Worked examples

Assignment 1 Q1: storage-floor beams and an annotated connection sketch

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For the source storage-floor plan, find ultimate slab intensity, B1 and B2 maximum shear/moment, both section classes, and B2 shear/bending/deflection adequacy. Then produce the requested annotated isometric slab–B1–B2 web-weld connection sketch. S355 beams are simply supported and fully laterally restrained by $170 mm$ concrete slabs, with brittle finishes.

Original source: Assignment/AY2627s 1-CON4334-Assignment 1.pdf — p. 1. 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.

Figure Q1 has two side-by-side bays of

8 m

. Each B1 lies horizontally on the centreline between the upper and lower B3 beams. The dimension chain on the right,

3.5 + 3.5 = 7 m

, is the span of the vertical B2 in the plan, supported at its upper and lower ends. The end reactions of the two B1 beams meet at the

3.5 m

midpoint of B2. The slab arrows show one-way spanning vertically in the plan to the horizontal beams. B1 receives half of the

3.5 m

bay on each side, giving a total tributary width of

3.5 m

. The original drawing specifies B1 as

457 \times 152 \times 60

UB and B2 as

533 \times 210 \times 92

UB; the last numbers in these designations are the respective nominal masses per metre,

60

92 \frac{kg}{m}

. The main explanation separately gives the exact tabulated masses; do not count the two descriptions as additional loads.Open full-size image

Read the diagram and collect the data

Input	Exact origin
Slab	Given thickness $170 mm$ ; unit weight $24.5 \frac{kN}{m^{3}}$ ; additional finish $0.6 \frac{kN}{m^{2}}$ ; imposed $4.5 \frac{kN}{m^{2}}$ .
Gravity	The question explicitly specifies the beam self-weight calculation using $g = 10 \frac{m}{s^{2}}$ ; use this value rather than another default.
$B1$	Given $457 \times 152 \times 60$ UB; Data File p.9 mass $59.8 \frac{kg}{m}$ (nominal designation $60$ ).
$B2$	Given $533 \times 210 \times 92$ UB; Data File p.9 mass $92.1 \frac{kg}{m}$ (nominal designation $92$ ).
Restraints/finish	Given simple supports, full slab compression-flange lateral restraint and brittle finish. LTB is suppressed in this particular check.
Connection sketch	Part(b) explicitly calls for secondary B1 supported by primary B2 with fillet welds at both sides of B1’s web; no weld leg, length, cope size or connection plate detail is specified.

Before calculating: recognition and strategy

Start at the slab, convert surface intensity to B1 line load using tributary width, and pass both B1 reactions into the middle of B2. End-joint reactions from B3 occur at B2’s vertical supports and do not create additional B2 span bending in the simple model. Strength and deflection use different load combinations.

(a)(i) Ultimate slab 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

Slab thickness:

170 mm = 0.170 m

Slab self-weight:

0.170 \times 24.5 = 4.165 \frac{kN}{m^{2}}

Characteristic dead load:

g_{k} = 4.165 + 0.6 = 4.765 \frac{kN}{m^{2}}

Characteristic imposed load:

q_{k} = 4.5 \frac{kN}{m^{2}}

Ultimate area load:

1.4 g_{k} + 1.6 q_{k} = 1.4 \times 4.765 + 1.6 \times 4.5

= 6.671 + 7.2 = 13.871 \frac{kN}{m^{2}}

The extra $0.6 \frac{kN}{m^{2}}$ finish is dead load; “brittle” controls the later deflection limit and does not change its load category.

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) B1 maximum factored shear and moment

Tributary width:

\frac{3.5}{2} + \frac{3.5}{2} = 3.5 m

B1 self-weight:

\frac{59.8 \frac{kg}{m} \times 10 \frac{m}{s^{2}}}{1,000} = 0.598 \frac{kN}{m}

Characteristic dead line load:

4.765 \times 3.5 + 0.598 = 17.2755 \frac{kN}{m}

Characteristic imposed line load:

4.5 \times 3.5 = 15.75 \frac{kN}{m}

Ultimate line load:

w = 1.4 \times 17.2755 + 1.6 \times 15.75

= 24.1857 + 25.2 = 49.3857 \frac{kN}{m}

L = 8 m

; by symmetry, each end reaction:

R = \frac{w L}{2} = \frac{49.3857 \times 8}{2} = 197.5428 kN

Maximum shear occurs at the supports:

V_{\max} = 197.5428 kN

V (x) = R - w x = 0

occurs at

x = 4 m

; hence maximum moment:

M_{\max} = \frac{w L^{2}}{8} = \frac{49.3857 \times 8^{2}}{8} = 395.0856 kN \cdot m

One B1’s full factored load is $395.0856 kN$ , but its reaction on B2 is half of that, $197.5428 kN$ . The numerical equality between this full load in $kN$ and its maximum moment in $kN \cdot m$ is coincidental for $8 m$ ; the quantities and units remain different.

Animation labFollow the floor load in 3D4 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 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) B2 maximum factored shear and moment

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

B2 receives two B1 reactions at midspan:

P = 2 \times 197.5428 = 395.0856 kN

B2 self-weight:

\frac{92.1 \times 10}{1,000} = 0.921 \frac{kN}{m}

Ultimate UDL:

w = 1.4 \times 0.921 = 1.2894 \frac{kN}{m}

B2 span:

L = 3.5 + 3.5 = 7 m

Support reaction at each span end:

R = \frac{P}{2} + \frac{w L}{2}

= \frac{395.0856}{2} + \frac{1.2894 \times 7}{2} = 202.0557 kN

Maximum shear:

V_{\max} = 202.0557 kN

Maximum moment at midspan:

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

= \frac{395.0856 \times 7}{4} + \frac{1.2894 \times 7^{2}}{8}

= 691.3998 + 7.897575 = 699.297375 kN \cdot m

Do not apply $1.4$ / $1.6$ to $395.0856$ again: it is already assembled from ultimate B1 reactions. Only B2’s newly introduced self-weight is factored at this step.

Animation labFollow the floor load in 3D4 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 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)(iv) Both section classes

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

Read each exact UB row; the weakest compression element governs. Pure bending has a web compression/tension stress gradient, so use the bending web limit $80 ε$ for Class 1, not the uniform-compression column limit.

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.

Both sections are Class 1, and both flange thicknesses lie in the $\leq 16 mm$ range, so use $p_{y} = 355 \frac{N}{{mm}^{2}}$ . Both web ratios are also below $70 ε$ , so the course trigger does not require a separate shear-buckling check.

Animation labWhy thin elements buckle locally1 concept · 20 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)(v) B2 strength and brittle-finish 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

Shear area:

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

Shear resistance:

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

V_{\max} = 202.055700 kN < V_{c}

Shear passes.

0.6 V_{c} = 662.138792 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 2360}{1,000}, \frac{1.2 \times 355 \times 2072}{1,000})

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

| M |_{\max} = 699.297375 kN \cdot m

Bending passes.

The slab fully restrains the compression flange by the question’s explicit assumption, so use the cross-section bending resistance. Do not claim that this assumption proves the weld detail in(b) supplies a rigid moment connection.

For serviceability, each B1 end reaction due only to imposed load is

\frac{15.75 \times 8}{2} = 63 kN

The two reactions form the central imposed point load

P = 126 kN

. Under the one-way slab model, B2 has no additional direct imposed UDL.

L = 7,000 mm

I_{x} = 55,230 {cm}^{4} = 55,230 \times 10^{4} {mm}^{4}

Maximum deflection:

δ_{\max} = \frac{P L^{3}}{48 E I}

= \frac{126,000 \times {7,000}^{3}}{48 \times 205,000 \times 55,230 \times 10^{4}} = 7.952332 mm

Brittle-finish limit:

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

7.952332 < 19.444444

: deflection passes.

B2 is adequate for the requested shear, bending and imposed-load deflection checks. This part does not supply the contact dimensions needed for separate support web-bearing detail checks.

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) Completed annotated isometric teaching sketch

Annotated isometric AI-authored teaching sketch: lifted170 mm slab, perpendicular B1/B2 beams, paired fillet welds on B1 web, loads and modelling assumptions. — AI-authored vector teaching sketch drawn for Assignment 1 Q1. Not to scale; the slab is lifted to expose the joint. The orange welds and flange cope are conceptual only: no unprovided dimension has been assumed.Open/download the local SVG.
Selectable equivalents of the diagram labels and formulas: slab thickness $170 mm$ ; primary beam B2 is $533 \times 210 \times 92$ UB, with a support-to-support span of $7 m$ ; secondary beam B1 is $457 \times 152 \times 60$ UB, with a support-to-support span of $8 m$ . For the slab, $g_{k} = 4.765 \frac{kN}{m^{2}}$ , $q_{k} = 4.5 \frac{kN}{m^{2}}$ , with ultimate load $13.871 \frac{kN}{m^{2}}$ . The tributary width of B1 is $3.5 m$ ; each B1 delivers an ultimate reaction to B2 of $197.5428 kN$ . Together with the B1 on the opposite side, which is not drawn, the total point load is $395.0856 kN$ . All steel is S355. The simply supported overall model and the fully restrained condition given in the question are unchanged; no increase in bending resistance from composite action is assumed. These are selectable equivalents of values already labelled in the drawing, not additional question data.

The drawing shows the requested slab, primary B2 and a perpendicular secondary B1. A matching B1 meets the far side in the floor plan; it is identified in the annotations to keep the joint visible. Orange marks the two fillets on the faces of B1’s web. The slab is lifted only for visibility; in the actual assembly it bears on the beam top flanges. A conceptual top-flange cope is identified to avoid drawing solid steel members unrealistically through each other.

The simple-support global model requires appropriate rotational behaviour in the actual connection. Welding both web faces does not, by itself, establish a rigid beam-end moment connection. The sketch deliberately makes no unsupported weld-size or cope-dimension claim. Those are connection-design inputs not provided in Q1; the supplied beam sizes, spans, slab depth, loads and assumptions are fully annotated.

AI-process evidence required by the assignment: the SVG above is an actual AI-authored output for this learning website. It is not a historical screenshot record of a student’s conversation. For an assessed submission, capture your own complete prompt/output sequence, including revisions; do not fabricate screenshots or present suggested prompts as past interactions.

An illustrative prompt sequence for learning	What to verify in the resulting output
“Create an isometric schematic from this exact floor plan: $170 mm$ slab; B1 $457 \times 152 \times 60$ spanning $8 m$ into B2 $533 \times 210 \times 92$ spanning $7 m$ ; two fillet welds on the B1 web; include only the provided dimensions.”	Correct beam orientation, both web welds, real slab support and no invented weld leg.
“Add the independently checked loads: $g_{k} = 4.765$ , $q_{k} = 4.5 \frac{kN}{m^{2}}$ ; reaction from each B1 $197.5428 kN$ , combining at B2 midspan to give $395.0856 kN$ . Identify the exploded slab display and any conceptual cope.”	Units, centroid/load positions and given-versus-assumed labels match the calculation.
“Audit the sketch for crossed solid flanges, absent welds, unsupported connection dimensions and an unjustified rigid-joint claim; revise and list changes.”	Keep both the flawed intermediate image and the corrected output in your genuine evidence trail.

These prompts are clearly labelled illustrations, not a fabricated transcript. The completed vector can be opened locally and printed without online services. The assignment’s student-specific screenshot trail remains separate evidence that this source package cannot supply.

Animation labFollow the floor load in 3D3 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.

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.

Compact exam answer

(a) Ultimate area load $13.871 \frac{kN}{m^{2}}$ . B1: $w = 49.3857 \frac{kN}{m}$ , $V = 197.5428 kN$ , $M = 395.0856 kN \cdot m$ . B2: $P = 395.0856 kN$ plus uniformly distributed load $1.2894 \frac{kN}{m}$ ， $V = 202.0557 kN$ , $M = 699.297375 kN \cdot m$ . Both are Class 1, $p_{y} = 355$ . B2 passes shear and low-shear bending; imposed-load deflection $δ = 7.952332 < 19.444444 mm$ . (b) An annotated local isometric SVG is provided. Connection dimensions absent from the source and a student-specific screenshot history have not been invented.

Mistakes to avoid

Do not use $8 m$ as B2’s span.
Do not send one B1’s full load as its end reaction.
Do not double-factor B1 reactions on B2.
Do not use $1.6 \times 126 kN$ in the imposed-only deflection.
Do not fabricate AI-process screenshot evidence.

Procedure for an unfamiliar variant

Read slab span direction and tributary width.
Calculate characteristic surface loads, then factored intensity.
Include each beam’s self-weight once.
Transfer reactions through the floor hierarchy.
Classify and run the requested strength/serviceability checks.
Annotate the sketch with verified values and clearly identify conceptual details.

Independent self-check

Try it yourself. Invented variant: only one of the two B1 beams meets B2 at midspan, with all other values unchanged. Find B2 maximum moment and imposed deflection.

Reveal answer and reasoning

Central $P$ becomes $197.5428 kN$ . B2 moment: $M = \frac{197.5428 \times 7}{4} + \frac{1.2894 \times 7^{2}}{8} = 353.597475 kN \cdot m$ Central imposed load $P$ halves to $63 kN$ , giving $δ = \frac{7.952332}{2} = 3.976166 mm$ B2 self-weight is unchanged, so the total moment used for the strength check does not halve exactly.

Animation labBalance reactions and moments4 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.