were applied at the top, at the bottom, or on either side of the box. In this case it is obvious that the internal pressures on the opposite sides of this closed vessel must exactly balance each other, so that on the whole the internal pressure would have no tendency to move the box in either direction.

If a small vessel is charged with hydraulic pressure under a great head, we shall introduce no serious error by assuming the pressure, in this case also, to have a uniform intensity, although in strict truth the assumption would be incorrect. To obtain a measurement of the error that would thus be involved, we might take a cubical box measuring 12 inches on each side, and placing it upon a weighing machine, let the weight of the empty box be registered. Then after charging the vessel with water under an accumulator pressure of 1000 lbs. per square inch, we should of course find the weighing machine to register a greater load than before, the difference being exactly the weight of the cubic foot of water, or say 62 lbs. But the box is now acted upon by three vertical forces (besides its own weight), and these forces balance each other. The upward exterior pressure or reaction at the base of the box is 62 lbs. as registered by the weighing machine; and it follows that the internal fluid pressure acting vertically downwards upon its floor must be greater than that which acts vertically upwards upon the lid, by exactly 62 lbs. The force exerted by the fluid upon each face of the cube under the assumed uniform pressure of 1000 lbs. per square inch, would amount to 144,000 lbs. But if that figure represents exactly the lifting force upon the lid, the floor must evidently be subjected to a downward pressure of 144,062 lbs., or to a fluid pressure of 1000-43 lbs. per square inch. The difference, in such a case as this, is too small to be worth considering, unless a high degree of accuracy is required in the calculations, and in practice we should seldom need to take account of so small a percentage of error.

As we could nowhere find a fluid without weight, it follows that we could hardly ever find a fluid pressure of really uniform intensity; but when the variations are so slight, we can neglect them in practice, and treat some of our problems as though p were a constant quantity; and when the surface is a horizontal plane, the pressure on that surface will in general be quite

uniform.

Art. 3. Hydrostatic Pressure of Varying Intensity.-A different

set of problems will come before us when we are dealing with standing water, open to the air at its upper surface, or subjected only to the pressure which is due to its own weight; for here the pressure-intensity p must be considered as a varying quantity proportional to the vertical head h. In other words, if we conceive the whole mass of the standing water to be divided into a number of horizontal layers, the fluid pressure in each layer will be proportional to the depth at which it lies below the open surface.

If this statement needed any demonstration, it would again be sufficient to refer to those commonplace facts which are familiar to the mind of every practical engineer. Thus, taking off the lid from the cubical vessel before referred to, and placing the open box upon a weighing machine, we may fill the vessel up to the brim with cold water, at a temperature of say 40° Fahr.; and we shall then find that, after deducting the weight of the empty vessel, the net weight of the cubic foot of water contained in it is just 62:425 lbs. As the fluid is now exerting no lifting force, and as the horizontal pressure on the sides of the cube cannot possibly affect the weighing machine, it follows that the load of 62:425 lbs., registered by the weighing machine, is simply the distributed downward pressure of the fluid upon the flat bottom of the vessel. In this horizontal layer of water, therefore, lying close to the bottom at a depth of 1 foot below the open surface, the pressure p is 62-425 lbs. on the square foot, or 0·433 lb. per square inch; and the same pressure will act in a horizontal direction upon the vertical sides of the vessel surrounding the layer.

To find the fluid pressure p in a layer lying at any greater depth, we may carry up the vertical sides of the box, and then fill it to a greater height h and weigh it again. If we fill it to a height of 10 feet, for example, the net load registered by the weighing machine will be the fluid pressure per square foot in a layer 10 feet below the open surface.

In the case of water, which is a practically incompressible fluid, the density is not altered by any increase of pressure within ordinary limits, and for this reason the net load registered by the weighing machine will be exactly ten times as great as in the first instance, or exactly 624.25 lbs., showing a fluid pressure of 4.33 lbs. per square inch.

Obviously, therefore, we have the following general rule for

the hydrostatic pressure p at any point whose vertical depth below the open surface is denoted by h, viz. :

where y is the weight of the incompressible fluid per unit of volume.

Measuring h in feet, and inserting for y its value for fresh water at the temperature of greatest density, we have

}

62.425hp in pounds per square foot
0.433hp in pounds per square inch S

la

and of course this pressure p is the same, at whatever angle of inclination the boundary-wall may present its surface to the contact of the fluid; while the direction of the pressure is always at right angles to that surface.

At the beginning of this article, we assumed that the standing water would be open to the air at its upper surface. At its upper surface, therefore, it would be exposed to the barometric pressure; and if the water surrounds the sides of a condenser we shall have to add the barometric pressure to the value of p as above determined; but in most other cases we may leave the atmospheric pressure out of account, because the structures we are concerned with are themselves surrounded by atmospheric pressure.

The forces, upsetting moments, and internal stresses, caused by the action of fluid pressure upon solid bodies, will of course be determined by the ordinary principles of static equilibrium.

CHAPTER II.

THE ACTION OF FLUID PRESSURE OF UNIFORM INTENSITY.

Art. 4. Centre of Action.-In considering the equilibrium of any structure under the action of fluid pressure, we shall always have to calculate the magnitude of that distributed force, and to find the position of its centre of action. When the pressure has a varying intensity, the calculations will be quite different from those which can be applied to a pressure of uniform intensity; and we may first confine our attention to the latter, assuming the pressure p to be a constant quantity.

If the surface of a level floor is covered with a load spread evenly, or unevenly, over it, the line of action of the distributed downward pressure must pass through the centre of gravity of the load in every case; and if the load is uniformly distributed, the centre of action will coincide with what is called the centre of gravity of the plane surface.

It is sufficiently obvious that, if we substitute for the level floor a plane surface of the same figure inclined at any angle with the horizon, and for the load a fluid pressure acting at right angles to that surface, the centre of action must still have the same position in the surface, and will coincide with its so-called centre of gravity if the pressure-intensity p is uniform.

For surfaces of regular geometrical outline we may here note down, for future reference, the following values of the superficial area A, and of the co-ordinates x and y, which fix the position of the centre of gravity.

1. For any rectangle, whose breadth is b and depth d, we have, of course, very simply

2. In a right-angled triangle, whose base and height are denoted by b and d, the area will be

while the ordinates measured from the square

corner in Fig. 2 will be—

3. In any triangle, ABC, the perpen

dicular distance of the centre of gravity G

FIG.2

from any axis OX or OY in Fig. 3, will always be the arithmetical mean of the three ordinates measured from the same axis to the three corners A, B, Y

FIG. 3

AA1 + BB1 + CC1
3

The moment of the wind-pressure pA (upon the whole sail) about any axis OX is, of course, the product pay.

4. In any quadrilateral, ABCD (Fig. 4), the area A will be

the sum of the two areas A, and

A2 of the two triangles ABC and BCD; and finding for these triangles the values of the co-ordinates x1, y1 and x2, y2 to the centre of gravity of each, we can easily calculate the co-ordinates x and y to the centre of gravity G of the whole quadrilateral by summating the moments of the two triangles. Thus if Fig. 4 repre

X

A

FIG. 4

Y

G

B

sents the mainsail of a cutter, the wind-pressure on the whole