Internal Friction in Metals

A tubular specimen of the metal is made the elastic control of a heavy pendulum bar. The system can be set into forced vibrations, either bending or twisting the specimen, by a measured periodic magnetic force-couple. The resonance amplitude and frequency, with the moment of the impressed force-couple, yield the ratio of the energy dissipated in one cycle to the energy of the vibration. The tubular form of the specimen enables the relation between dissipation and stress-amplitude to be studied in detail, at least for torsion. For normal (unfatigued) metal, the internal friction appears to be entirely associated with shear. This is not borne out in certain specimens of fatigued or overstrained metal; but these exceptions are easily accounted for. If the energy dissipated per cc. per cycle of stress-amplitude is represented by $F$, the relation of friction to stress-amplitude is given by the formula $F=8\mathrm{}\phi W(1-\frac{f_0}{f})$ where $W$ is the total strain energy, $\phi$ is the coefficient of internal friction, and $f_0$ is a "threshold" stress-level which is zero initially, but takes on larger values as the vibration history becomes longer. If $f_0=0\mathrm{}$, this formula is the same as that proposed by Kimball, viz. $F=\xi f^2$. The coefficient $\phi$ here introduced is independent of systems of units and to a certain extent independent of the vibration history. If $f$ is less than $f_0$ the dissipation is negligible; if $f$ is greater than another much higher level $f_1$, $F$ increases faster than any power of $f$. Repeated cycles of high amplitudes cause all the constants gradually to change; $f_0$ decreases again, $\phi$ increases many fold, the upper limit itself may change. These progressive changes are probably associated with the progress of fatigue.