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 = 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 = 6 mE = 200 GPa (steel)b × h = 20 × 100 mm
Pcr = kN
3.Medium
A column of length L made of aluminium is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 2.5 mE = 70 GPa (aluminium)b × h = 40 × 80 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 = 5 mE = 70 GPa (aluminium)b × h = 25 × 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 aluminium is fixed at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 3 mE = 70 GPa (aluminium)d = 20 mm
σcr = MPa
Answer key
- 1. Pcr = 17.58 kN
- 2. Pcr = 3.655 kN
- 3. Pcr = 11.79 kN
- 4. Pcr = 11.51 kN
- 5. σcr = 9.715 MPa
- 6. σcr = 7.676 MPa
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