Worked examples

2023 BQ3: column class and all three combined-action checks

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Answer all $30$ marks of BQ3. Check $305 \times 305 \times 198$ UC S355 with effective length $6 m$ and factored compression $2300 kN$ . Applied clockwise-positive top/bottom moments: $x$ -axis $+ 350$ / $+ 45 kN \cdot m$ ; $y$ -axis $+ 90$ / $- 35 kN \cdot m$ . Amplification factors $1.1$ / $1.15$ . The separately supplied amplified Mᴸᵀ is 110 kNm.

Original source: Pastpaper/22ENGTY033.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.

6 m

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

M^{LT} = 110 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} = 110 kN \cdot m

and the main explanation's superscript form

M^{LT}

denote the same already amplified quantity. The last section-designation number is nominal mass per metre

198 \frac{kg}{m}

. Open full-size image

Read the diagram and collect the data

Lookup: Data File p.11, exact row $305 \times 305 \times 198$ 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	$T = 31.4 mm$ 、 $t = 19.1 mm$ 、 $d = 246.7 mm$ 。
Local slenderness	$\frac{b}{T} = 5.01$ 、 $\frac{d}{t} = 12.9$ 。
Radii (converted from $cm$ )	$r_{x} = 14.2 \times 10 = 142 mm$ ； $r_{y} = 8.04 \times 10 = 80.4 mm$ 。
Area	$A = 252 {cm}^{2} = 25200 {mm}^{2}$ 。
Elastic moduli	$Z_{x} = 2995 {cm}^{3}$ ； $Z_{y} = 1037 {cm}^{3}$ 。
Plastic moduli	$S_{x} = 3440 {cm}^{3}$ ； $S_{y} = 1581 {cm}^{3}$ 。
LTB parameters	$u = 0.854$ , $x = 10.2$ ; both dimensionless

The 2023 question explicitly states a braced non-sway frame.

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 = 31.4 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} = 5.01

; 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{2300 \times 1,000}{246.7 \times 19.1 \times 345}

Using

0 \leq r_{1} \leq 1

is capped to give

1.000000

. For any

r_{1}

in this range, the conservative Class 1 web limit is

40 ε = 35.712215

\frac{d}{t} = 12.9 < 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 — 6 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{252 \times 100 \times 345}{1,000} = 8694 kN

M_{x, amp} = 1.1 \times 350 = 385 kN \cdot m

M_{y, amp} = 1.15 \times 90 = 103.5 kN \cdot m

M_{cx} = \min (\frac{345 \times 3440}{1,000}, \frac{1.2 \times 345 \times 2995}{1,000})

= \min (1186.8, 1239.93) = 1186.8 kN \cdot m

M_{cy} = \min (\frac{345 \times 1581}{1,000}, \frac{1.2 \times 345 \times 1037}{1,000})

= \min (545.445, 429.318) = 429.318 kN \cdot m

Section interaction ratio:

\frac{2300}{8694} + \frac{385}{1186.8} + \frac{103.5}{429.318} = 0.830032

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 — 20 printed marks

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

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

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}{350} = - 0.128571

Table 8.9:

m_{x} = 0.6 + 0.2 β_{x} = 0.574286

β_{y} = - \frac{(- 35)}{90} = 0.388889

Table 8.9:

m_{y} = 0.6 + 0.4 β_{y} = 0.755556

Table 8.4a:

m_{LT} = \max (0.44, 0.6 + 0.4 β_{x}) = 0.548571

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.

L_{Ex} = L_{Ey} = 6000 mm

is the given effective length.

λ_{x} = \frac{6000}{142} = 42.253521

λ_{y} = \frac{6000}{80.4} = 74.626866

x

axis, curve b:

42

row gives

306

;

44

row gives

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

. Interpolation fraction:

\frac{42.253521 - 42}{44 - 42} = 0.126761

p_{c} = 306 + 0.126761 \times (302 - 306) = 305.492958 \frac{N}{{mm}^{2}}

y

axis, curve c:

74

row gives

202

;

76

row gives

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

. Interpolation fraction:

\frac{74.626866 - 74}{76 - 74} = 0.313433

p_{c} = 202 + 0.313433 \times (196 - 202) = 200.119403 \frac{N}{{mm}^{2}}

P_{cx} = \frac{A p_{cx}}{10} = \frac{252 \times 305.492958}{10} = 7698.422535 kN

P_{cy} = \frac{252 \times 200.119403}{10} = 5043.008955 kN

P_{c} = \min (P_{cx}, P_{cy}) = 5043.008955 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 2995}{1,000} = 1033.28 kN \cdot m

M_{ey} = \frac{345 \times 1037}{1,000} = 357.765 kN \cdot m

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

= \frac{2300}{5043.008955} + \frac{0.574286 \times 385}{1033.28} + \frac{0.755556 \times 103.5}{357.765}

= 0.888636

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{6000}{80.4} = 74.626866

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

= {[1 + 0.05 {(\frac{74.626866}{10.2})}^{2}]}^{- \frac{1}{4}} = 0.722175

λ_{LT} = u v λ \sqrt{β_{W}} = 0.854 \times 0.722175 \times 74.626866 \times 1 = 46.025206

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

45

row gives

302

;

50

row gives

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

. Interpolation fraction:

\frac{46.025206 - 45}{50 - 45} = 0.205041

p_{b} = 302 + 0.205041 \times (285 - 302) = 298.514301 \frac{N}{{mm}^{2}}

M_{b} = p_{b} S_{x} = \frac{298.514301 \times 3440}{1,000} = 1026.889195 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{2300}{5043.008955} + \frac{0.548571 \times 110}{1026.889195} + \frac{0.755556 \times 90}{357.765}

= 0.704909

Passes.

$p_{y} = 345 \frac{N}{{mm}^{2}}$ ; $P_{cx} = 7698.423 kN$ , $P_{cy} = 5043.009 kN$ ; $M_{cx} = 1186.800 kN \cdot m$ , $M_{cy} = 429.318 kN \cdot m$ ; $M_{b} = 1026.889 kN \cdot m$ . Interaction ratios: section $0.830032$ , flexural buckling $0.888636$ , axial force/LTB $0.704909$ . 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 385,My 103.5 kNm. (c) mx 0.574286,my 0.755556,mLT0.548571; givenMLT110 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 $1.1 \times 0.888636 = 0.977500$ . 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.