid surface, and replace them with a field of pressure vectors that are:
compressive;
perpendicular to the surface geometry; and
computed with Pascal's Law.
In the interactive, you'll see how to calculate pressure prisms on horizontal surface ABCD and vertical surface EFGH.
This flipbook contains the same information as the 3D model.
Inspect it carefully and work each problem.
Key takeaways so far:
Pressure prisms for horizontal surfaces depict constant pressure.
Pressure prisms for vertical surfaces depict pressure that increases linearly with depth).
❏ Flipbook: horizontal and vertical surfaces
What about a vertical surface with an unusual shape? Below, inspect the pressure prism of surface ABCD. The pressure prism shows the effect of the fluid particles in contact with the surface. Be sure to visualize the process of cutting the fluid away from the surface, and replacing it with a vector at each point. The pressure distribution (or prism) is a field of vectors.
❏ An irregular vertical surface
❏ Use Pascal's Law to construct the prism
🟦 12.7 Pressure prisms: inward-leaning and outward-leaning surfaces
Sometimes we need to calculate the resultant force of the fluid pressure on a surface that is inclined (inward-leaning or outward-leaning) or even curved.
To do so, we can apply what we have already learned about vertical and horizontal pressure prisms.
Explore the interactive model to learn the basics.
This flipbook contains a general procedure that can be used to calculate the resultant force on any submerged surface (inclined, curved, etc.).
Carefully work through the flipbook.
❏ Flipbook: inward- and outward-leaning surfaces
🟦 12.8 Creating FBDs and solving reactions with the E.o.E.
Once you have determined the magnitude, inclination, and position of resultant forces on a submerged surface, you can solve unknowns using all of our regular tools (FBDs and E.o.E).
For example, consider the rotational and translational gates in this interactive:
The rotational gate pivots about the top hinge when the tank is to be drained.
The translational gate is pulled upwards when the tank is to be drained.
We're not studying the gate's motion in Statics, but we can determine the forces (and moments) on the gate components in the closed position.
Below is a sample calculation that combines loads from fluid statics with our usual Statics tools (FBDs and the E.o.E.).
🟦 12.9 A word of advice
Units are important throughout this course, but in fluid statics problems they are especially helpful. You start with the fluid density (force per volume). Use Pascal's Law to calculate pressures (force per area). Then, multiply by length to get line loads (force per distance). And finally, take the area under the curve to compute resultant force.
➜ Practice Problems
Problem 1.
A cylindrical glass jar is 12” tall. It is filled with water. What is the pressure at the top surface of the fluid? What is the pressure halfway down? What is the pressure at the bottom fluid particle?
Highlight the invisible text below to check your answers.
Solution: at A, p = 0. at B, p = 31.2 psf. at C, p = 62.4 psf
Problem 2.
An aquarium measures 8m x 20m in plan and is 6m deep. There are 2 tunnels going through the aquarium for people to walk through.
The aquarium contains water (and fish, but you can pretend that the density of a fish is the same as the density of water).
Solve for the resultant force on surface ABCD.
Report your answer in terms of the location of the force, magnitude of the force, and direction of the force.
Highlight the invisible text below to check your answers.
Answer: The force is located at (10m, 4m, 0m). The direction is negative z (or, as defined by unit vector <0,0,-1>). The magnitude is 3.77 MN.Remember to use engineering notation in intermediate steps and report final answers with appropriate prefixes (in this case, MN).Problem 3.
It's the same aquarium as before.
This time, you want to solve for the resultant force on a different surface (even though it's still labeled ABCD).
Report your answer in terms of the location of the force, magnitude of the force, and direction of the force.
Highlight the invisible text below to check your answers.
Location (10m, 8m, 2m). Magnitude is 353 kN. Direction is positive y, or in terms of a unit vector, <0,1,0>.
Problem 4.
It's still the same aquarium. Now you're interested in a different surface (even though it's still labeled ABCD).
Report your answer in terms of the location of the force, magnitude of the force, and direction of the force.
Highlight the invisible text below to check your answers.
Magnitude is 157 kN. Location is (10.95, 8, 2.961) meters. Direction is positive y, or unit vector <0,1,0>.
Problem 5.
Fascinated with this aquarium, you want to compute the resultant force on yet another surface (still labeled ABCD).
The top of the tunnel is a semi-circle with radius 1m.
For this one, just compute the magnitude of the force. (It's possible to also compute the location and direction, but not necessary.)
Highlight the invisible text below to check your answer.
The magnitude is140kN.
Problem 6.
You have found a different aquarium. This one has an inward-learning wall.
Note that the lengths are in meters.
Compute the resultant force on surface ABCD. (You do not have to compute the location and direction on this one.)
Highlight the invisible text below to check your answer.
The magnitude is 13.0 MN. How did you do with units?
Problem 7.
You encounter one last aquarium. This one has an outward-learning wall.
Note that the lengths are in meters.
Compute the resultant force on the 2m by 5m inclined surface. The top of the surface lies at z = 8m (or a depth of 4m below the surface of the water).
(You do not have to compute the location and direction on this one.)
Highlight the invisible text below to check your answer.
The magnitude is 588 kN.