Worked examples

Column example 4: tributary area and cumulative column loads

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Chinese–English terminology

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Design internal column A in S355 for axial load only. Roof dead/imposed intensities are $5$ / $1.5 \frac{kN}{m^{2}}$ ; floor intensities $7$ / $3 \frac{kN}{m^{2}}$ . Upper storey height $4 m$ ; lower $4.5 m$ . The foundation is fixed. Floor beams prevent translation but do not restrain column rotation. All given dead loads include beam and column self-weight.

Original source: LectureNotes/Ch 4_Column.pdf — p. 32, p. 33, p. 34, p. 35. 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

Input	Where to read it
Tributary dimensions	Plan labels $3.8 + 3.8 = 7.6 m$ and $3 + 3 = 6 m$ ; use plan area, not storey height.
Vertical dimensions	Elevation $4 m$ upper and $4.5 m$ lower.
Roof/floor loads	Question text; the same floor plan applies at both levels.
Axial-only scope	Explicitly requested; no eccentric moment is introduced into this exercise.
Imposed reduction	Lecture p.33 applies its 2011 load-code floor-count table; roof counts as a floor.

Before calculating: recognition and strategy

Multiply each level’s pressure by the area feeding the column. The upper segment carries only roof loading; the lower carries roof plus first floor. Reduce the cumulative imposed load only when the source’s permitted floor-count rule applies. After factoring, design each storey separately because loads, lengths and end conditions differ.

1. Derive the tributary area from the hatched plan

Simple explanation: How a floor load reaches a beam

Each beam collects the load from its own strip of floor.

Find the tributary width from the actual plan.
Area load × tributary width gives load per beam length.
A supporting beam receives the other beam’s end reaction.

Remember: A reaction becomes a point load, not automatically a UDL.

Related concept and full method

Tributary width:

\frac{7.6}{2} + \frac{7.6}{2} = 3.8 + 3.8 = 7.6 m

Tributary depth:

\frac{6}{2} + \frac{6}{2} = 3 + 3 = 6 m

A_{trib} = 7.6 \times 6 = 45.6 m^{2}

Half of each adjacent simply supported bay reaches column A. The hatch is a load-allocation area, not the column’s steel cross-sectional area. Both roof and floor use $45.6 m^{2}$ .

Animation labFollow the floor load in 3D1 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 floor carries pressure in $\frac{kN}{m ^{2}}$ . The highlighted strip belongs to one secondary beam.
Multiply pressure by tributary width: $w = q \times b$ . The illustration uses $q = 5 \frac{kN}{m ^{2}}$ .
A primary beam receives the secondary beam reaction at their connection, not a new full-span UDL.
Trace reactions down to columns and foundations. Count each loaded area once.

2. Read the source’s floor-count reduction

Floors carried	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	Over $8$
Imposed reduction	$0 %$	$5 %$	$10 %$	$15 %$	$20 %$	$25 %$	$30 %$	$35 %$	$40 %$ maximum

The upper member supports one roof level → $0 %$ reduction. The lower supports roof plus first floor →two levels → $5 %$ . Retain $95 %$ of the cumulative imposed load below the floor. This is the rule applied by this lecture example, attributed there to the 2011 Dead and Imposed Loads Code; do not extend it automatically to an occupancy or question where reduction is disallowed. Dead load is not reduced.

Animation labFollow the floor load in 3D3 concepts

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

The floor carries pressure in $\frac{kN}{m ^{2}}$ . The highlighted strip belongs to one secondary beam.
Multiply pressure by tributary width: $w = q \times b$ . The illustration uses $q = 5 \frac{kN}{m ^{2}}$ .
A primary beam receives the secondary beam reaction at their connection, not a new full-span UDL.
Trace reactions down to columns and foundations. Count each loaded area once.

3. Compute characteristic and ultimate loads at each level

Roof:

G = 5 \times 45.6 = 228 kN; Q = 1.5 \times 45.6 = 68.4 kN

Additional first-floor load:

G = 7 \times 45.6 = 319.2 kN; Q = 3 \times 45.6 = 136.8 kN

Lower-storey accumulated load:

G = 228 + 319.2 = 547.2 kN

Q = 68.4 + 136.8 = 205.2 kN

Upper-storey design force:

F = 1.4 \times 228 + 1.6 \times 68.4 = 319.2 + 109.44 = 428.64 kN

Reduced lower-storey imposed load:

Q = 0.95 \times 205.2 = 194.94 kN

Lower-storey design force:

F = 1.4 \times 547.2 + 1.6 \times 194.94

= 766.08 + 311.904 = 1,077.984 kN

All self-weight is already included. The lower load uses the whole accumulated load rather than adding an already factored/reduced upper force to a separately reduced floor force.

Animation labFrom characteristic to design load1 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.

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.

4. Select and check the upper column

Trial $152 \times 152 \times 30$ UC. Data File p.11 gives $T = 9.4 mm$ , $\frac{b}{T} = 8.13$ , $\frac{d}{t} = 19.0$ , $A = 38.3 {cm}^{2}$ , $r_{x} = 6.76 cm$ , $r_{y} = 3.83 cm$ . The original lecture uses the same area $38.3$ . $T \leq 16$ , giving $p_{y} = 355$ . Both ends are held in position but free to rotate, so $K = 1$ , with both axes having $L^{E} = 4,000 mm$ .

ε = \sqrt{\frac{275}{355}} = 0.880141

\frac{b}{T} = 8.13 < 13 ε = 11.4418; \frac{d}{t} = 19.0 < 40 ε = 35.2056

Non-slender.

λ_{x} = \frac{4,000}{67.6} = 59.171598

λ_{y} = \frac{4,000}{38.3} = 104.438642

Hot-rolled H-section thickness

\leq 40 mm

，

x

axis curve b;

y

axis curve c;

p_{y} = 355

. Curve b:

58

row gives

278

，

60

row gives

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

. Interpolation fraction:

\frac{59.171598 - 58}{60 - 58} = 0.585799

p_{c} = 278 + 0.585799 \times (272 - 278) = 274.485207 \frac{N}{{mm}^{2}}

Curve c:

104

row gives

133

，

106

row gives

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

. Interpolation fraction:

\frac{104.438642 - 104}{106 - 104} = 0.219321

p_{c} = 133 + 0.219321 \times (129 - 133) = 132.122715 \frac{N}{{mm}^{2}}

P_{cx} = \frac{3,830 \times 274.485207}{1,000} = 1051.278343 kN

P_{cy} = \frac{3,830 \times 132.122715}{1,000} = 506.030000 kN

P_{c} = 506.030000 kN > 428.64 kN

Adequate. Utilisation:

\frac{428.64}{506.030000} = 0.847064

The source uses rounded $p_{c} = 132$ , so $\frac{38.3 \times 132}{10} = 505.56 kN$ and prints $505.6$ . Exact interpolation gives $506.030 kN$ ; both pass. The previously omitted major-axis comparison is provided above.

Animation labEffective length and buckling axes5 concepts · 13 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 column can bow sideways before its section reaches the crushing resistance.
Each axis has its own radius of gyration and restraint spacing.
$λ = \frac{L _{E}}{r}$ , with compatible length units. A tie affects only the directions it actually restrains.
Select each buckling curve and compressive strength before forming Pc. The governing axis is determined by resistance, not slenderness alone.

5. Select and check the lower column

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

Trial $203 \times 203 \times 46$ UC. Data File p.11 gives $T = 11.0 mm$ , $\frac{b}{T} = 9.25$ , $\frac{d}{t} = 22.3$ , $A = 58.7 {cm}^{2}$ , $r_{x} = 8.82 cm$ , $r_{y} = 5.13 cm$ . The flange thickness again gives $p_{y} = 355$ . The base is fixed; the floor end is held in position but free to rotate, so use the recommended factor $K = 0.85$ .

L_{Ex} = L_{Ey} = 0.85 \times 4,500 = 3,825 mm

\frac{b}{T} = 9.25 < 13 ε = 11.4418; \frac{d}{t} = 22.3 < 40 ε = 35.2056

Non-slender.

λ_{x} = \frac{3,825}{88.2} = 43.367347

λ_{y} = \frac{3,825}{51.3} = 74.561404

Curve b,

p_{y} = 355

42

row gives

314

，

44

row gives

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

. Interpolation fraction:

\frac{43.367347 - 42}{44 - 42} = 0.683673

p_{c} = 314 + 0.683673 \times (310 - 314) = 311.265306 \frac{N}{{mm}^{2}}

Curve c,

p_{y} = 355

74

row gives

205

，

76

row gives

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

. Interpolation fraction:

\frac{74.561404 - 74}{76 - 74} = 0.280702

p_{c} = 205 + 0.280702 \times (200 - 205) = 203.596491 \frac{N}{{mm}^{2}}

P_{cx} = \frac{5,870 \times 311.265306}{1,000} = 1827.127347 kN

P_{cy} = \frac{5,870 \times 203.596491}{1,000} = 1195.111404 kN

P_{c} = 1195.111404 kN > 1,077.984 kN

Adequate. Utilisation:

\frac{1,077.984}{1195.111404} = 0.901995

The source rounds $p_{c}$ to $204$ , giving $\frac{58.7 \times 204}{10} = 1,197.48 kN$ (printed $1,197$ ). Exact interpolation gives $1,195.111 kN$ . Both selected columns pass the requested axial-only design; the lower weak-axis check is more highly utilised.

Animation labEffective length and buckling axes5 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.

A column can bow sideways before its section reaches the crushing resistance.
Each axis has its own radius of gyration and restraint spacing.
$λ = \frac{L _{E}}{r}$ , with compatible length units. A tie affects only the directions it actually restrains.
Select each buckling curve and compressive strength before forming Pc. The governing axis is determined by resistance, not slenderness alone.

Compact exam answer

$A_{trib} = 45.6 m^{2}$ . Roof $G$ ／ $Q$ is $228$ ／ $68.4 kN$ ; lower-storey accumulation $547.2$ ／ $205.2 kN$ . Upper-storey ultimate force $428.64 kN$ ; lower-storey imposed load after reduction $5 %$ gives $1,077.984 kN$ . Upper-storey $152 \times 152 \times 30$ UC: non-slender, $L^{E} = 4 m$ , $P_{c} = 506.030 kN$ . Lower-storey $203 \times 203 \times 46$ UC: non-slender, $L^{E} = 3.825 m$ , $P_{c} = 1,195.111 kN$ . Both are S355 with $p_{y} = 355$ , governed by minor-axis curve c; both are adequate.

Mistakes to avoid

Do not use $4$ or $4.5 m$ elevation heights when calculating tributary plan area.
Include roof load in the lower segment.
Reduce only imposed load and count the roof as instructed.
Do not add self-weight twice.
Use fixed/pinned recommended $0.85$ for the lower column, not the upper one.

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: imposed-load reduction is explicitly prohibited. Does the selected lower column still pass?

Reveal answer and reasoning

Lower force becomes $1.4 \times 547.2 + 1.6 \times 205.2 = 766.08 + 328.32 = 1,094.4 kN$ . Compare with $1,195.111 kN$ : utilisation $\approx 0.9157 < 1$ , so it still passes. The roof segment is unchanged because it already had zero reduction.

Animation labEffective length and buckling axes2 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.

A column can bow sideways before its section reaches the crushing resistance.
Each axis has its own radius of gyration and restraint spacing.
$λ = \frac{L _{E}}{r}$ , with compatible length units. A tie affects only the directions it actually restrains.
Select each buckling curve and compressive strength before forming Pc. The governing axis is determined by resistance, not slenderness alone.