Total Hydrostatic Force and Center of Pressure on Submerged Plane Areas

Derivation of the Total Hydrostatic Force on Plane Areas

• Differential Force Concept: The derivation begins by considering a differential force $df$ acting on a differential area $da$.

• Coordinate System:

  • For an inclined surface, the axes are tilted. The vertical depth is represented by $h$.
  • The axis along the inclined surface is the $y$ axis.
  • The $x$ axis is directed perpendicular to the screen (towards the viewer).
  • The distance from the water surface to the differential area along the incline is denoted as $y$, and the vertical distance from the water surface is $h$.

• Geometric Relationships:

  • By forming a right triangle, the relationship between vertical depth ($h$) and the distance along the incline ($y$) is expressed using the sine of the angle of inclination ($\theta$):     $\sin(\theta) = \frac{h}{y}$
  • Solving for $h$:     $h = y \sin(\theta)$

• Integrating for Total Force:

  • The differential force is defined as the product of the unit weight of the liquid ($\gamma$), the depth ($h$), and the differential area ($da$):     $df = \gamma \times h \times da$
  • Substituting the value of $h$:     $df = \gamma \times (y \sin(\theta)) \times da$
  • Integrating both sides of the equation while treating constants ($\gamma$ and $\sin(\theta)$) as external to the integral:     $F = \gamma \sin(\theta) \int y \,da$
  • From integral calculus, the first moment of area $\int y \,da$ is equal to $\bar{y} A$, where $\bar{y}$ is the distance from the reference axis to the centroid (CG) and $A$ is the total area.     $F = \gamma \sin(\theta) \bar{y} A$

• Final Simplified Formula:

  • Considering the centroid of the body at vertical depth $\bar{h}$, the same right triangle relationship applies:     $\sin(\theta) = \frac{\bar{h}}{\bar{y}} \implies \bar{h} = \bar{y} \sin(\theta)$
  • Substituting $\bar{h}$ into the force equation yields the definitive formula for total hydrostatic force on plane areas:     $F = \gamma \bar{h} A$

Location of the Total Hydrostatic Force: Center of Pressure

• Center of Pressure (CP): The total hydrostatic force $F$ does not act on the centroid (CG) of the body. Instead, it acts at a specific point called the Center of Pressure ($CP$).
• Eccentricity ($e$): This is the distance between the CG and the CP along the $y$ axis. It is calculated using the following formula:     $e = \frac{I_G}{A \bar{y}}$
• Variables involved:
  • $I_G$: This is the centroidal moment of inertia of the submerged shape (e.g., $\frac{1}{12} b h^3$ for rectangles, $\frac{1}{36} b h^3$ for triangles).
  • $A$: The total area of the submerged surface.
  • $\bar{y}$: The distance from the centroid of the body to the reference axis (the water surface) measured along the incline of the surface.
• Vertical Submersion Rule: For shapes that are submerged vertically (where $\theta = 90^{\circ}$), the vertical depth $\bar{h}$ is exactly equal to the distance along the surface $\bar{y}$.

Example 1: Vertical Rectangular Plate

• Problem Description: A vertical rectangular plate $1.0\,m$ wide and $3.0\,m$ high is submerged in water. Its top edge is flush with the water surface. Find the total pressure acting on one side and its location from the bottom of the plate.
• Given Data:
  • Width ($b$) = $1.0\,m$
  • Height ($h$) = $3.0\,m$
  • Unit weight of water ($\gamma$) = $9.81\,kN/m^3$
• Calculations for Total Force ($F$):
  • Since it is vertical, $\bar{h} = \bar{y}$.
  • The centroid of the rectangle is at half the height: $\bar{h} = \frac{3.0}{2} = 1.5\,m$.
  • Area ($A$) = $1.0\,m \times 3.0\,m = 3.0\,m^2$.
  • Force formula: $F = 9.81\,kN/m^3 \times 1.5\,m \times 3.0\,m^2$
  • Result: $F = 44.145\,kN$
• Calculations for Eccentricity ($e$):
  • Centroidal moment of inertia ($I_G$) for a rectangle: $I_G = \frac{1}{12} b h^3 = \frac{1}{12} \times 1 \times 3^3 = 2.25\,m^4$.
  • $e = \frac{I_G}{A \bar{y}} = \frac{2.25}{3.0 \times 1.5} = 0.5\,m$.
• Location from the Bottom:
  • The distance from the surface to the CP is $\bar{y} + e = 1.5 + 0.5 = 2.0\,m$.
  • The total height of the plate is $3.0\,m$.
  • Distance from the bottom: $3.0 - 2.0 = 1.0\,m$.
  • Result: The force acts $1.0\,m$ from the bottom.

Example 2: Vertical Triangular Gate

• Problem Description: A vertical triangular gate has a horizontal top base $1.5\,m$ wide and is $3.0\,m$ high. It is submerged in oil with a specific gravity (SG) of $0.82$. The top base is at a depth of $2.0\,m$ below the oil surface. Determine the magnitude of the force and its location measured from the CG.
• Given Data:
  • Width of base ($b$) = $1.5\,m$
  • Height ($h$) = $3.0\,m$
  • Depth to top of gate = $2.0\,m$
  • Specific Gravity ($SG$) of oil = $0.82$
  • Unit weight of liquid ($\gamma$) = $9.81 \times 0.82\,kN/m^3$
• Calculations for Total Force ($F$):
  • Centroid of the triangle: Since the base is at the top, the centroid is located at one-third of the height from the base: $\frac{1}{3} \times 3.0 = 1.0\,m$.
  • $\bar{h} = \text{depth of top} + \text{distance to centroid} = 2.0 + 1.0 = 3.0\,m$.
  • Area ($A$) = $\frac{1}{2} \times 1.5\,m \times 3.0\,m = 2.25\,m^2$.
  • Force formula: $F = (9.81 \times 0.82) \times 3.0 \times 2.25$
  • Result: $F = 54.298\,kN$
• Calculations for Location from CG ($e$):
  • Centroidal moment of inertia ($I_G$) for a triangle: $I_G = \frac{1}{36} b h^3 = \frac{1}{36} \times 1.5 \times 3^3 = 1.125\,m^4$.
  • $e = \frac{I_G}{A \bar{y}} = \frac{1.125}{2.25 \times 3.0}$
  • Result: $e = 0.167\,m$

Important Considerations in Calculations

• Specific Gravity (SG): It is critical to multiply the unit weight of water by the specific gravity of the liquid. For oil ($SG = 0.82$), failing to do this causes errors, though small. For liquids like mercury ($SG = 13.6$), the difference becomes massive.
• Shapes and Formulas:
  • Rectangular $I_G = \frac{1}{12} b h^3$
  • Triangular $I_G = \frac{1}{36} b h^3$
• Area Calculations: Use standard geometric formulas. For any triangle, use $A = \frac{1}{2} b h$. Do not double the area unless there are two distinct gates.
• Reference Point: Always measure $\bar{h}$ and $\bar{y}$ from the liquid surface, not the top of the submerged object.