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11.2 Mechanical Properties, Fatigue, Fracture, and Corrosion

Key Takeaways

  • True stress (σt=σe(1+ϵe)\sigma_t = \sigma_e(1 + \epsilon_e)) and true strain (ϵt=ln⁡(1+ϵe)\epsilon_t = \ln(1 + \epsilon_e)) account for instantaneous cross-sectional area changes up to the onset of necking in uniaxial tension testing.

  • Linear Elastic Fracture Mechanics (LEFM) defines failure via the stress intensity factor KI=YσπaK_I = Y \sigma \sqrt{\pi a}, where brittle fracture occurs when KI≥KIcK_I \ge K_{Ic} (plane-strain fracture toughness).

  • The Goodman equation (σaSe+σmSut=1\frac{\sigma_a}{S_e} + \frac{\sigma_m}{S_{ut}} = 1) modifies endurance limits to evaluate fatigue life under combined alternating stress (σa\sigma_a) and non-zero mean tensile stress (σm\sigma_m).

  • Steady-state creep behavior occurs at elevated homologous temperatures (T>0.4TmT > 0.4 T_m), where life prediction is modeled using the Larson-Miller parameter LMP=T[C+log⁡10(tr)]LMP = T[C + \log_{10}(t_r)].

  • Electrochemical corrosion involves simultaneous anodic oxidation (M→Mn++ne−M \rightarrow M^{n+} + n e^-) and cathodic reduction, mitigated using sacrificial cathodic protection or protective oxide film passivation.

Last updated: August 2026

11.2 Mechanical Properties, Fatigue, Fracture, and Corrosion

Exam Focus: Mechanical behavior and degradation topics on the FE exam require mastering engineering vs. true stress-strain relationships, elastic constants (E,G,νE, G, \nu), fracture mechanics stress intensity (KI=YσπaK_I = Y \sigma \sqrt{\pi a}), fatigue design with mean stress corrections (Goodman equation), high-temperature creep modeling via the Larson-Miller parameter, and electrochemical corrosion mechanisms.

Engineering vs. True Stress-Strain Relationships and Elastic Constants

Standard tension testing measures the response of a specimen under uniaxial loading. Uniaxial stress and strain are classified as either engineering or true parameters.

Metric Definitions

  • Engineering Stress (σe\sigma_e): Force divided by original cross-sectional area: σe=FA0\sigma_e = \frac{F}{A_0}
  • Engineering Strain (ϵe\epsilon_e): Change in length divided by original gage length: ϵe=ΔLL0=Li−L0L0\epsilon_e = \frac{\Delta L}{L_0} = \frac{L_i - L_0}{L_0}
  • True Stress (σt\sigma_t): Force divided by instantaneous cross-sectional area AiA_i: σt=FAi\sigma_t = \frac{F}{A_i}
  • True Strain (ϵt\epsilon_t): Instantaneous elongation strain integrated over gage length: ϵt=∫L0LidLL=ln⁡(LiL0)\epsilon_t = \int_{L_0}^{L_i} \frac{dL}{L} = \ln\left(\frac{L_i}{L_0}\right)

Mathematical Conversions (Prior to Necking)

Assuming constant plastic volume (A0L0=AiLiA_0 L_0 = A_i L_i), the relationship between true and engineering metrics prior to the onset of necking is:

LiL0=1+ϵe  ⟹  Ai=A01+ϵe\frac{L_i}{L_0} = 1 + \epsilon_e \implies A_i = \frac{A_0}{1 + \epsilon_e}

σt=FAi=σe(1+ϵe)\sigma_t = \frac{F}{A_i} = \sigma_e (1 + \epsilon_e)

ϵt=ln⁡(1+ϵe)\epsilon_t = \ln(1 + \epsilon_e)

Elastic Constants and Hooke's Law

Within the linear elastic region:

  • Hooke's Law: σ=Eϵ\sigma = E \epsilon
  • Poisson's Ratio (ν\nu): Ratio of lateral contracting strain to axial elongating strain: ν=−ϵlateralϵaxial\nu = -\frac{\epsilon_{\text{lateral}}}{\epsilon_{\text{axial}}}
  • Shear Modulus (GG): Governs shear stress-strain response (τ=Gγ\tau = G \gamma) and is related to EE and ν\nu by: G=E2(1+ν)G = \frac{E}{2(1 + \nu)}

Key Stress-Strain Curve Parameters

ParameterFormula / DefinitionEngineering Significance
Modulus of Elasticity (EE)E=σϵE = \frac{\sigma}{\epsilon}Slope of linear elastic region; material stiffness
0.2% Offset Yield Strength (σy\sigma_y)Intercept of 0.002 strain offset line parallel to elastic slopeBoundary between elastic and permanent plastic deformation
Ultimate Tensile Strength (σut\sigma_{ut})Peak engineering stress on stress-strain curveMaximum force capacity prior to localized necking
Modulus of Resilience (UrU_r)Ur=σy22EU_r = \frac{\sigma_y^2}{2E}Elastic strain energy absorbed per unit volume without permanent deformation
Toughness (UtU_t)Area under total stress-strain curve up to fractureMechanical energy absorbed per unit volume prior to fracture
Percent Elongation (%EL)%EL=(Lf−L0L0)×100%\%EL = \left(\frac{L_f - L_0}{L_0}\right) \times 100\%Metric of material ductility
Percent Reduction in Area (%RA)%RA=(A0−AfA0)×100%\%RA = \left(\frac{A_0 - A_f}{A_0}\right) \times 100\%Measure of localized necking ductility

Worked Example: True Stress and Strain Calculation

Problem: A cylindrical titanium alloy specimen with initial diameter d0=12.8 mmd_0 = 12.8 \text{ mm} (A0=128.68 mm2A_0 = 128.68 \text{ mm}^2) and gage length L0=50.00 mmL_0 = 50.00 \text{ mm} is loaded in tension. Under a load of F=110.0 kNF = 110.0 \text{ kN} prior to necking, the specimen length is measured as 54.20 mm54.20 \text{ mm}. Calculate the engineering stress, engineering strain, true stress, and true strain.

Solution:

  1. Engineering Stress (σe\sigma_e): σe=110,000 N1.2868×10−4 m2=8.548×108 Pa=854.8 MPa\sigma_e = \frac{110,000 \text{ N}}{1.2868 \times 10^{-4} \text{ m}^2} = 8.548 \times 10^8 \text{ Pa} = 854.8 \text{ MPa}
  2. Engineering Strain (ϵe\epsilon_e): ϵe=54.20−50.0050.00=4.2050.00=0.0840\epsilon_e = \frac{54.20 - 50.00}{50.00} = \frac{4.20}{50.00} = 0.0840
  3. True Stress (σt\sigma_t): σt=σe(1+ϵe)=854.8 MPa×(1+0.0840)=854.8×1.0840≈926.6 MPa\sigma_t = \sigma_e (1 + \epsilon_e) = 854.8 \text{ MPa} \times (1 + 0.0840) = 854.8 \times 1.0840 \approx 926.6 \text{ MPa}
  4. True Strain (ϵt\epsilon_t): ϵt=ln⁡(1+ϵe)=ln⁡(1.0840)≈0.0807\epsilon_t = \ln(1 + \epsilon_e) = \ln(1.0840) \approx 0.0807

Hardness Testing, Charpy Impact, and Ductile-to-Brittle Transition

Hardness Testing Methods

Hardness measures a material's resistance to localized surface indentation or plastic deformation:

  • Rockwell (HRB, HRC): Measures depth of penetration under fixed minor/major loads using a 1/16-inch steel ball (HRB) or a Brale diamond cone penetrator (HRC).
  • Brinell (HB): Uses a 10 mm hardened steel or tungsten carbide sphere pressed into the surface under a 3000 kgf load, measuring indentation diameter.
  • Empirical Tensile Strength Correlation for Steels: UTS (MPa)≈3.45×HB\text{UTS (MPa)} \approx 3.45 \times \text{HB} UTS (psi)≈500×HB\text{UTS (psi)} \approx 500 \times \text{HB}

Charpy Impact Energy and DBTT

Impact testing measures dynamic energy absorption under high strain rates using a swinging pendulum striker:

  • Charpy V-Notch (CVN) Test: Specimen supported as a simple beam with a central V-notch; energy absorbed during fracture is recorded in Joules (J\text{J}) or foot-pounds (ft⋅lbf\text{ft}\cdot\text{lbf}).
  • Ductile-to-Brittle Transition Temperature (DBTT): Body-Centered Cubic (BCC) and Hexagonal Close-Packed (HCP) metals exhibit a sharp reduction in impact energy absorption as temperature drops below DBTT. Failure transitions from ductile dimpled shear fracture to brittle cleavage fracture along crystallographic planes.
  • FCC Metals: Face-Centered Cubic (FCC) metals (such as aluminum, copper, and austenitic stainless steels) retain high toughness down to cryogenic temperatures and do not exhibit a DBTT.

Fracture Mechanics and Stress Intensity

Conventional stress analysis assumes flaw-free materials. Linear Elastic Fracture Mechanics (LEFM) evaluates component safety in the presence of pre-existing microcracks or defects.

Stress Concentration Factor (KtK_t)

Geometric discontinuities (notches, holes, fillets) intensify local stresses:

σmax=Ktσ0\sigma_{\text{max}} = K_t \sigma_0

For an internal elliptical crack of length (2a) with crack tip radius ρt\rho_t:

σmax=σ0(1+2aρt)\sigma_{\text{max}} = \sigma_0 \left(1 + 2\sqrt{\frac{a}{\rho_t}}\right)

Stress Intensity Factor (KIK_I) and Fracture Toughness (KIcK_{Ic})

Under Mode I (tensile opening) loading, the stress field at a crack tip is defined by the Stress Intensity Factor (KIK_I):

KI=YσπaK_I = Y \sigma \sqrt{\pi a}

where:

  • (Y) = dimensionless geometry factor (typically Y≈1.0−1.12Y \approx 1.0 - 1.12)
  • σ\sigma = nominal applied tensile stress (MPa\text{MPa})
  • (a) = crack depth for an edge crack, or half-length for an internal crack (m\text{m})

Catastrophic fast brittle fracture occurs when KIK_I reaches the critical material property known as Plane-Strain Fracture Toughness (KIcK_{Ic}):

KI≥KIcK_I \ge K_{Ic}

Critical Stress and Critical Flaw Size Equations

σc=KIcYπa\sigma_c = \frac{K_{Ic}}{Y \sqrt{\pi a}} ac=1π(KIcYσ)2a_c = \frac{1}{\pi} \left(\frac{K_{Ic}}{Y \sigma}\right)^2

Worked Example: Allowable Tensile Stress via Fracture Mechanics

Problem: A structural steel plate containing a through-thickness internal crack of total length 2a=16.0 mm2a = 16.0 \text{ mm} (a=8.0 mm=0.0080 ma = 8.0 \text{ mm} = 0.0080 \text{ m}) is subjected to a uniform tensile stress. The material has a plane-strain fracture toughness KIc=45.0 MPa⋅m1/2K_{Ic} = 45.0 \text{ MPa}\cdot\text{m}^{1/2} and a yield strength σy=550 MPa\sigma_y = 550 \text{ MPa}. Assuming (Y = 1.0):

  1. Calculate the critical tensile stress σc\sigma_c that causes brittle fracture.
  2. Determine whether the plate fails by yielding or brittle fracture first.

Solution:

  1. Calculate Critical Stress σc\sigma_c: σc=KIcYπa=45.0 MPa⋅m1/21.0×π×0.0080 m=45.00.025133=45.00.15853≈283.9 MPa\sigma_c = \frac{K_{Ic}}{Y \sqrt{\pi a}} = \frac{45.0 \text{ MPa}\cdot\text{m}^{1/2}}{1.0 \times \sqrt{\pi \times 0.0080 \text{ m}}} = \frac{45.0}{\sqrt{0.025133}} = \frac{45.0}{0.15853} \approx 283.9 \text{ MPa}
  2. Evaluate Failure Mode: Since σc=283.9 MPa\sigma_c = 283.9 \text{ MPa} is substantially less than the yield strength σy=550 MPa\sigma_y = 550 \text{ MPa}, the component will fail by sudden brittle fracture before yielding occurs.

Fatigue Mechanics, Goodman Diagram, and Creep Mechanics

Cyclic Fatigue Parameters

Fatigue represents progressive structural damage under fluctuating or cyclic stress state loading:

  • Mean Stress (σm\sigma_m): σm=σmax+σmin2\sigma_m = \frac{\sigma_{\text{max}} + \sigma_{\text{min}}}{2}
  • Stress Range (Δσ\Delta \sigma): Δσ=σmax−σmin\Delta \sigma = \sigma_{\text{max}} - \sigma_{\text{min}}
  • Alternating Stress Amplitude (σa\sigma_a): σa=σmax−σmin2\sigma_a = \frac{\sigma_{\text{max}} - \sigma_{\text{min}}}{2}
  • Stress Ratio ((R)): R=σminσmaxR = \frac{\sigma_{\text{min}}}{\sigma_{\text{max}}}

S-N Curves and Endurance Limit

Ferrous metals (steels) exhibit an Endurance Limit (SeS_e) (typically Se≈0.5SutS_e \approx 0.5 S_{ut} up to 100 ksi100 \text{ ksi}), below which the component can endure an infinite number of cycles without fatigue failure. Non-ferrous metals (e.g., aluminum) do not possess a true endurance limit; their fatigue strength continuously decreases with cycle count.

Mean Stress Corrections (Goodman Equation)

When cyclic stress is accompanied by a positive mean tensile stress (σm>0\sigma_m > 0), the allowable alternating stress is reduced according to the Modified Goodman Equation:

σaSe+σmSut=1nf\frac{\sigma_a}{S_e} + \frac{\sigma_m}{S_{ut}} = \frac{1}{n_f}

where SeS_e is the uncorrected endurance limit, SutS_{ut} is ultimate tensile strength, and nfn_f is the fatigue factor of safety.

Worked Example: Goodman Fatigue Safety Factor

Problem: A steel shaft with Sut=700 MPaS_{ut} = 700 \text{ MPa} and Se=280 MPaS_e = 280 \text{ MPa} experiences cyclic tensile loading between σmax=400 MPa\sigma_{\text{max}} = 400 \text{ MPa} and σmin=100 MPa\sigma_{\text{min}} = 100 \text{ MPa}. Calculate the factor of safety nfn_f using the Goodman relation.

Solution:

  1. Calculate Mean and Alternating Stresses: σm=400+1002=250 MPa\sigma_m = \frac{400 + 100}{2} = 250 \text{ MPa} σa=400−1002=150 MPa\sigma_a = \frac{400 - 100}{2} = 150 \text{ MPa}
  2. Apply Goodman Equation: σaSe+σmSut=150280+250700=0.5357+0.3571=0.8928\frac{\sigma_a}{S_e} + \frac{\sigma_m}{S_{ut}} = \frac{150}{280} + \frac{250}{700} = 0.5357 + 0.3571 = 0.8928 nf=10.8928≈1.12n_f = \frac{1}{0.8928} \approx 1.12

Cumulative Damage: Miner's Rule

Under multi-amplitude cyclic loading, fatigue damage accumulates linearly according to Miner's Rule:

∑i=1kniNi=C\sum_{i=1}^{k} \frac{n_i}{N_i} = C

where nin_i is the number of applied cycles at stress level σi\sigma_i, and NiN_i is the fatigue life at stress level σi\sigma_i. Fatigue failure is predicted when C≥1.0C \ge 1.0.

Creep Mechanics and Larson-Miller Parameter

Creep is time-dependent plastic deformation occurring under constant load at elevated temperatures (typically T>0.4TmT > 0.4 T_m in Kelvin).

  • Creep Curve Stages:
    1. Primary (Transient) Creep: Strain rate decreases over time due to work hardening.
    2. Secondary (Steady-State) Creep: Minimum constant creep rate (ϵ˙s\dot{\epsilon}_s) established by balance between work hardening and thermal recovery.
    3. Tertiary Creep: Strain rate accelerates rapidly due to internal necking and micro-void formation, leading to rupture.
  • Larson-Miller Parameter (LMP): Predicts rupture time (trt_r in hours) at absolute temperature (T): LMP=T[C+log⁡10(tr)]\text{LMP} = T [C + \log_{10}(t_r)] where (T) is temperature in Kelvin (K\text{K}) or Rankine (R\text{R}), and (C) is a material constant (typically C≈20C \approx 20 for steels).

Electrochemical Corrosion Mechanisms and Mitigation

Corrosion is the destructive electrochemical attack of a material by reaction with its surrounding environment.

Electrochemical Cell Components

An electrochemical corrosion cell requires four essential elements:

  1. Anode: The site of oxidation where metal atoms dissolve and release electrons.
  2. Cathode: The site of reduction where electrons are consumed.
  3. Electrolyte: An aqueous solution conducting ionic current.
  4. Electrical Connection: Metallic path conducting electron flow from anode to cathode.

Electrochemical Half-Cell Reactions

  • Anodic Oxidation Reaction (Metal Dissolution): M→Mn++ne−M \rightarrow M^{n+} + n e^-
  • Cathodic Reduction Reactions:
    • Acidic solutions (Hydrogen Evolution): 2H++2e−→H2(g)2H^+ + 2e^- \rightarrow H_2 (g)
    • Neutral/Basic aerated solutions (Oxygen Reduction): O2+2H2O+4e−→4OH−O_2 + 2H_2O + 4e^- \rightarrow 4OH^-

Faraday's Law for Corrosion Mass Loss

m=I⋅t⋅Mn⋅Fm = \frac{I \cdot t \cdot M}{n \cdot F}

where (I) is corrosion current (A\text{A}), (t) is time (s\text{s}), (M) is molar mass (g/mol\text{g/mol}), (n) is valence state, and F=96,485 C/molF = 96,485 \text{ C/mol}.

Forms of Corrosion

  • Galvanic Corrosion: Occurs when two dissimilar metals are electrically coupled in an electrolyte. The metal higher in the galvanic series (more active / anodic) corrodes rapidly to protect the cathodic metal.
  • Crevice Corrosion: Localized corrosion occurring in narrow gaps or stagnant fluid pockets driven by differential oxygen concentration cells.
  • Pitting Corrosion: Extremely localized breakdown of passive oxide films, creating deep pits in stainless steels exposed to chloride ions (Cl−\text{Cl}^-).
  • Intergranular Corrosion: Sensitization of austenitic stainless steels during welding between 500∘C500^\circ\text{C} and 800∘C800^\circ\text{C}, causing chromium carbide (Cr23C6\text{Cr}_{23}\text{C}_6) precipitation along grain boundaries and depleting local chromium below 12 wt% Cr12 \text{ wt\% Cr}.
  • Stress Corrosion Cracking (SCC): Rapid brittle cracking resulting from simultaneous tensile stress and a specific corrosive environment.

Corrosion Control Methods

  • Sacrificial Anode Cathodic Protection: Attaching sacrificial anodes (e.g., zinc or magnesium) to protect steel structures.
  • Impressed Current Cathodic Protection (ICCP): Applying external DC voltage to drive the structure cathodic.
  • Passivation: Maintaining a self-healing protective oxide film (e.g., Cr2O3\text{Cr}_2\text{O}_3 on stainless steel).
Test Your Knowledge

A cylindrical steel alloy specimen with initial cross-sectional area A₀ = 100 mm² and initial length L₀ = 50.0 mm is pulled in tension. At a tensile force of F = 50 kN prior to necking, the specimen length measures 52.5 mm. What are the true stress and true strain experienced by the specimen at this load?

A

True stress = 525 MPa, True strain = 0.0488

B

True stress = 500 MPa, True strain = 0.0500

C

True stress = 476 MPa, True strain = 0.0512

D

True stress = 550 MPa, True strain = 0.0450

Test Your Knowledge

A high-strength steel pressure vessel contains a surface edge crack of depth a = 4.0 mm (0.0040 m). The material has a plane-strain fracture toughness of K_Ic = 60.0 MPa·m^1/2 and a dimensionless geometry factor Y = 1.12. What is the maximum allowable tensile stress σ_max that can be applied without causing fast brittle fracture?

A

654.2 MPa

B

398.2 MPa

C

534.8 MPa

D

477.5 MPa

Test Your Knowledge

A structural steel component with an ultimate tensile strength S_ut = 600 MPa and an endurance limit S_e = 240 MPa is subjected to cyclic loading consisting of a maximum tensile stress of 300 MPa and a minimum tensile stress of 100 MPa. According to the modified Goodman relation, what is the maximum allowable alternating stress amplitude σ_a for infinite fatigue life under this mean stress condition?

A

192.0 MPa

B

120.0 MPa

C

160.0 MPa

D

144.0 MPa

Test Your Knowledge

An underground carbon steel gas pipeline is buried in moist soil. To prevent electrochemical corrosion, sacrificial anodes are electrically connected to the pipeline at regular intervals. Which metal is most suitable to act as a sacrificial anode to protect the steel pipeline?

A

Copper

B

Platinum

C

Lead

D

Magnesium

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