Worked examples

Tutorial 3 Q3: remove slab restraint and check B2/B3 LTB

Open “Animation lab” beside a teaching step for a visual explanation or a walkthrough of its original expressions. Models are illustrative; source answers remain unchanged.

Chinese–English terminology

Need a simpler picture? Open “Simple explanation” beside a difficult step. These optional notes do not replace the full solution.

Repeat the Q2 B2 and B3 assessment for lateral-torsional buckling when the slab does not provide full lateral restraint. B3 has lateral restraints at the supports and at B2. Use actual segment lengths as effective lengths and normal loads.

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

Reuse the load path fully derived in Q2: B2 $w = 48.525551 \frac{kN}{m}$ , $M_{\max} = 491.321208 kN \cdot m$ . B3 central $P = 218.364981 kN$ , self-weight $w = 0.921551 \frac{kN}{m}$ , reaction $R = 111.947145 kN$ , midspan $M = 331.694453 kN \cdot m$ . Removing restraints changes stability, not these simply supported vertical-load internal forces.

Before calculating: recognition and strategy

A section can pass its local bending resistance and still buckle sideways. For each unrestrained length, calculate its own equivalent uniform moment and its own λᴸᵀ. Do not use B3’s total $6 m$ where the midpoint is an effective restraint.

1. B2: the full $9 m$ is unrestrained

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

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.

For a simply supported full UDL the quarter-point moments are $0.75 M_{\max}$ , $M_{\max}$ and $0.75 M_{\max}$ . Table 8.4b therefore gives:

m_{LT} = 0.2 + 0.15 \times 0.75 + 0.5 \times 1 + 0.15 \times 0.75 = 0.925

Moment demand:

0.925 \times 491.321208 = 454.472117 kN \cdot m

Normal-load effective length is the given segment length. All quantities in $λ$ and λᴸᵀ are dimensionless. This Class 1 section has $β_{W} = 1$ .

λ = \frac{L_{E}}{r_{y}} = \frac{9000}{41.2} = 218.446602

v = {[1 + 0.05 {(\frac{λ}{x})}^{2}]}^{- \frac{1}{4}}

= {[1 + 0.05 {(\frac{218.446602}{37.9})}^{2}]}^{- \frac{1}{4}} = 0.782955

λ_{LT} = u v λ \sqrt{β_{W}}

= 0.872 \times 0.782955 \times 218.446602 \times 1 = 149.141533

Table 8.3a, Data File p.5, pᵧ355 column:

145

row gives

75

;

150

row gives

71 \frac{N}{{mm}^{2}}

. Interpolation fraction:

\frac{149.141533 - 145}{150 - 145} = 0.828307

p_{b} = 75 + 0.828307 \times (71 - 75) = 71.686774 \frac{N}{{mm}^{2}}

M_{b} = \frac{p_{b} S_{x}}{1,000} = \frac{71.686774 \times 1471}{1,000} = 105.451244 kN \cdot m

Equivalent uniform moment demand:

m_{LT} M_{\max} = 454.472117 kN \cdot m

Utilisation:

\frac{454.472117}{105.451244} = 4.309784

Fails.

B2 fails LTB severely. Its Q2 fully restrained bending pass does not apply once the restraint is removed. It needs an effective restraint arrangement or a larger verified section.

Animation labA beam bends sideways and twists4 concepts · 13 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 the illustrated sagging beam the top flange is compressed.
The unrestrained compression flange can move sideways while the complete cross-section twists.
Only effective restraints divide the member into unbraced segments. They do not automatically add vertical supports.
Use the segment effective length, section properties and matching moment factor. This is an exaggerated mode shape, not a calculated displacement.

2. B3: check one $3 m$ segment and use symmetry

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

The two segments are mirror images. On the left segment, $x$ is measured from the left support, before reaching the point load atx 3 m. Therefore $M (x) = R x - \frac{w x^{2}}{2}$ . Use local quarter points $0.75, 1.5, 2.25 m$ , not quarters of the total $6 m$ .

M_{2} = 111.947145 \times 0.75 - \frac{0.921551 \times {0.75}^{2}}{2} = 83.701172 kN \cdot m

M_{3} = 111.947145 \times 1.5 - \frac{0.921551 \times {1.5}^{2}}{2} = 166.883972 kN \cdot m

M_{4} = 111.947145 \times 2.25 - \frac{0.921551 \times {2.25}^{2}}{2} = 249.548399 kN \cdot m

M_{\max} = 331.694453 kN \cdot m

m_{LT} = 0.2 + \frac{0.15 M_{2} + 0.5 M_{3} + 0.15 M_{4}}{M_{\max}} = 0.602266 > 0.44

Equivalent moment demand:

199.768312 kN \cdot m

Normal-load effective length is the given segment length. All quantities in $λ$ and λᴸᵀ are dimensionless. This Class 1 section has $β_{W} = 1$ .

λ = \frac{L_{E}}{r_{y}} = \frac{3000}{41.2} = 72.815534

v = {[1 + 0.05 {(\frac{λ}{x})}^{2}]}^{- \frac{1}{4}}

= {[1 + 0.05 {(\frac{72.815534}{37.9})}^{2}]}^{- \frac{1}{4}} = 0.958541

λ_{LT} = u v λ \sqrt{β_{W}}

= 0.872 \times 0.958541 \times 72.815534 \times 1 = 60.862695

Table 8.3a, Data File p.5, pᵧ355 column:

60

row gives

257

;

65

row gives

239 \frac{N}{{mm}^{2}}

. Interpolation fraction:

\frac{60.862695 - 60}{65 - 60} = 0.172539

p_{b} = 257 + 0.172539 \times (239 - 257) = 253.894299 \frac{N}{{mm}^{2}}

M_{b} = \frac{p_{b} S_{x}}{1,000} = \frac{253.894299 \times 1471}{1,000} = 373.478514 kN \cdot m

Equivalent uniform moment demand:

m_{LT} M_{\max} = 199.768312 kN \cdot m

Utilisation:

\frac{199.768312}{373.478514} = 0.534886

Passes.

Both B3 segments pass. The small self-weight makes its BMD slightly curved; using the actual ordinates avoids silently replacing it by a straight line. The original shear, deflection and local support checks remain those in Q2.

Animation labA beam bends sideways and twists4 concepts · 5 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 the illustrated sagging beam the top flange is compressed.
The unrestrained compression flange can move sideways while the complete cross-section twists.
Only effective restraints divide the member into unbraced segments. They do not automatically add vertical supports.
Use the segment effective length, section properties and matching moment factor. This is an exaggerated mode shape, not a calculated displacement.

Compact exam answer

B2: LE9,000 mm, mLT0.925, demand $454.472 kN \cdot m$ , Mb 105.451 kNm → fails. B3: two LE3,000 mm segments, mLT0.602266, demand $199.768 kN \cdot m$ , Mb 373.479 kNm → both pass. Normal loads and the explicitly assumed effective midpoint restraint are essential.

Mistakes to avoid

Do not transfer B2’s $9 m$ effective length to B3.
Do not claim concrete restraint in a question that removes it.
Do not assume every supported beam inherently gives torsional restraint.

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 reasoning variant: B3’s midpoint connection ceases to provide lateral restraint. What must be redone?

Reveal answer and reasoning

Use one $6 m$ LTB segment, recompute its full-span quarter moments and $m_{LT}$ , and recompute $λ$ , $v$ , $λ_{LT}$ , $p_{b}$ and $M_{b}$ withLE6,000 mm. The existing $3 m$ result no longer proves adequacy. Vertical reactions, shear and bending from the same loads do not change.

Tutorial 3 Q3: remove slab restraint and check B2/B3 LTB

Read the diagram and collect the data

Before calculating: recognition and strategy

1. B2: the full 9m9m is unrestrained

2. B3: check one 3m3m segment and use symmetry

Compact exam answer

Mistakes to avoid

Procedure for an unfamiliar variant

Independent self-check

1. B2: the full $9 m$ is unrestrained

2. B3: check one $3 m$ segment and use symmetry