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Mock Q1: full solution and invented marking guide

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MOCK Q1 · slab → B1 → B2; all geometry repeated in the adjacent question. — Invented mock-question diagram. Slab → B1 → B2 shows slab load passing first to B1, then B2. The slab spans one way; interior denotes an internal beam. Squares mark corner columns providing vertical support. B1 spans $6 m$ ; the two slab spans on the right are each $3 m$ . Schematic, not to scale; stated dimensions govern.Open full-size image.

(a) Load intensity

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

G = 0.160 m \times 24.5 \frac{kN}{m^{3}} + 0.6 = 4.52 \frac{kN}{m^{2}}

U = 1.4 \times 4.52 + 1.6 \times 3 = 11.128 \frac{kN}{m^{2}}

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.

(b) Trace the load path and calculate actions

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

Interior B1 collects half of each $3 m$ slab bay: $1.5 + 1.5 = 3 m$ . The B2 midpoint receives one B1 end reaction; top/bottom edge B1 reactions arrive directly at columns. Those edge reactions do not add midpoint bending to B2.

B1 self-weight:

\frac{59.8 \times 10}{1000} = 0.598 \frac{kN}{m}

w_{G} = 4.52 \times 3 + 0.598 = 14.158

w_{Q} = 3 \times 3 = 9 \frac{kN}{m}

w_{u} = 1.4 \times 14.158 + 1.6 \times 9 = 34.2212 \frac{kN}{m}

R_{A} = R_{B} = \frac{w_{u} L}{2} = \frac{34.2212 \times 6}{2} = 102.664 kN

M_{\max} = \frac{w_{u} L^{2}}{8} = \frac{34.2212 \times 36}{8} = 153.995 kN \cdot m

At B2 midspan,

P = 102.664 kN

, already factored. B2 UDL:

\frac{1.4 \times 92.1 \times 10}{1000} = 1.2894 \frac{kN}{m}

Reaction at each B2 support:

\frac{P}{2} + \frac{w L}{2} = \frac{102.664}{2} + \frac{1.2894 \times 6}{2} = 55.2 kN

B2 maximum moment:

M_{\max} = \frac{P L}{4} + \frac{w L^{2}}{8} = \frac{102.664 \times 6}{4} + \frac{1.2894 \times 36}{8} = 159.798 kN \cdot m

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.

(c) Read exact rows and classify

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 $457 \times 152 \times 60$ 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	D454.6, web t 8.1, flange T13.3, root r 10.2, clear d 407.6, all $mm$
Local ratios	$\frac{b}{T} = 5.75$ ； $\frac{d}{t} = 50.3$ 。
Major-axis properties	$I_{x} = 25500 {cm}^{4} = 2.55 \times 10^{8} {mm}^{4}$ ; $Z_{x} = 1122 {cm}^{3}$ ; $S_{x} = 1287 {cm}^{3}$ .
LTB properties	$r_{y} = 3.23 cm = 32.3 mm$ ; $u = 0.868$ ; torsional index $x = 37.5$ .

S355, flange

T = 13.3 mm

, giving

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

ε = \sqrt{\frac{275}{355}} = 0.880141

Flange:

\frac{b}{T} = 5.75 < 9 ε = 7.921268

Web in bending:

\frac{d}{t} = 50.3 < 80 ε = 70.411267

Both are Class 1, so

β_{W} = 1

; plastic bending resistance is permitted.

\frac{d}{t} = 50.3 < 70 ε = 61.609858

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

Lookup: Data File pp.9–10, exact $533 \times 210 \times 92$ 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	D533.1, web t 10.1, flange T15.6, root r 12.7, clear d 476.5, all $mm$
Local ratios	$\frac{b}{T} = 6.71$ ； $\frac{d}{t} = 47.2$ 。
Major-axis properties	$I_{x} = 55230 {cm}^{4} = 5.523 \times 10^{8} {mm}^{4}$ ; $Z_{x} = 2072 {cm}^{3}$ ; $S_{x} = 2360 {cm}^{3}$ .
LTB properties	$r_{y} = 4.51 cm = 45.1 mm$ ; $u = 0.872$ ; torsional index $x = 36.5$ .

S355, flange

T = 15.6 mm

, giving

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

ε = \sqrt{\frac{275}{355}} = 0.880141

Flange:

\frac{b}{T} = 6.71 < 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.

Animation labWhy thin elements buckle locally2 concepts · 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.

(d) B1 resistance and deflection

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 = 8.1 \times 454.6 = 3682.26 {mm}^{2}

V_{c} = \frac{p_{y} A_{v}}{\sqrt{3}} = \frac{355 \times 3682.26}{\sqrt{3} \times 1,000} = 754.713600 kN

V_{\max} = 102.663600 kN < V_{c}

Shear passes.

0.6 V_{c} = 452.828160 kN > V_{\max}

Even maximum shear is low, so the low-shear bending formula applies throughout the beam.

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

= \min (\frac{355 \times 1287}{1,000}, \frac{1.2 \times 355 \times 1122}{1,000})

= \min (456.885, 477.972) = 456.885 kN \cdot m

| M |_{\max} = 153.995400 kN \cdot m

Bending passes.

Imposed line load:

w = 9 \frac{kN}{m} = 9 \frac{N}{mm}

I = 25500 \times 10000 = 255000000 {mm}^{4}

δ = \frac{5 w L^{4}}{384 E I} = \frac{5 \times 9 \times 6000^{4}}{384 \times 205000 \times 255000000} = 2.905308 mm

Limit

\frac{6000}{360} = 16.666667 mm

;

δ

: 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.

(e) B2 resistance and deflection

A_{v} = t D = 10.1 \times 533.1 = 5384.31 {mm}^{2}

V_{c} = \frac{p_{y} A_{v}}{\sqrt{3}} = \frac{355 \times 5384.31}{\sqrt{3} \times 1,000} = 1103.564654 kN

V_{\max} = 55.200000 kN < V_{c}

Shear passes.

0.6 V_{c} = 662.138792 kN > V_{\max}

Even maximum shear is low, so the low-shear bending formula applies throughout the beam.

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

= \min (\frac{355 \times 2360}{1,000}, \frac{1.2 \times 355 \times 2072}{1,000})

= \min (837.8, 882.672) = 837.8 kN \cdot m

| M |_{\max} = 159.797700 kN \cdot m

Bending passes.

B1 end reaction from imposed load only:

P_{q} = \frac{9 \times 6}{2} = 27 kN

I = 55230 \times 10000 = 552300000 {mm}^{4}

δ = \frac{P_{q} L^{3}}{48 E I} = \frac{27000 \times 6000^{3}}{48 \times 205000 \times 552300000} = 1.073118 mm

Limit

\frac{6000}{360} = 16.666667 mm

;

δ

: passes.

Full lateral restraint is explicitly supplied, so no LTB check is needed here. This does not establish an unspecified web-contact detail; the mock expressly excludes that scope.

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.

(f) Strength versus stiffness

Simple explanation: Strong enough and stiff enough are two questions

A shelf can avoid breaking yet still sag too much.

ULS checks safety against the relevant failure modes.
SLS checks the specified everyday-use limit.
Use the load case required for each check.

Remember: Passing bending resistance does not prove deflection passes.

Related concept and full method

Strength concerns resistance to yielding/failure, using design strength and capacities. Stiffness concerns elastic deformation, usingE andI. A higher grade can raise bending resistance without changing the courseE; changing section depth can substantially increaseI and reduce deflection. Therefore a member may pass strength and fail serviceability.

Animation labSee stiffness and deflection1 concept

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 script and invented marking guide

Part	Results / suggested credit
a /4	$G = 4.52$ , $U = 11.128 \frac{kN}{m^{2}}$ ; correct thickness units, each worth $2$ marks.
b /8	B1 w 34.2212,V102.664,M153.995; B2 P102.664,w 1.2894,V55. $2$ ,M159.798. Credit load path $2$ , line loads $2$ , reactions $2$ , moments $2$ .
c /4	BothClass 1; $2$ per correctly justified section.
d /10	Shear $2$ , low-shear/bending $3$ , service load $2$ , deflection/limit $3$ . $δ_{B 1} = 2.9053 mm$ .
e /10	Same distribution. $δ_{B 2} = 1.0731 mm$ ; use imposed reaction only.
f /4	$2$ per explained distinction, including one implication.

Animation labFollow the calculation sequence5 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.

Locate the load, supports, connection geometry and any stated assumptions.
Keep given values, table lookups and calculated values distinct; reconcile their units.
The calculation player steps through the existing expressions in their original order.
Compare demand with resistance or the relevant limit. Keep missing inputs and conditional conclusions explicit.