A column of length L made of aluminium is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
Column buckling worksheet with answers
1.Easy
L = 1.5 mE = 70 GPa (aluminium)d = 60 mm
Pcr = kN
2.Easy
A column of length L made of steel is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 2 mE = 200 GPa (steel)b × h = 50 × 80 mm
Pcr = kN
3.Medium
A column of length L made of steel is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 2 mE = 200 GPa (steel)b × h = 20 × 60 mm
Pcr = kN
4.Medium
A column of length L made of steel is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 1.5 mE = 200 GPa (steel)b × h = 30 × 80 mm
Pcr = kN
5.Hard
A column of length L made of aluminium is pinned at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 4 mE = 70 GPa (aluminium)d = 60 mm
σcr = MPa
6.Hard
A column of length L made of steel is pinned at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 1.5 mE = 200 GPa (steel)d = 50 mm
σcr = MPa
Answer key
- 1. Pcr = 195.3 kN
- 2. Pcr = 411.2 kN
- 3. Pcr = 78.96 kN
- 4. Pcr = 322.3 kN
- 5. σcr = 9.715 MPa
- 6. σcr = 137.1 MPa
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