Mechanical Engineering Design · Chapter 6
Mechanical Engineering Design · Part 2 — Failure Prevention

Fatigue Failure Resulting from Variable Loading

The complete stress-life workflow for cyclic loading: the S-N diagram and endurance limit, Marin modifying factors, fatigue notch sensitivity, and the failure criteria for fluctuating stress — Goodman, Gerber, ASME-elliptic, Soderberg — extended to combined loading, cumulative damage by Miner's rule, and surface (contact) fatigue.

Learning objectives

  • Recognize variable-amplitude, mean-stress, finite-life, and crack-growth fatigue problems.
  • Separate load histories into alternating and mean components with explicit signs.
  • Estimate part endurance strength from documented specimen data and modifying factors.
  • Apply S-N, Goodman, combined-stress, and cumulative-damage models within their limits.
  • Check first-cycle yield, units, plausibility, reliability assumptions, and manufacturing sensitivity.

Physical meaning

Fatigue is progressive local damage: a repeated stress range initiates and grows a crack even when every nominal stress is below static yield. Mean tension generally worsens that process, while surface condition and geometry control where it begins.

Problem-solving strategy

  1. Turn the service history into signed maxima, minima, counted cycles, and a critical-location free-body model.
  2. Compute nominal alternating and mean stresses, then apply documented fatigue concentration factors once.
  3. Select stress-life for elastic high-cycle work, strain-life for local cyclic plasticity, or LEFM when a crack is explicit.
  4. Correct laboratory endurance data for the actual surface, size, load mode, temperature, reliability, and environment.
  5. Check fatigue utilization, first-cycle yield, damage accumulation, and sensitivity before iterating geometry or process.

6–1Introduction to Fatigue in Metals

Why Fatigue Is Different

Most machine loads are not static but vary or repeat over time. Parts fail at stress levels far below the static strength when the load is applied and removed many times — a phenomenon called fatigue. Failure is sudden, complete, and dangerous: it begins with a microscopic crack at a stress raiser (a notch, fillet, keyway, tool mark, or inclusion), propagates a little with each cycle, and finally fractures the remaining section abruptly with no visible warning or gross yielding.

The Fatigue-Fracture Surface

A fatigue fracture shows two distinct zones: a smooth, often "beach-marked" region where the crack advanced slowly cycle by cycle, and a coarse, dull region where the part finally broke in a single overload. The size of the final-fracture zone relative to the section indicates how heavily the part was loaded.

6–2Approach to Fatigue Failure in Analysis and Design

Engineering Where Science Is Incomplete

Fatigue is not fully explained by science, yet aircraft must fly and cars must last. The chapter proceeds methodically: three life-prediction methods (Secs. 6–3 to 6–6), the S-N diagram and endurance limit (6–7, 6–8), modifying factors for real parts (6–9), notch sensitivity (6–10), fluctuating stresses (6–11 to 6–13), combined loading (6–14), cumulative damage (6–15), and surface fatigue (6–16). Beyond Sec. 6–6 only the stress-life method is carried forward for design.

6–3Fatigue-Life Methods

Three Approaches, Two Life Regimes

Life is measured in cycles to failure \(N\): low-cycle fatigue is \(1 \le N \le 10^3\); high-cycle fatigue is \(N > 10^3\). The stress-life method (stress only) is least accurate at low cycles but is the most traditional, easiest, and best-supported for high-cycle design. The strain-life method analyzes localized plastic strain and is best for low-cycle work. The linear elastic fracture mechanics (LEFM) method assumes a detectable crack already exists and predicts its growth.

6–4The Stress-Life Method and the S-N Diagram

The Rotating-Beam Test and the S-N Diagram

The R. R. Moore high-speed rotating-beam machine subjects a carefully polished specimen to pure reversed bending. Testing many specimens at descending stress levels and recording cycles to failure produces the S-N diagram (fatigue strength \(S_f\) vs. life \(N\), usually log-log). For steels and irons the curve develops a knee and goes horizontal: below the corresponding stress — the endurance limit — failure never occurs. Nonferrous metals (e.g., aluminum) show no knee and hence no true endurance limit, so a fatigue strength is quoted at a stated life (commonly \(5\times10^8\) cycles).

Regions of the Diagram

The S-N plot divides into a finite-life region (the sloped portion) and an infinite-life region (beyond the knee, between \(10^6\) and \(10^7\) cycles for steels), and separately into low-cycle (\(N \le 10^3\)) and high-cycle (\(N > 10^3\)) regimes. A given \(S_f\) is meaningless without stating its \(N\).

log N (cycles) log S_f S_ut (10⁰) knee → S_e 10⁶–10⁷ finite life infinite life low cycle | high cycle →
FIG. 6.1Schematic S-N diagram for steel: the sloped finite-life line bends at the knee into the horizontal endurance limit Se. (Original diagram, redrawn.)

6–5The Strain-Life Method

Localized Plastic Strain

The strain-life method is the best physical explanation of fatigue: a crack nucleates wherever a stress raiser drives repeated localized plastic strain, and each reversal cumulatively damages that material. Total strain amplitude combines elastic and plastic parts, captured by the Manson-Coffin relationship below, where \(\sigma_F'\) and \(\epsilon_F'\) are the fatigue strength and ductility coefficients and \(b\), \(c\) the corresponding exponents. Because applying it requires several idealizations (and local strain data that are seldom available), it is presented here for understanding rather than routine design.

Manson-Coffin Strain-Life Relationship
\[ \frac{\Delta\epsilon}{2} = \frac{\sigma_F'}{E}(2N)^{b} + \epsilon_F'\,(2N)^{c} \]Eq. (6–5)

6–6The Linear-Elastic Fracture Mechanics Method

Crack-Growth (Paris) Approach

LEFM assumes a crack of detectable size already exists and tracks its growth under cyclic loading. The stress-intensity range per cycle is \(\Delta K = \beta\,\Delta\sigma\sqrt{\pi a}\), and the crack advances per the Paris equation \(da/dN = C(\Delta K)^m\). Integrating from the initial crack size to the critical size (\(K\) reaching \(K_{Ic}\)) gives the remaining cycles of life. The method is most practical for large structures monitored by periodic inspection and computer codes.

Stress-Intensity Range
\[ \Delta K = \beta\,\Delta\sigma\,\sqrt{\pi a} \]Eq. (6–6)
Paris Crack-Growth Law
\[ \frac{da}{dN} = C\,(\Delta K)^{m} \]Eq. (6–7)

6–7The Endurance Limit

Estimating S′ₑ from Tensile Strength

Plotting rotating-beam endurance limits against tensile strength shows \(S_e'/S_{ut}\) clusters near 0.4–0.6 for steels up to about 210 kpsi, then levels off. The simple estimate below is used for steels when test data are unavailable; the prime denotes the polished laboratory specimen, reserved separately from \(S_e\), the endurance limit of an actual part. Because the estimate carries large scatter, larger design factors are warranted than for static design.

Endurance Limit Estimate — Steels
\[ S_e' = \begin{cases} 0.5\,S_{ut} & S_{ut} \le 200\ \text{kpsi (1400 MPa)} \\ 100\ \text{kpsi} & S_{ut} > 200\ \text{kpsi} \\ 700\ \text{MPa} & S_{ut} > 1400\ \text{MPa} \end{cases} \]Eq. (6–8)

6–8Fatigue Strength (Finite-Life Region)

The Power-Law S-N Equation

In the high-cycle finite-life band (\(10^3 \le N \le 10^6\)), the S-N data straighten on a log-log plot, giving \(S_f = a\,N^{b}\). The constants are fixed by two anchor points: the \(10^3\)-cycle strength \((S_f)_{10^3} = f\,S_{ut}\) and the \(10^6\)-cycle endurance limit \(S_e\), where the fatigue-strength fraction \(f\) is read from a chart (≈0.9 for \(S_{ut} < 70\) kpsi, dropping toward ~0.8 at higher strengths). Setting \(S_f = \sigma_{rev}\) inverts the relation to give life from a known completely reversed stress.

Finite-Life Fatigue Strength
\[ S_f = a\,N^{b} \]Eq. (6–13)
S-N Constants a and b
\[ a = \frac{(f\,S_{ut})^2}{S_e} \qquad\qquad b = -\frac{1}{3}\log\!\left(\frac{f\,S_{ut}}{S_e}\right) \]Eqs. (6–14, 6–15)
Cycles to Failure from a Reversed Stress
\[ N = \left(\frac{\sigma_{rev}}{a}\right)^{1/b} \]Eq. (6–16)

6–9Endurance-Limit Modifying Factors (Marin)

From Specimen to Real Part

A real component never matches the polished laboratory specimen. Marin's statistical study favored a multiplicative correction: the part endurance limit \(S_e\) is the specimen value \(S_e'\) scaled by six factors for surface, size, loading, temperature, reliability, and miscellaneous effects. When part-specific endurance tests are unavailable, these factors are the standard estimating route.

Marin Equation
\[ S_e = k_a\,k_b\,k_c\,k_d\,k_e\,k_f\,S_e' \]Eq. (6–18)
Surface Factor
\[ k_a = a\,S_{ut}^{\,b} \]Eq. (6–19)

Surface Coefficients (Eq. 6–19)

With \(S_{ut}\) in MPa (and the matching kpsi pair): Ground a = 1.58, b = −0.085; Machined / cold-drawn a = 4.51, b = −0.265; Hot-rolled a = 57.7, b = −0.718; As-forged a = 272, b = −0.995. (In kpsi the a-values are 1.34, 2.70, 14.4, 39.9 respectively.) Surface finish matters enormously — a forged surface can cut the endurance limit nearly in half versus a machined one.

Size Factor — Bending & Torsion (SI)
\[ k_b = \begin{cases} 1.24\,d^{-0.107} & 2.79 \le d \le 51\ \text{mm} \\ 1.51\,d^{-0.157} & 51 < d \le 254\ \text{mm} \end{cases} \qquad (k_b = 1\ \text{for axial}) \]Eqs. (6–20, 6–21)

Effective Diameter for Non-Rotating / Non-Round Sections

For a non-rotating round, \(d_e = 0.370d\); for a rectangle \(h\times b\), \(d_e = 0.808\sqrt{hb}\). These come from equating the volume stressed above 95% of the maximum to that of the rotating specimen, then substituting \(d_e\) into Eq. (6–20).

Loading Factor
\[ k_c = \begin{cases} 1 & \text{bending} \\ 0.85 & \text{axial} \\ 0.59 & \text{torsion} \end{cases} \]Eq. (6–26)

Temperature, Reliability, Miscellaneous

Temperature \(k_d = S_T/S_{RT}\) (a polynomial fit, Eq. 6–27, valid 70–1000 °F); below room temperature, check brittle fracture first. Reliability \(k_e = 1 - 0.08\,z_a\): 0.897 (90%), 0.868 (95%), 0.814 (99%), 0.753 (99.9%). Miscellaneous \(k_f\) is a reminder for residual stress (shot peening helps via compressive surface stress), corrosion (removes any true fatigue limit), plating (chromium/nickel can cut life up to 50%), directional grain, case hardening, and fretting (0.24–0.90).

Temperature Factor (curve fit, °F)
\[ k_d = 0.975 + 0.432(10^{-3})T_F - 0.115(10^{-5})T_F^2 + 0.104(10^{-8})T_F^3 - 0.595(10^{-12})T_F^4 \]Eq. (6–27)
Reliability Factor
\[ k_e = 1 - 0.08\,z_a \]Eq. (6–29)

6–10Stress Concentration and Notch Sensitivity

Fatigue Stress-Concentration Factor

Under fatigue, the full geometric factor \(K_t\) often overstates the damage because materials are only partly notch-sensitive. The reduced factor \(K_f\) (the fatigue stress-concentration factor) is the ratio of un-notched to notched fatigue strength at equal life. Notch sensitivity \(q\) runs 0 (no sensitivity, \(K_f = 1\)) to 1 (full sensitivity, \(K_f = K_t\)). Find \(K_t\) from geometry, then \(q\) from material/notch-radius charts, then combine. When in doubt, conservatively take \(K_f = K_t\).

Notch Sensitivity
\[ q = \frac{K_f - 1}{K_t - 1} \qquad\qquad q_{shear} = \frac{K_{fs} - 1}{K_{ts} - 1} \]Eq. (6–31)
Fatigue Stress-Concentration Factor
\[ K_f = 1 + q\,(K_t - 1) \qquad\qquad K_{fs} = 1 + q_{shear}\,(K_{ts} - 1) \]Eq. (6–32)
Neuber Equation for K_f
\[ K_f = 1 + \frac{K_t - 1}{1 + \sqrt{a}/\sqrt{r}} \qquad\qquad q = \frac{1}{1 + \sqrt{a}/\sqrt{r}} \]Eqs. (6–33, 6–34)
Neuber Constant √a — Steels (S_ut in kpsi, √a in √in)
\[ \text{Bending/axial: } \sqrt{a} = 0.246 - 3.08(10^{-3})S_{ut} + 1.51(10^{-5})S_{ut}^2 - 2.67(10^{-8})S_{ut}^3 \] \[ \text{Torsion: } \sqrt{a} = 0.190 - 2.51(10^{-3})S_{ut} + 1.35(10^{-5})S_{ut}^2 - 2.67(10^{-8})S_{ut}^3 \]Eq. (6–35a, b)

6–11Characterizing Fluctuating Stresses

Midrange and Alternating Components

For a periodic load with one peak and one valley per cycle, only the maximum and minimum matter (the wave shape does not). From \(\sigma_{max}\) and \(\sigma_{min}\) define the midrange (mean) and alternating (amplitude) components. The steady/static stress (from a fixed preload, e.g. a compression spring) is not the same as the midrange stress. The stress ratio \(R = \sigma_{min}/\sigma_{max}\) and amplitude ratio \(A = \sigma_a/\sigma_m\) are also used.

Midrange and Alternating Stress
\[ \sigma_m = \frac{\sigma_{max} + \sigma_{min}}{2} \qquad\qquad \sigma_a = \left|\frac{\sigma_{max} - \sigma_{min}}{2}\right| \]Eq. (6–36)
Stress Ratio and Amplitude Ratio
\[ R = \frac{\sigma_{min}}{\sigma_{max}} \qquad\qquad A = \frac{\sigma_a}{\sigma_m} \]Eqs. (6–37, 6–38)

Applying K_f to Both Components

In the absence of yielding, the fatigue stress-concentration factor multiplies both the alternating and midrange nominal stresses (\(\sigma_a = K_f\sigma_{ao}\), \(\sigma_m = K_f\sigma_{mo}\)) — a prescription of "no local yielding." If the steady component is high enough to yield the notch, the mean-stress factor \(K_{fm}\) (Dowling) is reduced; the prudent course is to keep \(\sigma_{max,o}\) below \(S_y\) so plastic strain is avoided altogether.

6–12Fatigue Failure Criteria for Fluctuating Stress

The Designer's Fatigue Diagram

Plotting alternating stress \(\sigma_a\) against midrange stress \(\sigma_m\), several criteria bound the safe region. Each connects the endurance limit \(S_e\) (on the \(\sigma_a\) axis) to a strength on the \(\sigma_m\) axis: Soderberg (to \(S_y\), the only one that also guards against yield, but conservative), modified Goodman (straight line to \(S_{ut}\)), Gerber (parabola to \(S_{ut}\), through the data's center), ASME-elliptic (ellipse to \(S_y\)), and the Langer first-cycle yield line. A load line of slope \(r = \sigma_a/\sigma_m\) locates the operating point; its first intersection with a criterion sets the safety factor.

Soderberg
\[ \frac{\sigma_a}{S_e} + \frac{\sigma_m}{S_y} = \frac{1}{n} \]Eq. (6–45)
Modified Goodman
\[ \frac{\sigma_a}{S_e} + \frac{\sigma_m}{S_{ut}} = \frac{1}{n} \]Eq. (6–46)
Gerber
\[ \frac{n\,\sigma_a}{S_e} + \left(\frac{n\,\sigma_m}{S_{ut}}\right)^2 = 1 \]Eq. (6–47)
ASME-Elliptic
\[ \left(\frac{n\,\sigma_a}{S_e}\right)^2 + \left(\frac{n\,\sigma_m}{S_y}\right)^2 = 1 \]Eq. (6–48)
Langer First-Cycle Yield
\[ \sigma_a + \sigma_m = \frac{S_y}{n} \]Eq. (6–49)
Goodman Fatigue Factor of Safety
\[ n_f = \frac{1}{\dfrac{\sigma_a}{S_e} + \dfrac{\sigma_m}{S_{ut}}} \]Table 6–6
Gerber Fatigue Factor of Safety (σ_m > 0)
\[ n_f = \frac{1}{2}\left(\frac{S_{ut}}{\sigma_m}\right)^2 \frac{\sigma_a}{S_e}\left[-1 + \sqrt{1 + \left(\frac{2\,\sigma_m S_e}{S_{ut}\,\sigma_a}\right)^2}\right] \]Table 6–7
ASME-Elliptic Fatigue Factor of Safety
\[ n_f = \sqrt{\frac{1}{(\sigma_a/S_e)^2 + (\sigma_m/S_y)^2}} \]Table 6–8
σ_m σ_a S_e S_y S_ut mod. Goodman Gerber ASME-elliptic Soderberg Langer (yield) load line r
FIG. 6.2Designer's fatigue diagram: the five failure criteria and a load line of slope r = σa/σm. Points on or beyond a line indicate failure by that criterion. (Original diagram, redrawn.)

6–13Torsional Fatigue Strength under Fluctuating Stresses

Smith's Results

For ductile, polished, notch-free round parts, a steady torsional stress up to the torsional yield strength has no effect on the torsional endurance limit. But for parts with notches or surface imperfections (i.e., nearly all real parts), the torsional fatigue limit falls steadily with mean stress — so Gerber, ASME-elliptic, and the modified Goodman relation apply. For pulsating torsion the ultimate shear strength is taken as \(S_{su} = 0.67\,S_{ut}\), and the shear yield as \(S_{sy} = 0.577\,S_y\).

Ultimate Shear Strength (pulsating torsion)
\[ S_{su} = 0.67\,S_{ut} \]Eq. (6–54)

6–14Combinations of Loading Modes

Von Mises Components for Combined Fatigue

When bending, torsion, and axial loads act together, build two stress elements — one of alternating components, one of midrange — apply the appropriate \(K_f\) to each stress, then collapse each element to a von Mises stress \(\sigma_a'\) and \(\sigma_m'\). Use bending Marin factors for \(S_e\); do not apply the torsion load factor \(k_c = 0.59\) (the von Mises calculation already accounts for shear), and divide the alternating axial stress by 0.85. Then apply any fluctuating-stress criterion. First-cycle yield is checked conservatively with \(\sigma_{max}' \approx \sigma_a' + \sigma_m'\).

Alternating Von Mises Component
\[ \sigma_a' = \left\{\left[(K_f)_b(\sigma_a)_b + (K_f)_{ax}\frac{(\sigma_a)_{ax}}{0.85}\right]^2 + 3\big[(K_{fs})_t(\tau_a)_t\big]^2\right\}^{1/2} \]Eq. (6–55)
Midrange Von Mises Component
\[ \sigma_m' = \left\{\big[(K_f)_b(\sigma_m)_b + (K_f)_{ax}(\sigma_m)_{ax}\big]^2 + 3\big[(K_{fs})_t(\tau_m)_t\big]^2\right\}^{1/2} \]Eq. (6–56)

6–15Varying, Fluctuating Stresses; Cumulative Fatigue Damage

Counting Cycles and the Hidden Cycle

Real stress histories are irregular, with many different peaks and valleys. The challenge is to identify what counts as a "cycle." Reordering the trace to begin and end at the largest magnitude exposes a "hidden" large cycle that simple peak-counting would miss — and it is often the most damaging. The rain-flow counting technique systematizes this, resolving the history into a set of completed \(\sigma_a,\sigma_m\) cycles.

Palmgren-Miner Linear Damage Rule
If a part sees \(n_i\) cycles at a stress level whose life to failure is \(N_i\), damage accumulates linearly and failure occurs when the cycle-ratio sum reaches the experimental constant \(c\) (range 0.7–2.2, average ≈ 1): \[ \sum \frac{n_i}{N_i} = c \approx 1 \]
Miner's Rule / Accumulated Damage
\[ \sum \frac{n_i}{N_i} = c \qquad\qquad D = \sum \frac{n_i}{N_i}\ \ (\text{failure at } D = 1) \]Eqs. (6–57, 6–58)

6–16Surface Fatigue Strength

Pitting from Repeated Contact

Where two bodies roll or roll-and-slide (gear teeth, cams, wheel-rail, bearings), repeated Hertzian contact eventually causes pitting. The subsurface maximum shear stress (Sec. 3–19) is widely thought to initiate a crack that propagates to the surface; lubricant then wedges loose a chip, leaving a pit. The contact pressure serves as the index of surface strength even though it is compressive. Hardened steel resists surface fatigue so well it shows essentially no limit, which is why it dominates contact applications.

Surface Endurance Strength and Buckingham's Load-Stress Factor

From the cylinder-contact equations, the surface endurance strength \(S_C\) (the contact pressure causing surface failure at a stated life) relates the controllable geometry/load on the left to a material load-stress factor \(K_1\) on the right. \(S_C\) can be written through an elastic coefficient \(C_p\) that bundles both materials' \(E\) and \(\nu\).

Surface Endurance Strength (contact pressure)
\[ S_C = \frac{2F}{\pi\,b\,w} \]Eq. (6–63)
Buckingham Load-Stress (Wear) Factor
\[ \frac{F}{w}\left(\frac{1}{r_1} + \frac{1}{r_2}\right) = \pi S_C^2\left[\frac{1-\nu_1^2}{E_1} + \frac{1-\nu_2^2}{E_2}\right] = K_1 \]Eq. (6–64)
Surface Strength via Elastic Coefficient C_p
\[ S_C = C_p\sqrt{\frac{F}{w}\left(\frac{1}{r_1} + \frac{1}{r_2}\right)}, \qquad \frac{1}{\pi C_p^2} = \frac{1-\nu_1^2}{E_1} + \frac{1-\nu_2^2}{E_2} \]Eq. (6–65 ff.)

6–17Road Map and Important Design Equations

The Stress-Life Design Procedure

Completely reversed loading: (1) estimate \(S_e'\) from Eq. (6–8); (2) apply the Marin factors (Eq. 6–18) to get \(S_e\); (3) find \(K_f\) (Eqs. 6–32/6–33); (4) apply \(K_f\) to the stress (or divide \(S_e\) by it — not both); (5) for finite life use the power-law S-N relation (Eqs. 6–13 to 6–16). Fluctuating simple loading: compute \(\sigma_a, \sigma_m\) (Eq. 6–36, with \(K_f\) on both), pick a criterion (Goodman, Gerber, ASME-elliptic, or Soderberg), and solve for \(n\) or size; follow with a Langer yield check. Combined loading: reduce to von Mises components \(\sigma_a', \sigma_m'\) (Eqs. 6–55/6–56), then apply a criterion as above.

Key Reminders

The deterministic results here are guides, not precise predictions — fatigue is inherently statistical, so testing remains essential and conservative design factors are appropriate. Most often fatigue governs; always verify whether first-cycle yielding (Langer) intervenes by comparing the load-line slope \(r\) with the critical slope \(r_{crit}\).

EXAMPLE 1Separating and checking a tensile cycle
A polished axial coupon cycles between tensile stresses of \(\sigma_{min}=30\ \text{MPa}\) and \(\sigma_{max}=150\ \text{MPa}\). Given part endurance strength \(S_e=240\ \text{MPa}\), ultimate strength \(S_{ut}=720\ \text{MPa}\), and yield strength \(S_y=600\ \text{MPa}\), find Goodman fatigue and first-cycle yield factors.

Recognition strategy

Two nonzero tensile endpoints signal fluctuating stress with positive mean. Modified Goodman evaluates elastic high-cycle fatigue, but it does not replace the separate maximum-stress yield check.

Given information

  • Signed stress endpoints, part value \(S_e=240\ \text{MPa}\), certified static strengths \(S_{ut}=720\ \text{MPa}\) and \(S_y=600\ \text{MPa}\), proportional axial cycling.

Model and interpretation

Take tension positive. The stated stresses are local elastic values with no additional notch factor; use \(\sigma_a\ge 0\) by definition and retain the sign of \(\sigma_m\).

Step-by-step solution
1. Plan. Resolve the history into \(\sigma_m=(\sigma_{max}+\sigma_{min})/2\) and \(\sigma_a=|\sigma_{max}-\sigma_{min}|/2\).
2. Formulate. The cycle gives \(\sigma_m=(150+30)/2=90\ \text{MPa}\), \(\sigma_a=(150-30)/2=60\ \text{MPa}\), and \(R=30/150=0.20\).
3. Calculate. Goodman utilization is \(1/n_f=\sigma_a/S_e+\sigma_m/S_{ut}=60/240+90/720=0.375\).
4. Resolve. Thus \(n_f=2.67\). The largest local stress is \(150\ \text{MPa}\), so \(n_y=S_y/\sigma_{max}=600/150=4.00\).

Verification

Units: All ratios compare MPa with MPa and are dimensionless; reconstructing \(90\pm60\ \text{MPa}\) returns the original endpoints.

Signs and direction: Both endpoints are tensile, so the positive mean correctly penalizes Goodman life rather than providing a compressive-mean benefit.

Plausibility: The fatigue factor is smaller than the yield factor, which is reasonable because repeated damage can govern well below static yielding.

Final result: \(\boxed{\sigma_a=60\ \text{MPa},\ \sigma_m=90\ \text{MPa},\ n_f=2.67,\ n_y=4.00}\).

Engineering interpretation

Fatigue controls this calculation, but the given endurance strength must match surface, size, environment, temperature, and reliability. Coupon results do not certify a manufactured component with scratches, fretting, residual stress, or load-spectrum uncertainty.

EXAMPLE 2Interpolating finite life from two anchors
A component S-N model is explicitly supplied by two fully reversed test anchors: \(S_f=600\ \text{MPa}\) at \(10^3\) cycles and \(S_f=300\ \text{MPa}\) at \(10^6\) cycles. Estimate life at a constant local alternating stress of \(420\ \text{MPa}\), with zero mean stress.

Recognition strategy

The stress lies between two high-cycle test anchors and has zero mean, so log-log power-law interpolation is appropriate. No Goodman correction or endurance-limit extrapolation is needed.

Given information

  • Two test-proven \(N,S_f\) pairs, local \(\sigma_a=420\ \text{MPa}\), \(\sigma_m=0\), fully reversed constant amplitude, and unknown cycles to failure \(N\).

Model and interpretation

Represent only the stated interval by \(S_f=aN^b\). Treat the anchors as median test data; do not infer a reliability factor or behavior outside \(10^3\le N\le 10^6\).

Step-by-step solution
1. Plan. Take logarithms so the exponent is \(b=\log_{10}(300/600)/\log_{10}(10^6/10^3)=-0.10034\).
2. Formulate. Use the first anchor to find \(a=600/(10^3)^b=1200\ \text{MPa}\), making the model \(S_f=1200N^{-0.10034}\).
3. Calculate. Set model strength equal to local alternating demand: \(420=1200N^{-0.10034}\), hence \(N=(420/1200)^{1/(-0.10034)}\).
4. Resolve. Evaluation gives \(N=3.50\times10^4\) cycles, which lies inside the fitted interval and therefore does not require extrapolation.

Verification

Units: The coefficient \(a\) carries MPa because cycle count is dimensionless; the ratio \(420/1200\) is dimensionless before exponentiation.

Signs and direction: Alternating stress is reported as a nonnegative amplitude and mean stress is zero, consistent with fully reversed endpoints of \(+420\) and \(-420\ \text{MPa}\).

Plausibility: The predicted life lies between \(10^3\) and \(10^6\), and 420 MPa lies between 600 and 300 MPa; both monotonic checks pass.

Final result: \(\boxed{a=1200\ \text{MPa},\ b=-0.10034,\ N\approx3.50\times10^4\ \text{cycles}}\).

Engineering interpretation

The steep life sensitivity means small stress, finish, or notch errors produce large life errors. A released design needs representative specimens, statistical allowables, spectrum effects, and a check that local cyclic plasticity does not invalidate stress-life analysis.

EXAMPLE 3Maintenance interval from a two-level duty block
One machine duty block contains 200 cycles at a level whose tested life is \(N_1=2.00\times10^5\) cycles and 20 cycles at a severe level whose tested life is \(N_2=2.00\times10^4\) cycles. Using Miner’s stated linear rule \(D=1\), find allowable blocks and total counted cycles.

Recognition strategy

A repeated spectrum with known constant-amplitude lives is a cumulative-damage problem. The supplied lives already embody stress and mean-stress corrections, so applying another fatigue criterion would double count them.

Given information

  • Per block, \(n_1=200\), \(N_1=200000\), \(n_2=20\), \(N_2=20000\); linear Miner threshold \(D=1\); block order neglected by assumption.

Model and interpretation

Each counted event consumes fraction \(n_i/N_i\) of median life. Add damage linearly per block, then invert; this is an empirical planning model, not a crack-growth law.

Step-by-step solution
1. Plan. Compute ordinary-level damage \(D_1=200/(2.00\times10^5)=0.00100\) per duty block.
2. Formulate. Compute severe-level damage \(D_2=20/(2.00\times10^4)=0.00100\); only one tenth as many severe cycles cause equal damage.
3. Calculate. Total block damage is \(D_b=D_1+D_2=0.00200\), so the continuous prediction is \(B=1/D_b=500\) complete blocks.
4. Resolve. Each block contains 220 counted cycles, giving \(500(220)=1.10\times10^5\) total cycles at the modeled damage threshold.

Verification

Units: Damage fractions and block count are dimensionless; multiplying blocks by cycles per block correctly returns total counted cycles.

Signs and direction: Cycle counts have no algebraic sign. Load sign and mean-stress effects must already be embedded in each supplied constant-amplitude life value.

Plausibility: Five hundred blocks produce \(500(0.002)=1\) exactly, and the equal damage contributions correctly reflect the tenfold life difference and tenfold count difference.

Final result: \(\boxed{B=500\ \text{blocks},\qquad n_{total}=1.10\times10^5\ \text{cycles}}\).

Engineering interpretation

Miner’s rule ignores sequence, overload interaction, corrosion, and scatter; the arithmetic threshold is not a safe maintenance interval by itself. A professional plan applies reliability policy, representative spectrum testing, inspection capability, and damage-tolerance requirements, then rounds retirement downward.

Chapter recap

  • Separate each cycle into alternating and mean stress before choosing a criterion.
  • Modify specimen endurance data once for the actual part, and distinguish finite-life interpolation from endurance design.
  • Check cumulative damage, first-cycle yield, crack growth, surface fatigue, and uncertainty as separate limit states.

Before moving on

You should be able to recognize the fatigue method, construct an auditable stress-life calculation, apply a mean-stress relation, sum spectrum damage, and state which manufacturing, test, code, and inspection evidence remains necessary.