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GERTC free review material · Fluid mechanics

Hydrostatic Forces on Plane and Curved Surfaces

Find the total force of water on gates and walls, and where it acts. Choose a plane gate (rectangle, triangle, circle, semicircle or trapezoid) at any angle, or a curved gate (quarter circle, semicircle, Tainter gate or cylinder). Add a hinge, water on both sides, the gate weight, or a gate lifted in its guides, and see the forces in 2D and 3D.

Plane and curved gatesPressure diagramsHinges and force to openLifted gate with friction7 worked examples12-item quiz
Open the gate lab
Topic 1

Force on a plane surface

Water pressure grows with depth: . On a flat gate, all the small pushes are square to the gate and add up to one force. It acts below the centroid, because the deeper part is pushed harder.

Total force

= vertical depth of the centroid, kN/m³.

Where it acts (centre of pressure)

and are measured along the slope from the water surface: .

Distance below the centroid

The deeper the gate, the smaller e.

Partly under water or water on both sides: use only the wetted part for each side. The pressure diagram (a triangle or a trapezoid for a rectangle) gives the same answer: force = volume of the pressure prism, acting through its centroid.
Topic 2

Area, centroid, Ig and Ig/A of common gates

Shape (height L measured along the slope)Area ACentroid from the bottom edgeIg (about the centroid)Ig/A
Rectangle b × L
Triangle, base b at the bottom
Triangle, base b at the top
Circle, diameter D
Semicircle, radius r from the flat side
Trapezoid, b₁ bottom, b₂ top
Shortcut: the distance of the centre of pressure below the centroid is . For a rectangle, ; for a circle, .
Topic 3

Hinged gates and lifted gates

Hinged gate: force P at the free edge

The water always tries to push the gate open (away from the water). P holds it closed, or opens it if the gate stays shut by itself.

Hinge reaction

Vertical gate lifted in guides

The water presses the gate on its guides, so friction μF resists the motion.

Topic 4

Force on a curved surface

Split the force into a horizontal part and a vertical part. They are easy to find separately.

Horizontal part

The force on the vertical projection of the curve. It acts at the centre of pressure of that projection.

Vertical part

The weight of the liquid directly above the curve, up to the surface. If the liquid is below the curve, use the imaginary liquid above it; FV then points up. It acts through the centroid of that volume.

Resultant

On a circular surface every push points to the centre, so the resultant passes through the centre. A Tainter gate hinged there feels no turning effect from the water.

Topic 5

Gate lab

Choose a gate, set the water depth above the bottom of the gate, and watch the pressure diagram and the forces. Drag the 3D view to look around.

WHAT IFGate lab
Plane gate
Water and gate weight Depths are measured vertically from the bottom of the gate. Set the other side to 0 if it is dry.
SECTION · PRESSURE AND FORCES
3D VIEW · DRAG TO TURN
WaterPressureResultant forceForce P / hinge

Solution

Topic 6

Worked examples

Example 1 · Vertical rectangular gate

A vertical gate 2 m wide and 3 m high has its bottom 5 m below the water surface. Find the total force and where it acts.

Show solutionHide solution
F = 206.01 kN at 3.7143 m below the surface
Example 2 · Inclined circular gate

A circular gate 1.2 m in diameter lies on a wall sloping at 60° from the horizontal. Its lowest point is 3 m below the water surface. Find the force and the centre of pressure.

Show solutionHide solution
F = 27.52 kN, 0.0314 m below the centroid (along the slope)
Example 3 · Gate hinged at the top

A vertical gate 1.5 m wide and 2 m high is hinged at its top edge. The water is 3 m deep above the bottom of the gate. Neglecting the gate weight, find the force P at the bottom edge needed to keep it closed.

Show solutionHide solution

The hinge is 1 m below the surface, so the arm of F is 1.1667 m.

P = 34.34 kN
Example 4 · Gate lifted in guides

A vertical gate 3 m wide and 2 m high weighs 20 kN. The water is 4 m deep above its bottom on one side and 1.5 m on the other. The friction coefficient on the guides is 0.2. What force is needed to raise the gate?

Show solutionHide solution
48.69 kN to raise the gate
Example 5 · Quarter-circle gate

A quarter-circle gate of radius 2 m and length 3 m bulges toward the water. The water surface is level with the top of the gate. Find the horizontal and vertical forces and the resultant.

Show solutionHide solution

The water directly above the curve fills the square 2 m × 2 m minus the quarter circle:

F_H = 58.86 kN, F_V = 25.26 kN, F = 64.05 kN, through the centre of the circle
Example 6 · Tainter gate

A Tainter gate has a radius of 6 m and is hinged 4 m above the sill. It holds water 6 m deep. Find the forces per metre of length and show that the water makes no moment about the hinge.

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Below the hinge level the curved face looks down, so the water under it pushes up. Above the hinge level it looks up, so the water on it pushes down. Adding the slices along the arc gives a net upward force:

Every push points to the centre of the circle, so the resultant passes through the hinge:

F_H = 176.58 kN/m, F_V = 68.87 kN/m up, F = 189.54 kN/m through the hinge
Example 7 · Cylinder with water on both sides

A cylinder 2 m in diameter holds back water 2 m deep on one side and 1 m deep on the other. Find the net forces per metre of length.

Show solutionHide solution
Net F_H = 14.72 kN/m, F_V = 23.11 kN/m up, through the centre
Topic 7

Quiz: test yourself

Pick an answer to check it right away. Open the solution when you want to see the steps.

Score: 0 / 12 (0 answered)

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