Mathematical Discussion. 
37 
(72 rp 
the member is in tension the value of T and that of , 
dz 2 
must have their signs changed. Take first the case when E K is 
in compression, so that equation (8) gives the correct value of 
In this case, since the right portion of the truss is free from 
loads, b' will be zero; and since n x is negative, the term involv¬ 
ing c' is the only one which can make the whole expression neg¬ 
ative. Hence a load will be at C when the compression in H K 
is a maximum. (When it is a minimum a load may be at A or 
at D.) 
d 2 T 
If H K is in tension, the value of - must be the negative 
of that given in (8), that is 
It ®- g - = -r l 2 a ' — l x b +— (w 8 c' + n x d') ..(14) 
The left portion of the truss being now unloaded, a ' is zero. 
Also, since l x and n x are negative and n 2 is positive, the only 
term in (14) which can be intrinsically negative is that contain¬ 
ing d '; that is, a load will be at D when the tension in E K is 
a maximum. (Loads at 5 or at C may correspond to minimum 
values of the tension.) 
12. Application of General Condition to Web Member — Second 
Case. —We now consider the case in which the point F falls 
between A and (7, so that l x is positive while n x is negative. 
This case is represented in Fig. 3, the member considered being 
marked E K , as in all the figures. It may be shown by the 
usual method that a load at any point of A C produces in II K 
a tension which is greater the nearer the load is to C ; and 
that a load at any point of D B produces in E K a tension which 
is greater the nearer the load is to D. We need further to com¬ 
pare the effects of equal loads at C and D. From equations (2) 
and (3) it is seen that a load P at C produces a tension of 
Ph(h~n 1 ) 
It 
while a load P at D produces a tension of 
Pl\{l 2 nf) 
It 
The former will be greater than the latter if 
Idd i n i 
that is if 
