Structural Buckling Study

Overview#

Buckling is a stability failure: a slender column, thin panel or shell suddenly deflects sideways under compressive load, often well below the material's strength. A static stress study can show low stress yet the part still collapses. The Structural Buckling study estimates the load at which this happens.

Fusion's buckling study is linear (eigenvalue-based) buckling. Autodesk notes that linear buckling gives nonconservative results, so avoid designing near the critical load. The result is the Buckling Load Factor (BLF), a multiplier of the applied load that causes buckling: critical load = load factor x applied load. Autodesk's study-type page says a multiplier below 1.0 means the structure buckles before the applied load is reached.

Licensing: Structural Buckling is an advanced study that costs 6 tokens per solve, or is covered by the Fusion Simulation Extension; only Linear Static Stress cloud solves are covered by the subscription (Autodesk token and extension topics). The Solve dialog shows the live cost. Personal-use availability is not stated in the Simulation help.

Where to find it#

• Simulation workspace > Setup tab > Study panel > New Simulation Study, then select Structural Buckling.
• Constraints, loads, materials, contacts and mesh follow the static stress workflow. Set Number of Modes in Setup tab > Manage panel > Settings > General.
• Results: Total Displacement for the first buckling mode is shown by default (normalized, no units), and Buckling Load Factor is shown in the Mode dropdown above the legend.

Step by step#

1. Create a Structural Buckling study.
2. Assign materials. Young's modulus and Poisson's ratio drive the result; yield strength does not enter linear buckling.
3. Constrain the part the way it is supported. Constraints affect the k constant in the Euler critical load equation. End conditions change the critical load massively (general Euler theory): a column fixed at both ends buckles at four times the load of a pinned-pinned column of the same size, and a fixed-free column at one quarter.
4. Apply a compressive load at a convenient reference value (for example 1000 N). Autodesk says applied loads should compress the model to avoid negative buckling load factors; the load factor scales your reference.
5. Mesh with enough elements through the thickness and along the length to capture the half-waves of the buckled shape. Check mesh settings under Settings > Mesh.
6. Set the number of modes to calculate (default 3, which Autodesk says is typically sufficient; technically a mode can be run for each degree of freedom).
7. Solve.
8. Read the load factor in the Mode dropdown (for example "Mode 1: 0.922 x Load"). The first positive load factor is the desired result. With a 3.2 load factor and a 1000 N reference, the predicted critical load is 3200 N.
9. View the mode shape animation to confirm it looks like a plausible instability.
10. Compare to the design load: the buckling safety factor equals the critical load divided by the real service load. How large it must be is your design decision; as general guidance (not Autodesk data) real structures buckle below the ideal prediction, so a margin well above 1 is used. Also run a Static Stress study: Autodesk notes buckling assumes all materials stay below yield, so a high load factor does not by itself mean the structure is safe.

Options reference#

Tips and pitfalls#

Warning: Autodesk says linear buckling results are nonconservative. General engineering practice (not Autodesk data) is to apply a knockdown factor for imperfections, especially for shells; the size of the factor depends on the structure and your standards.

Tip: Compare with Euler: P = pi^2 E I / (K L)^2. If your slender column result differs noticeably, check constraints and mesh.

Warning: If the buckling stress exceeds the yield strength, the part yields first, and the elastic buckling load is irrelevant. Compute the buckling stress (critical load divided by area) and compare with yield. Use nonlinear analysis for inelastic buckling.

Tip: Thin-walled sections can buckle locally (a flange wrinkles) before global column buckling. Higher modes may show these; examine more than one mode.

Tip: Shell elements are not supported in Fusion, so thin sheets are meshed as solids: use local mesh control to get enough elements through the thickness.

Practice exercise#

Create a 300 mm long, 10 mm diameter aluminium rod. Fix one end, apply 100 N axial compression at the free end. Solve and compute the Euler load P = pi^2 E I / (2L)^2 for the fixed-free case. Expected result: the first-mode load factor times 100 N should be close to the Euler value (the agreement depends on mesh). Then support both ends as pinned-pinned (leaving the axial direction free at the loaded end) and watch the critical load rise to about four times the fixed-free value.

End conditions and effective length#

Euler's critical load is P = pi^2 E I / (K L)^2, where K is the effective length factor.

Real joints sit between these ideals. A bolted end is neither pinned nor fixed; run the study with both extremes to bracket the real behavior, and design to the lower bound unless you have evidence of stiffness.

Worked example: a 3D-printed strut#

All values are illustrative assumptions (not Autodesk or measured data). A plastic strut, 6 mm diameter, 150 mm long, pinned at both ends, assumed E = 2.1 GPa and assumed yield about 45 MPa.

1. I = pi d^4 / 64 = 3.1416 x 1296 / 64 = 63.6 mm^4.
2. P_cr = pi^2 x 2100 x 63.6 / 150^2 = 9.8696 x 133560 / 22500 = 58.6 N.
3. Slenderness ratio: radius of gyration r = d / 4 = 1.5 mm, so L / r = 100. Euler is valid when the stress at buckling is below yield: sigma_cr = 58.6 / 28.3 mm^2 = 2.07 MPa, far below yield (about 45 MPa). Euler governs.
4. With an assumed knockdown factor of 3 for imperfections (your choice): allowable load = 58.6 / 3 = 19.5 N.
5. In Fusion: apply 10 N axial as reference; the load factor of mode 1 should be near 5.86. Check this against the hand value. Constrain one end with a Pin constraint (cylindrical faces only) and the other end so it can move axially, where the 10 N load acts; make sure rigid-body motion is prevented.

A fatter strut helps greatly: doubling the diameter increases I by 16 times, so P_cr becomes 937 N.

Thin walls and local buckling#

For a thin flat plate under compression, P_cr = k pi^2 E t^3 / (12 (1 - nu^2) b), with k = 4 for simply supported edges. For a 1.5 mm aluminium plate (E = 68.9 GPa, nu = 0.33) 50 mm wide: sigma_cr = k pi^2 E (t / b)^2 / (12 (1 - nu^2)) = 4 x 9.87 x 68900 x (0.03)^2 / (12 x 0.891) = 4 x 9.87 x 68900 x 0.0009 / 10.69 = 229 MPa. Plates therefore wrinkle at about 230 MPa when 50 mm wide and 1.5 mm thick; since 6061-T6 yields near 276 MPa the margin to wrinkling is small. A 0.8 mm plate (t / b = 0.016) would wrinkle at about 65 MPa, well below yield.

Mesh and mode settings#

General guidance, not Autodesk data (except the default number of modes):

If modes 1 and 2 have nearly the same load factor, the structure is likely sensitive to imperfections (general engineering knowledge); round tubes and domes behave this way.

Load factor interpretation#

Autodesk's interpretation of the Buckling Load Factor:

If a negative value appears, check the other modes to find the first positive one.

Compare the critical load with service load times a safety factor chosen for your application. Autodesk does not publish buckling knockdown values; use design codes or test data.

Beyond linear buckling#

Linear buckling assumes small displacements before buckling, a perfect structure and fully elastic behavior. For post-buckling behavior, imperfections or plastic buckling, a nonlinear static stress study ramps the load in increments and can include nonlinear material behavior. General practice is to build a small imperfection into the geometry and read the load at which displacement runs away; trust the nonlinear result when it is well below the linear prediction.

Failure modes and fixes#

Practice exercise#

Check three end conditions numerically.

1. Build a 6 mm diameter, 150 mm long rod with an assumed E = 2.1 GPa (a custom material). Apply a 10 N reference compressive load.
2. Pinned-pinned: expect a load factor near 5.86 (58.6 N). Record the Fusion value.
3. Fixed-free: expect 1.46 (14.6 N). Fixed-fixed: expect 23.4 (234 N). Record both.
4. Change diameter to 8 mm and predict P_cr = 58.6 x (8/6)^4 = 185 N before solving.
5. Report the percentage difference between Fusion and Euler for each case and state which end condition was hardest to reproduce, and why.

Related#

• Static Stress Study
• Nonlinear Static Stress
• Cloud Solve and Credits