Page:EB1911 - Volume 17.djvu/979

Rh moment is represented by twice the area of the polygon; this is proved by taking moments about any point. If the polygon intersects itself, care must be taken to attribute to the different parts of the area their proper signs.

Again, any coplanar system of forces can be replaced by a single force R acting at any assigned point O, together with a couple G. The force R is the geometric sum of the given forces, and the moment (G) of the couple is equal to the sum of the moments of the given forces about O. The value of G will in general vary with the position of O, and will vanish when O lies on the line of action of the single resultant.

The formal analytical reduction of a system of coplanar forces is as follows. Let (x1, y1), (x2, y2), be the rectangular co-ordinates of any points A1, A2, on the lines of action of the respective forces. The force at A1 may be replaced by its components X1, Y1, parallel to the co-ordinate axes; that at A2 by its components X2, Y2, and so on. Introducing at O two equal and opposite forces ±X1 in Ox, we see that X1 at A1 may be replaced by an equal and parallel force at O together with a couple −y1X1. Similarly the force Y1 at A1 may be replaced by a force Y1 at O together with a couple x1Y1. The forces X1, Y1, at O can thus be transferred to O provided we introduce a couple x1Y1 − y1X1. Treating the remaining forces in the same way we get a force X1 + X2 + or (X) along Ox, a force Y1 + Y2 +  or (Y) along Oy, and a couple (x1Y1 − y1X1) + (x2Y2 − y2X2) + or (xY − yX). The three conditions of equilibrium are therefore

(X) = 0, &emsp; (Y) = 0, &emsp; (xY − yX) = 0. (8)

If O′ be a point whose co-ordinates are, the moment of the couple when the forces are transferred to O′ as a new origin will be {(x − ) Y − (y − ) X}. This vanishes, i.e. the system reduces to a single resultant through O′, provided

−·(Y) + ·(X) + (xY − yX) = 0. (9)

If, be regarded as current co-ordinates, this is the equation of the line of action of the single resultant to which the system is in general reducible.

If the forces are all parallel, making say an angle with Ox, we may write X1 = P1 cos, Y1 = P1 sin , X2 = P2 cos , Y2 = P2 sin, The equation (9) then becomes

{(xP) − ·(P)} sin − {(yP) − ·(P)} cos  = 0. (10)

If the forces P1, P2, be turned in the same sense through the same angle about the respective points A1, A2, so as to remain parallel, the value of  is alone altered, and the resultant (P) passes always through the point

(11)

which is determined solely by the configuration of the points A1, A2, and by the ratios P1 : P2 :  of the forces acting at them respectively. This point is called the centre of the given system of parallel forces; it is finite and determinate unless (P) = 0. A geometrical proof of this theorem, which is not restricted to a two-dimensional system, is given later (§ 11). It contains the theory of the centre of gravity as ordinarily understood. For if we have an assemblage of particles whose mutual distances are small compared with the dimensions of the earth, the forces of gravity on them constitute a system of sensibly parallel forces, sensibly proportional to the respective masses. If now the assemblage be brought into any other position relative to the earth, without alteration of the mutual distances, this is equivalent to a rotation of the directions of the forces relatively to the assemblage, the ratios of the forces remaining unaltered. Hence there is a certain point, fixed relatively to the assemblage, through which the resultant of gravitational action always passes; this resultant is moreover equal to the sum of the forces on the several particles.

The theorem that any coplanar system of forces can be reduced to a force acting through any assigned point, together with a couple, has an important illustration in the theory of the distribution of shearing stress and bending moment in a horizontal beam, or other structure, subject to vertical extraneous forces. If we consider any vertical section P, the forces exerted across the section by the portion of the structure on one side on the portion on the other may be reduced to a vertical force F at P and a couple M. The force measures the shearing stress, and the couple the bending moment at P; we will reckon these quantities positive when the senses are as indicated in the figure.

If the remaining forces acting on the portion of the structure on either side of P are known, then resolving vertically we find F, and taking moments about P we find M. Again if PQ be any segment of the beam which is free from load, Q lying to the right of P, we find

FP = FQ, &emsp; MP − MQ = −F·PQ; (12)

hence F is constant between the loads, whilst M decreases as we travel to the right, with a constant gradient −F. If PQ be a short segment containing an isolated load W, we have

FQ − FP = −W, MQ = MP; (13)

hence F is discontinuous at a concentrated load, diminishing by an amount equal to the load as we pass the loaded point to the right, whilst M is continuous. Accordingly the graph of F for any system of isolated loads will consist of a series of horizontal lines, whilst that of M will be a continuous polygon.

To pass to the case of continuous loads, let x be measured horizontally along the beam to the right. The load on an element x of the beam may be represented by wx, where w is in general a function of x. The equations (12) are now replaced by

F = −wx, &emsp; M = −Fx,

whence FQ − FP = − &int; Q P wdx, &emsp; MQ − MP = − &int; Q P Fdx. (14)

The latter relation shows that the bending moment varies as the area cut off by the ordinate in the graph of F. In the case of uniform load we have

F = −wx + A, &emsp; M = wx2 − Ax + B, (15) where the arbitrary constants A,B are to be determined by the conditions of the special problem, e.g. the conditions at the ends of the beam. The graph of F is a straight line; that of M is a parabola with vertical axis. In all cases the graphs due to different distributions of load may be superposed. The figure shows the case of a uniform heavy beam supported at its ends.

§ 5. Graphical Statics.—A graphical method of reducing a plane system of forces was introduced by C. Culmann (1864). It involves the construction of two figures, a force-diagram and a funicular polygon. The force-diagram is constructed by placing end to end a series of vectors representing the given forces in