Mathematical Discussion. 
27 
2nd. Tension due to load on D B. — For a load P on D B, we 
compute the reaction at A, its value being 
P(z + l—x) m 
l * 
Considering the portion of the truss to the left of M W, we see 
that this reaction is in equilibrium with the internal forces pro¬ 
duced in the three members cut. Hence for the tension in UK 
we have (taking moments about F f as before and remembering 
that A F = 17) 
Pl x {z^l-x) 
It 
(3) 
3rd. Tension due to a load on C D. —A load P between C 
and D comes upon the truss partly at G and partly at D. The 
part coming at C is 
P(z + l 1 -4- n 2 —x) 
n 
and that coming at D is 
P ( x—z—l 1 + n-7) 
n 
The tension due to the former part is found from equation (2) 
by substituting 
P(z + l, +n^—x ) 
n 
for P and putting (z _p l x — n x ) instead of x in the factor (x — z). 
The tension due to the latter part is found from equation (3) 
by substituting 
• P{x—z-l x +n ± ) 
n 
for P and putting (z _p l x n. 2 ) for x in the factor (z-\-l — x). 
Combining the two values thus found, we get for the total ten¬ 
sion in H K due to a load P between C and P, 
P l % (li—nA 
n 11 
(z + lp-h n 2 —x)+ P - - n - s - {x—z—l x + n{) -(4) 
We may now determine the total tension due to all loads on 
the truss, by putting P=wdx in each of the expressions given 
in equations (2), (3) and (4), and integrating each between 
limits corresponding to that part of the span to which it applies. 
