Worked examples

Tutorial 4 Q1: compression resistance at a given effective length

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Find the compressive resistance of $305 \times 305 \times 118$ UC, Grade S355, with effective length $4.5 m$ .

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.

Ch 4 p.40 is a written question without a separate column sketch. The terms identify effective length and compressive resistance. The original website caption writes

L^{E}

, while the question prints

L_{E}

; both mean the given

4.5 m

effective length here, without multiplying by

0.85

0.7

again. The final section-designation number is nominal mass per metre

118 \frac{kg}{m}

; the precise mass per metre in the section table below is

117.9 \frac{kg}{m}

, not an additional imposed load. UC axes and flange/web dimensions are defined by the linked original section table.Open full-size image

Data File p.11,

305 \times 305 \times 118

dimensions and properties row. Universal Columns identifies the UC sections; the upper table gives dimensions and the lower properties. In the sketch,

D

is overall depth;

B

overall width;

t

web thickness;

T

flange thickness;

r

root radius;

d

clear web depth between fillets;

b

the flange outstand used by the table.

x - x

passes horizontally through the web;

y - y

passes vertically through its centre.

C

= end clearance;

N

／

n

= notch dimensions. Other dimensions in this row:

D = 314.5 mm

B = 307.4 mm

r = 15.2 mm

C = 8 mm

N = 158 mm

n = 34 mm

. The lower table gives second moments of area

I_{x} = 27670 {cm}^{4}

I_{y} = 9059 {cm}^{4}

; warping constant

H = 1.98 {dm}^{6}

, torsional constant

J = 161 {cm}^{4}

. Upper-table surface areas are per metre or per tonne, with respective row values

1.81 m^{2}

and

15.3 m^{2}

. These additional lookup values do not alter this question's resistance calculations. The ratios, area and radii of gyration used are individually typeset below. Do not read

cm

mm

. Open full-size image

Read the diagram and collect the data

Lookup: Data File p.11, exact row $305 \times 305 \times 118$ 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	$T = 18.7 mm$ 、 $t = 12 mm$ 、 $d = 246.7 mm$ 。
Local slenderness	$\frac{b}{T} = 8.22$ 、 $\frac{d}{t} = 20.6$ 。
Radii (converted from $cm$ )	$r_{x} = 13.6 \times 10 = 136 mm$ ； $r_{y} = 7.77 \times 10 = 77.7 mm$ 。
Area	$A = 150 {cm}^{2} = 15000 {mm}^{2}$ 。
Elastic moduli	$Z_{x} = 1760 {cm}^{3}$ ； $Z_{y} = 589 {cm}^{3}$ 。
Plastic moduli	$S_{x} = 1958 {cm}^{3}$ ； $S_{y} = 895 {cm}^{3}$ 。
LTB parameters	$u = 0.85$ , $x = 16.2$ ; both dimensionless

Before calculating: recognition and strategy

This is the same section as Column Example 1 but a shorter effective length. Compare both axis resistances. Because the task is pure compression, it only needs a non-slender local-section check before $A_{g} p_{c}$ .

1. Local section and material strength

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 = 18.7 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} = 8.22 < 13 ε = 11.606470

Web:

\frac{d}{t} = 20.6 < 40 ε = 35.712215

Non-slender under uniform compression, so gross area is permitted.

This proves Class 3 or better for axial design. It is not a claim that plastic bending capacity has been established.

Animation labWhy thin elements buckle locally2 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.

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. Both buckling axes and the answer

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.

L_{Ex} = L_{Ey} = 4500 mm

is the given effective length.

λ_{x} = \frac{4500}{136} = 33.088235

λ_{y} = \frac{4500}{77.7} = 57.915058

x

axis, curve b:

30

row gives

325

;

35

row gives

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

. Interpolation fraction:

\frac{33.088235 - 30}{35 - 30} = 0.617647

p_{c} = 325 + 0.617647 \times (318 - 325) = 320.676471 \frac{N}{{mm}^{2}}

y

axis, curve c:

56

row gives

252

;

58

row gives

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

. Interpolation fraction:

\frac{57.915058 - 56}{58 - 56} = 0.957529

p_{c} = 252 + 0.957529 \times (247 - 252) = 247.212355 \frac{N}{{mm}^{2}}

P_{cx} = \frac{A p_{cx}}{10} = \frac{150 \times 320.676471}{10} = 4810.147059 kN

P_{cy} = \frac{150 \times 247.212355}{10} = 3708.185328 kN

P_{c} = \min (P_{cx}, P_{cy}) = 3708.185328 kN

P_{c} = 3708.185328 kN

The weak axis controls. No applied axial force is supplied in Q1, so the answer is a resistance, not an adequacy statement for an invented load. This is lower than the squash resistance $5,175 kN$ because member buckling reduces the usable strength.

Animation labEffective length and buckling axes3 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.

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

$T = 18.7 mm$ , giving $p_{y} = 345$ . $\frac{b}{T} = 8.22 < 13 ε$ ; $\frac{d}{t} = 20.6 < 40 ε$ . $λ_{x} = \frac{4,500}{136} = 33.088235$ ; $λ_{y} = \frac{4,500}{77.7} = 57.915058$ . The hot-rolled H-section uses curves b/c respectively. $P_{cx} = 4810.147 kN$ ; $P_{cy} = 3708.185 kN$ ; compression resistance $P_{c} = 3708.185 kN$ .

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.

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: can the same column carry $4,000 kN$ in pure compression?

Reveal answer and reasoning

No. $\frac{F}{P_{c}} = \frac{4,000}{3708.185328} = 1.078695 > 1$ . The gross squash resistance is not the member resistance.

Animation labEffective length and buckling axes1 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.

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.