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.

New worksheetPractise online with hints

For teachers: free to print, copy and use in class. Every new worksheet has new values and its own answer key. A link to ExamProblems on your course page is appreciated.

Name: ____________________   Date: ____________

1.Easy
PL = 1 md = 25 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 = 1 mE = 200 GPa (steel)d = 25 mm
Pcr = kN
2.Easy
PL = 5 m25 × 80 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 = 5 mE = 70 GPa (aluminium)b × h = 25 × 80 mm
Pcr = kN
3.Medium
PL = 5 md = 80 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 = 5 mE = 200 GPa (steel)d = 80 mm
Pcr = kN
4.Medium
PL = 5 m50 × 80 mmcross-section

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

L = 5 mE = 200 GPa (steel)b × h = 50 × 80 mm
Pcr = kN
5.Hard
P = 65 kNL = 6 md = ?cross-section

A solid round column of length L made of steel is pinned at both ends. It must carry 65 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 65 kNn = 2L = 6 mE = 200 GPa (steel)
d = mm
6.Hard
PL = 3 md = 20 mmcross-section

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. 1. Pcr = 37.85 kN
  2. 2. Pcr = 2.879 kN
  3. 3. Pcr = 324 kN
  4. 4. Pcr = 16.45 kN
  5. 5. d = 83.37 mm
  6. 6. σcr = 7.676 MPa

Full step-by-step solutions: practise this topic on examproblems.com.