Worked examples

Column example 1: a pinned column in axial compression

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Select a Grade S355 UC for a $6 m$ column pinned at both ends, carrying a design axial compression of $2,500 kN$ . Check the selected section’s compression resistance.

Original source: LectureNotes/Ch 4_Column.pdf — p. 21, p. 22. Values tagged given are in the question or diagram; lookup values come from a named table; calculated values follow from the working; assumptions are stated explicitly.

Read the diagram and collect the data

Input	Source and meaning
Design load	Given $F_{c} = 2,500 kN$ , already factored; no dead/imposed breakdown is needed.
Restraints	Pinned-pinned, translation prevented at ends; Table 8.6 gives $K = 1$ in both axes.
Selected row	Data File p.11, $305 \times 305 \times 118$ UC: flange $T = 18.7 mm$ , $\frac{b}{T} = 8.22$ , $\frac{d}{t} = 20.6$ , $r_{x} = 13.6 cm$ , $r_{y} = 7.77 cm$ , $A = 150 {cm}^{2}$ .
Strength and curves	S355， $T = 18.7$ , giving $p_{y} = 345 \frac{N}{{mm}^{2}}$ . Hot-rolled H-section, $T \leq 40$ , so $x$ axis uses curve b; $y$ axis uses curve c.

Before calculating: recognition and strategy

There is no stated bending moment, so the core numerical task is axial member buckling. First ensure the local plates are not slender, allowing gross area. Then calculate both axis slendernesses and select the correct curve for each. The area alone is not enough: a long weak column can buckle well before reaching A𝗀pᵧ.

1. Choose a section and confirm its strength

The lecturer selects a trial $305 \times 305 \times 118$ UC. Its nominal mass $118 \frac{kg}{m}$ is part of the section designation. The strength calculation uses the tabulated gross area $150 {cm}^{2}$ . The flange thickness is $18.7 mm$ , falling within the S355 thickness band $16 < T \leq 40 mm$ .

p_{y} = 345 \frac{N}{{mm}^{2}}

A_{g} = 150 {cm}^{2} \times 100 = 15,000 {mm}^{2}

Gross-section squash resistance:

A_{g} p_{y} = \frac{15,000 \times 345}{1,000} = 5,175 kN

$5,175 kN$ exceeds the load but is not the member resistance. Proceed to local-slenderness and overall-buckling checks.

Animation labEffective length and buckling axes2 concepts · 2 source expressions

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

A column can bow sideways before its section reaches the crushing resistance.
Each axis has its own radius of gyration and restraint spacing.
$λ = \frac{L _{E}}{r}$ , with compatible length units. A tie affects only the directions it actually restrains.
Select each buckling curve and compressive strength before forming Pc. The governing axis is determined by resistance, not slenderness alone.

2. Establish a non-slender section

Simple explanation: A thin part can wrinkle first

A thin plate may wrinkle before the whole steel member reaches its intended resistance.

Check flange and web slenderness using their own definitions.
Compare each ratio with the correct class limits.
The less favourable element determines the section class.

Remember: Bending limits and uniform-compression limits are different.

Related concept and full method

ε = \sqrt{\frac{275}{345}} = 0.892805

Semi-compact limit for a uniformly compressed flange:

13 ε = 11.6065

\frac{b}{T} = 8.22 < 11.6065

Semi-compact limit for a uniformly compressed web:

40 ε = 35.7122

\frac{d}{t} = 20.6 < 35.7122

Both pass; the section is non-slender and the gross-section compression-resistance method applies.

This establishes Class 3 or better, not necessarily a plastic flange. The pure-compression example does not need plastic bending resistance; checking the non-slender boundary is sufficient for its stated method.

Animation labWhy thin elements buckle locally1 concept · 1 source expressions

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

The flange outstand and web have different widths, thicknesses and edge support conditions.
A thinner plate can wrinkle locally before the complete member loses stability.
Class 1 allows plastic rotation; Class 2 reaches plastic resistance; Class 3 reaches elastic resistance; Class 4 requires effective properties.
Check every relevant compression element with the supplied limits and stress distribution. The deformation shown is qualitative.

3. Calculate slenderness about both axes

Simple explanation: Why a long column can fail before crushing

Push a long thin ruler from both ends: it may bow sideways first.

Find effective length and radius of gyration for each axis.
Calculate slenderness for both directions.
Use the appropriate buckling curve before forming resistance.

Remember: Compare the final resistances; slenderness alone may not identify the controlling axis.

Related concept and full method

L_{Ex} = L_{Ey} = 1 \times 6,000 = 6,000 mm

r_{x} = 13.6 cm = 136 mm; r_{y} = 7.77 cm = 77.7 mm

λ_{x} = \frac{6,000}{136} = 44.117647

λ_{y} = \frac{6,000}{77.7} = 77.220077

The smaller weak-axis radius makes its slenderness much larger. Still read both curves rather than assuming the larger slenderness by itself proves which resistance controls.

Animation labEffective length and buckling axes1 concept · 1 source expressions

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

A column can bow sideways before its section reaches the crushing resistance.
Each axis has its own radius of gyration and restraint spacing.
$λ = \frac{L _{E}}{r}$ , with compatible length units. A tie affects only the directions it actually restrains.
Select each buckling curve and compressive strength before forming Pc. The governing axis is determined by resistance, not slenderness alone.

4. Read and interpolate the two compressive strengths

Simple explanation: A table lookup needs several keys

A table is like an address: knowing only the street is not enough.

Select the curve using section type, thickness and axis.
Select the material-strength column.
Bracket the slenderness and interpolate if required.

Remember: Do not jump between steel grades or extrapolate past supplied data.

Related concept and full method

Table 8.7, rolled H section with maximum thickness $18.7 \leq 40 mm$ : $x - x$ uses 8.8(b), $y - y$ uses 8.8(c). In each use the pᵧ=345 column of Data File p.8.

Major axis, curve b:

44

row gives

302

，

46

row gives

298 \frac{N}{{mm}^{2}}

. Interpolation fraction:

\frac{44.117647 - 44}{46 - 44} = 0.058824

p_{c} = 302 + 0.058824 \times (298 - 302) = 301.764706 \frac{N}{{mm}^{2}}

Minor axis, curve c:

76

row gives

196

，

78

row gives

191 \frac{N}{{mm}^{2}}

. Interpolation fraction:

\frac{77.220077 - 76}{78 - 76} = 0.610039

p_{c} = 196 + 0.610039 \times (191 - 196) = 192.949807 \frac{N}{{mm}^{2}}

The source rounds these to $302$ and $193 \frac{N}{{mm}^{2}}$ . The minor axis has the smaller $p_{c}$ and controls for the same area.

Animation labEffective length and buckling axes2 concepts · 2 source expressions

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

A column can bow sideways before its section reaches the crushing resistance.
Each axis has its own radius of gyration and restraint spacing.
$λ = \frac{L _{E}}{r}$ , with compatible length units. A tie affects only the directions it actually restrains.
Select each buckling curve and compressive strength before forming Pc. The governing axis is determined by resistance, not slenderness alone.

5. Form resistances and conclude

P_{cx} = \frac{15,000 \times 301.764706}{1,000} = 4526.471 kN

P_{cy} = \frac{15,000 \times 192.949807}{1,000} = 2894.247 kN

P_{c} = \min (P_{cx}, P_{cy}) = 2894.247 kN

F_{c} = 2,500 < P_{c}

; adequate under the specified axial compression. Utilisation:

\frac{2,500}{2894.247} = 0.8638

Using the lecturer’s rounded $193$ gives $\frac{150 \times 193}{10} = 2,895 kN$ , agreeing with the original answer. The governing failure mode is minor-axis overall buckling, not yielding of the full gross area.

Animation labEffective length and buckling axes1 concept · 2 source expressions

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

A column can bow sideways before its section reaches the crushing resistance.
Each axis has its own radius of gyration and restraint spacing.
$λ = \frac{L _{E}}{r}$ , with compatible length units. A tie affects only the directions it actually restrains.
Select each buckling curve and compressive strength before forming Pc. The governing axis is determined by resistance, not slenderness alone.

Compact exam answer

Select S355 $305 \times 305 \times 118$ UC; $T = 18.7$ , giving $p_{y} = 345$ . $\frac{b}{T} = 8.22 < 13 ε$ ， $\frac{d}{t} = 20.6 < 40 ε$ , non-slender. $L_{x}^{E} = L_{y}^{E} = 6 m$ ， $λ_{x} = 44.1176$ ， $λ_{y} = 77.2201$ . Curves b/c give respectively $p_{cx} = 301.765$ , $p_{cy} = 192.950 \frac{N}{{mm}^{2}}$ . $P_{c} = 2,894.247 kN > 2,500 kN$ , adequate. Rounded source result $2,895 kN$ .

Mistakes to avoid

Do not use $355 \frac{N}{{mm}^{2}}$ solely because the steel is S355.
Do not halve the effective length just because both ends are supported.
Do not use the same strut curve for both UC axes.
Do not treat the lower reaction arrow as a second applied load.

Procedure for an unfamiliar variant

Identify axial load, any moments, actual length and restraints in each axis.
Choose/read a section and confirm strength from the actual thickness.
Check the appropriate flange and web local-slenderness limits.
Find $λ_{x}$ and $λ_{y}$ separately; select each axis curve and interpolate $p_{c}$ .
Calculate both axial resistances and identify the governing axis.
If moments exist, complete section, flexural and axial/LTB interactions using their own capacities and moment definitions.

Independent self-check

Try it yourself. Invented variant: the same column now carries $3,000 kN$ design axial compression, with the same length and restraints. Does it pass?

Reveal answer and reasoning

No. Non-slender classification and buckling resistance are unchanged. Utilisation $= \frac{3,000}{2,894.247} \approx 1.0365 > 1$ . The gross squash resistance $5,175 kN$ cannot override the member buckling failure.

Animation labEffective length and buckling axes2 concepts · 1 source expressions

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

A column can bow sideways before its section reaches the crushing resistance.
Each axis has its own radius of gyration and restraint spacing.
$λ = \frac{L _{E}}{r}$ , with compatible length units. A tie affects only the directions it actually restrains.
Select each buckling curve and compressive strength before forming Pc. The governing axis is determined by resistance, not slenderness alone.