Reservoir Flip-Gate — Open or Remain Closed?: 04-BS-7, May 2017, Q2

Question 2: Reservoir Flip-Gate — Open or Remain Closed? (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Gate mass $m=800\ \text{kg}$; width (into page) $b=2.0\ \text{m}$; horizontal arm length $L=1.2\ \text{m}$; centre of gravity $e=0.3\ \text{m}$ right of and 0.3 m above pivot O; water depth above the pivot $x=2.0\ \text{m}$. From the figure: the vertical leaf projects above the water surface, reservoir water lies against its upstream face and beneath the horizontal arm, and the arm tip is prevented from rising by a lip on the spillway crest (see check note).

L-shaped gate pivoted at O. Water on the vertical leaf (acting x/3 above O) and the gate weight both turn the gate clockwise (open); uplift under the 1.2 m arm turns it anticlockwise, pressing the arm tip against the crest lip (closed).

Find. Whether the net moment about O opens the gate or keeps it closed.

Approach. Take moments about the pivot O for the three loads on the gate: the hydrostatic thrust on the vertical leaf (triangular pressure, resultant at x/3 above O), the uniform uplift pressure $\rho g x$ on the underside of the arm (resultant at L/2 from O), and the gate's weight (horizontal lever arm 0.3 m). Compare the clockwise (opening) and anticlockwise (closing) totals.

Thrust on the vertical leaf. The leaf is wetted from the surface down to O, so $y_c=x/2=1.0\ \text{m}$ and $A=xb=4.0\ \text{m}^2$: $$F_v=\rho g\,y_cA=1000\times9.81\times1.0\times4.0=39{,}240\ \text{N}$$ For a rectangle with its top edge at the surface, $y_p=y_c+I_c/(y_cA)=1.0+\tfrac{2.0\times2.0^3/12}{1.0\times4.0}=1.333\ \text{m}$ below the surface, i.e. $x/3=0.667\ \text{m}$ above O: $$M_v=39{,}240\times0.6667=\boxed{26{,}160\ \text{N}\cdot\text{m}\ \text{(clockwise, opens)}}$$
Uplift on the horizontal arm. The underside of the arm is at the pivot depth, so the pressure there is uniform, $p=\rho g x=19{,}620\ \text{Pa}$, over $A=Lb=2.4\ \text{m}^2$: $$F_u=19{,}620\times2.4=47{,}088\ \text{N},\qquad M_u=F_u\times\frac{L}{2}=47{,}088\times0.6=\boxed{28{,}253\ \text{N}\cdot\text{m}\ \text{(anticlockwise, closes)}}$$
Weight of the gate. $W=mg=800\times9.81=7848\ \text{N}$ acting 0.3 m to the right of O (the 0.3 m height of the CG gives no moment for a vertical force): $$M_W=7848\times0.3=\boxed{2354\ \text{N}\cdot\text{m}\ \text{(clockwise, opens)}}$$
Compare. $M_{open}=M_v+M_W=26{,}160+2354=28{,}514\ \text{N}\cdot\text{m}$ against $M_{close}=M_u=28{,}253\ \text{N}\cdot\text{m}$. The net moment is $+262\ \text{N}\cdot\text{m}$ clockwise: $\boxed{\text{the gate opens (only just)}}$.
How close to the trip depth? Setting the net moment to zero, $\rho g b\,x^3/6+mge-\rho g b L^2x/2=0$, gives $3270x^3-14{,}126x+2354=0$, whose upper root is $x_{crit}\approx1.99\ \text{m}$. The gate stays shut between about 0.17 m and 1.99 m of water above the pivot and flips open once the depth passes about 1.99 m, so 2.0 m is just past the trip point.

Check: the margin is small (262 N·m, about 1% of either moment), so the verdict rests on reading the figure correctly: full-depth wetting of the vertical leaf, water under the arm, and the stop on the crest lip that allows only clockwise opening. Ignoring the uplift under the arm would wrongly make the gate open at any depth.

Quantity	Value
Thrust on vertical leaf, F_v	39.24 kN
Opening moment of F_v about O	26.16 kN·m
Opening moment of weight about O	2.35 kN·m
Closing moment of arm uplift about O	28.25 kN·m
Net moment	0.26 kN·m clockwise
Trip depth, x_crit	≈ 1.99 m
Verdict	Gate opens (just)

The design logic is visible in how the moments scale. The uplift moment on the arm grows only linearly with depth ($\propto x$), while the thrust moment on the vertical leaf grows with the cube of depth ($\propto x^3$), because both the force and its lever arm increase with $x$. At moderate depths the linear uplift term wins and presses the arm tip against the stop. As the reservoir rises, the cubic term overtakes it, and just above 1.99 m the gate flips open to spill water.

Concept & Theory

Topic. Hydrostatic forces on plane surfaces; moment balance of a hinged, self-actuating gate.

Key relations. $F=\rho g y_c A$; $y_p=y_c+I_c/(y_cA)$ (so $2x/3$ below the surface for a surface-piercing rectangle); uniform pressure $\rho g x$ on a horizontal surface at depth $x$; moment of a vertical weight uses only its horizontal lever arm.

Why this works. Every wetted face of the gate is loaded, and each load is reduced to a resultant with its own lever arm about the hinge. The competition between an $x^3$ opening moment and an $x$ closing moment is what gives the gate a sharp, automatic trip depth.

Common pitfall. Forgetting the water beneath the horizontal arm. Its uplift is the only thing holding the gate shut, and leaving it out makes the gate open at every depth. The other common slip is placing the thrust on the vertical leaf at its centroid (x/2) instead of its centre of pressure (x/3 above the base).

Source. White, Fluid Mechanics, Ch. 2 (Hydrostatic Forces on Plane Surfaces).

Question 2 of 13: Reservoir Flip-Gate — Open or Remain Closed?

Question 2: Reservoir Flip-Gate — Open or Remain Closed? (5 marks)