Worked examples

Tutorial 4 Q4: eccentric roof reactions in simple construction

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Check the roof-storey S355 section using simple construction: $203 \times 203 \times 60$ UC. $L^{E} = L = 4.5 m$ . First-order ultimate loads: concentric $650 kN$ , eccentric $e_{y} = 105 mm$ at $140 kN$ , eccentric $e_{x} = 208 mm$ at $100 kN$ . All moment amplification factors are $1.15$ .

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 $203 \times 203 \times 60$ 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	$T = 14.2 mm$ 、 $t = 9.4 mm$ 、 $d = 160.8 mm$ 。
Local slenderness	$\frac{b}{T} = 7.25$ 、 $\frac{d}{t} = 17.1$ 。
Radii (converted from $cm$ )	$r_{x} = 8.96 \times 10 = 89.6 mm$ ； $r_{y} = 5.2 \times 10 = 52 mm$ 。
Area	$A = 76.4 {cm}^{2} = 7640 {mm}^{2}$ 。
Elastic moduli	$Z_{x} = 584 {cm}^{3}$ ； $Z_{y} = 201 {cm}^{3}$ 。
Plastic moduli	$S_{x} = 656 {cm}^{3}$ ； $S_{y} = 305 {cm}^{3}$ 。
LTB parameters	$u = 0.846$ , $x = 14.1$ ; both dimensionless

Before calculating: recognition and strategy

Sum every vertical load for compression, but use each force’s perpendicular lever arm for its bending axis. A roof joint has no upper column to share moment with in this stated model. Simple construction fixes the moment factors to $1$ and provides the special 0.5L/rᵧ LTB rule.

1. Axial force, moments and classification

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

F = 650 + 140 + 100 = 890 kN

M_{x, first} = F_{y} e_{x} = \frac{100 \times 208}{1,000} = 20.8 kN \cdot m

M_{y, first} = F_{x} e_{y} = \frac{140 \times 105}{1,000} = 14.7 kN \cdot m

M_{x, amp} = M_{LT} = 1.15 \times 20.8 = 23.92 kN \cdot m

M_{y, amp} = 1.15 \times 14.7 = 16.905 kN \cdot m

m_{x} = m_{y} = m_{LT} = 1

T = 14.2 mm

; use the S355 thickness band to select

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

ε = \sqrt{\frac{275}{355}} = 0.880141

Flange

\frac{b}{T} = 7.25

; Class 1 limit

9 ε = 7.921268

; Class 2 limit

10 ε = 8.801408

. Flange is Class 1. Web stress parameter:

r_{1} = \frac{F}{d t p_{y}} = \frac{890 \times 1,000}{160.8 \times 9.4 \times 355}

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 = 35.205633

\frac{d}{t} = 17.1 < 35.205633

. 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.

Animation labFollow the floor load in 3D4 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 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. Check the section

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{76.4 \times 100 \times 355}{1,000} = 2712.2 kN

M_{x, amp} = 1.15 \times 20.8 = 23.92 kN \cdot m

M_{y, amp} = 1.15 \times 14.7 = 16.905 kN \cdot m

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

Section ratio:

\frac{890}{2712.2} + \frac{23.92}{232.88} + \frac{16.905}{85.626} = 0.628289

Passes.

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.

3. Axial and flexural member resistance

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 $x$ use curve b; about $y$ use curve c. Read the Data File p.8 $p_{y} = 355 \frac{N}{{mm}^{2}}$ column. The two axes use different curves, so slenderness alone cannot determine the governing axis.

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

is the given effective length.

λ_{x} = \frac{4500}{89.6} = 50.223214

λ_{y} = \frac{4500}{52} = 86.538462

x

axis, curve b:

50

row gives

298

;

52

row gives

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

. Interpolation fraction:

\frac{50.223214 - 50}{52 - 50} = 0.111607

p_{c} = 298 + 0.111607 \times (293 - 298) = 297.441964 \frac{N}{{mm}^{2}}

y

axis, curve c:

86

row gives

173

;

88

row gives

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

. Interpolation fraction:

\frac{86.538462 - 86}{88 - 86} = 0.269231

p_{c} = 173 + 0.269231 \times (168 - 173) = 171.653846 \frac{N}{{mm}^{2}}

P_{cx} = \frac{A p_{cx}}{10} = \frac{76.4 \times 297.441964}{10} = 2272.456607 kN

P_{cy} = \frac{76.4 \times 171.653846}{10} = 1311.435385 kN

P_{c} = \min (P_{cx}, P_{cy}) = 1311.435385 kN

Member interaction uses elastic moment denominators pᵧZ, even when the cross-section check used plastic moduli. Use amplified moments here.

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

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

= \frac{890}{1311.435385} + \frac{1 \times 23.92}{207.32} + \frac{1 \times 16.905}{71.355}

= 1.030937

Fails.

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. Simple-construction LTB and conclusion

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

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 4500}{52} = 43.269231

Read Data File p.5 Table 8.3a, pᵧ355 column:

40

row gives

325

;

45

row gives

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

. Interpolation fraction:

\frac{43.269231 - 40}{45 - 40} = 0.653846

p_{b} = 325 + 0.653846 \times (309 - 325) = 314.538462 \frac{N}{{mm}^{2}}

M_{b} = p_{b} S_{x} = \frac{314.538462 \times 656}{1,000} = 206.337231 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{890}{1311.435385} + \frac{1 \times 23.92}{206.337231} + \frac{1 \times 14.7}{71.355}

= 1.000585

Fails.

$p_{y} = 355 \frac{N}{{mm}^{2}}$ ; $P_{cx} = 2272.457 kN$ , $P_{cy} = 1311.435 kN$ ; $M_{cx} = 232.880 kN \cdot m$ , $M_{cy} = 85.626 kN \cdot m$ . $M_{b} = 206.337 kN \cdot m$ . Ratios: section $0.628289$ , flexural buckling $1.030937$ , axial force/LTB $1.000585$ . At least one required ratio exceeds $1$ , so the column fails.

Animation labA strong slice can belong to an unstable member5 concepts · 9 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

F890 kN; first-order Mx 20.8/My 14.7; amplified $23.92$ / $16.905 kN \cdot m$ . No roof moment sharing; all $m = 1$ ; actualL4.5 m in λᴸᵀ=0.5L/rᵧ.

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.
Reaction position along $X$ creates moment about $y$ ; the force label alone is not the moment-axis label.
The $650 kN$ central load adds compression but no eccentric moment.

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: add an equal $140 kN$ vertical reaction at the opposite $105 mm$ position about $y$ . What changes?

Reveal answer and reasoning

Compression rises to $1,030 kN$ . The two $140 kN$ reactions cancel their $y$ -axis moments, so My becomes 0; $M_{x}$ remains $20.8 kN \cdot m$ first-order. It is unsafe to assume the column improves automatically: the axial term rises while the minor bending term disappears. Recompute all three ratios.