Art. 60. 



GRAPHICAL SOLUTION BY SECTIONS. 



81 



and L. R may be found by drawing force and string poly- 

 gons (see Fig. 37), and U, D, and L may be found by the 

 method of Art. 33 as shown in the force polygon of Fig. 59. 

 The resultant of U and R is R f and must pass through a; the 

 resultant of L and D must also equal R' ', but be opposite in 

 direction; and must pass through c. Then ac determines the 

 direction of R' . In the force polygon R is laid out to scale 

 and through its extremities lines are drawn parallel to the 

 directions of U and R', thus determining their amounts from 

 the force triangle 1-2-3 ; in like manner, from R' and lines par- 

 allel to the directions of D and L, the force triangle 1-3-4 is 

 formed. 



Three in dependent solutions are possible by means of the 

 following combinations : 



R with U and D with L (as shown). 



R with L and D with U. 



R with D and U with L. 



Combinations giving acute intersections, either in the force 

 or space diagrams should be avoided. When this is not possible, 

 it is possible to find the three stresses by using two convenient 

 components in place of R. 



By taking successive sections all the stresses may be found, 

 or any three stresses gotten by some other method may be readily 

 checked. In the former case, by combining the various force 

 polygons, a stress diagram is formed, but the construction is not 

 so simple as that of Art. 54. If, however, the sections are taken 

 in the order indicated in Fig. 59, there will be but two unknowns 

 in each case, and the construction reduces to that of Art. 54. 



Fig. 60. 



