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 = 2.5 mE = 70 GPa (aluminium)b × h = 40 × 80 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 = 1 mE = 200 GPa (steel)d = 40 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 = 5 mE = 200 GPa (steel)b × h = 25 × 50 mm
Pcr = kN
4.Medium
A column of length L made of aluminium is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 6 mE = 70 GPa (aluminium)d = 40 mm
Pcr = kN
5.Hard
A column of length L made of steel is fixed at the base and pinned at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 2 mE = 200 GPa (steel)d = 50 mm
σcr = MPa
6.Hard
A column of length L made of aluminium is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 2 mE = 70 GPa (aluminium)d = 50 mm
σcr = MPa
Answer key
- 1. Pcr = 47.16 kN
- 2. Pcr = 248.1 kN
- 3. Pcr = 20.56 kN
- 4. Pcr = 9.646 kN
- 5. σcr = 157.4 MPa
- 6. σcr = 6.747 MPa
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