, for stresses always
of the same kind, F = (t-u)/(t-f_{max.}) approximately agreed with
experiment. For stresses of different kinds Weyrauch found F =
(u-s)/(2u-s-f_{max.}) to be similarly approximate. Now let
f_{max.}/f_{min.} = [phi], where [phi] is + or - according as the stresses
are of the same or opposite signs. Putting the values of F in (1) and
solving for f_{max.}, we get for the breaking stress of a bar subjected to
repetition of varying stress,

f_{max.} = u(1+(t-u)[phi]/u) [Stresses of same sign.]
f_{max.} = u(1+(u-s)[phi]/u) [Stresses of opposite sign.]

The working stress in any case is f_{max.} divided by a factor of safety.
Let that factor be 3. Then Woehler's results for iron and Bauschinger's for
steel give the following equations for tension or thrust:--

Iron, working stress, f = 4.4 (1+1/2[phi])
Steel, working stress, f = 5.87 (1+1/2[phi]).

In these equations [phi] is to have its + or - value according to the case
considered. For shearing stresses the working stress may have 0.8 of its
value for tension. The following table gives values of the working stress
calculated by these equations:--

_Working Stress for Tension or Thrust by Launhardt and Weyrauch Formula._

------------------------+-------+-----------+--------------------+
| [phi] | [phi] | Working Stress f, |
| | 1 + ----- | tons per sq. in. |
| | 2 +--------------------+
| | | Iron. | Steel. |
------------------------+-------+-----------+--------------------+
All dead load | 1.0 | 1.5 | 6.60 | 8.80 |
| 0.75 | 1.375 | 6.05 | 8.07 |
| 0.50 | 1.25 | 5.50 | 7.34 |
| 0.25 | 1.125 | 4.95 | 6.60 |
All live load | 0.00 | 1.00 | 4.40 | 5.87 |
| -0.25 | 0.875 | 3.85 | 5.14 |
| -0.50 | 0.75 | 3.30 | 4.40 |
| -0.75 | 0.625 | 2.75 | 3.67 |
Equal stresses + and - | -1.00 | 0.500 | 2.20 | 2.93 |
------------------------+-------+-----------+--------------------+

[v.04 p.0550] To compare this with the previous table, [phi] = (A+B)/A =
1+[rho]. Except when the limiting stresses are of opposite