The formulas used in materials science courses and exams, grouped by topic. Each one lists symbols with SI units, conditions and range, common mistakes and a numerical example, and the dimensions of both sides have all been checked by machine. Links lead to drill types and calculators that use the same formula.
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Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Cubic lattices only (simple cubic, BCC, FCC, diamond, NaCl type, …). For tetragonal, hexagonal and other lattices use “Interplanar spacing (tetragonal, orthorhombic, hexagonal)” below. It gives the spacing of the plane family and holds numerically even for (hkl) whose reflection is extinct.
Common mistakes
Forgetting the square root or dividing by h + k + l. Confusing (100) and (200): in BCC the (100) reflection is extinct and appears as (200) with d/2. Mixing nm and Å (1 Å = 0.1 nm).
Example
Cu (FCC, a = 0.36146 nm), (111) planes: d = 0.36146/√3 = 0.2087 nm. For BCC Fe (a = 0.28665 nm), (110): 0.28665/√2 = 0.2027 nm.
1/d² = 4(h² + hk + k²)/(3a²) + l²/c²hexagonal[m⁻²]
Symbol
Meaning
SI unit
d
interplanar spacing
m
a, b, c
lattice parameters (orthorhombic a ≠ b ≠ c, tetragonal a = b)
m
h, k, l
Miller indices (for hexagonal, h, k, l of (hkil), i = −(h + k))
1
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Use the first form for lattices with orthogonal axes (cubic, tetragonal, orthorhombic) and the second for hexagonal (a axes at 120°, c perpendicular). Monoclinic and triclinic need the general form with the axial angles (reciprocal metric). In the hexagonal form h, k, l are the indices of (hkil) without i.
Common mistakes
Putting the i of the four-index (10−11) into the formula (i follows from h and k). Using the cubic form for hexagonal crystals. Swapping a and c in a tetragonal cell.
Example
Tetragonal β-Sn (a = 0.58318 nm, c = 0.31818 nm), (101): 1/d² = 1/a² + 1/c² gives d = 0.2793 nm. Hexagonal Mg (a = 0.32094 nm, c = 0.52107 nm), (101) (i.e. (10−11)): 1/d² = 4/(3a²) + 1/c² gives d = 0.2452 nm.
order of reflection (usually n = 1; the nth order is treated as first order from d/n)
1
λ
X-ray wavelength
m
d
interplanar spacing
m
θ
Bragg angle (between beam and planes; diffractometers report 2θ)
rad
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Condition for constructive interference from lattice planes (kinematic diffraction). A reflection exists only if λ ≤ 2d. It fixes peak positions only; intensities come from the structure factor and LP factor.
Common mistakes
Confusing θ and 2θ (instruments report 2θ); degrees vs radians; mixing Kα1 with the averaged Kα wavelength; assigning a peak to an extinct (hkl).
Example
Cu Kα1 (λ = 0.15406 nm) on Cu (111) (d = 0.2087 nm): sin θ = λ/(2d) = 0.3691, θ = 21.66°, peak at 2θ = 43.32°.
atoms per unit cell (SC 1, BCC 2, FCC 4, diamond 8, HCP hexagonal prism 6)
1
M
molar mass (atomic weight × 10⁻³ kg/mol)
kg/mol
NA
Avogadro constant
mol⁻¹
Vc
unit-cell volume
m³
a
lattice parameter
m
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Density of a perfect crystal; real materials differ slightly through vacancies, porosity and impurities. For compounds use n formula units and the formula mass (NaCl type: 4 Na⁺–Cl⁻ pairs).
Common mistakes
Cubing a in nm (convert nm → cm by × 10⁻⁷ for g/cm³); using M in g/mol with m³ (for kg/m³, M must be in kg/mol); swapping n for BCC and FCC.
Dimensions both sides agree on every line ([ ] in SI base units) · dimensionless: c/a
Conditions and range
Hard spheres touching along the cube edge (SC), body diagonal (BCC), face diagonal (FCC) or within the basal plane (HCP). The “radius” depends on bonding (metallic radii refer to coordination 12).
Common mistakes
Radius vs diameter (4r is two diameters); swapping the BCC and FCC relations.
Example
BCC Fe (a = 0.28665 nm): r = √3a/4 = 0.1241 nm. FCC Cu (a = 0.36146 nm): r = √2a/4 = 0.1278 nm.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Hard-sphere model. Metals: SC 0.524, BCC 0.680, FCC and HCP 0.740, diamond 0.340. In ionic crystals, a follows from the direction in which cation and anion touch (edge in NaCl type, body diagonal in CsCl type).
Common mistakes
Taking a from anion–anion contact (only right for small radius ratios); counting one NaCl pair per cell instead of four.
atoms centred on the plane or line (counted by the fraction inside the section)
1
A
plane area
m²
L
line length
m
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Count only atoms whose centres lie on the plane (line). When counting within a unit-cell section, add corner and edge atoms by the fraction inside the section. Slip planes are the densest planes and slip directions the densest directions (FCC {111}⟨110⟩, BCC {110}⟨111⟩).
Common mistakes
Counting atoms that merely intersect the plane; using a² for every plane ((110) has √2a², the (111) triangle (√3/2)a²).
Example
FCC Al (a = 0.40496 nm), (111): the section of the unit cell is an equilateral triangle of side √2a (area (√3/2)a² = 0.1420 nm²) holding 3 × 1/6 + 3 × 1/2 = 2 atoms. PD = 14.08 nm⁻². Linear density along [110]: 2/(√2a) = 3.492 nm⁻¹.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Count in the hexagonal prism (three primitive cells): 12 corners × 1/6, 2 faces × 1/2, 3 inside = 6 atoms. The primitive cell ((√3/2)a²c) holds 2. Use either in the density formula. Ideal c/a = √(8/3) = 1.633 (Mg 1.624, Zn 1.856, Ti 1.587).
Common mistakes
Pairing the prism volume with 2 atoms, or the primitive cell with 6 (density off by a factor of 3).
Example
Mg (a = 0.32094 nm, c = 0.52107 nm): c/a = 1.624 (close to the ideal 1.633). The hexagonal prism (3√3/2)a²c = 1.394 × 10⁻²² cm³ holds 6 atoms; with M = 24.305 g/mol, ρ = 1.737 g/cm³ (handbook: 1.74 g/cm³).
Dimensions both sides agree on every line ([ ] in SI base units) · dimensionless: r/R, r/R, r/R
Conditions and range
Octahedral and tetrahedral holes of close packing (FCC, HCP) and the cube centre of simple cubic. For ionic crystals it is the smallest radius ratio at which the cation still touches the anions, a guide to coordination (0.225–0.414: 4, 0.414–0.732: 6, ≥ 0.732: 8). BCC holes are distorted: octahedral 0.155, tetrahedral 0.291.
Common mistakes
Using FCC’s 0.414 for BCC octahedral holes. The radius-ratio rule is only a guide and fails for strongly covalent compounds.
Example
γ-Fe (FCC, a ≈ 0.3647 nm): R = √2a/4 = 0.1289 nm; the sphere that fits the octahedral hole is r = (√2 − 1)R = 0.0534 nm. A carbon atom (radius about 0.071 nm) is larger and pushes the neighbouring atoms apart.
atomic scattering factor of atom j (falls with sin θ/λ)
1
h, k, l
Miller indices
1
xj, yj, zj
fractional coordinates of atom j
1
I
integrated intensity (relative)
1
p
multiplicity (planes of equal d)
1
θ
Bragg angle
rad
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Kinematic intensities for a randomly oriented powder in Bragg–Brentano geometry with an infinitely thick sample, neglecting absorption and extinction. The LP factor (1 + cos²2θ)/(sin²θ cos θ) combines polarisation and Lorentz terms (its form changes with a monochromator). Multiplying by exp(−2B sin²θ/λ²) weakens high angles. Extinctions: BCC h + k + l odd; FCC h, k, l of mixed parity give F = 0.
Common mistakes
Comparing F instead of |F|²; leaving out the multiplicity; counting reflections of equal d (e.g. BCC (411) and (330)) separately.
Example
For FCC, F = 4f when h, k, l are all even or all odd, and 0 when mixed. Al with Cu Kα1: (111) at 2θ = 38.47°, f = 8.96, F = 35.8, LP = 15.7, multiplicity 8; (200) at 2θ = 44.72°, f = 8.51, F = 34.0, LP = 11.2, multiplicity 6. Intensity ratio I(200)/I(111) = 0.48 (without the temperature factor and anomalous dispersion).
crystallite size (volume-weighted, normal to the planes)
m
K
shape factor (0.89–0.94, usually 0.9)
1
λ
X-ray wavelength
m
β
size broadening (FWHM after removing the instrumental width)
rad
θ
Bragg angle
rad
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
When broadening comes from size alone (separate strain with Williamson–Hall). Useful up to about 100 nm; beyond that the size broadening is lost in the instrumental width. Remove the instrumental width βi as β² = βm² − βi² (Gaussian) or β = βm − βi (Lorentzian).
Common mistakes
Using β in degrees (convert to radians); putting 2θ into cos θ; not removing the instrumental width. The result is the coherent-domain size, not necessarily the grain size.
Example
Cu Kα1, 2θ = 44.5°, measured FWHM 0.55°, instrumental width 0.08° (Gaussian: β = √(0.55² − 0.08²) = 0.5442° = 0.009497 rad), K = 0.9: D = 0.9 × 0.15406/(0.009497 × cos 22.25°) = 15.8 nm.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Assumes size (1/cos θ) and strain (tan θ) broadening are both Lorentzian and simply add. Plot β cos θ against 4 sin θ: the intercept gives D and the slope ε. Elastically anisotropic materials scatter; use the modified W–H method.
Common mistakes
Plotting against sin θ and forgetting the factor 4 in the slope; reporting D when the intercept is negative or near zero (no measurable size broadening).
Example
With the five Ni peaks of the XRD workbench example (2θ = 44.5°–98.4°, instrumental width 0.08° removed as Gaussian), a least-squares line of β cos θ against 4 sin θ gives intercept 6.924 × 10⁻³ rad and slope ε = 1.50 × 10⁻³ (= 0.150 %). D = Kλ/intercept = 0.9 × 0.15406/6.924 × 10⁻³ = 20.0 nm.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Differential of Bragg’s law, valid for small Δd/d. Uniform strain (residual stress, solute, thermal expansion) shifts peaks; non-uniform strain broadens them. Shifts grow with angle, so precise lattice parameters use high-angle peaks.
Common mistakes
Sign: an expanded lattice moves 2θ down. The result is in radians (× 180/π for degrees). Using 2θ for θ. Sample displacement (zero error) also shifts peaks.
Example
Cu (111) (2θ = 43.32°, θ = 21.66°) with the lattice expanded by 0.1 % (Δd/d = 1.0 × 10⁻³): Δ(2θ) ≈ −2 × 1.0 × 10⁻³ × tan 21.66° = −7.94 × 10⁻⁴ rad = −0.0455° (towards lower angle).
lattice parameters of pure A and B (same structure)
m
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Empirical rule for continuous solid solutions of the same structure. Many alloys deviate (bowing up or down); precise compositions use measured calibrations. x is a mole fraction, not wt%.
Common mistakes
Using wt% for x; using lattice parameters of different structures.
Example
A Si–Ge solid solution (aSi = 0.54306 nm, aGe = 0.56574 nm) measured at a = 0.5500 nm: xGe = (0.5500 − 0.54306)/(0.56574 − 0.54306) = 0.306. Si–Ge deviates slightly below Vegard’s law, so precise compositions use a corrected relation.
F = C − P + 1constant pressure (condensed systems)[1]
Symbol
Meaning
SI unit
F
degrees of freedom (independent intensive variables)
1
C
number of components
1
P
number of coexisting phases
1
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Equilibrium only. Phase diagrams of metals and ceramics are usually at constant pressure, so use the +1 form. Phase amounts are not degrees of freedom.
Common mistakes
Counting microconstituents (pearlite) as phases; using +2 for a constant-pressure diagram; applying it to non-equilibrium structures (martensite).
Example
Pb–Sn eutectic point at constant pressure (C = 2; α, β and liquid, P = 3): F = 2 − 3 + 1 = 0, so temperature and all three compositions are fixed (183 °C). In a two-phase field F = 1 (fixing T fixes both compositions). Triple point of water (pressure variable, C = 1, P = 3): F = 1 − 3 + 2 = 0.
tie-line ends (phase compositions at that temperature, wt%)
1
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Inside a two-phase field at equilibrium. wt% inputs give mass fractions, at% give mole fractions; divide by phase densities for volume fractions. Fractions of eutectic or proeutectoid constituents also come from the lever rule just above or below the three-phase line.
Common mistakes
Using the arm on the side of the wanted phase (use the opposite arm); mixing wt% and at%; taking the alloy composition as a phase composition.
Example
Cu–35 wt% Ni held at 1250 °C (tie-line ends CL = 31.5 wt%, Cα = 42.5 wt%): WL = (42.5 − 35)/(42.5 − 31.5) = 0.682, Wα = 0.318.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Just below the eutectic temperature after equilibrium solidification. For hypereutectic alloys replace α by β and Cα by Cβ. The eutectic itself contains α, so the total α phase is a separate lever rule using Cβ.
Common mistakes
Confusing microconstituent fractions (primary, eutectic) with phase fractions (α, β). Fast (Scheil-like) solidification gives eutectic even where equilibrium predicts none.
Example
Pb–40 wt% Sn (hypoeutectic) just below 183 °C: eutectic microconstituent We = (40 − 18.3)/(61.9 − 18.3) = 0.498, primary α 0.502. As phases: β = (40 − 18.3)/(97.8 − 18.3) = 0.273, α = 0.727 (including the α inside the eutectic).
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Fe–Fe₃C cooled slowly to just below 727 °C. Values: eutectoid 0.76 wt% C, ferrite limit 0.022 wt%, Fe₃C 6.70 wt% (as in the site’s diagram and questions; some texts use 0.77 and 6.67).
Common mistakes
Confusing proeutectoid ferrite with total ferrite (including that in pearlite); using the hypo/hyper formula on the wrong side; applying it to rapid cooling.
Example
A 0.40 wt% C steel (hypoeutectoid) just below 727 °C: pearlite WP = (0.40 − 0.022)/(0.76 − 0.022) = 0.512, proeutectoid ferrite 0.488. Total ferrite phase (including that in pearlite): (6.70 − 0.40)/(6.70 − 0.022) = 0.943. For 1.00 wt% C (hypereutectoid) the proeutectoid cementite is (1.00 − 0.76)/(6.70 − 0.76) = 0.0404.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
No diffusion in the solid, complete mixing in the liquid, local equilibrium at the interface, constant k (straight solidus and liquidus). Estimates casting segregation and non-equilibrium eutectic. For k < 1 the last liquid is the most enriched.
Common mistakes
Swapping fs and fL; inverting k; expecting the equilibrium lever-rule result (equilibrium assumes a uniform solid).
Example
Al–4 wt% Cu (straight-line diagram, k = 5.65/33.2 = 0.170): at half solidified (fs = 0.5) the solid at the interface is Cs = 0.170 × 4 × 0.5^(−0.830) = 1.21 wt%. The liquid reaches the eutectic 33.2 wt% at fL = (33.2/4)^(1/(k − 1)) = 0.0781: about 7.8 % solidifies as eutectic although equilibrium predicts none.
ΔSmix = −R(xA ln xA + xB ln xB)[kg·m²·s⁻²·K⁻¹·mol⁻¹]
Tc = Ω/(2R)top of the miscibility gap[K]
Symbol
Meaning
SI unit
ΔGmix
Gibbs energy of mixing per mole
J/mol
ΔSmix
ideal entropy of mixing
J/(mol·K)
Ω
interaction parameter (0: ideal, > 0: tendency to separate)
J/mol
xA, xB
mole fractions
1
R
gas constant
J/(mol·K)
T
temperature
K
Tc
critical temperature
K
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Random mixing with heat of mixing ΩxA xB (regular solution). For Ω > 2RT the ΔGmix curve has two minima: phase separation (miscibility gap, spinodal decomposition).
Common mistakes
Using log₁₀ for ln; T in °C; dropping the minus sign (ΔSmix is positive).
Example
Ideal solution (Ω = 0), xB = 0.5, T = 1000 K: ΔGmix = RT(0.5 ln 0.5 × 2) = −5763 J/mol, ΔSmix = 5.763 J/(mol·K). With Ω = 20 kJ/mol the miscibility gap closes at Tc = Ω/(2R) = 1203 K.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Spherical nucleus, isotropic interfacial energy, small ΔT (linear ΔGv near Tm). Heterogeneous nucleation multiplies ΔG* by S(θ) = (2 + cos θ)(1 − cos θ)²/4.
Common mistakes
Sign conventions (Callister takes ΔHf negative: r* = −2γTm/(ΔHf ΔT)); using latent heat per mole (divide by molar volume); radius vs diameter.
Example
Pure gold (latent heat Lv = 1.16 × 10⁹ J/m³, γ = 0.132 J/m², Tm = 1337 K) undercooled by ΔT = 230 K: r* = 2 × 0.132 × 1337/(1.16 × 10⁹ × 230) = 1.32 nm, ΔG* = 9.68 × 10⁻¹⁹ J (the same numbers as the Callister chapter 10 example).
rate constant (temperature dependent; units depend on n)
s⁻ⁿ
t
time
s
n
Avrami exponent (1–4 depending on nucleation and growth)
1
Dimensions both sides agree on every line ([ ] in SI base units) · dimensionless: ktⁿ
Conditions and range
Isothermal transformations by random nucleation and growth (recrystallisation, pearlite). A plot of log ln[1/(1 − y)] against log t has slope n. The rate is often expressed as 1/t₀.₅.
Common mistakes
Assuming k is in s⁻¹ (it depends on n); applying it directly to continuous cooling.
Example
With n = 3 and half-transformation time t₀.₅ = 100 s, k = ln 2/t₀.₅³ = 6.93 × 10⁻⁷ s⁻³. At 150 s, y = 1 − exp(−k × 150³) = 0.904.
ln D = ln D₀ − (Q/R)(1/T)Arrhenius plot (slope −Q/R)[1]
Q = R ln(D₂/D₁)/(1/T₁ − 1/T₂)from two points[kg·m²·s⁻²·mol⁻¹]
Symbol
Meaning
SI unit
D
diffusion coefficient
m²/s
D₀
pre-exponential factor
m²/s
Q
activation energy per mole (per atom: replace R by k)
J/mol
R
gas constant 8.314
J/(mol·K)
T
absolute temperature
K
Dimensions both sides agree on every line ([ ] in SI base units) · dimensionless: Q/RT
Conditions and range
Within a temperature range controlled by one mechanism (lattice diffusion). At low temperature grain-boundary or pipe diffusion bends the Arrhenius plot; D₀ and Q change across magnetic or phase transitions (Fe).
Common mistakes
T in °C; dividing kJ/mol by R in J/(mol·K) (factor 1000); confusing eV/atom and kJ/mol (1 eV/atom = 96.49 kJ/mol); ln vs log₁₀ slopes (log₁₀ slope is −Q/(2.303R)).
Example
Carbon in γ-Fe (D₀ = 2.3 × 10⁻⁵ m²/s, Q = 148 kJ/mol) at 1000 °C = 1273.15 K: Q/RT = 13.98, D = 1.95 × 10⁻¹¹ m²/s. Conversely, D = 5.92 × 10⁻¹² at 900 °C and 5.39 × 10⁻¹¹ m²/s at 1100 °C give back Q = R ln(D₂/D₁)/(1/T₁ − 1/T₂) = 148.0 kJ/mol. Per atom, Q/NA = 1.53 eV.
concentration (mol/m³ or atoms/m³ also work; J follows)
kg/m³
x
position
m
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Steady state, where the profile does not change with time (e.g. hydrogen permeation through a membrane), with D independent of concentration. Flux runs down the gradient (minus sign).
Common mistakes
Not converting mm to m; using wt% (convert to amount per volume); using it for non-steady problems.
Example
Steel plate at 700 °C with 1.2 kg/m³ of carbon at 5 mm and 0.8 kg/m³ at 10 mm, D = 3 × 10⁻¹¹ m²/s (steady state): J = 3 × 10⁻¹¹ × (1.2 − 0.8)/(5 × 10⁻³) = 2.4 × 10⁻⁹ kg/(m²·s) (the same numbers as the Callister chapter 5 example).
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
General equation when the profile changes with time. Solutions depend on boundary conditions: erfc for a semi-infinite solid with constant surface concentration, Gaussian for a fixed amount at the surface, erf for a diffusion couple, Fourier series or numerics for a finite slab. Concentration-dependent D (interdiffusion) needs the second form.
Common mistakes
Using the constant-D form when D varies strongly; choosing erfc vs Gaussian without checking the boundary conditions.
Example
The time to smooth out concentration differences over L = 50 µm (e.g. dendrite arm spacing) with D = 1 × 10⁻¹³ m²/s is of order L²/D = 2.5 × 10⁴ s (about 6.9 h). Doubling the distance quadruples the time.
C(x, t) = (CA + CB)/2 − ((CA − CB)/2) erf(x/(2√(Dt)))diffusion couple (interface at x = 0)[kg·m⁻³]
Symbol
Meaning
SI unit
Cx
concentration at depth x and time t
kg/m³
C₀
initial uniform concentration
kg/m³
Cs
constant surface concentration
kg/m³
CA, CB
initial concentrations left (x < 0) and right of the couple
kg/m³
x
distance from surface (interface)
m
D
diffusion coefficient (constant)
m²/s
t
time
s
Dimensions both sides agree on every line ([ ] in SI base units) · dimensionless: x/(2√(Dt))
Conditions and range
Semi-infinite solid (√(Dt) much smaller than the thickness), constant D, constant surface concentration (or constant concentrations far on both sides of a couple). The ratio form also works with wt%.
Common mistakes
Mixing erf and erfc (= 1 − erf); dropping the 2 in 2√(Dt); time in hours; using Cs − Cx as the denominator.
Example
A 0.25 wt% C steel carburised at 950 °C (D = 1.10 × 10⁻¹¹ m²/s, same data as the diffusion simulator) with surface 1.20 wt%, to reach 0.80 wt% at 0.5 mm: (0.80 − 0.25)/(1.20 − 0.25) = 0.5789 = 1 − erf z, so erf z = 0.4211, z = 0.3924, t = x²/(4z²D) = 3.69 × 10⁴ s = 10.2 h. In a diffusion couple at x = √(Dt) (z = 0.5), C = (CA + CB)/2 − (CA − CB)/2 × erf 0.5, i.e. a fraction 0.760 of the way from CA to CB.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
For the same boundary conditions when the profile depends only on x/√(Dt) (erf, Gaussian). Doubling the depth needs four times the time.
Common mistakes
Making depth proportional to time (dropping the √); swapping T₁ and T₂ (higher T means shorter time).
Example
Cu in Al (D₀ = 6.5 × 10⁻⁵ m²/s, Q = 136 kJ/mol): D = 4.22 × 10⁻¹⁴ at 500 °C and 4.75 × 10⁻¹³ m²/s at 600 °C. To match 10 h at 500 °C, 600 °C needs t = 10 h × D₅₀₀/D₆₀₀ = 0.887 h. √(Dt) for 10 h at 500 °C is 39.0 µm.
amount per unit area initially at the surface (conserved)
m⁻²
D
diffusion coefficient
m²/s
t
time
s
x
depth
m
xj
depth where C equals background CB
m
CB
background concentration
m⁻³
Dimensions both sides agree on every line ([ ] in SI base units) · dimensionless: x²/(4Dt)
Conditions and range
A fixed amount M initially at the surface, no flux through the surface, spreading into a semi-infinite solid: semiconductor drive-in, tracer diffusion. For a thin layer inside an infinite body spreading both ways use M/(2√(πDt)).
Common mistakes
Mixing the semi-infinite and infinite prefactors (1/√(πDt) vs 1/(2√(πDt))); cm vs m (10¹⁴ cm⁻² = 10¹⁸ m⁻², 1 cm²/s = 10⁻⁴ m²/s).
Example
Boron in Si (D₀ = 0.76 cm²/s, Ea = 3.46 eV) driven in at 1100 °C (D = 1.52 × 10⁻¹⁷ m²/s) for 1 h with dose M = 5 × 10¹⁸ m⁻² (5 × 10¹⁴ cm⁻²): surface concentration M/√(πDt) = 1.21 × 10²⁵ m⁻³. Depth where it equals the background 10²¹ m⁻³ (junction depth): xj = √(4Dt ln(C(0)/CB)) = 1.43 µm.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Forward-time, centred-space solution of the 1-D second law with constant D. For r > 1/2 the solution oscillates and diverges. Used where no analytic solution exists (finite slabs, changing surface concentration).
Common mistakes
Refining the grid without reducing Δt (halving Δx needs a quarter of Δt).
Example
D = 1 × 10⁻¹¹ m²/s, Δx = 10 µm: Δt ≤ Δx²/(2D) = 5.0 s. The diffusion simulator uses r = 0.4 for margin (Δt = 4.0 s).
vacancy formation energy per vacancy (often given in eV)
J
k
Boltzmann constant (8.617 × 10⁻⁵ eV/K)
J/K
T
absolute temperature
K
NA
Avogadro constant
mol⁻¹
ρ
density
kg/m³
A
molar mass
kg/mol
Dimensions both sides agree on every line ([ ] in SI base units) · dimensionless: Qv/kT
Conditions and range
Thermal-equilibrium vacancies; the formation-entropy factor exp(Sv/k) (about 1–10) is omitted. Quenched samples retain high-temperature vacancies.
Common mistakes
Dividing eV by k in J/K (use k in eV/K); T in °C; the 10⁶ when converting g/cm³ to m⁻³.
Example
Cu at 1000 °C (ρ = 8.4 g/cm³, A = 63.5 g/mol, Qv = 0.9 eV): N = NA ρ/A = 7.97 × 10²⁸ m⁻³, Qv/kT = 8.203, Nv = 2.2 × 10²⁵ m⁻³ (the same numbers as the Callister chapter 4 example): one vacancy per 3653 atoms.
denominator (2 for perfect dislocations in FCC and BCC)
1
u, v, w
direction indices
1
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
For perfect dislocations b is the shortest lattice translation (FCC a/2⟨110⟩, BCC a/2⟨111⟩, HCP a/3⟨11−20⟩ with |b| = a). The FCC Shockley partial a/6⟨112⟩ has |b| = a/√6.
Common mistakes
Dropping the 1/2 and getting √2a; swapping the BCC and FCC directions.
Example
FCC Cu (a = 0.36146 nm), a/2⟨110⟩: |b| = a/√2 = 0.2556 nm (the nearest-neighbour distance). BCC Fe (a = 0.28665 nm), a/2⟨111⟩: |b| = √3a/2 = 0.2482 nm.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
ASTM E112 conversions; the original definition is N = 2G−1 grains per square inch at 100×. JIS G 0551 (ISO 643) uses m = 8 × 2G per mm², giving numbers about 0.05 lower for the same structure. Converting to a “diameter” (e.g. d ≈ ℓ̄) is a convention.
Common mistakes
Larger numbers mean finer grains; using counts from images not at 100× without converting to true area.
Example
Grain size number G = 6: 2⁵ = 32 grains per square inch at 100×, i.e. NA = 15.50 × 32 = 496 mm⁻² at 1×. Mean intercept ℓ̄ = 0.3200 × 2^(−3) = 40.0 µm.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Linear elasticity (below the proportional limit), small strain, uniaxial load on a uniform bar. It does not apply after yielding. For a stepped bar, find the extension of each section and add them.
Common mistakes
Using the diameter as the radius (πd² is four times too big); MPa vs GPa (N/mm² vs kN/mm²); applying it above the yield stress.
Example
A Cu bar pulled at σ = 276 MPa (l₀ = 305 mm, E = 110 GPa, elastic): ε = 276/110000 = 2.51 × 10⁻³, Δl = εl₀ = 0.765 mm. For a 10 mm diameter bar, F = σA = 276 × 78.54 = 21.68 kN.
Poisson’s ratio (metals 0.26–0.35, rubber near 0.5)
1
εz
axial strain
1
εx
lateral strain
1
d₀, Δd
original diameter and its change
m
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Isotropic linear elasticity. In plastic deformation volume is nearly constant, so the apparent ratio approaches 0.5.
Common mistakes
Sign (diameter shrinks in tension); treating diameter change as radius change.
Example
Load that reduces the diameter of a brass rod (d₀ = 10 mm, ν = 0.34, E = 97 GPa) by 2.5 µm: εx = −2.5 × 10⁻⁴, εz = −εx/ν = 7.35 × 10⁻⁴, σ = Eεz = 71.3 MPa, F = σπd₀²/4 = 5602 N (same numbers as the Callister Ch. 6 example).
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Isotropic solids only (two independent constants). Not valid for single crystals or fibre composites. Add αΔT for thermal strain. As ν → 0.5, K → ∞ (incompressible).
Common mistakes
Applying it to anisotropic materials (Si wafers, CFRP). Handbook G values that disagree with E/(2(1 + ν)) reflect texture or measurement differences.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
The (1 + ε) forms assume constant volume and hold only up to necking (uniform elongation). After necking use σT = F/Ai with the measured minimum area (Bridgman’s triaxiality correction applies further).
Common mistakes
Using σ(1 + ε) after necking; in compression ε is negative and σT smaller; computing true strain with log₁₀.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Polycrystals with grains above about 10 nm (finer grains may soften: inverse Hall–Petch). Compare annealed states with the same precipitation and texture.
Common mistakes
Using d in µm with ky in MPa·√m (convert to m); plotting against d rather than 1/√d; confusing grain size number with diameter.
Example
Steel with σ₀ = 70 MPa, ky = 0.74 MPa·√m: d = 20 µm gives σy = 70 + 0.74/√(2 × 10⁻⁵) = 235.5 MPa; d = 5 µm gives 400.9 MPa. Quartering the grain size doubles the ky term.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Single crystals under uniaxial load. The system with the largest Schmid factor cos φ cos λ (max 0.5) slips first. In cubic crystals cos is the normalised dot product of the indices.
Common mistakes
Assuming φ + λ = 90° (not true in general); choosing a direction that does not lie in the slip plane; using sin.
Example
A BCC Fe single crystal pulled along [010] at 52 MPa, slip system (110)[−111]: cos φ = 1/√2 = 0.7071 (φ = 45°), cos λ = 1/√3 = 0.5774 (λ = 54.7°), τR = 52 × 0.7071 × 0.5774 = 21.2 MPa. With τCRSS = 30 MPa the yield stress is 30/0.4082 = 73.5 MPa (same numbers as the Callister Ch. 7 example).
performance index (higher is lighter for the same duty; units per index)
—
E
Young’s modulus
Pa
σf
failure strength
Pa
ρ
density
kg/m³
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Minimum mass when the section size may change freely; the index depends on geometry, loading and what is fixed. On the material selection chart, candidates lie above the line of constant index (slopes 1, 2, 3).
Common mistakes
Using E/ρ for a beam (beams use √E/ρ); comparing indices with inconsistent units.
Example
Light, stiff beam (index √E/ρ, E in GPa, ρ in g/cm³): Al alloy √70/2.70 = 3.10, steel √210/7.85 = 1.85. For equal bending stiffness the Al beam weighs 1/1.68 of the steel one. For a tie (E/ρ) they are almost equal (25.9 vs 26.8).
surface crack depth (half length of an internal crack)
m
ρt
tip radius of curvature
m
Kt
stress concentration factor
1
Dimensions both sides agree on every line ([ ] in SI base units) · dimensionless: a/ρt
Conditions and range
Elliptical hole in an infinite plate loaded normal to the major axis, linear elastic. For sharp cracks (ρt → 0) the stress diverges, so use the stress intensity factor K. Use design charts for real geometries.
Common mistakes
Using the full internal crack length for a (use half); mixing mm and µm for a and ρt.
Example
Surface crack of depth a = 0.5 mm with tip radius ρt = 5 µm under σ₀ = 100 MPa: a/ρt = 100, Kt = 1 + 2√100 = 21.0, σmax = 2100 MPa (the approximation 2σ₀√(a/ρt) gives 2000 MPa).
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Linear elastic fracture mechanics: plastic zone small compared with crack and section (small-scale yielding). If the fracture stress approaches yield, the part yields first. KIC is measured in plane strain (E399).
Common mistakes
Using the full internal crack length; a in mm under the root (use m with MPa·√m); assuming Y = 1 always.
Example
Surface crack (Y = 1.12) in steel with KIC = 60 MPa·√m: at 300 MPa the critical depth is ac = (1/π)(60/(1.12 × 300))² = 10.2 mm. With a 5 mm crack the fracture stress is σc = 60/(1.12√(π × 0.005)) = 427 MPa.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Irwin’s first-order estimate (about one third in plane strain). 2.5(KIC/σy)² is the ASTM E399 validity requirement; thinner specimens approach plane stress and show higher apparent toughness.
Common mistakes
Treating rp as a diameter; calling a thin-sheet value KIC.
Example
K = 50 MPa·√m, σy = 500 MPa: plane-stress plastic zone rp = (1/2π)(50/500)² = 1.59 mm (about one third in plane strain). A valid KIC test needs thickness and crack length ≥ 2.5(KIC/σy)² = 25.0 mm.
Nf = [ac1−m/2 − a₀1−m/2]/[C(YΔσ√π)m(1 − m/2)]constant Y, m ≠ 2[1]
Symbol
Meaning
SI unit
da/dN
crack growth per cycle
m
C
material constant (units depend on m)
m·(Pa·√m)⁻ᵐ
ΔK
stress-intensity range
Pa·√m
m
Paris exponent (2–4 for metals)
1
Δσ
stress range (compressive part usually ignored)
Pa
Y
geometry factor
1
a, a₀, ac
crack size (initial, critical)
m
Nf
cycles to failure
1
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Stage II growth: ΔK above the threshold and Kmax well below KIC. This form ignores mean-stress (R) effects. Integrate numerically when Y varies with a.
Common mistakes
Using C without checking its unit basis (MPa·√m and m?); using Kmax for ΔK; applying the formula at m = 2 (it becomes logarithmic).
Example
Δσ = 100 MPa (compressive part ignored), initial surface crack 2.0 mm, KIC = 25 MPa·√m, Y = 1.0, C = 1.0 × 10⁻¹², m = 3 (MPa·√m and m units): final crack ac = (1/π)(25/100)² = 19.9 mm, Nf = [ac^(−1/2) − a₀^(−1/2)]/[C(Δσ√π)³(−1/2)] = 5.48 × 10⁶ cycles (same conditions as the Callister Ch. 8 example).
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Fully reversed constant amplitude on smooth specimens. The elastic term dominates at high cycles and the plastic term below about 10⁴ cycles. Correct for mean stress (e.g. Morrow) and combine with Miner’s rule for variable loads.
Common mistakes
Confusing reversals 2Nf with cycles Nf; amplitude vs range (factor 2).
Example
Assumed constants σf′ = 1000 MPa, b = −0.09, εf′ = 0.6, c = −0.6, E = 200 GPa: at Nf = 10⁴ cycles the strain amplitude is elastic 2.05 × 10⁻³ + plastic 1.58 × 10⁻³ = 3.63 × 10⁻³. Transition life (elastic = plastic): Nt = 5968 cycles. A stress amplitude of 400 MPa (high-cycle) gives Nf = (400/1000)^(1/b)/2 = 1.32 × 10⁴ cycles by Basquin.
Dimensions both sides agree on every line ([ ] in SI base units) · dimensionless: σa/σe, σm/σu
Conditions and range
Conservative straight-line estimate of how tensile mean stress lowers fatigue strength. For ductile metals Gerber’s parabola ((σm/σu)²) fits better; Soderberg uses σy. Compressive mean stress is usually not credited.
Common mistakes
Using the range as amplitude (halve it); confusing R = −1 (fully reversed) and R = 0 (zero to tension).
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Ignores load-sequence effects: high-then-low loading tends to fail at D < 1, low-then-high may exceed 1. Designs sometimes limit D to 0.3–0.5.
Common mistakes
Ignoring cycles below the fatigue limit or taking Ni = ∞ (once a crack grows, low stresses also damage).
Example
2 × 10⁴ cycles at a stress with life 10⁵ (0.2) and 3 × 10⁵ cycles at life 10⁶ (0.3): damage D = 0.50. At a stress with life 10⁷ the remaining life is (1 − 0.50) × 10⁷ = 5.0 × 10⁶ cycles.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Weakest-link brittle materials, written for uniform tension; for bending use the effective volume. m is the slope of ln ln[1/(1 − Pf)] against ln σ.
Common mistakes
Using bend strengths directly for tensile parts (different effective volume); confusing Pf with survival probability Ps = 1 − Pf.
Example
Ceramic with Weibull modulus m = 10 and σ₀ = 300 MPa (specimen volume V₀) at 200 MPa: Pf = 1 − exp[−(200/300)¹⁰] = 0.0172. A part ten times larger reaches the same probability at (1/10)^(1/10) = 0.794 times the stress.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
When α can be taken as constant over the range (otherwise integrate α(T)). The 3α relation is for isotropic materials only (anisotropic crystals and composites differ by direction).
Common mistakes
Dropping the 10⁻⁶; multiplying by an absolute temperature instead of a difference. A difference in °C equals a difference in K.
Example
Brass wire (α = 20 × 10⁻⁶ K⁻¹), 15 m, cooled from 40 °C to −9 °C: ΔL = 20 × 10⁻⁶ × 15 × (−49) = −1.47 × 10⁻² m = −14.7 mm (contraction). The volume decreases by 3α: 0.29 %.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Full constraint (zero net strain), linear elastic. Heating gives compression, cooling tension. For partial constraint or temperature gradients, see the thermal stress lab.
Common mistakes
Sign errors; using the uniaxial formula for a plate; using values above yield.
Example
Brass rod fixed at both ends (E = 100 GPa, α = 20 × 10⁻⁶ K⁻¹) heated from 20 °C: to keep the compressive stress below 172 MPa, ΔT = 172/(100000 × 20 × 10⁻⁶) = 86.0 K, T = 106 °C. A biaxially constrained plate (ν = 0.35) would see 1/(1 − ν) = 1.54 times the stress for the same ΔT.
film thermal stress (equibiaxial, tension positive)
Pa
Ef/(1 − νf)
biaxial modulus of the film
Pa
αs, αf
expansion coefficients of substrate and film
K⁻¹
ΔT
change from the stress-free temperature (negative on cooling)
K
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Film much thinner than the substrate (substrate undeformed), in-plane isotropic, linear elastic. Take the deposition temperature as stress-free (add intrinsic growth stress separately).
Common mistakes
Order of (αf − αs) vs (αs − αf); dropping 1 − ν (uniaxial form); ignoring anisotropy of single-crystal films.
Example
Al film (Ef = 70 GPa, νf = 0.33, αf = 23.1 × 10⁻⁶ K⁻¹) on Si (αs = 2.6 × 10⁻⁶ K⁻¹), stress-free at 400 °C, cooled to 25 °C: σf = (70000/0.67) × (2.6 − 23.1) × 10⁻⁶ × (−375) = 803 MPa (tensile). In practice Al yields long before reaching this elastic value.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
tf ≪ ts, deflection smaller than substrate thickness, isotropic substrate (use 180 GPa biaxial modulus for Si(100)). The film’s elastic constants are not needed.
Common mistakes
Swapping ts and tf (ts is squared); not subtracting the pre-deposition curvature; diameter vs radius of curvature.
Example
A (100) Si wafer (biaxial modulus Es/(1 − νs) = 180 GPa, 0.525 mm thick) with a 1 µm film bows to a 50 m radius: σf = 180 × 10⁹ × (0.525 × 10⁻³)²/(6 × 10⁻⁶ × 50) = 165 MPa.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Two parallel bars joined at both ends, no external load, uniform temperature, linear elastic, no bending. With an external load, put it on the right-hand side of the force balance.
Common mistakes
Using the fully constrained value (EαΔT) for one bar; ignoring different areas.
Example
Steel (E = 200 GPa, α = 12 × 10⁻⁶) and Al (70 GPa, 23 × 10⁻⁶) bars of equal section (100 mm²) joined at both ends and heated by 100 K: σ₁ (steel) = −(12 − 23) × 10⁻⁶ × 100/(1/200000 + 1/70000) = 57.0 MPa (tension), σ₂ (Al) = −57.0 MPa (compression).
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Two-layer beam, free in the width direction, linear elastic, uniform temperature. For wide plates replace E by E/(1 − ν).
Common mistakes
Wrong sense (the higher-expansion layer is on the convex side); taking h as one layer thickness.
Example
Two layers of 0.5 mm each with equal moduli, expansion mismatch 18 × 10⁻⁶ K⁻¹, heated by 50 K (Δε = 9.0 × 10⁻⁴): with m = n = 1, κ = 3Δε/(2h) = 1.35 m⁻¹ (radius 0.741 m).
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Onset of cracking when a brittle plate is suddenly cooled (or heated) at the surface. Large β (severe quench, thick, low conductivity) approaches R. Strength scatter and crack-propagation parameters (R‴ etc.) are separate.
Common mistakes
Using the full thickness for a (use half); dropping 10⁻⁶ in α; forgetting the surface stress reverses between heating and cooling.
Example
Alumina (data from the thermal expansion lab: σf = 310 MPa, ν = 0.23, E = 360 GPa, α = 7.2 × 10⁻⁶ K⁻¹, κ = 29 W/(m·K)): R = 92.1 K, R′ = κR = 2671 W/m. A 10 mm plate (a = 5 mm) quenched in water (h ≈ 10⁴ W/(m²·K)): β = ah/κ = 1.72, critical ΔTc ≈ R × (1.5 + 3.25/β − 0.5e^(−16/β)) = 312 K.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Governs the speed of transient conduction; the heat equation has the same form as Fick’s second law, so erfc solutions apply. Laser flash measures a, then κ = aρcp.
Common mistakes
Using cp per mole (use per kg); ρ in g/cm³.
Example
Cu (κ = 398 W/(m·K), ρ = 8960 kg/m³, cp = 385 J/(kg·K)): a = 1.15 × 10⁻⁴ m²/s. Heat penetrates roughly √(at) = 10.7 mm in 1 s.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Per mole of atoms in an element; multiply by atoms per formula for compounds. Metals add an electronic term (∝ T) at low temperature. Cp − Cv is small for solids.
Common mistakes
Per mole vs per kg; forgetting atoms per formula unit for compounds.
Example
Al (Debye temperature θD = 428 K): at 20 K, Cv = (12π⁴/5)R(20/428)³ = 0.198 J/(mol·K). The high-temperature limit 3R = 24.94 J/(mol·K), per mass 3R/M = 924 J/(kg·K) (measured cp 897 J/(kg·K)).
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Electronic conduction in metals; good at room temperature and above and at very low T (lower in between due to inelastic scattering). Excludes phonon conduction: κ = κe + κph in alloys and semiconductors.
Common mistakes
Confusing resistivity with conductivity; applying it to insulators and ceramics.
σθ = p(D² + d²)/(D² − d²)hoop stress at the hub bore[kg·m⁻¹·s⁻²]
ΔT = (δ + c)/(αd)heating needed for assembly[K]
Symbol
Meaning
SI unit
p
interface pressure
Pa
δ
diametral interference
m
d
interface diameter
m
D
outer diameter of the hub
m
di
bore of the inner member (0 if solid)
m
Eo, Ei
moduli of hub and shaft
Pa
νo, νi
Poisson’s ratios of hub and shaft
1
σθ
hoop stress
Pa
c
assembly clearance
m
α
expansion coefficient of the hub
K⁻¹
ΔT
temperature rise
K
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Thick-walled cylinders in plane stress, linear elastic, uniform pressure along the fit. Yielding lowers the pressure. The hoop stress at the hub bore is the largest.
Common mistakes
Using radial interference in the diametral formula (factor 2); the sign of ν in the shaft term (− νi).
Example
Steel hub (outer diameter 100 mm) shrunk onto a solid 50 mm steel shaft with 0.05 mm diametral interference (E = 200 GPa, ν = 0.3): p = 75.0 MPa; hoop stress at the hub bore p(D² + d²)/(D² − d²) = 125 MPa (tension). To assemble with 0.02 mm clearance, heat the hub by ΔT = (0.05 + 0.02)/(12 × 10⁻⁶ × 50) = 117 K.
Ecl = Vf Ef + Vm Emlongitudinal (isostrain, Voigt)[kg·m⁻¹·s⁻²]
Ff/Fm = Vf Ef/(Vm Em)load sharing[1]
Vf + Vm = 1no voids[1]
Symbol
Meaning
SI unit
Ecl
longitudinal modulus of the composite
Pa
Ef, Em
fibre and matrix moduli
Pa
Vf, Vm
fibre and matrix volume fractions
1
Ff, Fm
loads carried by fibre and matrix
N
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Continuous aligned fibres, perfect bonding so both phases share the same strain. Upper bound (Voigt). Use volume fractions (convert mass fractions first). Voids count as a phase of zero stiffness.
Common mistakes
Plugging in mass fractions; using it transverse to the fibres (use the inverse rule); confusing the load ratio with the load fraction Ff/(Ff + Fm).
Example
Unidirectional glass fibre 40 vol% (Ef = 69 GPa) in polyester (Em = 3.4 GPa), the same numbers as the Callister chapter 16 example: Ecl = 0.40 × 69 + 0.60 × 3.4 = 29.6 GPa. Load sharing Ff/Fm = (0.40 × 69)/(0.60 × 3.4) = 13.5 (fibres carry 93.1 %).
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Lower bound (Reuss), assuming both phases carry the same stress. Real transverse moduli are somewhat higher; Halpin–Tsai fits better. Carbon fibres are much softer radially (about 15 GPa for T300), so use the radial modulus for Ef.
Common mistakes
Forgetting to invert; using the axial modulus of carbon fibre for Ef.
Example
The same glass 40 vol%/polyester, transverse to the fibres: 1/Ect = 0.40/69 + 0.60/3.4 → Ect = 5.49 GPa (19 % of the longitudinal value).
Vf = (wf/ρf)/(wf/ρf + wm/ρm)mass to volume fraction[1]
wf = Vf ρf/ρcvolume to mass fraction[1]
ρc = Vf ρf + Vm ρmdensity (rule of mixtures)[kg·m⁻³]
Symbol
Meaning
SI unit
Vf, Vm
fibre and matrix volume fractions
1
wf, wm
fibre and matrix mass fractions
1
ρf, ρm, ρc
densities of fibre, matrix and composite
kg/m³
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
No voids. With voids Vf + Vm + Vv = 1; Vv follows from the measured density and the mass fraction (e.g. burn-off). Units of density cancel, so g/cm³ is fine.
Common mistakes
Using mass fractions in the rule of mixtures; using 1 − Vf instead of wm = 1 − wf.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Continuous fibres whose failure strain is below the matrix’s (carbon or glass/epoxy). Fibre strength scatter and misalignment usually bring measured values to 70–90 %. For matrix-first failure (ceramic matrices) another expression applies.
Common mistakes
Using the matrix strength σm* for σm′; applying it transverse to the fibres (governed by matrix and interface).
Example
T300 carbon fibre (σf* = 3530 MPa, Ef = 230 GPa) in 8552 epoxy (Em = 4.67 GPa, σm* = 121 MPa, failure strain 1.7 %), Vf = 0.60 (values from the composite designer): the fibre failure strain 3530/230000 = 1.53 % is below the matrix’s, so fibres break first. σm′ = 4670 × 0.0153 = 71.7 MPa, σcl* = 0.60 × 3530 + 0.40 × 71.7 = 2147 MPa. Vmin = 1.4 × 10⁻², so at any practical fraction the fibres govern.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Fibres aligned with the load; interfacial shear stress constant from the fibre ends (rigid-plastic). For random orientation multiply by an orientation efficiency (Krenchel, 3/8 for 2D random). With l above about 15 lc the fibres behave as continuous.
Common mistakes
Using the radius for d (factor 2); mixing µm and mm; using the long-fibre formula for l < lc.
Example
Carbon fibre (σf* = 3500 MPa, d = 7 µm) with interfacial shear strength τc = 40 MPa: lc = 3500 × 0.007/(2 × 40) = 0.306 mm. Aligned 3 mm fibres at 30 vol% with σm′ = 50 MPa: σcd* = 0.30 × 3500 × (1 − 0.306/6) + 0.70 × 50 = 1031 MPa. With 0.2 mm fibres (l < lc): 0.30 × 40 × 0.2/0.007 + 35 = 378 MPa.
longitudinal and transverse expansion coefficients
K⁻¹
αf, αm
fibre and matrix expansion coefficients
K⁻¹
Ef, Em
fibre and matrix moduli
Pa
νf, νm, ν₁₂
Poisson’s ratios
1
Vf, Vm
volume fractions
1
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Isotropic fibre and matrix (glass, basalt). Carbon and aramid fibres are anisotropic, so replace (1 + νf)αf by αf2 + νf αf1 (as in the composite designer). α₁ is dominated by the stiff fibres and lies far below the volume average.
Common mistakes
Using a plain volume average for α₁; forgetting that carbon fibre has negative axial α.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
A single orthotropic ply in plane stress pulled uniaxially at θ to the fibres. Laminates need classical laminate theory (A matrix). Because of the G₁₂ term, materials with G₁₂ below about E₂/2 dip below E₂ between 45° and 90° (for the material in the example, Ex stays at or above E₂ at every angle).
Common mistakes
Rotating E with cos²θ (it needs fourth powers and the shear term); forgetting to invert; degrees vs radians.
Example
T300/5208 ply (E₁ = 181 GPa, E₂ = 10.3 GPa, G₁₂ = 7.17 GPa, ν₁₂ = 0.28, values from the composite designer): Ex = 28.8 GPa at θ = 30° and 16.7 GPa at 45°. A 30° misalignment already drops the stiffness to 1/6.3 of the fibre direction.
R = 1/√ITHstrength ratio (proportional loading; ITH is the left-hand side)[1]
Symbol
Meaning
SI unit
σ₁, σ₂, τ₁₂
stresses in the ply axes
Pa
X, Y
longitudinal and transverse strengths (tension or compression by sign)
Pa
S
in-plane shear strength
Pa
ITH
value of the left-hand side (failure index)
1
R
strength ratio (load multiplier to failure)
1
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
In-plane (plane stress) loading of a unidirectional ply. For materials with different tension and compression strengths pick X and Y by the sign of the stress. For proportional loading the strength ratio is one over the square root of the left-hand side.
Common mistakes
Using global stresses without rotating them to the ply axes; taking the strength ratio as one over the left-hand side (needs a square root).
Example
T300/5208 (X = 1500, Y = 40, S = 68 MPa) pulled at 100 MPa, 30° from the fibres: σ₁ = 75.0, σ₂ = 25.0, τ₁₂ = −43.3 MPa. The left-hand side is 0.798 < 1, so no failure. Strength ratio R = 1/√0.798 = 1.12 (off-axis strength 112 MPa).
ρi = A ci(1 − ci)solute term (Nordheim)[kg·m³·s⁻³·A⁻²]
Symbol
Meaning
SI unit
ρ
total resistivity
Ω·m
ρt, ρi, ρd
thermal, impurity and deformation contributions
Ω·m
A
constant for the solute–solvent pair
Ω·m
ci
solute atom fraction
1
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Metals where the scattering mechanisms are independent. Only ρt depends on temperature; ρi and ρd remain at low temperature (residual resistivity). The Nordheim form is for alloys forming complete solid solutions, at dilute to moderate concentrations.
Common mistakes
Adding conductivities instead of resistivities; using at% instead of a fraction.
Example
Cu with 3 at% Ni in solid solution (room-temperature ρt of Cu 1.72 × 10⁻⁸ Ω·m; A for Cu–Ni read from the Callister figure, about 1.2 × 10⁻⁶ Ω·m): ρi = 1.2 × 10⁻⁶ × 0.03 × 0.97 = 3.49 × 10⁻⁸ Ω·m, ρ ≈ 5.21 × 10⁻⁸ Ω·m (3.03 times pure copper).
σ ≈ ND e μen-type (all donors ionised)[kg⁻¹·m⁻³·s³·A²]
ρ = 1/σresistivity[kg·m³·s⁻³·A⁻²]
Symbol
Meaning
SI unit
σ
conductivity
S/m
n, p
electron and hole concentrations
m⁻³
ND
donor concentration
m⁻³
e
elementary charge
C
μe, μh
electron and hole mobilities
m²/(V·s)
ρ
resistivity
Ω·m
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Near room temperature with all dopants ionised (extrinsic regime). Mobility falls with doping, so use a value for that concentration (empirical fits differ by a few per cent: this site’s drill uses the Caughey–Thomas form and the diffusion simulator the Arora fit, which differ by up to about 8 % for electrons in n-type Si at 10¹⁷–10¹⁸ cm⁻³).
Common mistakes
Not converting cm⁻³ and cm²/(V·s) to SI (or converting only one); confusing conductivity with resistivity.
ni ∝ T3/2 exp(−Eg/(2kT))temperature dependence[m⁻³]
np = ni²mass-action law[m⁻⁶]
Symbol
Meaning
SI unit
ni
intrinsic carrier concentration
m⁻³
Nc, Nv
effective densities of states
m⁻³
Eg
band gap (use k in eV/K for eV)
J
k
Boltzmann constant
J/K
T
temperature
K
n, p
electron and hole concentrations
m⁻³
Dimensions both sides agree on every line ([ ] in SI base units) · dimensionless: Eg/2kT
Conditions and range
Non-degenerate semiconductors (Fermi level several kT from the band edges). Nc and Nv scale as T3/2; Eg also varies slightly with T (1.12 eV for Si at 300 K).
Common mistakes
Dropping the factor 2 (Eg/kT); mixing eV and J.
Example
Si (Eg = 1.12 eV, ignoring the temperature dependence of Eg) from 300 K to 350 K: ni(350)/ni(300) = (350/300)^(1.5) × exp[−1.12/(2 × 8.617 × 10⁻⁵) × (1/350 − 1/300)] = 27.8. ni at 300 K is about 10¹⁰ cm⁻³ (Sze).
Rs = 1/(q∫₀xj μ(N) N(x) dx)with a doping profile[kg·m²·s⁻³·A⁻²]
Rs ≈ 1/(qμN xj)uniform layer[kg·m²·s⁻³·A⁻²]
Symbol
Meaning
SI unit
Rs
sheet resistance (Ω/□)
Ω
ρ
layer resistivity
Ω·m
xj
junction depth (layer thickness)
m
q
elementary charge
C
μ
carrier mobility (depends on N)
m²/(V·s)
N
net dopant concentration
m⁻³
x
depth
m
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Thin layers with current parallel to the surface. The four-point probe value Rs = 4.53 V/I holds for samples much wider than the probe spacing and thin. N is the net concentration (N − NB) for a layer of opposite type to the substrate.
Common mistakes
Using one mobility regardless of doping (it falls at high N); xj in µm.
Example
An n-type layer uniformly doped to 10¹⁹ cm⁻³ (= 10²⁵ m⁻³), 0.5 µm deep (electron mobility 114 cm²/(V·s) from Arora, as in the Diffusion and Surface-Treatment Simulator): Rs = 1/(qμNxj) = 1/(1.602 × 10⁻¹⁹ × 0.0114 × 10²⁵ × 0.5 × 10⁻⁶) = 109 Ω/□.
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
All charge goes to metal dissolution (100 % current efficiency) and attack is uniform. Localised attack such as pitting goes much deeper than the average rate.
Common mistakes
Dropping n; converting µA/cm² (1 µA/cm² = 10⁻² A/m²); mixing cm and µm with g/cm³.
Example
Iron (M = 55.845 g/mol, ρ = 7.87 g/cm³) dissolving uniformly as Fe²⁺ at 1 µA/cm²: per cm² and second 10⁻⁶ × 55.845/(2 × 96485) = 2.89 × 10⁻¹⁰ g, divided by density 3.68 × 10⁻¹¹ cm/s, so 11.6 µm per year (365 days). A current of 0.5 A for 30 days removes m = 0.5 × 2.592 × 10⁶ × 55.845/(2 × 96485) = 375 g.
E = E° − (0.0592/n) log₁₀Q25 °C (V)[kg·m²·s⁻³·A⁻¹]
Symbol
Meaning
SI unit
E, E°
cell potential, standard potential
V
R
gas constant
J/(mol·K)
T
temperature
K
n
electrons transferred
1
F
Faraday constant
C/mol
Q
reaction quotient
1
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Equilibrium potential (no net current). In dilute solutions use concentrations (mol/L) for activities. 0.0592 is (RT/F) ln 10 at 25 °C.
Common mistakes
Inverting Q relative to the reaction direction; mixing log₁₀ and ln; dropping n.
Example
Cell of Ni (Ni²⁺ 10⁻³ M) and Cd (Cd²⁺ 0.5 M) at 25 °C, the same numbers as the Callister chapter 17 example: for Ni²⁺ + Cd → Ni + Cd²⁺, E° = −0.250 − (−0.403) = 0.153 V. E = 0.153 − (RT/2F) ln(0.5/10⁻³) = 0.153 − 0.01285 × 6.215 = 0.073 V. Still positive, so the reaction runs as in the standard cell (Cd dissolves).
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
A rough guide to whether an oxide scale protects. Below 1 the scale is porous; 1–2 tends to be protective; well above 2 compressive stress causes spalling. Many exceptions (adhesion, diffusivity), so use as a guide only.
Common mistakes
Dropping n (2 for Al₂O₃); swapping the oxide and metal molar masses.
Example
Al₂O₃ (M = 101.96 g/mol, ρ = 3.95 g/cm³) on Al (26.982, 2.70): PBR = 101.96 × 2.70/(2 × 26.982 × 3.95) = 1.29 (between 1 and 2, protective). MgO (40.304, 3.58) on Mg (24.305, 1.74): 0.81 (below 1, porous and non-protective).
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Parabolic for high-temperature oxidation limited by ionic diffusion through a dense scale; not for very thin early films or cracking/spalling scales. In mass-gain form x becomes Δm/A and kp changes units.
Common mistakes
Assuming doubling the time doubles the thickness (√2 for parabolic); mixing thickness and mass-gain rate constants.
Example
Parabolic oxidation giving 2.0 µm in 10 h: kp = 2.0²/10 = 0.40 µm²/h. After 100 h, x = √(0.40 × 100) = 6.32 µm (ten times the time, only √10 times the thickness).
T₁(C + log₁₀t₁) = T₂(C + log₁₀t₂)two conditions at the same stress[K]
Symbol
Meaning
SI unit
P
Larson–Miller parameter (a function of stress)
K
T
absolute temperature
K
C
material constant (about 20 for many metals)
1
tr
rupture time (the number in hours goes into the log)
h
t₀
1 h (log₁₀tr means log₁₀(tr/t₀))
h
Dimensions both sides agree on every line ([ ] in SI base units)
Conditions and range
Trading temperature against time for creep rupture at the same stress (estimating long low-temperature lives from short hot tests). C refers to time in hours; 20 is typical (15–25). Keep extrapolations within about ten times the test duration.
Common mistakes
Using °C; using ln instead of log₁₀; time in seconds (C = 20 is for hours).
Example
An alloy that ruptured in 50 h at 900 °C under a given stress, used at 800 °C under the same stress (C = 20): P = 1173.15 × (20 + log₁₀50) = 25456 K, tr = 10^(25456/1073.15 − 20) = 5260 h. Lowering the temperature by 100 K extends the life about 105 times.
Dimensions both sides agree on every line ([ ] in SI base units) · dimensionless: Qc/RT
Conditions and range
Above about 0.4 Tm where dislocation creep dominates. At low stress and fine grains diffusion creep (n ≈ 1) takes over; at high stress the power law breaks down. Deformation-mechanism maps show which applies.
Common mistakes
Assuming doubling the stress doubles the rate (2ⁿ); mixing eV and J/mol for Q.
Example
A material with n = 5 and Q = 300 kJ/mol: raising the stress by 1.2 times multiplies the strain rate by 1.2⁵ = 2.49; going from 900 K to 950 K multiplies it by exp[(300000/8.314)(1/900 − 1/950)] = 8.25.
Type a formula name, symbol or term (“Bragg”, “KIC”, “diffusion length”) in the search box to keep only the matching formulas, or narrow them with the topic buttons. Each heading has a link (#f-…) that jumps straight to that formula.
Tick “Pick” on a formula to add it to your own formula sheet. “Print picked only” prints the formulas and symbol tables of the picked items on A4 in two columns (10–15 formulas fit on one or two pages). Tick “Also print conditions and examples” to include those as well.
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Choosing the right formula
Diffusion: if the concentration profile does not change with time (steady flow through a membrane), use Fick’s first law; the flux J is constant and the profile is linear across the thickness. For carburising and dopant diffusion, where concentration changes with time, use Fick’s second law: the erfc solution for a constant surface concentration, the Gaussian solution for a fixed total amount (drive-in after a short pre-deposition). Both assume the diffusion length √(Dt) is small compared with the specimen (semi-infinite).
Stress and strain: engineering stress F/A₀ and strain Δl/l₀ from a tensile test are used for design (yield and tensile strength). Work hardening (Hollomon) and large deformations need true stress σ(1 + ε) and true strain ln(1 + ε). The conversion assumes constant volume and uniform deformation, so it holds only up to necking (maximum load); after necking measure the cross-section.
Rule of mixtures: the longitudinal modulus follows the isostrain rule (upper bound, Voigt) and the transverse modulus roughly the isostress inverse rule (lower bound, Reuss). Real transverse moduli lie somewhat above the lower bound, which the Halpin–Tsai equations capture. For particle-filled materials the true value lies between the two bounds.
Fracture: stresses at holes and notches use the stress concentration factor Kt, but for a sharp crack Kt becomes unbounded. For cracks use the stress intensity factor K = Yσ√(πa) and compare it with the fracture toughness KIC. Crack growth under cyclic load follows the Paris law; fatigue of uncracked parts uses the S–N curve and Miner’s rule.
Thermal stress: a bar constrained in one direction gives EαΔT; a plate or thin film constrained in both in-plane directions gives EαΔT/(1 − ν). With partial constraint (two bars joined together, bimetals) solve from force balance and compatibility of elongation.
How the dimensions were checked
For all 85 formulas, the SI units in the symbol tables were reduced to base units (kg, m, s, K, mol, A) and a program checked that both sides of each equation have the same dimensions and that the arguments of exp, ln, erf and trigonometric functions, and symbolic exponents, are dimensionless. The [ ] to the right of each line shows the dimensions of both sides.
Some formulas expect numbers in a fixed unit, such as log₁₀tr in hours (Larson–Miller), the ASTM grain size number or the Paris constant C. The unit is given in the symbol table; with a different unit the constants change.
Unit conversions
Conversions for turning values into SI units before using a formula (after NIST SP 811).
Unit
In SI
Used for
1 MPa
10⁶ Pa = 1 N/mm²
stress
1 ksi
6.8948 MPa
stress (1000 lbf/in²)
1 ksi√in
1.0988 MPa√m
fracture toughness
1 Å
0.1 nm = 10⁻¹⁰ m
length (lattice parameter, wavelength)
1 eV
1.602176634 × 10⁻¹⁹ J
energy
1 eV/atom
96.485 kJ/mol
activation energy
1 cal
4.184 J
heat (thermochemical calorie)
T [K]
t [°C] + 273.15
temperature (a difference of 1 K equals 1 °C)
1 g/cm³
10³ kg/m³
density
1 cm²/s
10⁻⁴ m²/s
diffusion coefficient
1 cm⁻³
10⁶ m⁻³
carrier or dopant concentration
1 cm²/(V·s)
10⁻⁴ m²/(V·s)
mobility
1 Ω·cm
10⁻² Ω·m
resistivity (1 µΩ·cm = 10⁻⁸ Ω·m)
1 µA/cm²
10⁻² A/m²
current density
1 atm
101325 Pa
pressure (1 bar = 10⁵ Pa, 1 Torr = 133.322 Pa)
1 in
25.4 mm
length
1 lbf
4.448222 N
force
1 year (365 d)
3.1536 × 10⁷ s
time (corrosion rates)
Physical constants
CODATA 2022 recommended values (Mohr et al., Rev. Mod. Phys. 97, 025002 (2025); physics.nist.gov/cuu). “Exact” marks constants fixed by the 2019 SI and those derived from them. Figures in parentheses are the standard uncertainty in the last digits. The examples use these values (the drill uses rounded values such as R = 8.314).
Symbol
Constant
Value
Unit
Exact
NA
Avogadro constant
6.022 140 76 × 10²³
mol⁻¹
✓
k
Boltzmann constant
1.380 649 × 10⁻²³
J/K
✓
k
Boltzmann constant (eV)
8.617 333 262… × 10⁻⁵
eV/K
✓
R
molar gas constant NA k
8.314 462 618…
J/(mol·K)
✓
e
elementary charge
1.602 176 634 × 10⁻¹⁹
C
✓
F
Faraday constant NA e
96 485.332 12…
C/mol
✓
h
Planck constant
6.626 070 15 × 10⁻³⁴
J·s
✓
c
speed of light in vacuum
299 792 458
m/s
✓
σSB
Stefan–Boltzmann constant
5.670 374 419… × 10⁻⁸
W/(m²·K⁴)
✓
ε₀
vacuum electric permittivity
8.854 187 8188(14) × 10⁻¹²
F/m
μ₀
vacuum magnetic permeability
1.256 637 061 27(20) × 10⁻⁶
N/A²
me
electron mass
9.109 383 7139(28) × 10⁻³¹
kg
mu
atomic mass constant (1 u)
1.660 539 068 92(52) × 10⁻²⁷
kg
Main sources
Callister 10th ed.: W. D. Callister, Jr., D. G. Rethwisch, Materials Science and Engineering: An Introduction, 10th ed., Wiley (2018)
Cullity 3rd ed.: B. D. Cullity, S. R. Stock, Elements of X-Ray Diffraction, 3rd ed., Prentice Hall (2001)
Askeland 7th ed.: D. R. Askeland, W. J. Wright, The Science and Engineering of Materials, 7th ed., Cengage (2016)
Williamson & Hall 1953: G. K. Williamson, W. H. Hall, X-ray line broadening from filed aluminium and wolfram, Acta Metall. 1 (1953) 22
Porter 3rd ed.: D. A. Porter, K. E. Easterling, M. Y. Sherif, Phase Transformations in Metals and Alloys, 3rd ed., CRC Press (2009)
Scheil 1942: E. Scheil, Bemerkungen zur Schichtkristallbildung, Z. Metallkd. 34 (1942) 70
Shewmon 2nd ed.: P. G. Shewmon, Diffusion in Solids, 2nd ed., TMS (1989)
Crank 2nd ed.: J. Crank, The Mathematics of Diffusion, 2nd ed., Oxford University Press (1975)
Sze 3rd ed.: S. M. Sze, K. K. Ng, Physics of Semiconductor Devices, 3rd ed., Wiley (2007)
Hull & Bacon 5th ed.: D. Hull, D. J. Bacon, Introduction to Dislocations, 5th ed., Butterworth-Heinemann (2011)
ASTM E112: ASTM E112, Standard Test Methods for Determining Average Grain Size
Dowling 4th ed.: N. E. Dowling, Mechanical Behavior of Materials, 4th ed., Pearson (2013)
Ashby 5th ed.: M. F. Ashby, Materials Selection in Mechanical Design, 5th ed., Butterworth-Heinemann (2017)
Anderson 4th ed.: T. L. Anderson, Fracture Mechanics: Fundamentals and Applications, 4th ed., CRC Press (2017)
ASTM E399: ASTM E399, Standard Test Method for Linear-Elastic Plane-Strain Fracture Toughness of Metallic Materials
Paris & Erdogan 1963: P. Paris, F. Erdogan, A critical analysis of crack propagation laws, J. Basic Eng. 85 (1963) 528
Weibull 1951: W. Weibull, A statistical distribution function of wide applicability, J. Appl. Mech. 18 (1951) 293
Timoshenko & Goodier 3rd ed.: S. P. Timoshenko, J. N. Goodier, Theory of Elasticity, 3rd ed., McGraw-Hill (1970)
Stoney 1909: G. G. Stoney, The tension of metallic films deposited by electrolysis, Proc. R. Soc. Lond. A 82 (1909) 172
Timoshenko 1925: S. Timoshenko, Analysis of bi-metal thermostats, J. Opt. Soc. Am. 11 (1925) 233
Kingery 2nd ed.: W. D. Kingery, H. K. Bowen, D. R. Uhlmann, Introduction to Ceramics, 2nd ed., Wiley (1976)
Manson 1966: S. S. Manson, Thermal Stress and Low-Cycle Fatigue, McGraw-Hill (1966)
Kittel 8th ed.: C. Kittel, Introduction to Solid State Physics, 8th ed., Wiley (2005)
Hull & Clyne 2nd ed.: D. Hull, T. W. Clyne, An Introduction to Composite Materials, 2nd ed., Cambridge University Press (1996)
Halpin & Kardos 1976: J. C. Halpin, J. L. Kardos, The Halpin–Tsai equations: a review, Polym. Eng. Sci. 16 (1976) 344
Jones 2nd ed.: R. M. Jones, Mechanics of Composite Materials, 2nd ed., Taylor & Francis (1999)
Kelly & Tyson 1965: A. Kelly, W. R. Tyson, Tensile properties of fibre-reinforced metals, J. Mech. Phys. Solids 13 (1965) 329
Schapery 1968: R. A. Schapery, Thermal expansion coefficients of composite materials based on energy principles, J. Compos. Mater. 2 (1968) 380
IEC 60028: IEC 60028, International standard of resistance for copper
Pilling & Bedworth 1923: N. B. Pilling, R. E. Bedworth, The oxidation of metals at high temperatures, J. Inst. Met. 29 (1923) 529
Birks et al. 2nd ed.: N. Birks, G. H. Meier, F. S. Pettit, Introduction to the High-Temperature Oxidation of Metals, 2nd ed., Cambridge University Press (2006)
Larson & Miller 1952: F. R. Larson, J. Miller, A time-temperature relationship for rupture and creep stresses, Trans. ASME 74 (1952) 765
FAQ
What is the [ ] to the right of each formula?
The dimensions of both sides of that line in SI base units (kg, m, s, K, mol, A). Stress, for example, is kg·m⁻¹·s⁻² (= Pa); dimensionless quantities show 1. Use it to check the unit of your answer. Every formula has been checked by a program to have equal dimensions on both sides.
How many pages does printing the picked formulas take?
With formulas and symbol tables only, 10–15 formulas take roughly one or two A4 pages in two columns. Including conditions and examples makes it longer. The header, ads and search box are not printed, and the page reads fine with background graphics turned off.
Example values sometimes differ slightly from the drill or calculators.
Because constants are rounded differently. The examples here use CODATA 2022 values (R = 8.314462618 J/(mol·K) and so on), while drill questions state rounded values such as R = 8.314. The difference is in the fourth digit or beyond. Material properties are the same as in the calculators’ data.
Can I use this as a crib sheet for an exam?
Yes. Pick the formulas you need and use “Print picked only”. What you may bring in (examples, explanations) depends on the exam, so check the rules.
Are some formulas missing?
This sheet covers the formulas used by the drill and the calculators plus common textbook formulas. For crystal symmetry, elasticity of dislocations or detailed transformation kinetics, see a textbook.