less than a semicircle, will always lie between 0·4h and 0·424h. The exact position of G is given by

in which b denotes the width AB in the diagram.

Art. 5. The Pressure directly opposed by Contrary Forces or Resistances. In the plunger of a feed-pump and the ram of a hydraulic cylinder, we have examples of a direct opposition between the normal pressure at the base of the ram or plunger and the external resistances or forces; while the piston of a steam-engine works against an external resistance which is transmitted through the piston-rod to the centre of gravity of the plane circular surface, and is therefore directly opposed to the "resultant" of the fluid pressure. In the same way a pressure of uniform intensity acting normally upon any one of the plane surfaces, illustrated in the last article, may be directly opposed by an equal and opposite force or resistance P = pA applied at the point G in each figure.

More frequently, however, we have to consider structures in which the equilibrium is maintained by an indirect opposition of resistances to the normal fluid pressure.

Art. 6. Resolution of the Fluid Pressure into Components whose Direction is known or mechanically determined. The method described in this article will be found very useful for the solution of a large number of questions, for it very often happens that our calculations are concerned only with a particular component of the fluid pressure.

Suppose the hollow trunk, whose longitudinal section is sketched in Fig. 9, to have a square or rectangular cross-section

H

FIG. 9

B

D

FIG. 9A

as represented in Fig. 9a, and to be fitted with a square or rectangular piston sliding freely in the trunk. Let the piston

be formed with an oblique face AD which is presented to the fluid pressure, and let the angle contained between the plane AD and the normal AB be denoted by 0. Then the actual area of the plane surface AD exposed to fluid pressure will be S = b x d sec 0; and the pressure acting at right angles to this surface will be a uniformly distributed force having the magnitude pS = pbd sec 0. This force may be represented by the line OC meeting the plane at its centre of gravity C and at right angles to AD, and will be resolved into the components CH and CV, of which CH represents the force acting in the direction of the axis of the trunk. We have therefore

In other words, the force by which the piston is impelled along the trunk in the direction of its axis, is independent of the angle 0; and is always equal to the pressure p multiplied by the area of the rectangle bd of Fig. 9a, which is a projection of the inclined face of the piston in a plane at right angles to the axis of the trunk.

In like manner the vertical component will be

CV = pS sin 0 = pbd sec 0 sin 0 = pbd tan 0

in which expression the quantity d tan is equal to the horizontal length BD subtended by the inclined surface. The vertical component, therefore, is equal to the pressure p multiplied by the area of a horizontal projection of the inclined plane ; and it is obvious that whatever inclination the plane AD may have to the vertical plane AB, between 0 and 90°, the horizontal and the vertical components will always be measured by the areas of the projections of that surface on planes which are respectively at right angles to the directions of those component forces.

The force CH may evidently be determined by the same graphic method when the surface exposed to fluid pressure is made up of any number of elements or planes inclined at different angles to the axis, or when it consists of any cylindrical, conical, spherical, or other surface or surfaces of single or double curvature; for in all such cases we may conceive the whole surface to be made up of a large number of such elements, each having its own inclination to the axis CH, and each forming the oblique base of a prism whose axis is parallel to CH, like the prism sketched in Figs. 9 and 9a.

In general the action of fluid pressure in engineering structures and machines is equilibrated by forces, stresses, or resistances applied in certain definite directions, and the magnitude of the component which has thus to be opposed can be readily found by this method. A few examples will illustrate its application.

EXAMPLE 1.-The bridge cylinder sketched in section in Fig. 10 has a diameter AA, which is increased to the diameter BB, below the conical junction

B

B

FIG. 10

A

A,

A

A1

ring. The pneumatic pressure p, employed in the sinking of the cylinder, will act everywhere at right angles to the internal surface of cone and cylinder; but in connection with the requisite weighting of the cylinder, it may be desirable to calculate the vertical component of the fluid pressure, or the lifting force exerted by the compressed air.

As the direction of the required force is vertical, draw a horizontal projection or inverted plan of the surfaces as in Fig. 10a, where AA, represents the cover of the cylinder, while the annular area between the circles AA, and BB, will be the horizontal projection of the conical shoulder. This annular area multiplied by p will give us the vertical component of the pneumatic pressure acting upon the conical surface, and this vertical force added to the direct upward pressure upon the cover of the cylinder, will give us the total lifting force, which is therefore equal to the pressure p AQ A, B, multiplied by the whole area of the circle BB, in Fig. 10a.

FIG. IOA

EXAMPLE 2.-In the same bridge cylinder it may be necessary to calculate the resistance of the cylinder itself to the bursting pressure, and the requisite strength of the bolts by which the segments of the ring BC are united at their vertical joints. Suppose then that the diametral section BB, in Fig. 10a, at right angles to OX, passes through a pair of these joints and between the internal flanges by which the segments are united. At each of the joints B and B1, the bolts have to resist by their direct tensile strength, a force acting in a direction tangential to the circle and parallel to OX, tending to separate the semicircle BXB, from the semicircle BZB1; while the magnitude of that force will be found by summating all the components (in the direction OX) of the radial pressures acting upon the interior surface of the cylinder BXB, over the height BC. The rectangle BB1С1С in Fig. 10 is a projection of that surface on a plane at right angles to OX, and therefore the area of that rectangle multiplied by the pressure p will give us the force which is exerted in the direction OX, and which is equally divided between the two joints, so that one-half of the force

thus found will have to be borne in direct tension by the row of bolts uniting the joint BC.

The tangential stress in the metal of the cylinder itself is determined in the same way by taking any diametral section through the metal of the cylinder; and it is obvious that the sum of the tensile stresses in the metal cut through by the section is equal and opposite to the internal stress of the fluid acting across the plane of the same section.

In this example, as well as in the previous one, it is assumed that the cylinder BC and the conical junction-ring are standing above the water-level. If they were submerged, the tensile stress in the cylinder and the lifting force due to the internal pneumatic pressure would both be reduced by the contrary action of the external fluid pressure; but this point may be left for discussion at a later period.

FIG./I

Example 3.—The vessel shown in section in Fig. 11 may be taken to represent either the barrel of a plunger-pump or the cylinder of a hydraulic engine or ram; and it may be required to find, in either case, the vertical forces and stresses acting upon the vessel and upon the framing to which it is attached. The circle drawn in Fig. 11a to the diameter d2, which is the internal diameter of the cylinder, may be taken as a ground plan or horizontal projection of the vessel's bottom, which is visible from above; or if we regard the figure as an inverted plan, the inner circle will represent the base of the ram, or the circular opening in the stuffing box; so that the annular space between the two circles will be a horizontal projection of the internal shoulder which is visible from below.

The vertical force exerted by the fluid pressure upon the vessel as a whole, and borne by the framing which supports the outer flanges, will be the algebraical sum of all the vertical components, or the difference between the lifting pressure at the shoulder and the downward pressure upon the bottom of the cylinder. Therefore, subtracting the negative area of the annular space in Fig. 11a from the positive area of the larger circle, we have remaining the area of the inner circle, and this area multiplied by p will give us the downward pressure upon the engine-framing. This force is obviously equal to the load P1 upon the ram, and is independent of the diameter d2. At the same time it will be at once seen that the vertical tensile stress in the sides of the cylinder below the framing will have the greater value

1

FIG.I/A

dr

EXAMPLE 4. Suppose a cast-iron pipe to be formed as a bend curving through 180° as sketched in Fig. 12, and to be connected at each end to the straight vertical pipes by an ordinary flange-joint. It is required to find the stress in the flangebolts when the whole line of pipe is charged with a hydraulic pressure p.

The vertical lifting force which must be resisted by these bolts will evidently be the algebraical sum of all the vertical components of the radial pressures acting upon all the elements of area of the internal surface. In the straight vertical pipes all these vertical components will be zero, because normals drawn to all the elements of internal surface will be horizontal; but in the bend these normals will be inclined to the horizontal, some in one direction and some in the opposite

FIG. 12

FIG. 12A

A

E

G

H

A1

B

B1

F

direction. The dotted centre-line ACB in Fig. 12, if it were traced upon the sides of the pipe itself, would lie in vertical planes forming tangents to the pipe at all cross-sections of the bend; and we may conceive the centre-line as dividing the pipe into two troughs of semicircular section. The upper trough will contain all those elements of surface which are exposed to pressure having a vertical component in the upward (positive) direction; while the lower trough will contain all those parts of the surface on which the pressure has a downward vertical component. The horizontal projections of these surfaces are easily traced in the sectional plan of Fig. 12a. The figure ABFB11Е is the projection of the positive area, while the figure ABHB1ÁG is the projection of the negative area;

and subtracting the latter from the former, we have remaining the areas of the two circles which represent the open ends of the bend.

The lifting force which has to be resisted at each joint is therefore pd, where p is the fluid pressure and d the internal diameter of the pipe.

The same result might have been obtained still more simply by taking an imaginary section across the bend at the plane AB, and regarding the bend with its contained water, as forming a body whose equilibrium is to be considered. For the stress or pressure of fluid upon fluid, across this plane, must be balanced by the tensile resistance of the bolts.