# Statics

## 🟦  12.1  What is fluid statics?

Fluid statics is a topic that is covered in two different courses: Statics and Fluid Mechanics.

In Statics, we wish to investigate the structure (or container or 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. We aren't studying the fluid itself -- just its effect on the solid bodies around it.

Our analysis procedure typically breaks down into four general steps:

### ❏ Fluids and their quest for a low energy state ### ❏ 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 / m^3

weight density:  𝛄 = 𝜌g = 9.81 kN / m^3

U.S. customary units:

mass density: not commonly used

weight density:  𝛄 = 62.4 # / ft^3

### ❏ An alternate approach

As the example illustrates, a numeric analysis for problems involving the self-weight of water will generate decimals.

Some engineers use a workaround when working in SI units. Instead of using the more precise 𝛄 = 9.81 kN / m^3, an approximate value is used: 10 kN / m^3. That artificially inflates all numbers by (10/9.81), which is a ~1.9% increase.

Once you get to the final answer, you simply multiply it by (9.81/10). This is permissible because the density parameter has a linear effect on the analysis.

This is an optional math trick that you can use to speed up your work on fluids problems. If you choose to use it, be sure to effectively communicate your approach to your audience.

### ❏ Weight densities to memorize Conversationally, you might say: "A cubic meter of water weighs 9.81 kN" or "a cubic foot of water generates a body force of 62.4 pounds downwards."

### ❏ 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

Image shared via CC BY-SA 3.0 license: photo by Shawn Hartley, file: Hoover Dam HDR.jpg, created: 23 July 2013, location: 36° 0′ 58″ N, 114° 44′ 12″ W

### ❏ Tunnel at the Georgia Aquarium, USA

Image shared: By Diliff - Own work, CC BY 2.5, https://commons.wikimedia.org/w/index.php?curid=533725

## 🟦  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 pushes on 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.

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).

On the other hand, when we are talking about the fluid's container or any submerged surface (perhaps the outer wall or hull of a submarine), we say that the pressure is perpendicular to the geometry of that surface.

### ❏ A continuum of fluid particles Careful: this image does not show a pressure distribution, exactly. It shows differential resultant forces (which are computed by integrating the pressure function over the area of each surface). ## 🟦  12.5  Pascal's Law

We use Pascal's Law to compute the static fluid pressure at a specified depth. Pascal's Law 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 just 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 fluid column above the particle

and 𝛄 is the weight density (or unit weight) of the fluid

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 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.

### ❏ Definition of h (head) ### ❏ Units for Pascal's Law  In this example, Pascal's Law is used to compute the fluid pressure at points A, B, C, and D in four tanks of water. Be aware that some engineers incorrectly use the term hydrostatic pressure for both water and non-water fluids. Since “hydro” means water, “hydrostatic” refers specifically a reservoir of water in static equilibrium. It's best to use the more generic term static fluid pressure to refer to fluids in general, and hydrostatic pressure when you are specifically referring to water.

Fluid mechanics is the study of fluids in motion -- perhaps water that is pumped through a system of pipes. In Statics, our scope is limited to fluids at rest -- or static fluid pressure.

## 🟦  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 (1) compressive; (2) perpendicular to the surface geometry; and (3) computed with Pascal's Law.

As an overview of this concept, investigate horizontal surface ABCD and vertical surface EFGH in the interactive animation.

### ❏ 3D interactive animation: horizontal and vertical surfaces

In this example, a pressure prism of surface ABCD is illustrated. 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.  Pressure prisms for horizontal surfaces (with constant pressure) and vertical surfaces (with pressure that increases linearly with depth) are fairly easy to construct. This flipbook shows the entire process.

Inward-leaning or outward-leaning surfaces are a little trickier -- reference the next section for a useful and clever approach.

## 🟦  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 this interactive model to learn the basics.

### ❏ 3D interactive animation: inward- and outward-leaning surfaces

This flipbook contains a general procedure that can be used to calculate the resultant force on any submerged surface. Work through it slowly and carefully.

## 🟦  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 the interactive animation below. 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. Of course, in Statics, we are not studying motion, but we can determine the forces (and moments) on the gate components in the closed position.

### ❏ 3D interactive animation: rotational and translational gates

Below is a sample calculation for combining loads from fluid statics with our usual 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 Solution: at A, p = 0. at B, p = 31.2 psf. at C, p = 62.4 psf

2.  A tank of water is 12 feet high. What is the resultant force of the fluid on a unit width of vertical wall (a one foot strip)? The wall is perfectly vertical (it doesn't lean out or lean in).

Solution: 4.49 kip

3.  Consider the same FBD as the one you used in the prior problem. Let's model the unit width as a cantilever beam.

Use the bottom of the tank for the fixed connection. Solve for the force and moment reactions needed for static equilibrium on the one foot unit strip of wall.

Solution: force is 4.49 kip. the moment is 18.0 kip-feet

4. Determine the resultant force on surface ABCD.

Note the 60 degree angle... Solution: 13.0 MN

5. A weird tank has an illogical gate in the sloped surface.

The gate is held in static equilibrium with a dowel at E and a dowel at F.

You can think of dowels as types of pins.

Solve for the y-direction and z-direction reaction forces at dowel F when the gate is closed.

Note: the gate width in the x-direction is 2.5 m.  Solution: The z component of the reaction at dowel F is negative 89.7 kN. The y component of the reaction at dowel F is positive 155.3 kN.

6. Let’s revise our tank geometry slightly, so that the negative x face of the tank (the one at the left) is now slanting outward.

We wish to compute the resultant force on the 2m x 5m surface shown.

Say that the top edge of the 2m x 5m surface lies at z = 8m (or a depth of 4m below the surface of the water). Solution: 588 kN