Worked examples

Tutorial 3 Q2: fully restrained floor beams, including local web checks

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Chinese–English terminology

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Check B1, B2 and B3 for bending, shear and imposed-load deflection with brittle finishes. Also check B3 support web bearing and web buckling with the given stiff bearing and local restraints.

Original source: LectureNotes/Ch 3_Beam.pdf — p. 35, p. 36. 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

Given or adopted input	Exact origin
Floor geometry	FigureQ2: each horizontal B1/B2 spans $9 m$ ; vertical B3 spans $3 + 3 = 6 m$ . Slab spans vertically between horizontal beams.
Tributary widths	Top/bottom B1 takes half of one $3 m$ slab bay= $1.5 m$ . Middle B2 takes $1.5 m$ from each side= $3 m$ .
Slab thickness	The parenthesised $130$ in each slab bay denotes $130 mm$ RC slab; it is not a load intensity.
Surface loads	Q2 text: finishes $1$ , partitions $1$ , services $0.5$ , imposed $5$ , all $\frac{kN}{m^{2}}$ .
Members	B1 $406 \times 140 \times 46$ UB; B2 and B3 $457 \times 191 \times 67$ UB; S355, simple supports and full compression-flange restraint for Q2.
Additional B3 support data	Stiff bearing $150 mm$ , aₑ75 mm, bₑ0; both local flange restraints explicitly given.
Table mass for self-weight	$46.0$ kg/m and $67.1 \frac{kg}{m}$ from Data File p.9; multiply by the assumed $\frac{9.81}{1,000}$ for this numerical demonstration.

Keep $γ$ (concrete unit weight, $\frac{kN}{m^{3}}$ ) and $g$ (acceleration, $\frac{m}{s^{2}}$ ) symbolic until supplied. These equations are the source-supported general solution for the missing inputs:

Dead area load:

G = 0.130 γ + 1 + 1 + 0.5 = 0.130 γ + 2.5 \frac{kN}{m^{2}}

B1 ultimate UDL:

1.4 [1.5 (0.130 γ + 2.5) + \frac{46 g}{1000}] + 1.6 \times 7.5 \frac{kN}{m}

B2 ultimate UDL:

1.4 [3 (0.130 γ + 2.5) + \frac{67.1 g}{1000}] + 1.6 \times 15 \frac{kN}{m}

B3 midspan load = B2 UDL multiplied by

\frac{9}{2}

, giving

kN

; its own UDL =

\frac{1.4 \times 67.1 g}{1000} \frac{kN}{m}

. Substitute these loads into the equilibrium and strength formulas below. Imposed-load-only deflection is unaffected by

γ

g

Before calculating: recognition and strategy

Follow load paths before choosing formulas. Both B1 and B2 carry a floor UDL; B3 receives the central B2 end reaction plus its own weight. B1 end reactions are located at B3’s supports, so they increase column/joint reactions but not B3’s span bending. Full slab restraint suppresses LTB for Q2, but not shear, deflection or local web failure.

1. Convert slab thickness and collect surface loads

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

Slab dead load:

0.130 m \times 24 \frac{kN}{m^{3}} = 3.12 \frac{kN}{m^{2}}

g_{k} = 3.12 + 1 + 1 + 0.5 = 5.62 \frac{kN}{m^{2}}

q_{k} = 5 \frac{kN}{m^{2}}

Ultimate area load:

1.4 g_{k} + 1.6 q_{k} = 1.4 \times 5.62 + 1.6 \times 5 = 7.868 + 8 = 15.868 \frac{kN}{m^{2}}

B1 self-weight:

\frac{46.0 \times 9.81}{1,000} = 0.45126 \frac{kN}{m}

B2/B3 self-weight:

\frac{67.1 \times 9.81}{1,000} = 0.658251 \frac{kN}{m}

Self-weight is a dead load, so its $1.4$ factor is applied in the next lines. The imposed-only serviceability calculation does not use either strength factor.

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. Edge beam B1: loads, actions, section and deflection

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

B1 dead line load:

5.62 \times 1.5 + 0.45126 = 8.88126 \frac{kN}{m}

Imposed line load:

5 \times 1.5 = 7.5 \frac{kN}{m}

w_{u} = 1.4 \times 8.88126 + 1.6 \times 7.5 = 24.433764 \frac{kN}{m}

By symmetry:

R = \frac{w L}{2} = \frac{24.433764 \times 9}{2} = 109.951938 kN

V (x) = R - w x

, at

x = \frac{L}{2} = 4.5 m

V = 0

M (x) = R x - \frac{w x^{2}}{2}

Therefore:

M_{\max} = \frac{w L^{2}}{8} = \frac{24.433764 \times 9^{2}}{8} = 247.391861 kN \cdot m

Lookup: Data File pp.9–10, exact $406 \times 140 \times 46$ 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	D403.2, web t 6.8, flange T11.2, root r 10.2, clear d 360.4, all $mm$
Local ratios	$\frac{b}{T} = 6.35$ ； $\frac{d}{t} = 53$ 。
Major-axis properties	$I_{x} = 15690 {cm}^{4} = 1.569 \times 10^{8} {mm}^{4}$ ; $Z_{x} = 778 {cm}^{3}$ ; $S_{x} = 888 {cm}^{3}$ .
LTB properties	$r_{y} = 3.03 cm = 30.3 mm$ ; $u = 0.871$ ; torsional index $x = 38.9$ .

S355, flange

T = 11.2 mm

, so

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

ε = \sqrt{\frac{275}{355}} = 0.880141

Flange:

\frac{b}{T} = 6.35 < 9 ε = 7.921268

Web in bending:

\frac{d}{t} = 53 < 80 ε = 70.411267

Both are Class 1, so

β_{W} = 1

; plastic moment resistance is permitted.

\frac{d}{t} = 53 < 70 ε = 61.609858

No shear-buckling calculation is required under the course limit.

A_{v} = t D = 6.8 \times 403.2 = 2741.76 {mm}^{2}

V_{c} = \frac{p_{y} A_{v}}{\sqrt{3}} = \frac{355 \times 2741.76}{\sqrt{3} \times 1,000} = 561.949335 kN

V_{\max} = 109.951938 kN < V_{c}

: shear passes.

0.6 V_{c} = 337.169601 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{355 \times 888}{1,000}, \frac{1.2 \times 355 \times 778}{1,000})

= \min (315.24, 331.428) = 315.24 kN \cdot m

| M |_{\max} = 247.391861 kN \cdot m

: bending passes.

Deflection uses imposed UDL $7.5 \frac{kN}{m} = 7.5 \frac{N}{mm}$ , $L = 9,000 mm$ , $E = 205,000 \frac{N}{{mm}^{2}}$ and $I_{x} = 15,690 \times 10^{4} {mm}^{4}$ . The simply supported UDL formula follows by integrating the curvature and $E I$ relationship with zero displacement at both supports.

δ_{\max} = \frac{5 q L^{4}}{384 E I}

= \frac{5 \times 7.5 \times {9,000}^{4}}{384 \times 205,000 \times 15,690 \times 10^{4}} = 19.920181 mm

Brittle-finish limit:

\frac{L}{360} = \frac{9,000}{360} = 25 mm

19.920181 < 25

: deflection passes.

Animation labFollow the floor load in 3D7 concepts · 16 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.

3. Interior beam B2: twice the floor tributary width

B2 dead line load:

5.62 \times 3 + 0.658251 = 17.518251 \frac{kN}{m}

Imposed line load:

5 \times 3 = 15 \frac{kN}{m}

w_{u} = 1.4 \times 17.518251 + 1.6 \times 15 = 48.525551 \frac{kN}{m}

R = \frac{w L}{2} = \frac{48.525551 \times 9}{2} = 218.364981 kN

M_{\max} = \frac{w L^{2}}{8} = \frac{48.525551 \times 9^{2}}{8} = 491.321208 kN \cdot m

Lookup: Data File pp.9–10, exact $457 \times 191 \times 67$ 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	D453.4, web t 8.5, flange T12.7, root r 10.2, clear d 407.6, all $mm$
Local ratios	$\frac{b}{T} = 7.48$ ； $\frac{d}{t} = 48$ 。
Major-axis properties	$I_{x} = 29380 {cm}^{4} = 2.938 \times 10^{8} {mm}^{4}$ ; $Z_{x} = 1296 {cm}^{3}$ ; $S_{x} = 1471 {cm}^{3}$ .
LTB properties	$r_{y} = 4.12 cm = 41.2 mm$ ; $u = 0.872$ ; torsional index $x = 37.9$ .

S355, flange

T = 12.7 mm

, so

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

ε = \sqrt{\frac{275}{355}} = 0.880141

Flange:

\frac{b}{T} = 7.48 < 9 ε = 7.921268

Web in bending:

\frac{d}{t} = 48 < 80 ε = 70.411267

Both are Class 1, so

β_{W} = 1

; plastic moment resistance is permitted.

\frac{d}{t} = 48 < 70 ε = 61.609858

No shear-buckling calculation is required under the course limit.

A_{v} = t D = 8.5 \times 453.4 = 3853.9 {mm}^{2}

V_{c} = \frac{p_{y} A_{v}}{\sqrt{3}} = \frac{355 \times 3853.9}{\sqrt{3} \times 1,000} = 789.892822 kN

V_{\max} = 218.364981 kN < V_{c}

: shear passes.

0.6 V_{c} = 473.935693 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{355 \times 1471}{1,000}, \frac{1.2 \times 355 \times 1296}{1,000})

= \min (522.205, 552.096) = 522.205 kN \cdot m

| M |_{\max} = 491.321208 kN \cdot m

: bending passes.

Imposed-load deflection:

\frac{5 \times 15 \times {9,000}^{4}}{384 \times 205,000 \times 29,380 \times 10^{4}}

= 21.276218 mm < \frac{9,000}{360} = 25 mm

Passes. Characteristic imposed reaction delivered to each B3:

\frac{15 \times 9}{2} = 67.5 kN

Each B3 receives one half of B2’s whole load. Do not use its $218.364981 kN$ ultimate reaction in an imposed-only deflection calculation.

Animation labFollow the floor load in 3D7 concepts · 12 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.

4. Supporting beam B3: one midpoint reaction plus self-weight

Given span

L = 3 + 3 = 6 m

. Ultimate midspan point load

P = R_{B 2} = 218.364981 kN

. Ultimate self-weight:

w = 1.4 \times 0.658251 = 0.921551 \frac{kN}{m}

Support reaction due to loads within the beam span:

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

= \frac{218.364981}{2} + \frac{0.921551 \times 6}{2} = 111.947145 kN

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

= \frac{218.364981 \times 6}{4} + \frac{0.921551 \times 6^{2}}{8} = 331.694453 kN \cdot m

B3 has the same $457 \times 191 \times 67$ section as B2, so the already demonstrated Class 1 and no-shear-buckling results apply. At each corner the column additionally receives B1’s $109.951938 kN$ end reaction; that load at a support does not create span moment in B3.

A_{v} = t D = 8.5 \times 453.4 = 3853.9 {mm}^{2}

V_{c} = \frac{p_{y} A_{v}}{\sqrt{3}} = \frac{355 \times 3853.9}{\sqrt{3} \times 1,000} = 789.892822 kN

V_{\max} = 111.947145 kN < V_{c}

: shear passes.

0.6 V_{c} = 473.935693 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{355 \times 1471}{1,000}, \frac{1.2 \times 355 \times 1296}{1,000})

= \min (522.205, 552.096) = 522.205 kN \cdot m

| M |_{\max} = 331.694453 kN \cdot m

: bending passes.

This serviceability check uses only the midspan imposed point load

P = 67.5 kN

δ = \frac{P L^{3}}{48 E I} = \frac{67,500 \times {6,000}^{3}}{48 \times 205,000 \times 29,380 \times 10^{4}}

= 5.043252 mm

Limit

\frac{6,000}{360} = 16.666667 mm

: passes.

The midspan point-load expression contains $L^{3}$ . The point-load expression printed on Data File p.2, $L^{4}$ , is dimensionally incorrect; the course example uses $L^{3}$ . $\frac{N \times {mm}^{3}}{\frac{N}{{mm}^{2}} \times {mm}^{4}}$ gives $mm$ .

Animation labFollow the floor load in 3D7 concepts · 3 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.

5. B3 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

The question supplies $b_{1} = 150 mm$ stiff bearing, $a_{e} = 75 mm$ and $b_{e} = 0$ . At an end bearing, the course dispersal factor is $n = 2$ . $k = T + r$ is the flange-plus-root distance. The local load is the B3 span reaction $111.947145 kN$ under this model.

k = 12.7 + 10.2 = 22.9 mm

b_{1} + n k = 150 + 2 \times 22.9 = 195.8 mm

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

= \frac{195.8 \times 8.5 \times 355}{1,000} = 590.8265 kN

111.947145 < 590.8265

: web bearing passes.

The B1/column corner load-transfer detail is not shown. If the B1 reaction also enters the same B3 web-bearing zone, a conservative demand is $111.947145 + 109.951938 = 221.899083 kN$ . This alternative demand is also below the bearing and buckling resistances below. Do not invent a different actual bearing detail.

Animation labSpread a concentrated force into the web1 concept · 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.

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.

6. B3 support web buckling and conclusion

Simple explanation: What to do when one input is missing

A calculator cannot supply a dimension that the drawing never gave.

Separate given values, table values and calculated values.
Complete the checks whose required inputs are available.
State the missing input beside the remaining conditional result.

Remember: An illustrative assumption must not become an unstated exam given.

Related concept and full method

d = 407.6 mm

;

0.7 d = 285.32 mm

a_{e} = 75 < 285.32

; use the end-reduced web formula. End factor:

\frac{a_{e} + 0.7 d}{1.4 d} = \frac{75 + 285.32}{1.4 \times 407.6}

= \frac{360.32}{570.64} = 0.631431

Web factor:

\frac{25 ε t}{\sqrt{(b_{1} + n k) d}}

= \frac{25 \times \sqrt{\frac{275}{355}} \times 8.5}{\sqrt{195.8 \times 407.6}}

P_{x} = [\frac{360.32}{570.64}] \times [\frac{25 \times \sqrt{\frac{275}{355}} \times 8.5}{\sqrt{195.8 \times 407.6}}] \times 590.8265

= 246.986831 kN

Both local flange restraints specified in the question are present, so no additional $P_{xr}$ reduction is needed. $111.947145 kN$ is below 246.986831; even the conservative combined corner demand $221.899083 kN$ is below it. All requested Q2 beam checks pass for the explicitly adopted loading constants.

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.

Compact exam answer

Beam	Ultimate $V$ / $M$	Imposed $δ$ / limit	Result
$B1$	$109.952 kN$ ／ $247.392 kN \cdot m$	$19.920$ ／ $25 mm$	Pass
$B2$	$218.365 kN$ ／ $491.321 kN \cdot m$	$21.276$ ／ $25 mm$	Pass
$B3$	$111.947 kN$ ／ $331.694 kN \cdot m$	$5.043$ ／ $16.667 mm$	Pass

$g_{k} = 5.62$ , $q_{k} = 5 \frac{kN}{m^{2}}$ ; including tabulated beam weight and using $g = 9.81$ . Section is Class 1. B3: $P_{bw} = 590.8265 kN$ , $P_{x} = 246.9868 kN$ , with local restraint given. Full slab restraint means Q2 requires no LTB check.

Mistakes to avoid

The $9 m$ dimension is the B1/B2 span, not B3’s span.
B2 receives $3 m$ tributary width; B1 receives $1.5 m$ .
Do not apply ULS factors in imposed-load deflection.
Use $L^{3}$ for a central point, $L^{4}$ for a UDL.

Procedure for an unfamiliar variant

Read slab direction, tributary widths, beam spans and restraint positions from the source.
Separate characteristic dead/imposed loads; factor only strength loads.
Pass a supporting beam the reaction from the supported beam, not its full load.
Draw the beam free body, solve reactions and obtain shear/moment ordinates.
Read the exact section row, classify and check shear before selecting the bending formula.
Check imposed-load deflection with consistent $N$ and $mm$ .
Where requested, check bearing/web buckling and each unrestrained segment independently.

Independent self-check

Try it yourself. Invented variant: imposed surface load rises from $5$ to $6 \frac{kN}{m^{2}}$ . Which $9 m$ beam first fails its brittle-finish deflection limit?

Reveal answer and reasoning

Deflection increases linearly with imposed load. B1: $\frac{19.920181 \times 6}{5} = 23.904217 mm < 25 mm$ ; B2: $\frac{21.276218 \times 6}{5} = 25.531461 mm > 25 mm$ , fails. B3: $6.051902 mm < 16.666667 mm$ . Ultimate loads have changed; strength checks must be recalculated separately.

Animation labSee stiffness and deflection2 concepts · 3 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.