62 



REACTIONS FOR ANY NUMBER OF LOADS. Art. 47. 



The resultant of the loads passes through C, the intersection 

 of 1 and 4, since 1 and 4 are in equilibrium with ad (triangle 

 adO). The string polygon DEC is similar to that of Fig. 32. 



Fig. 43. 



Fig. 43 shows a cantilever beam, the part BC extending 

 beyond the support. The loads 20 and 16 will produce an up- 

 ward reaction at A, while the loads 8 and 15 produce a downward 

 reaction. The direction of the resultant R^ is in doubt. As- 

 suming R t to act upward, and with center at B, 



187^-20X13-16X7+8X4+15x9 = 

 Ji x = \ B 5 . Since this is plus as was assumed in writing the 

 equation of equilibrium, the assumption was correct. See rule 

 Art. 39. 



R 2 evidently acts upward. With center at A, 



18 R 2 = 20X5+16XH+4X18+8X22+15X27. 

 2 = sfg.. As a check Ri + R 2 = W + W = 63 = 20 +16+ 

 4 + 8 + 15. 



In drawing the string polygon, it is 

 necessary to have the lines of action of 

 those rays which form a triangle with a 

 load or reaction in the force polygon, 

 intersect on the line of action of that 

 load or reaction. Thus 5, 6, and the load 

 15 form the triangle efO and their lines 

 of action intersect at H. 



nO, parallel to the closing line, de- 

 termines R t and R 2 



Fig. 44 shows a trestle bent with 

 foundations at A and B, and wind loads 

 Pj, P 2 , and P 3 ; the resultant of the en- 

 tire weight is P w and acts in the axis of 

 symmetry. It is convenient to use only 



& 



Fie. 44. 



