Worked examples

Column example 2: a weak-axis tie permits a lighter section

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Repeat the $2,500 kN$ axial-column design with a $6 m$ pinned column and an effective midheight tie against weak-axis buckling. Select and check a Grade S355 UC.

Original source: LectureNotes/Ch 4_Column.pdf — p. 23, p. 24. 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	Origin
Axial force	Given $2,500 kN$ design compression.
Direction of tie	Question states weak direction: shorten $y$ -axis buckling length only.
Trial section	Data File p.11, $254 \times 254 \times 73$ UC: $T = 14.2 mm$ , $\frac{b}{T} = 8.96$ , $\frac{d}{t} = 23.3$ , $r_{x} = 11.1 cm$ , $r_{y} = 6.48 cm$ , $A = 93.1 {cm}^{2}$ .
Tables	pᵧ355 because $T \leq 16$ ; rolled H $\leq 40$ uses curves b/c.

Before calculating: recognition and strategy

Compare this to Example 1: restraints can be as important as area. The weak-axis tie reduces its unbraced length from $6$ to $3 m$ . A smaller column is now possible, but after changing section both radii, the area and the thickness-dependent strength must be reread. Do not reuse Example 1’s section properties.

1. Set each effective length

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

Major axis has no effective intermediate tie:

L_{Ex} = 6,000 mm

Minor axis has a mid-height tie:

L_{Ey} = 3,000 mm

Each of the two segments has length

3 m

and carries the full

2,500 kN

axial force.

A horizontal positional restraint suppresses a buckling displacement. It does not support half the vertical load or create a new $\frac{2,500}{2} kN$ compression case.

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.

2. Trial $254 \times 254 \times 73$ UC

T = 14.2 mm \leq 16 \to p_{y} = 355 \frac{N}{{mm}^{2}}

A_{g} = 93.1 {cm}^{2} = 9,310 {mm}^{2}

r_{x} = 11.1 cm = 111 mm; r_{y} = 6.48 cm = 64.8 mm

These values come from the same exact section row on Data File p.11. This lighter trial is the one selected in the lecture.

Animation labEffective length and buckling axes3 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.

3. Check local slenderness

ε = \sqrt{\frac{275}{355}} = 0.880141

13 ε = 11.4418; \frac{b}{T} = 8.96 < 11.4418

40 ε = 35.2056; \frac{d}{t} = 23.3 < 35.2056

Flange and web are both non-slender in uniform compression.

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.

4. Slenderness for the tied and untied directions

λ_{x} = \frac{6,000}{111} = 54.054054

λ_{y} = \frac{3,000}{64.8} = 46.296296

Now $λ_{x}$ exceeds $λ_{y}$ . That alone does not prove the $x$ axis controls because its curve b is more favourable than the weak-axis curve c.

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.

5. Interpolate each curve

Simple explanation: A value between two table rows

Move the same fraction of the way across both number ranges.

Choose the correct table, curve and strength column first.
Find how far your input lies between the two rows.
Apply that fraction to the change between their answers.

Remember: Teaching example: halfway between outputs 100 and 80 gives 90.

Related concept and full method

Use Table 8.8(b), pᵧ355 for $x$ and Table 8.8(c), pᵧ355 for $y$ .

Major axis:

54

row gives

288

，

56

row gives

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

. Interpolation fraction:

\frac{54.054054 - 54}{56 - 54} = 0.027027

p_{c} = 288 + 0.027027 \times (283 - 288) = 287.864865 \frac{N}{{mm}^{2}}

Minor axis:

46

row gives

286

，

48

row gives

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

. Interpolation fraction:

\frac{46.296296 - 46}{48 - 46} = 0.148148

p_{c} = 286 + 0.148148 \times (280 - 286) = 285.111111 \frac{N}{{mm}^{2}}

Despite its smaller slenderness, the weak axis gives the slightly smaller strength. The source values $288$ and $285 \frac{N}{{mm}^{2}}$ are rounded representations.

Animation labEffective length and buckling axes3 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.

6. Check the controlling resistance

P_{cx} = \frac{9,310 \times 287.864865}{1,000} = 2680.022 kN

P_{cy} = \frac{9,310 \times 285.111111}{1,000} = 2654.384 kN

P_{c} = 2654.384 kN > 2,500 kN

Adequate. Utilisation:

\frac{2,500}{2654.384} = 0.9418

The original rounds $p_{c}$ to $285$ , giving $\frac{93.1 \times 285}{10} = 2,653.35 kN$ , printed $2,653 kN$ . The exact interpolation gives $2,654.384 kN$ . The tie makes the $73 \frac{kg}{m}$ trial adequate where the untied example used $118 \frac{kg}{m}$ . This conclusion assumes the tie provides the stated effective restraint; tie design itself is not part of this question.

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

Use S355 $254 \times 254 \times 73$ UC， $p_{y} = 355$ ; under the $13 ε$ , $40 ε$ limits, it is non-slender. $L_{x}^{E} = 6 m$ , $L_{y}^{E} = 3 m$ ; $λ_{x} = 54.0541$ , $λ_{y} = 46.2963$ . Curves b/c give $p_{cx} = 287.865$ , $p_{cy} = 285.111 \frac{N}{{mm}^{2}}$ . $P_{c} = 2,654.384 kN > 2,500$ , adequate. Rounded source resistance $2,653 kN$ .

Mistakes to avoid

The tie changes only its effective buckling direction.
Use the full axial force in both $3 m$ portions.
Compare actual $p_{c}$ values, not slenderness alone.
Reclassify after selecting a lighter section.

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: remove the weak-axis tie but retain $254 \times 254 \times 73$ UC. Find the weak-axis slenderness and estimate its supplied-table resistance.

Reveal answer and reasoning

$λ_{y} = \frac{6,000}{64.8} = 92.59259$ . Curve c,pᵧ355: row $92$ gives $158$ and $94$ gives $153$ , so $p_{c} = 158 + (\frac{0.59259}{2}) (153 - 158) = 156.5185 \frac{N}{{mm}^{2}}$ . $P_{cy} = \frac{9,310 \times 156.5185}{1,000} = 1,457.187 kN$ , much less than $2,500$ . The untied version fails.

Animation labEffective length and buckling axes2 concepts · 3 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.

Column example 2: a weak-axis tie permits a lighter section

Read the diagram and collect the data

Before calculating: recognition and strategy

1. Set each effective length

2. Trial 254×254×73254×254×73 UC

3. Check local slenderness

4. Slenderness for the tied and untied directions

5. Interpolate each curve

6. Check the controlling resistance

Compact exam answer

Mistakes to avoid

Procedure for an unfamiliar variant

Independent self-check

2. Trial $254 \times 254 \times 73$ UC