2.3 Fatigue Analysis, S-N Curves & Fluctuating Stresses

Fatigue Analysis, S-N Curves & Fluctuating Stresses

Upwards of 90% of all mechanical failures in rotating machinery, ground vehicles, turbomachinery, and pressure systems are caused by fatigue—failure under repetitive, fluctuating, or reversed cyclic stresses. Fatigue occurs at stress amplitudes substantially lower than the static tensile yield strength ($S_y$) with little or no macroscopic warning.

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1. Fatigue Mechanisms & High-Cycle vs. Low-Cycle Regimes

+-----------------------------------------------------------------------------+
|                        STAGES OF FATIGUE FAILURE                            |
|                                                                             |
|   Stage I:   Micro-crack initiation at persistent slip bands, grain         |
|              boundaries, inclusions, or geometric notches.                  |
|                                                                             |
|   Stage II:  Transgranular crack propagation perpendicular to maximum       |
|              tensile stress; leaves microscopic striations and macroscopic  |
|              clamshell / beach marks.                                       |
|                                                                             |
|   Stage III: Sudden fast fracture when the remaining uncracked ligament     |
|              exceeds the plane strain fracture toughness (K_I \ge K_{Ic}).   |
+-----------------------------------------------------------------------------+

• Low-Cycle Fatigue (LCF, $N < 10^3$ to $10^4$ cycles): Associated with significant cyclic plastic strains (e.g. thermal cycling of turbine casings). Governed by the Manson-Coffin strain-life relationship ($\Delta \epsilon_p / 2 = \epsilon_f' (2N)^c$).
• High-Cycle Fatigue (HCF, $N > 10^3$ to $10^4$ cycles): Macroscopic strains remain purely elastic. Governed by stress-life ($S\text{-}N$) methods.

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2. S-N Diagrams & Uncorrected Endurance Limit

In standard laboratory R.R. Moore rotating-beam testing (pure reversed bending, polished specimen, room temperature), ferrous alloys display a true endurance limit ($S_e'$) below which the material can endure an infinite number of cycles ($N > 10^6$) without failing.

   Stress Amplitude S_f
     ^
     |  (f * S_ut) at 10^3 cycles
     |    \ 
     |     \   Finite Life Region (S_f = a * N^b)
     |      \ 
     |       \-------------------- Endurance Limit S_e' (Infinite Life)
     |        \
     +---------+-------------------+-----------------------------> Cycles N
              10^3                10^6

Uncorrected Endurance Limit for Steels ($S_e'$):

$$S_e' = \begin{cases} 0.50 S_{ut} & \text{for } S_{ut} \le 200\text{ kpsi } (1400\text{ MPa}) \\ 100\text{ kpsi } (700\text{ MPa}) & \text{for } S_{ut} > 200\text{ kpsi } (1400\text{ MPa}) \end{cases}$$

Finite Life Equation ($10^3 \le N \le 10^6$ cycles):

$$S_f = a N^b$$ $$b = -\frac{1}{3} \log_{10}\left(\frac{f S_{ut}}{S_e}\right), \quad a = \frac{(f S_{ut})^2}{S_e}$$

Where $f$ is the fatigue strength fraction at $10^3$ cycles (typically $f \approx 0.90$ for $S_{ut} \le 70\text{ kpsi}$ down to $f \approx 0.80$ for $S_{ut} \ge 200\text{ kpsi}$).

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3. Marin Modifying Factors for Real-World Components

The laboratory endurance limit $S_e'$ must be reduced using Marin modifying factors to account for surface finish, component size, loading mode, operating temperature, target reliability, and environmental effects:

$$S_e = k_a k_b k_c k_d k_e k_f S_e'$$

+-----------------------------------------------------------------------------+
|                          MARIN FACTOR REFERENCE                             |
|                                                                             |
|   1. Surface Factor:     k_a = a S_{ut}^b                                   |
|      - Ground:           a = 1.58 MPa, b = -0.085  (a = 1.34 kpsi)          |
|      - Machined/Cold-Drawn: a = 4.51 MPa, b = -0.265 (a = 2.70 kpsi)        |
|      - Hot-Rolled:       a = 57.7 MPa, b = -0.718 (a = 14.4 kpsi)          |
|      - As-Forged:        a = 272 MPa,  b = -0.995 (a = 39.9 kpsi)          |
|                                                                             |
|   2. Size Factor (Bending/Torsion):                                         |
|      - 2.79 mm \le d \le 51 mm:  k_b = 1.24 d^{-0.107}                      |
|      - 51 mm < d \le 254 mm:     k_b = 1.51 d^{-0.157}                      |
|      - Axial Loading:            k_b = 1.0 (size accounted for in k_c)      |
|                                                                             |
|   3. Load Factor:                                                           |
|      - Bending:                  k_c = 1.0                                  |
|      - Axial:                    k_c = 0.85                                 |
|      - Torsion / Shear:          k_c = 0.59                                 |
|                                                                             |
|   4. Temperature Factor:         k_d = S_T / S_{RT}                         |
|                                                                             |
|   5. Reliability Factor:                                                    |
|      - 50%: k_e = 1.000 | 90%: k_e = 0.897 | 99%: k_e = 0.814 | 99.9%: 0.753|
+-----------------------------------------------------------------------------+

Fatigue Stress Concentration Factor ($K_f$):

$$K_f = 1 + q (K_t - 1)$$

Where notch sensitivity $q = \frac{1}{1 + \frac{\sqrt{a}}{\sqrt{r}}}$ (Neuber equation, with material constant $\sqrt{a}$ and notch radius $r$).

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4. Fluctuating Multiaxial Stresses & Mean Stress Failure Criteria

A general fluctuating stress cycle varies between $\sigma_{\text{max}}$ and $\sigma_{\text{min}}$:

• Alternating Stress: $\sigma_a = \frac{|\sigma_{\text{max}} - \sigma_{\text{min}}|}{2}$
• Mean Stress: $\sigma_m = \frac{\sigma_{\text{max}} + \sigma_{\text{min}}}{2}$
• Stress Ratio: $R = \frac{\sigma_{\text{min}}}{\sigma_{\text{max}}}$

                                Alternating Stress \sigma_a
                                    ^
                                    |
                         Endurance  * (Soderberg Line)
                           Limit    |\ 
                            S_e ----+-\---* (Modified Goodman Line)
                                    |  \   \ 
                                    |   \   \---* (Gerber Parabola)
                                    |    \   \   \ 
                                    +-----+---+---+-----> Mean Stress \sigma_m
                                         S_y     S_{ut}

Classical Mean Stress Criteria Formulation

First-Cycle Static Yield Check (Langer Static Yield Line)

Regardless of which fatigue criterion is used, the component must be verified against gross yielding during startup transients:

$$n_y = \frac{S_y}{\sigma_a + \sigma_m} = \frac{S_y}{\sigma_{\text{max}}}$$

Combined Multiaxial Fluctuating Stresses (von Mises):

$$\sigma_a' = \sqrt{\sigma_{xa}^2 - \sigma_{xa}\sigma_{ya} + \sigma_{ya}^2 + 3\tau_{xya}^2}$$ $$\sigma_m' = \sqrt{\sigma_{xm}^2 - \sigma_{xm}\sigma_{ym} + \sigma_{ym}^2 + 3\tau_{xym}^2}$$

(Note: When evaluating multiaxial fatigue with von Mises effective stresses, set the Marin load factor $k_c = 1.0$, as the $\sqrt{3}$ factor already incorporates shear behavior).

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5. Cumulative Fatigue Damage & Palmgren-Miner's Rule

When a mechanical component is subjected to variable-amplitude multi-block loading spectra, cumulative damage is quantified using the Palmgren-Miner Linear Damage Hypothesis:

$$D = \sum_{i=1}^k \frac{n_i}{N_i} = \frac{n_1}{N_1} + \frac{n_2}{N_2} + \cdots + \frac{n_k}{N_k} = 1.0$$

Where $n_i$ is the number of applied cycles at stress level $\sigma_i$, and $N_i$ is the fatigue life (cycles to failure) at that same stress level from the component's $S\text{-}N$ curve.

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6. Step-by-Step Worked Engineering Problem

Problem Statement

A solid transmission shaft made of cold-drawn steel ($S_{ut} = 650\text{ MPa}, S_y = 530\text{ MPa}$) with diameter $d = 35\text{ mm}$ operates at $20^\circ\text{C}$ with 99% reliability. It contains a shoulder fillet with $K_t = 1.70$ and notch sensitivity $q = 0.85$. The shaft is subjected to a completely reversed bending moment $M_a = 320\text{ N}\cdot\text{m}$ and a steady torsional torque $T_m = 450\text{ N}\cdot\text{m}$.

Determine: (a) fully corrected endurance limit $S_e$, (b) fatigue factor of safety $n_f$ using Modified Goodman, and (c) static yield factor of safety $n_y$.

Step 1: Baseline Endurance Limit & Marin Factors

$$S_e' = 0.50 S_{ut} = 0.50(650) = 325\text{ MPa}$$

• Surface Factor ($k_a$, Cold-Drawn): $$k_a = 4.51 (650)^{-0.265} = 4.51 (0.1798) = 0.811$$
• Size Factor ($k_b$, Bending, $d = 35\text{ mm}$): $$k_b = 1.24 (35)^{-0.107} = 1.24 (0.6833) = 0.847$$
• Load Factor ($k_c$): $k_c = 1.0$ (bending/von Mises)
• Temperature Factor ($k_d$): $k_d = 1.0$
• Reliability Factor ($k_e$, 99%): $k_e = 0.814$
• Corrected Endurance Limit: $$S_e = (0.811)(0.847)(1.0)(1.0)(0.814)(325) = 181.7\text{ MPa}$$

Step 2: Fatigue Stress Concentration Factor ($K_f$)

$$K_f = 1 + q (K_t - 1) = 1 + 0.85(1.70 - 1) = 1 + 0.85(0.70) = 1.595$$

Step 3: Nominal and Peak Alternating / Mean Stresses

• Alternating Bending Stress (Reversed, $M_m = 0$): $$\sigma_{a} = K_f \frac{32 M_a}{\pi d^3} = 1.595 \frac{32(320)}{\pi (0.035)^3} = 1.595 \frac{10240}{1.347 \times 10^{-4}} = 121.26\text{ MPa}$$
• Mean Torsional Shear Stress (Steady, $T_a = 0$): $$\tau_m = \frac{16 T_m}{\pi d^3} = \frac{16(450)}{\pi (0.035)^3} = 53.45\text{ MPa}$$

Step 4: von Mises Alternating and Mean Stresses

$$\sigma_a' = \sqrt{\sigma_a^2 + 3(0)^2} = \sigma_a = 121.26\text{ MPa}$$ $$\sigma_m' = \sqrt{(0)^2 + 3\tau_m^2} = \sqrt{3}(53.45) = 92.58\text{ MPa}$$

Step 5: Safety Factors

• Modified Goodman Fatigue Safety Factor ($n_f$): $$\frac{1}{n_f} = \frac{\sigma_a'}{S_e} + \frac{\sigma_m'}{S_{ut}} = \frac{121.26}{181.7} + \frac{92.58}{650} = 0.6674 + 0.1424 = 0.8098 \implies n_f = 1.23$$
• First-Cycle Static Yield Safety Factor ($n_y$): $$n_y = \frac{S_y}{\sigma_a' + \sigma_m'} = \frac{530}{121.26 + 92.58} = \frac{530}{213.84} = 2.48$$

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7. Common Exam Traps & PE Pro-Tips

• Trap 1 — Double-Counting Shear Factor: When using the von Mises effective stress formulation for multiaxial fatigue, set Marin load factor $k_c = 1.0$. Do not multiply by $0.59$, as $\sqrt{3}\tau_m$ already performs the distortion energy conversion.
• Trap 2 — Neglecting First-Cycle Yielding: High mean stresses can cause catastrophic plastic deformation on the very first load cycle even when fatigue safety factor $n_f > 1.0$. Always check the Langer static line.
• Trap 3 — Using $K_t$ instead of $K_f$: Always calculate $K_f = 1 + q(K_t - 1)$. Using $K_t$ directly unnecessarily over-designs the component.