The beam _AB_ is supposed for the moment to have no weight. Consequently the only force acting upon the portion of the beam _AO_ is the reaction _R_, and, similarly, _Rʹ_ is the only force acting upon the portion _OB_. Obviously so far as the simple action of these two forces or reactions is concerned, the tendency of each is to cause vertical slices of the beam, so to speak, to slide over each other. In other words, in engineering language, the portion _AO_ of the beam is subjected to the shear _S = R_, while _OB_ is subjected to the shear _Sʹ = -Rʹ_. The cross-sectional area of the beam must be sufficient to resist the shear _S_ or _Sʹ_. The upper part of Fig. 13 shaded with broken vertical lines indicates this condition of shear. It is evident from this simple case that the total vertical shears at the ends of any beam will be the reactions or supporting forces exerted at those ends, and that each will remain constant for the adjoining portion of the beam. The third member of equation (13) shows that the greatest bending moment _M_₁ in the beam varies as the product _x_₁_x_₂ of the segments of the span. That product will have its greatest value when _x_₁ = _x_₂. Hence _a simple beam loaded by a single weight will be subjected to the greatest possible bending moment when the weight is placed at the middle of the span, at which point also that moment will be found_. =82. Bending Moments and Shears with any System of Loads.=—The general case of a simple beam loaded with any system of weights whatever may be represented in Fig. 13, in which the beam of Fig. 12 is supposed to carry three loads, _w_₁, _w_₂, _w_₃. The spacing of the loads is as shown. The reactions or supporting forces _Rʹ_ are determined precisely as in Fig. 12, each reaction in this case being the resultant of three loads instead of one. Applying the law of the lever as before, the reaction _R_ will have the value _d d + c d + c + b_ _R = W₃--- + W₂------- + W₁-----------_. (14) _l l l_ A similar value may be written for _Rʹ_, but it is probably simpler, after having found one reaction, to write _R′ = W₁ + W₂ + W₃ - R_. (15) As the beam is supposed to have no weight, no load will act upon the beam between the given weights. The bending moments at the points of application of the three weights or loads will be _M₁_ = _Ra_, } _M₂_ = _R_(_a + b_) - _W₁b_, } (16) _M₃_ = _R_(_a + b + c_) - _W₁_(_b + c_) - W₂c_.} After substituting the value of _R_ from equation (14) in equations (16) the values of the latter are at once known. [Illustration: FIG. 13.]