### Key Questions

- What is the basic relationship between depth and pressure?
- How are absolute and relative pressure different?
- How can use our knowledge of centroids to compute the equivalent point forces of fluids?

- What is the basic relationship between depth and pressure?
- How are absolute and relative pressure different?
- How can use our knowledge of centroids to compute the equivalent point forces of fluids?

Pressure is the term used for a force distributed over an area

\begin{equation}
P =\frac{F}{A}\text{.}\tag{7.9.1}
\end{equation}

We will consider the effect of fluid pressure on underwater surfaces, including slanted or curved objects. In all cases we will simply ask the question: what is the pressure at each point and how does it change along the surface?

Pressure can be measured in two different ways

- Absolute pressure is the pressure measured above an absolute or perfect vacuum. The absolute pressure of the surrounding atmosphere is approximately \(\kPa{101.3}\) or \(\psinch{14.7}\text{,}\) and a perfect vacuum is 0 psi or 0 kPa.
- Gage pressure is the pressure indicated by a standard pressure gage. The gage reads zero when exposed directly to the atmosphere, positive when the pressure is higher than atmospheric pressure, and negative pressure indicates a vacuum. In effect, pressure gages ignores the pressure of the atmosphere which surrounds us.

We will use gage pressure for the remainder of the chapter.

Commonly used pressure units include:

- 1 pascals (Pa) = 1 \(\text{N}/\text{m}^2\)
- 1 kilopascal (kPa) = 1000 \(\text{N}/\text{m}^2\)
- 1 pound per square inch (psi) = \(\text{lb}/\text{in}^2\)
- 1 kip per square inch (ksi) = 1000 \(\text{lb}/\text{in}^2\)
- 1 pound per square foot (psf) = 1 \(\text{lb}/\text{ft}^2\)

A fluid, like water or air exerts a pressure on its surroundings. This pressure applies a distributed load on surfaces surrounding the fluid, like the face of a dam, an irrigation control gate, a teakettle, or the drum of a steam boiler.

When you dive underwater, the pressure you feel in your ears increases with depth. At the surface, the gage pressure is zero no matter which unit system you use. As you descend, the fluid pressure \(P\) increases with depth according to the equation

\begin{equation}
P = \rho g h\text{,}\tag{7.9.2}
\end{equation}

where:

- \(\rho\) is the density of a fluid,
- \(g\) is gravitational acceleration, and
- \(h\) is the height of fluid above the point of interest.

Since fluid pressure increases linearly with depth, it behaves as a distributed load which increases linearly from 0 at the surface to \(\rho g h\) at depth \(h\text{,}\) acting normal to the surface. The pressure can be replaced with an equivalent force acting through the centroid of the triangular loading, with a magnitude equal to the triangular area. The pressure on horizontal surfaces is constant, and it is normal to all surfaces.

Some points to remember when solving fluid pressure problems.

- The pressure due to the fluid always acts perpendicular the surface.
- A particle underwater will feel the same pressure from all directions.
- Pressure increases linearly with depth. \(P = \rho g h\)
- \(P = \rho g h\) assumes a constant density and thus is valid only for incompressible fluids like water or oil, but not for compressible fluids like air.
- In English units, specific weight \(\gamma\) is often used instead of density \(\rho\) to describe fluids. Specific weight is the
*weight*per unit volume of a substance, while density is its*mass*per unit volume. The two properties are related by \(\gamma = \rho g\text{.}\) The specific weight of freshwater at room temperature is about \(\pqf{62.4}\text{.}\) - Gage pressure is the pressure above the surrounding atmospheric pressure. Atmospheric pressure is approximately \(\psinch{14.7}\) or \(101.3~\text{kPa}\text{,}\) but since this pressure acts on everything equally and from all directions, the pressure scale can be offset to make the pressure of the surroundings \(\psinch{0}\text{,}\) gage.

Does fluid pressure depend on the surface area of the container? For instance, is the pressure below the Atlantic Ocean less than the pressure below the Pacific Ocean since the Pacific is larger?

No. Fluid pressure is a function of density and depth only, so the surface area of an ocean or tank is insignificant.

\begin{equation*}
P = \rho g h\text{.}
\end{equation*}

Assuming that the density of seawater and \(g\) are the same everywhere under the ocean, the gage pressure depends on depth only.

Compare the pressure at three feet and thirty feet below the surface of freshwater to the atmospheric pressure.

The gage pressure at \(\ft{3}\) is

\begin{equation*}
p = \gamma d = \pqf{62.4} \times \ft{3} = \psf{187} = \psinch{1.30}\text{.}
\end{equation*}

This is

\begin{equation*}
\frac{ \psinch{14.7} + \psinch{1.3}} {\psinch{14.7}} = 1.088\text{,}
\end{equation*}

approximately 9% greater than atmospheric pressure.

At \(\ft{30}\) below the surface, the pressure is 10 times higher, \(\psinch{13.0}\) which is nearly twice atmospheric pressure.

An aquarium tank has a \(\m{3}\times \m{1.5}\) window AB for viewing the inhabitants. The tank contains water with density \(\rho = \kgqm{1000}\text{.}\)

Find the force of the water on the window, and the location of the equivalent point load.

\(F = \kN{155}\) acting \(\m{1.29}\) above point \(B\) or \(\m{3.71}\) below the surface of the water.

Begin by drawing a diagram of the window showing the load intensity and the equivalent concentrated force.

The pressure at the top and the bottom of the window are

\begin{align*}
P_A \amp = \rho\ g (\m{2})= \Nsm{19620}\\
P_B \amp = \rho\ g (\m{5}) = \Nsm{49050}
\end{align*}

Since the loading is linear, the average pressure acting on the window is

\begin{align*}
P_{ave}\amp = (P_A + P_B)/2\\
\amp = \Nsm{34300}
\end{align*}

The total force acting on the window is the average pressure times the area of the window

\begin{align*}
F \amp = (P_{ave})( \m{3} \times \m{1.5})\\
\amp = \kN{155}
\end{align*}

This force may also be visualized as the volume of a trapezoidal prism with a \(\m{1.5}\) depth into the page.

The line of action of the equivalent force passes through the centroid of the trapezoid, which may be calculated using composite areas, see Section 7.5.

Dividing the trapezoid into a triangle and a rectangle and measuring down from the surface of the tank, the distance to the equivalent force is

\begin{align*}
d \amp= \frac{\sum A_i \bar{y}_i}{\sum A_i}\\
d \amp = \frac{\big[P_A(\m{3})\big](\m{3.5}) + \left[\dfrac{1}{2}(P_B-P_A)(\m{3})\right](\m{4})} { \big[P_A(\m{3})\big] +\left[\dfrac{1}{2}(P_B-P_A)(\m{3})\right]}\\
d \amp = \m{3.71}
\end{align*}

If you prefer, you may use the formula from the Centroid table to locate the centroid of the trapezoid instead.

A concrete retaining wall \(\m{3}\) tall and \(\m{0.8}\) thick encloses wet clay mud.

Find the maximum depth \(h\) of mud that can be retained without tipping the wall.

Assume the density of mud is \(\kgqm{1760}\) and the density of concrete is \(\kgqm{2400}\text{.}\)

\begin{equation*}
h = \m{1.99}
\end{equation*}

Begin by drawing free-body diagram of the wall, showing the forces acting on it: the weight of the concrete \(W_c\text{,}\) acting at the center of gravity of the rectangular wall, the horizontal force \(F\) due to the mud, and the reaction components at \(A\text{.}\) Define symbol \(D_z\) for the unspecified length of the wall into the page.

The weight of the concrete is found by multiplying the weight density of concrete \(\lambda_c =\rho_c\ g\) by the volume of the rectangular wall.

\begin{align*}
W_c \amp= \lambda_c\ V\\
\amp= (\rho_c\ g)\ (D_x D_y D_z)
\end{align*}

The horizontal force is caused by the hydraulic pressure of the mud on the submerged face of the wall. The pressure is a distributed load that increases linearly from zero at the surface to \(P_\text{max} = \rho_m\ g\ h\) at depth \(h\text{.}\) Since the loading is triangular, the average pressure is one-half the maximum pressure, and the resulting force is:

\begin{align*}
F \amp= P_\text{ave} A = \frac{1}{2} P_\text{max} A\\
\amp = \frac{1}{2}(\rho_m\ g\ h) (h D_z)
\end{align*}

The wall will tip if the counter-clockwise moment due to \(F\) is greater than the clockwise moment caused by the concrete’s weight. By taking moments about point \(A\text{,}\) we can find the maximum allowable height of the mud, \(h\text{.}\) Note that the length of the dam and \(g\) appear on both sides and cancel out.

\begin{align*}
\Sigma M_A \amp= 0\\
(W_c) \left(\frac{D_x}{2}\right) \amp = F \left(\frac{h}{3}\right)\\
(\rho_c\ \cancel{g} D_x D_y \cancel{D_z}) \left(\frac{D_x}{\cancel{2}}\right) \amp = \left(\frac{1}{\cancel{2}}\rho_m\ \cancel{g}\ h^2 \cancel{D_z}\right)\left(\frac{h}{3}\right)\\
h^3 \amp= 3 \frac{\rho_c}{\rho_m} D_x^2 D_y\\
h^3 \amp= 3 \frac{2400}{1760}\!\cancel{\frac{\kgqm{}}{\kgqm{}}} (\m{0.8})^2(\m{3.0})\\
h \amp=\m{1.98}
\end{align*}

A sea gate is hinged at point \(A\) and is designed to rotate and release the water when the depth \(d\) exceeds a certain value.

The gate extends \(\m{2}\) into the page. The mass density of the water is \(\rho = \kgqm{1000}\text{.}\)

What depth will cause the gate to open?

\begin{equation*}
d \ge \m{1.50}
\end{equation*}

For the gate to tip, the equivalent force of the water must act at or above \(A\text{.}\) That happens when the centroid of the load intensity diagram of the water is at or above \(A\text{,}\) so

\begin{align*}
\frac{d}{3} \amp \ge \mm{500} \\
d \amp \ge \mm{1500}\text{.}
\end{align*}

A gate at the end of a freshwater channel is fabricated from three \(\kg{125}\text{,}\) \(\m{0.6}\times \m{1}\) rectangular steel plates. The gate is hinged at \(A\) and rests against a frictionless support at \(D\text{.}\) The depth of the water \(d\) is \(\m{0.75}\text{.}\)

Draw the free-body diagram and determine the reactions at \(A\) and \(D\text{.}\)

\begin{align*}
D_x \amp= \N{124} \rightarrow\\
A_x \amp = \N{2636} \leftarrow \\
A_y \amp = \N{2795} \uparrow
\end{align*}

A free-body diagram of a cross section of the gate is shown. For simplicity the thickness of the steel plates has been ignored. You should ensure that sufficient distances are provided to locate the loads.

The easiest way to solve this is to apply the principle of transmissibility: slide the lower trapezoid left until it aligns with the upper triangle and makes a triangular loading.

The total horizontal force from the water will be

\begin{align*}
F_x \amp = P_{ave}\ A\\
\amp = \left[\frac{1}{2} \rho\ g\ \m{0.75}\right](\m{0.75} \times \m{1})\\
\amp = \left[\frac{1}{2} (\kgqm{1000}) (\aSI{9.81})\ \m{0.75}\right](\m{0.75} \times \m{1})\\
\amp = \N{2760}
\end{align*}

acting to the right \(\m{0.25}\) above point \(A\text{.}\)

The total vertical load from the water is

\begin{align*}
F_y \amp = P_{ave}\ A\\
\amp = [\rho\ g\ (\m{0.15})] (\m{0.6} \times \m{1})\\
\amp = [( \kgqm{1000})( \aSI{9.81})\ (\m{0.15})] (\m{0.6} \times \m{1})\\
\amp = \N{882.9}
\end{align*}

acting upward \(\m{0.3}\) to the left of \(A\text{.}\)

Each plate weighs

\begin{align*}
W \amp = m g \\
\amp = (\kg{125})( \aSI{9.81}) \\
\amp = \N{1226}\text{.}
\end{align*}

From here solve the equilibrium equations to find the reactions. You should complete this for practice.

\begin{align*}
\Sigma M_A \amp = 0 \amp \amp \implies \amp D_x \amp = \N{239} \rightarrow\\
\Sigma F_x \amp = 0 \amp \amp \implies \amp A_x \amp = \N{2998} \leftarrow \\
\Sigma F_y \amp = 0 \amp \amp \implies \amp A_y \amp = \N{2795} \uparrow
\end{align*}