Worked examples

Beam example 1: fully restrained beam, from loads to deflection

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Select a Grade S355 section for a fully laterally restrained, simply supported $7 m$ beam. Check shear, bending, support web bearing, support web buckling and deflection. The central loads are $20 kN$ dead and $60 kN$ imposed; full-span UDLs are $7 \frac{kN}{m}$ dead and $35 \frac{kN}{m}$ imposed.

Original source: LectureNotes/Ch 3_Beam.pdf — p. 18, p. 19, p. 20, p. 21. 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

Quantity	Given, lookup or assumption
Span and loading	Given in the upper drawing: $L = 3.5 + 3.5 = 7 m$ , one central point load and two full-span UDL components.
Bearing	Given in the question: $b_{1} = 160 mm$ , $a_{e} = 80 mm$ , $b_{e} = 0$ . Loaded flange is restrained against both specified local movements.
Lateral restraint	Given: compression flange fully restrained, so no member LTB check is needed.
Deflection	Lecturer assumes brittle finishes and uses unfactored imposed loads only; limit $\frac{L}{360}$ . $E = 205,000 \frac{N}{{mm}^{2}}$ .
Section properties	Lookup Data File pp.9–10, row $533 \times 210 \times 82$ UB. Original solution also lists the values on p.19.

Before calculating: recognition and strategy

Learn the difference between a load and an action first: load is applied to the beam; shear $V$ and moment $M$ are internal actions needed to carry it. Symmetry makes the two reactions equal. The shear changes sign at the central point load, so that is the maximum-moment position. Full lateral restraint removes LTB from this example, but does not remove local web checks. Select a trial section from its plastic modulus, then verify every other requirement.

1. Factor each load component

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

P = 1.4 G_{point} + 1.6 Q_{point} = 1.4 \times 20 + 1.6 \times 60 = 28 + 96 = 124 kN

w = 1.4 g + 1.6 q = 1.4 \times 7 + 1.6 \times 35 = 9.8 + 56 = 65.8 \frac{kN}{m}

The given dead UDL is the prescribed design input. Do not add an extra assumed self-weight on top of it. Keep the unfactored imposed loads $60 kN$ and $35 \frac{kN}{m}$ for the separate deflection calculation.

Animation labFrom characteristic to design load2 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.

G is permanent load; Q is imposed load. A surface load and a line load also have different units.
This illustration uses the course gravity case $1.4 G + 1.6 Q$ . Other combinations in the original text retain their own factors.
For illustrative $G = 5 \frac{kN}{m ^{2}}$ , change Q and watch each separate contribution.
Do not carry this ULS total automatically into deflection. Follow the stated SLS load case.

2. Reactions, shear and maximum moment

Total ultimate load:

124 + 65.8 \times 7 = 584.6 kN

R_{A} = R_{B} = \frac{584.6}{2} = 292.3 kN

Immediately left of midspan:

V = 292.3 - 65.8 \times 3.5 = 62 kN

Immediately right of midspan:

V = 62 - 124 = - 62 kN

Immediately left of the right support:

V = - 62 - 65.8 \times 3.5 = - 292.3 kN

M_{mid} = R_{A} \times 3.5 - \frac{w \times {3.5}^{2}}{2}

= 292.3 \times 3.5 - \frac{65.8 \times 12.25}{2}

= 1,023.05 - 403.025 = 620.025 kN \cdot m

Equivalently $\frac{P L}{4} + \frac{w L^{2}}{8} = \frac{124 \times 7}{4} + \frac{65.8 \times 49}{8} = 217 + 403.025 = 620.025 kN \cdot m$ . The source labels $620$ after rounding. UDL makes shear linear and bending parabolic; the point load makes a shear jump of $124 kN$ but no moment jump.

Animation labBalance reactions and moments2 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.

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. Select a trial section and read the correct columns

Initially assume flange thickness

T \leq 16 mm

and trial

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

S_{x, required} = \frac{M}{p_{y}} = \frac{620.025 \times 10^{6}}{355}

= 1,746,549 {mm}^{3} = 1,746.549 {cm}^{3}

Trial

533 \times 210 \times 82

UB:

S_{x} = 2,059 {cm}^{3} > 1,746.549

Table column	Selected-row value
$D$ ; web $t$ ; flange $T$ ; root $r$ ; clear web $d$ ( $mm$ )	$528.3; 9.6; 13.2; 12.7; 476.5$
Flange $\frac{b}{T}$ ; web $\frac{d}{t}$	$7.91; 49.6$
Major-axis $I_{x}$ ; elastic $Z_{x}$ ; plastic $S_{x}$	$47,540 {cm}^{4}; 1,800 {cm}^{3}; 2,059 {cm}^{3}$

$533$ and $210$ are nominal section designations, not the exact $D$ and flange width. The table’s $T = 13.2 mm$ confirms the initial $p_{y} = 355$ assumption. Select the major-axis columns because the vertical load bends the beam about $x - x$ . This is the lecturer’s suitable trial section, not a demonstrated minimum-mass optimum.

Animation labRead a table without losing the keys1 concept · 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.

Name the required property: material strength, section property, buckling strength or a moment factor.
Keep section size, steel grade, thickness band, curve and axis as separate lookup keys.
$cm^{2}$ , $cm^{3}$ and $cm^{4}$ require different conversion powers. Do not use adjacent columns interchangeably.
Use bracketing rows within the same valid column. The original page remains the source of all table values.

4. Establish the section class and web-shear branch

Simple explanation: A thin part can wrinkle first

A thin plate may wrinkle before the whole steel member reaches its intended resistance.

Check flange and web slenderness using their own definitions.
Compare each ratio with the correct class limits.
The less favourable element determines the section class.

Remember: Bending limits and uniform-compression limits are different.

Related concept and full method

ε = \sqrt{\frac{275}{p_{y}}} = \sqrt{\frac{275}{355}} = 0.880141

Class 1 flange limit:

9 ε = 7.92127; \frac{b}{T} = 7.91 \leq 7.92127

Class 1 web limit in pure bending:

80 ε = 70.4113; \frac{d}{t} = 49.6 \leq 70.4113

Both elements are Class 1, giving a plastic section. Web shear-buckling threshold:

70 ε = 61.6099; \frac{d}{t} = 49.6 < 61.6099

The flange classification is close to its limit: use the table ratio and keep enough precision. The web’s $80 ε$ bending-class limit and $70 ε$ shear-buckling threshold are different tests. A separate shear-web-buckling calculation is not required; concentrated reaction web buckling is still checked below.

Animation labWhy thin elements buckle locally2 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.

The flange outstand and web have different widths, thicknesses and edge support conditions.
A thinner plate can wrinkle locally before the complete member loses stability.
Class 1 allows plastic rotation; Class 2 reaches plastic resistance; Class 3 reaches elastic resistance; Class 4 requires effective properties.
Check every relevant compression element with the supplied limits and stress distribution. The deformation shown is qualitative.

5. Resist the largest shear

A_{v} = t D = 9.6 \times 528.3 = 5,071.68 {mm}^{2}

V_{c} = \frac{p_{y} A_{v}}{\sqrt{3}}

= \frac{355 \times 5,071.68}{\sqrt{3}} = 1,039,488 N = 1,039.488 kN

V_{\max} = 292.3 kN < 1,039.488 kN

Passes.

Animation labSee shear in the web1 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.

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.

6. Check coexistent shear, then bending

At the maximum-moment section,

| V | = 62 kN

0.6 V_{c} = 0.6 \times 1,039.488 = 623.693 kN

62 < 623.693

; the low-shear bending formula applies.

p_{y} S_{x} = \frac{355 \times 2,059 \times 10^{3}}{10^{6}} = 730.945 kN \cdot m

1.2 p_{y} Z_{x} = \frac{1.2 \times 355 \times 1,800 \times 10^{3}}{10^{6}} = 766.8 kN \cdot m

M_{c} = \min (730.945, 766.8) = 730.945 kN \cdot m

620.025 < 730.945

: bending passes.

The largest shear actually also lies below $0.6 V_{c}$ , so every section is low-shear here. Do not use the larger of the two moment resistances: the second is a ceiling.

Animation labCompression and tension across a section2 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.

For sagging, the top flange is in compression and the bottom in tension; hogging reverses this.
Elastic bending stress varies with distance from the neutral axis: $σ = \frac{M y}{I}$ .
The section class governs whether elastic, plastic or effective properties may be used.
Use the shear at the section under examination; the largest shear elsewhere is not automatically coexistent.

7. Support web bearing

Simple explanation: A concentrated force can hurt one small region

A narrow contact presses much harder locally than the same force spread over a wider contact.

Find the actual stiff bearing length and end position.
Check local crushing and local web buckling separately.
Use the restraint conditions required by each expression.

Remember: Missing contact dimensions cannot be guessed from drawing scale.

Related concept and full method

k = T + r = 13.2 + 12.7 = 25.9 mm

n = \min (2 + \frac{0.6 b_{e}}{k}, 5) = \min (2 + 0, 5) = 2

b_{1} + n k = 160 + 2 \times 25.9 = 211.8 mm

P_{bw} = (b_{1} + n k) t p_{yw}

= 211.8 \times 9.6 \times 355 = 721,814.4 N = 721.8144 kN

721.8144 > 292.3

: support bearing passes.

Web thickness $9.6 mm$ also gives $p_{yw} = 355$ . This is an end bearing, hence $n = 2$ with $b_{e} = 0$ ; using the interior value $5$ would artificially increase resistance.

Animation labSpread a concentrated force into the web1 concept · 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.

A concentrated reaction first enters through the bearing/contact region.
The flange and root geometry spread the force before it enters the web.
A wider effective bearing region can reduce local stress for the same force.
End distance, stiff bearing length and restraint conditions must come from the original question. This slider is illustrative only.

8. Support web buckling

0.7 d = 0.7 \times 476.5 = 333.55 mm

a_{e} = 80 mm < 333.55 mm

; use the end-reduced branch. End factor:

\frac{80 + 333.55}{1.4 \times 476.5} = \frac{413.55}{667.1} = 0.619922

Web factor:

\frac{25 \times 0.880141 \times 9.6}{\sqrt{211.8 \times 476.5}}

= \frac{211.2338}{\sqrt{100,922.7}} \approx 0.664919

P_{x} = [\frac{413.55}{667.1}] \times [\frac{211.2338}{\sqrt{100,922.7}}] \times 721.8144

= 297.531 kN > 292.3 kN

Passes, but with a small margin.

Both local flange restraints are explicitly given, so no $P_{xr}$ reduction is needed. The reaction uses $98.24 %$ of this resistance: support web buckling is the tightest strength check. Keep the given $a_{e}$ and $b_{e}$ distinct.

Animation labThe web behaves like a short strut1 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.

A concentrated force compresses a local region between the flanges.
The thin web can buckle sideways before local crushing governs.
The source distinguishes restraint against rotation and relative lateral movement.
Web bearing and web buckling are separate checks. Select the original expression whose assumptions are satisfied.

9. Deflection under the specified serviceability case

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

P_{service} = 60 kN = 60,000 N; w_{service} = 35 \frac{kN}{m} = 35 \frac{N}{mm}

L = 7,000 mm; I_{x} = 47,540 \times 10^{4} = 475,400,000 {mm}^{4}

δ_{P} = \frac{P L^{3}}{48 E I}

= \frac{60,000 \times {7,000}^{3}}{48 \times 205,000 \times 475,400,000} = 4.39938 mm

δ_{w} = \frac{5 w L^{4}}{384 E I}

= \frac{5 \times 35 \times {7,000}^{4}}{384 \times 205,000 \times 475,400,000} = 11.22757 mm

δ_{total} = 4.39938 + 11.22757 = 15.62695 mm

Limit:

\frac{L}{360} = \frac{7,000}{360} = 19.44444 mm

15.62695 < 19.44444

: passes under the assumed brittle-finish condition.

Symmetry puts the maximum at midspan for both components, so adding these two maximum values is valid. The selected $533 \times 210 \times 82$ UB passes all requested checks under the stated model.

Animation labSee stiffness and deflection1 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.

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.

Compact exam answer

Use S355 $533 \times 210 \times 82$ UB, Class 1. Ultimate $P = 124 kN$ , $w = 65.8 \frac{kN}{m}$ , $R = V_{\max} = 292.3 kN$ , $M_{\max} = 620.025 kN \cdot m$ . $V_{c} = 1,039.49 kN$ . $M_{c} = \min (730.945, 766.8) = 730.945 kN \cdot m$ . $P_{bw} = 721.814 kN$ ; $P_{x} = 297.531 kN > 292.3$ . Imposed-load $δ = 15.627 mm < \frac{L}{360} = 19.444 mm$ . Adequate; support web buckling is the closest strength check.

Mistakes to avoid

Do not factor imposed loads again in the prescribed deflection check.
Use actual table dimensions, not $533 mm$ as $D$ .
Compare against the smaller moment-capacity expression.
Do not confuse shear-web slenderness with concentrated-load web buckling.

Procedure for an unfamiliar variant

Identify the vertical supports, span, lateral restraints and individual load positions.
Keep dead and imposed loads separate; form the required ultimate and serviceability cases.
Find reactions and the maximum shear/moment by equilibrium, with units.
Choose a section-table row, confirm thickness-dependent strength, and classify both flange and web.
Check shear and bending; add segment LTB where restraint is discrete.
Complete the requested web and deflection checks; state the governing result and any missing data.

Independent self-check

Try it yourself. Invented variant: keep the section and ultimate loads, but reduce the stiff bearing to $140 mm$ , centred at the member end ( $a_{e} = 70 mm$ , $b_{e} = 0$ ). Is the support web still adequate?

Hint

The change affects both the bearing resistance and the end-reduction factor.

Reveal answer and reasoning

Recalculate $n = 2$ and $b_{1} + n k = 191.8 mm$ . $P_{bw} = \frac{191.8 \times 9.6 \times 355}{1,000} = 653.6544 kN$ . $P_{x} = [\frac{70 + 333.55}{667.1}] \times [\frac{25 \times \sqrt{\frac{275}{355}} \times 9.6}{\sqrt{191.8 \times 476.5}}] \times 653.6544 \approx 276.3 kN$ , below $292.3 kN$ . Bearing passes but web buckling fails; the original section conclusion cannot simply be reused.

Animation labSpread a concentrated force into the web2 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.

A concentrated reaction first enters through the bearing/contact region.
The flange and root geometry spread the force before it enters the web.
A wider effective bearing region can reduce local stress for the same force.
End distance, stiff bearing length and restraint conditions must come from the original question. This slider is illustrative only.