In the case of heavy beams, either built or rolled, as in railroad structures, it is of the greatest importance to determine both the bending moments and the shears, as represented in the preceding equations and diagrams, and to provide sufficient metal to resist them. The case of Fig. 13 is perfectly general for moments and shears, and the methods developed are applicable to any amount or any system of loading whatever. =83. Bending Moments and Shears with Uniform Loads.=—Fig. 14 represents what is really a special case of Fig. 13, in which the loading is uniform for each unit of length of the beam throughout the whole span _l_. Inasmuch as the load is uniformly distributed, it is evident that the reaction at each end of the beam will be one half the total load, or _wl_ _R = R_ʹ = ----. (17) 2 [Illustration: FIG. 14.] The general expression for the bending moment at any point _G_ in the span, and located at the distance _x_ from the end _A_, will take the form _x w_ _M_ = _Rx_ - _wx_.--- = --- _x_(l-_x_). (18) 2 2 This equation, giving the value of _M_, is the equation of a parabola with the vertex over the middle of the span. The bending moment at the latter point will be found by placing _x = l/2_ in equation (18), which will give _wl²_ _M_ = -----. (19) 8 Hence, in Fig. 14, if the vertical line _DC_ be erected at _D_, so as to represent the value of _M_ in equation (19) to a convenient scale, the parabola _ACB_ may be at once drawn. Any vertical intercept, as _GF_ between _AB_ and the curve _AFCB_, will represent by the same scale the bending moment in the beam at the point indicated by the intercept. Equation (19), giving the greatest external bending moment in a simple beam due to a uniform load, is constantly employed in structural work, and shows that that moment is equal to the total load multiplied by one eighth of the span. It has already been shown, in connection with Fig. 12, that when a single centre weight rests on a beam the centre bending moment is equal to that weight multiplied by one fourth the span. If the total uniform load in the one case is equal to the single load in the other, these equations show that the single centre load will produce just double the bending moment due to the same load uniformly distributed over the span. Wherever it is feasible, therefore, the load should be distributed rather than concentrated at the centre of the span. That portion of Fig. 14 shaded with vertical lines shows the shear existing in the beam. Evidently the shear at each end is equal to the reaction, or one half the total load on the span. The expression for the shear at any point, as _G_, distant _x_ from _A_ will be ( _l_ ) _S = R_ - _wx_ = _w_( ---- - _x_). (20) ( 2 )