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Mock Q3: full solution and invented marking guide

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MOCK Q3 · braced non-sway column; all geometry repeated in the adjacent question. — Invented mock-question diagram: braced non-sway column. The top arrow denotes design compression $F_{c} = 1000 kN$ . Section $254 \times 254 \times 89$ UC; effective lengths about the two axes $L_{E, x} = L_{E, y} = 3500 mm$ . Uniform first-order moments $M_{x} = 80$ , $M_{y} = 40 kN \cdot m$ ; amplification factors about the two axes $1.1$ . Given $m_{x} = m_{y} = m_{LT} = 1$ and already amplified $M_{LT} = 88 kN \cdot m$ . Schematic, not to scale; numerical labels govern.Open full-size image.

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

(a) Classify before choosing resistance

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{1000 \times 1,000}{200.3 \times 10.3 \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} = 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.

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 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

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 80 = 88 kN \cdot m

M_{y, amp} = 1.1 \times 40 = 44 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{1000}{3898.5} + \frac{88}{422.28} + \frac{44}{156.906} = 0.745324

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.

(c) 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.

Given effective length:

L_{Ex} = L_{Ey} = 3500 mm

λ_{x} = \frac{3500}{112} = 31.250000

λ_{y} = \frac{3500}{65.5} = 53.435115

x

axis uses curve b:

30

gives

325

，

35

gives

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

. Interpolation fraction:

\frac{31.250000 - 30}{35 - 30} = 0.250000

p_{c} = 325 + 0.250000 \times (318 - 325) = 323.250000 \frac{N}{{mm}^{2}}

y

axis uses curve c:

52

gives

263

，

54

gives

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

. Interpolation fraction:

\frac{53.435115 - 52}{54 - 52} = 0.717557

p_{c} = 263 + 0.717557 \times (258 - 263) = 259.412214 \frac{N}{{mm}^{2}}

P_{cx} = \frac{A p_{cx}}{10} = \frac{113 \times 323.250000}{10} = 3652.725000 kN

P_{cy} = \frac{113 \times 259.412214}{10} = 2931.358015 kN

P_{c} = \min (P_{cx}, P_{cy}) = 2931.358015 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{1000}{2931.358015} + \frac{1 \times 88}{378.12} + \frac{1 \times 44}{130.755}

= 0.910376

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.

(c) 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

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{3500}{65.5} = 53.435115

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

= {[1 + 0.05 {(\frac{53.435115}{14.5})}^{2}]}^{- \frac{1}{4}} = 0.878487

λ_{LT} = u v λ \sqrt{β_{W}} = 0.85 \times 0.878487 \times 53.435115 \times 1 = 39.900763

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

35

gives

332

，

40

gives

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

. Interpolation fraction:

\frac{39.900763 - 35}{40 - 35} = 0.980153

p_{b} = 332 + 0.980153 \times (317 - 332) = 317.297710 \frac{N}{{mm}^{2}}

M_{b} = p_{b} S_{x} = \frac{317.297710 \times 1224}{1,000} = 388.372397 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{1000}{2931.358015} + \frac{1 \times 88}{388.372397} + \frac{1 \times 40}{130.755}

= 0.873641

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.

Compact script and invented marking guide

$p_{y} = 345 \frac{N}{{mm}^{2}}$ ; $P_{cx} = 3652.725 kN$ , $P_{cy} = 2931.358 kN$ ; $M_{cx} = 422.280 kN \cdot m$ , $M_{cy} = 156.906 kN \cdot m$ ; $M_{b} = 388.372 kN \cdot m$ . Interaction ratios: section $0.745324$ , flexural buckling $0.910376$ , axial force/LTB $0.873641$ . All three requested strength checks pass.

Invented credit: classification $4$ ; section properties/amplification $3$ and interaction $3$ ; curves/slenderness $4$ , compression interpolation $4$ , flexural equation/result $4$ , accurate LTB calculation $5$ , second interaction/conclusion $3$ . Do not multiply the supplied $88 kN \cdot m$ by $1.1$ again.

Animation labFollow the calculation sequence2 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.

Locate the load, supports, connection geometry and any stated assumptions.
Keep given values, table lookups and calculated values distinct; reconcile their units.
The calculation player steps through the existing expressions in their original order.
Compare demand with resistance or the relevant limit. Keep missing inputs and conditional conclusions explicit.