Case Study 2 — Sizing a Trapezoidal Flood Gate

Field: Hydraulic and structural engineering (flood control) Calculus used: Hydrostatic force (§18.8), plus a second moment integral to locate the center of pressure (built from the §18.9 moment idea)

The problem a flood-control district must answer

A river town protects itself with a levee, and where a road or canal must pass through the levee, engineers install a flood gate — a steel plate that swings or slides shut when the river rises. When the gate is closed and the river is high, the water on the upstream side pushes against it with enormous force. Two numbers decide whether the gate and its hinges survive: the total hydrostatic force the water exerts, and the height at which that force effectively acts (the center of pressure). Underestimate either and the gate buckles or its lower hinge tears out under flood. Both numbers are integrals, and this case study works them end to end for a realistic gate.

Flood gates are rarely simple rectangles. To fit a channel that is wider at the waterline than at its bed, a common gate is a trapezoid: wide at the top, narrower at the bottom. That varying width is exactly the complication §18.8 warns about — force is not pressure times total area, both because the pressure changes with depth and because the width $w(h)$ changes with depth. We must read $w(h)$ off the geometry before we can integrate.

The geometry and the slice

Take a vertical trapezoidal gate that is $8\ \text{m}$ wide at the top, $4\ \text{m}$ wide at the bottom, and $H = 6\ \text{m}$ tall. Assume the worst design case: the river has risen so the water just reaches the top edge of the gate.

Measure depth $h$ downward from the water surface, so $0 \le h \le 6$. The gate's width shrinks linearly from $8\ \text{m}$ at $h = 0$ to $4\ \text{m}$ at $h = 6$. A line through those two points gives the width function

$$w(h) = 8 - \frac{8 - 4}{6}\,h = 8 - \frac{2}{3}h.$$

Now apply the slice-and-sum recipe. Cut a thin horizontal strip at depth $h$, of height $dh$ and width $w(h)$, so its area is $w(h)\,dh$. The strip is so thin that the pressure is essentially constant across it, equal to $\rho g h$. Pressure times area gives the force on that one strip:

$$dF = \underbrace{\rho g h}_{\text{pressure at depth } h}\cdot\underbrace{w(h)\,dh}_{\text{strip area}} = \rho g\, h\left(8 - \tfrac{2}{3}h\right)dh.$$

Notice the integrand carries two sources of $h$-dependence: the pressure $\rho g h$ grows with depth, and the width $8 - \tfrac23 h$ shrinks with depth. A rectangular gate would have constant width; the trapezoid's taper is what makes this a genuine modeling problem rather than a plug-in.

Total force on the gate

Sum the strips from the surface to the bottom:

$$F = \rho g \int_0^6 h\left(8 - \tfrac{2}{3}h\right)dh = \rho g \int_0^6 \left(8h - \tfrac{2}{3}h^2\right)dh.$$

Integrate:

$$F = \rho g\left[4h^2 - \tfrac{2}{9}h^3\right]_0^6 = \rho g\left[4(36) - \tfrac{2}{9}(216)\right] = \rho g\,[144 - 48] = 96\,\rho g.$$

With $\rho g = (1000)(9.81) = 9810\ \text{N/m}^3$:

$$F = 96 \times 9810 \approx 9.42 \times 10^5\ \text{N} \approx 0.94\ \text{MN}.$$

Nearly a meganewton — the weight of about $96{,}000\ \text{kg}$, pressing on a gate barely larger than a garage door. That is why flood gates are thick steel and why their anchorages are engineered with the same care as a bridge.

A quick units sanity check: $96$ has units of $\text{m}^4$ (from $h \cdot w \cdot dh \sim \text{m}\cdot\text{m}\cdot\text{m}$ integrated, giving $\text{m}^4$ ... actually $h\,w\,dh$ has units $\text{m}\cdot\text{m}\cdot\text{m} = \text{m}^3$, and $\rho g$ carries $\text{N/m}^3$, so $F$ comes out in newtons). The dimensional bookkeeping confirms we have a force, not a pressure.

Where does the force act? The center of pressure

Total force alone is not enough. A hinge or anchor bolt feels a torque, and torque depends on where the resultant force acts. Because the pressure piles up toward the bottom — every strip's force is weighted by its depth — the effective point of application, the center of pressure $\bar h$, sits below the gate's geometric center. Finding it is a second integral, and it is the §18.9 moment idea transplanted from mass to force: the center of pressure is the force-weighted average depth,

$$\bar h = \frac{\displaystyle\int_0^6 h\,dF}{\displaystyle\int_0^6 dF} = \frac{\rho g\displaystyle\int_0^6 h^2\left(8 - \tfrac{2}{3}h\right)dh}{F}.$$

Just as $\bar x = \int x\,\rho\,dx / \int \rho\,dx$ weights each slice of a rod by its position, here we weight each strip by its depth $h$, using the strip's force $dF$ as the "mass." Compute the numerator integral:

$$\int_0^6 h^2\left(8 - \tfrac{2}{3}h\right)dh = \int_0^6\left(8h^2 - \tfrac{2}{3}h^3\right)dh = \left[\tfrac{8}{3}h^3 - \tfrac{1}{6}h^4\right]_0^6 = \tfrac{8}{3}(216) - \tfrac{1}{6}(1296) = 576 - 216 = 360.$$

So the force-moment about the surface is $\rho g \cdot 360$, and dividing by the total force's coefficient $96\,\rho g$:

$$\bar h = \frac{360\,\rho g}{96\,\rho g} = \frac{360}{96} = 3.75\ \text{m}.$$

The resultant force acts at a depth of $3.75\ \text{m}$ — well below the halfway mark and below the gate's centroid. To see how much lower, compute the gate's geometric centroid depth, which weights each strip by area instead of by force:

$$\bar h_{\text{centroid}} = \frac{\int_0^6 h\,w(h)\,dh}{\int_0^6 w(h)\,dh} = \frac{96}{\int_0^6 (8 - \tfrac23 h)\,dh} = \frac{96}{[8h - \tfrac13 h^2]_0^6} = \frac{96}{48 - 12} = \frac{96}{36} \approx 2.67\ \text{m}.$$

The center of pressure ($3.75\ \text{m}$) lies more than a meter deeper than the centroid ($2.67\ \text{m}$). This is a general law: because pressure increases with depth, the line of action of hydrostatic force is always below the area centroid. It is precisely why a dam or gate is overturned about its base, not its middle, and why the lower hinge of this gate carries the lion's share of the load.

What the two numbers tell the engineer

With $F \approx 0.94\ \text{MN}$ acting at $\bar h = 3.75\ \text{m}$, the designer can now do the structural work the integrals were for:

• Hinge/anchor reactions. Taking moments about the top hinge, the force at depth $3.75\ \text{m}$ determines how hard the bottom anchor is pulled. The deeper center of pressure means the bottom anchorage, not the top, is the critical fastener — a non-obvious result that falls straight out of the moment integral.
• Overturning check. Multiply $F$ by the height of $\bar h$ above the gate's base to get the overturning moment, then compare against the restoring moment from the gate's own weight and its sill. This is the same overturning logic the chapter flagged for gravity dams in §18.8's Real-World Application box.
• Plate thickness. The bending stress on the steel plate is largest near the center of pressure, so the gate is reinforced most heavily across its lower third — not its middle.

Every one of these design decisions traces back to two definite integrals: one for the force, one for its location. Get the width function $w(h)$ wrong, or forget that pressure must stay inside the integral, and the whole structural calculation is built on sand.

Verifying numerically (theme 4)

Hand calculus gives the exact $F = 96\,\rho g$ and $\bar h = 3.75\ \text{m}$; numerical quadrature confirms both without touching an antiderivative. The values in the comments are computed by hand; the listing is for the reader to run.

import numpy as np
from scipy.integrate import quad

rho_g = 9810.0          # N/m^3 for water
w = lambda h: 8 - (2/3)*h            # gate width at depth h
F, _  = quad(lambda h: rho_g * h * w(h), 0, 6)      # total force
Mom,_ = quad(lambda h: rho_g * h*h * w(h), 0, 6)    # moment about surface
print(f"force            : {F:.3e} N")   # 9.418e+05
print(f"center of pressure: {Mom/F:.3f} m")  # 3.750

The numbers shake hands with the hand computation: $9.42\times10^5\ \text{N}$ and a center of pressure at $3.75\ \text{m}$.

Discussion Questions

1. We assumed the river just reached the top of the gate. Suppose the flood crests $2\ \text{m}$ above the gate, fully submerging it. Re-derive $F$ with depth running from $h = 2$ to $h = 8$ (the gate now spans depths $2$ to $8$). Does the force more than double? Reason from the linear pressure law before computing.

2. A rectangular gate of the same height and the same area (width $6\ \text{m}$ throughout) would feel what total force? Is it larger or smaller than the trapezoid's $0.94\ \text{MN}$, and why does the taper matter?

3. Show in general that the center of pressure of any vertical plate is always at or below its area centroid. (Hint: compare the force-weighted average $\int h^2 w\,dh / \int h\,w\,dh$ to the area-weighted average $\int h\,w\,dh / \int w\,dh$, and use that the extra factor of $h$ shifts weight downward.)

4. The §18.9 center-of-mass formula and this section's center-of-pressure formula are the same "moment over total" structure with different weights (mass density vs. pressure-area). Name a third quantity in the chapter that fits the same template.

5. Real gates leak, and a thin film of water on the downstream side reduces the net force. Sketch how you would modify the integral to subtract a downstream hydrostatic load.

A short annotated reading

• Stewart, Calculus: Early Transcendentals (9th ed.), §6.5 "Average Value of a Function" and §8.3 "Applications to Physics and Engineering." §8.3 is the home of hydrostatic-force and centroid problems; the trapezoidal-plate setup here is a direct descendant of Stewart's worked dam and trough examples.
• OpenStax, Calculus Volume 2, §6.5 "Physical Applications." Free, and works the force-on-a-plate integral with the same slice-and-sum framing this chapter uses; good for a second pass.
• Munson, Young & Okiishi, Fundamentals of Fluid Mechanics, ch. 2 "Fluid Statics." The engineering reference for hydrostatic force and center of pressure on plane and curved surfaces, including the parallel-axis formula that gives $\bar h$ directly — a shortcut equivalent to our second integral.

A flood gate is held shut by steel and held standing by calculus. One integral says how hard the river pushes; a second says exactly where — and the bottom bolt that never fails is the one an engineer sized from a moment integral.