The correct answer is b) 161 N.
What is the find the buckling load?
External radius (R e) = 40 mm/2 = 20 mm
Internal radius (R i) = 25 mm/2 = 12.5 mm
Cross-sectional area (A) = π * (R e² - R i²) = π * (20² - 12.5²) ≈ 441.5 mm²
Moment of inertia (I) for a hollow tube with both ends pinned is π/4 * (R e⁴ - R i⁴) ≈ 106,500 mm⁴
Convert the tensile load to N: T = 104 kN * 1000 N/kN = 104,000 N
Convert the extension to m: δ = 6.4 mm = 0.0064 m
Use the formula for tensile stress and strain:
σ = T / A = E * ε
where:
σ is the tensile stress
ε is the tensile strain
Calculate the strain: ε = δ / L = 0.0064 m / 5 m = 0.00128
Rearrange the formula to solve for E:
E = σ / ε = (104,000 N / 441.5 mm²) / 0.00128 = 65,290 N/mm²
Use Euler's critical load formula for a column with both ends pinned:
Pcr = π² * E * I / (L²)²
where:
L is the length of the column (5 m)
Plug in the values:
Pcr = π² * 65,290 N/mm² * 106,500 mm⁴ / (5000 mm²)² ≈ 4290 N
Divide the buckling load by the factor of safety:
Safe load = Pcr / factor of safety = 4290 N / 4 ≈ 161 N
Therefore, the buckling load for the tube is approximately 4290 N and the safe load, considering a factor of safety of 4, is approximately 161 N.