Worked examples

2023 AQ1: compulsory floor-beam design and theory

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Chinese–English terminology

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Complete the $40$ -mark compulsory question. Slab thickness $125 mm$ , concrete unit weight $24.5 \frac{kN}{m^{3}}$ , finishes $0.5 \frac{kN}{m^{2}}$ , imposed load $2.5 \frac{kN}{m^{2}}$ . B1 is $356 \times 127 \times 39$ UB and B2 as $356 \times 171 \times 51$ UB, all S355. Include self-weight using $g = 10 \frac{m}{s^{2}}$ . The beams are simply supported, fully laterally restrained and carry brittle finishes.

Original source: Pastpaper/22ENGTY033.pdf — p. 2. 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 type	Exact source feature
Given slab data	Question paragraph: $125 mm$ thickness, $24.5$ unit weight,finish $0.5$ ,imposed $2.5$ .
Given supports/spans	Dimension chains and column symbols inFigureAQ1; use the explanation alongside the original crop.
Calculated widths	Interior B1 tributary width $\frac{2.5}{2} + \frac{2.5}{2} = 2.5 m$ .
Lookup self-weight	Data File p.9 gives B1 mass $39.1 \frac{kg}{m}$ ;B2 mass $51 \frac{kg}{m}$ . Nominal section labels may round the mass.
Method applicability	Slab is expressly a full lateral restraint; no composite-section enhancement is assumed.

Before calculating: recognition and strategy

This compulsory family tests the same chain repeatedly: slab intensity → tributary beam line load → reaction onto supporting beam → section class → shear/bending → imposed-only deflection. Work from the actual plan orientation; the names B1/B2 alone do not tell you which way a beam spans.

(a)(i) Slab factored intensity —4 printed marks

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

Slab thickness:

125 mm = 0.125 m

Concrete dead load:

0.125 \times 24.5 = 3.0625 \frac{kN}{m^{2}}

g_{k} = 3.0625 + 0.5 = 3.5625 \frac{kN}{m^{2}}

q_{k} = 2.5 \frac{kN}{m^{2}}

Ultimate area load:

1.4 \times 3.5625 + 1.6 \times 2.5 = 8.9875 \frac{kN}{m^{2}}

Concrete unit weight is in $\frac{kN}{m^{3}}$ , so multiply by slab thickness in metres to obtain $\frac{kN}{m^{2}}$ . A surface intensity is not yet a beam line load; the next step supplies the tributary width.

Animation labFrom characteristic to design load1 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.

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.

(a)(ii) Both beam actions —8 printed marks

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

B1:

Tributary width:

\frac{2.5}{2} + \frac{2.5}{2} = 2.5 m

B1 self-weight:

\frac{39.1 \frac{kg}{m} \times 10 \frac{m}{s^{2}}}{1,000} = 0.391 \frac{kN}{m}

Dead line load:

3.5625 \times 2.5 + 0.391 = 9.297250 \frac{kN}{m}

Imposed line load:

2.5 \times 2.5 = 6.25 \frac{kN}{m}

Ultimate line load:

w = 1.4 \times 9.297250 + 1.6 \times 6.25 = 23.016150 \frac{kN}{m}

Reaction at each end:

R = \frac{w L}{2} = \frac{23.016150 \times 7}{2} = 80.556525 kN

V_{\max} = R = 80.556525 kN

Shear

V (x) = R - w x

lie on opposite sides of the

x = 3.5 m

is zero at this position.

M_{\max} = \frac{w L^{2}}{8} = \frac{23.016150 \times 7^{2}}{8} = 140.973919 kN \cdot m

Here $39 \frac{kg}{m}$ is the nominal designation; the exact section-table mass is $39.1 \frac{kg}{m}$ . Use one stated convention consistently. B1’s vertical support spacing is $7 m$ even though the slab tributary width is $2.5 m$ .

B2:

Each B2 receives one B1 end reaction at midspan,

P = 80.556525 kN

. B2 self-weight:

\frac{51 \times 10}{1,000} = 0.51 \frac{kN}{m}

B2 ultimate UDL:

w = 1.4 \times 0.51 = 0.714 \frac{kN}{m}

Span:

L = 2.5 + 2.5 = 5 m

Reaction at each end:

R = \frac{P}{2} + \frac{w L}{2} = \frac{80.556525}{2} + \frac{0.714 \times 5}{2} = 42.063262 kN

V_{\max} = 42.063262 kN

M_{\max} = \frac{P L}{4} + \frac{w L^{2}}{8}

= \frac{80.556525 \times 5}{4} + \frac{0.714 \times 5^{2}}{8} = 102.926906 kN \cdot m

The transferred $P$ is already factored. Do not factor it again. The full-span UDL on B2 here is only its own weight: the one-way slab loads pass through B1 or arrive at supported end joints, rather than being spread arbitrarily along B2.

Animation labFollow the floor load in 3D4 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.

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.

(a)(iii) Both section classes —4 printed marks

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

Lookup: Data File pp.9–10, exact $356 \times 127 \times 39$ 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	D353.4, web t 6.6, flange T10.7, root r 10.2, clear d 311.6, all $mm$
Local ratios	$\frac{b}{T} = 5.89$ ； $\frac{d}{t} = 47.2$ 。
Major-axis properties	$I_{x} = 10170 {cm}^{4} = 1.017 \times 10^{8} {mm}^{4}$ ; $Z_{x} = 576 {cm}^{3}$ ; $S_{x} = 659 {cm}^{3}$ .
LTB properties	$r_{y} = 2.68 cm = 26.8 mm$ ; $u = 0.871$ ; torsional index $x = 35.2$ .

S355, flange

T = 10.7 mm

, giving

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

ε = \sqrt{\frac{275}{355}} = 0.880141

Flange:

\frac{b}{T} = 5.89 < 9 ε = 7.921268

Web in bending:

\frac{d}{t} = 47.2 < 80 ε = 70.411267

Both are Class 1, so

β_{W} = 1

; plastic bending resistance is permitted.

\frac{d}{t} = 47.2 < 70 ε = 61.609858

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

Lookup: Data File pp.9–10, exact $356 \times 171 \times 51$ 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	D355, web t 7.4, flange T11.5, root r 10.2, clear d 311.6, all $mm$
Local ratios	$\frac{b}{T} = 7.46$ ； $\frac{d}{t} = 42.1$ 。
Major-axis properties	$I_{x} = 14140 {cm}^{4} = 1.414 \times 10^{8} {mm}^{4}$ ; $Z_{x} = 797 {cm}^{3}$ ; $S_{x} = 896 {cm}^{3}$ .
LTB properties	$r_{y} = 3.86 cm = 38.6 mm$ ; $u = 0.881$ ; torsional index $x = 32.1$ .

S355, flange

T = 11.5 mm

, giving

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

ε = \sqrt{\frac{275}{355}} = 0.880141

Flange:

\frac{b}{T} = 7.46 < 9 ε = 7.921268

Web in bending:

\frac{d}{t} = 42.1 < 80 ε = 70.411267

Both are Class 1, so

β_{W} = 1

; plastic bending resistance is permitted.

\frac{d}{t} = 42.1 < 70 ε = 61.609858

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

Both beams are Class 1. No Class 4 effective-property calculation is needed. Full slab lateral restraint, given in the paper, removes the LTB requirement for these floor-beam questions; it does not remove local shear/bending or deflection checks.

Animation labWhy thin elements buckle locally1 concept · 18 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.

(a)(iv) B1 adequacy —10 printed marks

Simple explanation: Why one flange squeezes and the other stretches

Bending makes opposite sides of the section do opposite jobs.

Find the moment from the actual loads and supports.
Choose the resistance formula allowed by the section class.
Check whether the coexistent shear changes that formula.

Remember: Use shear at the location being checked, not an unrelated maximum.

Related concept and full method

A_{v} = t D = 6.6 \times 353.4 = 2332.44 {mm}^{2}

V_{c} = \frac{p_{y} A_{v}}{\sqrt{3}} = \frac{355 \times 2332.44}{\sqrt{3} \times 1,000} = 478.055376 kN

V_{\max} = 80.556525 kN < V_{c}

Shear passes.

0.6 V_{c} = 286.833226 kN > V_{\max}

Even peak shear is low, so the low-shear bending expression applies throughout the beam.

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

= \min (\frac{355 \times 659}{1,000}, \frac{1.2 \times 355 \times 576}{1,000})

= \min (233.945, 245.376) = 233.945 kN \cdot m

| M |_{\max} = 140.973919 kN \cdot m

Bending passes.

B1 imposed line load:

q = 2.5 \times 2.5 = 6.25 \frac{kN}{m} = 6.25 \frac{N}{mm}

L = 7,000 mm

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

I_{x} = 10,170 \times 10^{4} {mm}^{4}

δ_{B 1} = \frac{5 q L^{4}}{384 E I}

= \frac{5 \times 6.25 \times {7,000}^{4}}{384 \times 205,000 \times 10,170 \times 10^{4}} = 9.372083 mm

Brittle-finish limit:

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

9.372083 < 19.444444

: passes.

Animation labSee shear in the web4 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.

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.

(a)(v) B2 adequacy —10 printed marks

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 = 7.4 \times 355 = 2627 {mm}^{2}

V_{c} = \frac{p_{y} A_{v}}{\sqrt{3}} = \frac{355 \times 2627}{\sqrt{3} \times 1,000} = 538.428201 kN

V_{\max} = 42.063262 kN < V_{c}

Shear passes.

0.6 V_{c} = 323.056920 kN > V_{\max}

Even peak shear is low, so the low-shear bending expression applies throughout the beam.

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

= \min (\frac{355 \times 896}{1,000}, \frac{1.2 \times 355 \times 797}{1,000})

= \min (318.08, 339.522) = 318.08 kN \cdot m

| M |_{\max} = 102.926906 kN \cdot m

Bending passes.

B1 reaction delivered to B2, imposed load only:

P_{q} = \frac{(2.5 \times 2.5) \times 7}{2} = 21.875 kN

L = 5,000 mm

I_{x} = 14,140 \times 10^{4} {mm}^{4}

δ_{B 2} = \frac{P L^{3}}{48 E I}

= \frac{21875 \times {5,000}^{3}}{48 \times 205,000 \times 14,140 \times 10^{4}} = 1.965231 mm

Brittle-finish limit:

\frac{5,000}{360} = 13.888889 mm

1.965231 < 13.888889

: passes.

Animation labSee shear in the web4 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.

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.

(b) Two corrosion-protection methods —4 printed marks

Protective paint/coating system: prepare the steel surface, then apply and maintain a suitable protective coating to isolate steel from water/oxygen. Explain the barrier mechanism; merely saying “paint” omits why it helps.

Galvanizing: apply a zinc coating, which provides a barrier and sacrificial protection because the zinc preferentially corrodes. Identify zinc explicitly; it is not a generic colour finish.

These two methods correspond to the course corrosion discussion in Ch 1 p.2. Keep drainage/moisture-trap avoidance as a useful detailing addition, not an unsupported substitute for explaining the two requested methods. No coating thickness or maintenance interval is specified in the paper.

Animation labHow corrosion protection works1 concept

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

Corrosion can consume steel and reduce its effective thickness.
Paint and metallic coatings protect the exposed surface; coating integrity matters.
Edges, damaged coatings and water-trapping details deserve attention. The model shows layers schematically.
Weathering steel and cathodic protection have specific environmental applications; they are not universal substitutes for paint.

Compact exam answer

Result	$B1$	$B2$
Ultimate line self/floor model	$w = 23.016150 \frac{kN}{m}$	$P = 80.556525 kN$ ； $w = 0.714 \frac{kN}{m}$
Maximum shear	$80.556525 kN$	$42.063262 kN$
Maximum moment	$140.973919 kN \cdot m$	$102.926906 kN \cdot m$
Section class	$1$	$1$
Imposed deflection / limit	$9.372083$ ／ $19.444444 mm$	$1.965231$ ／ $13.888889 mm$
Requested adequacy	Pass	Pass

Slab $g_{k} = 3.5625$ , $q_{k} = 2.5$ ,ultimate= $8.9875 \frac{kN}{m^{2}}$ . Use the theory explanations above for the $4$ -mark final part. A concise script should still show units, the plastic ceiling and the serviceability load choice.

Mistakes to avoid

Do not read a slab bay width as the supporting beam span.
Do not double-factor a transferred reaction.
Do not use design load in an imposed-only deflection equation.
Do not confuseI( ${cm}^{4}$ ), $Z$ ( ${cm}^{3}$ ) andS( ${cm}^{3}$ ).
Passing bending does not prove that deflection passes.

Procedure for an unfamiliar variant

Mark slab arrows, support points and member spans.
Find dead/imposed surface and line loads separately.
Transfer each supported beam’s reaction exactly once.
Find ULS actions and classify each exact section.
Check shear/low-shear bending and then imposed-only deflection.
End with an explicit pass/fail statement and answer the short theory part.

Independent self-check

Try it yourself. Invented variant: increase only the imposed surface load $20 %$ . What happens to the two deflection checks?

Reveal answer and reasoning

Multiply only the imposed-load deflection by $1.2$ : B1 becomes $11.246500 mm$ , and B2 becomes $2.358277 mm$ . Compare respectively with $19.444444$ and $13.888889 mm$ ; both remain below their limits. Recalculate strength actions using $1.4 G + 1.6 \times 1.2 Q$ . Dead load is unchanged, so do not multiply everything directly by $1.2$ .