— 96 — 
the billige is covered by the moving loaa. This horizontal component will be zero 
for a truss with an odd number of panels, and ^ W 1 tan 0 for a truss with an even 
number of panels. 
The stress in the next panel of the bottom chord towards the nearest end of 
the bridge is found by subtracting from the one already determined the horizontal 
component of the stress in the main diagonal at the panel point between tbo two 
p «au el a considered ; the bridge, as before, being fully loaded. This component is a 
multiple of W n tan 0. In this way can bo found all the stresses in tbo panels of the 
bottom chord, the correctness of the work being checked by seeing if the stress in 
the end panel be equal to the re- action multipliod by tan Q. If so, the remaining 
upper-chord stresses may bo at once written by inspection ; for tlio stress iu tlio nth 
panel of the top chord, counting from the nearest pier or abutment, and supplying 
the missing pauel at tlio end, is numerically equal to that in the (n + l)th panel 
of the bottom chord. 
Next let us consider the double-intersection truss. 
The formulas for this case are so complicated that it is better not to employ 
tlicm. The simplest method is to draw a skeleton diagram, and number tlio panel 
points, as in the single- intersection truss. The double -intersection truss really con- 
sists of two trusses, as may bo seen in the accompanying diagram. 
Sucli a division is necessary in order to calculate the chord stresses when the 
ti.uss contains an odd number of panels. This is accomplished by finding by the 
method of momonts already explained, the chord stresses in each of the trusses shown 
in Figs. 2 and 8, and then combining them. Thus the stress in panel 9-10 of the 
lower chord in Fig. 1 is equal to tliat in panel 9-11 of Fig. 2, plus that of panel 8 - 10 
of Fig. 8. 
Tlio live-load stress in any diagonal sloping upward from right to left is found 
by noting whether the number at its foot be odd or even, then taking tlio sum 
of tlio odd or oven numbers, from one or two up to tlio number at tlio foot of 
the diagonal, and multiplying the sum by ^ sec a, or sec ß f as the case may be. 
The stress due to the dead load is found by taking the sum of tho same numb- 
ers, and from it subtracting tlio sum of the odd or even numbers from one or two up 
to (n 一 n f — 2), wliere n is the number of panels in the span, and n f is the number 
at the foot of the diagonal considered. Whether tho odd or even numbers should be 
