Worked examples

Tutorial 4 Q2: the same column with biaxial bending

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Check the Q1 column in a non-sway frame for first-order ultimate $F = 1,450 kN$ , $M_{x} = 220 kN \cdot m$ , Mᵧ=35 kNm and Mᴸᵀ=200 kNm. Amplification factors are $1.1$ about $x$ and $1.15$ about $y$ .

Original source: LectureNotes/Ch 4_Column.pdf — p. 40. 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

Lookup: Data File p.11, exact row $305 \times 305 \times 118$ UC. Read the dimensions/local ratios table and the properties table separately. The $x - x$ axis crosses the web horizontally; $y - y$ passes vertically through its centre in the table sketch.

Property	Value and units
Flange/web/root-to-root web depth	$T = 18.7 mm$ 、 $t = 12 mm$ 、 $d = 246.7 mm$ 。
Local slenderness	$\frac{b}{T} = 8.22$ 、 $\frac{d}{t} = 20.6$ 。
Radii (converted from $cm$ )	$r_{x} = 13.6 \times 10 = 136 mm$ ； $r_{y} = 7.77 \times 10 = 77.7 mm$ 。
Area	$A = 150 {cm}^{2} = 15000 {mm}^{2}$ 。
Elastic moduli	$Z_{x} = 1760 {cm}^{3}$ ； $Z_{y} = 589 {cm}^{3}$ 。
Plastic moduli	$S_{x} = 1958 {cm}^{3}$ ； $S_{y} = 895 {cm}^{3}$ 。
LTB parameters	$u = 0.85$ , $x = 16.2$ ; both dimensionless

Before calculating: recognition and strategy

Adding moments changes the task: local classification must now justify the bending resistance, and all three combined-action checks are required. The flange is compact, not plastic, but Class 2 still allows the course capped plastic resistance. Conservatively take the listed axis maxima to coexist.

1. Classify, amplify and check the cross section

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

T = 18.7 mm

; use the S355 thickness band to select

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

ε = \sqrt{\frac{275}{345}} = 0.892805

Flange

\frac{b}{T} = 8.22

; Class 1 limit

9 ε = 8.035248

; Class 2 limit

10 ε = 8.928054

, so the flange is Class 2. Web stress parameter:

r_{1} = \frac{F}{d t p_{y}} = \frac{1450 \times 1,000}{246.7 \times 12 \times 345}

Limit to

0 \leq r_{1} \leq 1

, then use

1.000000

. For any

r_{1}

within this range, a conservative Class 1 web limit is

40 ε = 35.712215

\frac{d}{t} = 20.6 < 35.712215

. The web satisfies the stricter bound. Overall Class 2; use capped plastic section resistance.

The lecture’s Class 1 combined-stress web limit $\frac{80 ε}{1 + r_{1}}$ cannot be below $40 ε$ because $r_{1} \leq 1$ . This check therefore avoids an unjustified plastic classification while remaining conservative. The flange is checked independently.

M_{LT, amp} = 1.1 \times 200 = 220 kN \cdot m

;

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

At a cross section, compression and bending share the material. Use amplified moments and capped plastic resistances. The total must not exceed $1$ ; all terms below are dimensionless.

A_{g} p_{y} = \frac{150 \times 100 \times 345}{1,000} = 5175 kN

M_{x, amp} = 1.1 \times 220 = 242 kN \cdot m

M_{y, amp} = 1.15 \times 35 = 40.25 kN \cdot m

M_{cx} = \min (\frac{345 \times 1958}{1,000}, \frac{1.2 \times 345 \times 1760}{1,000})

= \min (675.51, 728.64) = 675.51 kN \cdot m

M_{cy} = \min (\frac{345 \times 895}{1,000}, \frac{1.2 \times 345 \times 589}{1,000})

= \min (308.775, 243.846) = 243.846 kN \cdot m

Section ratio:

\frac{1450}{5175} + \frac{242}{675.51} + \frac{40.25}{243.846} = 0.803504

Passes.

Animation labA strong slice can belong to an unstable member4 concepts · 5 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.

2. Flexural member interaction

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

Table 8.7: hot-rolled H-section (UC), maximum thickness $\leq 40 mm$ , about the $x$ axis use curve $b$ , about the $y$ axis use curve $c$ . Use the Data File p.8 $p_{y} = 345 \frac{N}{{mm}^{2}}$ column. The two axes use different curves, so slenderness alone cannot identify the governing axis.

L_{Ex} = L_{Ey} = 4500 mm

is the given effective length.

λ_{x} = \frac{4500}{136} = 33.088235

λ_{y} = \frac{4500}{77.7} = 57.915058

x

axis, curve b:

30

row gives

325

;

35

row gives

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

. Interpolation fraction:

\frac{33.088235 - 30}{35 - 30} = 0.617647

p_{c} = 325 + 0.617647 \times (318 - 325) = 320.676471 \frac{N}{{mm}^{2}}

y

axis, curve c:

56

row gives

252

;

58

row gives

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

. Interpolation fraction:

\frac{57.915058 - 56}{58 - 56} = 0.957529

p_{c} = 252 + 0.957529 \times (247 - 252) = 247.212355 \frac{N}{{mm}^{2}}

P_{cx} = \frac{A p_{cx}}{10} = \frac{150 \times 320.676471}{10} = 4810.147059 kN

P_{cy} = \frac{150 \times 247.212355}{10} = 3708.185328 kN

P_{c} = \min (P_{cx}, P_{cy}) = 3708.185328 kN

Member interaction uses elastic moment denominators pᵧZ, even when the cross-section check used plastic moduli. Use amplified moments here.

M_{ex} = \frac{345 \times 1760}{1,000} = 607.2 kN \cdot m

M_{ey} = \frac{345 \times 589}{1,000} = 203.205 kN \cdot m

\frac{F}{P_{c}} + \frac{m_{x} M_{x, amp}}{M_{ex}} + \frac{m_{y} M_{y, amp}}{M_{ey}}

= \frac{1450}{3708.185328} + \frac{1 \times 242}{607.2} + \frac{1 \times 40.25}{203.205}

= 0.987653

Passes.

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.

3. Combined compression and lateral-torsional buckling

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

Continuous/non-sway member: Class 1/2 gives $β_{W} = 1$ . Use the full $u$ – $v$ calculation with the stated effective length.

λ = \frac{L_{E}}{r_{y}} = \frac{4500}{77.7} = 57.915058

v = {[1 + 0.05 {(\frac{λ}{x})}^{2}]}^{- \frac{1}{4}}

= {[1 + 0.05 {(\frac{57.915058}{16.2})}^{2}]}^{- \frac{1}{4}} = 0.883798

λ_{LT} = u v λ \sqrt{β_{W}} = 0.85 \times 0.883798 \times 57.915058 \times 1 = 43.507440

Read Data File p.5 Table 8.3a, pᵧ345 column:

40

row gives

317

;

45

row gives

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

. Interpolation fraction:

\frac{43.507440 - 40}{45 - 40} = 0.701488

p_{b} = 317 + 0.701488 \times (302 - 317) = 306.477681 \frac{N}{{mm}^{2}}

M_{b} = p_{b} S_{x} = \frac{306.477681 \times 1958}{1,000} = 600.083299 kN \cdot m

Course Eq.8.81 uses first-order minor-axis moment in its last term. Mᴸᵀ is the specified amplified major-axis value; do not amplify it twice. The axial denominator is $P_{cy}$ .

\frac{F}{P_{cy}} + \frac{m_{LT} M_{LT}}{M_{b}} + \frac{m_{y} M_{y, first}}{M_{ey}}

= \frac{1450}{3708.185328} + \frac{1 \times 220}{600.083299} + \frac{1 \times 35}{203.205}

= 0.929882

Passes.

$p_{y} = 345 \frac{N}{{mm}^{2}}$ ; $P_{cx} = 4810.147 kN$ , $P_{cy} = 3708.185 kN$ ; $M_{cx} = 675.510 kN \cdot m$ , $M_{cy} = 243.846 kN \cdot m$ . $M_{b} = 600.083 kN \cdot m$ . Ratios: section $0.803504$ , flexural buckling $0.987653$ , axial force/LTB $0.929882$ . All three requested strength checks pass.

Animation labA strong slice can belong to an unstable member5 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.

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

Conservative unknown-distribution factors are all $1$ ; first-order Mᴸᵀ200 becomes $220 kN \cdot m$ . This is a checked conservative solution, not a reconstruction of an absent end-moment diagram.

Mistakes to avoid

Do not use $355 \frac{N}{{mm}^{2}}$ for a flange thicker than $16 mm$ .
Do not substitute plastic moduli in the member elastic denominators.
Do not amplify an already amplified Mᴸᵀ twice.
Do not treat an axial resistance pass as proof of combined-load adequacy.
Do not automatically call this $8.22$ flange Class 1: $9 ε \approx 8.035$ , so it is Class 2.

Procedure for an unfamiliar variant

Identify construction type, axis, actual length and effective length.
Read thickness, grade and the exact UC row. Classify before using plastic moduli.
Sum vertical forces and derive signed eccentric moments; share moments only when the joint has two columns.
Apply specified amplification once. Check the capped section interaction.
Select curves $b$ / $c$ for these rolled H sections, interpolate both strengths and check flexural interaction with elastic moduli.
Use the appropriate simple/continuous LTB slenderness rule and the course first-order minor-axis term.
Report every utilisation and let any failed check govern.

Independent self-check

Try it yourself. Invented variant: a revised question says “Mᴸᵀ=200 kNm already amplified”. Which calculation changes?

Reveal answer and reasoning

Use $200$ instead of $220$ only in the LTB numerator. The new axial/LTB ratio is $0.896554$ . Section and flexural checks still use $1.1 \times 220$ and $1.15 \times 35$ . The wording determines whether amplification is needed.