50 MOMENT OF DOTATION PARALLEL FORCES. [dlAP. V. 



parallel to itself. Since at the point of maximum strain the 

 sum of the forces either side is zero, this point is given by the 

 intersection of the broken line thus found with the axis. 



Thus in PI. 7. Fig. 32, let A B be the beam sustaining a uni- 

 form load, and also the concentrated loads P x P 2 P 8 P 4 . The 

 reaction of the uniform load at the supports is equal to hall 

 that load. To find the reactions for the concentrated loads we 

 draw the force polygon 01234, choose a pole C, then con- 

 struct the equilibrium polygon A' 1 2 3 4 B', and parallel to 

 A' B' draw C L. L and L 4: are the reactions at A and B. 

 Now through L draw AQ L horizontal, make it equal to the 

 length of the beam, and take it as axis of abscissas. [It is of 

 course advantageous here to lay off the forces along the verti- 

 cal through B, as done in the Fig. Then A falls in the vertical 

 through A and 1 2 3 4 are directly under the forces them- 

 selves.] 



The ordinate to be laid off at A is equal to L + half the 

 uniform load. Between A and 1 the line A.\ V is inclined to 

 the axis at an angle depending upon the uniform load. Lay off 

 L U equal to this load and draw AQ U. A' t 1' must be parallel 

 to this line. At 1' the line A\ V is shifted to 1", so that 11" 

 is the load P!. Then 1" 2' is parallel as before to A U, and 

 2' 2" is the load P 2 , and so on. The intersection 2 with A L 

 gives the point of maximum moment or cross-section of rup- 

 ture. The force P 2 at this point in our Fig. is divided, as shown 

 by L in the force polygon, into two portions, one of which is to 

 be added to the forces left, the other to the forces right. The 

 ordinate ?/ y' at any point gives the shear or sum of the forces 

 acting at that point. This force acts up or down according as 

 the ordinate is above or below the axis. 



Moreover, the area between the broken line and axis A L, 

 limited by this ordinate, gives the moment of rotation of the 

 forces beyond the section y, areas below the axis being nega- 

 tive. For a section at z, therefore, we have area A A.\ 1' 1" 

 2' 2 , minus 2 2" 3' 3" z' z, or what is the same thing, the area 

 ZQ z' 4:' 4" B'x L, since the sum of the moments of all the forces 

 is zero. 



46. Influence of a Concentrated Load, passing over the 

 Beam. If in addition to the already existing uniform and 

 concentrated loads, a new force operates, we have by (44) simply 

 to ernstruct for this new force its force and equilibrium poly- 



