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 = 2 m25 × 50 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 = 2 mE = 200 GPa (steel)b × h = 25 × 50 mm
Pcr = kN
2.Easy
PL = 1 md = 20 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 mE = 70 GPa (aluminium)d = 20 mm
Pcr = kN
3.Medium
PL = 2.5 m30 × 40 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 = 2.5 mE = 70 GPa (aluminium)b × h = 30 × 40 mm
Pcr = kN
4.Medium
PL = 6 m40 × 100 mmcross-section

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 = 6 mE = 200 GPa (steel)b × h = 40 × 100 mm
Pcr = kN
5.Hard
P = 100 kNL = 3 md = ?cross-section

A solid round column of length L made of steel is pinned 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 = 3 mE = 200 GPa (steel)
d = mm
6.Hard
P = 70 kNL = 2.5 md = ?cross-section

A solid round column of length L made of aluminium is fixed at the base and pinned at the top. It must carry 70 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 70 kNn = 2L = 2.5 mE = 70 GPa (aluminium)
d = mm

Answer key

  1. 1. Pcr = 32.13 kN
  2. 2. Pcr = 5.426 kN
  3. 3. Pcr = 2.487 kN
  4. 4. Pcr = 59.68 kN
  5. 5. d = 69.42 mm
  6. 6. d = 59.63 mm

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