Worked examples

Assignment 2 Q1: simple-construction UC with eccentric reactions

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Check the Class $1$ $356 \times 406 \times 287$ UC S355 column below the shown floor joint. Actual height $4 m$ ; effective length $3.4 m$ about both axes; upper/lower stiffness equal; all moment factors $1$ . First-order factored centralP4,500 kN, Pₓ₁600 kN at $220 mm$ from $y - y$ , Pᵧ₁900 kN at $320 mm$ from $x - x$ and Pᵧ₂850 kN on the opposite side at $280 mm$ . Amplification $1.1$ .

Original source: Assignment/AY2627s 1-CON4334-Assignment 2.pdf — p. 1. 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.

The drawing labelled Figure Q3 in the source actually belongs to Q1. It is a plan of the UC section; the black dots mark vertical compressive reactions.

P_{y 1}

and

P_{y 2}

lie on opposite sides of the

x - x

axis, so their moments act in opposite directions. The only

600 kN

reaction, measured relative to the

y - y

axis, acts to its left by

220 mm

P

acts at the centroid and produces no eccentric moment.Open full-size image

2023 paper, physical page 22 / Data 16: the exact dimensions row for

356 \times 406 \times 287

. The headings identify UNIVERSAL COLUMNS and their DIMENSIONS.

D

is overall depth;

B

overall width;

t

web thickness;

T

flange thickness;

r

root radius;

d

clear depth between root fillets;

\frac{b}{T}

and

\frac{d}{t}

are the width/thickness ratios for local buckling.

C

denotes end clearance. The notch columns use

N

and

n

, with different letter cases. In the sketch,

b

is the flange outstand width. The selected row gives a mass per metre of

287.1 \frac{kg}{m}

; its other dimensions are

D = 393.6 mm

B = 399.0 mm

r = 15.2 mm

C = 12 mm

N = 200 mm

n = 52 mm

. Surface areas are listed per metre and per tonne; the respective values in this row are

2.31 m^{2}

and

8.05 m^{2}

. Thicknesses, clear depth and width/thickness ratios are listed individually in the selected-row table below.Open full-size image

2023 paper, physical page 23 / Data 17: the exact properties row for

356 \times 406 \times 287

. Under PROPERTIES, each paired set of columns gives values about the

x - x

／

y - y

axes: second moment of area, radius of gyration, elastic section modulus and plastic section modulus. These are followed by buckling parameter

u

, torsional index

x

, warping constant

H

, torsional constant

J

and area

A

. Read the units printed in each column. The selected-row table and calculations below provide selectable equivalents. Original table units: second moment of area

{cm}^{4}

, radius of gyration

cm

, both section moduli

{cm}^{3}

, warping constant

{dm}^{6}

, torsional constant

{cm}^{4}

, area

{cm}^{2}

. The buckling parameter and torsional index are dimensionless. Additional values from this row are

I_{x} = 99880 {cm}^{4}

I_{y} = 38680 {cm}^{4}

H = 19.3 {dm}^{6}

J = 1441 {cm}^{4}

; the remaining properties used in design are in the table below.Open full-size image

Read the diagram and collect the data

Lookup: 2023 paper, physical pp.22–23 (printed Data Pages 16–17), exact row $356 \times 406 \times 287$ 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	$T = 36.5 mm$ 、 $t = 22.6 mm$ 、 $d = 290.2 mm$ 。
Local slenderness	$\frac{b}{T} = 5.47$ 、 $\frac{d}{t} = 12.8$ 。
Radii (converted from $cm$ )	$r_{x} = 16.5 \times 10 = 165 mm$ ； $r_{y} = 10.3 \times 10 = 103 mm$ 。
Area	$A = 366 {cm}^{2} = 36600 {mm}^{2}$ 。
Elastic moduli	$Z_{x} = 5075 {cm}^{3}$ ； $Z_{y} = 1939 {cm}^{3}$ 。
Plastic moduli	$S_{x} = 5812 {cm}^{3}$ ； $S_{y} = 2949 {cm}^{3}$ 。
LTB parameters	$u = 0.835$ , $x = 10.2$ ; both dimensionless

Before calculating: recognition and strategy

Separate the axial sum from the joint moment balance. All vertical reactions increase the lower column’s compression. Opposite eccentric reactions subtract for moment, then the equal column stiffnesses split that net moment equally. Use $3.4 m$ for axial buckling and actual $4 m$ for simple-construction LTB.

(a) Total axial force, joint moments and column moments

F_{lower} = 4,500 + 600 + 900 + 850 = 6,850 kN

M_{x, joint} = \frac{900 \times 320}{1,000} - \frac{850 \times 280}{1,000}

= 288 - 238 = 50 kN \cdot m

| M_{y, joint} | = \frac{600 \times 220}{1,000} = 132 kN \cdot m

The upper and lower column

\frac{I}{L}

values are equal; each takes the following share of joint moment:

\frac{1}{2}

M_{x, first} = \frac{50}{2} = 25 kN \cdot m

M_{y, first} = \frac{132}{2} = 66 kN \cdot m

M_{x, amp} = M_{LT} = 1.1 \times 25 = 27.5 kN \cdot m

M_{y, amp} = 1.1 \times 66 = 72.6 kN \cdot m

Give both first-order $25$ / $66$ and amplified $27.5$ / $72.6 kN \cdot m$ values so the answer to(a) is unambiguous. Do not split the $6,850 kN$ axial sum between columns. The problem says centralP includes self-weight; no additional self-weight is added.

Animation labEccentric reactions and stiffness sharing3 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.

(b) Cross-section capacity

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

The question permits Class 1. The supplied property row also supports it; verify the material band and conservative local limits:

T = 36.5 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} = 5.47

; 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{6850 \times 1,000}{290.2 \times 22.6 \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} = 12.8 < 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.

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{366 \times 100 \times 345}{1,000} = 12627 kN

M_{x, amp} = 1.1 \times 25 = 27.5 kN \cdot m

M_{y, amp} = 1.1 \times 66 = 72.6 kN \cdot m

M_{cx} = \min (\frac{345 \times 5812}{1,000}, \frac{1.2 \times 345 \times 5075}{1,000})

= \min (2005.14, 2101.05) = 2005.14 kN \cdot m

M_{cy} = \min (\frac{345 \times 2949}{1,000}, \frac{1.2 \times 345 \times 1939}{1,000})

= \min (1017.4, 802.746) = 802.746 kN \cdot m

Section ratio:

\frac{6850}{12627} + \frac{27.5}{2005.14} + \frac{72.6}{802.746} = 0.646643

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) Both member-buckling interactions

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

Simple construction: $m_{x} = m_{y} = m_{LT} = 1$ as given. Effective lengths $3,400 mm$ are already specified, so apply no further $K$ factor.

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} = 3400 mm

(known effective length).

λ_{x} = \frac{3400}{165} = 20.606061

λ_{y} = \frac{3400}{103} = 33.009709

x

axis, curve b:

20

row gives

339

;

25

row gives

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

. Interpolation fraction:

\frac{20.606061 - 20}{25 - 20} = 0.121212

p_{c} = 339 + 0.121212 \times (332 - 339) = 338.151515 \frac{N}{{mm}^{2}}

y

axis, curve c:

30

row gives

315

;

35

row gives

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

. Interpolation fraction:

\frac{33.009709 - 30}{35 - 30} = 0.601942

p_{c} = 315 + 0.601942 \times (305 - 315) = 308.980583 \frac{N}{{mm}^{2}}

P_{cx} = \frac{A p_{cx}}{10} = \frac{366 \times 338.151515}{10} = 12376.345455 kN

P_{cy} = \frac{366 \times 308.980583}{10} = 11308.689320 kN

P_{c} = \min (P_{cx}, P_{cy}) = 11308.689320 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 5075}{1,000} = 1750.88 kN \cdot m

M_{ey} = \frac{345 \times 1939}{1,000} = 668.955 kN \cdot m

\frac{F}{P_{c}} + \frac{m_{x} M_{x, amp}}{M_{ex}} + \frac{m_{y} M_{y, amp}}{M_{ey}}

= \frac{6850}{11308.689320} + \frac{1 \times 27.5}{1750.88} + \frac{1 \times 72.6}{668.955}

= 0.729963

Passes.

Simple-construction special rule: use actual storey length $L$ in 0.5L/rᵧ; the axial effective length is a different quantity.

λ_{LT} = \frac{0.5 \times 4000}{103} = 19.417476

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

Calculated

λ_{LT} = 19.417476

, below the first printed row

25

. Conservatively use the greater slenderness

25

without extrapolation:

p_{b} = 345 \frac{N}{{mm}^{2}}

, also not exceeding material design strength.

M_{b} = p_{b} S_{x} = \frac{345.000000 \times 5812}{1,000} = 2005.140000 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{6850}{11308.689320} + \frac{1 \times 27.5}{2005.140000} + \frac{1 \times 66}{668.955}

= 0.718105

Passes.

$p_{y} = 345 \frac{N}{{mm}^{2}}$ ; $P_{cx} = 12376.345 kN$ , $P_{cy} = 11308.689 kN$ ; $M_{cx} = 2005.140 kN \cdot m$ , $M_{cy} = 802.746 kN \cdot m$ . $M_{b} = 2005.140 kN \cdot m$ . Ratios: section $0.646643$ ; flexural buckling $0.729963$ ; axial force/LTB $0.718105$ . All three requested strength checks pass.

The calculated simple-construction value $λ_{LT} = 19.417$ is below the first printed Table 8.3a row, $25$ . Conservatively read the more slender $25$ row to obtain $p_{b} = 345 \frac{N}{{mm}^{2}}$ , equal to the material ceiling, without extrapolation. All three requested checks pass.

Animation labA strong slice can belong to an unstable member6 concepts · 16 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

(a) F6,850 kN; net jointMx 50/My 132; lower first-order $25$ / $66$ and amplified $27.5$ / $72.6 kN \cdot m$ . Class 1, pᵧ345; curvesb/c; LE3.4 m, actualL4 m; allm 1.

Mistakes to avoid

Do not replace the missing current-data row by a similarly named $305$ UC.
Do not add opposite eccentric moments.
Do not halve the axial force.
Do not apply $0.85$ to an effective length that is already given.

Procedure for an unfamiliar variant

Find the exact property row, including supplied supplementary data if necessary.
Sum compression, balance signed joint moments and apply the stiffness fractions.
Amplify once and check the section.
Use each axis’s radius/curve and the correct moment denominators.
Apply the simple-construction LTB length rule and report every check.

Independent self-check

Try it yourself. Invented variant: Pᵧ₂ rises from $850$ to $950 kN$ . Does the $x$ -axis moment necessarily increase?

Reveal answer and reasoning

Not necessarily. Net joint moment $M_{x} = 900 \times 0.320 - 950 \times 0.280 = 22 kN \cdot m$ , so the lower column's first-order $M_{x}$ becomes $11 kN \cdot m$ ; after amplification, $M_{x} = 12.1 kN \cdot m$ . Axial force increases to $6,950 kN$ . Opposing reactions reduce the net moment but increase compression; recalculate every interaction check.

Animation labEccentric reactions and stiffness sharing4 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 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.