Mathematical Discussion. 
85 
guished; first, that in which F is to the left of A (Fig. 1), and. 
second, that in which F is between A and C (Fig. 3), 
In the first of these two cases it may be shown without diffi¬ 
culty that loads on A C and on D B produce opposite kinds of 
stress in the member E K. (This is seen by comparing equa¬ 
tions (2) and (3), remembering that l x is now negative.) Either 
tension or compression may therefore be produced in H K by the 
live loads, depending upon the position of the train. 
Treating H K as a tension member, we see that all loads on 
D B produce tension, and the effect of such a load is greater the 
nearer it is to D ; while a load on A G causes a compression 
which is greater the nearer the load is to C. Hence for the 
greatest tension in H K we must load D B as completely as pos¬ 
sible, putting heavy loads as near D as possible, and leave A G 
free from loads. These requirements may in some cases be some¬ 
what contradictory, as when the foremost loads are light com¬ 
pared with those that follow. But they serve as a general 
guide, and must be satisfied as nearly as possible, at the same 
time that equation (7) is exactly satisfied. 
To get the greatest compression in E K , the load must occupy 
A C as fully as possible, D B being free from loads (if this is 
possible), at the same time that equation (7) is satisfied. 
Equation (7) may conveniently be put in another form for use 
in the present case, as follows: Adding the numerators and 
denominators respectively of the two equal fractions in equation 
(7), and putting W— the total load on the truss, we get 
h ~ h ~ h+h ~ l . 1 ' 
Now if AT AT is treated as a compression member, P 2 should be 
zero if this is possible consistently with equation (7). Hence 
from equation (9), 
n 2 Q _ W 
71/ ? 2 l 
or 
Q-n-T- W 
( 10 ) 
