CHAP. V.] MOMENT OF ROTATION PARALLEL FORCES. 43 



corresponding polygons A V D and A c" D coincide. This is 

 easily done, as if the closing line of the second polygon for any 

 assumed position of O 2 (O 2 H being equal to G O t ) does not co- 

 incide with A D, the ordinate at c" can be laid off from C and 

 A c" D thus found in proper position, and then the pole O a can 

 be located. It will evidently be at the intersection of the ver- 

 tical O 2 O' 2 with c" D. 



The two force polygons being thus formed, we construct the 

 polygon A C" D by drawing lines B B", E E", C C", etc., so 

 that their angles with the vertical shall be equal to the angle 

 between the planes of the forces, and making them equal tc 

 the ordinates B b", E e", C c", etc., respectively. Join b' B", 

 e > E", /' F", d C", etc., and lay off the ordinates B b, E e, F/, 

 C c, etc., respectively equal. The ordinates to the polygon 

 thus obtained, viz. : A b efc D multiplied by the pole distance 

 Oj G or O 2 H, give the moments at any point.- A b and c D 

 are straight lines, b efc is a curve (hyperbola). If we drop 

 verticals through O x and O 2 , and draw the perpendiculars O/ M, 

 O' 2 K ; A M is the reaction R 1? and D K the reaction R 2 , both 

 measured to the scale of the force polygon. Their directions 

 are found by the composition of A G and H 2 and D H and 

 G M respectively, under the angle of the forces. 



4TH. COMBINED TWISTING AND BENDING MOMENTS. 



In many constructions pieces occur which are subjected at 

 the same time to both bending and twisting moments. Both 

 can be represented and given by moment areas. Thus, Fig. 

 28, PL 7, represents an axle turning upon supports at A and B 

 and having at C a wheel upon which the force P acts tangenti- 

 ally. We have then a moment of torsion M t = P R and reac- 



o /7 



tions P, = P and P 2 = P - : s being the distance of P 



' 



a+s 

 from B, and a of P from A. 



Let the bending moments be represented by the ordinates to 

 the polygon a C b ; then laying off a o equal to P and drawing 

 o O parallel to b c, we find the corresponding pole distance 

 O &, and the reactions P! and P a equal to k a and o Jc respec- 

 tively. 



Now, in the force polygon O a o thus found, at a distance 

 from O equal to R, draw a line m n parallel to P. This line 

 m n evidently gives for the same pole distance the moment of 



