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Materials Science · High temperature, cryogenic conditions and high pressure

Thermal shock and expansion mismatch

Derive how a nonuniform temperature field creates self-equilibrating stress, distinguish diffusion and surface heat-transfer timescales, and calculate uniform-temperature mismatch in a bonded film. Keep fracture and residual-stress conclusions tied to the actual field and history.

Useful background: Energy, stability and environmental context; Heat transport, expansion and thermal response; Fatigue, creep and time-dependent damage. Each worked case explains the quantities it uses.

Back to the chapter

After reading, you can

  • Use diffusivity, Biot number and a stated length scale to decide whether a spatial temperature model is needed.
  • Derive the sign and magnitude of in-plane thermal stress in a symmetric, freely contracting elastic plate.
  • Distinguish a surface-to-average temperature difference from the environmental temperature jump.
  • Combine a supplied geometry-matched crack-driving model with a residual-stress contribution.
  • Calculate a bonded film’s expansion mismatch and explain why slow temperature change need not remove it.
01

One region changes temperature first

A hot surface or rapidly cooled exterior can prefer a different strain from the interior. Internal compatibility creates stress. Thermal conductivity, diffusivity, expansion, stiffness, geometry and fracture resistance all influence the outcome.

For a small temperature change from a stress-free reference state, a freely expanding isotropic material has thermal strain ε_th = αΔT. Here α is the linear expansion coefficient, ΔT = T − T_ref and tensile strain is positive. If α varies appreciably over the interval, replace αΔT by the integral of α(T) with respect to temperature. The same temperature difference in kelvins or Celsius degrees gives the same strain.

The actual strain need not equal that free thermal strain. In a simple uniaxial elastic bar, ε_total = σ/E + ε_th. Free uniform cooling permits ε_total = ε_th and σ = 0. Holding its original length fixes ε_total = 0, giving σ = −EαΔT: cooling produces tension and heating compression, while this elastic constraint remains applicable.

Thermal shock concerns a rapid temperature change that generates a transient field and associated stress. A cold surface initially wants to contract more than its warmer interior; attachment to that interior leaves it in tension. Rapid surface heating can reverse that pattern. Which region is tensile, how large the peak becomes and whether a crack grows require the actual thermal and mechanical boundary conditions.

02

The temperature field has its own model

Consider a broad plate of total thickness 2L, cooled equally on both large faces. Let z measure distance through the thickness, with z = 0 at the centre and surfaces at z = ±L. Away from edges, take one-dimensional conduction, constant conductivity k, density ρ and specific heat c, with no internal heat generation or phase change. Thermal diffusivity is a = k/(ρc), in m²/s. The symbol a here is diffusivity, distinct from the expansion coefficient α.

Energy conservation and Fourier conduction give ∂T/∂t = a∂²T/∂z². Scaling distance by L shows that the diffusion timescale is t_D = L²/a. The dimensionless time Fo = at/L² compares elapsed time with that scale. Fo = 1 is a scale comparison, not a guarantee that every point has reached the environment’s temperature.

If heat exchange at each face is represented by a coefficient h and an environment at T_∞, the outward conductive heat flux equals h(T_s − T_∞). Equivalently, −k∂T/∂n = h(T_s − T_∞), where n points outward and T_s is surface temperature. This boundary model requires a suitable effective h; strong radiation, boiling, contact changes or temperature-dependent exchange may require another relation.

The Biot number Bi = hL/k compares internal conduction resistance with surface exchange resistance for this plate. Bi much smaller than one supports an approximately uniform internal temperature. Large Bi permits the surface to change much faster than the centre; a single average temperature then loses the field needed for stress. Bi is not itself a failure index.

When the uniform-temperature approximation is justified, an energy balance gives τ = ρcV/(hA_s), with A_s the total exchanging area. For a broad plate exchanging through both faces, V/A_s = L, so τ = ρcL/h and (T − T_∞)/(T_i − T_∞) = exp(−t/τ). A spatially uniform lumped model cannot calculate the gradient it deliberately neglects. Even a small Bi does not remove stress from external restraint or a bonded expansion mismatch.

03

A free plate can carry internal stress

Now specify the mechanical model. Use a homogeneous isotropic plate, small strains, linear elasticity, traction-free large faces, and a symmetric temperature field T(z) = T(−z). Examine its broad interior, away from free-edge corrections. Temperature varies through thickness, not along the plate. Mechanical equilibration is assumed fast compared with the thermal change. No applied in-plane force or moment is present, and no plasticity, creep, new crack growth or phase change has yet occurred.

Symmetry removes thermal bending in this model. Bonded layers share a common in-plane total strain ε₀ in both directions. Their free thermal strains differ through thickness. The equal in-plane stresses σ_x = σ_y = σ, with negligible through-thickness stress, satisfy the plane-stress relation ε₀ = (1 − ν)σ/E + αΔT(z). Here ν is Poisson’s ratio. It follows that σ(z) = E[ε₀ − αΔT(z)]/(1 − ν).

The plate is free to contract, so the stress resultant across its thickness must vanish: ∫₋ᴸᴸσ(z) dz = 0. Define the thickness average ⟨ΔT⟩ = (1/2L)∫₋ᴸᴸΔT(z) dz. With constant E, ν and α, force balance gives ε₀ = α⟨ΔT⟩, hence σ(z) = [Eα/(1 − ν)][⟨ΔT⟩ − ΔT(z)]. Symmetry also makes the moment resultant ∫₋ᴸᴸzσ(z) dz zero.

This result explains the signs. A surface colder than the thickness average has positive tensile stress; a warmer centre has compression. The two regions balance internally, even though the whole plate contracts. A uniform temperature change makes the bracket zero everywhere. An asymmetric field can instead produce curvature; an external clamp changes the force and displacement conditions. Neither case can be solved merely by reusing the symmetric free-plate expression.

04

Keep the stress-driving temperature difference

The surface stress in this model depends on ⟨T⟩ − T_s at that instant. It does not generally equal Eα times the difference between the initial specimen and its environment. The environment influences stress through the evolving temperature field and surface exchange. A strong ideal restraint that keeps a cooling surface near its original in-plane dimensions gives the scale Eα|ΔT_s|/(1 − ν), but the freely contracting plate generally has a different ε₀.

For a concrete original snapshot, prescribe ΔT(z) = −60(z/L)² K. This is an illustrative symmetric field supplied for a mechanical calculation, not a solved cooling history or measured temperature record. Its centre has not cooled, its surfaces have cooled by 60 K and its thickness average is −20 K. The average follows because the average of (z/L)² over −L to L is 1/3.

With E = 100 GPa, α = 8 × 10⁻⁶ K⁻¹ and ν = 0.25, the stress factor Eα/(1 − ν) is 1.06667 MPa/K. The common total strain is −160 microstrain. Surface free contraction would be −480 microstrain, leaving +320 microstrain of elastic strain; the centre instead has −160 microstrain of elastic strain. The in-plane stresses are therefore +42.67 MPa at the surfaces and −21.33 MPa at the centre.

The table and graph show this compatibility account. These stresses are a snapshot, not the maximum over an unknown history. If the field later becomes uniform and the plate remained homogeneous, free and fully elastic, the thermal stress vanishes. Permanent residual stress requires some additional retained incompatibility, such as plastic strain, creep, transformation, damage or bonding to a different material.

Calculated symmetric-plate snapshot for the invented field and material inputs. Negative strain is contraction; stress is one of the two equal in-plane components. One microstrain is 10⁻⁶.
Position z/LΔT (K)Free thermal strain (microstrain)Common total strain (microstrain)In-plane stress (MPa)
−1 and +1−60−480−160+42.67
−0.5 and +0.5−15−120−160−5.33
000−160−21.33
Compatible temperature, strain and stress profilesThree aligned calculated profiles across the symmetric free plate, with surfaces at q = −1 and +1 and centre at 0. Supplied temperature change is −60 q² K, with mean −20 K. Free thermal strain is −480 q² microstrain, while common actual total strain is −160 microstrain. One of the equal in-plane stresses is 64 q² − 21.333… MPa: +42.67 MPa at each surface, −21.33 MPa at the centre and zero near q = ±0.577. The stress and moment resultants vanish. Solid blue curves are calculated profiles; dashed orange lines mark the mean temperature or common total strain. This is an original mechanical snapshot, not measured data or a solved cooling history.A contracting plate can have surface tensionInvented symmetric snapshot · common position q = z/LA · Temperature change ΔT (K)-60-40-200-1-0.50+0.5+1Mean −20 KB · In-plane strain (microstrain)-480-320-1600-1-0.50+0.5+1Free thermal strainCommon total strainC · One in-plane stress component (MPa)-2002040-1-0.50+0.5+1Position q = z/LCold surfaces: tension. Warmer centre: compression.Stress and moment resultants are zero in this plate model.
Three aligned calculated profiles across the symmetric free plate, with surfaces at q = −1 and +1 and centre at 0. Supplied temperature change is −60 q² K, with mean −20 K. Free thermal strain is −480 q² microstrain, while common actual total strain is −160 microstrain. One of the equal in-plane stresses is 64 q² − 21.333… MPa: +42.67 MPa at each surface, −21.33 MPa at the centre and zero near q = ±0.577. The stress and moment resultants vanish. Solid blue curves are calculated profiles; dashed orange lines mark the mean temperature or common total strain. This is an original mechanical snapshot, not measured data or a solved cooling history.
05

A stress prediction is not yet a fracture prediction

For a brittle material, tensile regions can open suitably oriented flaws. A tensile-initiation stress taken from a relevant specimen population is conditional on surface condition, stressed volume, environment and loading timescale; it is not automatically transferable to another part. For a known crack in an appropriate linear-elastic, small-process-zone regime, compare its mode-I stress intensity K_I with a resistance established at the relevant temperature, rate and constraint.

Under a geometry-matched uniform remote stress, an expression such as K_I = Yσ√(πa_c) uses crack size a_c and a dimensionless geometry factor Y. A thermal stress varying along the prospective crack path generally needs its own field-dependent solution. Substituting the surface peak into a uniform-stress formula can misrepresent the crack’s driving force. A geometry-matched calculation can instead supply how a stated stress-profile amplitude maps to K_I.

Linear-elastic contributions can be added while the crack geometry and model remain valid: K_total = K_thermal + K_residual + K_applied. A tensile residual contribution can move a crack above its initiation resistance even when the thermal contribution alone is lower. A compressive contribution can reduce opening, but contact or crack closure then needs appropriate treatment; algebraic superposition is not a complete contact model.

Elastic stress calculations also cease to describe the whole response after substantial yielding, creep or damage. Relaxation can reduce an instantaneous peak while leaving a changed strain state for the next reversal. Thermal cycling can therefore accumulate fatigue or crack growth. Surviving one event establishes neither long-cycle life nor an undamaged specimen; post-exposure crack evidence and retained properties answer additional questions.

06

Uniform temperature can still stress a bonded assembly

A film and substrate can be at the same spatially uniform temperature and still prefer different in-plane contractions. Consider a thin, perfectly bonded isotropic film on a much thicker substrate. Take a stress-free bonding reference, constant properties, no slip or inelastic relaxation, and a substrate whose film-induced change of in-plane strain and curvature is negligible in the region considered. This is a substrate-dominated limit, not every bilayer.

Let α_f and α_s be the film and substrate expansion coefficients. The film follows the substrate’s in-plane strain, approximately α_sΔT, while its own free strain would be α_fΔT. The elastic mismatch is (α_s − α_f)ΔT in each in-plane direction. Using the same biaxial plane-stress relation gives σ_f ≈ [E_f/(1 − ν_f)](α_s − α_f)ΔT.

If α_f is larger than α_s, cooling from the bonding reference produces film tension: the film is prevented from contracting as much as it would freely. Reversing the temperature change reverses the elastic sign. A uniform film stress is not the interfacial shear distribution at its edges, and it does not by itself decide delamination. Finite substrate compliance, curvature, patterned edges, anisotropy, yielding and imperfect adhesion require an extended model.

Slowing a temperature change can reduce a transient gradient while leaving this equilibrium mismatch at the final temperature. It may also give creep or other relaxation time, so path and dwell matter if those mechanisms are active. Distinguish a homogeneous free plate’s disappearing elastic gradient stress from an assembly’s retained mismatch under uniform temperature.

Free contraction versus bonded film compatibilityOriginal strain comparison for the specified uniform 100 K cooling of a perfectly bonded film on a dominant thick substrate. Film free thermal strain is −1200 microstrain. Substrate free strain, approximately the bonded film’s actual total strain, is −400 microstrain. Their difference is +800 microstrain of film elastic mismatch, giving +106.67 MPa in each in-plane direction for E_f = 100 GPa and nu_f = 0.25. Horizontal coordinates use a common strain scale; line lengths are strain differences, not a physically scaled drawing of the assembly. No real material data are plotted.Uniform cooling: the film cannot contract freelySubstrate-dominated elastic model · strain is dimensionless-1200-800-4000In-plane strain (microstrain)Free film-1200 microstrainFree substrate ≈ bonded film total-400 microstrainElastic mismatch: +800 microstrainFilm contracts less than freely: +106.67 MPa tension.No temperature gradient is required for this mismatch.
Original strain comparison for the specified uniform 100 K cooling of a perfectly bonded film on a dominant thick substrate. Film free thermal strain is −1200 microstrain. Substrate free strain, approximately the bonded film’s actual total strain, is −400 microstrain. Their difference is +800 microstrain of film elastic mismatch, giving +106.67 MPa in each in-plane direction for E_f = 100 GPa and nu_f = 0.25. Horizontal coordinates use a common strain scale; line lengths are strain differences, not a physically scaled drawing of the assembly. No real material data are plotted.
07

A temperature jump alone is incomplete

Specify heat-transfer rate, geometry, starting state and boundary conditions. Different surface environments can create different gradients for the same nominal temperature difference. Cracks and residual stress influence how close a component is to failure.

For a defensible comparison, connect the quantities causally: surface exchange and conduction determine T(z,t); compatibility and restraints determine stress; crack geometry and resistance or an appropriate inelastic model determine the damage prediction. Confirm temperatures at relevant locations and times, rather than validating only a final average. Check fixture motion or bonding assumptions where they set compatibility.

Low expansion, lower elastic stiffness or greater fracture resistance can reduce a stated cracking tendency, but changing conductivity, thickness or h changes the temperature field too. There is no context-free thermal-shock ranking from one property or one quoted “critical temperature difference”. The models here explain which comparison is being made; they do not prescribe an extreme-temperature experiment or establish a component operating limit.

Worked examples

Same endpoint, different thermal models

Two otherwise identical broad plates reach the same final temperature. Compare two idealised surface-exchange cases using k = 2 W/(m·K), ρ = 2500 kg/m³, c = 800 J/(kg·K), half-thickness L = 5 mm and h = 20 or 2000 W/(m²·K). Compute the diffusion scale and Bi. Which case admits an approximately lumped calculation, and what does that establish about peak stress?

  1. The volumetric heat capacity is ρc = 2500 × 800 = 2.0 × 10⁶ J/(m³·K). Thus a = k/(ρc) = 1.0 × 10⁻⁶ m²/s. Convert L to 0.005 m: t_D = L²/a = 25 s.
  2. At h = 20, Bi = 20 × 0.005/2 = 0.05. Internal conduction is fast compared with surface exchange, so a nearly uniform temperature is a useful first approximation. The lumped time constant is τ = ρcL/h = 500 s.
  3. At t = 500 s in this approximate lumped model, Fo = 20 and (T − T_∞)/(T_i − T_∞) = e⁻¹ ≈ 0.368. For an initial specimen–environment difference of 100 K, about 36.8 K remains. The number describes an approximately uniform temperature, not a calculated local stress.
  4. At h = 2000, Bi = 5. Merely substituting this h into a lumped expression would give 5 s, shorter than the 25 s internal diffusion scale. Because its spatial-uniformity premise is now poor, that 5 s is not a justified whole-plate cooling prediction. A spatial transient solution is required.
  5. The large-Bi case can develop a stronger early surface-to-interior difference, but neither Bi alone nor the common final temperature supplies its peak tensile stress. Use the evolving field with the actual restraints. A small-Bi temperature model also cannot dismiss assembly mismatch.

a = 10⁻⁶ m²/s and t_D = 25 s. Bi is 0.05 or 5. The first case supports approximate lumped cooling with τ = 500 s; the second needs a spatial field. Matching endpoints does not match transient peak stress.

From a temperature profile to a conditional crack prediction

Use ΔT(z) = −60(z/L)² K and E = 100 GPa, α = 8 × 10⁻⁶ K⁻¹, ν = 0.25 for the symmetric free-plate model. Find its average, strain and stress signs. Then assess a stipulated opening crack: a geometry-matched linear-elastic solution supplies K_thermal = C_T σ_s, with C_T = 0.020 √m and σ_s the positive thermal surface-stress amplitude. A specified self-equilibrating residual field of amplitude σ_r = 10 MPa supplies K_residual = C_R σ_r, C_R = 0.025 √m. These original teaching coefficients encode two different field shapes for that crack. Take its matched initiation resistance as K_R = 1.00 MPa√m.

  1. Write q = z/L. Its average q² is (1/2)∫₋₁¹q² dq = 1/3, so ⟨ΔT⟩ = −20 K. The common total in-plane strain is α⟨ΔT⟩ = −160 × 10⁻⁶.
  2. The biaxial thermal factor is Eα/(1 − ν) = (100 000 MPa)(8 × 10⁻⁶ K⁻¹)/0.75 = 1.066666… MPa/K. Thus σ(q) = 1.066666…(−20 + 60q²) MPa.
  3. At q = ±1, σ_s = +42.6667… MPa. At q = 0, σ = −21.3333… MPa. Stress crosses zero at q² = 1/3, or q ≈ ±0.577. Its thickness average is zero; symmetry gives zero moment. Contraction and internal surface tension are compatible.
  4. For the supplied crack solution, K_thermal = 0.020 × 42.6667… = 0.85333… MPa√m, below the stipulated 1.00 MPa√m resistance. This is a field-matched coefficient, not a universal conversion from a surface peak.
  5. The second field contributes K_residual = 0.025 × 10 = 0.250 MPa√m. While linear superposition and the unchanged opening crack remain applicable, K_total = 1.10333… MPa√m. The combined model crosses the supplied initiation resistance.
  6. The residual field is separately specified and self-equilibrating; it is not a claim that a freely cooling elastic homogeneous plate creates a uniform residual tension. The calculation flags initiation for this stipulated crack/model at this snapshot. It does not predict later crack growth, measured life or an unknown component’s capacity.

The average is −20 K and the common strain −160 microstrain. Thermal surface stress is +42.67 MPa and centre stress −21.33 MPa. The supplied crack model gives K_thermal ≈ 0.853 and K_total ≈ 1.103 MPa√m; only the combined case exceeds the stated initiation resistance.

No temperature gradient, but a film in tension

In the substrate-dominated film model, take E_f = 100 GPa, ν_f = 0.25, α_f = 12 × 10⁻⁶ K⁻¹ and α_s = 4 × 10⁻⁶ K⁻¹. The entire bonded assembly finishes 100 K below its stress-free bonding reference. Calculate the film stress. Would a slower change remove it if all behaviour remained elastic and the other assumptions held?

  1. Use the signed change ΔT = −100 K. The film’s free thermal strain would be α_fΔT = −0.0012, or −1200 microstrain.
  2. The thick substrate instead imposes approximately α_sΔT = −0.0004, or −400 microstrain. The film contracts, but by less than it would freely. Its elastic mismatch is −400 − (−1200) = +800 microstrain in each in-plane direction.
  3. The biaxial modulus is E_f/(1 − ν_f) = 100/0.75 = 133.333… GPa. Multiplying by 0.0008 gives σ_f = +0.106666… GPa = +106.67 MPa, a tensile in-plane stress.
  4. Slowing the change can reduce gradients during the path, but the final mismatch is unchanged in this elastic substrate-dominated model. If creep, slip, curvature or damage occurs, this calculation must be extended and the retained state assessed.
  5. No interfacial fracture energy or edge field has been supplied. The computed film stress alone therefore does not establish whether the film delaminates or survives repeated cycles.

Approximately +106.67 MPa in each in-plane direction. Slow, uniform cooling does not by itself remove elastic expansion mismatch at the stated final temperature.

Try the reasoning

  1. Two invented brittle materials have identical α, ν, temperature profile and applicable tensile-initiation conditions. Material A has E = 100 GPa and a matched initiation stress 80 MPa; B has E = 200 GPa and initiation stress 120 MPa. For the profile above, does B’s higher strength alone imply a better margin?

    Show the answer and reasoning

    No. The thermal surface stress doubles from 42.67 to 85.33 MPa. Stress divided by the stipulated initiation value rises from about 0.533 to 0.711.

    This controlled comparison holds the field fixed to isolate stiffness and the conditional strength criterion. The 50% strength rise is smaller than the 100% stress rise. Different transport, flaws or temperature-dependent properties would require a fresh comparison; neither ratio is a real design safety factor.

  2. Use E = 100 GPa and α = 8 × 10⁻⁶ K⁻¹ for a uniform 60 K cooling. Compare an unrestrained homogeneous bar with the same bar held at its original length, assuming small elastic strain. Does either calculation automatically equal the symmetric plate snapshot stress?

    Show the answer and reasoning

    The free bar has zero stress. The length-fixed bar has σ = −EαΔT = +48 MPa in axial tension. Neither is automatically the plate’s +42.67 MPa surface stress.

    The bar constraints and uniaxial relation differ from a plate with two equal in-plane stresses, free contraction and a temperature gradient. A uniform biaxially length-fixed plate would instead have +64 MPa in each in-plane direction for these inputs; naming the constraint selects the formula.

  3. Halve L to 2.5 mm while keeping k, ρ, c and h = 2000 W/(m²·K) from the first worked case. Find t_D and Bi. Does this prove thermal stress becomes one quarter as large?

    Show the answer and reasoning

    t_D becomes 6.25 s and Bi becomes 2.5. Neither number proves a quarter-stress result; a lumped field is still not justified by small Bi.

    Diffusion time scales with L² and Bi with L. The symmetric-plate stress depends on the actual temperature differences. If exactly the same ΔT(z/L) snapshot were prescribed, its stresses would be unchanged; a real cooling history changes that profile and its time of occurrence.

  4. For the same free-plate model, replace the snapshot by ΔT(z) = +60(z/L)² K. Find the mean and surface/centre stresses. Is compression at the surface evidence that no thermal fracture is possible?

    Show the answer and reasoning

    The mean is +20 K; the surfaces have −42.67 MPa compression and the centre +21.33 MPa tension. No general no-fracture conclusion follows.

    The sign reverses because the surfaces now prefer more expansion than the average. Interior flaws, other stress components, geometry and later transients need consideration. The result is specific to the symmetric heating field, not all heating events.

  5. The environment is 100 K below the initial plate temperature, but the measured snapshot is the −60(z/L)² K field. A worksheet calculates surface tension as Eα(100 K)/(1 − ν). Diagnose the error and compare its number with the plate calculation.

    Show the answer and reasoning

    It uses the specimen–environment jump as if it were the stress-driving constraint difference. It predicts 106.67 MPa instead of 42.67 MPa for the stated free-plate snapshot, a factor of 2.5.

    The plate formula uses ⟨ΔT⟩ − ΔT_s = −20 − (−60) = 40 K. The environment belongs to the heat-transfer boundary problem; the plate is also free to contract. One cannot replace both facts with a fully restrained 100 K surface drop.

  6. Interchange the film/substrate expansion coefficients in the uniform-film example, leaving E_f, ν_f and ΔT unchanged. Predict the stress and identify what extra evidence a claim of delamination needs.

    Show the answer and reasoning

    The film stress becomes −106.67 MPa, compressive. Delamination needs an interfacial/geometry and energy assessment, including any defect or buckling mode; the sign alone does not establish it.

    Now the substrate contracts more than the film would freely. Perfect bonding forces extra contraction. Compression can enable buckling-assisted separation in a suitable configuration, but adhesion, edge fields, defect size and substrate compliance remain consequential.

  7. After a transient, a homogeneous unrestrained plate is uniformly cooler than its initial reference. A report attributes a remaining stress profile solely to the earlier elastic gradient. What mechanism or model assumption must be investigated?

    Show the answer and reasoning

    Purely elastic gradient stress vanishes in the stated free homogeneous model once temperature is uniform. Investigate retained incompatible strain or a changed constraint, such as plasticity, creep, transformation, damage, a bonded layer or a fixture.

    An earlier gradient is a loading history, not by itself a permanent strain mechanism. In contrast, a bonded two-material assembly can retain mismatch at a uniform temperature away from its bonding reference. A measured residual field calls for a mechanism and a compatible force balance.

What to carry forward

  • Thermal stress comes from incompatible strain, which can arise from gradients, external restraint or bonded expansion mismatch.
  • Use a = k/(ρc), a consistent length scale and Bi to choose the thermal model before calculating stress.
  • A symmetric free plate has σ = Eα[⟨T⟩ − T(z)]/(1 − ν) under the stated elastic assumptions.
  • Fracture and residual-property conclusions need the actual stress field, flaw or damage model and temperature history.

References

Teaching resources and research supporting this lesson’s explanations and examples.

  1. University Physics Volume 2 · 1.3 Thermal ExpansionSamuel J. Ling, Jeff Sanny and William Moebs · OpenStaxTextbook sectionopenstax.orgRead source
  2. Massachusetts Institute of Technology4 sources
    1. Introduction to Fracture MechanicsDavid Roylance / MIT OpenCourseWareCourse notes, Fall 1999 · PDFocw.mit.eduRead source
    2. Thermodynamics and Propulsion · 18.3 Transient Heat TransferMIT Unified EngineeringUniversity teaching chapterweb.mit.eduRead source
    3. Thermal Properties · Lecture 9, transcribed notesLorna Gibson / MIT OpenCourseWareCourse notes, Spring 2015 · PDFThermal Properties · Lecture 9, transcribed notesSpoken lecture transcriptocw.mit.edu
    4. 3.032 Mechanical Behavior of Materials · Problem Set 3MIT Department of Materials Science and Engineering / MIT OpenCourseWareProblem set and solutions, Fall 2006 · PDFocw.mit.eduRead source
  3. Behavior of Materials Under Conditions of Thermal StressS. S. Manson / NACANACA Report 1170, 1954 · PDFntrs.nasa.govRead source