Learn

Columns: buckling, eccentric reactions and combined forces

Open “Animation lab” beside a teaching step for a visual explanation or a walkthrough of its original expressions. Models are illustrative; source answers remain unchanged.

Chinese–English terminology

Need a simpler picture? Open “Simple explanation” beside a difficult step. These optional notes do not replace the full solution.

A compressed column can fail by local plate buckling, cross-section yielding, or overall member buckling. Passing one does not prove the others. This chapter follows LectureNotes/Ch 4_Column.pdf and the supplied HK2011 tables. The numerical design scope is non-sway columns; sway theory is explained for recognition because it appears in the lectures.

Lecture examples are included immediately after the relevant concepts. Expand an example to read its complete diagram, working, exam answer and self-check here.

1. Effective length belongs to an axis and a restraint model

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

Slenderness measures how long and weak a member is in a particular buckling direction: λ=Lᴱ/ $r$ . Radius of gyration $r = \sqrt{\frac{I}{A}}$ describes how far area lies from that bending axis. Larger $r$ gives a smaller slenderness; a weak-axis tie can improve resistance without changing the steel section. It only helps the direction it actually restrains.

L_{Ex} = K_{x} L_{x}; λ_{x} = \frac{L_{Ex}}{r_{x}}

L_{Ey} = K_{y} L_{y}; λ_{y} = \frac{L_{Ey}}{r_{y}}

Idealized restraint pattern in Table 8.6	Theoretical $K$	Recommended $K$ when ideal conditions are approximated
Both ends fixed against rotation and translation	$0.5$	$0.70$
One end fixed, other pinned; translation prevented	$0.7$	$0.85$
One end fixed, other rotation-fixed but translation-free	$1.0$	$1.20$
Both ends pinned, translation prevented	$1.0$	$1.00$
Fixed-free cantilever	$2.0$	$2.10$
Source sixth pattern with partial rotational restraint and free translation at top	Not specified	$1.5$

Original Table8.6, Ch4 p.2: use the pictured end conditions and distinguish the theoretical and recommended rows. — Original Table 8.6, Ch 4 p.2: effective lengths of idealised columns. Dashed lines indicate the buckled shape. The six end-condition patterns, read left to right, and both rows of factors are reproduced in the table above. From top to bottom, the legend means: rotation and translation fixed; rotation free and translation fixed; rotation fixed and translation free; both free; rotation partially restrained and translation free. The source word “Transition” means translation here. The page formula is $λ = \frac{L_{E}}{r}$ , and $L_{E} = K L$ . Open full-size image

Use the question’s explicitly given effective length if present. Otherwise choose the source’s recommended factor for the modeled conditions, as the examples do. Do not use textbook theoretical $0.5$ where the worked course method uses $0.70$ . Pin symbols indicate free rotation, not free lateral translation. Keep $r$ and $L$ in the same units.

Try it yourself. A6 m pinned column has a midheight tie effective only against weak-axis buckling. Find Lᴱx and Lᴱy.

Reveal answer and reasoning

$L_{x}^{E} = 6 m$ ; $L_{y}^{E} = 3 m$ . Major-axis buckling can still extend over the full height.

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.

Animation labEffective length and buckling axes1 concept · 4 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. Select the buckling curve before reading compressive strength

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

First establish that the section is not slender. For uniform compression the lecture checks flange $\frac{b}{T} \leq 13 ε$ and web $\frac{d}{t} \leq 40 ε$ , the semi-compact limits. That permits the gross-area resistance $P_{c} = A_{g} p_{c}$ ; it does not assert Class 1. For combined bending and axial force, use the appropriate class limits before claiming plastic bending resistance.

Section; maximum thickness	Curve about $x - x$	Curve about $y - y$
Rolled I; $\leq 40 mm$	$a$	$b$
Rolled I; $> 40 mm$	$b$	$c$
Rolled H (UC); $\leq 40 mm$	$b$	$c$
Rolled H; $> 40 mm$	$c$	$d$
Welded I or H; $\leq 40 mm$	$b$	$c$
Welded I or H; $> 40 mm$	$b$	$d$

These rows are Table 8.7 in the Data File p.7. The full source on Ch 4 p.4 also covers hollow sections, bars, angles, channels, cover plates and welded boxes. The full table’s notes matter: thickness $40$ – $50 mm$ permits the specified averaging treatment; fabrication and cover-plate geometry can alter the curve. Do not select a curve solely from steel grade.

Full Table8.7, Ch4 p.4: section type, maximum thickness, buckling-axis columns and special notes. — Full Table 8.7, Ch 4 p.4: section type, maximum thickness, buckling axis and special notes. Each curve pair below is in the order $x - x$ ／ $y - y$ ; thickness boundaries follow the source table.
Hot-finished structural hollow sections in steel above S460, or hot-finished seamless hollow sections: $a_{0}$ ／ $a_{0}$ . Hot-finished hollow sections in steel below S460: $a$ ／ $a$ . These two source rows say “above” and “below”; neither explicitly assigns exactly S460, so do not invent an equality boundary. Cold-formed hollow sections with longitudinal or spiral welds: $c$ ／ $c$ .
Hot-rolled I-sections: thickness $\leq 40 mm$ uses $a$ ／ $b$ ; $> 40 mm$ uses $b$ ／ $c$ . Hot-rolled H-sections: for the same two thickness intervals, use respectively $b$ ／ $c$ and $c$ ／ $d$ . Welded I/H-sections: use respectively $b$ ／ $c$ and $b$ ／ $d$ ; also see note 2.
Hot-rolled sections with welded flange cover plates: select the row using the Figure 8.4 ratio $\frac{U}{B}$ . When $0.25 < \frac{U}{B} < 0.80$ : for I-sections in the two thickness intervals above, use respectively $a$ ／ $b$ and $b$ ／ $c$ ; for H-sections, use respectively $b$ ／ $c$ and $c$ ／ $d$ . When $\frac{U}{B} \geq 0.80$ : for I/H-sections, use respectively $b$ ／ $a$ and $c$ ／ $b$ . When $\frac{U}{B} \leq 0.25$ : for I/H-sections, use respectively $b$ ／ $c$ and $b$ ／ $d$ .
Welded box sections: for the two thickness intervals above, use respectively $b$ ／ $b$ and $c$ ／ $c$ ; also see note 3. Round, square or flat bars: use respectively $b$ ／ $b$ and $c$ ／ $c$ . Hot-rolled angles, channels and tees; two hot-rolled sections connected by lacing, battens or back to back; and built-up hot-rolled sections: for either axis, use $c$ .
Note 1: for thickness $40 mm$ to $50 mm$ , for the same relevant $p_{y}$ , average the values for thickness not exceeding $40 mm$ and thickness exceeding $40 mm$ of $p_{c}$ . Note 2: for welded I/H-section flanges machine flame-cut without subsequent edge grinding or machining, when buckling about the $y - y$ axis, flange thickness not exceeding $40 mm$ may use curve $b$ ; above that thickness, use $c$ . Note 3: “welded box sections” includes boxes assembled from plates or hot-rolled sections with all longitudinal welds near the corners, but excludes boxes with longitudinal stiffeners. Note 4: the source permits curves from other recognised codes when differences in load/material partial factors are addressed and the curves are calibrated against Tables 8.8(a₀), (a) to (h); the Table 8.8 footnote also applies. This does not supply numerical values that can simply be mixed with the course tables.Open full-size image

Identify section family and fabrication, maximum thickness and buckling axis; choose its curve.
Confirm pᵧ from grade and thickness. Select that strength column within the appropriate Table 8.8 curve.
Calculate $λ$ for that axis and locate bracketing rows. Interpolate $p_{c}$ ; do not extrapolate beyond supplied tables.
Compute $P_{cx} = A_{g} p_{cx}$ and $P_{cy} = A_{g} p_{cy}$ ; use the smaller for the flexural compression check.

A

uses

{cm}^{2}

，

p_{c}

uses

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

P_{c} (kN) = \frac{A \times 100 \times p_{c}}{1,000} = \frac{A p_{c}}{10}

Check both axes even if rᵧ is smaller: different effective lengths and different curves may change which controls. Data File p.8 supplies S355 curves only over its printed rows; the lecture and 2023 extra data contain further tables. Choose a supplied applicable table and cite it; do not invent a value from the trend.

Column example 1: a pinned column in axial compression

Open this example on its own page

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.

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

Open this example on its own page

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.

Animation labEffective length and buckling axes3 concepts · 17 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. Cross-section capacity: axial force consumes some of the same resistance

Simple explanation: The column needs more than one pass

A slice can be strong while the whole member still buckles.

Check cross-section compression plus bending.
Then check the separate member-buckling expressions.
Keep each moment, factor and resistance in its specified expression.

Remember: The three checks do not share interchangeable denominators.

Related concept and full method

\frac{F_{c}}{A_{g} p_{y}} + \frac{| M_{x} |}{M_{cx}} + \frac{| M_{y} |}{M_{cy}} \leq 1

Class 1／2:

M_{cx} = \min (p_{y} S_{x}, 1.2 p_{y} Z_{x})

M_{cy} = \min (p_{y} S_{y}, 1.2 p_{y} Z_{y})

$F_{c}$ is design compression at the critical section; $M_{x}$ and Mᵧ are design moments there, including the specified amplification. $A_{g}$ is gross area. Each ratio is dimensionless, so keep force units consistent and moment units consistent. Use magnitudes for the conservative linear interaction; moments about perpendicular axes do not cancel. If using maxima from different locations, label the check as a conservative envelope. This lecture expression excludes Class 4 without an appropriate local-buckling treatment.

The minor-axis 1.2pᵧZᵧ ceiling frequently controls UC sections: calculating pᵧSᵧ alone overstates the allowed moment. The axial denominator here is A𝗀pᵧ, not the reduced member resistance $A_{g} p_{c}$ .

Animation labA strong slice can belong to an unstable member2 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.

Combine axial compression and the two bending demands using the specified section resistances.
The whole member adds effective-length and buckling-curve effects.
This uses its own moment factor and bending resistance; it is not a copy of the section check.
Elastic, plastic and buckling resistances are not interchangeable. Read the three original expressions and their first-order/amplified moments.

4. Member buckling: keep the source’s barred quantities distinct

Simple explanation: The column needs more than one pass

A slice can be strong while the whole member still buckles.

Check cross-section compression plus bending.
Then check the separate member-buckling expressions.
Keep each moment, factor and resistance in its specified expression.

Remember: The three checks do not share interchangeable denominators.

Related concept and full method

The source uses bars to distinguish first-order moments and some resistance definitions. Small overbars can disappear in extracted text, so read the original image. Here “Mᵧ,first” explicitly means the source’s barred Mᵧ. “ $M_{x, amp}$ ” and “Mᵧ,amp” include the required amplification. Elastic moment denominators in these member equations are pᵧZ, even for a Class 1 section.

Elastic denominator:

M_{ex} = p_{y} Z_{x}

;

M_{ey} = p_{y} Z_{y}

.
Non-sway flexural-buckling interaction (course Eq.8.80):

\frac{F_{c}}{P_{c}} + \frac{m_{x} | M_{x, amp} |}{M_{ex}} + \frac{m_{y} | M_{y, amp} |}{M_{ey}} \leq 1

Combined axial force/LTB (Eq.8.81):

\frac{F_{c}}{P_{cy}} + \frac{m_{LT} | M_{LT} |}{M_{b}} + \frac{m_{y} | M_{y, first} |}{M_{ey}} \leq 1

For the non-sway worked method, $P_{c}$ is the smaller axis resistance based on the stated/computed effective lengths; $P_{cy}$ specifically uses weak-axis buckling. Mᴸᵀ is the amplified major-axis moment governing lateral-torsional resistance, often given separately by an exam question. Do not amplify a supplied “already amplified” Mᴸᵀ twice. The last term of Eq.8.81 uses first-order minor-axis moment in the supplied source; the flexural equation uses the amplified one. This explains the apparently different $30$ and $34.5 kN \cdot m$ values in Example 5.

Ch4 p.15: original Eqs.8.79–8.81 and definitions; inspect the overbars. — Ch 4 p.15: original Eqs.8.79–8.81 and their definitions. Overbars are retained as printed. These are the source symbols; the renamed teaching symbol $M_{ex}$ , $M_{ey}$ used above is not substituted into the original equations.
Eqn 8.79: $\frac{F_{c}}{P_{c}} + \frac{m_{x} {\bar{M}}_{x}}{M_{cx}} + \frac{m_{y} {\bar{M}}_{y}}{M_{cy}} \leq 1$ Eqn 8.80: $\frac{F_{c}}{{\bar{P}}_{c}} + \frac{m_{x} M_{x}}{M_{cx}} + \frac{m_{y} M_{y}}{M_{cy}} \leq 1$ Eqn 8.81: $\frac{F_{c}}{P_{cy}} + \frac{m_{LT} M_{LT}}{M_{b}} + \frac{m_{y} {\bar{M}}_{y}}{M_{cy}} \leq 1$ Source definitions: $F_{c}$ is the design compressive force at the critical location; $P_{cx}$ , $P_{cy}$ are the compression resistances about the two axes in sway mode; $P_{c}$ is the smaller of these; ${\bar{P}}_{c}$ is the smaller of the two non-sway compression resistances, obtained by second-order analysis or by taking effective length equal to actual member length. $M_{b}$ is the buckling moment resistance under clause 8.3.5.2. $M_{x}$ , $M_{y}$ are the maximum major- and minor-axis design moments already amplified for $P - Δ - δ$ effects; ${\bar{M}}_{x}$ , ${\bar{M}}_{y}$ are the corresponding first-order linear-analysis moments. $M_{LT}$ is the amplified major-axis moment governing $M_{b}$ . In this source figure, $M_{cx} = p_{y} Z_{x}$ , $M_{cy} = p_{y} Z_{y}$ denotes elastic moment resistance. $m_{x}$ , $m_{y}$ comes from Table 8.9; $m_{LT}$ comes from Table 8.4.Open full-size image

For sway recognition only, Eq.8.79 uses sway-mode compression resistance with first-order end moments, while Eq.8.80 uses the alternative non-sway-mode compression resistance with amplified moments. Eq.8.81 adds axial/LTB interaction. The source explains that different lateral-load/initial-curvature effects can control. Its p.17 explicitly excludes numerical sway-column design from the course; the worked pages therefore use the non-sway branches rather than constructing an unsupported sway analysis.

Animation labA strong slice can belong to an unstable member1 concept · 6 source expressions

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

Combine axial compression and the two bending demands using the specified section resistances.
The whole member adds effective-length and buckling-curve effects.
This uses its own moment factor and bending resistance; it is not a copy of the section check.
Elastic, plastic and buckling resistances are not interchangeable. Read the three original expressions and their first-order/amplified moments.

5. Non-sway, sway and ultra-sensitive frames

Source analysis route	Non-sway	Sway	Ultra-sensitive
Ordinary course route	$λ_{cr} \geq 10$	$5 \leq λ_{cr} < 10$	$λ_{cr} < 5$
Advanced-analysis classification quoted by lecture	$λ_{cr} \geq 15$	$5 \leq λ_{cr} < 15$	$λ_{cr} < 5$

$λ_{cr}$ is the elastic critical load factor: the multiple of the applied loading at which an elastic instability mode occurs. It is not the member slenderness λ=Lᴱ/r. Non-sway classification permits the stated global $P - Δ$ simplification; local member $P - δ$ amplification may still be needed. For ultra-sensitive frames the lecture requires second-order $P - Δ - δ$ or advanced analysis. Bracing is the physical system restraining sway; the classification checks how sensitive the modeled structure is.

Animation labRestraints, sway and imperfections2 concepts · 6 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 frame needs a defined path for horizontal force as well as gravity.
Pinned joints alone do not provide frame moment resistance; sway restraint needs a real structural system.
Diagonal bracing carries horizontal action through axial forces. This sketch does not assign a numerical frame classification.
Use the specified imperfection/notional-force model and critical-load criteria. Avoid counting alternative imperfection models twice.

6. Why moments increase under compression

Simple explanation: Moving sideways gives the load a new lever arm

Once a compressed member moves sideways, the same compression creates extra moment.

P–Δ concerns overall frame or storey movement.
P–δ concerns bowing relative to the member’s end line.
Amplify only the first-order moments specified by the method.

Remember: Do not amplify a moment that the question already gives as amplified.

Related concept and full method

A compression force acting through a lateral displacement creates additional moment. Capital $Δ$ usually denotes frame/storey sway; small $δ$ denotes member curvature. Second-order analysis can include these effects directly. If the exercise instead gives amplification factors, multiply the appropriate first-order moment once.

Lecture Eq.6.1:

λ_{cr} = (\frac{F_{N}}{F_{V}}) (\frac{h}{δ_{N}})

This lecture normally takes

F_{N} = 0.005 F_{V}

.
Global amplification factor:

\frac{1}{1 - \frac{F_{V} δ_{N}}{F_{N} h}}

Member amplification factor:

\frac{1}{1 - \frac{F_{c} L_{E}^{2}}{π^{2} E I}}

$F_{v}$ is the factored vertical load at and above the storey; Fᴺ is the notional horizontal load; $h$ is storey height; δᴺ is relative storey displacement caused by Fᴺ. The lecture uses the larger relevant global/member factor for its general discussion. For non-sway member design it specifies the local member factor. Use a consistent force/length system so both denominator ratios are dimensionless. A nonpositive denominator does not mean a negative useful amplification: the approximation has reached/breached its instability condition.

Try it yourself. Invented arithmetic: $h = 4,000 mm$ , Fᴺ/Fᵥ=0.005, δᴺ=2 mm. Find $λ_{cr}$ and the global factor.

Reveal answer and reasoning

$λ_{cr} = \frac{0.005 \times 4,000}{2} = 10$ . The ordinary route is at the non-sway boundary. The displayed global factor is $\frac{10}{10 - 1} = 1.1111$ ; a separate member effect may still govern.

Animation labSeparate P–Δ from P–δ1 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.

The dashed line marks the original frame position; axial compression acts downwards.
Global displacement Δ creates the additional moment PΔ.
Local displacement δ is measured from the line joining the displaced ends, and adds Pδ.
Follow the original analysis route. Do not amplify an already amplified moment or confuse critical-load factor with member slenderness.

Animation labSeparate P–Δ from P–δ1 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.

The dashed line marks the original frame position; axial compression acts downwards.
Global displacement Δ creates the additional moment PΔ.
Local displacement δ is measured from the line joining the displaced ends, and adds Pδ.
Follow the original analysis route. Do not amplify an already amplified moment or confuse critical-load factor with member slenderness.

7. Flexural moment factors are not the LTB factors

Simple explanation: Similar-looking moment factors come from different tables

Two recipes can use the same ingredients but different amounts.

Identify flexural buckling or lateral-torsional buckling first.
Read the actual moment diagram, including any change of sign.
Use that check’s table and its own sign and minimum-factor rules.

Remember: Table 8.9 flexural factors are not Table 8.4 LTB factors.

Related concept and full method

For an end-moment-only segment, read Table 8.9 with the signed internal end-moment ratio $β$ . $β$ positive means the BMD stays on the same side; negative means it reverses. The lecture’s column-end applied-arrow convention gives $β = - \frac{M_{bottom}}{M_{top}}$ ; confirm the actual arrow/BMD rather than applying that minus sign blindly to internal BMD ordinates.

Table 8.9, end moments only:
When

0 \leq β \leq 1

m = 0.6 + 0.4 β

.
When

- 1 \leq β \leq 0

m = 0.6 + 0.2 β

.
For example,

β = - 0.3 \to m = 0.54

;

β = + 0.7 \to m = 0.88

.
But Table 8.4a for LTB, at

β = - 0.3

, gives

m_{LT} = 0.48

The two tables differ especially for reversing bending. They also have different general quarter-point formulas. For Table 8.9:

m = 0.2 + \frac{0.1 M_{2} + 0.6 M_{3} + 0.1 M_{4}}{M_{\max}}

but require

m \geq \frac{0.8 M_{24}}{M_{\max}}

$M_{2}$ and $M_{4}$ are quarter-point moments; $M_{3}$ is mid-length moment. $M_{24}$ is the maximum moment magnitude in the central half. If the three ordinates lie on one side, take them positive; if they straddle the axis, choose the positive side that gives the larger $m$ . $M_{\max}$ and $M_{24}$ are always positive. This sign rule differs from Table 8.4b, which uses all magnitudes. The lecture’s special flexural cases give $0.90$ for one central point, $0.95$ for a UDL, $0.95$ for three equal equally spaced points and $0.80$ for the pictured two equal points; match the drawing before using them.

Table8.9, Ch4 p.18, including special cases and the general-case sign rule. — Table 8.9, Ch 4 p.18: special cases and sign rules for the general case. X marks lateral restraint; $L$ ／ $L_{LT}$ marks segment length. The two end moments above are $M$ and $β M$ . In the left sketch, $β$ is positive; in the right sketch, $β$ is negative. The end-moment table gives $β \to m$ row by row as follows:
$1.0 \to 1.00$ ， $0.9 \to 0.96$ ， $0.8 \to 0.92$ ， $0.7 \to 0.88$ ， $0.6 \to 0.84$ ， $0.5 \to 0.80$ ， $0.4 \to 0.76$ ， $0.3 \to 0.72$ ， $0.2 \to 0.68$ ， $0.1 \to 0.64$ ， $0.0 \to 0.60$ ， $- 0.1 \to 0.58$ ， $- 0.2 \to 0.56$ ， $- 0.3 \to 0.54$ ， $- 0.4 \to 0.52$ ， $- 0.5 \to 0.50$ ， $- 0.6 \to 0.48$ ， $- 0.7 \to 0.46$ ， $- 0.8 \to 0.44$ ， $- 0.9 \to 0.42$ ， $- 1.0 \to 0.40$ .
The special cases at lower left, from top to bottom, are: one point load at midspan, $m = 0.90$ ; UDL, $m = 0.95$ ; three equal, equally spaced point loads, $m = 0.95$ ; two equal point loads at the third points, $m = 0.80$ . In the lower-right sketch, $M_{1}$ , $M_{5}$ are the ends; $M_{2}$ , $M_{4}$ are the quarter points; $M_{3}$ is the midpoint. The source expression is: $m = 0.2 + \frac{0.1 M_{2} + 0.6 M_{3} + 0.1 M_{4}}{M_{\max}}$ but $m \geq \frac{0.8 M_{24}}{M_{\max}}$ . If all three sampled moments lie on the same side, take them positive. If they lie on different sides, choose as positive the side producing the larger $m$ . $M_{\max}$ , $M_{24}$ are both positive: respectively the maximum moment over the whole segment and over its central half.Open full-size image

Column example 5: axial load plus biaxial end moments

Open this example on its own page

Check a $254 \times 254 \times 89$ UC S355 column in a braced non-sway frame, $4.5 m$ long, fixed at one end and pinned at the other. Design compression is $1,650 kN$ . First-order factored end moments are $M_{x}$ (top/bottom)= $+ 80$ / $+ 24 kN \cdot m$ and Mᵧ(top/bottom)= $+ 30$ / $- 21 kN \cdot m$ . Amplification factors are $1.08$ about $x$ and $1.15$ about $y$ . The given amplified Mᴸᵀ=86.4 kNm has the same distribution as $M_{x}$ .

Original source: LectureNotes/Ch 4_Column.pdf — p. 36, p. 37, p. 38. 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.

Ch 4 p.36 supplies a load table, not a structural sketch. Clockwise moments are positive and anticlockwise moments negative. Top/bottom identify member ends; factored loads already include partial load factors; the amplification factor increases the moment.

4.5 m

length and the fixed/pinned-end wording establish the model. The source table writes the amplification effect as

P - Δ - δ

; its original expression for the already amplified moment is

M_{LT} = 80 \times 1.08 = 86.4 kN \cdot m

, with the same distribution as

M_{x}

. The main explanation also uses

M^{LT}

for that same given moment, not a separate unamplified input. The source load table and sign convention are retained; no new dimensions are inferred. End moments and section properties are typeset step by step below.Open full-size image

Read the diagram and collect the data

Input	Meaning/value source
End moments	Given applied-arrow convention: $+ 80$ / $+ 24$ about $x$ ; $+ 30$ / $- 21$ about $y$ . Convert to the Table 8.9 internal-diagram ratio.
Amplification	Given $1.08$ and $1.15$ ; Mᴸᵀ86.4 already includes amplification.
Section row	Data File p.11, $254 \times 254 \times 89$ UC: $T = 17.3$ , $t = 10.3$ , $d = 200.3 mm$ ; $\frac{b}{T} = 7.41$ , $\frac{d}{t} = 19.4$ ; $r_{x} = 11.2 cm$ , $r_{y} = 6.55 cm$ ; $A = 113 {cm}^{2}$ ; $Z_{x} = 1,096$ , $Z_{y} = 379$ , $S_{x} = 1,224$ , $S_{y} = 575 {cm}^{3}$ ; $u = 0.850$ , $x = 14.5$ .
Restraint/curve choices	Table 8.6 recommended fixed/pinned K0.85; rolled H $\leq 40 mm$ uses b about $x$ ,c about $y$ .

Before calculating: recognition and strategy

This is continuous-frame end-moment design, so use the full $u - v$ LTB procedure and moment factors from the separate flexural/LTB tables. Check three things: local section interaction, flexural member interaction, and axial/LTB interaction. The section moment denominator is capped plastic resistance, whereas the member denominators are elastic pᵧZ. The source’s LTB minor-axis term uses first-order Mᵧ; its flexural term uses amplified Mᵧ.

1. Strength, combined-stress classification and moments

T = 17.3 mm

lies within

16 < T \leq 40

, so

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

ε = \sqrt{\frac{275}{345}} = 0.892805

Class 1 flange limit

9 ε = 8.03525

;

\frac{b}{T} = 7.41 < 8.03525

. Web compression ratio:

r_{1} = \frac{F_{c}}{d t p_{y}}

= \frac{1,650,000}{200.3 \times 10.3 \times 345} \approx 2.318

Under the supplied classification rule,

r_{1}

is limited to

1

. Class 1 web limit:

\frac{80 ε}{1 + r_{1}} = \frac{80 \times 0.892805}{2} = 35.7122

\frac{d}{t} = 19.4 < 35.7122

, so the web is Class 1 and the overall section is Class 1.

M_{x, amp} = 1.08 \times 80 = 86.4 kN \cdot m

M_{y, amp} = 1.15 \times 30 = 34.5 kN \cdot m; M_{y, first} = 30 kN \cdot m

Given

M_{LT} = 86.4 kN \cdot m

; do not amplify it again.

The raw $r_{1}$ exceeds $1$ because the axial load is divided by the web’s $d t$ area, not total section area. The course classification parameter is bounded; this does not mean gross-area compression exceeds capacity.

Animation labWhy thin elements buckle locally2 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.

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.

2. Cross-section interaction

Simple explanation: The column needs more than one pass

A slice can be strong while the whole member still buckles.

Check cross-section compression plus bending.
Then check the separate member-buckling expressions.
Keep each moment, factor and resistance in its specified expression.

Remember: The three checks do not share interchangeable denominators.

Related concept and full method

A_{g} p_{y} = \frac{113 \times 100 \times 345}{1,000} = 3,898.5 kN

Major axis:

p_{y} S_{x} = \frac{345 \times 1,224}{1,000} = 422.28 kN \cdot m

Upper cap:

1.2 p_{y} Z_{x} = \frac{1.2 \times 345 \times 1,096}{1,000} = 453.744 kN \cdot m

M_{cx} = 422.28 kN \cdot m

Minor axis:

p_{y} S_{y} = \frac{345 \times 575}{1,000} = 198.375 kN \cdot m

Upper cap:

1.2 p_{y} Z_{y} = \frac{1.2 \times 345 \times 379}{1,000} = 156.906 kN \cdot m

M_{cy} = 156.906 kN \cdot m

Utilisation:

\frac{1,650}{3,898.5} + \frac{86.4}{422.28} + \frac{34.5}{156.906}

= 0.847720 < 1

Section passes.

Both maximum moment magnitudes occur at the top in the given load schedule, so they coexist with the stated compression there. The minor-axis ceiling is the governing branch for $M_{cy}$ .

Animation labA strong slice can belong to an unstable member2 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.

Combine axial compression and the two bending demands using the specified section resistances.
The whole member adds effective-length and buckling-curve effects.
This uses its own moment factor and bending resistance; it is not a copy of the section check.
Elastic, plastic and buckling resistances are not interchangeable. Read the three original expressions and their first-order/amplified moments.

3. Effective length and both axial buckling resistances

L_{Ex} = L_{Ey} = 0.85 \times 4,500 = 3,825 mm

r_{x} = 112 mm; r_{y} = 65.5 mm

λ_{x} = \frac{3,825}{112} = 34.151786

λ_{y} = \frac{3,825}{65.5} = 58.396947

Select curve b for $x$ and c for $y$ ; both use pᵧ345. The source shows only the controlling $y$ result; the $x$ lookup below completes the omitted comparison.

Curve b,

x

axis:

30

row gives

325

，

35

row gives

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

. Interpolation fraction:

\frac{34.151786 - 30}{35 - 30} = 0.830357

p_{c} = 325 + 0.830357 \times (318 - 325) = 319.187500 \frac{N}{{mm}^{2}}

Curve c,

y

axis:

58

row gives

247

，

60

row gives

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

. Interpolation fraction:

\frac{58.396947 - 58}{60 - 58} = 0.198473

p_{c} = 247 + 0.198473 \times (241 - 247) = 245.809160 \frac{N}{{mm}^{2}}

P_{cx} = \frac{113 \times 319.187500}{10} = 3606.818750 kN

P_{cy} = \frac{113 \times 245.809160}{10} = 2777.643511 kN

P_{c} = P_{cy} = 2777.643511 kN

Animation labEffective length and buckling axes3 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.

4. Read the two flexural factors and the LTB factor

Simple explanation: Similar-looking moment factors come from different tables

Two recipes can use the same ingredients but different amounts.

Identify flexural buckling or lateral-torsional buckling first.
Read the actual moment diagram, including any change of sign.
Use that check’s table and its own sign and minimum-factor rules.

Remember: Table 8.9 flexural factors are not Table 8.4 LTB factors.

Related concept and full method

The question defines signs for applied clockwise/anticlockwise end moments. The internal end-moment diagram reverses the bottom-end sign relative to that schedule. Thus the same positive applied signs about $x$ create a reversing BMD, while the opposite applied signs about $y$ create a same-side BMD.

β_{x} = \frac{- 24}{80} = - 0.3

Table 8.9:

m_{x} = 0.54

β_{y} = \frac{- (- 21)}{30} = + 0.7

Table 8.9:

m_{y} = 0.88

M_{LT}

and

x

axis has the same moment-diagram shape;

β = - 0.3

. Table 8.4a:

m_{LT} = 0.48

Do not read $0.48$ as $m_{x}$ : that belongs to the different LTB table. Uniform amplification about an axis multiplies both ends equally and leaves the end-moment ratio unchanged.

Animation labRead the moment shape within one segment3 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.

Moment ordinates must belong to the same effective unbraced segment.
Same-side and reverse-curvature diagrams have different signed end ratios.
The markers show ¼, ½ and ¾ of this segment, not of the entire beam.
LTB and column flexural factors are different. Preserve their individual bounds and coefficient sets.

5. Flexural member interaction with elastic denominators

M_{ex} = p_{y} Z_{x} = \frac{345 \times 1,096}{1,000} = 378.12 kN \cdot m

M_{ey} = p_{y} Z_{y} = \frac{345 \times 379}{1,000} = 130.755 kN \cdot m

U_{flex} = \frac{1,650}{2777.643511} + \frac{0.54 \times 86.4}{378.12} + \frac{0.88 \times 34.5}{130.755}

= 0.594029 + 0.123389 + 0.232190

= 0.949608 < 1

Passes.

Animation labA strong slice can belong to an unstable member2 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.

Combine axial compression and the two bending demands using the specified section resistances.
The whole member adds effective-length and buckling-curve effects.
This uses its own moment factor and bending resistance; it is not a copy of the section check.
Elastic, plastic and buckling resistances are not interchangeable. Read the three original expressions and their first-order/amplified moments.

6. Axial/LTB interaction

Simple explanation: A beam can escape sideways

The compressed flange can move sideways while the section twists.

Divide the beam at effective lateral restraints.
Use each segment’s effective length to obtain its buckling resistance.
Compare that resistance with the segment’s equivalent moment demand.

Remember: A section bending check alone does not check lateral-torsional buckling.

Related concept and full method

λ = 58.396947

; section torsional index

x = 14.5

v = {[1 + 0.05 {(\frac{58.396947}{14.5})}^{2}]}^{- 0.25} = 0.862028

β_{w} = 1; λ_{LT} = 0.850 \times 0.862028 \times 58.396947 = 42.788808

Table 8.3a，

p_{y} = 345

column:

40

row gives

317

，

45

row gives

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

. Interpolation fraction:

\frac{42.788808 - 40}{45 - 40} = 0.557762

p_{b} = 317 + 0.557762 \times (302 - 317) = 308.633576 \frac{N}{{mm}^{2}}

M_{b} = \frac{308.633576 \times 1,224}{1,000} = 377.767497 kN \cdot m

U_{LT} = \frac{1,650}{2777.643511} + \frac{0.48 \times 86.4}{377.767497} + \frac{0.88 \times 30}{130.755}

= 0.594029 + 0.109782 + 0.201904

= 0.905715 < 1

Passes.

The $30 kN \cdot m$ in the last term is intentional: source Eq.8.81 has barred, first-order Mᵧ. The $86.4 kN \cdot m$ Mᴸᵀ is already amplified. All three interactions pass; flexural member interaction is closest to $1$ .

Animation labA strong slice can belong to an unstable member5 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.

Combine axial compression and the two bending demands using the specified section resistances.
The whole member adds effective-length and buckling-curve effects.
This uses its own moment factor and bending resistance; it is not a copy of the section check.
Elastic, plastic and buckling resistances are not interchangeable. Read the three original expressions and their first-order/amplified moments.

Compact exam answer

S355 $254 \times 254 \times 89$ UC: $p_{y} = 345$ , Class 1. Section utilisation $0.84772$ . $L^{E} = 3,825 mm$ ; $λ_{x} = 34.1518$ , $λ_{y} = 58.3969$ ; curves b/c give $p_{cx} = 319.1875$ , $p_{cy} = 245.8092 \frac{N}{{mm}^{2}}$ . $P_{c} = P_{cy} = 2,777.6435 kN$ . $m_{x} = 0.54$ , $m_{y} = 0.88$ ; elastic resistance $378.12$ ／ $130.755 kN \cdot m$ ; flexural-buckling utilisation $0.94961$ . $v = 0.862028$ , $λ^{LT} = 42.7888$ , $p_{b} = 308.6336$ , $M_{b} = 377.7675 kN \cdot m$ , $m^{LT} = 0.48$ ; axial force/LTB utilisation $0.90571$ . All pass.

Mistakes to avoid

Use $345$ , not $355$ , for the $17.3 mm$ flange.
Apply the 1.2pᵧZ ceiling on the minor axis.
Do not use capped plastic capacities in the member interaction.
Distinguish applied end signs from internal moment-diagram signs.
Keep first-order Mᵧ in the source Eq.8.81 term.

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: only the imposed major-axis end moments in the given factored schedule are replaced by $+ 88$ / $+ 26.4 kN \cdot m$ ; treat these as a $10 %$ increase of the stated total first-order $x$ moments, with axial load and $y$ moments unchanged. How do the checks change?

Reveal answer and reasoning

$x$ -axis end-moment ratio remains $- 0.3$ , so $m_{x} = 0.54$ , $m^{LT} = 0.48$ are unchanged. $M_{x, amp} = M^{LT} = 1.08 \times 88 = 95.04 kN \cdot m$ . Section utilisation increases by $\frac{8.64}{422.28} \approx 0.02046$ to $0.86818$ . Flexural-buckling utilisation increases by $\frac{0.54 \times 8.64}{378.12} \approx 0.01234$ to $0.96195$ . LTB utilisation increases by $\frac{0.48 \times 8.64}{377.7675} \approx 0.01098$ to $0.91669$ . All still pass, with resistance properties unchanged.

Animation labA strong slice can belong to an unstable member3 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.

Combine axial compression and the two bending demands using the specified section resistances.
The whole member adds effective-length and buckling-curve effects.
This uses its own moment factor and bending resistance; it is not a copy of the section check.
Elastic, plastic and buckling resistances are not interchangeable. Read the three original expressions and their first-order/amplified moments.

Animation labRead the moment shape within one segment1 concept · 31 source expressions

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

Moment ordinates must belong to the same effective unbraced segment.
Same-side and reverse-curvature diagrams have different signed end ratios.
The markers show ¼, ½ and ¾ of this segment, not of the entire beam.
LTB and column flexural factors are different. Preserve their individual bounds and coefficient sets.

8. Simple construction: eccentric reactions still bend columns

A simply supported beam-to-column joint does not transmit a rigid-frame end moment, but its vertical reaction may act away from the column centre. The nominal eccentricity is measured from the column centroid to the reaction line. For a beam resting on a column face, take the reaction position as the larger of $100 mm$ from that face and the centre of stiff bearing. Add the centroid-to-face distance (for the illustrated major-axis case, $\frac{D}{2}$ ).

M = R e

With columns both above and below the joint, distribute net moment in proportion to

\frac{I}{L}

M_{lower} = M \frac{\frac{I_{lower}}{L_{lower}}}{(\frac{I_{lower}}{L_{lower}}) + (\frac{I_{upper}}{L_{upper}})}

For identical sections above and below, $I$ cancels. A3 m lower column is twice as stiff by $\frac{I}{L}$ as a $6 m$ upper one, so shares are $\frac{2}{3}$ and $\frac{1}{3}$ . Opposite equal reactions cancel their nominal moments about that axis but still add to axial compression. The upper column carries roof loads; the lower carries roof plus intermediate-floor loads. Do not distribute accumulated axial load using the moment-stiffness fractions.

Special simple-construction rule:

m_{x} = m_{y} = m_{LT} = 1

When looking up bending strength:

λ_{LT} = \frac{0.5 L}{r_{y}}

where

L

is actual storey length. For axial buckling, use the applicable end-restraint conditions:

λ = \frac{L_{E}}{r}

The source allows all beams fully loaded without pattern loading for this simplified column procedure. It also says a beam loaded above $90 %$ of its capacity should not be assumed to afford column rotational restraint. These are applicability conditions, not universal properties of every beam connection. The special 0.5L/rᵧ rule uses actual $L$ , not the reduced effective column length.

Column example 3: loads and eccentric moments in two storeys

Open this example on its own page

Check $203 \times 203 \times 60$ UC S355 in a two-storey simple-construction column. The lower storey is $3 m$ , the upper $6 m$ . Use the supplied reaction table and stated construction assumptions, including a pinned base and $1.15$ moment amplification.

Original source: LectureNotes/Ch 4_Column.pdf — p. 25, p. 26, p. 27, p. 28, p. 29, p. 30, p. 31. 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.

Ch 4 p.25: the left elevation labels levels

1, 2, 3

. The lower storey height is

L_{1} = 3,000 mm

, and the upper storey height is

L_{2} = 6,000 mm

. At right are the column sections and three beam reactions at each level. Opposing secondary-beam reactions

R_{1 j}

／

R_{3 j}

join symmetrically. The one-sided primary-beam reaction

R_{2 j}

acts at the column flange and creates a nominal major-axis moment.

j

denotes the level number.Open full-size image

Original reaction table, Ch 4 p.26, listing each given characteristic load and the lecturer's assumed self-weight. Headings distinguish load factors, dead loads, imposed loads, unfactored and factored values, assumed self-weight and total accumulated axial load. The original load factors are

γ_{f}

: dead load

1.4

, imposed load

1.6

. The table below recalculates each row. Distinguish the rounded source-table values from the higher-precision results here. At level 3, the original dead-load entry uses

34 kN

instead of

33.6 kN

, so its dead-load subtotal is

28 + 28 + 34 + 28 = 118 kN

and its accumulated dead load is

118 + 154 = 272 kN

. The respective recalculated values here are

117.6

and

271.6 kN

. The source imposed-load subtotal/accumulation is

192

640

and

832 kN

, consistent with the main explanation. Below the source table, the

203 \times 203 \times 60

UC properties are individually reproduced under “Section data”. The original classification step is

T = 14.2 < 16 mm

, giving

p_{y} = 355 \frac{N}{{mm}^{2}}

(Table 3.2). The main explanation uses the tabulated interval

\leq 16 mm

to express the same thickness selection; it does not reinterpret the less-than sign as a different section thickness.Open full-size image

Read the diagram and collect the data

Given model	Implication
Continuous column, simple beam construction	Use nominal eccentricity and $\frac{I}{L}$ moment sharing; all moment factors are $1$ .
Base effectively pinned	Lower segment has no base rotational fixity. Substantial level 2 members give the source’s upper restraint assumption.
Roof secondary beams are small ties	Provide positional but not rotational restraint for the upper column’s weak-axis buckling.
Main beams connect by web and seating cleats	Source adopts reaction line $100 mm$ beyond the column face.
Self-weight	Source assumes $20 kN$ characteristic dead self-weight at each level; include it once per level.
Section data	Data p.11: $D = 209.6$ , $t = 9.4$ , $T = 14.2$ , $d = 160.8 mm$ , $\frac{b}{T} = 7.25$ , $\frac{d}{t} = 17.1$ , $r_{x} = 8.96 cm$ , $r_{y} = 5.20 cm$ , $A = 76.4 {cm}^{2}$ , $S_{x} = 656$ , $S_{y} = 305$ , $Z_{x} = 584$ , $Z_{y} = 201 {cm}^{3}$ .

Before calculating: recognition and strategy

There are two separate accounting tasks. Axial load accumulates down the column. Eccentric reactions create a net joint moment which is shared between the upper and lower column in proportion to $\frac{I}{L}$ . Equal opposite secondary reactions cancel their moments, but their vertical loads still add. Keep these operations separate. This solution retains arithmetic precision; a short comparison also explains the source’s rounded values.

1. Recalculate every reaction-table entry and storey total

Level/reaction	Given $G$ ( $kN$ )	$1.4 G$	Given $Q$ ( $kN$ )	$1.6 Q$	Ultimate total
3: $R_{13}$	$20$	$28$	$20$	$32$	$60$
3: $R_{33}$	$20$	$28$	$20$	$32$	$60$
3: $R_{23}$	$24$	$33.6$	$80$	$128$	$161.6$
3: assumed self-weight	$20$	$28$	$0$	$0$	$28$
3 totals	$84$	$117.6$	$120$	$192$	$309.6$
2: $R_{12}$	$30$	$42$	$100$	$160$	$202$
2: $R_{32}$	$30$	$42$	$100$	$160$	$202$
2: $R_{22}$	$30$	$42$	$200$	$320$	$362$
2: assumed self-weight	$20$	$28$	$0$	$0$	$28$
2 added totals	$110$	$154$	$400$	$640$	$794$

Upper column segment

2 - 3

F_{c} = 309.6 kN

. Lower column segment

1 - 2

F_{c} = 309.6 + 794 = 1,103.6 kN

. Check using accumulated components:

G = 84 + 110 = 194 kN; Q = 120 + 400 = 520 kN

1.4 \times 194 + 1.6 \times 520 = 271.6 + 832 = 1,103.6 kN

Animation labFrom characteristic to design load3 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.

G is permanent load; Q is imposed load. A surface load and a line load also have different units.
This illustration uses the course gravity case $1.4 G + 1.6 Q$ . Other combinations in the original text retain their own factors.
For illustrative $G = 5 \frac{kN}{m ^{2}}$ , change Q and watch each separate contribution.
Do not carry this ULS total automatically into deflection. Follow the stated SLS load case.

2. Strength and section classification

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

ε = \sqrt{\frac{275}{355}} = 0.880141

Flange:

\frac{b}{T} = 7.25 < 9 ε = 7.92127

, Class 1. Lower-storey web:

r_{1} = \frac{1,103,600}{160.8 \times 9.4 \times 355} \approx 2.057

Use the cap

r_{1} = 1

. Class 1 web limit:

\frac{80 ε}{1 + r_{1}} = 40 ε = 35.2056

\frac{d}{t} = 17.1 < 35.2056

; the web is Class 1. The upper storey has smaller axial force and also passes using this same conservative web limit.

The lower section’s larger axial compression is sufficient for checking this section’s worst compression/bending web limit. The selected section has the same geometry throughout both storeys.

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 section and elastic member capacities

A_{g} = 76.4 \times 100 = 7,640 {mm}^{2}

A_{g} p_{y} = \frac{7,640 \times 355}{1,000} = 2,712.2 kN

M_{cx} = \min (\frac{355 \times 656}{1,000}, \frac{1.2 \times 355 \times 584}{1,000})

= \min (232.88, 248.784) = 232.88 kN \cdot m

M_{cy} = \min (\frac{355 \times 305}{1,000}, \frac{1.2 \times 355 \times 201}{1,000})

= \min (108.275, 85.626) = 85.626 kN \cdot m

Elastic denominator for member checks:

M_{ex} = \frac{355 \times 584}{1,000} = 207.32 kN \cdot m

M_{ey} = \frac{355 \times 201}{1,000} = 71.355 kN \cdot m

The linear section interaction uses $232.88$ and $85.626$ . The member interaction uses $207.32$ and $71.355$ . The source reuses similar $M_{c}$ notation for different definitions; keep a clearly labelled capacity list.

Animation labCompression and tension across a section2 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.

For sagging, the top flange is in compression and the bottom in tension; hogging reverses this.
Elastic bending stress varies with distance from the neutral axis: $σ = \frac{M y}{I}$ .
The section class governs whether elastic, plastic or effective properties may be used.
Use the shear at the section under examination; the largest shear elsewhere is not automatically coexistent.

5. Lower-storey section interaction

Simple explanation: The column needs more than one pass

A slice can be strong while the whole member still buckles.

Check cross-section compression plus bending.
Then check the separate member-buckling expressions.
Keep each moment, factor and resistance in its specified expression.

Remember: The three checks do not share interchangeable denominators.

Related concept and full method

M_{x, amp} = 1.15 \times 49.4250667 = 56.838827 kN \cdot m

M_{LT} = M_{x, amp}; M_{y} = 0

U_{section} = \frac{1,103.6}{2,712.2} + \frac{56.838827}{232.88} + 0

= 0.650971 < 1

Lower-storey section passes.

Using the rounded source loads and moments gives $\frac{1,104}{2,712.2} + \frac{1.15 \times 49.5}{232.88} \approx 0.6515$ . Both precision levels agree with the source result of approximately $0.65$ .

Animation labA strong slice can belong to an unstable member3 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.

Combine axial compression and the two bending demands using the specified section resistances.
The whole member adds effective-length and buckling-curve effects.
This uses its own moment factor and bending resistance; it is not a copy of the section check.
Elastic, plastic and buckling resistances are not interchangeable. Read the three original expressions and their first-order/amplified moments.

6. Lower-storey member buckling

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

The source assumes rotational and positional restraint at level 2 in both axes; the base is pinned. Table 8.6 recommended fixed/pinned factor is $0.85$ . Use that assumed model explicitly, rather than inferring a rigid base from the drawing’s plate line.

L_{Ex} = L_{Ey} = 0.85 \times 3,000 = 2,550 mm

λ_{x} = \frac{2,550}{89.6} = 28.459821

λ_{y} = \frac{2,550}{52} = 49.038462

Hot-rolled H-section, thickness

\leq 40 mm

p_{y} = 355

x

axis uses curve b;

y

axis uses curve c. Curve b:

25

row gives

342

，

30

row gives

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

. Interpolation fraction:

\frac{28.459821 - 25}{30 - 25} = 0.691964

p_{c} = 342 + 0.691964 \times (335 - 342) = 337.156250 \frac{N}{{mm}^{2}}

Curve c:

48

row gives

280

，

50

row gives

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

. Interpolation fraction:

\frac{49.038462 - 48}{50 - 48} = 0.519231

p_{c} = 280 + 0.519231 \times (275 - 280) = 277.403846 \frac{N}{{mm}^{2}}

P_{cx} = \frac{7,640 \times 337.156250}{1,000} = 2575.873750 kN

P_{cy} = \frac{7,640 \times 277.403846}{1,000} = 2119.365385 kN

P_{c} = P_{cy} = 2119.365385 kN

Simple construction sets all $m$ factors to $1$ . For bending buckling use actual storey length $3,000 mm$ , not $2,550 mm$ :

λ_{LT} = \frac{0.5 L}{r_{y}} = \frac{0.5 \times 3,000}{52} = 28.846154

Table 8.3a，

p_{y} = 355

25

and

30

rows all give

355

, giving

p_{b} = 355

M_{b} = \frac{355 \times 656}{1,000} = 232.88 kN \cdot m

Flexural-buckling interaction:

\frac{1,103.6}{2119.365385} + \frac{1 \times 56.838827}{207.32} + 0

= 0.794882 < 1

Axial force/LTB interaction:

\frac{1,103.6}{2119.365385} + \frac{1 \times 56.838827}{232.88} + 0

= 0.764791 < 1

Animation labA strong slice can belong to an unstable member5 concepts · 4 source expressions

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

Combine axial compression and the two bending demands using the specified section resistances.
The whole member adds effective-length and buckling-curve effects.
This uses its own moment factor and bending resistance; it is not a copy of the section check.
Elastic, plastic and buckling resistances are not interchangeable. Read the three original expressions and their first-order/amplified moments.

7. Upper-storey governing moment and section interaction

Roof reaction:

R_{23} = 1.4 \times 24 + 1.6 \times 80 = 161.6 kN

Roof nominal moment:

M_{3} = 161.6 \times 0.2048 = 33.09568 kN \cdot m

Moment allocated to the bottom of the upper column segment:

24.712533 kN \cdot m

. Governing first-order moment magnitude:

\max (33.09568, 24.712533) = 33.09568 kN \cdot m

M_{x, amp} = M_{LT} = 1.15 \times 33.09568 = 38.060032 kN \cdot m

F_{c} = 309.6 kN; M_{y} = 0

U_{section} = \frac{309.6}{2,712.2} + \frac{38.060032}{232.88}

= 0.277583 < 1

Upper-storey section passes.

The upper-storey moment is governed by the roof reaction, not automatically the moment inherited at level 2. The source uses $162 \times 0.205 \approx 33.2$ , giving amplified $38.18 \approx 38.2$ . Its printed $0.27$ comes from early rounding of the separate utilisation terms; full arithmetic is about $0.28$ at two decimal places.

Animation labA strong slice can belong to an unstable member3 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.

Combine axial compression and the two bending demands using the specified section resistances.
The whole member adds effective-length and buckling-curve effects.
This uses its own moment factor and bending resistance; it is not a copy of the section check.
Elastic, plastic and buckling resistances are not interchangeable. Read the three original expressions and their first-order/amplified moments.

8. Upper-storey buckling about both axes

The main roof beam supplies the source’s major-axis rotational restraint. The small secondary roof ties only restrain position in the weak direction. With level 2 restrained, the lecture uses $0.70 L$ for $x$ and $0.85 L$ for $y$ .

L_{Ex} = 0.70 \times 6,000 = 4,200 mm; L_{Ey} = 0.85 \times 6,000 = 5,100 mm

λ_{x} = \frac{4,200}{89.6} = 46.875

λ_{y} = \frac{5,100}{52} = 98.076923

Curve b,

355

column:

46

row gives

306

，

48

row gives

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

. Interpolation fraction:

\frac{46.875000 - 46}{48 - 46} = 0.437500

p_{c} = 306 + 0.437500 \times (302 - 306) = 304.250000 \frac{N}{{mm}^{2}}

Curve c,

355

column:

98

row gives

145

，

100

row gives

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

. Interpolation fraction:

\frac{98.076923 - 98}{100 - 98} = 0.038462

p_{c} = 145 + 0.038462 \times (140 - 145) = 144.807692 \frac{N}{{mm}^{2}}

P_{cx} = 7.64 \times 304.250000 = 2324.470000 kN

P_{cy} = 7.64 \times 144.807692 = 1106.330769 kN

Minor axis governs. Simple construction:

λ_{LT} = \frac{0.5 \times 6,000}{52} = 57.692308

Table 8.3a，

355

column:

55

row gives

274

，

60

row gives

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

. Interpolation fraction:

\frac{57.692308 - 55}{60 - 55} = 0.538462

p_{b} = 274 + 0.538462 \times (257 - 274) = 264.846154 \frac{N}{{mm}^{2}}

M_{b} = \frac{264.846154 \times 656}{1,000} = 173.739077 kN \cdot m

Flexural-buckling interaction:

\frac{309.6}{1106.330769} + \frac{38.060032}{207.32}

= 0.463425 < 1

Axial force/LTB interaction:

\frac{309.6}{1106.330769} + \frac{38.060032}{173.739077}

= 0.498908 < 1

Both storeys pass all three demonstrated interaction checks. The lower-storey flexural interaction, approximately $0.795$ , is the largest. The original source-rounded results are approximately lower $0.65$ / $0.79$ / $0.76$ and upper $0.28$ / $0.46$ / $0.50$ .

Animation labA strong slice can belong to an unstable member6 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.

Combine axial compression and the two bending demands using the specified section resistances.
The whole member adds effective-length and buckling-curve effects.
This uses its own moment factor and bending resistance; it is not a copy of the section check.
Elastic, plastic and buckling resistances are not interchangeable. Read the three original expressions and their first-order/amplified moments.

Compact exam answer

Using exact arithmetic: upper/lower compression $309.6$ / $1,103.6 kN$ . pᵧ355,Class 1; A𝗀pᵧ2,712.2 kN; section capacities $232.88$ / $85.626 kN \cdot m$ ; elastic $207.32$ / $71.355$ . $e = 204.8 mm$ ; level 2 moment $74.1376 kN \cdot m$ shares $49.4251$ / $24.7125$ . Amplified lower moment $56.8388$ ; upper governing $38.0600$ . Lower $P_{c} = 2,119.365 kN$ ,Mᵦ232.88; interactions $0.6510$ / $0.7949$ / $0.7648$ . Upper $P_{c} = 1,106.331 kN$ ,Mᵦ173.7391; interactions $0.2776$ / $0.4634$ / $0.4989$ . Both storeys adequate under the stated simple-construction restraint assumptions.

Mistakes to avoid

Accumulate both levels for lower axial load.
Cancel opposite moments, not the underlying compressive reactions.
Share moment by $\frac{I}{L}$ , not equally and not by axial-force ratios.
Use actual storey length in 0.5L/rᵧ.
Do not treat a small roof tie as rotational restraint.

Procedure for an unfamiliar variant

Trace the loads above each column segment and keep dead/imposed components separate.
Read actual storey heights, section properties and end restraints in both axes.
For simple construction derive nominal eccentric moments and share them by $\frac{I}{L}$ .
Classify the section and form the correct section and member resistances.
Apply only the prescribed load reductions and moment amplification rules.
Check every storey segment and report both the governing check and source assumptions.

Independent self-check

Try it yourself. Invented variant: change only the upper-storey length from $6 m$ to $3 m$ , retaining the same column section and level 2 moment $74.1376 kN \cdot m$ . How is that joint moment shared?

Reveal answer and reasoning

The upper and lower $\frac{I}{L}$ values now match, so each receives $\frac{74.1376}{2} = 37.0688 kN \cdot m$ . The lower amplified share becomes $42.62912 kN \cdot m$ . This redistribution does not alter the accumulated axial loads. Upper buckling lengths and 0.5L/rᵧ also change and must be recalculated; the original upper resistance cannot be reused.

Animation labEccentric reactions and stiffness sharing2 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 beam reaction can act away from the column centre even at a nominally pinned beam connection.
$M = R e$ . Opposing reactions can cancel part of the signed moment, while both still add compression.
The course simple model distributes the joint moment in proportion to $\frac{I}{L}$ of the columns above and below.
Equal relevant stiffness gives half each. A roof joint with no upper column is a different case.

Column example 4: tributary area and cumulative column loads

Open this example on its own page

Design internal column A in S355 for axial load only. Roof dead/imposed intensities are $5$ / $1.5 \frac{kN}{m^{2}}$ ; floor intensities $7$ / $3 \frac{kN}{m^{2}}$ . Upper storey height $4 m$ ; lower $4.5 m$ . The foundation is fixed. Floor beams prevent translation but do not restrain column rotation. All given dead loads include beam and column self-weight.

Original source: LectureNotes/Ch 4_Column.pdf — p. 32, p. 33, p. 34, p. 35. 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	Where to read it
Tributary dimensions	Plan labels $3.8 + 3.8 = 7.6 m$ and $3 + 3 = 6 m$ ; use plan area, not storey height.
Vertical dimensions	Elevation $4 m$ upper and $4.5 m$ lower.
Roof/floor loads	Question text; the same floor plan applies at both levels.
Axial-only scope	Explicitly requested; no eccentric moment is introduced into this exercise.
Imposed reduction	Lecture p.33 applies its 2011 load-code floor-count table; roof counts as a floor.

Before calculating: recognition and strategy

Multiply each level’s pressure by the area feeding the column. The upper segment carries only roof loading; the lower carries roof plus first floor. Reduce the cumulative imposed load only when the source’s permitted floor-count rule applies. After factoring, design each storey separately because loads, lengths and end conditions differ.

1. Derive the tributary area from the hatched plan

Simple explanation: How a floor load reaches a beam

Each beam collects the load from its own strip of floor.

Find the tributary width from the actual plan.
Area load × tributary width gives load per beam length.
A supporting beam receives the other beam’s end reaction.

Remember: A reaction becomes a point load, not automatically a UDL.

Related concept and full method

Tributary width:

\frac{7.6}{2} + \frac{7.6}{2} = 3.8 + 3.8 = 7.6 m

Tributary depth:

\frac{6}{2} + \frac{6}{2} = 3 + 3 = 6 m

A_{trib} = 7.6 \times 6 = 45.6 m^{2}

Half of each adjacent simply supported bay reaches column A. The hatch is a load-allocation area, not the column’s steel cross-sectional area. Both roof and floor use $45.6 m^{2}$ .

Animation labFollow the floor load in 3D1 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 floor carries pressure in $\frac{kN}{m ^{2}}$ . The highlighted strip belongs to one secondary beam.
Multiply pressure by tributary width: $w = q \times b$ . The illustration uses $q = 5 \frac{kN}{m ^{2}}$ .
A primary beam receives the secondary beam reaction at their connection, not a new full-span UDL.
Trace reactions down to columns and foundations. Count each loaded area once.

2. Read the source’s floor-count reduction

Floors carried	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	Over $8$
Imposed reduction	$0 %$	$5 %$	$10 %$	$15 %$	$20 %$	$25 %$	$30 %$	$35 %$	$40 %$ maximum

The upper member supports one roof level → $0 %$ reduction. The lower supports roof plus first floor →two levels → $5 %$ . Retain $95 %$ of the cumulative imposed load below the floor. This is the rule applied by this lecture example, attributed there to the 2011 Dead and Imposed Loads Code; do not extend it automatically to an occupancy or question where reduction is disallowed. Dead load is not reduced.

Animation labFollow the floor load in 3D3 concepts

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

The floor carries pressure in $\frac{kN}{m ^{2}}$ . The highlighted strip belongs to one secondary beam.
Multiply pressure by tributary width: $w = q \times b$ . The illustration uses $q = 5 \frac{kN}{m ^{2}}$ .
A primary beam receives the secondary beam reaction at their connection, not a new full-span UDL.
Trace reactions down to columns and foundations. Count each loaded area once.

3. Compute characteristic and ultimate loads at each level

Roof:

G = 5 \times 45.6 = 228 kN; Q = 1.5 \times 45.6 = 68.4 kN

Additional first-floor load:

G = 7 \times 45.6 = 319.2 kN; Q = 3 \times 45.6 = 136.8 kN

Lower-storey accumulated load:

G = 228 + 319.2 = 547.2 kN

Q = 68.4 + 136.8 = 205.2 kN

Upper-storey design force:

F = 1.4 \times 228 + 1.6 \times 68.4 = 319.2 + 109.44 = 428.64 kN

Reduced lower-storey imposed load:

Q = 0.95 \times 205.2 = 194.94 kN

Lower-storey design force:

F = 1.4 \times 547.2 + 1.6 \times 194.94

= 766.08 + 311.904 = 1,077.984 kN

All self-weight is already included. The lower load uses the whole accumulated load rather than adding an already factored/reduced upper force to a separately reduced floor force.

Animation labFrom characteristic to design load1 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.

G is permanent load; Q is imposed load. A surface load and a line load also have different units.
This illustration uses the course gravity case $1.4 G + 1.6 Q$ . Other combinations in the original text retain their own factors.
For illustrative $G = 5 \frac{kN}{m ^{2}}$ , change Q and watch each separate contribution.
Do not carry this ULS total automatically into deflection. Follow the stated SLS load case.

4. Select and check the upper column

Trial $152 \times 152 \times 30$ UC. Data File p.11 gives $T = 9.4 mm$ , $\frac{b}{T} = 8.13$ , $\frac{d}{t} = 19.0$ , $A = 38.3 {cm}^{2}$ , $r_{x} = 6.76 cm$ , $r_{y} = 3.83 cm$ . The original lecture uses the same area $38.3$ . $T \leq 16$ , giving $p_{y} = 355$ . Both ends are held in position but free to rotate, so $K = 1$ , with both axes having $L^{E} = 4,000 mm$ .

ε = \sqrt{\frac{275}{355}} = 0.880141

\frac{b}{T} = 8.13 < 13 ε = 11.4418; \frac{d}{t} = 19.0 < 40 ε = 35.2056

Non-slender.

λ_{x} = \frac{4,000}{67.6} = 59.171598

λ_{y} = \frac{4,000}{38.3} = 104.438642

Hot-rolled H-section thickness

\leq 40 mm

，

x

axis curve b;

y

axis curve c;

p_{y} = 355

. Curve b:

58

row gives

278

，

60

row gives

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

. Interpolation fraction:

\frac{59.171598 - 58}{60 - 58} = 0.585799

p_{c} = 278 + 0.585799 \times (272 - 278) = 274.485207 \frac{N}{{mm}^{2}}

Curve c:

104

row gives

133

，

106

row gives

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

. Interpolation fraction:

\frac{104.438642 - 104}{106 - 104} = 0.219321

p_{c} = 133 + 0.219321 \times (129 - 133) = 132.122715 \frac{N}{{mm}^{2}}

P_{cx} = \frac{3,830 \times 274.485207}{1,000} = 1051.278343 kN

P_{cy} = \frac{3,830 \times 132.122715}{1,000} = 506.030000 kN

P_{c} = 506.030000 kN > 428.64 kN

Adequate. Utilisation:

\frac{428.64}{506.030000} = 0.847064

The source uses rounded $p_{c} = 132$ , so $\frac{38.3 \times 132}{10} = 505.56 kN$ and prints $505.6$ . Exact interpolation gives $506.030 kN$ ; both pass. The previously omitted major-axis comparison is provided above.

Animation labEffective length and buckling axes5 concepts · 13 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. Select and check the lower column

Simple explanation: The column needs more than one pass

A slice can be strong while the whole member still buckles.

Check cross-section compression plus bending.
Then check the separate member-buckling expressions.
Keep each moment, factor and resistance in its specified expression.

Remember: The three checks do not share interchangeable denominators.

Related concept and full method

Trial $203 \times 203 \times 46$ UC. Data File p.11 gives $T = 11.0 mm$ , $\frac{b}{T} = 9.25$ , $\frac{d}{t} = 22.3$ , $A = 58.7 {cm}^{2}$ , $r_{x} = 8.82 cm$ , $r_{y} = 5.13 cm$ . The flange thickness again gives $p_{y} = 355$ . The base is fixed; the floor end is held in position but free to rotate, so use the recommended factor $K = 0.85$ .

L_{Ex} = L_{Ey} = 0.85 \times 4,500 = 3,825 mm

\frac{b}{T} = 9.25 < 13 ε = 11.4418; \frac{d}{t} = 22.3 < 40 ε = 35.2056

Non-slender.

λ_{x} = \frac{3,825}{88.2} = 43.367347

λ_{y} = \frac{3,825}{51.3} = 74.561404

Curve b,

p_{y} = 355

42

row gives

314

，

44

row gives

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

. Interpolation fraction:

\frac{43.367347 - 42}{44 - 42} = 0.683673

p_{c} = 314 + 0.683673 \times (310 - 314) = 311.265306 \frac{N}{{mm}^{2}}

Curve c,

p_{y} = 355

74

row gives

205

，

76

row gives

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

. Interpolation fraction:

\frac{74.561404 - 74}{76 - 74} = 0.280702

p_{c} = 205 + 0.280702 \times (200 - 205) = 203.596491 \frac{N}{{mm}^{2}}

P_{cx} = \frac{5,870 \times 311.265306}{1,000} = 1827.127347 kN

P_{cy} = \frac{5,870 \times 203.596491}{1,000} = 1195.111404 kN

P_{c} = 1195.111404 kN > 1,077.984 kN

Adequate. Utilisation:

\frac{1,077.984}{1195.111404} = 0.901995

The source rounds $p_{c}$ to $204$ , giving $\frac{58.7 \times 204}{10} = 1,197.48 kN$ (printed $1,197$ ). Exact interpolation gives $1,195.111 kN$ . Both selected columns pass the requested axial-only design; the lower weak-axis check is more highly utilised.

Animation labEffective length and buckling axes5 concepts · 10 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

$A_{trib} = 45.6 m^{2}$ . Roof $G$ ／ $Q$ is $228$ ／ $68.4 kN$ ; lower-storey accumulation $547.2$ ／ $205.2 kN$ . Upper-storey ultimate force $428.64 kN$ ; lower-storey imposed load after reduction $5 %$ gives $1,077.984 kN$ . Upper-storey $152 \times 152 \times 30$ UC: non-slender, $L^{E} = 4 m$ , $P_{c} = 506.030 kN$ . Lower-storey $203 \times 203 \times 46$ UC: non-slender, $L^{E} = 3.825 m$ , $P_{c} = 1,195.111 kN$ . Both are S355 with $p_{y} = 355$ , governed by minor-axis curve c; both are adequate.

Mistakes to avoid

Do not use $4$ or $4.5 m$ elevation heights when calculating tributary plan area.
Include roof load in the lower segment.
Reduce only imposed load and count the roof as instructed.
Do not add self-weight twice.
Use fixed/pinned recommended $0.85$ for the lower column, not the upper one.

Procedure for an unfamiliar variant

Trace the loads above each column segment and keep dead/imposed components separate.
Read actual storey heights, section properties and end restraints in both axes.
For simple construction derive nominal eccentric moments and share them by $\frac{I}{L}$ .
Classify the section and form the correct section and member resistances.
Apply only the prescribed load reductions and moment amplification rules.
Check every storey segment and report both the governing check and source assumptions.

Independent self-check

Try it yourself. Invented variant: imposed-load reduction is explicitly prohibited. Does the selected lower column still pass?

Reveal answer and reasoning

Lower force becomes $1.4 \times 547.2 + 1.6 \times 205.2 = 766.08 + 328.32 = 1,094.4 kN$ . Compare with $1,195.111 kN$ : utilisation $\approx 0.9157 < 1$ , so it still passes. The roof segment is unchanged because it already had zero reduction.

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.

Animation labEccentric reactions and stiffness sharing1 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 beam reaction can act away from the column centre even at a nominally pinned beam connection.
$M = R e$ . Opposing reactions can cancel part of the signed moment, while both still add compression.
The course simple model distributes the joint moment in proportion to $\frac{I}{L}$ of the columns above and below.
Equal relevant stiffness gives half each. A roof joint with no upper column is a different case.

9. A repeatable exam sequence

Identify simple or continuous construction, sway status, column segment, effective lengths in both axes, section and grade.
Assemble factored axial load at each level; derive any eccentric moments or read given end moments.
Read exact section properties and thickness-dependent pᵧ. Classify local elements under the applicable stress pattern.
Compute capped section moment resistances and check the axial/biaxial section interaction.
Calculate $λ_{x}$ and $λ_{y}$ , choose the proper curves, interpolate $p_{c}$ and form both axis resistances.
For continuous construction obtain $m_{x}$ , $m_{y}$ and mᴸᵀ from their separate tables. For simple construction set all three to $1$ .
Use the correct amplified/first-order moments and elastic pᵧZ denominators for flexural interaction.
Compute λᴸᵀ using the appropriate simple/continuous rule; read pᵦ, calculate Mᵦ and check Eq.8.81.
State all utilisation ratios and whether every required check is $\leq 1$ .

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.

Practise this chapter: tutorials, assignments and past-paper answers →

Columns: buckling, eccentric reactions and combined forces

1. Effective length belongs to an axis and a restraint model

2. Select the buckling curve before reading compressive strength

Read the diagram and collect the data

Before calculating: recognition and strategy

1. Choose a section and confirm its strength

2. Establish a non-slender section

3. Calculate slenderness about both axes

4. Read and interpolate the two compressive strengths

5. Form resistances and conclude

Compact exam answer

Mistakes to avoid

Procedure for an unfamiliar variant

Independent self-check

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

3. Cross-section capacity: axial force consumes some of the same resistance

4. Member buckling: keep the source’s barred quantities distinct

5. Non-sway, sway and ultra-sensitive frames

6. Why moments increase under compression

7. Flexural moment factors are not the LTB factors

Read the diagram and collect the data

Before calculating: recognition and strategy

1. Strength, combined-stress classification and moments

2. Cross-section interaction

3. Effective length and both axial buckling resistances

4. Read the two flexural factors and the LTB factor

5. Flexural member interaction with elastic denominators

6. Axial/LTB interaction

Compact exam answer

Mistakes to avoid

Procedure for an unfamiliar variant

Independent self-check

8. Simple construction: eccentric reactions still bend columns

Read the diagram and collect the data

Before calculating: recognition and strategy

1. Recalculate every reaction-table entry and storey total

2. Strength and section classification

3. Calculate section and elastic member capacities

4. Nominal eccentricity and stiffness sharing at level 2

5. Lower-storey section interaction

6. Lower-storey member buckling

7. Upper-storey governing moment and section interaction

8. Upper-storey buckling about both axes

Compact exam answer

Mistakes to avoid

Procedure for an unfamiliar variant

Independent self-check

Read the diagram and collect the data

Before calculating: recognition and strategy

1. Derive the tributary area from the hatched plan

2. Read the source’s floor-count reduction

3. Compute characteristic and ultimate loads at each level

4. Select and check the upper column

5. Select and check the lower column

Compact exam answer

Mistakes to avoid

Procedure for an unfamiliar variant

Independent self-check

9. A repeatable exam sequence

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