2.1
Obtain an expression for the depth of the centre of pressure of a plane surface wholly submerged in a fluid and inclined at an angle to the free surface of the liquid.
A horizontal circular pipe, 1.25m diameter, is closed by a butterfly disk which rotates about a horizontal axis through its centre. Determine the torque which would have to be applied to the disk spindle to keep the disk closed in a vertical position when there is a 3m head of fresh water above the axis.
[1176 Nm]

Answer:

The question asks what is the moment you have to apply to the spindle to keep the disc vertical i.e. to keep the valve shut?

So you need to know the resultant force exerted on the disc by the water and the distance x of this force from the spindle.

We know that the water in the pipe is under a pressure of 3m head of water (to the spindle)

, the depth to the centroid of the disc

h' = depth to the centre of pressure (or line of action of the force)

Calculate the force:

Calculate the line of action of the force, h'.

By the parallel axis theorem 2^nd moment of area about O (in the surface) where I_GG is the 2nd moment of area about a line through the centroid of the disc and I_GG = pr⁴/4.

So the distance from the spindle to the line of action of the force is

And the moment required to keep the gate shut is

2.2
A dock gate is to be reinforced with three horizontal beams. If the water acts on one side only, to a depth of 6m, find the positions of the beams measured from the water surface so that each will carry an equal load. Give the load per meter.
[58 860 N/m, 2.31m, 4.22m, 5.47m]

First of all draw the pressure diagram, as below:

The resultant force per unit length of gate is the area of the pressure diagram. So the total resultant force is

Alternatively the resultant force is, R = Pressure at centroid Area , (take width of gate as 1m to give force per m)

This is the resultant force exerted by the gate on the water.

The three beams should carry an equal load, so each beam carries the load f, where

If we take moments from the surface,

Taking the first beam, we can draw a pressure diagram for this, (ignoring what is below),

We know that the resultant force, , so

And the force acts at 2H/3, so this is the position of the 1^st beam,

Taking the second beam into consideration, we can draw the following pressure diagram,

The reaction force is equal to the sum of the forces on each beam, so as before

The reaction force acts at 2H/3, so H=3.27m. Taking moments from the surface,

For the third beam, from before we have,

2.3
The profile of a masonry dam is an arc of a circle, the arc having a radius of 30m and subtending an angle of 60 at the centre of curvature which lies in the water surface. Determine (a) the load on the dam in N/m length, (b) the position of the line of action to this pressure.
[4.28 10⁶ N/m length at depth 19.0m]

Draw the dam to help picture the geometry,

Calculate F_v = total weight of fluid above the curved surface (per m length)

Calculate F_h = force on projection of curved surface onto a vertical plane

The resultant,

acting at the angle

As this force act normal to the surface, it must act through the centre of radius of the dam wall. So the depth to the point where the force acts is,

y = 30sin 39.31=19m

2.4
The arch of a bridge over a stream is in the form of a semi-circle of radius 2m. the bridge width is 4m. Due to a flood the water level is now 1.25m above the crest of the arch. Calculate (a) the upward force on the underside of the arch, (b) the horizontal thrust on one half of the arch.
[263.6 kN, 176.6 kN]

The bridge and water level can be drawn as:

1. The upward force on the arch = weight of (imaginary) water above the arch.

b)

The horizontal force on half of the arch, is equal to the force on the projection of the curved surface onto a vertical plane.

2.5
The face of a dam is vertical to a depth of 7.5m below the water surface then slopes at 30 to the vertical. If the depth of water is 17m what is the resultant force per metre acting on the whole face?
[1563.29 kN]

h₂ = 17.0 m, so h₁ = 17.0 - 7.5 = 9.5 . x = 9.5/tan 60 = 5.485 m.

Vertical force = weight of water above the surface,

The horizontal force = force on the projection of the surface on to a vertical plane.

The resultant force is

And acts at the angle

2.6
A tank with vertical sides is square in plan with 3m long sides. The tank contains oil of relative density 0.9 to a depth of 2.0m which is floating on water a depth of 1.5m. Calculate the force on the walls and the height of the centre of pressure from the bottom of the tank.
[165.54 kN, 1.15m]

Consider one wall of the tank. Draw the pressure diagram:

density of oil r_oil = 0.9r_water = 900 kg/m³.

Force per unit length, F = area under the graph = sum of the three areas = f₁ + f₂ + f₃

To find the position of the resultant force F, we take moments from any point. We will take moments about the surface.

Information provided by: http://www.efm.leeds.ac.uk