Describing a load cycle

The simplest load history is a constant-amplitude cycle between the largest stress \(\sigma_{\max}\) and the smallest \(\sigma_{\min}\), Figure 8.21.3. Fatigue tests and design rules describe it by its mean value and its amplitude,

\[ \sigma_m = \frac{\sigma_{\max} + \sigma_{\min}}{2}, \qquad \sigma_a = \frac{\sigma_{\max} - \sigma_{\min}}{2} \tag{8.21.1}\]

or by the stress range \(\Delta\sigma = \sigma_{\max} - \sigma_{\min} = 2\sigma_a\) together with the stress ratio

\[ R = \frac{\sigma_{\min}}{\sigma_{\max}} \tag{8.21.2}\]

The fully reversed cycle is the one of a rotating shaft under a constant bending moment, where every point of the surface passes from tension to compression and back on every revolution. The repeated cycle is that of a gear tooth, loaded once per revolution and then unloaded. A bolt that is pretensioned and then loaded further in service follows a fluctuating cycle with a large mean stress. Swedish texts call the fully reversed cycle växlande and the repeated one pulserande. Real load histories, such as the stress in a suspension arm on a rough road, are irregular. They are reduced to a list of cycles with known amplitudes and means by counting methods, of which rainflow counting is the standard, and the methods of this chapter are then applied cycle by cycle. Figure 8.21.4 contrasts the two kinds of history.

How a fatigue crack starts and grows

A fatigue failure passes through three stages. In the first, a crack initiates. Even when the nominal stress is elastic, single grains at the surface, where they are least constrained, deform plastically by slip along favourably oriented crystal planes. Repeated slip back and forth roughens the surface into tiny intrusions and extrusions, and a microscopic crack forms. Initiation is fastest where the local stress is highest and the surface is weakest: at notches, fillets and holes, at scratches and machining marks, at inclusions in the material, at weld toes, and in a 3D-printed part at the boundaries between layers.

In the second stage the crack grows, by a small amount on every cycle, in the direction perpendicular to the largest principal stress. Each change in the loading, a stop or a period of higher load, leaves a mark on the crack face, and the crack front is preserved as a set of curved beach marks, Figure 8.21.5. The crack faces rub against each other as the load reverses, which makes this part of the fracture surface smooth and often dark. In the third stage the crack has grown so long that the remaining section can no longer carry the peak load, and it breaks suddenly in a single overload, leaving a rough, bright surface.

The surface of Figure 8.21.5 shows what makes fatigue dangerous. The part gave no warning by visible deformation, the final fracture looks brittle although aluminium is a ductile metal, and most of the life of the part was spent growing a crack that nobody saw.

How much plastic deformation each cycle causes divides fatigue into two regimes, Figure 8.21.6. If the stress amplitude exceeds the yield strength, every cycle deforms the material plastically and traces a hysteresis loop in the stress-strain diagram, whose width is the plastic strain range \(\Delta\varepsilon_p\). Such low-cycle fatigue breaks a part within roughly \(10^3\) to \(10^4\) cycles, and it is analysed in terms of the strain, since the strain is what the loop controls; it concerns, e.g., parts heated and cooled in every start of a machine. In high-cycle fatigue the nominal stress stays elastic, the cycle runs back and forth along the straight part of the curve, and the plastic slip is confined to single grains at the surface. Most machine parts, which must last millions of cycles, are in this regime, and it is analysed in terms of the stress.

The strain-based description of low-cycle fatigue goes back to Manson and Coffin, who studied parts cracked by repeated heating and cooling [2,3]. They split the strain amplitude \(\Delta\varepsilon/2\) of the loop into its elastic and plastic parts and found that each falls as a power of the number of load reversals \(2N\), two per cycle,

\[ \frac{\Delta\varepsilon}{2} = \frac{\sigma_f'}{E}(2N)^{b} + \varepsilon_f'(2N)^{c} \tag{8.21.3}\]

The first term is the elastic strain of the S-N line of the next section, 8.21.4, divided by Young’s modulus \(E\) and written for reversals, with the fatigue strength coefficient \(\sigma_f'\) and the same exponent \(b\). The second term is the Coffin-Manson relation for the plastic strain, with the fatigue ductility coefficient \(\varepsilon_f'\), close to the true strain at fracture, and an exponent \(c\) between about \(-0.5\) and \(-0.7\). The plastic term falls much more steeply, so it dominates at short lives and the elastic term at long ones. The two are equal at the transition life, which is longer for a soft, ductile steel than for a hard one. The rest of the chapter treats high-cycle fatigue, where only the elastic term is left.

The S-N curve

The fatigue strength of a material is measured by loading specimens with constant-amplitude cycles until they break. The classical test, due to Wöhler, bends a polished cylindrical specimen while it rotates, so that every point of its surface sees a fully reversed cycle once per revolution, and counts the revolutions to failure. Repeating the test at several amplitudes gives the S-N curve, also called the Wöhler curve: the stress amplitude \(\sigma_a\) against the number of cycles to failure \(N\), on logarithmic axes. Figure 8.21.7 shows the test.

The shaft between the load bearings is in four-point bending, with no shear force and the same moment along the whole specimen, so the specimen breaks where it is thinnest. At its smallest diameter \(d\) the surface carries the bending stress of 8.16.11,

\[ \sigma_a = \frac{32M}{\pi d^3} = \frac{16Fa}{\pi d^3} \]

and as the shaft turns, every point of the surface passes from the tension side to the compression side and back once per revolution, so this stress is the amplitude of a fully reversed cycle. Each specimen gives one point of the curve, the amplitude it was tested at and the number of revolutions it survived, Figure 8.21.9. A specimen that has not broken after a set number of cycles, often \(10^7\), is stopped and recorded as a run-out.

rng = np.random.default_rng(7)
R_m_d, f_d, se_d = 700, 0.85, 350                    # a polished steel specimen, @eq-endurance-estimate
a_d = (f_d*R_m_d)**2/se_d
b_d = -np.log10(f_d*R_m_d/se_d)/3
N_runout = 1e7

levels = np.repeat([560, 520, 480, 440, 400, 375, 360, 345, 330], 3)
N_med = (levels/a_d)**(1/b_d)                         # the median life on the sloping line
N_test = N_med*10**(0.25*rng.standard_normal(levels.size))    # the scatter of real tests
se_own = se_d*(1 + 0.04*rng.standard_normal(levels.size))     # each specimen's own limit
broke = (levels > se_own) & (N_test < N_runout)

N_line = np.logspace(3, 8, 300)
fig, ax = plt.subplots(figsize=(6, 3.8))
ax.loglog(N_line, np.where(N_line < 1e6, a_d*N_line**b_d, se_d), color='C0', lw=1.5,
          label='fitted curve')
ax.loglog(N_test[broke], levels[broke], 'o', color='C3', ms=6, label='broken specimen')
ax.loglog(np.full((~broke).sum(), N_runout), levels[~broke], '>', color='C2', ms=7,
          mfc='white', mew=1.5, label='run-out, stopped unbroken')
ax.set_xlabel('cycles to failure $N$')
ax.set_ylabel(r'stress amplitude $\sigma_a$ [MPa]')
ax.set_xlim(1e3, 1e8)
ax.set_ylim(300, 650)
ax.set_yticks([300, 350, 400, 500, 650], ['300', '350', '400', '500', '650'])
ax.minorticks_off()
ax.grid(True, which='both', alpha=0.4)
ax.legend(loc='upper right')
plt.show()

Data of this kind is expensive. A rotating-bending machine runs at around 50 revolutions per second, so a single specimen taken to \(10^7\) cycles occupies it for more than two days, and one taken to \(10^8\) cycles for more than three weeks. The scatter means that every amplitude needs several specimens, so one S-N curve with its knee takes a few dozen specimens, all machined and polished alike, and months of machine time. The curve then holds only for that material, that surface, that size and a zero mean stress. Yet there is no way around it: every fatigue calculation in this chapter starts from a measured curve, and even the estimate below is a fit to such tests. This is why fatigue data is collected in handbooks and guarded by the companies that paid for it.

A fatigue test at Jönköping University

The rotating-bending machine is the classical test, but most fatigue tests today run in a servohydraulic machine such as the one in Figure 8.21.10 (a), in the materials and manufacturing department of the School of Engineering at Jönköping University. Two columns carry a crosshead, from which the upper grip hangs below a load cell that measures the force. The lower grip sits on a hydraulic actuator in the base, which pushes and pulls the specimen. Oil at high pressure comes through the blue hoses from the hydraulic power unit on the right, and it also clamps the specimen in the two hydraulic wedge grips of Figure 8.21.10 (b). A controller compares the measured force with the commanded one and adjusts a servo valve on the actuator continuously, so the machine follows any prescribed force history: a sine for a constant-amplitude test, or a recorded history from a vehicle for a variable-amplitude one, Figure 8.21.4.

The specimen is loaded axially, in tension and compression, and the whole of its waist carries the same stress. In rotating bending only a thin layer at the surface is highly stressed, so an axial test finds a defect to start a crack from more easily. This is why the load factor of the previous section is \(k_c = 0.85\) for an axial load.

The machine stands in a former cleaning closet. Its reinforced concrete floor is the roof of a bomb shelter built during the Cold War, and that floor, able to take the weight of the machine and the forces it shakes into the ground, is why the room was chosen.

When the photographs were taken, the controller was cycling the force between \(+20.7\) and \(-21.3~\text{kN}\), a fully reversed cycle with \(R \approx -1\), and it had counted \(N = 1\,859\,164\) cycles after the running time \(t = 25~\text{h}\;49~\text{min}\;27~\text{s}\). A sinusoidal force of amplitude \(F_a\) and frequency \(f\), and the number of cycles it has applied, are

\[ F(t) = F_a \sin(2\pi f t), \qquad N = f t \]

so the record gives the test frequency \(f = N/t\), and the time a run-out at \(10^7\) cycles takes is \(t_{10^7} = 10^7/f\).

Code

N_count = 1859164
t_run = 25*3600 + 49*60 + 27                 # s
f_test = N_count/t_run
t_runout = 1e7/f_test/(24*3600)              # days

ltx(r"f = \frac{N}{t} =", f_test, r"~\text{Hz}, \qquad t_{10^7} = \frac{10^7}{f} =", t_runout,
    r"~\text{days}", precision=3)

\[ f = \frac{N}{t} =20.0~\text{Hz}, \qquad t_{10^7} = \frac{10^7}{f} =5.79~\text{days} \]

The machine runs at \(20~\text{Hz}\), and the specimen, a day into its test, has covered less than a fifth of the way to a run-out, which would take almost six days. Figure 8.21.11 redraws two of the records the controller displays, the force over a few cycles and the force against the displacement of the actuator. The displacement follows the force through the stiffness \(k\) of the specimen and the grips, \(\delta = F/k\), and the measured peaks, \(\pm 21~\text{kN}\) at about \(\pm 0.13~\text{mm}\), give \(k \approx 160~\text{kN/mm}\).

Code

F_a, k_train = 21.0, 21.0/0.13             # kN, kN/mm, from the controller's peak readings
t = np.linspace(0, 0.2, 400)                  # s, four cycles at 20 Hz
F = F_a*np.sin(2*np.pi*f_test*t)

fig, (ax1, ax2) = plt.subplots(1, 2, figsize=(9, 3.4))
ax1.plot(1e3*t, F, color='C0', lw=2)
ax1.axhline(0, color='0.35', lw=0.8)
ax1.set_xlabel('time $t$ [ms]')
ax1.set_ylabel('axial force $F$ [kN]')
ax1.set_xticks([0, 50, 100, 150, 200])
ax1.set_yticks([-21, -10, 0, 10, 21])
ax1.set_xlim(0, 200)
ax1.grid(True, alpha=0.4)
ax2.plot(F/k_train, F, color='C3', lw=2)
ax2.axhline(0, color='0.35', lw=0.8)
ax2.axvline(0, color='0.35', lw=0.8)
ax2.set_xlabel(r'axial displacement $\delta$ [mm]')
ax2.set_ylabel('axial force $F$ [kN]')
ax2.set_xticks([-0.13, -0.05, 0, 0.05, 0.13])
ax2.set_yticks([-21, -10, 0, 10, 21])
ax2.set_xlim(-0.16, 0.16)
ax2.grid(True, alpha=0.4)
fig.tight_layout()
plt.show()

The force-displacement record has no loop. Every cycle runs up and down the same straight line, the elastic line of Figure 8.21.6, so the specimen is in high-cycle fatigue: no plastic strain can be measured, and the damage builds up in single grains at the surface until a crack starts. The record will look the same until shortly before the specimen breaks, when the growing crack lowers its stiffness and the line starts to tilt.

Figure 8.21.12 shows the whole curve of a steel, from a single cycle, where the amplitude that breaks the specimen equals the tensile strength \(R_m\), to \(10^8\) cycles. Below about \(10^3\) cycles lies the low-cycle fatigue of the previous section. Between \(10^3\) and \(10^6\) cycles the part has a finite life that the curve predicts, and beyond the knee its life is infinite as long as the amplitude stays below the endurance limit.

Between about \(10^3\) and \(10^6\) cycles the S-N curve of a steel is close to a straight line on the double logarithmic axes, as Basquin observed in 1910 [4],

\[ \sigma_a = a N^b \tag{8.21.4}\]

with a negative exponent \(b\), typically between \(-0.05\) and \(-0.15\). Near \(10^6\) to \(10^7\) cycles the curve of most steels bends over into a horizontal line, and below this endurance limit \(\sigma_e'\) a polished specimen lasts indefinitely. Aluminium alloys, most other non-ferrous metals and polymers show no such knee: their curve keeps falling, and their fatigue strength is quoted at a stated number of cycles, often \(5 \cdot 10^8\), Figure 8.21.13.

When no test data is available, Shigley gives an estimate for steels from the tensile strength alone [5]. The endurance limit of the polished specimen is about half the tensile strength,

\[ \sigma_e' \approx \begin{cases} 0.5\,R_m & R_m \le 1400~\text{MPa} \\ 700~\text{MPa} & R_m > 1400~\text{MPa} \end{cases} \tag{8.21.5}\]

and the line 8.21.4 is drawn from the strength \(f R_m\) at \(N = 10^3\) cycles to the endurance limit \(\sigma_e\) at \(N = 10^6\), where \(f\) falls from about \(0.9\) for \(R_m = 500~\text{MPa}\) to \(0.8\) for very strong steels. The two points fix the constants,

\[ a = \frac{(f R_m)^2}{\sigma_e}, \qquad b = -\frac{1}{3}\log_{10}\frac{f R_m}{\sigma_e} \tag{8.21.6}\]

and solving 8.21.4 for \(N\) gives the life at a given amplitude above the endurance limit,

\[ N = \left(\frac{\sigma_a}{a}\right)^{1/b} \tag{8.21.7}\]

Figure 8.21.14 draws the estimate for a steel with \(R_m = 700~\text{MPa}\), for which \(f \approx 0.85\), together with the curve of the shaft of Example 1 below, whose endurance limit has been reduced by the corrections of the next section.

Code

def sn_curve(N, R_m, sigma_e, f):
    """The stress amplitude at the life N: @eq-basquin above 10^6 cycles, sigma_e beyond."""
    a = (f*R_m)**2/sigma_e
    b = -np.log10(f*R_m/sigma_e)/3
    return np.where(N < 1e6, a*N**b, sigma_e)

R_m, f_sn = 700, 0.85
N = np.logspace(3, 8, 300)

fig, ax = plt.subplots(figsize=(6, 3.8))
ax.loglog(N, sn_curve(N, R_m, 0.5*R_m, f_sn), color='C0', lw=2,
          label=r"polished specimen, $\sigma_e' = 0.5\,R_m$")
ax.loglog(N, sn_curve(N, R_m, 238, f_sn), color='C1', lw=2,
          label=r'machined shaft, $\sigma_e = 238$ MPa')
ax.axvline(1e6, color='0.35', lw=0.8, ls=':')
ax.set_xlabel('cycles to failure $N$')
ax.set_ylabel(r'stress amplitude $\sigma_a$ [MPa]')
ax.set_xlim(1e3, 1e8)
ax.set_ylim(150, 700)
ax.set_yticks([150, 238, 350, 595, 700], ['150', '238', '350', '595', '700'])
ax.minorticks_off()
ax.grid(True, which='both', alpha=0.4)
ax.legend()
plt.show()

fit = broke & (levels >= 400)                    # broken on the sloping line
x, y = np.log10(levels[fit]), np.log10(N_test[fit])
slope, A_fit = np.polyfit(x, y, 1)                # the least-squares line of log N on log sigma_a
m_fit = -slope
s_fit = np.sqrt(np.sum((y - (A_fit - m_fit*x))**2)/(x.size - 2))
k_dc = 2

ltx(r"n_s =", int(fit.sum()), r", \quad m =", m_fit, r", \quad s =", s_fit,
    r", \quad 10^{-ks} =", 10**(-k_dc*s_fit), r", \quad 10^{-ks/m} =", 10**(-k_dc*s_fit/m_fit),
    precision=3)

sig = np.linspace(395, 600, 100)
fig, ax = plt.subplots(figsize=(6, 3.8))
ax.loglog(10**(A_fit - m_fit*np.log10(sig)), sig, color='C0', lw=2, label='mean curve, 50 % fail')
ax.loglog(10**(A_fit - k_dc*s_fit - m_fit*np.log10(sig)), sig, color='C3', lw=2, ls='--',
          label=r'design curve, $k = 2$, 2.3 % fail')
ax.loglog(N_test[fit], levels[fit], 'o', color='0.3', ms=5, label='broken specimens')
ax.set_xlabel('cycles to failure $N$')
ax.set_ylabel(r'stress amplitude $\sigma_a$ [MPa]')
ax.set_xlim(1e3, 1e6)
ax.set_ylim(380, 620)
ax.set_yticks([400, 480, 560, 620], ['400', '480', '560', '620'])
ax.minorticks_off()
ax.grid(True, which='both', alpha=0.4)
ax.legend(loc='upper right')
fig.tight_layout()
plt.show()

From the test specimen to the part

A real part differs from the polished specimen of the rotating-bending test in its surface, its size and the way it is loaded. Marin collected these differences into correction factors that multiply the endurance limit of the specimen [5],

\[ \sigma_e = k_a k_b k_c k_d k_e k_f\,\sigma_e' \tag{8.21.9}\]

The surface factor \(k_a\) accounts for the roughness of the surface, where cracks start. It is fitted as \(k_a = a_s R_m^{b_s}\) with \(R_m\) in MPa and the constants of Table 8.21.1. A rough surface harms a strong steel most, since a strong steel is more sensitive to small notches, and Figure 8.21.16 shows that a forged surface can throw away almost the whole advantage of a high tensile strength.

R = np.linspace(300, 1400, 200)
finishes = [('ground', 1.58, -0.085), ('machined or cold-drawn', 4.51, -0.265),
            ('hot-rolled', 57.7, -0.718), ('as-forged', 272, -0.995)]

fig, ax = plt.subplots(figsize=(6, 3.6))
for (name, a_s, b_s), c in zip(finishes, ('C0', 'C1', 'C2', 'C3')):
    ax.plot(R, np.minimum(a_s*R**b_s, 1), color=c, lw=2, label=name)
ax.set_xlabel(r'tensile strength $R_m$ [MPa]')
ax.set_ylabel(r'surface factor $k_a$')
ax.set_xticks([300, 600, 900, 1200, 1400])
ax.set_yticks([0, 0.25, 0.5, 0.75, 1])
ax.set_xlim(300, 1400)
ax.set_ylim(0, 1.05)
ax.grid(True, alpha=0.4)
ax.legend()
plt.show()

The size factor \(k_b\) accounts for the larger volume of highly stressed material in a thick part, which is more likely to contain a defect from which a crack can start. For round parts in bending or torsion

\[ k_b = 1.24\,d^{-0.107}, \qquad 2.79 \le d \le 51~\text{mm} \]

with \(d\) in millimetres, and \(k_b = 1\) for an axial load, which stresses the whole section equally. The load factor \(k_c\) converts the endurance limit measured in bending to other loads: \(k_c = 1\) in bending, \(0.85\) for an axial load and \(0.59\) for torsion. The temperature factor \(k_d\) falls below one at high temperatures. The reliability factor \(k_e\) moves the median curve down to a stated probability of survival, \(k_e = 1\) for \(50~\%\), \(0.897\) for \(90~\%\) and \(0.814\) for \(99~\%\). The miscellaneous-effects factor \(k_f\) collects everything else, e.g., corrosion, residual stresses and plating, and is one unless data says otherwise. It is unrelated to the fatigue notch factor \(K_f\) below, despite the similar symbol.

A notch is not part of 8.21.9. Its effect is applied to the stress, by multiplying the nominal stress amplitude with the fatigue notch factor \(K_f = 1 + q(K_t - 1)\) of 8.19.8, which accounts for the notch sensitivity \(q\) of the material.

The mean stress and the Haigh diagram

The S-N curve is measured with fully reversed cycles, \(\sigma_m = 0\). A tensile mean stress holds a crack open and lets it grow faster, so it lowers the amplitude a part can endure, while a compressive mean stress closes the crack and helps. Tested at the same amplitude and different mean stresses, the specimens give the family of S-N curves of Figure 8.21.17.

The combinations of \(\sigma_a\) and \(\sigma_m\) that a part endures indefinitely are drawn in the Haigh diagram, with the mean stress on the horizontal axis and the amplitude on the vertical, Figure 8.21.20 in Example 1 below. A failure line runs from the endurance limit \(\sigma_e\) on the amplitude axis, where the load is fully reversed, to a static strength on the mean axis, where there is no amplitude left at all.

The straight line to the tensile strength is the Goodman line [10]. With a factor of safety \(n\) applied to both stresses, a load cycle is safe if

\[ \frac{\sigma_a}{\sigma_e} + \frac{\sigma_m}{R_m} = \frac{1}{n} \tag{8.21.11}\]

defines an \(n\) above the required value. Test data fall mostly between the Goodman line and the Gerber parabola, \(n\sigma_a/\sigma_e + (n\sigma_m/R_m)^2 = 1\), so Goodman is the somewhat conservative choice and the common one in design. The Soderberg line replaces \(R_m\) by the yield strength and is more conservative still. A part must also not yield on its first cycle, which requires \(\sigma_a + \sigma_m < R_{p0.2}\), the Langer line from \(R_{p0.2}\) on one axis to \(R_{p0.2}\) on the other.

8.21.11 also converts a cycle with a mean stress into a fully reversed cycle that is equally damaging, with the amplitude

\[ \sigma_{ar} = \frac{\sigma_a}{1 - \sigma_m/R_m} \tag{8.21.12}\]

which can be entered into the S-N curve to estimate a finite life.

When the stress at the critical point is multiaxial, as at the surface of a shaft in bending and torsion, Shigley combines the components into two von Mises stresses, one from the amplitudes and one from the means,

\[ \sigma_a' = \sqrt{\left(K_f\sigma_a\right)^2 + 3\left(K_{fs}\tau_a\right)^2}, \qquad \sigma_m' = \sqrt{\left(K_f\sigma_m\right)^2 + 3\left(K_{fs}\tau_m\right)^2} \tag{8.21.13}\]

and uses them in 8.21.11 in place of \(\sigma_a\) and \(\sigma_m\) [5]. Since the combination already accounts for the type of load, the load factor is then \(k_c = 1\).

Varying amplitude and the Palmgren-Miner rule

Most parts see cycles of many amplitudes. The simplest way to combine them is the linear damage rule proposed by Palmgren for ball bearings in 1924 and by Miner for aircraft structures in 1945 [11,12]. A cycle at an amplitude whose life is \(N_i\) consumes the fraction \(1/N_i\) of the life, so \(n_i\) such cycles consume \(n_i/N_i\), and the part fails when the fractions add up to one,

\[ D = \sum_i \frac{n_i}{N_i} = 1 \tag{8.21.14}\]

The rule ignores the order of the cycles, although a few large cycles early in life start a crack that the later small cycles then grow, and it counts cycles below the endurance limit as harmless, which they are not once a crack exists. Measured damage sums at failure scatter widely around one, and design codes often set the limit lower, or continue the S-N line below the knee for variable-amplitude loading. Taken with these reservations, 8.21.14 is the standard tool for an estimate.

def rainflow(S):
    """Rainflow cycles of the turning points S, as (index, index, count) triples.

    The stack holds the turning points that are not yet paired. Whenever the inner
    range of its last four is no larger than both outer ranges, the inner pair has
    closed a loop: it is a full cycle and leaves the stack. The residue left at the
    end counts as half cycles between neighbours.
    """
    stack, cycles = [], []
    for i in range(len(S)):
        stack.append(i)
        while len(stack) >= 4:
            i1, i2, i3, i4 = stack[-4:]
            inner = abs(S[i3] - S[i2])
            if inner <= abs(S[i2] - S[i1]) and inner <= abs(S[i4] - S[i3]):
                cycles.append((i2, i3, 1.0))
                del stack[-3:-1]
            else:
                break
    cycles += [(i, j, 0.5) for i, j in zip(stack[:-1], stack[1:])]
    return cycles, stack

S_block = np.array([340, -100, 180, 20, 260, -240, 120, -40, 340])     # MPa
cycles, residue = rainflow(S_block)
pairs = [(i, j) for i, j, n in cycles if n == 1.0] + [(residue[0], residue[1])]

fig, ax = plt.subplots(figsize=(7, 3.6))
turn = np.arange(S_block.size)
ax.plot(turn, S_block, color='0.55', lw=1.2, zorder=1)
for c, (num, (i, j)) in zip(('C0', 'C1', 'C2', 'C3'), enumerate(pairs, 1)):
    shown = [i, j] if num < len(pairs) else residue
    lo, hi = sorted(S_block[[i, j]])
    ax.plot(shown, S_block[shown], 'o', color=c, ms=9, zorder=3, label=f'cycle {num}: {lo} to {hi} MPa')
    if num < len(pairs):                       # the loop closes where the history next passes S_i
        for p in range(j, S_block.size - 1):
            s_lo, s_hi = sorted(S_block[p:p+2])
            if s_lo <= S_block[i] <= s_hi:
                x_close = p + (S_block[i] - S_block[p])/(S_block[p+1] - S_block[p])
                break
        ax.plot([i, x_close], [S_block[i]]*2, color=c, lw=1.5, ls=':', zorder=2)
ax.axhline(0, color='0.35', lw=0.8)
ax.set_xticks(turn)
ax.set_xlim(-0.3, 8.3)
ax.set_xlabel('turning point')
ax.set_ylabel(r'stress $\sigma$ [MPa]')
ax.set_yticks([-240, -100, 0, 180, 340])
ax.grid(True, alpha=0.4)
ax.legend(loc='center left', bbox_to_anchor=(1.01, 0.5), fontsize=9)
fig.tight_layout()
plt.show()

from IPython.display import Markdown

R_mb, sigma_eb, f_b = 700, 200, 0.85
a_b = (f_b*R_mb)**2/sigma_eb                                     # @eq-basquin-constants
b_b = -np.log10(f_b*R_mb/sigma_eb)/3

def sci(x):
    """A number as LaTeX in powers of ten, 9.04e+05 -> 9.04 \\cdot 10^{5}."""
    mantissa, exponent = f"{x:.2e}".split("e")
    return f"{float(mantissa):.3g} \\cdot 10^{{{int(exponent)}}}"

rows, D_b = [], 0
for num, (i, j) in enumerate(pairs, 1):
    S_min, S_max = sorted(S_block[[i, j]])
    s_a, s_m = (S_max - S_min)/2, (S_max + S_min)/2
    s_ar = s_a/(1 - s_m/R_mb)                                     # @eq-equivalent-amplitude
    N_i = (s_ar/a_b)**(1/b_b) if s_ar > sigma_eb else np.inf     # @eq-life, harmless below sigma_e
    D_b += 1/N_i                                                  # @eq-miner, one of each cycle per block
    rows.append(f"| {num} | {S_min} | {S_max} | {s_a:.0f} | {s_m:.0f} | {s_ar:.0f} | "
                + (f"${sci(N_i)}$ | ${sci(1/N_i)}$ |" if np.isfinite(N_i) else "$\\infty$ | $0$ |"))

Markdown("| cycle | $S_{\\min}$ [MPa] | $S_{\\max}$ [MPa] | $\\sigma_a$ [MPa] | $\\sigma_m$ [MPa] "
         "| $\\sigma_{ar}$ [MPa] | $N_i$ | $n_i/N_i$ |\n|---|---|---|---|---|---|---|---|\n" + "\n".join(rows))

ltx(r"a =", a_b, r"~\text{MPa}, \quad b =", b_b, r", \quad D_b =", D_b,
    r", \quad \frac{1}{D_b} =", 1/D_b, r"~\text{blocks}", precision=3)

Example 1: A rotating shaft with a shoulder

We return to the shaft of Example 1 in Chapter 8.19, which steps from \(D = 40~\text{mm}\) to \(d = 32~\text{mm}\) at a shoulder with a fillet of radius \(r = 2~\text{mm}\), Figure 8.21.19, and carries the bending moment \(M = 200~\text{N\,m}\) and the torque \(M_t = 300~\text{N\,m}\). That example found \(K_t = 1.88\) in bending and \(K_{ts} = 1.51\) in torsion, and a factor of safety of about \(3\) against yielding. The shaft now rotates, the moment acting always in the same plane and the torque constant. It is machined from a steel with \(R_m = 700~\text{MPa}\) and \(R_{p0.2} = 500~\text{MPa}\), for which Shigley’s charts give the notch sensitivities \(q = 0.82\) in bending and \(q_s = 0.86\) in torsion at \(r = 2~\text{mm}\) [5]. The shaft turns at \(n_r = 1000~\text{rpm}\). Determine the factor of safety against fatigue, and the life in revolutions and in hours if the bending moment doubles and if the doubled moment acts during \(5~\%\) of the revolutions.

As the shaft turns, a point on its surface passes from the tension side to the compression side and back, so the bending stress is fully reversed, while the torque gives a constant shear stress. With the nominal stresses of 8.19.5,

\[ \sigma_a = \frac{32M}{\pi d^3}, \quad \sigma_m = 0, \qquad \tau_a = 0, \quad \tau_m = \frac{16M_t}{\pi d^3} \]

The notch factors follow from 8.19.8, \(K_f = 1 + q(K_t - 1)\) and \(K_{fs} = 1 + q_s(K_{ts} - 1)\), and 8.21.13 reduces to \(\sigma_a' = K_f\sigma_a\) and \(\sigma_m' = \sqrt{3}K_{fs}\tau_m\). The endurance limit of the shaft is 8.21.9 with \(\sigma_e' = 0.5R_m\) from 8.21.5, the machined surface factor \(k_a = 4.51R_m^{-0.265}\), the size factor \(k_b = 1.24d^{-0.107}\) and all other factors equal to one, for a median life. The factor of safety follows from 8.21.11, and the first-cycle check against yielding from the largest von Mises stress, \(\sigma_{\max}' = \sqrt{(K_f\sigma_a)^2 + 3(K_{fs}\tau_m)^2}\).

d_s, M, M_t = 32, 200e3, 300e3             # mm, N mm, N mm, so stresses in MPa
R_m, R_p = 700, 500
K_t, K_ts, q, q_s = 1.88, 1.51, 0.82, 0.86

sigma_a = 32*M/(np.pi*d_s**3)
tau_m = 16*M_t/(np.pi*d_s**3)
K_f, K_fs = 1 + q*(K_t - 1), 1 + q_s*(K_ts - 1)                  # @eq-kf
sa_vm, sm_vm = K_f*sigma_a, np.sqrt(3)*K_fs*tau_m               # @eq-fatigue-von-mises

k_a, k_b = 4.51*R_m**-0.265, 1.24*d_s**-0.107
sigma_e = k_a*k_b*0.5*R_m                                       # @eq-marin

n_fatigue = 1/(sa_vm/sigma_e + sm_vm/R_m)                       # @eq-goodman
n_yield = R_p/np.sqrt(sa_vm**2 + sm_vm**2)

ltx(r"K_f &=", K_f, r", \quad K_{fs} =", K_fs,
    r", \quad \sigma_a' =", sa_vm, r"~\text{MPa}, \quad \sigma_m' =", sm_vm, r"~\text{MPa}",
    r"\\ k_a &=", k_a, r", \quad k_b =", k_b, r", \quad \sigma_e =", sigma_e, r"~\text{MPa}",
    r"\\ n &=", n_fatigue, r", \quad n_y =", n_yield,
    aligned=True, precision=3)

The surface and the size together reduce the endurance limit from \(350\) to \(238~\text{MPa}\), and the shoulder raises the bending amplitude from \(62\) to \(107~\text{MPa}\). The factor of safety against fatigue is about \(1.6\), half the static factor of safety of the same shaft. The point \((\sigma_m', \sigma_a')\) lies inside the Goodman line in Figure 8.21.20, and the shaft has an infinite life.

If the bending moment doubles, the amplitude doubles while the mean stays. 8.21.12 converts the cycle to an equally damaging fully reversed one, and if that lies above the endurance limit, the line of 8.21.6 with \(f = 0.85\) and 8.21.7 give the life,

\[ \sigma_{ar} = \frac{2\sigma_a'}{1 - \sigma_m'/R_m}, \qquad N = \left(\frac{\sigma_{ar}}{a}\right)^{1/b} \]

and at \(n_r\) revolutions per minute the shaft completes them in \(t = N/(60\,n_r)\) hours.

sa_over = 2*sa_vm
s_ar = sa_over/(1 - sm_vm/R_m)                                  # @eq-equivalent-amplitude
a_sn = (f_sn*R_m)**2/sigma_e                                    # @eq-basquin-constants
b_sn = -np.log10(f_sn*R_m/sigma_e)/3
N_over = (s_ar/a_sn)**(1/b_sn)                                  # @eq-life
n_r = 1000                                                      # rpm
t_over = N_over/(60*n_r)                                        # h

ltx(r"\sigma_{ar} =", s_ar, r"~\text{MPa}, \quad a =", a_sn, r"~\text{MPa}, \quad b =", b_sn,
    r", \quad N =", N_over/1e5, r"\cdot 10^5, \quad t =", t_over, r"~\text{h}", precision=3)

The doubled moment gives a life of about \(5.7 \cdot 10^5\) revolutions, under ten hours of running. Since the endurance limit is exceeded by only \(8~\%\), the life is sensitive to every number in the calculation, and the result should be read as an order of magnitude.

If the doubled moment acts only during \(5~\%\) of the revolutions, the remaining \(95~\%\) lie below the endurance limit and, by 8.21.14, cause no damage. A total of \(N_{\text{tot}}\) revolutions then gives the damage \(D = 0.05\,N_{\text{tot}}/N\), and the shaft lasts

\[ N_{\text{tot}} = \frac{N}{0.05}, \qquad t_{\text{tot}} = \frac{N_{\text{tot}}}{60\,n_r} \]

N_tot = N_over/0.05
ltx(r"N_{\text{tot}} =", N_tot/1e7, r"\cdot 10^7, \quad t_{\text{tot}} =", N_tot/(60*n_r), r"~\text{h}",
    precision=3)

The shaft lasts about \(1.1 \cdot 10^7\) revolutions, or \(190\) hours. The reservations of the previous section apply with force here: once the overloads have started a crack, the normal cycles grow it too, and the true life may be considerably shorter. Figure 8.21.20 collects the two load cycles of Example 1 in the Haigh diagram.

sm = np.linspace(0, R_m, 200)

fig, ax = plt.subplots(figsize=(6, 4.0))
ax.plot([0, R_m], [sigma_e, 0], color='C0', lw=2, label='Goodman')
ax.plot(sm, sigma_e*(1 - (sm/R_m)**2), color='C2', lw=1.5, ls='--', label='Gerber')
ax.plot([0, R_p], [sigma_e, 0], color='C4', lw=1.5, ls='-.', label='Soderberg')
ax.plot([0, R_p], [R_p, 0], color='C3', lw=1.5, ls=':', label='Langer (yield)')
ax.plot([0, 3*sm_vm], [0, 3*sa_vm], color='0.35', lw=0.8)
ax.plot(sm_vm, sa_vm, 'o', color='black', ms=6, label=f'$M = 200$ N m, $n = {n_fatigue:.2f}$')
ax.plot(sm_vm, sa_over, 's', color='C3', ms=6, label='$M = 400$ N m')
ax.set_xlabel(r"mean stress $\sigma_m'$ [MPa]")
ax.set_ylabel(r"stress amplitude $\sigma_a'$ [MPa]")
ax.set_xticks([0, 116, 500, 700])
ax.set_yticks([0, 107, 214, 238, 500])
ax.set_xlim(0, R_m)
ax.set_ylim(0, 520)
ax.grid(True, alpha=0.4)
ax.legend(fontsize=9)
plt.show()

The example shows where the design leverage lies. A fillet radius of \(4~\text{mm}\) would lower \(K_t\) and with it the amplitude, a ground surface would raise \(k_a\) from \(0.80\) to about \(0.91\), and both act on the term \(\sigma_a'/\sigma_e\) that dominates 8.21.11, while a stronger steel helps less than its tensile strength suggests, because \(k_a\) falls as \(R_m\) rises.

Designing against fatigue

The rules that follow from this chapter are few. Keep the stress amplitude low where the stress concentrations are, which means generous fillet radii and discontinuities placed where the nominal stress is small, as Chapter 8.19 recommends. Make the surface at the critical points smooth, since cracks start at the surface. Introduce compressive residual stresses there, by shot peening, by rolling the fillets and threads, or by surface hardening, since a compressive mean stress closes cracks. Avoid welding or attaching anything to a highly stressed member: a weld toe is a sharp notch with tensile residual stresses, and the Kielland platform failed from a small fillet weld on a member that carried the main loads. Where cyclic loads cannot be avoided, use a material with an endurance limit, and for an aluminium part, which has none, design for a finite and stated life.

For 3D-printed polymer parts no endurance limit can be assumed, and the boundaries between layers act as built-in notches. A cyclic tension that pulls the layers apart is the most damaging, so a part loaded repeatedly should be oriented on the print bed so that the main stresses run along the layers, and its fillets should be rounded as generously as for a metal part. Published fatigue data for printed parts depend strongly on the material and the print settings, and a part that matters should be tested.

The finite element method delivers the stress field from which a fatigue check starts. Fatigue post-processors take the stresses of each load case, or a measured or simulated load history, and apply the methods of this chapter point by point: they count the cycles, compute amplitudes and means, correct for the surface and the mean stress, and sum the damage with 8.21.14. They inherit the scatter of the material data, and their result is an estimate of the life, not a guarantee.

A crack that has already been found falls outside the S-N curve, whose lives include the initiation that has already happened, and it is assessed with fracture mechanics. The stress field at the tip of a crack of length \(a\) under the nominal stress \(\sigma\) is scaled by the stress intensity factor \(K = Y\sigma\sqrt{\pi a}\), where the factor \(Y\) depends on the geometry of the part and the crack. The part breaks when \(K\) reaches the fracture toughness \(K_{Ic}\) of the material, which fixes a critical crack length \(a_c\). Below it the crack grows on every cycle by an amount that Paris and Erdogan found to depend on the range \(\Delta K = Y\Delta\sigma\sqrt{\pi a}\) alone [15],

\[ \frac{da}{dN} = C\,\Delta K^{m} \]

with the material constants \(C\) and \(m\), where \(m\) is about three for steels. Integrating from the crack that was found, \(a_0\), to \(a_c\) gives the cycles left, and since a short crack grows slowly, most of them are spent while the crack is still small. For a constant \(Y\) the integral gives a life proportional to \(\Delta\sigma^{-m}\), an S-N line of slope \(m \approx 3\), the slope of the design curves for welded joints, whose life is almost all crack growth. The same calculation sets the inspection intervals of aircraft structures, so that a crack too small to be found at one inspection cannot reach \(a_c\) before the next. The wheels at Eschede were inspected by eye, which finds only a long crack. Fracture mechanics is a subject of its own [16].

Further reading

Shigley’s chapter on fatigue failure covers the stress-life method of this chapter in full, including the charts for the notch sensitivity and many worked examples of shafts [5]. DNV-RP-C203 is free to download and shows how a design code turns the methods of this chapter into rules for welded structures [6].