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 = 6 mE = 70 GPa (aluminium)b × h = 30 × 50 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 = 4 mE = 200 GPa (steel)d = 25 mm
Pcr = kN
3.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 = 2 mE = 70 GPa (aluminium)d = 40 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 = 6 mE = 200 GPa (steel)b × h = 20 × 60 mm
Pcr = kN
5.Hard
A solid round column of length L made of aluminium is fixed at both ends. It must carry 115 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 115 kNn = 2.5L = 4 mE = 70 GPa (aluminium)
d = mm
6.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 = 4 mE = 200 GPa (steel)d = 60 mm
σcr = MPa
Answer key
- 1. Pcr = 2.159 kN
- 2. Pcr = 2.366 kN
- 3. Pcr = 86.82 kN
- 4. Pcr = 8.773 kN
- 5. d = 76.31 mm
- 6. σcr = 111 MPa
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