breaking strength. Let k_{max.} be the breaking strength of
the same bar when subjected to stresses varying from k_{max.} to k_{min.}
alternately and repeated an indefinitely great number of times; k_{min.} is
to be reckoned + if of the same kind as k_{max.} and - if of the opposite
kind (tension or thrust). The range of stress is therefore
k_{max.}-k_{min.}, if the stresses are both of the same kind, and
k_{max.}+k_{min.}, if they are of opposite kinds. Let [Delta] = k_{max.} +-
k_{min.} = the range of stress, where [Delta] is always positive. Then
Woehler's results agree closely with the rule,

k_{max.} = 1/2[Delta]+[root](K squared-n[Delta]K),

where n is a constant which varies from 1.3 to 2 in various qualities of
iron and steel. For ductile iron or mild steel it may be taken as 1.5. For
a statical load, range of stress nil, [Delta] = 0, k_{max.} = K, the
statical breaking stress. For a bar so placed that it is alternately loaded
and the load removed, [Delta] = k_{max.} and k_{max.} = 0.6 K. For a bar
subjected to alternate tension and compression of equal amount, [Delta] = 2
f_{max.} and k_{max.} = 0.33 K. The safe working stress in these different
cases is k_{max.} divided by the factor of safety. It is sometimes said
that a bar is "fatigued" by repeated straining. The real nature of the
action is not well understood, but the word fatigue may be used, if it is
not considered to imply more than that the breaking stress under repetition
of loading diminishes as the range of variation increases.

It was pointed out as early as 1869 (Unwin, _Wrought Iron Bridges and
Roofs_) that a rational method of fixing the working stress, so far as
knowledge went at that time, would be to make it depend on the ratio of
live to dead load, and in such a way that the factor of safety for the live
load stresses was double that for the dead load stresses. Let A be the dead
load and B the live load, producing stress in a bar; [rho] = B/A the ratio
of live to dead load; f_1 the safe working limit of stress for a bar
subjected to a dead load only and f the safe working stress in any other
case. Then

f_1 (A+B)/(A+2B) = f_1(1+[rho])/(1+2[rho]).

The following table gives values of f so computed on the assumption that
f_1 = 71/2 tons per sq. in. for iron and 9 tons per sq. in. for steel.

_Working Stress for combined Dead and Live Load. Factor of Safety twice as
great for Live Load as for Dead Load._

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