Thrust on a plane surface

Figure 2.15 shows a plane surface of arbitrary shape, wholly submerged in a liquid in equilibrium. The plane of the surface makes an angleθ with the horizontal, and the intersection of this plane with the plane of the free

Hydrostatic thrusts on submerged surfaces 61 surface (where the pressure is atmospheric) is taken as the x-axis. The y-axis

is taken down the sloping plane. Every element of the area is subjected to a force due to the pressure of the liquid. At any element of areaδA, at a depth h below the free surface, the (gauge) pressure is p = gh and the corresponding force is

δF = pδA = ghδA = gy sin θδA (2.14) As the fluid is not moving relative to the plane there are no shear stresses.

Thus the force is perpendicular to the element, and since the surface is plane all the elemental forces are parallel. The total force on one side of the plane is therefore

But ∫AydA is the first moment of the area about the x-axis and may be represented by Ay where A represents the total area and(x, y) is the position of its centroid C. Therefore

F= g sin θAy = gAh (2.15)

Nowgh is the pressure at the centroid, so, whatever the slope of the plane, the total force exerted on it by the static fluid is given by the product of the area and the pressure at the centroid. Whether the fluid actually has a free surface in contact with the atmosphere is of no consequence: for a fluid of uniform density in equilibrium the result is true however the pressure is produced.

In addition to the magnitude of the total force we need to know its line of Centre of pressure for a plane surface

action. Since all the elemental forces are perpendicular to the plane, their total is also perpendicular to the plane. It remains to determine the point at which its line of action meets the plane. This point is known as the centre of pressure (although centre of thrust might be a better term).

For the resultant force to be equivalent to all the individual forces its moment about any axis must be the same as the sum of the moments of the individual forces about the axis. The x- and y-axes are most suitable to our purpose. From eqn 2.14 the force on an element of the area isgy sin θδA and the moment of this force about Ox is thereforegy²sinθδA. Let the centre of pressure P be at (x^,y^). Then the total moment about Ox is

Fy^=

Agy²sinθdA

Substituting for the total force F from eqn 2.15 we obtain y^=g sin θ

Ay²dA

g sin θAy = (Ak²)Ox

Ay (2.16)

where(Ak²)Oxis the second moment of the area about Ox. In other words, the slant depth (i.e. measured down the plane) of the centre of pressure equals the second moment of the area about the intersection of its plane with that

of the free (i.e. atmospheric) surface divided by the first moment of the area about the intersection of its plane with that of the free surface.

The centre of pressure is always lower than the centroid (except when the surface is horizontal) as the following calculation demonstrates.

From the parallel axes theorem:

(Ak²)Ox= (Ak²)C+ Ay² so eqn 2.16 becomes

y^= (Ak²)C+ Ay²

Ay = y + (Ak²)C/Ay (2.17) Since a second moment of area is always positive it follows that y^> y. QED.

We see also that the more deeply the surface is submerged, that is, the greater the value of y, the smaller is the contribution made by the last term in eqn 2.17 and the closer is the centre of pressure to the centroid. This is because, as the pressure becomes greater with increasing depth, its vari-ation over a given area becomes proportionately smaller, so making the distribution of pressure more uniform. Thus where the variation of pressure is negligible the centre of pressure may be taken as approximately at the centroid. This is justifiable in gases, because in them the pressure changes very little with depth, and also in liquids provided the depth is very large and the area small.

The expressions 2.16 and 2.17, it is re-emphasized, give the distance to the centre of pressure measured down the plane from the level of the free surface and not vertically.

The x-coordinate of the centre of pressure may be determined by taking moments about Oy. Then the moment ofδF is gy sin θδAx and the total moment is

When the area has an axis of symmetry in the y direction, this axis may be taken as Oy and then

AxydA is zero, that is, the centre of pressure lies on the axis of symmetry. It will be noted from eqns 2.16 and 2.18 that the position of the centre of pressure is independent of the angleθ and of the density of the fluid. However, a constant value of was used; the relations are therefore valid only for a single homogeneous fluid.

For the plane lamina of negligible thickness illustrated in Fig. 2.15, the force on one face would exactly balance the force on the other if both faces were in contact with the fluid. In most cases of practical interest, however, there is no continuous path in the fluid from one face of the plane to the other and therefore the pressures at corresponding points on the two faces are not necessarily the same. For example, the surface may be that of a plate covering a submerged opening in the wall of a reservoir, or a canal lock-gate

Hydrostatic thrusts on submerged surfaces 63 which has different depths of water on the two sides. The surface may extend

up through the liquid and into the atmosphere; only the part below the free surface then has a net hydrostatic thrust exerted on it.

The pressures we have considered have been expressed as gauge pressures.

It is unnecessary to use absolute pressures because the effect of atmospheric pressure at the free surface is to provide a uniform addition to the gauge pressure throughout the liquid, and therefore to the force on any surface in contact with the liquid. Normally atmospheric pressure also provides a uniform force on the other face of the plane, and so it has no effect on either the magnitude or position of the resultant net force.

It should be particularly noted that, although the total force acts at the centre of pressure, its magnitude is given by the product of the area and the pressure at the centroid.

Example 2.2 A cylindrical tank 2 m diameter and 4 m long, with its axis horizontal, is half filled with water and half filled with oil of density 880 kg· m⁻³. Determine the magnitude and position of the net hydrostatic force on one end of the tank.

Solution

We assume that the tank is only just filled, that is, the pressure in the fluids is due only to their weight, and thus the (gauge) pressure at the top is zero. Since two immiscible fluids are involved we must consider each separately. In equilibrium conditions the oil covers the upper semicircular half of the end wall.

Since the centroid of a semicircle of radius a is on the central radius and 4a/3π from the bounding diameter, the centroid Co of the upper semicircle is 4(1 m)/3π = 0.4244 m above the centre of

the tank, that is, (1 − 0.4244) m = 0.5756 m from the top. The pressure of the oil at this point is

gh = (880 kg · m⁻³)(9.81 N · kg⁻¹)(0.5756 m) = 4969 Pa

and thus the force exerted by the oil on the upper half of the wall

By eqn 2.17 the centre of pressure is(AK²)C/Ay below the centroid.

Now Ak² about the bounding diameter= πa⁴/8 (see Fig. 2.14). So, by the parallel axes theorem, for a horizontal axis through Co,

(AK²)C= πa⁴ (0.5756 m) = 0.1214 m below the centroid, that is, (0.5756 + 0.1214) m = 0.6970 m below the top.

For the lower semicircle, in contract with water, the centroid C_wis 0.4244 m below the central diameter. The pressure here is that due to 1 m of oil together with 0.4244 m of water, that is,

(800 kg · m⁻³)(9.81 N · kg⁻¹)(1 m)

+ (1000 kg · m⁻³)(9.81 N · kg⁻¹)(0.4244 m) = 12, 796 Pa Thus the force on the lower semicircle is (12 796 Pa)

1

2π1²m²

= 20 100 N.

(AK²)C is again 0.1098 m⁴but we must be very careful in calcul-ating y since there is not a single fluid between this centroid and the zero-pressure position. However, conditions in the water are the same as if the pressure at the oil–water interface [(880 kg · m⁻³) (9.81 N · kg⁻¹)(1 m)] were produced instead by 0.88 m of water [(1000 kg · m⁻³)(9.81 N · kg⁻¹)(0.88 m)]. In that case the vertical distance from the centroid Cw to the zero-pressure position would be 0.4244 m+ 0.88 m = 1.3044 m.

∴ Centre of pressure Pw for lower semicircle is 0.1098 m⁴

The total force on the circular end is(7805+20 100) N = 27 905 N, acting horizontally. Its position may be determined by taking moments about, for example, a horizontal axis at the top of the cylinder:

(7805 N)(0.697 m) + (20 100 N)(1.478 m) = (27 905 N)x

where x= distance of line of action of total force from top of cylinder

= 1.260 m

Hydrostatic thrusts on submerged surfaces 65 By symmetry, the centre of pressure is on the vertical diameter of the

circle.

An alternative, though algebraically more tiresome, technique would be to consider horizontal strips of the surface of vertical thick-ness, say,δy; and then to integrate, over each semicircle, expressions for forces on the strips and their moments about a horizontal axis at, say, the top of the cylinder. However, there is no escape from dealing separately with the surfaces in contact with each fluid. 2