—— 1G4 — 
Comparing the sections of tlie members given in Table XIII. with those found 
in this chapteL* we do not in all cases find an exact agreement. This is because tlie 
t.iblo w.i-s m tde l>e *OL*e the exact wind pressures per lineal loot were determinea, 
nevertheless with ono exception the agreement is sufficiently close. The exception 
is in tlie sizes oi* tlie ujjpor lateral rods, whicli tlio t ible gives ns greater than those 
found in this clia[>t«r. This was done advisedly in preparing the table, in order to 
insure sufficient lateral stiffness to tlie bridges ami thus prevent undue vibration 
under passing loads. It is only lor sp ms under '2UÜ feet that this allowance is 
necessary. 
It is now a convenient time to test the size of tlie middle post to see it it be 
strong enongli to resist the bemling ami trail Hferrecl load stresses clue to a wind pres- 
sure ot* thirty pounds per square foot, in addition to tlie live and dead load stresses. 
Chapter IX. gives the stress produced by bending as 
(/，+/•') (ゴ ー/) = c 
2 //A 
Hei.e /， and / 7 have values equal to of those used when calculating the 
stresses in tlie vertical sway bracing, while tlie values of d and / are the same as be- 
foi.e. The value of m is nearly 1.1 feet. Tlie following is then the table of data 
P + Pf = 0.735 
d = 25 
/ = 10 
m = 1.1 
a 0.735 x 15 _ , , 】 ヽ 
There；ore C = 2 x 1.1 = 5 toi 】 s (^eavly). 
Using an intensity of five tons makes tlie section required lor one channel, to 
resist bending, 1.00 square inch. 
The simplest way to find the value of the transferred panel load V is to multiply 
the panel length by tlie wind pressure per lineal loot, multiply the product by tlie 
deptli of truss ami divide by tlie perpendicular distance between centres of trusses, 
reducing the result to tons. 
Thus V = 
21 X 320 x 25 
15.4 X 2UÜU 
= about 2 tons. 
Using the intensity for which the post was proportioned viz. 1.565 tons, and 
remembering tliat V must be equally divided between the two channels, we find the 
additional area of each channe], necessary to resist the stress considered, to be O.u* 
square inches, making the total area of one chamiel, needed to resist the effect 
of wind pressure, 1.00 + 0.64 = 1.64 square inches. This is slightly greater than 
one halt* tlie area ot* one channel, or 1.59 square inches, as previously determinea, 
but the tlilierence is so small that we will conclude that the post is strong enough. 
Next let ns proportion the pins, bogi lining witli the middle one of the bottom 
cliofd. 
The value ol* T in the formula M = ^ of Chapter XIII. is most easily obtained 
