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Queensferry

Module 07

Compression members and flexural buckling

The route from an ideal column to a design resistance — five steps, three of them derivable and two of them not, and the join is where most misunderstanding lives.

What this module covers

  • Follow the chain from the Euler load to a design buckling resistance
  • Compute relative slenderness and say which failure mode it points to
  • Explain what the buckling curves represent and why there are five of them
  • Identify which axis governs, and why it is usually the weak one
  • Calculate the effect of an intermediate restraint
  • Say when a higher steel grade helps a column and when it does not

Lessons

  1. 7.1

    From Euler to a design resistance

    Five steps. The first three are mechanics, the last two are calibration, and knowing where the join falls is the point of the lesson.

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  2. 7.2

    Buckling length, axis, and what restraint buys

    Everything a designer can actually change is in Lcr and I. One restraint in the right place can be worth more than a third more steel.

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Module checkpoint

Check what you have taken in

6 questions

  1. Question 1

    A column has I = 11 240 cm⁴ and a buckling length of 4.0 m. What is its Euler load, in kN? Take E = 210 000 N/mm².

  2. Question 2

    A section has A = 9172 mm² in S355 and an Euler load of 14 560 kN. What is λ̄?

  3. Question 3

    A 5.0 m column is fixed at the base and free at the top. What is its buckling length, in m?

  4. Question 4

    By what factor does the Euler load of a fixed–fixed column exceed that of a fixed–free one of the same height and section?

  5. Question 5

    At λ̄ = 1.5 on curve b, χ = 0.342 and the Euler ratio is 0.444. By what percentage is the design curve below the Euler curve?

  6. Question 6

    A column's design resistance is 2152 kN and its squash load is 3256 kN. What percentage of the cross-section resistance has been lost to buckling?