SIMPLE STRUCTURES 



191 



FIG. 129 



shown in Fig. 128. Let the depth of truss and panel length be 

 each 15 ft, and the loads carried at the joints of the upper chord 

 be 7, 10, 9, and 15 tons respectively. The reactions at B and J 

 are found by taking moments about J and B to be 17| tons mid 

 23| tons respectively. 



Since this form of truss has parallel chords and a single web 

 system, it is not necessary to begin at any particular point, but a 

 section may be taken anywhere, provided it cuts both chords and 

 a single web member. Taking any 

 section xy, and considering only the 

 portion of the structure on one side of 

 the section, the external forces acting 

 on this portion will be in equilibrium 

 with the stresses P, $, 7?, in the mem- 

 bers cut (Fig. 129). Since Q is the 

 only stress having a vertical compo- 

 nent, it must equilibrate the external 

 forces at B and C. That is to say, 

 from the condition of equilibrium 



V vertical forces 0, we have Q sin 63 26' = 17| 7 ; whence 

 Q = 11.854 tons and is compressive. 



To find P take moments about D. Then since Q and R both 

 pass through D, their moments about this point are zero; therefore 



Pxl5 = 17f x 15- 7 x 7.5; 



whence P = 14i tons. By observing the signs of the moments of 

 the external forces at B and C about D, P is found to act in the 

 direction shown by the arrow, that is, in compression.. 



Similarly, to find the stress R in Dl\ take the section xy just to 

 the left of E, then take moments about E. Since P and Q pass 

 through E, their moments about this point are zero, and hence 



J Kxl5=17f X 22.5-7x15; 



whence R = 19.44 tons. 



Since the loads are vertical, R might also have been found 

 from P and Q by the condition V horizontal forces = ; that is, 

 p _|_ Q cos 63 26' = R ; whence R = 19.427 tons. 



