The trapezium ABCD of Fig. 5 may be treated in the way; but as the sides AB and CD are parallel, we can denote by h the perpendicular h_b1 + 2b2 y = 3 b1 + b2 while the area of the figure is C he centre of gravity G is perhaps most easily found by In every regular polygon, as in every circle, the centre G2 B C er is the radius OC. The area of the semicircle is where d is the diameter; so that under a uniform pressure p, the moment of the distributed force about the axis AB will be The quadrant OBC in the same diagram will have its centre of gravity at the point G2, whose co-ordinates measured from the axes OC and OB will evidently be— while the area of the quadrant, and its moment about either axis, will have one-half of the values given for the semicircle. 7. The parabolic segment ACB in Fig. 7 has an area equal to two-thirds of the rectangle AB × CE; or if b denotes the width and h the central height of the segment, then FIG. 7 gravity G, is y = zh FIG. 8 8. The segment of a circle differs very little from the parabolic segment so long as the ratio of h to b is a small fraction, as it is, for example, in the invert of a culvert of ordinary design. To find its exact area A, we may subtract the area A2 of the triangle OAB, from the area A1 of the sector OACB in Fig. 8. To make the calculation, the angle 0 may first be found from a table of sines by means of the expression— If the height EC is denoted by h, the ordinate EG, in segments less than a semicircle, will always lie between 04h 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 FIG. 9A d 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 CH = pS cos 0 = pbd sec 0 cos 0 = pbd 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 FIG. 10 A A, A A1 B A B C AB1 FIG. IOA 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 AA1 and BB1 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 multiplied by the whole area of the circle BB, in Fig. 10a. 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 BXB1 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 BB1C1C 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 |