*0 MOMENT OF ROTATION PARALLEL FORCK8. [CHAP. V. 



Construct the force polygon 012. Choose a pole C and draw 

 C 0, C 1, C 2. Parallel to C 0, draw the first side of the equi- 

 librium polygon to intersection \v\\hfirstforce Pj ; then paral- 

 lel to C 1 to second force P 2 ; then parallel to C 2 to d. Join d 

 and 0. Parallel to this draw C a in force polygon. Then a 

 is the vertical reaction at A, which acts upwards, since it must 

 with C a replace C ; and C 0, when we follow round from o to 

 1 and 1 to C, acts from C to 0. 



We have the same vertical reaction at B, but here, since we 

 must follow from 1 to 2 and 2 to C, C 2 acts from 2 to C, hence 

 following round, the reaction at B is downward. The shaded 

 area gives the moments to pole distance H, as before. 



We see at once that at a certain point e the moment is zero. 

 Left and right of this point the moment is positive and nega- 

 tive. At the point itself we have a point of inflection, and 

 here, since the moment is zero, there is no longitudinal strain. 

 At b and c the moments are greatest ; here the beam is most 

 strained, and at these points, therefore, are the "cross sections 

 of rupture." Here again, if we had taken the pole C in the 

 perpendicular through a, the closing line of the polygon o d 

 would have been horizontal. It is, however, indifferent at 

 what inclination a d may lie, but we may if we wish make it 

 horizontal now, and then layoff from its new intersections with 

 P! and P 2 along the directions of these forces, the ordinates 

 already found at b and c, and join the points thus obtained with 

 the ends of od(i.e., with its intersections with the verticals 

 through the supports). The ordinatest of the new polygon 

 thus found will be for any point the same as beforehand will 

 also be perpendicular to the beam. 



[NOTE. Had we taken the forces precisely as above but in reverse order, 

 the force line would be reversed, and we should have and 2 in place of 1, 

 and 1 in place of and 2 ; that is. in place of O 1 we should have O and 

 O 2. Constructing then the equilibrium polygon by drawing a line paral- 

 lel to new O to intersection with new Pi, then parallel to new O 1 to in 

 tersection with new P 3 , then parallel with new O 2 to intersection with 

 vertical through B, and finally joining this last point with intersection of 

 the first lin drawn (O 0) with vertical through A, we have at first sight a 

 very different equilibrium polygon. This new polygon will consist of two 

 parts. If the ordinates in one of these parts are considered positive, those 

 in the other must be negative. The difference of the ordinates in these two 

 portions for any point, will give the same result as above. This, by mak- 

 ing the above construction, the reader can easily prove.] 



