— 157 — 
hus the stress in the second panel is 
0 W 2 tan ゲ •= G x 4.440 = 26.640. 
The next step is to find the reduced dead load stresses when the bridge is em* 
1%: they will he multiples of }V 6 tanQ, aucl the coefficients are given in the 
Co l_n f or Ir, i n Table III. 
They are to be written on the same diagram under tlio other stresses, and are 
to l)e followed by T to denote that they are tensile. Thus in the second panel the 
stl， ess is 
JV 6 tan 0 = 3 x 3.661 + 1.830 = 12.81 3. 
へ The next step is to subtract the values of T from the corresponding values of 
’ m order to find the greatest actual compressive stresses that can ever come upon 
le c h or <l. The largest of these differences is 18.82 7, and it is well to proportion 
le w h 0 l e bottom chord to resist this compression, so as to simplify tlie drawings 
! Uu ^ S ^ 10 P W0l ， k. There will be but little waste in so doing, for fclio different panel 
en gths ot the strut will have to act as tension members, and tlieir effective areas 
re to l> e subtracted from the total sections required, when finding the necessary 
areas for the chord bars. 
The next step is to find the wind stresses in tlie leeward bottom chord when the 
Ul( ge is covered by tlie train : they will be multiples of Tr：* tanQ^ and the cooffici- 
Gnts are given in Table V. 
Thus in the third panel tlie stress is 
^ tan 6 x 6.302 = 37.81 2. 
he sti esses thus determined are to be written on the principal diagram beneath 
6 live and dead load stresses already found. 
The next step is to find the stresses in the leeward bottom chord due to the 
^ansfeired load, when the bridge is covered by tlie train : they will be multiples of 
° 没 ， an d tlie coefficients are given in Table III. under the heading 
hus in the third panel tho stress is 
ö W^taii 0 =6 x 2.734 = 16.404. 
or こ S t 1GSSeS t】 Uls detemiued are to be written under those last found, and tlie two 
placeTb S ^ lesses ^ or eac h lower chord panel are to be added together, the sum being 
mi 
e 〒 ex t step is to find the sections required for the tension members. 
⑽ d tl 6 仏 6 am l l oa d stresses of tlie bottom chord are to be divided by five, 
ins 10 stresses by seven and a half, and tlie greater result taken. By 
tio ^ le ^ a o ram it will bo seen that tlio combined stresses determine the sec- 
US 111 evei ’y case but that of the end panels. The intensities for the main dia* 
aie 5, 4 普 and 4i tons, which divided iuto the proper stresses give the 
ßonals 
s ©ction g 
ieqiui’ecl as marked on the diagram. 
0r l no portion the counters we must first decide whether they are 
e, then consult Table VI. With channel struts iu the bottom c 
to be single 
the bottom chords, single 
