From ee73b7cf2f4e5733261faeea2acab1970feb67fa Mon Sep 17 00:00:00 2001
From: Alvie Rahman <alvierahman90@gmail.com>
Date: Wed, 20 Oct 2021 11:27:23 +0100
Subject: [PATCH] Lecture 3 - submerged surfaces

---
 mechanical/mmme1048_fluid_mechanics.md | 132 +++++++++++++++++++++++++
 1 file changed, 132 insertions(+)

diff --git a/mechanical/mmme1048_fluid_mechanics.md b/mechanical/mmme1048_fluid_mechanics.md
index 4c2cfea..d136710 100755
--- a/mechanical/mmme1048_fluid_mechanics.md
+++ b/mechanical/mmme1048_fluid_mechanics.md
@@ -243,3 +243,135 @@ p_1 - p_2 &= \rho_wg(z_C-z_B-z_C+z_A) + \rho_mg\Delta z \\
 - If there is no flow along the tube, then
 
   $$p_1 = p_2 + \rho_wg(z_{C2} - z_{C1})$$
+
+# Lecture 3 // Submerged Surfaces
+
+## Prepatory Maths
+
+### Integration as Summation
+
+### Centroids
+
+- For a 3D body, the centre of gravity is the point at which all the mass can be considered to act
+- For a 2D lamina (thin, flat plate) the centroid is the centre of area, the point about which the
+  lamina would balance
+
+To find the location of the centroid, take moments (of area) about a suitable reference axis:
+
+$$moment\,of\,area = moment\,of\,mass$$
+
+(making the assumption that the surface has a unit mass per unit area)
+
+$$moment\,of\,mass = mass\times distance\,from\,point\,acting\,around$$
+
+Take the following lamina:
+
+![](./images/vimscrot-2021-10-20T10:01:30,080819382+01:00.png)
+
+1. Split the lamina into elements parallel to the chosen axis
+2. Each element has area $\delta A = w\delta y$
+3. The moment of area ($\delta M$) of the element is $\delta Ay$
+4. The sum of moments of all the elements is equal to the moment $M$ obtained by assuing all the
+   area is located at the centroid or:
+
+   $$Ay_c = \int_{area} \! y\,\mathrm{d}A$$
+
+   or:
+
+   $$y_c = \frac 1 A \int_{area} \! y\,\mathrm{d}A$$
+
+   - $\int y\,\mathrm{d}A$ is known as the first moment of area
+
+<details>
+<summary>
+
+#### Example 1
+
+Determine the location of the centroid of a rectangular lamina.
+
+</summary>
+
+##### Determining Location in $y$ direction
+
+![](./images/vimscrot-2021-10-20T10:14:17,688774145+01:00.png)
+
+1. Take moments for area about $OO$
+
+   $$\delta M = y\delta A = y(b\delta y)$$
+
+2. Integrate to find all strips
+
+   $$M = b\int_0^d\! y \,\mathrm{d}y = b\left[\frac{y^2}2\right]_0^d = \frac{bd^2} 2$$
+
+   ($b$ can be taken out the integral as it is constant in this example)
+
+   but also $$M = (area)(y_c) = bdy_c$$
+
+   so $$y_c = \frac 1 {bd} \frac {bd} 2 = \frac d 2$$
+
+##### Determining Location in $x$ direction
+
+![](./images/vimscrot-2021-10-20T10:24:48,372189101+01:00.png)
+
+1. Take moments for area about $O'O'$:
+
+   $$\delta M = x\delta A = x(d\delta x)$$
+
+2. Integrate
+
+   $$M_{O'O'} = d\int_0^b\! x \,\mathrm{d}x = d\left[\frac{x^2} 2\right]_0^b = \frac{db^2} 2$$
+
+   but also $$M_{O'O'} = (area)(x_c) = bdx_c$$
+
+   so $$x_c = \frac{db^2}{2bd} = \frac b 2$$
+
+</details>
+
+## Horizontal Submereged Surfaces
+
+![](./images/vimscrot-2021-10-20T10:33:16,783724117+01:00.png)
+
+Assumptions for horizontal lamina:
+
+- Constant pressure acts over entire surface of lamina
+- Centre of pressure will coincide with centre of area
+- $total\,force = pressure\times area$
+
+![](./images/vimscrot-2021-10-20T10:36:12,520683729+01:00.png)
+
+## Vertical Submerged Surfaces
+
+![](./images/vimscrot-2021-10-20T11:05:33,235642932+01:00.png)
+
+- A vertical submerged plate does experience uniform pressure
+- Centroid of pressure and area are not coincident
+- Centroid of pressure is always below centroid of area for a vertical plate
+- No shear forces, so all hydrostatic forces are perpendicular to lamina
+
+![](./images/vimscrot-2021-10-20T11:07:52,929126609+01:00.png)
+
+Force acting on small element:
+
+\begin{align*}
+\delta F &= p\delta A \\
+&= \rho gh\delta A \\
+&= \rho gh w\delta h
+\end{align*}
+
+Therefore total force is
+
+$$F_p = \int_{area}\! \rho gh \,\mathrm{d}A = \int_{h_1}^{h_2}\! \rho ghw\,\mathrm{d}h$$
+
+### Finding Line of Action of the Force
+
+![](./images/vimscrot-2021-10-20T11:15:51,200869760+01:00.png)
+
+\begin{align*}
+\delta M_{OO} &= \delta Fh = (\rho gh\delta A)h \\
+&= \rho gh^2\delta A = \rho gh^2w\delta h \\
+\\
+M_{OO} &= F_py_p = \int_{area}\! \rho gh^2 \,\mathrm{d}A \\
+&= \int_{h1}^{h2}\! \rho gh^2w \,\mathrm{d}h \\
+\\
+y_p = \frac{M_{OO}}{F_p}
+\end{align*}