Structural Steel Design / module 7
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 you will understand
- 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
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.
Buckling length, axis, and what restraint buys
Everything a designer can actually change is in L_cr and I. One restraint in the right place can be worth more than a third more steel.
Module summary
- Ncr = π²EI/L² and Npl = A fy are both exact; λ̄ = √(Npl/Ncr) is their ratio
- λ̄ ≪ 1 squashes, λ̄ ≫ 1 buckles elastically, λ̄ ≈ 1 is the worst place to be
- At λ̄ = 1 the design curve sits 40% below both ideal curves
- χ is CALIBRATED — α stands for bow, residual stress and progressive yielding at once
- The curve depends on axis and fabrication because those change the residual-stress pattern
- The weak axis governs unless something stops it — and one restraint can gain 36%
- The best column is one where both axes govern at once
- A higher grade helps a stocky column and does almost nothing for a slender one
Teaching basis and review status
These are educational models and exercises. Check their assumptions before using the results. The full technical content and every diagram have not yet been independently verified.
Concrete and steel calculations retain their stated first-generation Eurocode teaching basis. Use the standards and amendments specified for your project; do not mix generations without an appropriate design basis.
Read the lesson register and outstanding review notesCheck your understanding
Practice
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.
Practice
A section has A = 9172 mm² in S355 and an Euler load of 14 560 kN. What is λ̄?
Practice
A 5.0 m column is fixed at the base and free at the top. What is its buckling length, in m?
Practice
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?
Practice
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?
Practice
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?