Worked examples

Column example 3: loads and eccentric moments in two storeys

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Chinese–English terminology

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Check $203 \times 203 \times 60$ UC S355 in a two-storey simple-construction column. The lower storey is $3 m$ , the upper $6 m$ . Use the supplied reaction table and stated construction assumptions, including a pinned base and $1.15$ moment amplification.

Original source: LectureNotes/Ch 4_Column.pdf — p. 25, p. 26, p. 27, p. 28, p. 29, p. 30, p. 31. 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.

Ch 4 p.25: the left elevation labels levels

1, 2, 3

. The lower storey height is

L_{1} = 3,000 mm

, and the upper storey height is

L_{2} = 6,000 mm

. At right are the column sections and three beam reactions at each level. Opposing secondary-beam reactions

R_{1 j}

／

R_{3 j}

join symmetrically. The one-sided primary-beam reaction

R_{2 j}

acts at the column flange and creates a nominal major-axis moment.

j

denotes the level number.Open full-size image

Original reaction table, Ch 4 p.26, listing each given characteristic load and the lecturer's assumed self-weight. Headings distinguish load factors, dead loads, imposed loads, unfactored and factored values, assumed self-weight and total accumulated axial load. The original load factors are

γ_{f}

: dead load

1.4

, imposed load

1.6

. The table below recalculates each row. Distinguish the rounded source-table values from the higher-precision results here. At level 3, the original dead-load entry uses

34 kN

instead of

33.6 kN

, so its dead-load subtotal is

28 + 28 + 34 + 28 = 118 kN

and its accumulated dead load is

118 + 154 = 272 kN

. The respective recalculated values here are

117.6

and

271.6 kN

. The source imposed-load subtotal/accumulation is

192

640

and

832 kN

, consistent with the main explanation. Below the source table, the

203 \times 203 \times 60

UC properties are individually reproduced under “Section data”. The original classification step is

T = 14.2 < 16 mm

, giving

p_{y} = 355 \frac{N}{{mm}^{2}}

(Table 3.2). The main explanation uses the tabulated interval

\leq 16 mm

to express the same thickness selection; it does not reinterpret the less-than sign as a different section thickness.Open full-size image

Read the diagram and collect the data

Given model	Implication
Continuous column, simple beam construction	Use nominal eccentricity and $\frac{I}{L}$ moment sharing; all moment factors are $1$ .
Base effectively pinned	Lower segment has no base rotational fixity. Substantial level 2 members give the source’s upper restraint assumption.
Roof secondary beams are small ties	Provide positional but not rotational restraint for the upper column’s weak-axis buckling.
Main beams connect by web and seating cleats	Source adopts reaction line $100 mm$ beyond the column face.
Self-weight	Source assumes $20 kN$ characteristic dead self-weight at each level; include it once per level.
Section data	Data p.11: $D = 209.6$ , $t = 9.4$ , $T = 14.2$ , $d = 160.8 mm$ , $\frac{b}{T} = 7.25$ , $\frac{d}{t} = 17.1$ , $r_{x} = 8.96 cm$ , $r_{y} = 5.20 cm$ , $A = 76.4 {cm}^{2}$ , $S_{x} = 656$ , $S_{y} = 305$ , $Z_{x} = 584$ , $Z_{y} = 201 {cm}^{3}$ .

Before calculating: recognition and strategy

There are two separate accounting tasks. Axial load accumulates down the column. Eccentric reactions create a net joint moment which is shared between the upper and lower column in proportion to $\frac{I}{L}$ . Equal opposite secondary reactions cancel their moments, but their vertical loads still add. Keep these operations separate. This solution retains arithmetic precision; a short comparison also explains the source’s rounded values.

1. Recalculate every reaction-table entry and storey total

Level/reaction	Given $G$ ( $kN$ )	$1.4 G$	Given $Q$ ( $kN$ )	$1.6 Q$	Ultimate total
3: $R_{13}$	$20$	$28$	$20$	$32$	$60$
3: $R_{33}$	$20$	$28$	$20$	$32$	$60$
3: $R_{23}$	$24$	$33.6$	$80$	$128$	$161.6$
3: assumed self-weight	$20$	$28$	$0$	$0$	$28$
3 totals	$84$	$117.6$	$120$	$192$	$309.6$
2: $R_{12}$	$30$	$42$	$100$	$160$	$202$
2: $R_{32}$	$30$	$42$	$100$	$160$	$202$
2: $R_{22}$	$30$	$42$	$200$	$320$	$362$
2: assumed self-weight	$20$	$28$	$0$	$0$	$28$
2 added totals	$110$	$154$	$400$	$640$	$794$

Upper column segment

2 - 3

F_{c} = 309.6 kN

. Lower column segment

1 - 2

F_{c} = 309.6 + 794 = 1,103.6 kN

. Check using accumulated components:

G = 84 + 110 = 194 kN; Q = 120 + 400 = 520 kN

1.4 \times 194 + 1.6 \times 520 = 271.6 + 832 = 1,103.6 kN

Animation labFrom characteristic to design load3 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.

G is permanent load; Q is imposed load. A surface load and a line load also have different units.
This illustration uses the course gravity case $1.4 G + 1.6 Q$ . Other combinations in the original text retain their own factors.
For illustrative $G = 5 \frac{kN}{m ^{2}}$ , change Q and watch each separate contribution.
Do not carry this ULS total automatically into deflection. Follow the stated SLS load case.

2. Strength and section classification

T = 14.2 \leq 16 mm \to p_{y} = 355 \frac{N}{{mm}^{2}}

ε = \sqrt{\frac{275}{355}} = 0.880141

Flange:

\frac{b}{T} = 7.25 < 9 ε = 7.92127

, Class 1. Lower-storey web:

r_{1} = \frac{1,103,600}{160.8 \times 9.4 \times 355} \approx 2.057

Use the cap

r_{1} = 1

. Class 1 web limit:

\frac{80 ε}{1 + r_{1}} = 40 ε = 35.2056

\frac{d}{t} = 17.1 < 35.2056

; the web is Class 1. The upper storey has smaller axial force and also passes using this same conservative web limit.

The lower section’s larger axial compression is sufficient for checking this section’s worst compression/bending web limit. The selected section has the same geometry throughout both storeys.

Animation labWhy thin elements buckle locally1 concept · 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.

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.

3. Calculate section and elastic member capacities

A_{g} = 76.4 \times 100 = 7,640 {mm}^{2}

A_{g} p_{y} = \frac{7,640 \times 355}{1,000} = 2,712.2 kN

M_{cx} = \min (\frac{355 \times 656}{1,000}, \frac{1.2 \times 355 \times 584}{1,000})

= \min (232.88, 248.784) = 232.88 kN \cdot m

M_{cy} = \min (\frac{355 \times 305}{1,000}, \frac{1.2 \times 355 \times 201}{1,000})

= \min (108.275, 85.626) = 85.626 kN \cdot m

Elastic denominator for member checks:

M_{ex} = \frac{355 \times 584}{1,000} = 207.32 kN \cdot m

M_{ey} = \frac{355 \times 201}{1,000} = 71.355 kN \cdot m

The linear section interaction uses $232.88$ and $85.626$ . The member interaction uses $207.32$ and $71.355$ . The source reuses similar $M_{c}$ notation for different definitions; keep a clearly labelled capacity list.

Animation labCompression and tension across a section2 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.

For sagging, the top flange is in compression and the bottom in tension; hogging reverses this.
Elastic bending stress varies with distance from the neutral axis: $σ = \frac{M y}{I}$ .
The section class governs whether elastic, plastic or effective properties may be used.
Use the shear at the section under examination; the largest shear elsewhere is not automatically coexistent.

5. Lower-storey 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

M_{x, amp} = 1.15 \times 49.4250667 = 56.838827 kN \cdot m

M_{LT} = M_{x, amp}; M_{y} = 0

U_{section} = \frac{1,103.6}{2,712.2} + \frac{56.838827}{232.88} + 0

= 0.650971 < 1

Lower-storey section passes.

Using the rounded source loads and moments gives $\frac{1,104}{2,712.2} + \frac{1.15 \times 49.5}{232.88} \approx 0.6515$ . Both precision levels agree with the source result of approximately $0.65$ .

Animation labA strong slice can belong to an unstable member3 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.

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.

6. Lower-storey member buckling

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

The source assumes rotational and positional restraint at level 2 in both axes; the base is pinned. Table 8.6 recommended fixed/pinned factor is $0.85$ . Use that assumed model explicitly, rather than inferring a rigid base from the drawing’s plate line.

L_{Ex} = L_{Ey} = 0.85 \times 3,000 = 2,550 mm

λ_{x} = \frac{2,550}{89.6} = 28.459821

λ_{y} = \frac{2,550}{52} = 49.038462

Hot-rolled H-section, thickness

\leq 40 mm

p_{y} = 355

x

axis uses curve b;

y

axis uses curve c. Curve b:

25

row gives

342

，

30

row gives

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

. Interpolation fraction:

\frac{28.459821 - 25}{30 - 25} = 0.691964

p_{c} = 342 + 0.691964 \times (335 - 342) = 337.156250 \frac{N}{{mm}^{2}}

Curve c:

48

row gives

280

，

50

row gives

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

. Interpolation fraction:

\frac{49.038462 - 48}{50 - 48} = 0.519231

p_{c} = 280 + 0.519231 \times (275 - 280) = 277.403846 \frac{N}{{mm}^{2}}

P_{cx} = \frac{7,640 \times 337.156250}{1,000} = 2575.873750 kN

P_{cy} = \frac{7,640 \times 277.403846}{1,000} = 2119.365385 kN

P_{c} = P_{cy} = 2119.365385 kN

Simple construction sets all $m$ factors to $1$ . For bending buckling use actual storey length $3,000 mm$ , not $2,550 mm$ :

λ_{LT} = \frac{0.5 L}{r_{y}} = \frac{0.5 \times 3,000}{52} = 28.846154

Table 8.3a，

p_{y} = 355

25

and

30

rows all give

355

, giving

p_{b} = 355

M_{b} = \frac{355 \times 656}{1,000} = 232.88 kN \cdot m

Flexural-buckling interaction:

\frac{1,103.6}{2119.365385} + \frac{1 \times 56.838827}{207.32} + 0

= 0.794882 < 1

Axial force/LTB interaction:

\frac{1,103.6}{2119.365385} + \frac{1 \times 56.838827}{232.88} + 0

= 0.764791 < 1

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.

7. Upper-storey governing moment and section interaction

Roof reaction:

R_{23} = 1.4 \times 24 + 1.6 \times 80 = 161.6 kN

Roof nominal moment:

M_{3} = 161.6 \times 0.2048 = 33.09568 kN \cdot m

Moment allocated to the bottom of the upper column segment:

24.712533 kN \cdot m

. Governing first-order moment magnitude:

\max (33.09568, 24.712533) = 33.09568 kN \cdot m

M_{x, amp} = M_{LT} = 1.15 \times 33.09568 = 38.060032 kN \cdot m

F_{c} = 309.6 kN; M_{y} = 0

U_{section} = \frac{309.6}{2,712.2} + \frac{38.060032}{232.88}

= 0.277583 < 1

Upper-storey section passes.

The upper-storey moment is governed by the roof reaction, not automatically the moment inherited at level 2. The source uses $162 \times 0.205 \approx 33.2$ , giving amplified $38.18 \approx 38.2$ . Its printed $0.27$ comes from early rounding of the separate utilisation terms; full arithmetic is about $0.28$ at two decimal places.

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.

8. Upper-storey buckling about both axes

The main roof beam supplies the source’s major-axis rotational restraint. The small secondary roof ties only restrain position in the weak direction. With level 2 restrained, the lecture uses $0.70 L$ for $x$ and $0.85 L$ for $y$ .

L_{Ex} = 0.70 \times 6,000 = 4,200 mm; L_{Ey} = 0.85 \times 6,000 = 5,100 mm

λ_{x} = \frac{4,200}{89.6} = 46.875

λ_{y} = \frac{5,100}{52} = 98.076923

Curve b,

355

column:

46

row gives

306

，

48

row gives

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

. Interpolation fraction:

\frac{46.875000 - 46}{48 - 46} = 0.437500

p_{c} = 306 + 0.437500 \times (302 - 306) = 304.250000 \frac{N}{{mm}^{2}}

Curve c,

355

column:

98

row gives

145

，

100

row gives

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

. Interpolation fraction:

\frac{98.076923 - 98}{100 - 98} = 0.038462

p_{c} = 145 + 0.038462 \times (140 - 145) = 144.807692 \frac{N}{{mm}^{2}}

P_{cx} = 7.64 \times 304.250000 = 2324.470000 kN

P_{cy} = 7.64 \times 144.807692 = 1106.330769 kN

Minor axis governs. Simple construction:

λ_{LT} = \frac{0.5 \times 6,000}{52} = 57.692308

Table 8.3a，

355

column:

55

row gives

274

，

60

row gives

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

. Interpolation fraction:

\frac{57.692308 - 55}{60 - 55} = 0.538462

p_{b} = 274 + 0.538462 \times (257 - 274) = 264.846154 \frac{N}{{mm}^{2}}

M_{b} = \frac{264.846154 \times 656}{1,000} = 173.739077 kN \cdot m

Flexural-buckling interaction:

\frac{309.6}{1106.330769} + \frac{38.060032}{207.32}

= 0.463425 < 1

Axial force/LTB interaction:

\frac{309.6}{1106.330769} + \frac{38.060032}{173.739077}

= 0.498908 < 1

Both storeys pass all three demonstrated interaction checks. The lower-storey flexural interaction, approximately $0.795$ , is the largest. The original source-rounded results are approximately lower $0.65$ / $0.79$ / $0.76$ and upper $0.28$ / $0.46$ / $0.50$ .

Animation labA strong slice can belong to an unstable member6 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.

Compact exam answer

Using exact arithmetic: upper/lower compression $309.6$ / $1,103.6 kN$ . pᵧ355,Class 1; A𝗀pᵧ2,712.2 kN; section capacities $232.88$ / $85.626 kN \cdot m$ ; elastic $207.32$ / $71.355$ . $e = 204.8 mm$ ; level 2 moment $74.1376 kN \cdot m$ shares $49.4251$ / $24.7125$ . Amplified lower moment $56.8388$ ; upper governing $38.0600$ . Lower $P_{c} = 2,119.365 kN$ ,Mᵦ232.88; interactions $0.6510$ / $0.7949$ / $0.7648$ . Upper $P_{c} = 1,106.331 kN$ ,Mᵦ173.7391; interactions $0.2776$ / $0.4634$ / $0.4989$ . Both storeys adequate under the stated simple-construction restraint assumptions.

Mistakes to avoid

Accumulate both levels for lower axial load.
Cancel opposite moments, not the underlying compressive reactions.
Share moment by $\frac{I}{L}$ , not equally and not by axial-force ratios.
Use actual storey length in 0.5L/rᵧ.
Do not treat a small roof tie as rotational restraint.

Procedure for an unfamiliar variant

Trace the loads above each column segment and keep dead/imposed components separate.
Read actual storey heights, section properties and end restraints in both axes.
For simple construction derive nominal eccentric moments and share them by $\frac{I}{L}$ .
Classify the section and form the correct section and member resistances.
Apply only the prescribed load reductions and moment amplification rules.
Check every storey segment and report both the governing check and source assumptions.

Independent self-check

Try it yourself. Invented variant: change only the upper-storey length from $6 m$ to $3 m$ , retaining the same column section and level 2 moment $74.1376 kN \cdot m$ . How is that joint moment shared?

Reveal answer and reasoning

The upper and lower $\frac{I}{L}$ values now match, so each receives $\frac{74.1376}{2} = 37.0688 kN \cdot m$ . The lower amplified share becomes $42.62912 kN \cdot m$ . This redistribution does not alter the accumulated axial loads. Upper buckling lengths and 0.5L/rᵧ also change and must be recalculated; the original upper resistance cannot be reused.

Animation labEccentric reactions and stiffness sharing2 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.

A beam reaction can act away from the column centre even at a nominally pinned beam connection.
$M = R e$ . Opposing reactions can cancel part of the signed moment, while both still add compression.
The course simple model distributes the joint moment in proportion to $\frac{I}{L}$ of the columns above and below.
Equal relevant stiffness gives half each. A roof joint with no upper column is a different case.

Column example 3: loads and eccentric moments in two storeys

Read the diagram and collect the data

Before calculating: recognition and strategy

1. Recalculate every reaction-table entry and storey total

2. Strength and section classification

3. Calculate section and elastic member capacities

4. Nominal eccentricity and stiffness sharing at level 2

5. Lower-storey section interaction

6. Lower-storey member buckling

7. Upper-storey governing moment and section interaction

8. Upper-storey buckling about both axes

Compact exam answer

Mistakes to avoid

Procedure for an unfamiliar variant

Independent self-check