4:8 MOMENT OF KOTATION PARALLEL FORCES. [CliAP. V. 



BO also its area is proportional to the moments of the moments, 

 or the moment of inertia of the load area. 



As to the shearing force at any point of a beam submitted 

 to the action of parallel forces, the reactions at the ends being 

 easily found as above by a line parallel to the closing line in 

 the force polygon, we have only to remember that the shear at 

 any point is equal to the reaction at one end, minus all the 

 weights between that end and the point in question. 



Thus for a uniformly distributed load we have simply to lay 

 off the reactions which are equal to one-half the load, above 

 and below the ends, and draw a straight line, which thus passes 

 through the centre of the span. The ordinates to this line are 

 evidently then the shearing forces. If we have a series of con- 

 centrated loads, we have a broken line similar to A.\ 1' 1" 2', 

 etc., Fig. 32, PI 7, where each successive weight as we arrive at 

 it, is subtracted from the preceding shear. 



44. Beam continuously Loaded and alo Subjected to 

 the Action of Concentrated Loads. In practice we have 

 to consider not only a continuously distributed load, such as the 

 weight of the truss or beam itself, but also concentrated forces, 

 such as the weight of cars, locomotives, etc., standing upon or 

 passing over the truss. 



In PI. 8, Fig. 31, we have a continuous loading represented 

 by the load area A a b B, and in addition four forces P' w . 

 Now, since the total moment about any point is equal to the 

 sum of the several moments, we can treat each method of load- 

 ing separately and then combine the results. Thus with the 

 force polygon (b) we obtain the equilibrium polygon A' 1 2 3 

 .... B' for the continuous loading, and with the force polygon 

 (a) the equilibrium polygon A' 1" 2" 3" B" for the con- 

 centrated loads. If now in (b) we draw C L parallel to the 

 closing line A' B', and in (a) C" it' parallel to the closing line 

 A' B", we obtain at once the reactions at the supports for each 

 case. 



Thus for continuous loading we have L for reaction at A, 

 and 10 I for reaction at B ; for the concentrated loads, L' 0' 

 at A and 4' L' at B. These reactions hold the beam in equi- 

 librium. 



For any cross-section y, the shear to the right is composed of 

 the two components .L 7 and I/ 3' (i.e., is equal to the reactions 

 minus the forces between cross-section and support). The mo- 



