²t²/(12α(1-ν²)b²). The parameters are plate thickness (t), coefficient of thermal expansion (α), Poisson's ratio (ν), and the width of the short side (b)."}}, {"@type": "Question", "name": "How is thermal buckling evaluated using FEM?", "acceptedAnswer": {"@type": "Answer", "text": "In Step 1, a thermal stress analysis is performed to obtain the stress field due to the temperature load. In Step 2, this stress field is applied as an initial stress condition in a linear buckling eigenvalue analysis. The resulting eigenvalue (λ) represents the critical temperature multiplier."}}, {"@type": "Question", "name": "What are common pitfalls to avoid in thermal buckling analysis?", "acceptedAnswer": {"@type": "Answer", "text": "Typical issues include unrealistic buckling modes due to incorrect constraint conditions, neglecting temperature-dependent material properties, and ignoring geometric initial imperfections. Since FEM eigenvalue analysis assumes an ideal structure, applying a knockdown factor is necessary for real-world structures."}}]}]}

熱座屈解析

Category: 連成解析 > 熱-構造連成 | 更新 2026-04-12
Thermal buckling analysis showing critical temperature difference and buckled plate mode shape with compressive stress contour
熱座屈解析 — 拘束された薄板が加熱により座屈する際の臨界モード形状と圧縮応力分布

Theory and Physics

What is Thermal Buckling?

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Does thermal buckling happen just because the temperature rises? Even without any load?

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Yes. Roughly speaking, when a constrained structure is heated, it wants to expand but cannot, so compressive stress builds up. The moment that compressive stress exceeds a critical value, buckling occurs even with zero external load.

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Huh, buckling with zero load goes against intuition, doesn't it? In what real-world situations does this actually happen?

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Let me give you three familiar examples:

  • Railway track "sun kink": Rails heat up in summer but are bolted to sleepers (ties), preventing longitudinal expansion. Compressive force accumulates, and when a certain temperature is exceeded, the rails buckle in a wavy pattern sideways. This causes train service suspensions.
  • Aircraft skin panels: During supersonic flight, aerodynamic heating raises the skin temperature, but the skin is constrained by frames (skeleton), generating compressive stress and causing panel buckling.
  • Thermal expansion of piping: High-temperature pipes in chemical plants, constrained between supports, can become compressed and buckle, leading to pipe bending accidents.

All follow the same mechanism: Constraint + Temperature Rise → Compressive Stress → Buckling.

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I see! So "constraint" is the key. If it could expand freely, it wouldn't buckle.

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Exactly. With free ends, it would simply expand by $\Delta L = \alpha L \Delta T$ and that's it. But if both ends are fixed, the expansion is constrained to zero, so that amount is entirely converted into compressive stress $\sigma = E \alpha \Delta T$. This is the starting point of thermal buckling.

Thermal Stress Generation Mechanism

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Can you explain the thermal stress formula in a bit more detail?

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Consider a bar constrained in one axis. When the temperature rises by $\Delta T$, the free thermal strain is:

$$ \varepsilon_{\text{th}} = \alpha \Delta T $$
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But if both ends are fixed, the total strain is zero ($\varepsilon_{\text{total}} = 0$), so elastic strain arises to counteract the thermal strain:

$$ \varepsilon_{\text{elastic}} = -\alpha \Delta T $$
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From Hooke's law, the compressive stress generated in the constrained structure is:

$$ \sigma = E \varepsilon_{\text{elastic}} = -E \alpha \Delta T $$
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The negative sign means compression. For example, with steel ($E = 200\,\text{GPa}$, $\alpha = 12 \times 10^{-6}\,/\text{K}$) and $\Delta T = 50\,\text{K}$...

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$\sigma = 200 \times 10^3 \times 12 \times 10^{-6} \times 50 = 120\,\text{MPa}$. A mere 50-degree temperature rise generates 120 MPa of compressive stress. For a thin plate, this stress is sufficient to cause buckling.

Theoretical Formula for Critical Temperature Difference

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So, how do you specifically determine "at what temperature rise does buckling occur"?

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Consider the most basic case: a rectangular plate simply supported on all four sides under uniform heating. This is the thermal version of Euler buckling theory.

The critical buckling stress from the plate buckling formula is:

$$ \sigma_{cr} = k \frac{\pi^2 E}{12(1-\nu^2)} \left(\frac{t}{b}\right)^2 $$
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Here, $k$ is the buckling coefficient (for simply supported four sides under uniaxial compression, $k = 4$), $t$ is the plate thickness, and $b$ is the short side width. Equating this with the thermal stress $\sigma = E \alpha \Delta T$ gives:

$$ \boxed{\Delta T_{cr} = \frac{k \pi^2 t^2}{12 \alpha (1 - \nu^2) b^2}} $$
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Wow, Young's modulus $E$ cancels out!

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Good observation. Since thermal stress is $E \alpha \Delta T$ and the critical stress is also proportional to $E$, the $E$ factors cancel out. This means the critical temperature difference is determined solely by plate thickness, width, thermal expansion coefficient, and Poisson's ratio, and does not depend on Young's modulus. This is a characteristic unique to thermal buckling.

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I'd like to get a feel for it with specific numbers. For example, for an aluminum plate ($\alpha = 23 \times 10^{-6}$, $\nu = 0.33$) with $t = 2\,\text{mm}$, $b = 200\,\text{mm}$, what happens?

🎓

Calculating with $k = 4$ (simply supported on four sides):

$\Delta T_{cr} = \dfrac{4 \times \pi^2 \times (0.002)^2}{12 \times 23 \times 10^{-6} \times (1 - 0.33^2) \times (0.2)^2} \approx 1.6\,\text{K}$

It buckles at just 1.6 degrees. This shows how susceptible thin aluminum panels are to thermal buckling. This is why thermal buckling is so important in aircraft skin design.

Physical Meaning of Each Term
  • $\alpha$ (Coefficient of linear thermal expansion): Expansion rate per 1K temperature rise. Aluminum has about twice the value of steel, making aluminum structures more susceptible to thermal buckling.
  • $t/b$ (Thickness/Short side width ratio): Equivalent to the slenderness ratio of the plate. The smaller this ratio (thinner and wider plate), the lower $\Delta T_{cr}$ becomes, making buckling easier. It is proportional to $t^2$, so doubling the thickness makes $\Delta T_{cr}$ four times larger.
  • $\nu$ (Poisson's ratio): The $(1-\nu^2)$ term accounts for the biaxial stress state effect. A larger Poisson's ratio slightly reduces $\Delta T_{cr}$.
  • $k$ (Buckling coefficient): Varies with boundary conditions and aspect ratio. For a plate clamped on all four sides, $k \approx 10.67$, which is about 2.7 times that of the simply supported case.
Assumptions and Applicability Limits
  • Linear elastic material (valid within the pre-yield range)
  • Small deflection theory (deflection is sufficiently small compared to plate thickness)
  • Uniform temperature distribution (uniform across the plate surface and through the thickness)
  • No initial imperfections (assumes a perfect flat plate)
  • Temperature-dependent material properties are ignored
  • In real structures, a knockdown factor (typically 0.5 to 0.8) needs to be applied
Dimensional Analysis and Unit System
VariableSI UnitRepresentative Value / Note
$\alpha$ (Coefficient of linear thermal expansion)1/KSteel: 12×10⁻⁶, Aluminum: 23×10⁻⁶, Titanium: 8.6×10⁻⁶
$E$ (Young's modulus)PaSteel: 200 GPa, Aluminum: 70 GPa (does not directly affect $\Delta T_{cr}$)
$t$ (Plate thickness)mAircraft skin: 1–3 mm, Pressure vessel: 10–50 mm
$\sigma_{cr}$ (Critical buckling stress)PaCan be back-calculated from $E \alpha \Delta T_{cr}$

Formulation as an Eigenvalue Problem

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How is it formulated in FEM?

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Linear buckling analysis is formulated as an eigenvalue problem. Using the stress stiffness matrix $[K_\sigma]$ due to thermal stress:

$$ \bigl([K_0] + \lambda [K_\sigma(T)]\bigr) \{\phi\} = \{0\} $$
🎓

The meaning of each symbol is as follows:

  • $[K_0]$: Standard elastic stiffness matrix
  • $[K_\sigma(T)]$: Stress stiffness matrix (also called geometric stiffness matrix) calculated from thermal stresses due to temperature $T$
  • $\lambda$: Eigenvalue = critical temperature load multiplier
  • $\{\phi\}$: Buckling mode shape

If $\lambda = 1$, buckling occurs exactly at the applied temperature load. If $\lambda = 0.5$, it buckles at half the temperature, which is dangerous. If $\lambda > 1$, it's on the safe side.

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I see, so $\lambda$ acts like a safety factor.

Relationship Between Boundary Conditions and Critical Temperature

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How much does the critical temperature change if you change the boundary conditions?

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The buckling coefficient $k$ changes significantly with boundary conditions, so the critical temperature changes proportionally. Let's compare for a square plate ($a/b = 1$):

Boundary ConditionBuckling Coefficient $k$$\Delta T_{cr}$ Multiplier (SS reference)
Simply Supported on All Sides (SSSS)4.01.0
Clamped on All Sides (CCCC)10.672.67
2 Sides Clamped + 2 Sides Free (CCFF)≈2.20.55
All Edges Free (with in-plane constraint)1.00.25

Simply clamping all four sides makes the critical temperature 2.7 times higher. Increasing fixed supports in design is a fundamental thermal buckling countermeasure. Conversely, structures with many free edges are weak.

🧑‍🎓

Having the critical temperature differ by nearly a factor of 4 just by changing the fixation conditions is significant for design.

Numerical Methods and Implementation

FEM Two-Step Analysis Procedure

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What are the specific steps for analyzing thermal buckling with FEM?

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The basic approach is a two-step process:

Step 1: Thermal Stress Analysis (Static Analysis with Thermal Load)

  • Input temperature distribution $\Delta T(\mathbf{x})$ as a load
  • Convert thermal strain $\varepsilon_{\text{th}} = \alpha \Delta T$ into equivalent nodal forces
  • Solve $[K]\{u\} = \{F_{\text{th}}\}$ to obtain the stress field $\sigma_{ij}$

Step 2: Linear Buckling Eigenvalue Analysis

  • Construct the stress stiffness matrix $[K_\sigma]$ from the stress field in Step 1
  • Solve the eigenvalue problem $([K_0] + \lambda [K_\sigma]) \{\phi\} = \{0\}$
  • The smallest positive eigenvalue $\lambda_1$ is the critical temperature multiplier
🧑‍🎓

So if I input a temperature of 100 degrees in Step 1 and get $\lambda_1 = 0.6$, does that mean it buckles at 60 degrees?

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Exactly. Since $\Delta T_{cr} = \lambda_1 \times \Delta T_{\text{ref}}$, the critical temperature difference is $0.6 \times 100 = 60$ degrees. In practice, a safety factor of 1.5 to 2.0 is applied, so the design allowable temperature becomes $\Delta T_{\text{allow}} = 30 \sim 40$ degrees.

Case of Non-Uniform Temperature Distribution

For non-uniform heating like one-sided or localized heating, first perform a steady-state/transient heat conduction analysis in Step 1 to obtain the temperature distribution, then map that result as a thermal load for the structural analysis in Step 1. This essentially becomes a three-step process (Heat Conduction → Thermal Stress → Buckling Eigenvalue).

Element Selection and Formulation

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What kind of elements should be used for thermal buckling analysis?

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It depends on the structure's shape:

Structure TypeRecommended ElementNotes
Thin plate panelsShell elements (S4R, SHELL181, etc.)Use 5–7 integration points through thickness to capture bending gradients
Thick plates / Block bodiesSolid elements (C3D20R, SOLID186, etc.)Mesh with at least 3–4 layers through the thickness
Rails / BeamsBeam elements (B31, BEAM188, etc.)Cannot capture local buckling of thin-walled parts; use in conjunction with shell elements
PipingPipe/Shell elementsSufficient circumferential divisions are needed to capture circumferential buckling modes
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Shell elements are used a lot. What's the effect of the number of integration points through the thickness?

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If there's a temperature gradient through the thickness, too few integration points cannot accurately capture the bending stress distribution. For example, with one-sided heating, a linear temperature distribution through the thickness occurs. If there are only 3 integration points, the bending thermal stress might be underestimated. In practice, at least 5, preferably 7 points are recommended.

Nonlinear Post-Buckling Analysis

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Is linear buckling analysis sufficient? What if I want to know the post-buckling behavior?

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Good question. Linear buckling tells you "when it buckles" but not "what happens after buckling." To track post-buckling behavior, nonlinear static analysis (geometric nonlinearity) is required.

The procedure is as follows:

  1. Obtain the buckling mode shape $\{\phi\}$ from linear buckling analysis.
  2. Scale this mode shape and add it to the geometry as an initial imperfection (typically 10–50% of the plate thickness).
  3. Perform a nonlinear static analysis (NLGEOM=ON) with the temperature load increased incrementally.
  4. Track the load-displacement curve (Riks method or Arc-length method).
🧑‍🎓

What happens if you don't include an initial imperfection?

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With perfect geometry, the solution is not unique at the bifurcation point, and the solver may fail to capture the buckling. In nonlinear analysis, it's standard practice to include an initial imperfection based on the premise that "perfect structures do not exist in reality." Real structures have tiny imperfections due to manufacturing errors, so a model with initial imperfections is actually more realistic.

Key Points for Solver Settings

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Are there any important points to watch for in solver settings?

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Written by NovaSolver Contributors
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About the Authors
Setting ItemRecommended ValueReason
Number of eigenvalues to find10–20Check influence of higher modes, not just the lowest
Eigenvalue solverLanczos methodStable and fast for large-scale models
Temperature reference temperatureStress-free temperatureMany accidents occur due to misconfiguration causing all stresses to be offset
NLGEOM (Nonlinear)ONMandatory for post-buckling tracking