When temperature changes, materials expand or contract, and if that movement is prevented, thermal stress appears. Pick a material and calculate free expansion, the stress for each type of constraint, the stress from temperature differences inside plates and cylinders, bonded pairs of materials (bimetal, thin film, solder), the temperature drop that cracks a part on quenching, and expansion matching for shrink fits and glass seals. The default is a structural steel bar fixed at both ends and heated from 20 °C to 70 °C, which gives a compressive stress of about 130 MPa.
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Results
Material values are typical values. Real values change with composition, processing and temperature, so for design enter the values from the data sheet of your material. The calculations are elastic and do not include stress relaxation by plasticity or creep.
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Why solids expand: the asymmetric potential
The potential energy between two atoms rises steeply when they come closer and only gently when they move apart, so the well is asymmetric. As temperature rises and atoms vibrate with larger amplitude, they swing further toward the gentle (far) side, and the mean interatomic distance grows. That is thermal expansion. If the potential were a perfect parabola (harmonic vibration), larger vibrations would not move the mean position and there would be no expansion. By the Grüneisen relation the volume expansion coefficient is proportional to the heat capacity at constant volume, so at low temperature it falls steeply like the heat capacity. For the copper in the library (NMIJ certified values), α is 0.28×10⁻⁶ K⁻¹ at 20 K, 10.5×10⁻⁶ K⁻¹ at 100 K and 16.7×10⁻⁶ K⁻¹ at 300 K. Silicon has negative α below about 120 K (about −0.47×10⁻⁶ K⁻¹ near 80 K) and expands on cooling, because some lattice vibrations stiffen when the volume grows (negative Grüneisen parameters).
Why α differs between materials
Materials with strong bonds and deep potential wells stretch less at a given temperature and tend to have high melting points. Near room temperature α is 4.5 for tungsten, 11.8 for iron, 16.5 for copper, 23.1 for aluminum and 28.9 for lead (×10⁻⁶ K⁻¹): the higher the melting point, the smaller α. Covalent ceramics are low (silicon nitride 2.8, silicon carbide 3.7, mean over 40–400 °C), and polymers, whose chains are held together by weak forces, are 60–200. Silica glass is as low as 0.5 because the network of SiO₄ tetrahedra can absorb the stretching by rotating its bonds. In Invar (Fe-36Ni) the volume expansion from magnetism (magnetovolume effect) shrinks as temperature rises and cancels the normal expansion, so below the Curie temperature (about 279 °C) α is only 1–2. Some materials have different α in different directions: an FR-4 type circuit board is 13–14 in-plane but 45×10⁻⁶ K⁻¹ through the thickness (below Tg).
Mean and instantaneous α; length, area and volume
Since α changes with temperature, check which meaning a data source uses. The instantaneous α is the fractional elongation per kelvin at a temperature, (1/L)(dL/dT). The mean α is the elongation from a reference temperature (20 °C or 25 °C) to a given temperature divided by the temperature difference; data sheet values such as "30–400 °C" are mean values. The length change is ΔL = L·∫α dT, which is αLΔT for constant α. This page integrates tables of instantaneous α with the trapezoidal rule and converts tables of mean α to thermal strain at each point. Alumina, for example, has an instantaneous α of 5.4 at 20 °C and 9.4×10⁻⁶ K⁻¹ at 827 °C. For isotropic materials the area changes by (1 + ε)² − 1 ≈ 2ε and the volume by (1 + ε)³ − 1 ≈ 3ε, so the volume expansion coefficient is about 3α.
Constraint and thermal stress: uniaxial, biaxial, triaxial and partial
A free body simply grows by the thermal strain αΔT with no stress. If the growth is prevented, an elastic strain appears that cancels the thermal strain, and that is the thermal stress. For a bar between rigid walls σ = −EαΔT (compression on heating). For a plate or a thin film on a substrate held in two in-plane directions, the stress in one direction partly cancels the strain in the other through the Poisson effect, so a larger stress is needed and σ = −EαΔT/(1 − ν). With all three directions held, the stress is hydrostatic, σ = −EαΔT/(1 − 2ν), which becomes very large as ν approaches 0.5. If the constraint also expands (a frame of another material) use α − αs instead of α; if the constraint is compliant like a spring, the bar and spring act in series and the stress is multiplied by η = 1/(1 + EA/(kL)). For full constraint the thermal stress does not depend on length. Length decides the amount of movement, which is why rails and bridges have expansion gaps (joints).
Stresses from temperature differences: plates and cylinders
Even without external constraint, a temperature difference inside a part creates stress, because hot regions are held back by cold ones. A plate with a linear temperature profile through its thickness has no stress if its edges are free; it simply bows with curvature κ = αΔT/h. If it is held flat, the faces carry ±EαΔT/(2(1 − ν)), with the hot face in compression and the cold face in tension. A non-linear profile (for example during quenching) leaves stress even in a free plate. For a long hollow cylinder with a steady inside–outside difference (temperature linear in ln r), the Timoshenko & Goodier solution gives the stresses: with the inside hotter, the inner surface is in hoop and axial compression and the outer surface in tension. For a thin wall the surface stresses approach ±EαΔT/(2(1 − ν)). These stresses cause cracking and fatigue in heat-exchanger tubes and furnace and engine parts.
Bimetal strips: the Timoshenko formula
Two bonded strips with different expansion coefficients bend uniformly when heated, with the higher-expansion side on the outside (convex); on cooling they bend the other way. From the force and moment balance of the two layers, Timoshenko (1925) obtained 1/ρ = 6(α2 − α1)ΔT(1 + m)²/(h[3(1 + m)² + (1 + mn)(m² + 1/(mn))]), where m = t1/t2, n = E1/E2 and h is the total thickness. For equal thickness and modulus, 1/ρ = 3ΔαΔT/(2h). For a fixed total thickness the curvature is largest at m = 1/√n, and with matched thicknesses the modulus ratio has little effect. Curvature is proportional to ΔαΔT and inversely proportional to the thickness, so thin strips bend more. Bimetals are used in thermostats, thermometers and circuit breakers, often with Invar or a similar alloy as the low-expansion side.
Thermal stress in thin films and the Stoney equation
When a film deposited at high temperature is cooled to room temperature, the mismatch in expansion leaves a biaxial stress in the film. For a thick substrate σf = Ef/(1 − νf)·(αs − αf)ΔT (ΔT = room temperature − deposition temperature): a film with larger α (Al on Si) ends in tension and a film with smaller α (SiO₂ on Si) in compression. The film stress bends the substrate, and the curvature is given by the Stoney equation κ = 6σf·tf/(Ms·ts²), where Ms = Es/(1 − νs) is the biaxial modulus of the substrate. The equation does not contain the film modulus, so measuring the wafer curvature by laser before and after deposition gives the film stress without knowing the film properties. This is the standard way to measure film stress in semiconductor processing. The biaxial modulus of a Si (100) wafer is about 180 GPa, a different quantity from the in-plane E (130 GPa) and ν (0.28). For thicker films (tf/ts above a few per cent), the Stoney equation departs from the bilayer formula.
Thermal strain and life of solder joints
Each temperature cycle, the expansion mismatch between a component (ceramic or silicon) and a circuit board (about 13×10⁻⁶ K⁻¹ in-plane for FR-4 types) shears the solder joints. The displacement at a distance LD from the center (the neutral point) is LD·|Δα|·ΔT, and for solder of height h the shear strain is Δγ ≈ F·LD·|Δα|·ΔT/h. Corners of large components and low joints see the largest strain and the shortest life. From fatigue tests of eutectic SnPb solder, Engelmaier proposed a Coffin–Manson type estimate, Nf = ½(Δγ/(2εf′))^(1/c), with εf′ = 0.325 and a negative exponent c that depends on the mean temperature and the cycle frequency. It is an empirical fit to SnPb tests and does not apply directly to lead-free solders, underfilled joints or components whose leads flex. Use it as an estimate for comparisons.
Thermal shock: R, R′ and the Biot number
When a hot body is quenched, the surface tries to shrink first but is held back by the interior, so the surface goes into tension. If the surface cooled instantly and was fully constrained, the stress would be EαΔT/(1 − ν), so the temperature difference that cracks it is R = σf(1 − ν)/(Eα) (Kingery's first thermal shock resistance parameter, 1955). In practice the surface cools at a rate set by heat transfer, and the stress is lower when the Biot number β = ah/κ (a half-thickness, h heat transfer coefficient) is small. Manson (1953) approximated the fraction of the maximum stress in an infinite plate as 1/(1.5 + 3.25/β − 0.5e^(−16/β)). For large β, ΔTc → R; for small β, ΔTc ≈ 3.25R′/(ah), where R′ = Rκ includes the thermal conductivity. Fused silica has a very small α and a large R; soda-lime glass, with a large α and low strength, has a small R. Silicon nitride, strong and low in α, is among the most shock-resistant ceramics. Cooling puts the surface in tension and heating in compression; brittle materials are strong in compression, so quenching is the more dangerous direction.
Expansion matching: shrink fits and clearances
In a shrink fit the part with the hole is heated to open it, the shaft is inserted, and the interference is locked in by pressure after cooling. The required temperature rise is (δ + c)/(αd), with δ the interference and c the clearance needed for insertion. In a cryogenic (expansion) fit the shaft is shrunk instead, often with liquid nitrogen (−196 °C). The contact pressure follows from the Lamé equations for thick-walled cylinders, and the bore of the outer part carries a hoop tension p(D² + d²)/(D² − d²). With dissimilar materials the interference changes at the service temperature: an aluminum part shrunk onto a steel shaft loses interference when heated, because the aluminum expands more. Clearances behave the same way: pairs with different α, such as an aluminum piston in a cast iron cylinder, are dimensioned for the clearance at operating temperature.
Glass-to-metal seals, Invar and Kovar
In a fused glass-to-metal seal, the difference in contraction from the temperature where the glass sets (the strain point or Tg is a guide) down to room temperature becomes stress. Glass is weak in tension, so the expansions are matched as closely as possible (matched seals), and any mismatch is arranged to leave the glass slightly in compression (compression seals). Kovar (Fe-29Ni-17Co) is specified in ASTM F15 with a mean α of 4.60–5.20×10⁻⁶ K⁻¹ over 30–400 °C, matched to hard (borosilicate) sealing glasses with α near 5. Even among borosilicates, BOROFLOAT 33 (3.25) differs a lot, so the choice of glass matters. Platinum (8.8) is close to soda-lime glass (9.0). Invar and Kovar owe their low α to the magnetovolume effect, and α rises above the Curie temperature (about 279 °C for Invar, about 435 °C for Kovar). A seal has to be matched over the whole temperature range, including this bend.
Practice: rails and bridges, semiconductor packages, glass, engines
A 50 m steel girder grows or shrinks by 12×10⁻⁶ × 50 m × 60 K = 36 mm when the temperature changes by 60 K, and expansion joints take up this movement. Continuously welded rail is instead restrained so it cannot grow, and the resulting axial force is designed for (with attention to buckling in hot weather). In semiconductor packages, the mismatch between the silicon die (about 2.6), molding compound, copper lead frame (about 17) and board causes warpage and delamination of solder and adhesive interfaces. Window glass can crack when sunlight heats the center while the shaded edge stays cool, putting the edge in tension (thermal breakage); makers specify an allowable temperature difference (80–175 K for BOROFLOAT 33 depending on the test). In engines, designers handle the expansion mismatch between aluminum pistons and cast iron or steel parts, thermal stress in exhaust systems, and the mismatch between thermal barrier coatings and metal on turbine blades, a cause of spallation.
Limits of these calculations
All formulas here are elastic. If the thermal stress exceeds the yield stress, a metal deforms plastically, and returning to the original temperature leaves residual stress of the opposite sign. At high temperatures (above about 0.4 of the absolute melting point) creep relaxes stress, so stresses held for a long time become smaller than calculated. E, yield strength and thermal conductivity also change with temperature; this page includes only the temperature dependence of α. Calculations that cross temperatures where properties change abruptly, such as the steel transformation (about 750–860 °C), the zirconia phase transformation or a polymer glass transition, cannot be used as they are. Composites, single crystals and rolled products have direction-dependent α and E (anisotropy), while most formulas here assume isotropic materials. The strength of ceramics and glasses scatters widely (Weibull statistics) and falls for larger parts and deeper flaws, so the thermal shock ΔTc is an estimate.
Common mistakes
Forgetting to divide by (1 − ν) for a thin film or an in-plane constrained plate underestimates the stress by about 30 %. The sign is easy to get right by asking which way the body would move without the constraint: heating with growth prevented gives compression, cooling with shrinkage prevented gives tension. Mixing mean and instantaneous α, or mean values with different reference temperatures, makes non-negligible errors over wide ranges. Using the flexural strength in R overestimates the strength of large parts. Putting the substrate E directly into the Stoney equation is wrong; use the biaxial modulus E/(1 − ν) (about 180 GPa for Si (100)). For checking orders of magnitude, remember 200 GPa × 12×10⁻⁶ K⁻¹ = 2.4 MPa/K.
Values and sources in the materials library
For pure metals, α, ν and thermal conductivity come from tables compiling CRC Handbook and similar values (near 25 °C), and E and density are the same values as in this site's periodic table for materials. Temperature data are for Al (Hidnert 1924 equation, 20–600 °C), structural steel (EN 1993-1-2 formula, 20–1200 °C), Cu (NMIJ CRM 5805-a, 20–300 K), Si (Swenson 1983 recommended values, 100–800 K), alumina (NMIJ CRM 5801-a, 293–1100 K), silica glass (NMIJ CRM 5802-a, 293–800 K), and Invar and Kovar (makers' tables of mean α). Ceramics are from Kyocera data sheets, glasses from ISO 16293-1, SCHOTT and Corning, and the FR-4 type board from the Isola 370HR data sheet. Polymers, CFRP and GFRP use the geometric mean of the same ranges as the material selection chart (Ashby's tables); ν is not given there and is left blank. The full list is in the "Materials library" tab.
AThe interatomic potential is asymmetric (anharmonic), so as the vibration amplitude grows, the average distance between atoms increases
BThe interatomic potential is parabolic (harmonic), and both the vibration amplitude and the average interatomic distance grow with temperature
CAs temperature rises, the atoms themselves become larger
Show answer and explanation
Answer: A. The interatomic potential is asymmetric (anharmonic), so as the vibration amplitude grows, the average distance between atoms increases
The interatomic potential has an asymmetric shape: repulsion rises steeply as atoms approach, while on the separating side the attraction weakens gradually. As temperature rises and the vibration amplitude grows, atoms swing farther toward the separating side, so the average interatomic distance, the center of the vibration, increases, and this is thermal expansion. If the potential were parabolic (a harmonic oscillator), the vibration would remain symmetric about the equilibrium position however large the amplitude, so the average distance would not change and no thermal expansion would occur. The size of the atoms themselves (the extent of their electron clouds) hardly changes with temperature. Anharmonicity is also the cause of phonon-phonon scattering, which is connected to thermal conductivity having a finite value.
Near room temperature, the linear thermal expansion coefficient is about 100–200×10⁻⁶ K⁻¹ for polyethylene, about 23×10⁻⁶ K⁻¹ for aluminum and about 4.5×10⁻⁶ K⁻¹ for tungsten. Materials with stronger interatomic bonds and deeper potential wells tend to show a smaller increase in interatomic distance from vibration at a given temperature, a higher melting point and a smaller linear expansion coefficient. Tungsten, with a melting point above 3400℃, has the smallest value, and polymers, whose molecular chains are held together by weak van der Waals forces, have the largest. For this reason, parts that combine metals with polymers or ceramics develop thermal stress and warping from the difference in expansion when the temperature changes.
ANi suppresses the lattice vibrations of Fe and raises the Debye temperature greatly, so at room temperature the atoms hardly vibrate
BBelow the Curie temperature, the magnetic volume expansion shrinks as magnetization weakens on heating, canceling normal expansion
COn heating, a gradual transformation from BCC to the more densely packed FCC phase proceeds, and the resulting shrinkage cancels the expansion
Show answer and explanation
Answer: B. Below the Curie temperature, the magnetic volume expansion shrinks as magnetization weakens on heating, canceling normal expansion
Invar is a ferromagnet with a large magnetovolume effect (spontaneous volume magnetostriction), in which spontaneous magnetization lengthens the interatomic distance and enlarges the volume. As temperature rises, the magnetization weakens, so this magnetic expansion decreases and nearly cancels the ordinary thermal expansion from lattice vibration. As a result, the linear expansion coefficient is as small as about 1×10⁻⁶ K⁻¹ below the Curie temperature, and Invar is used in precision measuring instruments, containers for liquefied natural gas and the like. Above the Curie temperature the cancellation disappears and it expands like an ordinary metal. Invar is single-phase FCC at room temperature; atomic vibration grows with temperature as usual, and no phase transformation that offsets the expansion takes place.
AIt keeps nearly the same value as at room temperature even as the temperature falls
BIt grows as the temperature falls and is largest at absolute zero
CIt decreases with temperature just as the heat capacity does, approaching 0 at absolute zero
Show answer and explanation
Answer: C. It decreases with temperature just as the heat capacity does, approaching 0 at absolute zero
By the Grüneisen relation β = γC_v/(BV) (β is the volume expansion coefficient, γ the Grüneisen parameter, C_v the constant-volume molar heat capacity, B the bulk modulus and V the molar volume), the thermal expansion coefficient follows nearly the same temperature dependence as the heat capacity. At temperatures well below the Debye temperature the heat capacity shrinks in proportion to T³, so the linear expansion coefficient of an insulator also approaches 0 in proportion to T³. Thermal expansion arises from the anharmonicity of vibrations, so at very low temperatures, where vibrations are hardly excited, expansion also hardly occurs. Using the room-temperature value all the way down to very low temperatures greatly overestimates the shrinkage on cooling. Above the Debye temperature the heat capacity is nearly constant, so the linear expansion coefficient is also nearly constant.
AIts linear expansion coefficient is very small, about 0.5×10⁻⁶ K⁻¹, so the thermal stress from a temperature difference is small
BIts thermal conductivity is as high as a metal's, so the surface–interior temperature difference vanishes quickly
CIts fracture toughness is as high as a metal's, so even if thermal stress starts surface cracks they hardly grow
Show answer and explanation
Answer: A. Its linear expansion coefficient is very small, about 0.5×10⁻⁶ K⁻¹, so the thermal stress from a temperature difference is small
On quenching, only the surface tries to contract first and is held back by the interior, creating a tensile thermal stress σ ≈ EαΔT/(1 − ν) at the surface. The linear expansion coefficient of quartz glass is about 0.5×10⁻⁶ K⁻¹, only about one twentieth of the roughly 9×10⁻⁶ K⁻¹ of soda-lime glass, so the thermal stress is far smaller for the same temperature difference and it does not crack. The thermal conductivity of quartz glass is low, about 1.4 W/(m·K), and its fracture toughness of about 0.8 MPa·√m is on par with soda-lime glass, both far below metals. Thermal conductivity acts through how the temperature difference between surface and interior develops; under mild heat transfer, materials with higher thermal conductivity are more resistant to thermal shock.
Δα = 23×10⁻⁶ − 2.6×10⁻⁶ = 20.4×10⁻⁶ K⁻¹ and ΔT = 100 K, so σ = 70×10⁹ × 20.4×10⁻⁶ × 100/(1 − 0.35) ≈ 2.2×10⁸ Pa, about 220 MPa. On cooling, aluminum tries to shrink more than silicon, but constrained by the substrate it cannot, so the film is pulled into tension. Calling it compressive gets the sign wrong. About 143 MPa forgets to divide by (1 − ν). The film is in a biaxial stress state, constrained simultaneously in both in-plane directions, so the biaxial modulus E/(1 − ν) must be used. Thermal stresses like this cause stress migration and voids in interconnects.
On cooling, silicon shrinks more than SiO₂. The SiO₂ film is made to shrink more than it would on its own, so it is squeezed in plane and develops compressive stress. The sign of thermal stress can be judged this way: if the film's thermal expansion coefficient is larger than the substrate's, the film is in tension after cooling; if smaller, in compression. The relative expansion coefficients are the reverse of those for an aluminum film on silicon, so the stress is of the opposite sign. As long as the expansion coefficients differ, stress develops whenever the temperature changes from the deposition temperature, so the choice that no stress develops is wrong. The compressive stress in thermal oxide reaches a few hundred MPa and can be determined by measuring wafer bow.
Eα = 380×10⁹ × 8.0×10⁻⁶ = 3.04×10⁶ Pa/K and σf(1 − ν) = 300×10⁶ × 0.78 = 2.34×10⁸ Pa, so ΔTc = 2.34×10⁸/(3.04×10⁶) ≈ 77 K. About 99 K forgets to multiply by (1 − ν), and about 127 K divides by (1 − ν) instead. Thermal shock resistance is greater the higher the strength and the lower the Young's modulus and thermal expansion coefficient. Fused silica, with a very small thermal expansion coefficient of about 0.5×10⁻⁶ K⁻¹, is extremely resistant to thermal shock. A high thermal conductivity also helps, because it keeps the temperature difference between the surface and the interior small.
ABeneath a low-conductivity zirconia (YSZ) top coat lies a bond coat such as MCrAlY, and a thin Al₂O₃ thermally grown oxide (TGO) forming at their interface protects the substrate from oxidation
BThe top coat is dense, highly conductive alumina that quickly carries heat out of the blade to cool the metal, while the bond coat absorbs the thermal expansion mismatch with the substrate
CThe YSZ top coat lets no oxygen through at all, so no bond coat is needed, and YSZ sprayed directly onto the substrate blocks both oxidation and heat
Show answer and explanation
Answer: A. Beneath a low-conductivity zirconia (YSZ) top coat lies a bond coat such as MCrAlY, and a thin Al₂O₃ thermally grown oxide (TGO) forming at their interface protects the substrate from oxidation
A TBC is a coating that lowers the metal temperature with an insulating surface layer while the inside of the blade is cooled by cooling air. The YSZ (yttria partially stabilized zirconia) top coat has a low thermal conductivity of about 1 to 2 W/(m·K) and, for a ceramic, a large thermal expansion coefficient. It is made as a porous sprayed coating or a columnar vapor-deposited coating to relieve the strain from the thermal expansion mismatch with the substrate. YSZ conducts oxide ions well and cannot stop oxygen, so Al in the bond coat oxidizes to form a thin Al₂O₃ TGO, which protects the substrate from oxidation. When the TGO thickens or undergoes thermal cycling, the top coat spalls from near it, which is the main factor limiting the life. A material with high thermal conductivity cannot serve as insulation.
FAQ
Does thermal stress depend on the length of a bar?
Not for full constraint. The thermal strain αΔT does not depend on length, so neither does the stress EαΔT. Length decides the movement ΔL = αLΔT, so longer members need larger expansion gaps. With a gap or a spring-like constraint, the length enters the degree of constraint and the stress changes.
Why divide by (1 − ν) for thin films and plates?
Because both in-plane directions are held at once. The stress in x partly cancels the strain in y through the Poisson effect, so cancelling the thermal strain in both directions needs more stress than in the uniaxial case. The effective modulus rises from E to E/(1 − ν), the biaxial modulus: about 1.43 times for ν = 0.3.
How do I tell the sign of the stress on heating and cooling?
Ask which way the body would move without the constraint. Heating with growth prevented gives compression; cooling with shrinkage prevented gives tension. For two materials, the higher-expansion side ends in tension after cooling and the lower-expansion side in compression: an Al film on Si is in tension after cooling, a thermal oxide on Si in compression.
How much does the temperature dependence of α matter?
It depends on the material and range. For aluminum α is about 23 at 20 °C and about 32×10⁻⁶ K⁻¹ at 500 °C (derivative of the Hidnert equation), so heating to 500 °C with the room-temperature α underestimates the expansion by more than 10 %. Invar's α rises several times above its Curie temperature, and silicon's becomes negative at low temperature. With "Use it" selected, materials with data are calculated along their curves, and the chart compares the result with constant α.
Why does the Stoney equation not include the film modulus?
When the film is much thinner than the substrate, the bending is set only by the film force σf·tf per width and by the bending stiffness of the substrate. The film stiffness matters for how large the film stress is, but not for converting stress into curvature. So the film stress can be found from the wafer curvature measured before and after deposition, without knowing the film modulus. For thick films the film's own bending stiffness causes deviations.
Which bimetal combinations bend most?
Curvature is proportional to the expansion mismatch Δα and the temperature change and inversely proportional to the total thickness. For a fixed total thickness it is largest at a thickness ratio m = 1/√n (n the modulus ratio), and for equal thickness and modulus 1/ρ = 3ΔαΔT/(2h). Using a low-expansion alloy such as Invar on the low side increases Δα.
What should I look at to choose a thermal-shock-resistant material?
For severe quenches (high heat transfer) choose a large R = σf(1 − ν)/(Eα); for mild heat transfer choose a large R′ = Rκ, which includes thermal conductivity. High strength and low α and E help; fused silica and silicon nitride are typical examples. The Biot number β = ah/κ tells which case applies.
How hot must the outer part be for a shrink fit?
Hot enough to open the bore by the interference δ plus the insertion clearance c. For constant α, ΔT = (δ + c)/(αd): for steel at 50 mm diameter with 50 µm interference and 20 µm clearance, about 117 K (from 20 °C to about 137 °C). Overheating can change the heat-treated condition of the material, so check the maximum temperature in its data as well.
What is the difference between Invar and Kovar?
Invar (Fe-36Ni) is an alloy with a very small α near room temperature (1–2×10⁻⁶ K⁻¹), used in precision instruments, molds and the low side of bimetals. Kovar (Fe-29Ni-17Co) is a sealing alloy whose α is matched to hard borosilicate-type glasses, with a mean α of about 5×10⁻⁶ K⁻¹ over 30–400 °C. Both show a much larger α above their Curie temperatures.
Can I use these values directly for design?
The material values are typical and vary with composition, processing and temperature. They are fine for understanding and for estimating magnitudes, but for design enter the values from the data sheet or standard of your material. The calculations are elastic and do not include stress relaxation by plasticity or creep or stress concentration from part geometry.