Worked examples

Tutorial 4 Q3: large UC with signed end moments

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Check $305 \times 305 \times 198$ UC S355, Lᴱ=6 m, in a non-sway frame. First-order $F = 2,500 kN$ . Top moments are $+ 350$ / $+ 90 kN \cdot m$ about $x$ / $y$ ; bottom $+ 150$ / $- 70 kN \cdot m$ . Amplification factors $1.1$ / $1.15$ . Given Mᴸᵀ=350 kNm is already amplified. Positive means clockwise.

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 $305 \times 305 \times 198$ 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

Before calculating: recognition and strategy

Translate the applied-arrow signs into internal moment ratios before reading two different factor tables. Q3 deliberately differs from Q2: Mᴸᵀ is explicitly amplified already, and end moments allow non-uniform-moment factors below $1$ .

1. Establish Class 1 and the moment factors

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{2500 \times 1,000}{246.7 \times 19.1 \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} = 12.9 < 35.712215

. The web satisfies the stricter bound. 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.

β_{x} = \frac{- 150}{350} = - 0.428571

Table 8.9, reverse-moment branch:

m_{x} = 0.6 + 0.2 (\frac{- 150}{350}) = 0.514286

β_{y} = \frac{- (- 70)}{90} = 0.777778

Table 8.9, same-side moment branch:

m_{y} = 0.6 + 0.4 (\frac{70}{90}) = 0.911111

LTB Table 8.4a:

m_{LT} = \max (0.44, 0.6 + 0.4 (\frac{- 150}{350}))

= \max (0.44, 0.428571) = 0.44

M_{LT} = 350 kN \cdot m

, already amplified.

For Mᴸᵀ the question provides a value but no separate distribution; the course end-moment method uses the stated major-axis pattern. The lower limit $0.44$ matters here.

Animation labWhy thin elements buckle locally3 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.

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 at the top

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 ratio:

\frac{2500}{8694} + \frac{385}{1186.8} + \frac{103.5}{429.318} = 0.853036

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.

3. Flexural buckling about both axes

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

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{2500}{5043.008955} + \frac{0.514286 \times 385}{1033.28} + \frac{0.911111 \times 103.5}{357.765}

= 0.950940

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.

4. Lateral-torsional buckling 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{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{2500}{5043.008955} + \frac{0.44 \times 350}{1026.889195} + \frac{0.911111 \times 90}{357.765}

= 0.874904

Passes.

$p_{y} = 345 \frac{N}{{mm}^{2}}$ ; $P_{cx} = 7698.423 kN$ , $P_{cy} = 5043.009 kN$ ; $M_{cx} = 1186.800 kN \cdot m$ , $M_{cy} = 429.318 kN \cdot m$ . $M_{b} = 1026.889 kN \cdot m$ . Ratios: section $0.853036$ , flexural buckling $0.950940$ , axial force/LTB $0.874904$ . 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

Factors: mₓ0.514286, mᵧ0.911111, mᴸᵀ0.44. Amplified moments $385$ / $103.5 kN \cdot m$ for section/flexural checks; Mᴸᵀ remains 350 and the LTB minor term uses $90 kN \cdot m$ .

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 use the raw $0.428571$ LTB factor below its $0.44$ floor.
Do not use the same signed ratio for both axes.

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: change the bottom major applied moment from $+ 150$ to $- 150 kN \cdot m$ , retaining top $+ 350$ . What factors change?

Reveal answer and reasoning

$β_{x} = \frac{- (- 150)}{350} = + 0.428571$ . Tables 8.9 and 8.4a now both give $0.6 + 0.4 \times 0.428571 = 0.771429$ . Peak major-axis moment is unchanged, so the section check is unchanged. Both member checks become less favourable; recalculate before concluding.

Animation labRead the moment shape within one segment3 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.

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.