What is buckling?
It is a sudden sideways bow of a slender rod under push. The danger is that it can happen at a stress far below the yield strength of the steel, so a stress check on the rod alone will not find it. K03-Q1-28 shows the same effect on a cylinder rod.
What is the formula?
For a long, slender rod, the Euler formula gives the critical load:
critical load = π² × E × I ÷ (k × L)²
E is the stiffness of the steel, I is the second moment of area of the section (π × d⁴ ÷ 64 for a solid round rod), L is the length, and k is a factor for the end conditions. The usual textbook values:
| Ends | k |
|---|---|
| Pivoted at both ends | 1 |
| One end fixed, the other pivoted | 0.7 |
| Both ends fixed | 0.5 |
| One end fixed, the other free | 2 |
A linkage rod with a rod end or ball joint at each end is pivoted at both ends, so use k = 1. The textbook values for fixed ends assume the ends are perfectly rigid, which a pinned joint with clearance never is, so do not count on a smaller k.
How do you find the diameter you need?
Turn the formula around. For a solid round steel rod, with a safety factor S on the push force F:
d⁴ = 64 × (F × S) × (k × L)² ÷ (π³ × E)
Example, with illustrative numbers: a push of 2,000 N, a safety factor of 3 (an example, not a rule), pivoted ends (k = 1), a length of 800 mm between the pin centres, and E = 200,000 N/mm².
- d⁴ = 64 × 6,000 × 640,000 ÷ (31.0 × 200,000) ≈ 39,600 mm⁴
- d ≈ 14.1 mm
Halve the length and the required diameter drops by about 30%. Double it and the diameter rises by about 40%, because the diameter goes with the square root of the length.
If the rod is threaded along its length, use the thread root diameter (the diameter at the bottom of the thread), not the outside diameter. That is the safer course.
When can you use the Euler formula?
When the rod is slender enough. The measure is the slenderness ratio L/r, where r is the radius of gyration of the section. For a solid round rod r = d ÷ 4, so L/r = 4 × L ÷ d. Euler holds above a transition value:
L/r > π × √(2E ÷ σy)
σy is the yield strength. For E = 200,000 N/mm² the transition is 106 for a yield strength of 350 MPa and 126 for 250 MPa. Below it the rod is too stocky for Euler, which then overestimates the load, and the Johnson formula applies instead: σcr = σy − σy² × (L/r)² ÷ (4π²E). K07-Q1-10 works both cases for a tube.
In the example, r = 14.1 ÷ 4 = 3.53 mm and L/r = 800 ÷ 3.53 ≈ 227, well above the transition, so Euler is the right formula. A check on the stress agrees: 6,000 N on a 14.1 mm rod is about 38 MPa, far below yield.
What does the formula not cover?
Real joints and real rods. A real mounting is not an ideal column: clevises, spherical bearings, clearance, stiff or flexible frames, and the position in the stroke all change the end conditions. Bends in the rod, threads and cross holes, side loads, shock and looseness at the pins all reduce the real margin. Treat the result as a first screening, not a rating, and add a margin.
How can you reduce the risk?
- Use a larger diameter. The effect is large, because diameter counts to the fourth power.
- Shorten the unsupported length, or add a guide in the middle.
- Use a tube instead of a solid bar where weight matters: for the same weight a tube is much stiffer (K07-Q1-10 gives a worked comparison).
- Keep the pins aligned and free of play, so the rod only sees push and pull (Q1-10).
Where does the push come from on a vehicle link, and how much tube does it need?
A radius rod, trailing link or lower link of a four-link axle is pushed or pulled by braking and drive, depending on the layout. An illustrative case: an axle carrying 6 kN of vertical load during braking, with a grip coefficient of 1.0, has a 6 kN braking force at the axle. If two links in line with it share the force equally, each sees 3 kN from the tyre force alone (more if the link is angled, as in Q1-02). On a solid axle the brake torque is also reacted through the upper and lower links as a couple, which can make the link loads considerably larger, so take the link loads from the suspension geometry. Now take a steel tube link with a pin at each end, E = 200 GPa and an assumed yield of 350 MPa. A 25.4 × 1.65 mm tube has a critical load of about 25.8 kN at 800 mm and 17.2 kN at 1,000 mm. A 31.75 × 1.65 mm tube at 1,000 mm carries about 33.3 kN, nearly twice as much for 27% more steel, because at the same wall thickness the stiffness of a tube grows roughly with the cube of its diameter (Johnson below the transition slenderness, Euler above it, as in K07-Q1-10). These are ratios for comparing tubes. A real link also meets impact loads far above braking, and dents or bent tubes lose strength, so the margin must come from measured loads or the vehicle maker's design data. The tube buckling calculator runs the same method, and the radius rod buyer's guide covers the link forms.