then 
— 131 — 
and 
wd\T = tension in bar, 
tv dG = compression on pin and eye. 
^lieso, of course, are equal ; and, as (7 = G tons when T = 5 tons, there results 
tlle equation, • 
d = = 0.838 rfi, 
'vhicli shows that the diameter of the pin should never be less than eighty- three per 
cent of the depth of the bar. It is possible, though, that good iron of twenty-five 
tons tensile strength will resist more than thirty tons per square inch in compression : 
consequently d may be taken at 0.8rfi as a matter of convenience. 
lo find the proportion between width and depth of bars for the smallest allow- 
able PiQ in an iron bridge,— 
Let the notation be as before, and first let us suppose that there be but one pair 
ブ bars acting at each end of the pin, and that the total tension be a fixed quantity. 
” stress in one bar is ivdiT, and its moment is u 2 di T. This must be equal to the 
lesi8 ting- moment of the pin, w-liicli is given by the well-known equation. 
M 
Here R ^ ^ ^ j 
RI 
万 • 
著 7H* 4 , and = r = 菩 , substituting wliich gives 
M = AttW 3 . 
•^luating the two values of the moments gives 
i^iTx = ^7rJ ( / 3 , • 
or 
w 2 
37T ds 
64 d\ 
Now, to make the diameter of tho pin as small as possible, the moment of the 
Ss be made as small as possible ; and, as the stress is constant, the lever - 
^ must be made as small as possible. But tlie product of w and di is a con- 
11 • so when w is smallest, must bo greatest. But the greatest value of d\ is 
が ； su ^stituting which gives 
w J 
and 
w が 
： 0.7 5 4 rf, s , 
W : 
0.2 7 4 ^, 
01 ' ab I ° ut °»e-fourth of the depth of the bars. 
於 . 1 ere わ 0 two pairs of similar bars acting at each end of tlie pin, instead of one 
1 1 lr, the equation of moments will be 
or 
2 iv 2 di 為 7T Td 3 f 
to 2 ; 
87T 
(P 
128 . di 
