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Materials science models

Link atomic structure and microstructure to diffusion, mixtures, thermal response, fracture and fatigue through twenty worked examples.

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Matter pathway: atom → solid → liquid → gas → plasma. Quantum mechanics and quantum field theory provide foundations across the pathway; they are not additional phases. This is a connected modeling route, not a universal heating curve. Actual phases depend on pressure, composition, and kinetics.

Material families and model-selection map

Choose the family, processing route, scale and service environment before choosing a constitutive equation. The examples below identify familiar applications, not grade-specific design allowables.

Metals and alloys
FCC/BCC/HCP structure, dislocations, solid solutions, precipitation, recovery and recrystallization. Examples: aluminum airframes, steel vehicle bodies, nickel-superalloy turbine components. Resolve texture and temperature when isotropic elasticity or a single yield stress is inadequate.
Ceramics and glasses
Ionic/covalent bonding, brittle fracture statistics, sintering and slow crack growth. Examples: alumina electrical insulators, zirconia dental components, silica optical fibers. Distinguish crystalline grains from amorphous glass and surface-flaw-controlled failure.
Polymers and elastomers
Chain structure, crystallinity, glass transition, viscoelasticity, rubber elasticity and aging. Examples: polyethylene packaging, epoxy adhesives, silicone seals. Time, temperature, humidity and loading rate affect response.
Composites and cellular materials
Fibers, particles, laminates, interfaces, foam cells, anisotropy and damage. Examples: carbon-fiber aircraft panels, glass-fiber wind blades, sandwich cores. Homogenized mixture bounds do not predict delamination or every directional property.
Semiconductors and functional materials
Band structure, charge carriers, dielectric polarization, ferroelectricity, magnetic domains and piezoelectric coupling. Examples: silicon processors, SiC power electronics, PZT actuators, ferrite cores. Couple electrical, thermal and mechanical fields where necessary.
Energy and environmental materials
Ion diffusion, electrochemical potentials, phase separation, corrosion, passivation and swelling. Examples: lithium-ion electrodes, fuel-cell membranes and corrosion-resistant piping. Transport and reaction models must share consistent species and charge balances.
Interfaces, nanoscale and biomaterials
Surface energy, adhesion, size effects and interaction with biological environments. Examples: thin-film coatings, implant alloys and hydrogel devices. Bulk continuum coefficients can fail at small scales.

Characterization connects the model to evidence

X-ray diffraction tests structure and phases; microscopy and EBSD reveal grains and texture; calorimetry identifies transitions; tensile, creep and dynamic-mechanical tests calibrate mechanical laws; spectroscopy and transport measurements test electronic models. Report specimen orientation, processing history, uncertainty and test temperature alongside fitted coefficients.

Plastic and viscoelastic deformation · Continuum mechanics · Solid-state physics · Reaction kinetics

1. Atomic structure and density

Definitions & inputs. n atoms/unit cell,M molar mass,NA Avogadro constant,a cubic lattice parameter.

  1. Convert a count of atoms into cell mass.

    mcell=nM/NAm_{cell}=nM/N_A

  2. Divide cell mass by cell volume.

    ρ=nM/(NAa3)\rho=nM/(N_Aa^3)

  3. Cubic reciprocal-lattice geometry gives plane spacing.

    dhkl=a/h2+k2+l2d_{hkl}=a/\sqrt{h^2+k^2+l^2}

Interpretation. Vacancies, alloy composition and thermal expansion alter density and diffraction positions.

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2. Diffusion and kinetics

Definitions & inputs. J molar flux,D diffusivity,c concentration,Dt time-diffusion product.

  1. Concentration gradients drive down-gradient transport.

    J=−D∇cJ=-D\nabla c

  2. Combine the flux law with conservation.

    ∂tc=D∇2c\partial_tc=D\nabla^2c

  3. Random-walk spreading in one coordinate and activated hopping set distance and temperature dependence.

    ⟨x2⟩=2Dt,D=D0e−Q/(RT)\langle x^2\rangle=2Dt,\quad D=D_0e^{-Q/(RT)}

Interpretation. Short-circuit paths, stress and composition dependence can require a more detailed transport model.

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3. Mixtures and microstructure

Definitions & inputs. Vf constituent volume fraction,E1,E2 elastic moduli,d grain size,kHP Hall–Petch coefficient.

  1. Shared strain adds constituent stresses, producing the Voigt estimate.

    EV=VfE1+(1−Vf)E2E_V=V_fE_1+(1-V_f)E_2

  2. Shared stress adds constituent strains, producing the Reuss estimate.

    ER−1=Vf/E1+(1−Vf)/E2E_R^{-1}=V_f/E_1+(1-V_f)/E_2

  3. An empirical grain-size relation describes strengthening in an appropriate regime.

    σy=σ0+kHPd−1/2\sigma_y=\sigma_0+k_{HP}d^{-1/2}

Interpretation. These ideal bounds and correlations do not replace a measured anisotropic composite response.

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4. Failure and time dependence

Definitions & inputs. KI crack intensity,Y geometry factor,a crack length,KIC toughness,σa fatigue stress amplitude.

  1. The near-tip elastic singularity scales with remote stress and crack size.

    KI=YσπaK_I=Y\sigma\sqrt{\pi a}

  2. Set the intensity equal to the fracture toughness and solve for crack size.

    ac=(KIC/(Yσ))2/πa_c=(K_{IC}/(Y\sigma))^2/\pi

  3. A Basquin fit relates elastic fatigue amplitude to reversals to failure.

    σa=σf′(2Nf)b\sigma_a=\sigma_f'(2N_f)^b

Interpretation. Toughness, crack geometry, plasticity, environment and fatigue scatter must be checked independently.

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5. Point defects and site occupancy

Definitions & inputs. cv vacancy fraction,ΔGf formation free energy,kB Boltzmann constant.

  1. Separate energetic and entropic contributions to forming a defect.

    ΔGf=ΔHf−TΔSf\Delta G_f=\Delta H_f-T\Delta S_f

  2. Minimize the dilute defect free energy including configurational entropy.

    cv≃e−ΔGf/(kBT)c_v\simeq e^{-\Delta G_f/(k_BT)}

  3. Substitute kB=8.617333262×10⁻⁵eV/K for a checkable toy value.

    ΔSf=0, ΔHf=1 eV, T=1000 K⇒cv≃9.125×10−6\Delta S_f=0,\ \Delta H_f=1\,\mathrm{eV},\ T=1000\,\mathrm K\Rightarrow c_v\simeq9.125\times10^{-6}

Interpretation. Vacancies enable diffusion; dislocations accommodate slip; grain boundaries introduce distinct mobility and segregation pathways.

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6. Phase equilibrium and tie lines

Definitions & inputs. G Gibbs energy,μi chemical potential,C composition on a single consistent mass or mole basis,f phase fraction.

  1. Transfer of a species cannot lower total Gibbs energy at coexistence.

    μiα=μiβ\mu_i^\alpha=\mu_i^\beta

  2. Conserve composition across two phases.

    C0=fαCα+fβCβ,fα+fβ=1C_0=f_\alpha C_\alpha+f_\beta C_\beta,\quad f_\alpha+f_\beta=1

  3. Derive the lever rule and apply the existing worked example.

    fβ=(C0−Cα)/(Cβ−Cα)=(40−20)/(80−20)=1/3f_\beta=(C_0-C_\alpha)/(C_\beta-C_\alpha)=(40-20)/(80-20)=1/3

Interpretation. Phase fraction is not generally volume fraction unless densities and composition conventions support that conversion. Equilibrium phase diagrams do not alone predict transformation speed.

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7. Nucleation and transformation

Definitions & inputs. γ interface energy,Δgv negative bulk free-energy change per volume,r nucleus radius,X transformed fraction.

  1. Competing surface cost and bulk benefit create a barrier.

    ΔG(r)=4πr2γ+43πr3Δgv\Delta G(r)=4\pi r^2\gamma+\tfrac43\pi r^3\Delta g_v

  2. Differentiate with respect to r and substitute the stationary radius.

    r∗=2γ/∣Δgv∣,ΔG∗=16πγ3/(3Δgv2)r_*=2\gamma/|\Delta g_v|,\quad\Delta G_*=16\pi\gamma^3/(3\Delta g_v^2)

  3. A numerical critical size and an independent Avrami transformation law illustrate the difference between nucleation and bulk kinetics.

    γ=0.1 J/m2, ∣Δgv∣=108 J/m3⇒r∗=2 nm;X=1−e−ktn\gamma=0.1\,\mathrm{J/m^2},\ |\Delta g_v|=10^8\,\mathrm{J/m^3}\Rightarrow r_*=2\,\mathrm{nm};\quad X=1-e^{-kt^n}

Interpretation. Heterogeneous sites reduce barriers; martensitic, diffusion-controlled and glass transitions require different models. An atomic-scale nucleus challenges the continuum approximation.

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8. Dislocations and slip systems

Definitions & inputs. τRSS resolved shear stress,σ uniaxial stress,φ angle to slip-plane normal,λ angle to slip direction,ρd dislocation line length per volume,b Burgers-vector magnitude,v glide speed.

  1. Project traction onto the slip plane and slip direction.

    τRSS=σcos⁡ϕcos⁡λ\tau_{RSS}=\sigma\cos\phi\cos\lambda

  2. Swept slip area per volume gives Orowan’s plastic shear rate.

    γ˙p=ρdbv\dot\gamma_p=\rho_dbv

  3. Check that the line density, Burgers vector and velocity produce inverse-time units.

    ρd=1012 m−2, b=2.5×10−10 m, v=10−6 m/s⇒γ˙p=2.5×10−4 s−1\rho_d=10^{12}\,\mathrm{m^{-2}},\ b=2.5\times10^{-10}\,\mathrm m,\ v=10^{-6}\,\mathrm{m/s}\Rightarrow\dot\gamma_p=2.5\times10^{-4}\,\mathrm{s^{-1}}

Interpretation. Work hardening, precipitates, solid-solution strengthening and grain boundaries impede different parts of this motion; one empirical strength law cannot identify all mechanisms.

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9. Creep and high-temperature deformation

Definitions & inputs. εdot steady creep rate,A prefactor,σstress,n stress exponent,Q activation energy,R gas constant.

  1. A Norton–Arrhenius law combines stress dependence and thermal activation.

    ϵ˙=Aσne−Q/(RT)\dot\epsilon=A\sigma^ne^{-Q/(RT)}

  2. Linearize to identify slopes only while the deformation mechanism remains unchanged.

    ln⁡ϵ˙=ln⁡A+nln⁡σ−Q/(RT)\ln\dot\epsilon=\ln A+n\ln\sigma-Q/(RT)

  3. Doubling stress can sharply increase the creep rate in this illustrative regime.

    ϵ˙(2σ)/ϵ˙(σ)=2n;n=4⇒16\dot\epsilon(2\sigma)/\dot\epsilon(\sigma)=2^n;\quad n=4\Rightarrow16

Interpretation. Diffusional creep, dislocation climb and grain-boundary sliding have different exponents and grain-size sensitivity. Creep rupture life is not obtained from a steady rate alone.

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10. Electronic and thermal transport

Definitions & inputs. n electron density,p hole density,μe andμh mobilities,σel electrical conductivity,k thermal conductivity.

  1. Sum carrier current responses to an electric field.

    σel=q(nμe+pμh)\sigma_{el}=q(n\mu_e+p\mu_h)

  2. Fourier transport and the metallic electronic Lorenz relation connect different response coefficients.

    q=−k∇T,ke/(σelT)≃L0\mathbf q=-k\nabla T,\quad k_e/(\sigma_{el}T)\simeq L_0

  3. Use q=1.602176634×10⁻¹⁹C and neglect holes for this example.

    n=1022 m−3, μe=0.1 m2/(Vs)⇒σel≃160.22 S/mn=10^{22}\,\mathrm{m^{-3}},\ \mu_e=0.1\,\mathrm{m^2/(Vs)}\Rightarrow\sigma_{el}\simeq160.22\,\mathrm{S/m}

Interpretation. Band structure, doping, defects and phonon scattering set the coefficients. Si logic, SiC power devices and thermoelectric materials optimize different combinations.

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11. Polymers and relaxation spectra

Definitions & inputs. E(t) relaxation modulus,E∞ long-time modulus,Ek modal stiffness,τk relaxation time.

  1. A generalized Maxwell model resolves multiple relaxation processes.

    E(t)=E∞+∑kEke−t/τkE(t)=E_\infty+\sum_kE_ke^{-t/\tau_k}

  2. Superpose responses to increments of strain history.

    σ(t)=∫−∞tE(t−s)ϵ˙(s) ds\sigma(t)=\int_{-\infty}^tE(t-s)\dot\epsilon(s)\,ds

  3. A one-mode example separates instantaneous and relaxed stiffness.

    E∞=1 GPa, E1=2 GPa, t=τ1⇒E=1+2/e≃1.736 GPaE_\infty=1\,\mathrm{GPa},\ E_1=2\,\mathrm{GPa},\ t=\tau_1\Rightarrow E=1+2/e\simeq1.736\,\mathrm{GPa}

Interpretation. Thermoplastics, thermosets and elastomers differ in chain mobility and crosslinking; a glass-transition temperature depends on the measurement time scale.

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12. Processing, porosity and material selection

Definitions & inputs. P porosity fraction,ρs fully dense density,E/ρ specific stiffness,σallow allowable stress under specified conditions.

  1. Neglect pore-gas mass in a porous solid.

    ρ=(1−P)ρs\rho=(1-P)\rho_s

  2. Combine density with component geometry to compare structural function rather than raw modulus alone.

    m=ρAL,kaxial=EA/Lm=\rho AL,\quad k_{axial}=EA/L

  3. For fixed axial stiffness and length, minimize ρ/E under the stated geometry model.

    m=kaxialL2ρ/Em=k_{axial}L^2\rho/E

Interpretation. Additive manufacture, casting, forging, heat treatment, sintering, layup and machining change defects, texture, residual stress and tolerances. A different load mode leads to a different material index.

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Graphical worked example

Fickian random-walk length √(2Dt), D=10⁻¹⁰ m²/s; not a finite-body concentration profile. X axis: Diffusion time (s). Y axis: One-coordinate RMS displacement (mm).
Fickian random-walk length √(2Dt), D=10⁻¹⁰ m²/s; not a finite-body concentration profile. Related worked calculation · Download SVG · Plot data

Twenty worked examples

Open a problem to see its defined inputs, assumptions, equation, numerical substitution, result, and interpretation. Values are illustrative analytical exercises.

Example 01. Cubic crystal density

Definitions & inputs. FCC n=4,M=.063546 kg/mol,a=.3615 nm,NA=6.02214076×10²³/mol.

  1. Choose the governing model and isolate the requested quantity.

    ρ=nM/(NAa3)\rho=nM/(N_Aa^3)

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    ρ=4(0.063546)/[(6.02214076×1023)(0.3615×10−9)3]\rho=4(0.063546)/[(6.02214076\times10^{23})(0.3615\times10^{-9})^3]

  3. Evaluate the expression; the result uses the units shown.

    Result=8934.544 kg m−3\mathrm{Result}=8934.544\ {\rm kg\,m}^{-3}

Interpretation. Ideal site occupancy and the stated lattice parameter are assumed.

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Example 02. Cubic (110) plane spacing

Definitions & inputs. a=.4 nm.

  1. Choose the governing model and isolate the requested quantity.

    d110=a/2d_{110}=a/\sqrt2

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    d110=0.4/2d_{110}=0.4/\sqrt2

  3. Evaluate the expression; the result uses the units shown.

    Result=0.2828427 nm\mathrm{Result}=0.2828427\ {\rm nm}

Interpretation. This is a cubic-lattice spacing formula.

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Example 03. First-order Bragg angle

Definitions & inputs. λ=.154 nm,d=.2 nm.

  1. Choose the governing model and isolate the requested quantity.

    θ=arcsin⁡[λ/(2d)]\theta=\arcsin[\lambda/(2d)]

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    θ=arcsin⁡(0.154/0.4)\theta=\arcsin(0.154/0.4)

  3. Evaluate the expression; the result uses the units shown.

    Result=22.64374 deg\mathrm{Result}=22.64374\ {\rm deg}

Interpretation. A diffraction instrument often reports 2θ rather than θ.

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Example 04. Vacancy equilibrium fraction

Definitions & inputs. Formation energy 1 eV,T=1000 K,kB=8.617333262×10⁻⁵ eV/K,prefactor one.

  1. Choose the governing model and isolate the requested quantity.

    cv=e−Ef/(kBT)c_v=e^{-E_f/(k_BT)}

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    cv=e−1/(8.617333262×10−5⋅1000)c_v=e^{-1/(8.617333262\times10^{-5}\cdot1000)}

  3. Evaluate the expression; the result uses the units shown.

    Result=9.124768×10−6 \mathrm{Result}=9.124768\times10^{-6}\ {}

Interpretation. Formation entropy is neglected in this illustrative prefactor.

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Example 05. Fickian flux

Definitions & inputs. D=10⁻¹⁰ m²/s,dc/dx=1000 mol/m⁴.

  1. Choose the governing model and isolate the requested quantity.

    J=−D dc/dxJ=-D\,dc/dx

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    J=−10−10(1000)J=-10^{-10}(1000)

  3. Evaluate the expression; the result uses the units shown.

    Result=−1×10−7 mol m−2s−1\mathrm{Result}=-1\times10^{-7}\ {\rm mol\,m}^{-2}{\rm s}^{-1}

Interpretation. The negative sign points toward lower concentration.

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Example 06. One-dimensional diffusion length

Definitions & inputs. D=10⁻¹⁰ m²/s,t=1000 s.

  1. Choose the governing model and isolate the requested quantity.

    xrms=2Dtx_{rms}=\sqrt{2Dt}

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    xrms=2(10−10)(1000)x_{rms}=\sqrt{2(10^{-10})(1000)}

  3. Evaluate the expression; the result uses the units shown.

    Result=0.0004472136 m\mathrm{Result}=0.0004472136\ {\rm m}

Interpretation. A three-dimensional radial RMS distance would use 6Dt.

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Example 07. Diffusion time estimate

Definitions & inputs. Desired one-dimensional RMS length 1 mm,D=10⁻¹⁰ m²/s.

  1. Choose the governing model and isolate the requested quantity.

    t=xrms2/(2D)t=x_{rms}^2/(2D)

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    t=(10−3)2/(2⋅10−10)t=(10^{-3})^2/(2\cdot10^{-10})

  3. Evaluate the expression; the result uses the units shown.

    Result=5000 s\mathrm{Result}=5000\ {\rm s}

Interpretation. This random-walk scale is not a complete finite-body concentration solution.

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Example 08. Activated diffusivity

Definitions & inputs. D0=10⁻⁵ m²/s,Q=100 kJ/mol,T=1000 K,R=8.314462618.

  1. Choose the governing model and isolate the requested quantity.

    D=D0e−Q/(RT)D=D_0e^{-Q/(RT)}

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    D=10−5e−100000/(8.314462618⋅1000)D=10^{-5}e^{-100000/(8.314462618\cdot1000)}

  3. Evaluate the expression; the result uses the units shown.

    Result=5.97913×10−11 m2s−1\mathrm{Result}=5.97913\times10^{-11}\ {\rm m}^2{\rm s}^{-1}

Interpretation. Mechanism and prefactor are assumed fixed over this temperature range.

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Example 09. Voigt composite modulus

Definitions & inputs. V1=.6,E1=200 GPa,E2=3 GPa.

  1. Choose the governing model and isolate the requested quantity.

    EV=V1E1+(1−V1)E2E_V=V_1E_1+(1-V_1)E_2

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    EV=0.6(200)+0.4(3)E_V=0.6(200)+0.4(3)

  3. Evaluate the expression; the result uses the units shown.

    Result=121.2 GPa\mathrm{Result}=121.2\ {\rm GPa}

Interpretation. Iso-strain loading is an ideal upper estimate under the stated scalar mixture model.

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Example 10. Reuss composite modulus

Definitions & inputs. Same V1=.6,E1=200,E2=3 GPa.

  1. Choose the governing model and isolate the requested quantity.

    ER=[V1/E1+(1−V1)/E2]−1E_R=[V_1/E_1+(1-V_1)/E_2]^{-1}

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    ER=[0.6/200+0.4/3]−1E_R=[0.6/200+0.4/3]^{-1}

  3. Evaluate the expression; the result uses the units shown.

    Result=7.334963 GPa\mathrm{Result}=7.334963\ {\rm GPa}

Interpretation. Iso-stress response gives a contrasting lower ideal estimate.

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Example 11. Mixture density

Definitions & inputs. V1=.6,ρ1=1800 kg/m³,ρ2=1200 kg/m³,no voids.

  1. Choose the governing model and isolate the requested quantity.

    ρ=V1ρ1+(1−V1)ρ2\rho=V_1\rho_1+(1-V_1)\rho_2

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    ρ=0.6(1800)+0.4(1200)\rho=0.6(1800)+0.4(1200)

  3. Evaluate the expression; the result uses the units shown.

    Result=1560 kg m−3\mathrm{Result}=1560\ {\rm kg\,m}^{-3}

Interpretation. Fractions are by volume, not mass.

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Example 12. Hall–Petch strength

Definitions & inputs. σ0=50 MPa,kHP=.5 MPa√m,d=25 μm.

  1. Choose the governing model and isolate the requested quantity.

    σy=σ0+kHP/d\sigma_y=\sigma_0+k_{HP}/\sqrt d

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    σy=50+0.5/25×10−6\sigma_y=50+0.5/\sqrt{25\times10^{-6}}

  3. Evaluate the expression; the result uses the units shown.

    Result=150 MPa\mathrm{Result}=150\ {\rm MPa}

Interpretation. Extrapolation to extreme nanoscale grains is not justified by this example.

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Example 13. Lever-rule phase fraction

Definitions & inputs. Overall composition C0=40%,phase α Cα=20%,phase β Cβ=80%; same composition basis.

  1. Choose the governing model and isolate the requested quantity.

    fβ=(C0−Cα)/(Cβ−Cα)f_\beta=(C_0-C_\alpha)/(C_\beta-C_\alpha)

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    fβ=(40−20)/(80−20)f_\beta=(40-20)/(80-20)

  3. Evaluate the expression; the result uses the units shown.

    Result=0.3333333 \mathrm{Result}=0.3333333\ {}

Interpretation. The tie line and phase compositions must correspond to equilibrium at the stated temperature.

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Example 14. Thermal free strain

Definitions & inputs. α=20×10⁻⁶/K,ΔT=100 K.

  1. Choose the governing model and isolate the requested quantity.

    ϵth=αΔT\epsilon_{th}=\alpha\Delta T

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    ϵth=20×10−6(100)\epsilon_{th}=20\times10^{-6}(100)

  3. Evaluate the expression; the result uses the units shown.

    Result=0.002 \mathrm{Result}=0.002\ {}

Interpretation. Restraint converts part of the free strain into mechanical stress.

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Example 15. Specific stiffness

Definitions & inputs. E=70 GPa,ρ=2700 kg/m³.

  1. Choose the governing model and isolate the requested quantity.

    E/ρE/\rho

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    E/ρ=70×109/2700E/\rho=70\times10^9/2700

  3. Evaluate the expression; the result uses the units shown.

    Result=2.592593×107 m2s−2\mathrm{Result}=2.592593\times10^{7}\ {\rm m}^2{\rm s}^{-2}

Interpretation. Component performance also depends on geometry and load mode.

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Example 16. Crack intensity

Definitions & inputs. Y=1,σ=100 MPa,a=.001 m.

  1. Choose the governing model and isolate the requested quantity.

    KI=YσπaK_I=Y\sigma\sqrt{\pi a}

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    KI=100π(0.001)K_I=100\sqrt{\pi(0.001)}

  3. Evaluate the expression; the result uses the units shown.

    Result=5.604991 MPam\mathrm{Result}=5.604991\ {\rm MPa}\sqrt{\rm m}

Interpretation. Crack-length convention is part of the selected geometry factor.

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Example 17. Critical crack length

Definitions & inputs. KIC=30 MPa√m,Y=1,σ=100 MPa.

  1. Choose the governing model and isolate the requested quantity.

    ac=(KIC/(Yσ))2/πa_c=(K_{IC}/(Y\sigma))^2/\pi

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    ac=(30/100)2/πa_c=(30/100)^2/\pi

  3. Evaluate the expression; the result uses the units shown.

    Result=0.02864789 m\mathrm{Result}=0.02864789\ {\rm m}

Interpretation. Small-scale yielding and the relevant toughness state are required.

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Example 18. Basquin fatigue amplitude

Definitions & inputs. σf′=1000 MPa,b=−.1,Nf=10⁶ cycles.

  1. Choose the governing model and isolate the requested quantity.

    σa=σf′(2Nf)b\sigma_a=\sigma_f'(2N_f)^b

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    σa=1000(2×106)−0.1\sigma_a=1000(2\times10^6)^{-0.1}

  3. Evaluate the expression; the result uses the units shown.

    Result=234.3673 MPa\mathrm{Result}=234.3673\ {\rm MPa}

Interpretation. The law uses reversals 2Nf and ignores mean-stress correction here.

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Example 19. Linear creep extension

Definitions & inputs. Constant steady creep rate 10⁻⁸/s,t=10⁶ s,L=1 m.

  1. Choose the governing model and isolate the requested quantity.

    δ=Lϵ˙t\delta=L\dot\epsilon t

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    δ=1(10−8)(106)\delta=1(10^{-8})(10^6)

  3. Evaluate the expression; the result uses the units shown.

    Result=0.01 m\mathrm{Result}=0.01\ {\rm m}

Interpretation. Primary and tertiary creep are excluded.

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Example 20. Avrami transformed fraction

Definitions & inputs. X=1−exp(−ktⁿ),n=2,k=.01 s⁻²,t=10 s.

  1. Choose the governing model and isolate the requested quantity.

    X=1−e−ktnX=1-e^{-kt^n}

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    X=1−e−0.01(10)2X=1-e^{-0.01(10)^2}

  3. Evaluate the expression; the result uses the units shown.

    Result=0.6321206 \mathrm{Result}=0.6321206\ {}

Interpretation. The fitted nucleation/growth model is limited to its calibrated transformation regime.

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Symbols and units

Each derivation and problem defines its own symbols and inputs. Symbols may be reused with different meanings in other subjects. Keep units consistent, retain sufficient precision during calculation, and apply the stated validity limits.