Example: A solid steel push rod with pinned ends
Given
- A solid round steel push rod, 16 mm diameter and 900 mm long, with pinned ends (K = 1).
- Assume E = 200 GPa (typical for steel) and a yield strength Sy = 350 MPa (an assumed value for this example).
- The design uses a factor of safety of 3 on buckling.
Find
- The critical buckling load, whether Euler’s formula is valid, and the allowable axial load.
Solution
- Moment of inertia of a solid circle (same about every diameter):
I = π·d⁴ / 64 = π(16)⁴ / 64 = 3217 mm⁴ - Euler critical load, with E = 200 000 MPa and KL = 900 mm:
Pcr = π² × 200 000 × 3217 / 900² = 7840 N = 7.84 kN - Radius of gyration and slenderness (for a solid circle r = d/4):
A = π(16)²/4 = 201.1 mm², r = 4 mm, λ = 900 / 4 = 225 - Transition slenderness for this steel:
λc = √(2π² × 200 000 / 350) = 106.2 → λ = 225 > 106.2, so Euler applies - Critical stress, well below yield:
σcr = 7840 / 201.1 = 39.0 MPa < 350 MPa - Allowable load with the factor of safety:
P_allow = 7.84 / 3 = 2.61 kN
Answer: Pcr = 7.84 kN, Euler applies (λ = 225 against λc = 106), and the allowable load is 2.61 kN. At buckling the rod is stressed to only 39 MPa, about a ninth of its yield strength: stability, not strength, governs.
Example: A rectangular bar: weak axis and end conditions
Given
- A steel bar 30 mm × 10 mm in section and 600 mm long is fixed at its base and free at the top (K = 2). It is not braced.
- Assume E = 200 GPa (typical for steel).
Find
- The critical load about the weak axis, the strong-axis value for comparison, and the critical load if the top is pinned instead (fixed–pinned, K = 0.699).
Solution
- Moments of inertia about the two axes:
I_weak = 30(10)³ / 12 = 2500 mm⁴, I_strong = 10(30)³ / 12 = 22 500 mm⁴ - Effective length for a fixed–free column:
K·L = 2 × 600 = 1200 mm - Critical load about the weak axis, which governs:
Pcr = π² × 200 000 × 2500 / 1200² = 3427 N = 3.43 kN - About the strong axis it would be nine times higher (I ratio 22 500 / 2500 = 9):
Pcr,strong = π² × 200 000 × 22 500 / 1200² = 30.8 kN (does not govern) - Slenderness check about the weak axis:
r = √(2500 / 300) = 2.89 mm, λ = 1200 / 2.89 = 416, far above typical λc - Pinning the top changes the effective length:
K·L = 0.699 × 600 = 419.4 mm; Pcr = π² × 200 000 × 2500 / 419.4² = 28.1 kN
Answer: As a fixed–free column the bar buckles about its weak axis at 3.43 kN; the strong-axis value of 30.8 kN is irrelevant unless the weak axis is braced. Pinning the top raises the critical load to 28.1 kN, about 8.2 times higher, from end conditions alone.