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 F_P = F_Q, M_P - M_Q = -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 F_Q - F_P = -W, M_Q = M_P; (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. [Illustration: FIG. 24.] To pass to the case of continuous loads, let x be measured horizontally along the beam to the right. The load on an element [delta]x of the beam may be represented by w[delta]x, where w is in general a function of x. The equations (12) are now replaced by [delta]F = -w[delta]x, [delta]M = -F[delta]x, whence _ _ / Q / Q F_Q - F_P = - | w dx, M_Q - M_P = - | F dx. (14) _/P _/P 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, M = ½wx² - 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. [Illustration: FIG. 25.] [Illustration: FIG. 26.]