Figure 15 — Loop 3 (Shear Flow K,) 



q 



OOP 



1 



q 



oop 



2 



q 



OOP 



3 



q 



oop 



4 



q 



oop 



5 1 



q 



oop 



6 



q 



oop 



7 



q 



oop 



9 



= [L,^]. 



+ 1 



-1 



0" 



+ 1 



-1 



-1 



+ 1 



-1 



-1 



+ 1 











+ 1 











+ 1 















+ 1 



+ 1 



'loop j 



0+10 



rl 



[151 



4 



tor shear flows in the plates 



It has been shown that the particular solution Iq ^ : 

 arises from the shear flows out of the nodes in , .j. We will now derive an expression for 



'Tout 1 



the terms q , . in terms of V and V by differentiating Equation [9] with respect to x. 



"out I y z '' o 1 



Figure 16 shows how the rate of change of tensile stress in a node is related to the 

 shear flows out of that node via all connecting plates. The arrows indicate forces exerted 

 on the members with which they are associated. The sum of q. of all the plates connected 

 to the node (two plates are shown in Figure 16) equals q^^^ . of the node. Equilibrium in 

 the longitudinal direction of the small tensile element of cross section A. and length Ax 



. A . = : 



requires that (since, a 



(da ) 



d\ 



^out i 



[16] 



67 



