A long, slender shaft under axial compression buckles sideways before it crushes. Euler's formula gives the critical load the shaft can carry.
⚙ Open the Hydraulic Calculator →| Quantity | Symbol | Unit |
|---|---|---|
| Elastic modulus | E | N/mm² (steel ≈ 210,000) |
| Second moment of area | I | mm⁴ |
| Length | L | mm |
| Critical load | Pcr | N |
Given: Solid shaft d = 40 mm, length L = 1000 mm, pinned-pinned (K = 1), E = 210,000 N/mm².
1) Second moment: I = π × 40⁴ / 64 = 125,664 mm⁴
2) Critical load: Pcr = π² × 210000 × 125664 / (1 × 1000)² = ≈ 260 kN
3) Safety factor S = 3.5 → allowable load: ≈ 74 kN
It captures how the shaft ends are supported. Pinned-pinned K=1, fixed-free K=2, fixed-pinned K=0.7, fixed-fixed K=0.5. A larger K lowers the buckling load.
Because buckling is sudden and dangerous, a factor of 3–5 is common; the critical load is divided by this to get the working load.
No. Euler applies only to slender (long/thin) members. Short/thick shafts are governed by yielding (crushing) and need a different check.