Hydrostatics

Introduction

Pascal’s law states that a fluid at rest creates a pressure at a point that is the same in all directions. Pressure acts normal to a surface and can be drawn as a distributed load. The equation for pressure is:

$p = \rho gz = \gamma z$

Where: $P =$ Pressure, $\rho =$ density, $g =$ gravity, $z =$ depth, and $\gamma =$specific weight

Flat Plate with Constant Width

• Loading is measured in intensity, which is $\frac{force}{length}$

• Intensity varies linearly between $w_{1} = bP_{1} = b\gamma z_{1}$and $w_{2} = bP_{s} = b\gamma z_{2}$, where b is the width of the plate.

• The magnitude is the trapezoidal area, and the line of action going through the area's centroid.

Curved Plate with Constant Width

• Can be solved by integrating the line of the curve

• Can also be solved by separating $F_{R}$ into its vertical and horizontal components where:

  • $W_{f} = \gamma b$, representing the weight of BDA and acts through $C_{BDA}$

  • $F_{AD}$ is the area of the trapezoid and acts through $C_{AD}$

  • $F_{AB}$ is the area of the rectangle and acts at the midpoint

  • $\overrightarrow{F_{R}} = \Sigma\overrightarrow{F_{AD}} + \overrightarrow{F_{AB}} + \overrightarrow{W_{f}}$

  • Location of $\overrightarrow{F_{R}}$ is found using $M_{R} = \Sigma M$ about a point

Steps to Solve

1. Find pressure using $p = \rho gz = \gamma z$

2. Find intensity by multiplying pressure by width of the plate ($w = bp = b\gamma z$)

3. Find $F_{R}$ using area and centroid using $\underline{y} = \frac{\Sigma\underline{y}A}{\Sigma A}$

Example 1: Problem F9-19

Determine the magnitude of the hydrostatic force acting on gate AB, which has a width of 1.5 m. Water has a density of $\rho = 1\ Mg/m^{3}$.

$w = \rho gz_{A}b$

$w_{A} = 1000\frac{kg}{m^{3}} \cdot 9.81\frac{m}{s^{2}} \cdot 0m \cdot 1.5m \Rightarrow w_{A} = 0\frac{kN}{m}$

$w_{B} = 1000\frac{kg}{m^{3}} \cdot 9.81\frac{m}{s^{2}} \cdot 2m \cdot 1.5m \Rightarrow w_{B} = 29.43\frac{kN}{m}$

To find the force of the water, multiply the intensity by the length of the gate. The intensity acts as a triangle on the surface of the gate.

$L = \sqrt{(1.5m)^2+(2m)^2}\Rightarrow L = 2.5m$

$F = \frac{1}{2} \cdot 29.43\frac{kN}{m} \cdot 2.5m \Rightarrow F = 36.79\ kN$

The resultant force acts halfway down the vertical at $h = 1m$

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