Why does a link that pushes need a different check from one that pulls?
A pulling link fails when the steel reaches its strength. A pushing link can bow sideways long before that: this is buckling. Global buckling is an immediate loss of strength of the whole cross-section and can end in collapse. So for any link that can be pushed, the buckling check governs. The solid-rod version, with the diameter formula, is in K06-Q1-03; this article does the same for a tube.
How do you work out the buckling load of a tube?
With four steps, using the section of the tube.
- Section. Area A = π ÷ 4 × (D² − Di²), second moment of area I = π ÷ 64 × (D⁴ − Di⁴), radius of gyration r = √(I ÷ A), where D and Di are the outside and inside diameters.
- Slenderness. λ = L ÷ r, with L the length between the pin centres for a pin at each end (K = 1). K = 1 is the ideal pinned–pinned case; a real frame needs a justified effective length.
- Which regime. The transition is λ_t = √(2π² × E ÷ σy). Above it, use Euler: σcr = π² × E ÷ λ². Below it, use Johnson: σcr = σy − σy² × λ² ÷ (4π² × E).
- Load. Pcr = σcr × A. Divide by a design factor that you choose.
The result is a comparison ratio, not a certificate of compliance. Real joints are not ideal columns (K06-Q1-03).
How much better is a tube than a solid rod?
Take steel with E = 200 GPa and a yield of 350 MPa, a pin-to-pin length of 800 mm and a design area of about 314 mm² in both cases. The transition is λ_t = 106.2.
| Solid rod, 20 mm | Tube, 32 × 3.5 mm | |
|---|---|---|
| Area | 314.2 mm² | 313.4 mm² |
| Second moment I | 7 854 mm⁴ | 32 297 mm⁴ |
| Radius of gyration r | 5.00 mm | 10.15 mm |
| Slenderness λ | 160 | 78.8 |
| Regime | Euler (λ > 106) | Johnson (λ < 106) |
| Critical stress | 77.1 MPa | 253.7 MPa |
| Critical load | 24.2 kN | 79.5 kN |
The two use the same amount of steel. The tube carries 3.3 times the load, because moving the metal outwards raises I by four times while the area stays the same. The tube's slenderness is lower, so it moves into the Johnson regime, where the load is limited by yield as well as stiffness. A pure Euler calculation would have given 99.6 kN for the tube, which is a critical stress of 318 MPa. Euler's curve holds only while that stress stays below about half the yield stress (175 MPa here), so it would have overstated the load, and that is why the regime matters.
The numbers are illustrations. Use the real length, material and section.
What about a thin wall?
A tube can fail locally before it fails as a column. Local buckling depends on the width-to-thickness ratio of each element of the section: when the ratio is above the limit set in structural steel design specifications, the element is slender, and its local buckling has to be considered. Hollow sections often use thin steel, so local buckling is more likely to control. A round tube's perimeter is a stiffened element, and has some strength after buckling. The example above has D ÷ t = 32 ÷ 3.5 = 9.1, a thick wall. The higher D ÷ t is, the sooner local buckling can come before column buckling. The limit for a round section is given in the structural steel design specification; take it from the current edition.
A thin-walled tube is also more easily dented by a clamp or a pin hole than a thick-walled one (Q1-02).
Where do bending and torsion come from?
A link with a rod end at each end carries axial load only, apart from its own weight and bearing friction: the spherical bearings cannot pass a torque, and with pins at both ends they cannot pass a bending moment. Bending and torsion enter when the load is not on the tube's axis, or when something other than the end fittings holds the tube.
An offset e between the line of the load and the tube axis gives a moment M = P × e. Take the 32 × 3.5 mm tube with a 30 kN load and a 3 mm offset. M = 90 N·m, and the bending stress is 90 000 ÷ (32 297 ÷ 16) = 44.6 MPa. The axial stress is 30 000 ÷ 313.4 = 95.7 MPa, so the offset adds nearly half as much again, and in a buckling-prone member it also reduces the load at which it bows. Keep the load on the axis (K06-Q1-10).
A tube used as a torque member, joined rigidly at both ends, is a different design. Its torsion constant is J = π ÷ 32 × (D⁴ − Di⁴), twice I.
How much do the adapter ends change the answer?
Mainly by adding length and a change of section. The length that counts runs from pin centre to pin centre, and the rod ends and adapters add to it. The solid adapter region is short and stiff, but it is not part of the tube's free section, so for a first estimate use the full pin-to-pin length with the tube's section. The abrupt change at the adapter is a stress concentration, which is a fatigue question rather than a buckling one (Q1-04).
How does a vehicle link change the tube choice?
Three things. First, the diameter-versus-wall trade-off is easier to see. Take the illustrative case above (E = 200 GPa, yield 350 MPa, pin to pin 800 mm). A 25.4 × 2.4 mm tube (1.36 kg/m) has a critical load of about 34.9 kN. A 31.75 × 1.65 mm tube is lighter at 1.22 kg/m and carries about 41.0 kN, because the metal sits further out. Its wall is thinner relative to its diameter (D ÷ t is 19.2 against 10.6), so local buckling and dents from stones and clamps become the next concern. Second, a tube joined to bushings or rigid mounts, not to rod ends with pins, may have a different effective length, which needs a justified factor. Third, rules for chassis and roll-cage tubing set tube size and material, and the calculation does not replace them. The tube buckling calculator uses the same method, and the radius rod buyer's guide covers link forms.