Norton’s power-law creep (1929)

Steam turbines of the 1920s ran hot enough for their steel to stretch slowly under constant load. Frederick Norton’s The Creep of Steel at High Temperatures (1929) found that the steady creep rate rises as a power of stress, \(\dot\varepsilon = A\sigma^n\), with \(n\) around 5. Its temperature dependence is thermally activated, \(e^{-Q/RT}\), with an activation energy close to that of self-diffusion (Dorn, 1950s) (arrhenius_creep_rate()). The example recovers \(n\) and \(Q\) from simulated creep tests.

import matplotlib.pyplot as plt
import numpy as np

from engineeringkit.materials import arrhenius_creep_rate
from engineeringkit.materials.utils import power_law_fit

Creep tests at three temperatures

Minimum creep rates of a ferritic steel with \(n = 5\) and \(Q = 280\) kJ/mol, measured with 15% scatter at five stresses and three temperatures.

A, Q, R, n = 1.8e-32, 280e3, 8.314, 5.0
stresses = np.geomspace(60e6, 200e6, 5)
temperatures = np.array([850.0, 900.0, 950.0])
rng = np.random.default_rng(11)
rates = arrhenius_creep_rate(A, Q, R, temperatures[:, None], stresses[None, :], n) * rng.lognormal(0, 0.15, (3, 5))

The stress exponent and the activation energy

At each temperature the slope of \(\log\dot\varepsilon\) against \(\log\sigma\) is \(n\). At fixed stress the slope of \(\ln\dot\varepsilon\) against \(1/T\) is \(-Q/R\).

exponents = [power_law_fit(stresses, row)[1] for row in rates]
slope = np.polyfit(1 / temperatures, np.log(rates).mean(axis=1), 1)[0]
print(f"stress exponents {np.round(exponents, 2).tolist()}; activation energy {-slope * R / 1e3:.0f} kJ/mol")
assert np.allclose(exponents, n, atol=0.4) and abs(-slope * R / Q - 1) < 0.05

life = 0.01 / arrhenius_creep_rate(A, Q, R, 850.0, 100e6, n)
print(f"time to 1% creep strain at 850 K and 100 MPa: {life / 3600:.0f} h")

fig, ax = plt.subplots(figsize=(6, 3.8))
for T, row in zip(temperatures, rates):
    ax.loglog(stresses / 1e6, row, "o", label=f"{T:.0f} K")
    ax.loglog(stresses / 1e6, arrhenius_creep_rate(A, Q, R, T, stresses, n), "k-", lw=0.8)
ax.set(xlabel="stress [MPa]", ylabel="minimum creep rate [1/s]", title="Norton's law")
ax.legend()
fig.tight_layout()

plt.show()

stress exponents [4.87, 4.88, 4.8]; activation energy 279 kJ/mol
time to 1% creep strain at 850 K and 100 MPa: 2487 h