Column buckling: practice problems with step-by-step solutions

Also called: Euler buckling · critical load

Free Column buckling practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: pinned ends, other end conditions, design and critical stress.

Free practice problemEasy
PL = 2.5 m40 × 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 = 2.5 mE = 200 GPa (steel)b × h = 40 × 80 mm
kN
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What you practise

  1. Easypinned ends
  2. Mediumother end conditions
  3. Harddesign and critical stress

Key formulas

Euler load
Pcr = π²EI/(KL)²
Effective length factor
K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
Critical stress
σcr = Pcr/A
Slenderness
λ = KL/r, r = √(I/A)

Learn Column buckling from scratch

Before this: Centroid & second moment of area

1What buckling is

Press on both ends of a plastic ruler. At first nothing seems to happen. Then, quite suddenly, it bows out sideways. That is buckling.

A column is a member loaded in compression along its length: a building column, a strut in a roof, a piston rod.

PL

A short, thick column fails by crushing: the stress P/A reaches the strength of the material. A long, slender column buckles at a much lower load, while the stress is still far below that strength.

The load at which a slender column buckles is its critical load Pcr. Above it the column is unstable and gives way.

Self-check Two steel columns have the same cross-section. One is 2 m long, the other 4 m. Which buckles at the lower load?
  1. Both at the same load
  2. Neither: steel columns cannot buckle
  3. The 2 m column
  4. The 4 m column

Answer: The 4 m column. A longer column is more slender and buckles more easily. Length is the most important factor in buckling.

2Euler’s formula

Leonhard Euler found the critical load of a slender column:

Pcr = π² · E · I / (K·L)²
  • E is the Young’s modulus of the material: 200 GPa for steel.
  • I is the second moment of area of the cross-section. Use the smallest one: the column bends the easy way.
  • L is the length, and K is the effective length factor, which depends on how the ends are held.
K = 1K = 2K = 0.7K = 0.5

A pinned end can rotate; a fixed end cannot. Holding the ends more firmly gives a shorter effective length KL and a much higher Pcr, because KL is squared.

For a round bar I = π·d⁴/64. For a rectangle b × h with b the smaller side, Imin = h·b³/12.

Self-check A pinned–pinned column has Pcr = 100 kN. The same column is made fixed at the base and free at the top (K = 2). What is Pcr now?
  1. 200 kN
  2. 25 kN
  3. 400 kN
  4. 50 kN

Answer: 25 kN. Pcr is divided by K² = 4: 100/4 = 25 kN. 50 kN forgets to square K; 200 and 400 kN multiply instead of divide.

3Slenderness and critical stress

Dividing Pcr by the area gives the critical stress. It is neatest written with the radius of gyration r and the slenderness ratio λ (lambda):

r = √(I / A) λ = K·L / r σcr = π² · E / λ²

For a solid round bar, r = d/4.

Euler’s formula only holds while σcr stays below the yield stress of the material. For structural steel that means a slenderness of roughly λ > 100. Stockier columns fail by yielding or by a mix of both, and need other design rules.

Self-check A rectangular column section is 20 mm × 40 mm. What is the smallest second moment of area?
  1. 106 667 mm⁴
  2. 26 667 mm⁴
  3. 320 000 mm⁴
  4. 7854 mm⁴

Answer: 26 667 mm⁴. The weak axis cubes the smaller side: 40·20³/12 = 26 667 mm⁴. 106 667 is the strong axis; 320 000 forgets the /12; 7854 uses the formula for a circle.

4The method step by step

  1. Read the end conditions and choose K: 1, 2, 0.7 or 0.5.
  2. Find the smallest I of the section, in mm⁴.
  3. Convert to N and mm: L in mm, E in N/mm² (GPa × 1000).
  4. Effective length: KL.
  5. Critical load: Pcr = π²EI/(KL)², in N. Divide by 1000 for kN.
  6. If asked, check the slenderness KL/r or the critical stress Pcr/A.

For a design question with a factor of safety n, the column must have Pcr = n·P. Solve Euler’s formula for I, then for the diameter.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

A solid round steel column (E = 200 GPa) of diameter 40 mm and length 2 m is pinned at both ends. Find its critical buckling load Pcr. Then find it again for the same column fixed at the base and free at the top.

1

Look at the column

PL = 2 m

Pinned at both ends, so K = 1 and KL = 2000 mm. E = 200 000 N/mm².

2

Second moment of area

A round section has the same I about every axis:

I = π · 40⁴ / 64 = 125 664 mm⁴
3

Critical load

Pcr = π² · 200 000 · 125 664 / 2000² = 62 013 N ≈ 62.0 kN
4

Check that Euler applies

r = d/4 = 10 mm λ = 2000 / 10 = 200
σcr = π² · 200 000 / 200² = 49.3 MPa

That is far below the yield stress of steel (about 250 MPa), so the column is slender and Euler’s formula is valid.

5

Fixed base, free top

Now K = 2, so the effective length doubles to 4000 mm and Pcr drops by a factor of 2² = 4:

Pcr = 62.0 / 4 = 15.5 kN
Answer

Pcr = 62.0 kN pinned at both ends, and 15.5 kN when fixed at the base and free at the top.

6Common mistakes

  • Choosing the wrong K, or forgetting to square it.
  • Using the larger I of a rectangle. A column always buckles about its weak axis.
  • Mixing units: L in metres with E in N/mm². Work in N and mm throughout.
  • Using π·d⁴/32 (the polar moment J from torsion) instead of π·d⁴/64.
  • Applying Euler to a short, stocky column whose critical stress would be above the yield stress.
Self-check A column pinned at both ends has Pcr = 40 kN. The same column is fixed at both ends (K = 0.5). What is Pcr?
  1. 10 kN
  2. 160 kN
  3. 20 kN
  4. 80 kN

Answer: 160 kN. Pcr is divided by K² = 0.25, so it becomes four times larger: 160 kN. 80 kN forgets to square K; 10 kN multiplies by K² instead of dividing.

7How exam questions look

  • Easy: pinned ends. Find Pcr of a round or rectangular steel or aluminium column.
  • Medium: other end conditions. The same question with K = 2, 0.7 or 0.5.
  • Hard: design and critical stress. Find the smallest diameter that carries a load with a factor of safety, or find the slenderness and σcr.

For the design case with a round bar: d = (64·n·P·(KL)² / (π³·E))1/4.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
PL = 5 md = 50 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)d = 50 mm

Find Pcr (kN)

Show hints
  1. Euler: Pcr = π²EI/(KL)².
  2. Effective length factor: K = 1 for a column pinned at both ends.
  3. I = 306796 mm⁴.
Show the answer and the step-by-step solution

Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·50⁴/64 = 306796 mm⁴

Effective length

KL = 1·5000 = 5000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·306796/5000² = 8478 N = 8.478 kN

Answer

Pcr = 8.478 kN
Example 2Medium
PL = 2.5 md = 30 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 = 2.5 mE = 200 GPa (steel)d = 30 mm

Find Pcr (kN)

Show hints
  1. Euler: Pcr = π²EI/(KL)².
  2. Effective length factor: K = 2 for a column fixed at the base and free at the top.
  3. I = 39761 mm⁴.
Show the answer and the step-by-step solution

Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·30⁴/64 = 39761 mm⁴

Effective length

KL = 2·2500 = 5000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·39761/5000² = 3139 N = 3.139 kN

Answer

Pcr = 3.139 kN
Example 3Hard
PL = 2.5 md = 25 mmcross-section

A column of length L made of aluminium is pinned at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.

L = 2.5 mE = 70 GPa (aluminium)d = 25 mm

Find σcr (MPa)

Show hints
  1. σcr = Pcr/A = π²E/(KL/r)², with r = √(I/A) the radius of gyration.
  2. Use the smallest I, because the column buckles about its weakest axis.
  3. r = 6.25 mm.
Show the answer and the step-by-step solution

Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·25⁴/64 = 19175 mm⁴

Radius of gyration and slenderness

A = 490.9 mm²
r = √(I/A) = √(19175/490.9) = 6.25 mm
KL = 1·2500 = 2500 mm
KL/r = 2500/6.25 = 400

Critical stress

σcr = π²E/(KL/r)² = π²·70000/400² = 4.318 MPa

Answer

σcr = 4.318 MPa

Questions and answers

What is Column buckling?

Press on both ends of a plastic ruler. At first nothing seems to happen. Then, quite suddenly, it bows out sideways. That is buckling.

A column is a member loaded in compression along its length: a building column, a strut in a roof, a piston rod.

How do I solve Column buckling problems step by step?

Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.

What do the Column buckling exercises cover?
  • Easy: pinned ends
  • Medium: other end conditions
  • Hard: design and critical stress

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Topics before and after this one

Learn first

Centroid & second moment of area

Calculators

Euler buckling calculator

Courses with this topic

Mechanics of MaterialsStructural AnalysisMachine DesignFE Mechanical exam practiceFE Civil exam practice

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Shear force & bending momentTruss: method of jointsBeam deflectionCentroid & second moment of areaEffective stress in soilConsolidation settlementHydrostatic forceOpen-channel flow (Manning)

Printable worksheet

Column buckling worksheet with answers
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