Column buckling worksheet with answers

Free printable Column buckling worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

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1.Easy
PL = 4 m20 × 80 mmcross-section

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)b × h = 20 × 80 mm
Pcr = kN
2.Easy
PL = 1.5 md = 30 mmcross-section

A column of length L made of aluminium is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.

L = 1.5 mE = 70 GPa (aluminium)d = 30 mm
Pcr = kN
3.Medium
PL = 4 md = 60 mmcross-section

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 = 4 mE = 70 GPa (aluminium)d = 60 mm
Pcr = kN
4.Medium
PL = 2 m40 × 60 mmcross-section

A column of length L made of aluminium is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.

L = 2 mE = 70 GPa (aluminium)b × h = 40 × 60 mm
Pcr = kN
5.Hard
PL = 2.5 md = 60 mmcross-section

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 = 2.5 mE = 200 GPa (steel)d = 60 mm
σcr = MPa
6.Hard
P = 10 kNL = 2 md = ?cross-section

A solid round column of length L made of aluminium is fixed at the base and free at the top. It must carry 10 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 10 kNn = 2L = 2 mE = 70 GPa (aluminium)
d = mm

Answer key

  1. 1. Pcr = 6.58 kN
  2. 2. Pcr = 12.21 kN
  3. 3. Pcr = 6.867 kN
  4. 4. Pcr = 112.8 kN
  5. 5. σcr = 145 MPa
  6. 6. d = 55.42 mm

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