— 132 — 
As before, to make d a minimum, w must be made a minimum, or di a maximum ; 
therefore d = |</i, which, substituted, gives 
w = 0.194 c?!, 
or about one-fifth of the depth of the bars. 
For three pairs of similar bars at each end of the pin, the equation of moments 
will be 
Siv^dx T = ^ 
substituting in which for a gives 
= 0.159 ん 
or about one-sixtli of the depth of the bars. 
Finally, if there be four pairs of similar bars at each end of tlie pin, tlie equa- 
tion oT moments will be 
T = ^7T Td\ 
which gives 
w = 0.137 も， 
or about one- seventh of tlie depth of bars. 
To find the greatest working sliearing-stress ( supposed to bo uniformly dis- 
tributed) in terms of tho working resistance to tension, — 
Let S = actual varying resistance to shearing, considered uniformly distributed. 
The greatest value of S will correspond to a value of w equal to 0.274^ ; for suppose 
tlie moment to remain at its maximum value, and tlie dimensions of tlie bar to vary 
(consequently the stress therein also), tlie tension in the bar will be greatest for tlie 
value of w corresponding with the greatest value of d t : therefore the shear will also 
be greatest for that value. 
Equating tlie tension to tlie shear gives 
wd, Tz 
~~ 4 ~ 
Substituting \d for^ , and 0.274 (Jrf) for iv 9 gives 
0.274(| ci) 2 T ： 
% d ' 2 S 
4 J 
and 
S = 0.545 T ； 
for T = 5 tons, S = 2.725 tons. But the greatest allowable value for S is, accord- 
ing to Bender, 2.91 tons. This proves, that, if an iron pin bo properly propor- 
tioned for crushing and bending, it will bo strong enough to resist shear, and in 
fact, that， before the pin could shear, it would either break by bending or crushiug, or 
tlie eye of the bar would give way. A similar investigation for steel bridges, where 
T = 8.85 tons, (7 = J T, and II (tlie intensity of working bending- stress) = 1.82 7 , 
gives d = 0.5714 w = 0.1816^, and S = 5.912 tons = tlie actual intensity of 
slieariiig- stress when the pin is strained up to the bending- limit, and the ratio Ü 
• • ( d\ 
for that condition of stress is at its minimum, and consequently the area of the bar, 
the tension therein, and the shear on the pin, at their maxima. But the greatest 
allowable shear is, according to Bender, 普 x ☆ x r = 普署 x 8.35 = 4.858 tons ; so 
