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)b × h = 40 × 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 = 3 mE = 200 GPa (steel)d = 30 mm
Pcr = kN
3.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 = 2.5 mE = 200 GPa (steel)d = 30 mm
Pcr = kN
4.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)d = 25 mm
Pcr = kN
5.Hard
A column of length L made of steel is fixed at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 2.5 mE = 200 GPa (steel)d = 50 mm
σcr = MPa
6.Hard
A solid round column of length L made of steel is fixed at both ends. It must carry 100 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 100 kNn = 2.5L = 2.5 mE = 200 GPa (steel)
d = mm
Answer key
- 1. Pcr = 8.843 kN
- 2. Pcr = 8.721 kN
- 3. Pcr = 25.63 kN
- 4. Pcr = 37.85 kN
- 5. σcr = 197.4 MPa
- 6. d = 44.81 mm
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