Worked examples

Tutorial 3 Q5: design a beam with a $3 m$ overhang

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Chinese–English terminology

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Select a suitable S355 UB for the source beam: supports at $0$ and $7 m$ , free end at $10 m$ ; characteristic point loads at $2$ and $5 m$ , each dead $60$ and imposed $35 kN$ ; full-length UDL dead $10$ and imposed $5 \frac{kN}{m}$ , including beam self-weight. Supports provide torsional restraint, no intermediate lateral support, and effective lengths equal actual segment lengths.

Original source: LectureNotes/Ch 3_Beam.pdf — p. 36, p. 37. 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

Input	Origin/type
Characteristic point loads	Given eachP: dead $60$ , imposed $35 kN$ ; two points total.
Characteristic UDL	Given dead $10$ , imposed $5 \frac{kN}{m}$ over $10 m$ ; self-weight included, so do not add the selected mass again.
LTB lengths	Given actual segment lengths: main span $7,000 mm$ ; overhang $3,000 mm$ .
Selected result	$533 \times 210 \times 122$ UB is a checked trial from the supplied Data File; it is not claimed to be the globally lightest possible design.

Before calculating: recognition and strategy

First solve the full $10 m$ free body; using $\frac{w L^{2}}{8}$ for this overhanging beam would be wrong. Peak sagging lies between the two point loads where shear crosses zero; the roller also has hogging moment. After deriving the LTB demand, test a plausible section and reject it if necessary. For the overhang use conservative $m_{LT} = 1$ ; do not apply a simply supported triangular-moment shortcut to a free end.

1. Factor the actual loads once

Simple explanation: Why dead and imposed loads stay separate

Keep two shopping baskets until their different multipliers are applied.

Put self-weight and permanent finishes in the dead-load basket.
Put the specified use load in the imposed-load basket.
For this course’s stated gravity combination: $1.4 G + 1.6 Q$ .

Remember: That ULS combination is not the imposed-load deflection load.

Related concept and full method

Each point load:

P = 1.4 \times 60 + 1.6 \times 35 = 84 + 56 = 140 kN

w = 1.4 \times 10 + 1.6 \times 5 = 14 + 8 = 22 \frac{kN}{m}

Total downward load:

2 \times 140 + 22 \times 10 = 500 kN

Full-length UDL resultant

220 kN

, acting at distance

\frac{10}{2} = 5 m

Self-weight is already included in dead $10 \frac{kN}{m}$ . All subsequent strength results use these ultimate loads. The serviceability calculation later returns to imposed $35 kN$ and $5 \frac{kN}{m}$ .

Animation labSee stiffness and deflection3 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.

Use the specified SLS load, span and support arrangement. The demonstrator has a full-span UDL.
The loaded beam bends; the deformation is exaggerated so its shape can be seen.
For a simply supported full-span UDL, $δ = \frac{5 w L ^{4}}{384 E I}$ . Double L with w, E and I unchanged: δ becomes 16 times as large.
The readout uses $w = 10 \frac{N}{mm}$ , $E = 205000 \frac{N}{mm ^{2}}$ and $I = 200000000 mm^{4}$ . Select the finish/support-specific limit from the original table.

2. Reactions, shear and bending over the full beam

Set A at

0 m

, support B at

7 m

, and free end C at

10 m

Σ M_{A} = 0 : 7 R_{B} - 140 \times 2 - 140 \times 5 - 220 \times 5 = 0

R_{B} = \frac{280 + 700 + 1,100}{7} = 297.142857 kN

R_{A} = 500 - 297.142857 = 202.857143 kN

x

is measured in

m

, define

⟨ x - a ⟩ = \max (x - a, 0)

M (x) = 202.857143 x - 11 x^{2} - 140 ⟨ x - 2 ⟩ - 140 ⟨ x - 5 ⟩ + 297.142857 ⟨ x - 7 ⟩ kN \cdot m

V

is the slope

\frac{d M}{d x}

, with a jump at each point force.

Interval	Shear equation ( $kN$ )	End shear levels
$0 < x < 2$
$2 < x < 5$
$5 < x < 7$
$7 < x < 10$

In the interval

2

5 m

V = 0

gives

x = \frac{62.857143}{22} = 2.857143 m

M_{\max} = 202.857143 \times 2.857143 - 11 \times {2.857143}^{2} - 140 \times (2.857143 - 2)

= 369.795918 kN \cdot m

is sagging moment.

M (2) = 361.714286

;

M (5) = 319.285714 kN \cdot m

M (7) = 202.857143 \times 7 - 11 \times 49 - 140 \times 5 - 140 \times 2 = - 99 kN \cdot m

Overhang:

M (x) = \frac{- 22 {(10 - x)}^{2}}{2}

;

M (10) = 0

. Immediately left of B,

| V |_{\max} = 231.142857 kN

Support reaction $297.142857 kN$ is local bearing demand, not the beam's maximum internal shear. The shear diagram has four sloping straight segments, with jumps $140$ , $140$ , $297.142857 kN$ . The moment diagram is piecewise parabolic, continuous at point loads, with the above peak and the roller moment $- 99 kN \cdot m$ .

Animation labBalance reactions and moments2 concepts · 6 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.

3. Obtain LTB demand and reject an insufficient trial

There is no lateral restraint at either point load. The main span is one $7 m$ LTB segment. Its local quarter positions are $1.75$ , $3.5$ and $5.25 m$ , and the maximum magnitude in that segment is $369.795918 kN \cdot m$ .

M_{2} = M (1.75) = 202.857143 \times 1.75 - 11 \times {1.75}^{2} = 321.3125 kN \cdot m

M_{3} = M (3.5) = 202.857143 \times 3.5 - 11 \times {3.5}^{2} - 140 \times 1.5 = 365.25 kN \cdot m

M_{4} = M (5.25) = 202.857143 \times 5.25 - 11 \times {5.25}^{2} - 140 \times 3.25 - 140 \times 0.25 = 271.8125 kN \cdot m

Table 8.4b:

m_{LT} = 0.2 + \frac{0.15 \times 321.3125 + 0.5 \times 365.25 + 0.15 \times 271.8125}{369.795918}

= 0.934442 > 0.44

Demand:

m_{LT} M_{\max} = 345.552934 kN \cdot m

Try $533 \times 210 \times 109$ UB. Its large plastic moment capacity is not enough evidence; compute its $7 m$ LTB resistance.

Lookup: Data File pp.9–10, exact $533 \times 210 \times 109$ 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	D539.5, web t 11.6, flange T18.8, root r 12.7, clear d 476.5, all $mm$
Local ratios	$\frac{b}{T} = 5.61$ ； $\frac{d}{t} = 41.1$ 。
Major-axis properties	$I_{x} = 66820 {cm}^{4} = 6.682 \times 10^{8} {mm}^{4}$ ; $Z_{x} = 2477 {cm}^{3}$ ; $S_{x} = 2828 {cm}^{3}$ .
LTB properties	$r_{y} = 4.6 cm = 46 mm$ ; $u = 0.875$ ; torsional index $x = 30.9$ .

S355, flange

T = 18.8 mm

, so

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

ε = \sqrt{\frac{275}{345}} = 0.892805

. Flange

\frac{b}{T} = 5.61 < 9 ε = 8.035248

; web in bending

\frac{d}{t} = 41.1 < 80 ε = 71.424431

. Both are Class 1, so

β_{W} = 1

; plastic bending resistance is permitted.

\frac{d}{t} = 41.1 < 70 ε = 62.496377

; no shear-buckling calculation is required under the course limit.

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{7000}{46} = 152.173913

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

= {[1 + 0.05 {(\frac{152.173913}{30.9})}^{2}]}^{- \frac{1}{4}} = 0.819921

λ_{LT} = u v λ \sqrt{β_{W}}

= 0.875 \times 0.819921 \times 152.173913 \times 1 = 109.174285

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

105

row gives

128

;

110

row gives

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

. Interpolation fraction:

\frac{109.174285 - 105}{110 - 105} = 0.834857

p_{b} = 128 + 0.834857 \times (119 - 128) = 120.486287 \frac{N}{{mm}^{2}}

M_{b} = \frac{p_{b} S_{x}}{1,000} = \frac{120.486287 \times 2828}{1,000} = 340.735220 kN \cdot m

Equivalent uniform demand:

m_{LT} M_{\max} = 345.552934 kN \cdot m

Utilisation:

\frac{345.552934}{340.735220} = 1.014139

Fails.

The trial fails: $340.735220 < 345.552934 kN \cdot m$ . Choose the next heavier row in this depth series, $533 \times 210 \times 122$ UB, and re-read all its properties; do not reuse the109 kg/m thickness or radius.

Lookup: Data File pp.9–10, exact $533 \times 210 \times 122$ 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	D544.5, web t 12.7, flange T21.3, root r 12.7, clear d 476.5, all $mm$
Local ratios	$\frac{b}{T} = 4.97$ ； $\frac{d}{t} = 37.5$ 。
Major-axis properties	$I_{x} = 76040 {cm}^{4} = 7.604 \times 10^{8} {mm}^{4}$ ; $Z_{x} = 2793 {cm}^{3}$ ; $S_{x} = 3196 {cm}^{3}$ .
LTB properties	$r_{y} = 4.67 cm = 46.7 mm$ ; $u = 0.877$ ; torsional index $x = 27.6$ .

S355, flange

T = 21.3 mm

, so

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

ε = \sqrt{\frac{275}{345}} = 0.892805

. Flange

\frac{b}{T} = 4.97 < 9 ε = 8.035248

; web in bending

\frac{d}{t} = 37.5 < 80 ε = 71.424431

. Both are Class 1, so

β_{W} = 1

; plastic bending resistance is permitted.

\frac{d}{t} = 37.5 < 70 ε = 62.496377

; no shear-buckling calculation is required under the course limit.

Animation labA beam bends sideways and twists3 concepts · 23 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.

4. Check both segments of the selected $533 \times 210 \times 122$ UB

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

Main $7 m$ span: use the same load-derived demand but the new section properties.

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{7000}{46.7} = 149.892934

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

= {[1 + 0.05 {(\frac{149.892934}{27.6})}^{2}]}^{- \frac{1}{4}} = 0.797293

λ_{LT} = u v λ \sqrt{β_{W}}

= 0.877 \times 0.797293 \times 149.892934 \times 1 = 104.808994

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

100

row gives

137

;

105

row gives

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

. Interpolation fraction:

\frac{104.808994 - 100}{105 - 100} = 0.961799

p_{b} = 137 + 0.961799 \times (128 - 137) = 128.343811 \frac{N}{{mm}^{2}}

M_{b} = \frac{p_{b} S_{x}}{1,000} = \frac{128.343811 \times 3196}{1,000} = 410.186820 kN \cdot m

Equivalent uniform demand:

m_{LT} M_{\max} = 345.552934 kN \cdot m

Utilisation:

\frac{345.552934}{410.186820} = 0.842428

Passes.

Overhang $3 m$ : its maximum hogging magnitude is $99 kN \cdot m$ at B. Use conservative $m_{LT} = 1$ , hence demand $99 kN \cdot m$ . The problem expressly permits effective length equal to the actual $3 m$ ; do not silently impose a different cantilever factor.

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}{46.7} = 64.239829

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

= {[1 + 0.05 {(\frac{64.239829}{27.6})}^{2}]}^{- \frac{1}{4}} = 0.941835

λ_{LT} = u v λ \sqrt{β_{W}}

= 0.877 \times 0.941835 \times 64.239829 \times 1 = 53.061400

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

50

row gives

285

;

55

row gives

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

. Interpolation fraction:

\frac{53.061400 - 50}{55 - 50} = 0.612280

p_{b} = 285 + 0.612280 \times (268 - 285) = 274.591241 \frac{N}{{mm}^{2}}

M_{b} = \frac{p_{b} S_{x}}{1,000} = \frac{274.591241 \times 3196}{1,000} = 877.593606 kN \cdot m

Equivalent uniform demand:

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

Utilisation:

\frac{99.000000}{877.593606} = 0.112808

Passes.

Both selected-section LTB checks pass. Restraint at B must be effective for the specified hogging/compression-flange condition; the question grants the torsional restraint model.

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.

5. Local strength, imposed-load deflection and remaining scope

Simple explanation: How much does the beam sag?

Strength asks whether it fails; deflection asks how far it moves.

Use the serviceability load case specified by the course question.
Choose the expression matching the support and load positions.
Use consistent units for load, length, E and I.

Remember: The largest deflection is not always at midspan.

Related concept and full method

A_{v} = t D = 12.7 \times 544.5 = 6915.15 {mm}^{2}

V_{c} = \frac{p_{y} A_{v}}{\sqrt{3}} = \frac{345 \times 6915.15}{\sqrt{3} \times 1,000} = 1377.399981 kN

V_{\max} = 231.142857 kN < V_{c}

: shear passes.

0.6 V_{c} = 826.439989 kN > V_{\max}

; even peak shear is low, so low-shear bending applies throughout.

M_{c} = \min (p_{y} S_{x}, 1.2 p_{y} Z_{x})

= \min (\frac{345 \times 3196}{1,000}, \frac{1.2 \times 345 \times 2793}{1,000})

= \min (1102.62, 1156.3) = 1102.62 kN \cdot m

| M |_{\max} = 369.795918 kN \cdot m

: bending passes.

Serviceability uses each imposed point load $P = 35,000 N$ , $q = 5 \frac{N}{mm}$ . Treat the whole beam as a continuous elastic member passing over two simple supports. A fixed-cantilever formula would omit rotation of the section at the roller.

Imposed-load reaction:

R_{B, q} = \frac{35 \times 2 + 35 \times 5 + 50 \times 5}{7} = 70.714286 kN

R_{A, q} = 120 - 70.714286 = 49.285714 kN

E = 205,000 \frac{N}{{mm}^{2}}

;

I = 76,040 \times 10^{4} {mm}^{4}

E I = 155,882,000,000,000 N \cdot {mm}^{2}

Take downward deflection $v$ positive, $x$ in $mm$ . Integrating $E I v^{″} = - M$ twice gives the following expression. The support atx 0 forces the constant displacement term to zero; $v (7,000) = 0$ determines the remaining constant $C$ .

E I v = C x - \frac{49,285.714286 x^{3}}{6} + \frac{5 x^{4}}{24}

+ \frac{35,000 {⟨ x - 2,000 ⟩}^{3}}{6} + \frac{35,000 {⟨ x - 5,000 ⟩}^{3}}{6}

- \frac{70,714.285714 {⟨ x - 7,000 ⟩}^{3}}{6}

In the interval

x = 7,000

C = \frac{[\frac{49,285.714286 \times {7,000}^{3}}{6} - \frac{5 \times {7,000}^{4}}{24} - \frac{35,000 \times {5,000}^{3}}{6} - \frac{35,000 \times {2,000}^{3}}{6}]}{7,000}

= \frac{[2.8175 \times 10^{15} - 5.002083333 \times 10^{14} - 7.291666667 \times 10^{14} - 4.666666667 \times 10^{13}]}{7,000}

= 2.202083333 \times 10^{11} N \cdot {mm}^{2}

Differentiate once and set slope to zero in each interval. The only internal stationary deflection along the beam is in $2,000 < x < 5,000 mm$ , where only the first point-load bracket is active:

E I v^{'} = C - \frac{49,285.714286 x^{2}}{2} + \frac{5 x^{3}}{6} + \frac{35,000 {(x - 2,000)}^{2}}{2} = 0

Collecting like terms:

\frac{5}{6} x^{3} - 7,142.857143 x^{2} - 70,000,000 x + 290,208,333,333.333 = 0

Root within the applicable interval:

x = 3,426.666453 mm

Substitute into

E I v

to obtain maximum downward deflection

v = 3.013388 mm

v (0) = v (7,000) = 0

v (10,000) = - 3.408027 mm

The free end moves upwards. The other intervals have no internal zero-slope root; their boundaries must still be checked. Even the stricter main-span limit for brittle finishes,

\frac{7,000}{360} = 19.444444 mm

, is satisfied. Overhang limit

\frac{3,000}{180} = 16.666667 mm

;

| - 3.408027 | < 16.666667

The source does not specify finishes. Demonstrating the stricter span limit avoids inventing a finish type. Loading on the main span rotates the support section, so the free end can move upwards. This calculation covers the simultaneous load pattern drawn; it does not claim an envelope from separate imposed-load patterns.

Animation labSee shear in the web4 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.

Internal shear keeps the two sides of the cut in vertical equilibrium.
For the course’s common I-section case, the web provides the principal shear area; use the specified definition.
A slender web may require a different shear-buckling route before a simple shear-resistance formula is used.
Use $V_{c} = \frac{p _{y} A _{v}}{3}$ where applicable in the course. Convert N to kN before comparing with design shear.

Compact exam answer

P140 kN,w 22 kN/m; RA202.857143,RB297.142857 kN. Peak internal shear $231.142857 kN$ ; saggingM369.795918 kNm atx 2.857143 m, hogging−99 atB. Main-span LTB demand $345.552934 kN \cdot m$ . Trial $533 \times 210 \times 109$ fails(Mb 340.735220); select $533 \times 210 \times 122$ : mainMb 410.186820, overhangMb 877.593606 kNm with conservative overhangmLT1. ShearVc 1377.400 kN and Mc 1102.620 kNm pass. Imposed deflection $3.013388 mm$ downward in span and $3.408027 mm$ upward at tip. Bearing/local-web detail checks remain incomplete because source inputs are absent.

Mistakes to avoid

Do not read the upper $5 m$ dimension as distance to the roller.
Do not add self-weight again.
Do not use the support reaction as the maximum beam internal shear.
Do not split LTB length at an unrestrained point load.
Do not use a fixed-cantilever deflection formula for a rotating support.

Procedure for an unfamiliar variant

Name supports and load positions using the source dimension chains.
Choose a sign convention; include applied couples in moment equilibrium.
Find reactions, then piecewise shear and bending moment.
Locate extrema where shear is zero and at load/support discontinuities.
Divide only at actual effective lateral restraints; classify and read the exact section row.
Calculate segment moment factors and LTB resistance, retaining UDL curvature where present.
Check shear, local bending and serviceability; state any missing bearing/detail inputs.

Independent self-check

Try it yourself. Invented variant: can the rejected $533 \times 210 \times 109$ section be accepted by rounding its $340.735 kN \cdot m$ resistance to $350 kN \cdot m$ ?

Reveal answer and reasoning

No. The demand $345.553 kN \cdot m$ exceeds the unrounded resistance $340.735 kN \cdot m$ by $4.818 kN \cdot m$ , a utilisation of about $1.01414$ . Rounding must not change a failed limit-state check into a pass. Change the section or restraint arrangement and recalculate.

Tutorial 3 Q5: design a beam with a 3m3m overhang

Read the diagram and collect the data

Before calculating: recognition and strategy

1. Factor the actual loads once

2. Reactions, shear and bending over the full beam

3. Obtain LTB demand and reject an insufficient trial

4. Check both segments of the selected 533×210×122533×210×122 UB

5. Local strength, imposed-load deflection and remaining scope

Compact exam answer

Mistakes to avoid

Procedure for an unfamiliar variant

Independent self-check

Tutorial 3 Q5: design a beam with a $3 m$ overhang

4. Check both segments of the selected $533 \times 210 \times 122$ UB