Worked examples

2024 BQ4: uneven point loads and segment-by-segment LTB

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Chinese–English terminology

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For the $11 m$ simply supported $610 \times 229 \times 140$ UB S355, assumed Class 1, find support reactions and draw annotated SFD/BMD, then check LTB inBC andCD using accurateu andv. Design loads: $100 kN$ at B3.5 m, $200 kN$ atC7.5 m, and full-span $1.4 \frac{kN}{m}$ UDL including beam self-weight. Lateral restraints atA/B/C/D.

Original source: Pastpaper/23ENGTY004.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 $610 \times 229 \times 140$ 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	D617.2, web t 13.1, flange T22.1, root r 12.7, clear d 547.6, all $mm$
Local ratios	$\frac{b}{T} = 5.21$ ； $\frac{d}{t} = 41.8$
Major-axis properties	$I_{x} = 111800 {cm}^{4} = 1.118 \times 10^{9} {mm}^{4}$ ; $Z_{x} = 3622 {cm}^{3}$ ; $S_{x} = 4142 {cm}^{3}$
LTB properties	$r_{y} = 5.03 cm = 50.3 mm$ ; $u = 0.875$ ; torsional index $x = 30.6$

Before calculating: recognition and strategy

Use global equilibrium for the complete $11 m$ beam. Then determine the critical section from shear changes and evaluate the BMD over each specified lateral segment. The full UDL creates curvature even between the point loads; retain it when computing quarter-point moment factors.

(a) Support reactions —3 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

UDL resultant:

1.4 \times 11 = 15.4 kN

Acts at

x = 5.5 m

Σ M_{A} = 0

11 R_{D} = 100 \times 3.5 + 200 \times 7.5 + 15.4 \times 5.5

= 350 + 1,500 + 84.7 = 1,934.7 kN \cdot m

R_{D} = \frac{1,934.7}{11} = 175.881818 kN

R_{A} = 100 + 200 + 15.4 - 175.881818 = 139.518182 kN

Check:

R_{A} + R_{D} = 315.4 kN

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) Annotated shear and bending diagrams —4 printed marks

x

starts at A and is measured in

m

. Define

⟨ x - a ⟩ = \max (x - a, 0)

M (x) = 139.518182 x - 0.7 x^{2} - 100 ⟨ x - 3.5 ⟩ - 200 ⟨ x - 7.5 ⟩ kN \cdot m

Segment AB:

V = 139.518182 - 1.4 x

Segment BC:

V = 39.518182 - 1.4 x

Segment CD:

V = - 160.481818 - 1.4 x

Position	Shear just before/after ( $kN$ )	Moment ( $kN \cdot m$ )
$A$	$0$ ／ $+ 139.518182$	$0$
$B$	$+ 134.618182$ ／ $+ 34.618182$	$479.738636$
$C$	$+ 29.018182$ ／ $- 170.981818$	$607.011364$
$D$	$- 175.881818$ ／ $0$	$0$

M (B) = 139.518182 \times 3.5 - 0.7 \times {3.5}^{2} = 479.738636 kN \cdot m

M (C) = 139.518182 \times 7.5 - 0.7 \times {7.5}^{2} - 100 \times 4 = 607.011364 kN \cdot m

Shear is positive throughout AB/BC and negative throughout CD, so maximum moment occurs at C, where the

200 kN

load changes the sign of shear. Shear magnitude is greatest beside D:

| V |_{\max} = 175.881818 kN

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, C and D correspond to the original beam positions. All values are also listed in the solution steps.Open full-size image.

Animation labBuild the shear and moment diagrams1 concept · 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.

For the illustrative 60 kN point load on a 6 m simply supported beam, balance the reactions before making a cut.
$V = R_{A}$ and $M = R_{A} x$ before the point load. Moment grows linearly.
Shear jumps down by P. To its right, $V = R_{A} - P$ and $M = R_{A} x - P (x - a)$ .
For this case, moment is continuous and returns to zero at B. A concentrated applied couple instead creates a moment jump.

(c) BC segment LTB —13 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 = 22.1 mm

, giving

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

ε = \sqrt{\frac{275}{345}} = 0.892805

Flange:

\frac{b}{T} = 5.21 < 9 ε = 8.035248

Web in bending:

\frac{d}{t} = 41.8 < 80 ε = 71.424431

Both are Class 1, therefore

β_{W} = 1

; plastic bending resistance is permitted.

\frac{d}{t} = 41.8 < 70 ε = 62.496377

No separate shear-buckling calculation is needed under the course limit.

Flange thickness $22.1 mm$ means that, despite the S355 grade name, use $p_{y} = 345 \frac{N}{{mm}^{2}}$ . In the model above, BC’s effective length is $4,000 mm$ . The quarter-point calculation is shown explicitly below:

In the interval

x = 4.5

M = 139.518182 \times 4.5 - 0.7 \times {4.5}^{2} - 100 \times 1

In the interval

x = 5.5

M = 139.518182 \times 5.5 - 0.7 \times {5.5}^{2} - 100 \times 2

In the interval

x = 6.5

M = 139.518182 \times 6.5 - 0.7 \times {6.5}^{2} - 100 \times 3

Local quarter positions are $4.5$ , $5.5$ , $6.5 m$ measured fromA. Use the full beam $M (x)$ evaluated there; segment distance and global $x$ are different quantities.

M_{2} = M (4.5) = 513.656818 kN \cdot m

M_{3} = M (5.5) = 546.175000 kN \cdot m

M_{4} = M (6.5) = 577.293182 kN \cdot m

For this segment,

M_{\max} = 607.011364 kN \cdot m

m_{LT} = 0.2 + \frac{0.15 | M_{2} | + 0.5 | M_{3} | + 0.15 | M_{4} |}{M_{\max}}

= 0.919476 > 0.44

Equivalent demand

558.132273 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{4000}{50.3} = 79.522863

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

= {[1 + 0.05 {(\frac{79.522863}{30.6})}^{2}]}^{- \frac{1}{4}} = 0.929847

λ_{LT} = u v λ \sqrt{β_{W}}

= 0.875 \times 0.929847 \times 79.522863 \times 1 = 64.701095

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

60

gives

251

，

65

gives

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

. Interpolation fraction:

\frac{64.701095 - 60}{65 - 60} = 0.940219

p_{b} = 251 + 0.940219 \times (234 - 251) = 235.016276 \frac{N}{{mm}^{2}}

M_{b} = \frac{p_{b} S_{x}}{1,000} = \frac{235.016276 \times 4142}{1,000} = 973.437414 kN \cdot m

Equivalent uniform moment demand:

m_{LT} M_{\max} = 558.132273 kN \cdot m

Utilisation:

\frac{558.132273}{973.437414} = 0.573362

Passes.

Animation labA beam bends sideways and twists4 concepts · 7 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.

(d) CD segment LTB —10 printed marks

Simple explanation: Why the shape of the moment diagram matters

The same peak moment is more demanding when a long region stays near that peak.

Use the diagram of this unrestrained segment.
Choose the applicable course sketch or quarter-point rule.
Apply its factor to the demand, following the stated check.

Remember: Column flexural-buckling factors and LTB factors are not interchangeable.

Related concept and full method

CD is $3.5 m$ long. Its UDL means the BMD is slightly curved, so do not replace the full calculation by an exact triangularβ0 case. At all three quarter positions, subtract both $100$ and $200 kN$ point-load moment contributions.

M (x) = 139.518182 x - 0.7 x^{2} - 100 (x - 3.5) - 200 (x - 7.5)

Substitute respectively

x = 8.375

9.25

10.125 m

Local quarter positions are $8.375$ , $9.25$ , $10.125 m$ measured fromA. Use the full beam $M (x)$ evaluated there; segment distance and global $x$ are different quantities.

M_{2} = M (8.375) = 456.866335 kN \cdot m

M_{3} = M (9.25) = 305.649432 kN \cdot m

M_{4} = M (10.125) = 153.360653 kN \cdot m

For this segment,

M_{\max} = 607.011364 kN \cdot m

m_{LT} = 0.2 + \frac{0.15 | M_{2} | + 0.5 | M_{3} | + 0.15 | M_{4} |}{M_{\max}}

= 0.602560 > 0.44

Equivalent demand

365.761037 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{3500}{50.3} = 69.582505

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

= {[1 + 0.05 {(\frac{69.582505}{30.6})}^{2}]}^{- \frac{1}{4}} = 0.944133

λ_{LT} = u v λ \sqrt{β_{W}}

= 0.875 \times 0.944133 \times 69.582505 \times 1 = 57.483259

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

55

gives

268

，

60

gives

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

. Interpolation fraction:

\frac{57.483259 - 55}{60 - 55} = 0.496652

p_{b} = 268 + 0.496652 \times (251 - 268) = 259.556920 \frac{N}{{mm}^{2}}

M_{b} = \frac{p_{b} S_{x}}{1,000} = \frac{259.556920 \times 4142}{1,000} = 1075.084763 kN \cdot m

Equivalent uniform moment demand:

m_{LT} M_{\max} = 365.761037 kN \cdot m

Utilisation:

\frac{365.761037}{1075.084763} = 0.340216

Passes.

Both requested segments pass under the stated restraint model. The paper does not ask forAB buckling in this question; do not label the whole beam fully checked merely becauseBC/CD pass.

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.

Compact exam answer

RA139.518182,RD175.881818 kN. MB479.738636,MC607.011364 kNm; maximum shear $175.881818 kN$ . py 345. BC:LE4 m,mLT0.919476,demand $558.132273$ ,λLT64.701095,pb 235.016276,Mb 973.437414 kNm → passes. CD:LE3.5 m,mLT0.602560,demand $365.761037$ ,λLT57.483259,pb 259.556920,Mb 1075.084763 kNm → passes. Retain UDL curvature and the stated normal-load/ $L_{E}$ assumption.

Mistakes to avoid

Do not factor $1.4 \frac{kN}{m}$ again.
Do not make B/C vertical supports.
Use $345 \frac{N}{{mm}^{2}}$ forT22.1 mm.
Use $4 m$ and $3.5 m$ segment lengths, not the full $11 m$ in both checks.
Quarter positions must be local to the chosen segment.

Procedure for an unfamiliar variant

Solve full-beam reactions including the UDL resultant.
Write piecewise shear/moment equations.
Use shear signs to locate the actual maximum.
For each requested segment, evaluate its own quarter moments and factor.
Read exact section properties and thickness-dependent strength, then calculate $u$ / $v$ and interpolatepb.
Report each segment result and scope.

Independent self-check

Try it yourself. Invented variant: replace the $1.4 \frac{kN}{m}$ design UDL by zero, keeping both point loads. Can you keep the sameBC/CD moment factors?

Reveal answer and reasoning

No. Reactions and allBMD ordinates change, and the segment diagrams become straight between point loads. Recompute both end moments and use the applicable straight-line moment factor. The section properties and the stated restraint lengths remain unchanged.