CHAP. V.] MOMENT OF ROTATION PARALLEL FORCES. 39 



polygon from to 1, and 1 to 2 or 0. Next choose a pole C, 

 and draw S S t and S 2 . Draw then a parallel to S till intersec- 

 tion with first force, P 1? then parallel with S t to second force, 

 P 2 , then parallel to S 2 or S to intersection with vertical through 

 support B, and finally draw the closing line I*. A line through 

 C, parallel to L, gives as before the vertical reactions. Follow- 

 ing round the force polygon, we find at A the reaction down- 

 wards, since S acts from C to and is to be replaced (Art. 4) 

 by L and V\; at B reaction upwards, since P 2 acts up, and fol- 

 lowing round, S 2 acts from to C. -Both reactions are equal to 

 a 0. At A then the support must be above, and at B below the 

 beam. The shaded area gives the moments to pole distance H. 

 Had we taken the pole in the perpendicular through o, S would 

 have been parallel with the beam itself. This is, however, a 

 matter of indifference. The moment area may lie at any in- 

 clination to the beam. We also see here again the effect of a 

 couple (Art. 23). S is simply shifted through a certain distance 

 to S 2 , parallel to S , and therefore the moment at any point be- 

 tween P 2 and B is constant. This is generally true of any 

 couple, as we have already seen, Article 21, and may be proved 

 analytically as follows : 



Let the distance between the forces be a A B, Fig. 22. 

 Then for any point o, we have P x(-f-B o) P xBo=P [a + 'Eo 

 B o]=P a. For o' between A and B, P x A o'+P x o' B= 

 P [A (/ + <?' B] =Pa. 



So also for any point to the left, the same holds true. 



Graphically the proof is as follows : 



Decompose both forces into parallel components, Fig. 23. 

 Then for any point, as o, we have the moment M = H x m n 

 Hxwj9 or M = H x np. But n p is the constant ordinate 

 between the parallel components A n and A. p. 



We see, therefore, by simple inspection, that the distance of 

 P! and P 2 from the support B, Fig. 21, has no influence what- 

 ever upon the moment or strain in A B, provided the distance 

 between the points of application remains the same, and that 

 the moment at all points between P 2 and the support B is con- 

 stant and a maximum. From B and P 2 the moments decrease 

 left and right, and become zero at A and P t . 



4O. Beam with Two Equal and Opposite Forces Ic- 

 tween the Two Supports. Let the beam A B, Fig. 24, PL 6, 

 be acted upon by the two equal and opposite forces P t P 2 . 



