Worked examples

2024 BQ3: column class and all three combined-action checks

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.

Answer all $30$ marks of BQ3. Check $254 \times 254 \times 89$ UC S355 with effective length $4 m$ and factored compression $650 kN$ . Applied clockwise-positive top/bottom moments: $x$ -axis $+ 75$ / $+ 45 kN \cdot m$ ; $y$ -axis $+ 65$ / $- 39 kN \cdot m$ . Amplification factors $1.1$ / $1.1$ . The separately supplied amplified Mᴸᵀ is 96 kNm.

Original source: Pastpaper/23ENGTY004.pdf — p. 5. 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.

Original BQ3 provides a written load table without a frame drawing. Its sign convention takes clockwise as positive and anticlockwise as negative. Applied top/bottom arrow moments must be converted into the end-moment ratio of the internal bending-moment diagram.

4 m

is already the effective length; the question does not require another end-restraint factor.

M^{LT} = 96 kN \cdot m

is explicitly amplified. For the source axis

x - x

／

y - y

, the moment is written

M_{x - x}

／

M_{y - y}

, abbreviated in the main explanation as

M_{x}

／

M_{y}

. The source term

P - Δ - δ

denotes second-order effects. The source subscript form

M_{LT} = + 96 kN \cdot m

and the main explanation's superscript form

M^{LT}

denote the same already amplified quantity, retaining the source's explicit positive sign. The last section-designation number is nominal mass per metre

89 \frac{kg}{m}

. Open full-size image

Read the diagram and collect the data

Lookup: Data File p.11, exact row $254 \times 254 \times 89$ 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	$T = 17.3 mm$ 、 $t = 10.3 mm$ 、 $d = 200.3 mm$ 。
Local slenderness	$\frac{b}{T} = 7.41$ 、 $\frac{d}{t} = 19.4$ 。
Radii (converted from $cm$ )	$r_{x} = 11.2 \times 10 = 112 mm$ ； $r_{y} = 6.55 \times 10 = 65.5 mm$ 。
Area	$A = 113 {cm}^{2} = 11300 {mm}^{2}$ 。
Elastic moduli	$Z_{x} = 1096 {cm}^{3}$ ； $Z_{y} = 379 {cm}^{3}$ 。
Plastic moduli	$S_{x} = 1224 {cm}^{3}$ ； $S_{y} = 575 {cm}^{3}$ 。
LTB parameters	$u = 0.85$ , $x = 14.5$ ; both dimensionless

Before calculating: recognition and strategy

The three checks answer different questions. Local classification determines the section resistance model. Cross-section interaction checks yielding/local resistance; flexural and axial/LTB interactions check member instability. Keep plastic versus elastic denominators and first-order versus amplified moments distinct. The top has both listed maximum moment magnitudes, so those maxima coexist.

(a) Classify the column section — 4 printed marks

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

T = 17.3 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} = 7.41

; Class 1 limit

9 ε = 8.035248

; Class 2 limit

10 ε = 8.928054

. Flange is Class 1.
Web stress parameter:

r_{1} = \frac{F}{d t p_{y}} = \frac{650 \times 1,000}{200.3 \times 10.3 \times 345}

Using

0 \leq r_{1} \leq 1

is capped to give

0.913221

. For any

r_{1}

in this range, the conservative Class 1 web limit is

40 ε = 35.712215

\frac{d}{t} = 19.4 < 35.712215

. The web satisfies this stricter limit. Overall Class 1; 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.

Use the flange’s $9 ε$ Class 1 bound and the conservative combined-compression web bound explained above. The given compression affects web stress distribution but does not change the actual dimension ratios. The result is Class 1.

Animation labWhy thin elements buckle locally1 concept · 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.

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.

(b) Cross-section capacity — 7 printed marks

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

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{113 \times 100 \times 345}{1,000} = 3898.5 kN

M_{x, amp} = 1.1 \times 75 = 82.5 kN \cdot m

M_{y, amp} = 1.1 \times 65 = 71.5 kN \cdot m

M_{cx} = \min (\frac{345 \times 1224}{1,000}, \frac{1.2 \times 345 \times 1096}{1,000})

= \min (422.28, 453.744) = 422.28 kN \cdot m

M_{cy} = \min (\frac{345 \times 575}{1,000}, \frac{1.2 \times 345 \times 379}{1,000})

= \min (198.375, 156.906) = 156.906 kN \cdot m

Section interaction ratio:

\frac{650}{3898.5} + \frac{82.5}{422.28} + \frac{71.5}{156.906} = 0.817786

Passes.

For this UC, the minor axis is governed by the $1.2 p_{y} Z_{y}$ cap. Using only $p_{y} S_{y}$ would overestimate allowable minor-axis moment. The denominator of the section axial-force term is gross squash resistance $A_{g} p_{y}$ , not member compression resistance $P_{c}$ .

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.

(c) Member buckling — 19 printed marks

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

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

First read the factors from the two separate moment-gradient tables. The opposite-face convention makes internal $β = - \frac{M_{bottom}}{M_{top}}$ for this clockwise-positive applied schedule.

β_{x} = - \frac{45}{75} = - 0.600000

Table 8.9:

m_{x} = 0.6 + 0.2 β_{x} = 0.480000

β_{y} = - \frac{(- 39)}{65} = 0.600000

Table 8.9:

m_{y} = 0.6 + 0.4 β_{y} = 0.840000

Table 8.4a:

m_{LT} = \max (0.44, 0.6 + 0.4 β_{x}) = 0.440000

Use the major-axis end pattern for the LTB moment factor in this course solution. The separate Mᴸᵀ magnitude must remain the value specified by the paper; do not replace it with the largest amplified $M_{x}$ simply because the values differ.

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.

Given effective length:

L_{Ex} = L_{Ey} = 4000 mm

λ_{x} = \frac{4000}{112} = 35.714286

λ_{y} = \frac{4000}{65.5} = 61.068702

x

axis uses curve b:

35

gives

318

，

40

gives

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

. Interpolation fraction:

\frac{35.714286 - 35}{40 - 35} = 0.142857

p_{c} = 318 + 0.142857 \times (310 - 318) = 316.857143 \frac{N}{{mm}^{2}}

y

axis uses curve c:

60

gives

241

，

62

gives

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

. Interpolation fraction:

\frac{61.068702 - 60}{62 - 60} = 0.534351

p_{c} = 241 + 0.534351 \times (236 - 241) = 238.328244 \frac{N}{{mm}^{2}}

P_{cx} = \frac{A p_{cx}}{10} = \frac{113 \times 316.857143}{10} = 3580.485714 kN

P_{cy} = \frac{113 \times 238.328244}{10} = 2693.109160 kN

P_{c} = \min (P_{cx}, P_{cy}) = 2693.109160 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 1096}{1,000} = 378.12 kN \cdot m

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

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

= \frac{650}{2693.109160} + \frac{0.48 \times 82.5}{378.12} + \frac{0.84 \times 71.5}{130.755}

= 0.805418

Passes.

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{4000}{65.5} = 61.068702

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

= {[1 + 0.05 {(\frac{61.068702}{14.5})}^{2}]}^{- \frac{1}{4}} = 0.853224

λ_{LT} = u v λ \sqrt{β_{W}} = 0.85 \times 0.853224 \times 61.068702 \times 1 = 44.289500

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

40

gives

317

，

45

gives

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

. Interpolation fraction:

\frac{44.289500 - 40}{45 - 40} = 0.857900

p_{b} = 317 + 0.857900 \times (302 - 317) = 304.131499 \frac{N}{{mm}^{2}}

M_{b} = p_{b} S_{x} = \frac{304.131499 \times 1224}{1,000} = 372.256955 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{650}{2693.109160} + \frac{0.44 \times 96}{372.256955} + \frac{0.84 \times 65}{130.755}

= 0.772402

Passes.

$p_{y} = 345 \frac{N}{{mm}^{2}}$ ; $P_{cx} = 3580.486 kN$ , $P_{cy} = 2693.109 kN$ ; $M_{cx} = 422.280 kN \cdot m$ , $M_{cy} = 156.906 kN \cdot m$ ; $M_{b} = 372.257 kN \cdot m$ . Interaction ratios: section $0.817786$ , flexural buckling $0.805418$ , axial force/LTB $0.772402$ . All three requested strength checks pass.

Suggested method-credit checkpoints (inferred, not an official allocation): correct strength/curve choice; both slendernesses; interpolation; correct signed factors; elastic member denominators; $u$ – $v$ calculation; both interaction ratios and a conclusion. Show these operations even if a final number differs slightly by rounding.

Animation labA strong slice can belong to an unstable member6 concepts · 16 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

(a) Class 1, pᵧ345. (b) amplifiedMx 82.5,My 71.5 kNm. (c) mx 0.480000,my 0.840000,mLT0.440000; givenMLT96 remains unchanged.

Mistakes to avoid

Do not apply an additionalK to the given effective height.
Do not use the same end-moment factor formula for reversing flexural and LTB patterns.
Do not amplify Mᴸᵀ twice.
The final LTB minor-axis term uses the course first-order $M_{y}$ ; the flexural one uses amplifiedMy.

Procedure for an unfamiliar variant

Read the sign convention and exact section row.
Determine strength and class.
Amplify each first-order moment once, preserving a separately given amplifiedMLT.
Check the capped section resistances.
Find both compression resistances and the signed moment factors.
Complete flexural and axial/LTB interactions separately.

Independent self-check

Try it yourself. Invented variant: all first-order actions and the supplied amplifiedMᴸᵀ increase $10 %$ proportionally while section, effective length and amplification factors remain unchanged. What is the largest of the three new ratios?

Reveal answer and reasoning

With the same class and moment ratios, resistances/factors stay unchanged and all interaction demands scale by $1.1$ . The largest becomes $1.1 \times 0.817786 = 0.899564$ . It remains below $1$ . Reconfirm classification after any changed axial load; the conservative web check here still covers boundedr₁.

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.