88 Hoskins—Maximum Stresses in Bridge Members. 
As in the present case n Y is negative, while l 2 and n 2 are 
positive, the above inequality always holds; hence a load at C 
produces a greater effect than one at D. 
The general principle reached for the present case is, then, 
the following: 
To produce the greatest stress in the member HK, the loads 
should cover.the truss as completely as possible, with the heaviest 
loads near the point C. 
This principle will usually enable us to choose readily which 
one, among several positions which may be found to satisfy 
equation (7), corresponds to the greatest tension in H K. 
If the member considered is L H instead of H K (Fig. 3), the 
stress is a compression; otherwise the same results apply. 
It remains to apply the criterion for distinguishing between 
positions giving maximum values and those giving minimum 
values of the stress. Since n l is negative, while l u l 2 and 
d 2 T 
n 2 are positive, the only term in the value of 
(equation (8) ) which can be really negative is that containing 
c'. Hence a load must be at C in order to make the stress a 
maximum. (Minimum values of the stress may occur when 
loads are at A, B or D. ) 
13. Web member — case in which lf= 0.—If the point F coin¬ 
cides with A, we have a limiting case between the two preced¬ 
ing. No figure is given for this case; but by referring to Fig. 1 
or Fig. 3, and assuming that F falls at A , it is easily seen that 
loads on D B produce no stress in II K , and that a load on A G 
produces a stress which is greater the nearer the load is to C.. 
Hence, for the greatest stress in IIK , the part A C of the truss 
should be loaded as fully as possible, the heaviest loads being 
near G\ while no attention need be paid to the loads on D B . 
(They must of course be considered in applying the general 
condition expressed by equation (7).) In the present case it 
will be convenient to employ equation (11); and since l 2 =l the 
equation becomes 
(W-P s ) 
( 16 ) 
