Worked examples

2023 BQ4: the same beam with two restraint arrangements

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For a $6 m$ simply supported $406 \times 140 \times 46$ UB S355, assumed Class 1, carrying a central $110 kN$ design point load and negligible self-weight: (a) find maximum shear/moment; (b) check LTB with support restraints only; (c) repeat with an additional effective midpoint lateral restraint.

Original source: Pastpaper/22ENGTY033.pdf — p. 6. 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 pp.9–10, exact $406 \times 140 \times 46$ UB row. Dimensions on p.9; properties on p.10. $D$ is overall depth; $d$ is the clear web depth between root fillets, not the nominal designation.

Property	Lookup value / conversion
Dimensions	D403.2, web t 6.8, flange T11.2, root r 10.2, clear d 360.4, all $mm$
Local ratios	$\frac{b}{T} = 6.35$ ； $\frac{d}{t} = 53$ 。
Major-axis properties	$I_{x} = 15690 {cm}^{4} = 1.569 \times 10^{8} {mm}^{4}$ ; $Z_{x} = 778 {cm}^{3}$ ; $S_{x} = 888 {cm}^{3}$ .
LTB properties	$r_{y} = 3.03 cm = 30.3 mm$ ; $u = 0.871$ ; torsional index $x = 38.9$ .

Before calculating: recognition and strategy

Vertical loading is identical in both cases, so the reactions and BMD stay identical. What changes is the unrestrained segment: one $6 m$ length becomes two $3 m$ lengths. Both the segment moment factor and the segment buckling resistance must be recalculated.

(a) Maximum shear and moment —2 printed marks

Simple explanation: Why reactions balance the loads

Think of a seesaw that must neither fall nor turn.

Upward and downward forces must balance.
Take moments about a support to eliminate its reaction.
Use perpendicular distance from the point to the force line.

Remember: A lateral restraint is not automatically a vertical support.

Related concept and full method

By symmetry:

R_{A} = R_{C} = \frac{110}{2} = 55 kN

Maximum shear magnitude:

| V |_{\max} = 55 kN

Maximum midspan moment:

M_{\max} = \frac{P L}{4} = \frac{110 \times 6}{4} = 165 kN \cdot m

Calculated shear and bending diagrams with labelled support/load ordinates; positive values upward. All values are repeated in the working. — Teaching diagrams drawn from the calculation, not original source images. Labels identify calculated diagrams, positive ordinates upwards, shear force and bending moment. Positions are in $m$ , $V$ is measured in $kN$ , $M$ is measured in $kN \cdot m$ ; A, B and C denote the left end, midspan and right end. All values are also listed in the solution steps.Open full-size image.

Animation labBalance reactions and moments1 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.

This demonstrator is a simply supported 6 m beam with a 60 kN point load; it is not the page’s original loading diagram.
$R_{B} \times L = P \times a$ . Moving the load towards B increases $R_{B}$ .
$R_{A} = P - R_{B}$ . The two upward reactions must sum to P.
With the load at midspan, $R_{A} = R_{B} = 30 kN$ . End couples, UDLs and overhangs require their own equilibrium terms.

(b) End restraints only —14 printed marks

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

S355, flange

T = 11.2 mm

, so

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

ε = \sqrt{\frac{275}{355}} = 0.880141

Flange:

\frac{b}{T} = 6.35 < 9 ε = 7.921268

Web in bending:

\frac{d}{t} = 53 < 80 ε = 70.411267

Both are Class 1, so

β_{W} = 1

; plastic moment resistance is permitted.

\frac{d}{t} = 53 < 70 ε = 61.609858

No shear-buckling calculation is required under the course limit.

For the full $6 m$ segment the BMD is triangular with its peak in the middle. At local quarter points $1.5$ , $3$ , $4.5 m$ , the moments are $82.5$ , $165$ , $82.5 kN \cdot m$ .

Table 8.4b:

m_{LT} = 0.2 + \frac{0.15 \times 82.5 + 0.5 \times 165 + 0.15 \times 82.5}{165}

= 0.2 + 0.075 + 0.5 + 0.075 = 0.85

Equivalent uniform moment demand:

0.85 \times 165 = 140.25 kN \cdot m

Normal-load effective length is the given segment length. All quantities in $λ$ and λᴸᵀ are dimensionless. This Class 1 section has $β_{W} = 1$ .

λ = \frac{L_{E}}{r_{y}} = \frac{6000}{30.3} = 198.019802

v = {[1 + 0.05 {(\frac{λ}{x})}^{2}]}^{- \frac{1}{4}}

= {[1 + 0.05 {(\frac{198.019802}{38.9})}^{2}]}^{- \frac{1}{4}} = 0.812407

λ_{LT} = u v λ \sqrt{β_{W}}

= 0.871 \times 0.812407 \times 198.019802 \times 1 = 140.120058

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

140

gives

80

，

145

gives

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

. Interpolation fraction:

\frac{140.120058 - 140}{145 - 140} = 0.024012

p_{b} = 80 + 0.024012 \times (75 - 80) = 79.879942 \frac{N}{{mm}^{2}}

M_{b} = \frac{p_{b} S_{x}}{1,000} = \frac{79.879942 \times 888}{1,000} = 70.933388 kN \cdot m

Equivalent uniform moment demand:

m_{LT} M_{\max} = 140.250000 kN \cdot m

Utilisation:

\frac{140.250000}{70.933388} = 1.977207

Fails.

The beam fails this LTB check. Its section bending resistance $315.24 kN \cdot m$ exceeds $165$ , but that local-section fact cannot override the much smaller lateral-buckling resistance $70.933388 kN \cdot m$ .

Animation labA beam bends sideways and twists4 concepts · 5 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 the illustrated sagging beam the top flange is compressed.
The unrestrained compression flange can move sideways while the complete cross-section twists.
Only effective restraints divide the member into unbraced segments. They do not automatically add vertical supports.
Use the segment effective length, section properties and matching moment factor. This is an exaggerated mode shape, not a calculated displacement.

(c) Add an effective midpoint restraint —14 printed marks

Simple explanation: Which length belongs in the calculation?

A sideways tie can shorten the buckling region without shortening the beam’s vertical span.

Separate vertical-support spacing from lateral-restraint spacing.
Apply the course rule for the actual end restraint and loading.
Keep the original vertical load analysis unless its supports change.

Remember: Restraints in one direction may not restrain the other direction.

Related concept and full method

The left $3 m$ segment now has moments $0$ atA and $165$ atB; the right segment mirrors it. The BMD over each segment is a straight triangle with its maximum at an end, not in the middle.

End-moment ratio:

β = \frac{0}{165} = 0

Table 8.4a:

m_{LT} = 0.6 + 0.4 β = 0.6

Equivalent demand:

0.6 \times 165 = 99 kN \cdot m

New effective length of each half-segment

L_{E} = 3,000 mm

Normal-load effective length is the given segment length. All quantities in $λ$ and λᴸᵀ are dimensionless. This Class 1 section has $β_{W} = 1$ .

λ = \frac{L_{E}}{r_{y}} = \frac{3000}{30.3} = 99.009901

v = {[1 + 0.05 {(\frac{λ}{x})}^{2}]}^{- \frac{1}{4}}

= {[1 + 0.05 {(\frac{99.009901}{38.9})}^{2}]}^{- \frac{1}{4}} = 0.932256

λ_{LT} = u v λ \sqrt{β_{W}}

= 0.871 \times 0.932256 \times 99.009901 \times 1 = 80.395537

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

80

gives

190

，

85

gives

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

. Interpolation fraction:

\frac{80.395537 - 80}{85 - 80} = 0.079107

p_{b} = 190 + 0.079107 \times (175 - 190) = 188.813390 \frac{N}{{mm}^{2}}

M_{b} = \frac{p_{b} S_{x}}{1,000} = \frac{188.813390 \times 888}{1,000} = 167.666290 kN \cdot m

Equivalent uniform moment demand:

m_{LT} M_{\max} = 99.000000 kN \cdot m

Utilisation:

\frac{99.000000}{167.666290} = 0.590459

Passes.

Both halves pass. A point load alone is not a lateral restraint; part(c) grants a separate effective restraint at that location. Suggested method-credit evidence: show the changedLE, moment factor, $λ$ / $v$ / $λ_{LT}$ , table interpolation, resistance and conclusion. Do not merely state “restraint makes it safe.”

Animation labA beam bends sideways and twists4 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.

For the illustrated sagging beam the top flange is compressed.
The unrestrained compression flange can move sideways while the complete cross-section twists.
Only effective restraints divide the member into unbraced segments. They do not automatically add vertical supports.
Use the segment effective length, section properties and matching moment factor. This is an exaggerated mode shape, not a calculated displacement.

Compact exam answer

(a)V55 kN,M165 kNm. (b)LE6 m,mLT0.85,demand $140.25$ ;λ198.019802,v 0.812407,λLT140.120058,pb 79.879942,Mb 70.933388 kNm → fails. (c)TwoLE3 m segments,mLT0.6,demand $99$ ;λ99.009901,v 0.932256,λLT80.395537,pb 188.813390,Mb 167.666290 kNm → passes. Normal-load/ $L_{E}$ assumptions as stated.

Mistakes to avoid

Do not factor the $110 kN$ design load.
Do not usemLT0.85 for the half-span segment.
Do not double the bolt/beam load merely because there are two segments.
A local bending pass does not rule out LTB.

Procedure for an unfamiliar variant

Keep the same vertical-load analysis.
Redraw only the restraint boundaries for each case.
Calculate a moment factor for each actual segment BMD.
Recompute slenderness and resistance for that length.
Compare equivalent demand withMb and state each case separately.

Independent self-check

Try it yourself. Invented variant: with the midpoint restraint present, increase the central design load to $180 kN$ . Does LTB still pass?

Reveal answer and reasoning

New maximum moment $M = \frac{180 \times 6}{4} = 270 kN \cdot m$ ; for each half-span, $m_{LT}$ remains $0.6$ , giving demand $162 kN \cdot m$ . Resistance remains $167.666290 kN \cdot m$ , so LTB passes with only a small margin. Other checks also need updating; this conclusion covers only the stated LTB model.

Animation labA beam bends sideways and twists1 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.

For the illustrated sagging beam the top flange is compressed.
The unrestrained compression flange can move sideways while the complete cross-section twists.
Only effective restraints divide the member into unbraced segments. They do not automatically add vertical supports.
Use the segment effective length, section properties and matching moment factor. This is an exaggerated mode shape, not a calculated displacement.