L r )ß — 
cl =5 
25 
TVi sec 0 = 
9.805 
h = 
15.4 
i TV 1 sec 6 — 
4.900 
ding = 
82.65 
W ,f tan 0 = 
11.379 
sec 0 = 
1.306 
i W ff tan 0 = 
5.689 
tan 0 = 
0.84 
E — 
n 
1.3375 
fan 0 1 — 
1.304- 
sec 0 = 
1.7468 
IV = 
6 .0；38 
盔 tan 0 = 
1.1235 
^\ = 
7.508 
W.j tan 6 f = 
4.440* 
Wf = 
2.500 
\ W. z tan 0 f = 
2.220 
W f ，= 
13.5 4(] 
W 3 tan 0 f = 
6.302 
JV 2 = 
3.255 
\ W 3 tan 0 f — 
8.151 
4.020 
W rt tan 0 ==■ 
2.7；H 
n 
：}.150 
i W rt tan 0 = 
1.807 
i 匕 = 
3.255 
W t] tan 0 = 
3.001 
!〆(!■ = 
4.358 
\ W Cl tan 6 = 
1.880 
E = 
10.700 
The next stop is to draw the skeleton diagram shown on Plate XIII” number- 
ing tlio panel points from light to left, bo^inning witli zero at the end. Next by re- 
forring to Table III we find the stresBes tlue to the uniform live load, tlie dead load 
iui(l Uie engine excesses on each member of the truss, add tliem together and write 
the sum iu itft proper place on the cliagraiu. 
Thus the stress in tlie second panel of the top chord is 
1V.W' 1 tun G + ^ E tan ¢=74 x 11.379 + 27 x 1 .1230 = 1 15.(577. 
That in the end main diagonal is 
^ W sec 6 + ^ 0 + ^ E sec 6 
— 21 x 0,9858 + 2 x 9.805 + 4.003 + ]1 x 1.7408 =： 64.430 
find that- in tlie middle post is 
IW - \ W l + + W 
= ö x 0.7548 - 3.754 + 5 x 1.3375 + 2.500 = 9.962. 
T]ie next step is to fiiul the wind stresses on tlie windward bottom chord wlien 
Uie bridge is empty : they will bo multiples of JV 2 tan0 f , and tlie coefficients are 
giyen in the Table Y. There is no need to enter these stresses oil the diagram, but 
tliej’ should bo written upon a temponiry diagnuu, as in the accompanying figure, 
followed l>y the letter C to denote that tlie stress is compressive. 
