Learn

Beams: loads, section checks and lateral stability

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.

A beam turns transverse loads into shear and bending, then carries them to its supports. First analyse the loading; only then compare the resulting actions with resistances. Sources: LectureNotes/Ch 3_Beam.pdf, pp.2–17, and the supplied Data File. Read equilibrium and section properties first.

Lecture examples are included immediately after the relevant concepts. Expand an example to read its complete diagram, working, exam answer and self-check here.

1. Identify what is restrained

Simple explanation: A support stops only certain movements

Holding a ruler up does not necessarily stop it sliding sideways or twisting.

Identify vertical support, lateral restraint and torsional restraint separately.
Find which flange is in compression in each region.
Use only restraints actually provided by the question.

Remember: An extra side restraint does not create a new vertical reaction.

Related concept and full method

A vertical support stops vertical movement. A lateral restraint stops sideways movement of the compression flange. Torsional restraint prevents the cross-section twisting. These are different conditions: a beam can stand on two supports yet buckle sideways between them. In sagging bending the top flange is compressed; in hogging bending the bottom flange is compressed. Follow the moment sign when identifying the flange requiring restraint.

The course accepts full lateral restraint when the restraint can resist $2.5 %$ of the maximum factored compression-flange force. Concrete slabs can generally supply this in the stated lecture models. If the question explicitly says “fully restrained”, carry out section, shear, local web and serviceability checks. If restraint is only at discrete points, additionally check LTB over every segment. A point load is not automatically a restraint.

Original Ch3 illustration: sideways movement and twist of a beam. — Original Ch 3 diagram: lateral movement and twist of a steel beam. Read its restraint labels as follows: “Restraint only”; “At supports—no movement”; “At intermediate restraints—vertical movement only”; and “Between restraints—vertical, lateral and torsional movement”. The label $L$ marks the spacing between restraints.Open full-size image

Try it yourself. A $9 m$ beam has lateral restraints at $0$ , $3$ , $6$ and $9 m$ . Is it three independently simply supported vertical beams?

Reveal answer and reasoning

No. Vertical bending analysis still uses the actual vertical supports. The three $3 m$ lengths are the LTB segments; internal moments at the lateral restraints are generally nonzero.

Animation labA beam bends sideways and twists2 concepts

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. Shear capacity and web slenderness

Simple explanation: What shear is trying to do

Imagine cutting the beam: shear stops the two cut faces sliding past each other.

Find the design shear from the load analysis.
Use the section’s applicable shear area and strength.
Compare demand with resistance in the same force units.

Remember: The web usually carries much of the vertical shear in an I-section.

Related concept and full method

V_{c} = \frac{p_{y} A_{v}}{\sqrt{3}}

p_{y}

is measured in

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

is used,

A_{v}

is measured in

{mm}^{2}

is used, divide the result by

1,000

to obtain

kN

Section / load direction	Shear area $A_{v}$
Rolled I, H or channel, load parallel to web	$t D$
Welded I, load parallel to web	$t d$
Rectangular hollow section	$2 t d$
T section	$t (D - T)$
Circular hollow section	$0.6 A$
Solid rectangle	$0.9 A$
Other section	$0.9 A_{0}$ , where $A_{0}$ is the relevant rectilinear element area

$D$ is overall depth; $d$ is clear web depth; $t$ is web thickness; $T$ is flange thickness. Rolled and welded formulas differ. For the rolled sections used in the worked beam examples, $\frac{d}{t} \leq 70 ε$ avoids a separate shear-web-buckling check, where $ε = \sqrt{\frac{275}{p_{y}}}$ . This does not waive concentrated-load web bearing or web buckling. Compare the largest absolute design shear with $V_{c}$ .

Animation labSee shear in the web1 concept · 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.

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.

3. Bending capacity at low shear

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

Use the section class from flange and web limits; take the less favourable class. At the location being checked, if $| V | \leq 0.6 V_{c}$ , use the low-shear resistance below. $S$ is plastic modulus, $Z$ elastic modulus. Properties in ${cm}^{3}$ multiply by $1,000$ to become ${mm}^{3}$ ; $p \times S$ in $N \cdot mm$ divides by $10^{6}$ for $kN \cdot m$ .

Class	Low-shear moment resistance
1 plastic or 2 compact	$M_{c} = \min (p_{y} S, 1.2 p_{y} Z)$
3 semi-compact	pᵧZ, or pᵧSₑff where the permitted effective plastic property is established
4 slender	pᵧZₑff or pᵧᵣZ, using an established effective property/reduced strength

The $1.2 p_{y} Z$ ceiling limits working-load plasticity in this course method. Check $M \leq M_{c}$ . Never replace $M$ by the reduced equivalent LTB moment here: the cross-section sees the actual moment. An unknown Class 4 effective modulus is missing design data, not permission to use gross $Z$ .

Animation labCompression and tension across a section2 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.

For sagging, the top flange is in compression and the bottom in tension; hogging reverses this.
Elastic bending stress varies with distance from the neutral axis: $σ = \frac{M y}{I}$ .
The section class governs whether elastic, plastic or effective properties may be used.
Use the shear at the section under examination; the largest shear elsewhere is not automatically coexistent.

4. What changes when shear is high?

Simple explanation: The web is doing two jobs

Heavy shear leaves less of the web’s strength available for bending.

Check the course trigger using the coexistent shear.
If triggered, use the appropriate reduced bending expression.
Recheck moment demand against the reduced resistance.

Remember: The reduction is conditional; do not apply it to every beam.

Related concept and full method

ρ = {(\frac{2 V}{V_{c}} - 1)}^{2}

Class 1／2:

M_{c} = \min [p_{y} (S - ρ S_{v}), 1.2 p_{y} (Z - \frac{ρ S_{v}}{1.5})]

Class 3:

p_{y} (Z - \frac{ρ S_{v}}{1.5})

Or use the supplied alternative effective-plastic method. Class 4:

p_{y} (Z_{eff} - \frac{ρ S_{v}}{1.5})

Apply this branch only when the coexistent shear exceeds $0.6 V_{c}$ . $S_{v}$ is the plastic modulus of the shear area about the bending axis, not the whole section $S$ . For a rectangular shear area of breadth $t$ and depth $h$ , derive $S_{v} = 2 (\frac{t h}{2}) (\frac{h}{4}) = \frac{t h^{2}}{4}$ . Establish the appropriate shear area before calculating it. If $V > V_{c}$ , the shear check already fails. The lecture examples have low shear, but this branch explains what to change in an unfamiliar heavily loaded variant.

Try it yourself. $\frac{V}{V_{c}} = 0.8$ . What is $ρ$ ?

Reveal answer and reasoning

$ρ = {(2 \times 0.8 - 1)}^{2} = 0.36$ . Reduce the bending contribution associated with the shear area; do not reduce the entire section capacity by $36 %$ .

Animation labHigh shear reduces bending resistance1 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.

Part of the web contributes to bending while also carrying shear.
The course low-shear route applies at $∣ V ∣ \leq 0.6 V_{c}$ . Above it, follow the applicable reduction expression.
The highlighted web contribution is reduced, not the complete cross-section by an arbitrary percentage.
Retain the section-class-specific formula and all bounds shown in the original calculation.

Animation labHigh shear reduces bending resistance1 concept · 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.

Part of the web contributes to bending while also carrying shear.
The course low-shear route applies at $∣ V ∣ \leq 0.6 V_{c}$ . Above it, follow the applicable reduction expression.
The highlighted web contribution is reduced, not the complete cross-section by an arbitrary percentage.
Retain the section-class-specific formula and all bounds shown in the original calculation.

5. A concentrated force can crush the web locally

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 reaction or point load spreads through the flange before entering the web. The stiff bearing length $b_{1}$ is the portion of the bearing that cannot appreciably bend; it is not automatically the whole support plate width. The source sketches show how load disperses through solid parts. Read the applicable sketch and its dimensions, or use $b_{1}$ explicitly supplied by the problem.

Ch3 p.5: four alternative stiff-bearing geometries; each formula belongs to its own sketch. — Ch 3 p.5: four alternative stiff-bearing geometries. Each formula belongs to its own sketch. Selectable formulas in left-to-right source order: $b_{1} = t + 1.6 r + 2 T$ $b_{1} = t + 1.6 s + 2 T$ $b_{1} = t + T + 0.8 r - g$ $b_{1} = 0.5 D_{c} + t + 0.8 s - g$ Legend: $t$ , $T$ , $r$ , $s$ , $g$ and $D_{c}$ are located by the dimension arrows in each individual sketch; do not borrow a dimension from another sketch. The source assumes a spread through solid material at $45 °$ and defines $b_{1}$ as the bearing length that cannot deform appreciably in bending (clause 8.4.10.2). The web-buckling formula at the bottom of the source page is typeset in the next section of this lesson.Open full-size image

Hot-rolled I/H-section:

k = T + r

; welded I/H-section:

k = T

. At a member end:

n = \min (2 + \frac{0.6 b_{e}}{k}, 5)

Within a member:

n = 5

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

$r$ is the root radius; $b_{e}$ is the distance from the near edge of the bearing to the member end. $p_{yw}$ is the design strength of the web at its actual thickness. All lengths in $mm$ give $P_{bw}$ in $N$ . Compare with the factored concentrated load or reaction at this location. Missing $b_{1}$ or $b_{e}$ prevents an unconditional numerical bearing result.

Animation labSpread a concentrated force into the web1 concept · 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.

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. The web can buckle as a short compression member

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

First establish whether the loaded flange is restrained against rotation relative to the web and lateral movement relative to the other flange. The following $P_{x}$ applies when both conditions hold. $a_{e}$ is the distance of the load/reaction line from the nearer member end; it differs from $b_{e}$ , measured from the bearing edge.

C = \frac{25 ε t}{\sqrt{(b_{1} + n k) d}}

a_{e} \geq 0.7 d

P_{x} = C P_{bw}

a_{e} < 0.7 d

P_{x} = [\frac{a_{e} + 0.7 d}{1.4 d}] C P_{bw}

If either flange-restraint condition is unsatisfied:

P_{xr} = (\frac{0.7 d}{L_{E, web}}) P_{x}

Both multipliers are dimensionless. $L_{E, web}$ is the effective length of the web as a compression member, not the beam’s lateral unbraced length. The lecture does not give a universal value for every unrestrained web geometry. Establish the restraint model and effective length before using the reduction. Compare resistance with the same concentrated force used for bearing. A good shear-capacity result does not prove this check.

Try it yourself. $b_{1} = 160 mm$ is centred on a bearing that starts at the member end. What are $b_{e}$ and $a_{e}$ ?

Reveal answer and reasoning

$b_{e} = 0$ ; $a_{e} = \frac{160}{2} = 80 mm$ . The first is measured to the bearing edge, the second to the centred reaction line.

Animation labThe web behaves like a short strut1 concept · 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.

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.

Animation labThe web behaves like a short strut1 concept · 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.

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.

7. Select and check a fully restrained beam

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

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

Draw the load path. Convert slab pressure $q$ ( $\frac{kN}{m^{2}}$ ) using tributary width $b$ ( $m$ ): $w = q b$ ( $\frac{kN}{m}$ ). Transfer supporting-beam reactions as point loads.
Keep characteristic dead and imposed loads separate. Include beam self-weight once, using mass×gravity/1,000, unless already included.
Use the stated ultimate factors; derive reactions, shear and bending from equilibrium. Locate maxima at load jumps or where $V = 0$ .
Estimate Srequired=M/pᵧ, assuming a thickness band only for the first trial. Select an actual section-table row; recheck pᵧ with its flange thickness.
Classify flange and web. Check shear, coexistent-shear bending, and requested local bearing/buckling.
Use the specified serviceability load combination and limit to check deflection. If a check fails, change section and repeat dependent checks.

For these lecture examples the deflection load is the unfactored imposed load, $E = 205,000 \frac{N}{{mm}^{2}}$ , and brittle-finish limit $\frac{L}{360}$ . The finish condition in Examples 1 and 2 is explicitly an assumption in the lecturer’s solution. Other conditions need their stated limit.

Simply supported beam, full-span UDL:

δ_{mid} = \frac{5 w L^{4}}{384 E I}

Simply supported beam, midspan point load:

δ_{mid} = \frac{P L^{3}}{48 E I}

For simultaneous loads, add deflections using elastic superposition.

Use $N$ , $mm$ and ${mm}^{4}$ throughout: $1 \frac{kN}{m} = 1 \frac{N}{mm}$ , $1 {cm}^{4} = 10^{4} {mm}^{4}$ . These two formulas are not valid unchanged for off-centre loads, cantilevers or continuous beams. See Example 3 for an off-centre-load derivation.

Beam example 1: fully restrained beam, from loads to deflection

Open this example on its own page

Select a Grade S355 section for a fully laterally restrained, simply supported $7 m$ beam. Check shear, bending, support web bearing, support web buckling and deflection. The central loads are $20 kN$ dead and $60 kN$ imposed; full-span UDLs are $7 \frac{kN}{m}$ dead and $35 \frac{kN}{m}$ imposed.

Original source: LectureNotes/Ch 3_Beam.pdf — p. 18, p. 19, p. 20, p. 21. 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

Quantity	Given, lookup or assumption
Span and loading	Given in the upper drawing: $L = 3.5 + 3.5 = 7 m$ , one central point load and two full-span UDL components.
Bearing	Given in the question: $b_{1} = 160 mm$ , $a_{e} = 80 mm$ , $b_{e} = 0$ . Loaded flange is restrained against both specified local movements.
Lateral restraint	Given: compression flange fully restrained, so no member LTB check is needed.
Deflection	Lecturer assumes brittle finishes and uses unfactored imposed loads only; limit $\frac{L}{360}$ . $E = 205,000 \frac{N}{{mm}^{2}}$ .
Section properties	Lookup Data File pp.9–10, row $533 \times 210 \times 82$ UB. Original solution also lists the values on p.19.

Before calculating: recognition and strategy

Learn the difference between a load and an action first: load is applied to the beam; shear $V$ and moment $M$ are internal actions needed to carry it. Symmetry makes the two reactions equal. The shear changes sign at the central point load, so that is the maximum-moment position. Full lateral restraint removes LTB from this example, but does not remove local web checks. Select a trial section from its plastic modulus, then verify every other requirement.

1. Factor each load component

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

P = 1.4 G_{point} + 1.6 Q_{point} = 1.4 \times 20 + 1.6 \times 60 = 28 + 96 = 124 kN

w = 1.4 g + 1.6 q = 1.4 \times 7 + 1.6 \times 35 = 9.8 + 56 = 65.8 \frac{kN}{m}

The given dead UDL is the prescribed design input. Do not add an extra assumed self-weight on top of it. Keep the unfactored imposed loads $60 kN$ and $35 \frac{kN}{m}$ for the separate deflection calculation.

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. Reactions, shear and maximum moment

Total ultimate load:

124 + 65.8 \times 7 = 584.6 kN

R_{A} = R_{B} = \frac{584.6}{2} = 292.3 kN

Immediately left of midspan:

V = 292.3 - 65.8 \times 3.5 = 62 kN

Immediately right of midspan:

V = 62 - 124 = - 62 kN

Immediately left of the right support:

V = - 62 - 65.8 \times 3.5 = - 292.3 kN

M_{mid} = R_{A} \times 3.5 - \frac{w \times {3.5}^{2}}{2}

= 292.3 \times 3.5 - \frac{65.8 \times 12.25}{2}

= 1,023.05 - 403.025 = 620.025 kN \cdot m

Equivalently $\frac{P L}{4} + \frac{w L^{2}}{8} = \frac{124 \times 7}{4} + \frac{65.8 \times 49}{8} = 217 + 403.025 = 620.025 kN \cdot m$ . The source labels $620$ after rounding. UDL makes shear linear and bending parabolic; the point load makes a shear jump of $124 kN$ but no moment jump.

Animation labBalance reactions and moments2 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.

3. Select a trial section and read the correct columns

Initially assume flange thickness

T \leq 16 mm

and trial

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

S_{x, required} = \frac{M}{p_{y}} = \frac{620.025 \times 10^{6}}{355}

= 1,746,549 {mm}^{3} = 1,746.549 {cm}^{3}

Trial

533 \times 210 \times 82

UB:

S_{x} = 2,059 {cm}^{3} > 1,746.549

Table column	Selected-row value
$D$ ; web $t$ ; flange $T$ ; root $r$ ; clear web $d$ ( $mm$ )	$528.3; 9.6; 13.2; 12.7; 476.5$
Flange $\frac{b}{T}$ ; web $\frac{d}{t}$	$7.91; 49.6$
Major-axis $I_{x}$ ; elastic $Z_{x}$ ; plastic $S_{x}$	$47,540 {cm}^{4}; 1,800 {cm}^{3}; 2,059 {cm}^{3}$

$533$ and $210$ are nominal section designations, not the exact $D$ and flange width. The table’s $T = 13.2 mm$ confirms the initial $p_{y} = 355$ assumption. Select the major-axis columns because the vertical load bends the beam about $x - x$ . This is the lecturer’s suitable trial section, not a demonstrated minimum-mass optimum.

Animation labRead a table without losing the keys1 concept · 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.

Name the required property: material strength, section property, buckling strength or a moment factor.
Keep section size, steel grade, thickness band, curve and axis as separate lookup keys.
$cm^{2}$ , $cm^{3}$ and $cm^{4}$ require different conversion powers. Do not use adjacent columns interchangeably.
Use bracketing rows within the same valid column. The original page remains the source of all table values.

4. Establish the section class and web-shear branch

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

ε = \sqrt{\frac{275}{p_{y}}} = \sqrt{\frac{275}{355}} = 0.880141

Class 1 flange limit:

9 ε = 7.92127; \frac{b}{T} = 7.91 \leq 7.92127

Class 1 web limit in pure bending:

80 ε = 70.4113; \frac{d}{t} = 49.6 \leq 70.4113

Both elements are Class 1, giving a plastic section. Web shear-buckling threshold:

70 ε = 61.6099; \frac{d}{t} = 49.6 < 61.6099

The flange classification is close to its limit: use the table ratio and keep enough precision. The web’s $80 ε$ bending-class limit and $70 ε$ shear-buckling threshold are different tests. A separate shear-web-buckling calculation is not required; concentrated reaction web buckling is still checked below.

Animation labWhy thin elements buckle locally2 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 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.

5. Resist the largest shear

A_{v} = t D = 9.6 \times 528.3 = 5,071.68 {mm}^{2}

V_{c} = \frac{p_{y} A_{v}}{\sqrt{3}}

= \frac{355 \times 5,071.68}{\sqrt{3}} = 1,039,488 N = 1,039.488 kN

V_{\max} = 292.3 kN < 1,039.488 kN

Passes.

Animation labSee shear in the web1 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.

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.

6. Check coexistent shear, then bending

At the maximum-moment section,

| V | = 62 kN

0.6 V_{c} = 0.6 \times 1,039.488 = 623.693 kN

62 < 623.693

; the low-shear bending formula applies.

p_{y} S_{x} = \frac{355 \times 2,059 \times 10^{3}}{10^{6}} = 730.945 kN \cdot m

1.2 p_{y} Z_{x} = \frac{1.2 \times 355 \times 1,800 \times 10^{3}}{10^{6}} = 766.8 kN \cdot m

M_{c} = \min (730.945, 766.8) = 730.945 kN \cdot m

620.025 < 730.945

: bending passes.

The largest shear actually also lies below $0.6 V_{c}$ , so every section is low-shear here. Do not use the larger of the two moment resistances: the second is a ceiling.

Animation labCompression and tension across a section2 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.

For sagging, the top flange is in compression and the bottom in tension; hogging reverses this.
Elastic bending stress varies with distance from the neutral axis: $σ = \frac{M y}{I}$ .
The section class governs whether elastic, plastic or effective properties may be used.
Use the shear at the section under examination; the largest shear elsewhere is not automatically coexistent.

7. 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

k = T + r = 13.2 + 12.7 = 25.9 mm

n = \min (2 + \frac{0.6 b_{e}}{k}, 5) = \min (2 + 0, 5) = 2

b_{1} + n k = 160 + 2 \times 25.9 = 211.8 mm

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

= 211.8 \times 9.6 \times 355 = 721,814.4 N = 721.8144 kN

721.8144 > 292.3

: support bearing passes.

Web thickness $9.6 mm$ also gives $p_{yw} = 355$ . This is an end bearing, hence $n = 2$ with $b_{e} = 0$ ; using the interior value $5$ would artificially increase resistance.

Animation labSpread a concentrated force into the web1 concept · 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.

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.

8. Support web buckling

0.7 d = 0.7 \times 476.5 = 333.55 mm

a_{e} = 80 mm < 333.55 mm

; use the end-reduced branch. End factor:

\frac{80 + 333.55}{1.4 \times 476.5} = \frac{413.55}{667.1} = 0.619922

Web factor:

\frac{25 \times 0.880141 \times 9.6}{\sqrt{211.8 \times 476.5}}

= \frac{211.2338}{\sqrt{100,922.7}} \approx 0.664919

P_{x} = [\frac{413.55}{667.1}] \times [\frac{211.2338}{\sqrt{100,922.7}}] \times 721.8144

= 297.531 kN > 292.3 kN

Passes, but with a small margin.

Both local flange restraints are explicitly given, so no $P_{xr}$ reduction is needed. The reaction uses $98.24 %$ of this resistance: support web buckling is the tightest strength check. Keep the given $a_{e}$ and $b_{e}$ distinct.

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.

9. Deflection under the specified serviceability case

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

P_{service} = 60 kN = 60,000 N; w_{service} = 35 \frac{kN}{m} = 35 \frac{N}{mm}

L = 7,000 mm; I_{x} = 47,540 \times 10^{4} = 475,400,000 {mm}^{4}

δ_{P} = \frac{P L^{3}}{48 E I}

= \frac{60,000 \times {7,000}^{3}}{48 \times 205,000 \times 475,400,000} = 4.39938 mm

δ_{w} = \frac{5 w L^{4}}{384 E I}

= \frac{5 \times 35 \times {7,000}^{4}}{384 \times 205,000 \times 475,400,000} = 11.22757 mm

δ_{total} = 4.39938 + 11.22757 = 15.62695 mm

Limit:

\frac{L}{360} = \frac{7,000}{360} = 19.44444 mm

15.62695 < 19.44444

: passes under the assumed brittle-finish condition.

Symmetry puts the maximum at midspan for both components, so adding these two maximum values is valid. The selected $533 \times 210 \times 82$ UB passes all requested checks under the stated model.

Animation labSee stiffness and deflection1 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.

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 exam answer

Use S355 $533 \times 210 \times 82$ UB, Class 1. Ultimate $P = 124 kN$ , $w = 65.8 \frac{kN}{m}$ , $R = V_{\max} = 292.3 kN$ , $M_{\max} = 620.025 kN \cdot m$ . $V_{c} = 1,039.49 kN$ . $M_{c} = \min (730.945, 766.8) = 730.945 kN \cdot m$ . $P_{bw} = 721.814 kN$ ; $P_{x} = 297.531 kN > 292.3$ . Imposed-load $δ = 15.627 mm < \frac{L}{360} = 19.444 mm$ . Adequate; support web buckling is the closest strength check.

Mistakes to avoid

Do not factor imposed loads again in the prescribed deflection check.
Use actual table dimensions, not $533 mm$ as $D$ .
Compare against the smaller moment-capacity expression.
Do not confuse shear-web slenderness with concentrated-load web buckling.

Procedure for an unfamiliar variant

Identify the vertical supports, span, lateral restraints and individual load positions.
Keep dead and imposed loads separate; form the required ultimate and serviceability cases.
Find reactions and the maximum shear/moment by equilibrium, with units.
Choose a section-table row, confirm thickness-dependent strength, and classify both flange and web.
Check shear and bending; add segment LTB where restraint is discrete.
Complete the requested web and deflection checks; state the governing result and any missing data.

Independent self-check

Try it yourself. Invented variant: keep the section and ultimate loads, but reduce the stiff bearing to $140 mm$ , centred at the member end ( $a_{e} = 70 mm$ , $b_{e} = 0$ ). Is the support web still adequate?

Hint

The change affects both the bearing resistance and the end-reduction factor.

Reveal answer and reasoning

Recalculate $n = 2$ and $b_{1} + n k = 191.8 mm$ . $P_{bw} = \frac{191.8 \times 9.6 \times 355}{1,000} = 653.6544 kN$ . $P_{x} = [\frac{70 + 333.55}{667.1}] \times [\frac{25 \times \sqrt{\frac{275}{355}} \times 9.6}{\sqrt{191.8 \times 476.5}}] \times 653.6544 \approx 276.3 kN$ , below $292.3 kN$ . Bearing passes but web buckling fails; the original section conclusion cannot simply be reused.

Animation labSpread a concentrated force into the web2 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.

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.

Beam example 2: trace a library-floor load into beam B3

Open this example on its own page

Design the fully restrained supporting beam B3 in the library floor plan. Characteristic dead and imposed floor loads are $6$ and $4 \frac{kN}{m^{2}}$ ; the dead load includes the self-weight allowances used by the example. Use S355 and the lecturer’s simply supported member model.

Original source: LectureNotes/Ch 3_Beam.pdf — p. 22, p. 23, p. 24, p. 25, p. 26. 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

Feature	Meaning for loading
Slab span marks crossing the B1 lines	Slab load reaches the parallel B1 beams over a $1.25 m$ tributary width.
B1 above and below each interior B2	Each B2 loading point receives two B1 end reactions.
B2 on both sides of the interior B3	B3 receives two B2 end reactions at its centre.
B3 also directly borders slab strips	Interior B3 carries its own $1.25 m$ tributary slab strip, represented as a UDL.
Given and assumed	All self-weight is already allowed for in the supplied dead intensity. B3 fully restrained; lecturer assumes brittle finishes for serviceability.

Before calculating: recognition and strategy

Work upstream before downstream: slab → B1 → B2 → B3. A supporting beam receives a point load equal to the end reaction of the supported beam, not that beam’s entire load. Keep dead and imposed reactions separate until the B3 loading is established. Do not treat the $5 m$ horizontal bay width as the B3 span.

1. Slab to B1

B1 span

= 3 m

; tributary width

= 1.25 m

g_{B 1} = 6 \frac{kN}{m^{2}} \times 1.25 m = 7.5 \frac{kN}{m}

q_{B 1} = 4 \times 1.25 = 5 \frac{kN}{m}

Dead-load reaction at each end:

\frac{7.5 \times 3}{2} = 11.25 kN

Imposed-load reaction at each end:

\frac{5 \times 3}{2} = 7.5 kN

The division by two comes from a uniformly loaded simply supported span. These are characteristic reactions, before the ultimate factors.

Animation labFollow the floor load in 3D2 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.

2. B1 reactions to B2

Each internal B1 line has two spans joining B2:

G_{point, B 2} = 2 \times 11.25 = 22.5 kN

Q_{point, B 2} = 2 \times 7.5 = 15 kN

In the interval

5 m

B2 span has three equally spaced point loads at

1.25, 2.5, 3.75 m

. B2 dead-load reaction at each end:

\frac{3 \times 22.5}{2} = 33.75 kN

B2 imposed-load reaction at each end:

\frac{3 \times 15}{2} = 22.5 kN

There are three internal point loads, not four: four slab strips are bounded by three interior B1 lines and the two B3 boundaries. Symmetry of the three loads gives half the total at each B2 end. Loads arriving directly at column-supported beam ends do not add span bending.

Animation labFollow the floor load in 3D2 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.

3. Assemble B3 point load and direct slab UDL

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

Midspan dead point load:

2 \times 33.75 = 67.5 kN

Midspan imposed point load:

2 \times 22.5 = 45 kN

Direct dead UDL:

6 \times 1.25 = 7.5 \frac{kN}{m}

Direct imposed UDL:

4 \times 1.25 = 5 \frac{kN}{m}

Ultimate point load:

P = 1.4 \times 67.5 + 1.6 \times 45 = 94.5 + 72 = 166.5 kN

Ultimate UDL:

w = 1.4 \times 7.5 + 1.6 \times 5 = 10.5 + 8 = 18.5 \frac{kN}{m}

The factor of two at the central point represents the B2 on each side of B3. It does not double B3’s $6 m$ span or its $1.25 m$ tributary strip.

Animation labFollow the floor load in 3D2 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.

4. Reactions, shear and moment of B3

Simple explanation: Why reactions balance the loads

Think of a seesaw that must neither fall nor turn.

Upward and downward forces must balance.
Take moments about a support to eliminate its reaction.
Use perpendicular distance from the point to the force line.

Remember: A lateral restraint is not automatically a vertical support.

Related concept and full method

Characteristic dead-load reaction:

\frac{67.5 + 7.5 \times 6}{2} = 56.25 kN

Characteristic imposed-load reaction:

\frac{45 + 5 \times 6}{2} = 37.5 kN

Ultimate reaction:

1.4 \times 56.25 + 1.6 \times 37.5 = 138.75 kN

Direct check:

\frac{166.5 + 18.5 \times 6}{2} = 138.75 kN

Shear immediately left of midspan:

138.75 - 18.5 \times 3 = 83.25 kN

Immediately right of midspan:

83.25 - 166.5 = - 83.25 kN

M_{\max} = \frac{166.5 \times 6}{4} + \frac{18.5 \times 6^{2}}{8}

= 249.75 + 83.25 = 333 kN \cdot m

Maximum shear occurs at the supports:

| V |_{\max} = 138.75 kN

Animation labBalance reactions and moments2 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.

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.

5. Select $457 \times 152 \times 52$ UB

Trial

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

S_{required} = \frac{333 \times 10^{6}}{355} = 938,028.17 {mm}^{3} = 938.028 {cm}^{3}

Selected

S_{x} = 1,096 {cm}^{3} > 938.028

Data File p.9 row $457 \times 152 \times 52$ gives $D = 449.8$ , $t = 7.6$ , $T = 10.9$ , $r = 10.2$ , $d = 407.6 mm$ , $\frac{b}{T} = 6.99$ and $\frac{d}{t} = 53.6$ . Page 10, the same designation, gives $I_{x} = 21,370 {cm}^{4}$ , $Z_{x} = 950 {cm}^{3}$ , $S_{x} = 1,096 {cm}^{3}$ . $T = 10.9 \leq 16$ confirms pᵧ=355. No interpolation is used for section properties: read the exact row.

Animation labRead a table without losing the keys3 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.

Name the required property: material strength, section property, buckling strength or a moment factor.
Keep section size, steel grade, thickness band, curve and axis as separate lookup keys.
$cm^{2}$ , $cm^{3}$ and $cm^{4}$ require different conversion powers. Do not use adjacent columns interchangeably.
Use bracketing rows within the same valid column. The original page remains the source of all table values.

6. Classify the trial section

ε = \sqrt{\frac{275}{355}} = 0.880141

\frac{b}{T} = 6.99 < 9 ε = 7.92127

The flange is Class 1.

\frac{d}{t} = 53.6 < 80 ε = 70.4113

The web in bending is Class 1; overall Class 1.

\frac{d}{t} = 53.6 < 70 ε = 61.6099

No separate web shear-buckling check is needed.

Animation labWhy thin elements buckle locally2 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 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.

7. Check shear resistance

A_{v} = t D = 7.6 \times 449.8 = 3,418.48 {mm}^{2}

V_{c} = \frac{355 \times 3,418.48}{\sqrt{3} \times 1,000} = 700.649 kN

138.75 < 700.649

: shear passes.

Animation labSee shear in the web1 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.

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.

8. Check moment resistance

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

Shear at the same midspan section

= 83.25 kN

0.6 V_{c} = 420.390 kN > 83.25

is low shear.

p_{y} S_{x} = \frac{355 \times 1,096}{1,000} = 389.08 kN \cdot m

1.2 p_{y} Z_{x} = \frac{1.2 \times 355 \times 950}{1,000} = 404.7 kN \cdot m

M_{c} = 389.08 kN \cdot m > 333 kN \cdot m

Passes.

The source’s requested B3 design demonstrates shear, moment and deflection. It does not specify a bearing geometry for a numerical local-web design here. Do not import the $160 mm$ bearing from Example 1.

Animation labCompression and tension across a section1 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.

For sagging, the top flange is in compression and the bottom in tension; hogging reverses this.
Elastic bending stress varies with distance from the neutral axis: $σ = \frac{M y}{I}$ .
The section class governs whether elastic, plastic or effective properties may be used.
Use the shear at the section under examination; the largest shear elsewhere is not automatically coexistent.

9. Check B3 serviceability

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

Use imposed load

P = 45,000 N

w = 5 \frac{N}{mm}

, and

L = 6,000 mm

I = 21,370 \times 10^{4} = 213,700,000 {mm}^{4}

δ_{P} = \frac{45,000 \times {6,000}^{3}}{48 \times 205,000 \times 213,700,000} = 4.62239 mm

δ_{w} = \frac{5 \times 5 \times {6,000}^{4}}{384 \times 205,000 \times 213,700,000} = 1.92600 mm

δ = 4.62239 + 1.92600 = 6.54839 mm

\frac{L}{360} = \frac{6,000}{360} = 16.66667 mm

6.54839 < 16.66667

: passes under the assumed brittle-finish condition.

Animation labSee stiffness and deflection1 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.

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 exam answer

B1 characteristic reactions $G$ / $Q$ = $11.25$ / $7.5 kN$ . B2 has three $22.5$ / $15 kN$ points; end reactions $33.75$ / $22.5 kN$ . B3 central $G$ / $Q$ = $67.5$ / $45 kN$ plus $7.5$ / $5 \frac{kN}{m}$ . Ultimate $P = 166.5 kN$ , $w = 18.5 \frac{kN}{m}$ , $V_{\max} = 138.75 kN$ , $M_{\max} = 333 kN \cdot m$ . Select $457 \times 152 \times 52$ UB S355, Class 1: $V_{c} = 700.649 kN$ , $M_{c} = 389.08 kN \cdot m$ , $δ = 6.548 mm < 16.667 mm$ . The demonstrated checks pass.

Mistakes to avoid

Do not make floor pressure a beam UDL without multiplying by tributary width.
Do not transfer a supported beam’s entire load to each end.
Include B3’s directly supported slab strip once.
Do not add self-weight already included by the problem.

Procedure for an unfamiliar variant

Identify the vertical supports, span, lateral restraints and individual load positions.
Keep dead and imposed loads separate; form the required ultimate and serviceability cases.
Find reactions and the maximum shear/moment by equilibrium, with units.
Choose a section-table row, confirm thickness-dependent strength, and classify both flange and web.
Check shear and bending; add segment LTB where restraint is discrete.
Complete the requested web and deflection checks; state the governing result and any missing data.

Independent self-check

Try it yourself. Invented variant: only the characteristic floor imposed pressure rises from $4$ to $5 \frac{kN}{m^{2}}$ . Find the revised B3 ultimate $P$ and $w$ , and imposed deflection; retain the original dead loading.

Reveal answer and reasoning

All imposed effects multiply by $\frac{5}{4}$ . B3 imposed point $= 56.25 kN$ ; imposed UDL $= 6.25 \frac{kN}{m}$ . $P = 1.4 \times 67.5 + 1.6 \times 56.25 = 184.5 kN$ ; $w = 1.4 \times 7.5 + 1.6 \times 6.25 = 20.5 \frac{kN}{m}$ . $M = \frac{184.5 \times 6}{4} + \frac{20.5 \times 36}{8} = 369 kN \cdot m < 389.08$ . Imposed deflection= $6.54839 \times 1.25 = 8.18549 mm < 16.66667$ . Maximum shear= $\frac{184.5 + 20.5 \times 6}{2} = 153.75 kN < 700.649$ ; these checks still pass.

Animation labFollow the floor load in 3D5 concepts · 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.

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.

Animation labFollow the calculation sequence10 concepts · 6 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.

8. Normal versus destabilizing loads

The source defines the destabilizing case by dominant loads applied to the top flange with both load and flange free to deflect and rotate relative to the shear centre. That movement increases the overturning effect as the beam twists. Merely seeing a downward arrow on the top of a drawing does not establish this condition. Use the stated “normal load” assumption when given; otherwise inspect load application and restraints.

Animation labA beam bends sideways and twists1 concept

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.

9. Select the LTB effective length from restraints

Simple explanation: Which length belongs in the calculation?

A sideways tie can shorten the buckling region without shortening the beam’s vertical span.

Separate vertical-support spacing from lateral-restraint spacing.
Apply the course rule for the actual end restraint and loading.
Keep the original vertical load analysis unless its supports change.

Remember: Restraints in one direction may not restrain the other direction.

Related concept and full method

Source condition, Ch 3 p.9	Effective length for normal load
End compression-flange lateral restraint, free rotation in plan, nominal torsional restraint	$L_{E} = L^{LT}$
Compression flange fully restrained against rotation in plan at end supports	$L_{E} = 0.8 L^{LT}$
End compression flanges not laterally restrained; both flanges free to rotate in plan	$L_{E} = 1.2 L^{LT} + 2 D$
Adequate intermediate lateral restraints	Lᴇ=segment length Lᴸᵀ

For the destabilizing cases described on p.9 multiply the relevant length by $1.2$ ; intermediate segments use 1.2Lᴸᵀ. A question may explicitly direct that effective and actual segment lengths are equal; follow that instruction. Cantilevers need Table 8.1: match the support condition, tip restraint and normal/destabilizing column, rather than borrowing the simply supported rule.

Original Table 8.1, Ch3 p.10. The four tip conditions are illustrated at the bottom. — Original Table 8.1, Ch 3 p.10. Four tip conditions are sketched at the bottom. Support groups: (a) continuous member with lateral restraint to the top flange; (b) continuous member with partial torsional restraint; (c) continuous member with lateral and torsional restraint; (d) restrained against lateral displacement, twist and rotation in the plane of bending. Within every group, tip conditions 1–4 are: 1 free; 2 top flange laterally restrained; 3 torsionally restrained; 4 laterally and torsionally restrained. Tip sketches 1 and 3 have no plan bracing; sketches 2 and 4 show plan bracing in one or more bays. The following values transcribe the source row by row, each pair in the order normal loading / destabilizing loading:
(a1) $3.0 L$ ／ $7.5 L$ ; (a2) $2.7 L$ ／ $7.5 L$ ; (a3) $2.4 L$ ／ $4.5 L$ ; (a4) $2.1 L$ ／ $3.6 L$ .
(b1) $2.0 L$ ／ $5.0 L$ ; (b2) $1.8 L$ ／ $5.0 L$ ; (b3) $1.6 L$ ／ $3.0 L$ ; (b4) $1.4 L$ ／ $2.4 L$ .
(c1) $1.0 L$ ／ $2.5 L$ ; (c2) $0.9 L$ ／ $2.5 L$ ; (c3) $0.8 L$ ／ $1.5 L$ ; (c4) $0.7 L$ ／ $1.2 L$ .
(d1) $0.8 L$ ／ $1.4 L$ ; (d2) $0.7 L$ ／ $1.4 L$ ; (d3) $0.6 L$ ／ $0.6 L$ ; (d4) $0.5 L$ ／ $0.5 L$ . Open full-size image

To read it: choose support block (a–d), then tip row (1–4), then the load-condition column. Example of reading, not an extra code rule: block (c), free tip row 1, normal loading gives $1.0 L$ ; the same geometry under destabilizing loading gives $2.5 L$ . The explicit overhang effective-length assumption in Tutorial 3 Q5 takes precedence for that exercise.

Animation labA beam bends sideways and twists1 concept · 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.

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.

10. Convert unbraced length to a resistance

Simple explanation: A beam can escape sideways

The compressed flange can move sideways while the section twists.

Divide the beam at effective lateral restraints.
Use each segment’s effective length to obtain its buckling resistance.
Compare that resistance with the segment’s equivalent moment demand.

Remember: A section bending check alone does not check lateral-torsional buckling.

Related concept and full method

λ = \frac{L_{E}}{r_{y}}

v = {[1 + 0.05 {(\frac{λ}{x})}^{2}]}^{- 0.25}

λ_{LT} = u v λ \sqrt{β_{w}}

Using

λ_{LT}

and

p_{y}

, look up

p_{b}

in Table 8.3a. Class 1/2 examples use:

M_{b} = p_{b} S_{x}

$r_{y}$ is weak-axis radius of gyration, in the same units as $L_{E}$ . $u$ is the section buckling parameter and $x$ the torsional index, both read from the selected section’s property row. This $x$ is not the beam position coordinate used in equilibrium. $β_{w} = 1$ for Classes 1 and 2. Do not assume that value for an unestablished Class 3/4 treatment. The source allows conservative $u = 0.9$ and $v = 1$ for rolled-section preliminary checks; accurate values matter in tight examples.

Select the design strength column within the steel grade: S355 with $T = 20 mm$ uses $p_{y} = 345$ , not the $355$ column. Find the two $λ_{LT}$ rows surrounding the calculated value and interpolate linearly. Never use $λ$ directly as the table row. A larger slenderness normally lowers $p_{b}$ .

Try it yourself. λᴸᵀ=121.6 and pᵧ=355. Rows $120$ and $125$ give $104$ and $97 \frac{N}{{mm}^{2}}$ . Find pᵦ.

Reveal answer and reasoning

Interpolation fraction $= \frac{121.6 - 120}{125 - 120} = 0.32$ . $p_{b} = 104 + 0.32 (97 - 104) = 101.76 \frac{N}{{mm}^{2}}$ , conventionally written $101.8$ .

Animation labA beam bends sideways and twists2 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.

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.

Animation labA beam bends sideways and twists2 concepts · 6 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.

11. Use the moment shape of each unbraced segment

Simple explanation: Why the shape of the moment diagram matters

The same peak moment is more demanding when a long region stays near that peak.

Use the diagram of this unrestrained segment.
Choose the applicable course sketch or quarter-point rule.
Apply its factor to the demand, following the stated check.

Remember: Column flexural-buckling factors and LTB factors are not interchangeable.

Related concept and full method

A short region at the maximum moment is less severe for LTB than the whole segment at that moment. The equivalent uniform moment factor $m_{LT}$ accounts for this. Choose Table 8.4a only when the actual load/moment pattern matches its sketch. For a straight end-moment diagram, use the signed ratio $β$ of the smaller to the larger internal end moment: $m_{LT} = \max (0.44, 0.6 + 0.4 β)$ . Check the source sketch’s sign convention: applied end arrows and internal BMD signs are not interchangeable.

For a general beam segment, Table 8.4b:

m_{LT} = \max [0.44, 0.2 + \frac{0.15 | M_{2} | + 0.5 | M_{3} | + 0.15 | M_{4} |}{M_{\max}}]

M_{2}

M_{3}

M_{4}

: the moments at the

\frac{1}{4}

\frac{1}{2}

\frac{3}{4}

positions within that segment.

M_{\max}

: the greatest absolute moment anywhere within the segment.

The table instructs that the ordinates are taken positive; use their magnitudes even if the BMD reverses sign. Quarter points refer to the current segment, not the full beam. Calculate their moments from equilibrium, retaining UDL curvature. For a cantilever without intermediate restraints the source sets $m_{LT} = 1.00$ . If a segment has zero moment throughout, there is no bending demand; do not divide by zero.

Table 8.4b, Ch3 p.16: quarter-point positions and both single-curvature and reversing diagrams. — Table 8.4b, Ch 3 p.16: quarter-point positions and moment diagrams with single curvature or a change of sign. The source formula is: $m_{LT} = 0.2 + \frac{0.15 M_{2} + 0.5 M_{3} + 0.15 M_{4}}{M_{\max}}$ It must also satisfy $m_{LT} \geq 0.44$ . The source specifies that all moments are taken as positive. $M_{2}$ , $M_{4}$ are the two quarter-point values; $M_{3}$ is the midpoint value; $M_{\max}$ is the maximum within the segment. The labels $M_{1}$ , $M_{5}$ are at the segment ends. The left-hand diagram retains one sign; the right-hand diagram crosses zero and changes sign. A cantilever without an intermediate lateral restraint uses $m_{LT} = 1.00$ . Open full-size image

Animation labRead the moment shape within one segment1 concept · 6 source expressions

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

Moment ordinates must belong to the same effective unbraced segment.
Same-side and reverse-curvature diagrams have different signed end ratios.
The markers show ¼, ½ and ¾ of this segment, not of the entire beam.
LTB and column flexural factors are different. Preserve their individual bounds and coefficient sets.

12. Put the checks together

Finish whole-beam reactions and SFD/BMD using the actual vertical supports.
Mark every lateral restraint and divide into unbraced segments.
For each segment establish Lᴇ, $λ$ , $v$ , λᴸᵀ and pᵦ; compute Mᵦ.
Find the segment’s maximum moment and its matching mᴸᵀ. Check mᴸᵀMmax≤Mᵦ.
Separately check actual $M_{\max} \leq M_{c}$ and $V \leq V_{c}$ , plus high-shear interaction if necessary.
Complete required bearing, web buckling and deflection checks. Report the governing segment and any missing inputs.

Do not assume the segment with the highest moment always governs: a longer, more weakly restrained segment may have a much smaller resistance. Keep a table of demand, resistance and utilisation for all segments.

Beam example 3: three restrained segments and an omitted deflection solution

Open this example on its own page

Check/design the S355 simply supported $9 m$ beam for shear and bending with lateral restraints at A, B, C and D. $AB = BC = CD = 3 m$ . At B: dead $40 kN$ and imposed $60 kN$ ; at C: dead $20 kN$ and imposed $30 kN$ . Dead self-weight is $3 \frac{kN}{m}$ . Effective segment lengths equal actual lengths. Also work through the lecturer’s requested follow-up web and deflection checks.

Original source: LectureNotes/Ch 3_Beam.pdf — p. 27, p. 28, p. 29, p. 30. 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.

Ch 3 p.27: only A and D have vertical support symbols. Crosses at A, B, C and D identify lateral restraints. B and C carry downward point loads; the small arrows over the entire span represent the specified

3 \frac{kN}{m}

self-weight. DL denotes dead load, LL imposed load, S/W self-weight, Shear shear force, and BM bending moment. From left to right, the original shear diagram prints

146, 133, 19, 32, 108, 120 kN

. The first two labels are above zero and the last four below. The source prints only magnitudes for those last four; under this lesson's sign convention they are negative shear values, not four positive loads. At B and C, the original moment diagram labels

419

and

342 kN \cdot m

. These are rounded values on the lecturer's diagram. The adjacent working recalculates from the given loads at higher precision and does not treat the diagram results as additional inputs.Open full-size image

Read the diagram and collect the data

Input	Exact diagram or table source
Vertical span	Three dimension strings of $3 m$ give $A = 0, B = 3, C = 6, D = 9 m$ .
Lateral model	Crosses define three $3 m$ unbraced segments; question sets Lᴇ=3,000 mm.
Trial section	The source selects $457 \times 191 \times 74$ UB. Data pp.9–10 give $D = 457$ , $t = 9$ , $T = 14.5$ , $r = 10.2$ , $d = 407.6 mm$ ; $\frac{b}{T} = 6.57$ , $\frac{d}{t} = 45.3$ ; $r_{y} = 4.20 cm$ ; $I_{x} = 33,320 {cm}^{4}$ ; $Z_{x} = 1,458$ , $S_{x} = 1,653 {cm}^{3}$ ; $u = 0.877$ , $x = 33.9$ .
Missing local-web data	No stiff bearing lengths, distances to member ends or complete local flange-restraint details are given. Crosses alone identify lateral restraint, not both web-buckling conditions.

Before calculating: recognition and strategy

Analyse the continuous $9 m$ beam on its two vertical supports. Use that BMD to examine the three separate lateral-buckling segments. First reproduce the lecturer’s rounded, straight-line BC approximation, then retain the self-weight curvature with Table 8.4b to check every segment. Deflection uses the imposed loads $60$ and $30 kN$ at their actual off-centre positions; a central-point formula cannot be substituted.

1. Factored loads

P_{B} = 1.4 \times 40 + 1.6 \times 60 = 56 + 96 = 152 kN

P_{C} = 1.4 \times 20 + 1.6 \times 30 = 28 + 48 = 76 kN

w = 1.4 \times 3 = 4.2 \frac{kN}{m}

Total load:

152 + 76 + 4.2 \times 9 = 265.8 kN

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.

2. Reactions without premature rounding

Simple explanation: Why reactions balance the loads

Think of a seesaw that must neither fall nor turn.

Upward and downward forces must balance.
Take moments about a support to eliminate its reaction.
Use perpendicular distance from the point to the force line.

Remember: A lateral restraint is not automatically a vertical support.

Related concept and full method

Take moments about D: each downward load contributes its distance to D; the UDL resultant $4.2 \times 9$ acts at midspan, $4.5 m$ from D.

R_{A} \times 9 = 152 \times 6 + 76 \times 3 + (4.2 \times 9) \times 4.5

= 912 + 228 + 170.1 = 1,310.1 kN \cdot m

R_{A} = \frac{1,310.1}{9} = 145.566667 kN

R_{D} = 265.8 - 145.566667 = 120.233333 kN

Check by moments about A: $120.233333 \times 9 = 152 \times 3 + 76 \times 6 + 37.8 \times 4.5 = 1,082.1 kN \cdot m$ .

Animation labBalance reactions and moments1 concept · 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.

3. Construct the SFD and BMD from the left-hand free body

For shear (

kN

), start with

V (x) = 145.566667 - 4.2 x

; when

x > 3

, add

- 152

; when

x > 6

, also add

- 76

. Negative signs mean the point loads cause downward shear jumps.
For moment (

kN \cdot m

), start with

M (x) = 145.566667 x - \frac{4.2 x^{2}}{2}

; when

x > 3

, add

- 152 (x - 3)

; when

x > 6

, also add

- 76 (x - 6)

. Each negative term is a point load times its lever arm to the section; include it only after the section has passed that load.

Read each “when” term as zero before its load position and include it afterwards. $x$ is in metres from A. The shorthand is often written with positive-part brackets $⟨ x - a ⟩$ ; it is a beam-position coordinate, unrelated to the torsional index.

Position	Shear just left / right ( $kN$ )	Moment ( $kN \cdot m$ )
$A$	$0$ ／ $+ 145.5667$	$0$
B, $x = 3$	$+ 132.9667$ ／ $- 19.0333$	$145.566667 \times 3 - 2.1 \times 9 = 417.8$
C, $x = 6$	$- 31.6333$ ／ $- 107.6333$	$145.566667 \times 6 - 2.1 \times 36 - 152 \times 3 = 341.8$
$D$	$- 120.2333$ ／ $0$	$0$

B is the maximum because shear changes from positive to negative there. BC and CD have negative shear throughout, so their bending moment decreases. Draw straight sloping shear segments, vertical jumps at the point loads, and curved bending segments continuous at B and C.

Animation labBuild the shear and moment diagrams1 concept · 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 illustrative 60 kN point load on a 6 m simply supported beam, balance the reactions before making a cut.
$V = R_{A}$ and $M = R_{A} x$ before the point load. Moment grows linearly.
Shear jumps down by P. To its right, $V = R_{A} - P$ and $M = R_{A} x - P (x - a)$ .
For this case, moment is continuous and returns to zero at B. A concentrated applied couple instead creates a moment jump.

4. Confirm strength and section class

T = 14.5 mm \leq 16 \to p_{y} = 355 \frac{N}{{mm}^{2}}

ε = \sqrt{\frac{275}{355}} = 0.880141

\frac{b}{T} = 6.57 < 9 ε = 7.92127; \frac{d}{t} = 45.3 < 80 ε = 70.4113

The section is Class 1;

\frac{d}{t} = 45.3 < 70 ε = 61.6099

, so no separate web shear-buckling calculation is needed.

Animation labWhy thin elements buckle locally1 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.

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.

5. Shear resistance

A_{v} = t D = 9 \times 457 = 4,113 {mm}^{2}

V_{c} = \frac{355 \times 4,113}{\sqrt{3} \times 1,000} = 842.9978 kN

V_{\max} = 145.5667 < 842.9978

Passes.

Animation labSee shear in the web1 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.

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.

6. Actual section moment resistance

Maximum shear coexisting with the moment at B

= 132.9667 kN

0.6 V_{c} = 505.7987 kN > 132.9667

Use the low-shear branch.

p_{y} S_{x} = \frac{355 \times 1,653}{1,000} = 586.815 kN \cdot m

1.2 p_{y} Z_{x} = \frac{1.2 \times 355 \times 1,458}{1,000} = 621.108 kN \cdot m

M_{c} = 586.815 kN \cdot m > 417.8

Section bending resistance passes.

Animation labCompression and tension across a section1 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.

For sagging, the top flange is in compression and the bottom in tension; hogging reverses this.
Elastic bending stress varies with distance from the neutral axis: $σ = \frac{M y}{I}$ .
The section class governs whether elastic, plastic or effective properties may be used.
Use the shear at the section under examination; the largest shear elsewhere is not automatically coexistent.

7. Check every unbraced segment

Simple explanation: A beam can escape sideways

The compressed flange can move sideways while the section twists.

Divide the beam at effective lateral restraints.
Use each segment’s effective length to obtain its buckling resistance.
Compare that resistance with the segment’s equivalent moment demand.

Remember: A section bending check alone does not check lateral-torsional buckling.

Related concept and full method

All three segments have the same section and $3 m$ effective length, hence the same buckling resistance. Use the weak-axis radius $4.20 cm = 42 mm$ .

λ = \frac{3,000}{42} = 71.428571

\frac{λ}{x} = \frac{71.428571}{33.9} = 2.10704

v = {[1 + 0.05 {(\frac{71.428571}{33.9})}^{2}]}^{- 0.25} = 0.951117

For Class 1,

β_{w} = 1

λ_{LT} = 0.877 \times 0.951117 \times 71.428571 \times \sqrt{1} = 59.580686

Table 8.3a,

p_{y} = 355

column:

55

row gives

274

，

60

row gives

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

f = \frac{59.580686 - 55}{60 - 55} = 0.916137

p_{b} = 274 + 0.916137 (257 - 274) = 258.425666 \frac{N}{{mm}^{2}}

M_{b} = \frac{258.425666 \times 1,653}{1,000} = 427.177626 kN \cdot m

Lecturer route: approximate BC as straight, $β = \frac{342}{419} \approx 0.816$ ; read mᴸᵀ≈0.93, giving $0.93 \times 419 = 389.67 kN \cdot m$ . Rounded λᴸᵀ≈59.5 gives pᵦ≈259 and Mᵦ≈428 kNm. It passes. To include the UDL exactly, use quarter-point magnitudes from $M (x)$ above:

Segment	Quarter positions from A ( $m$ )	$M_{2}$ 、 $M_{3}$ 、 $M_{4}$ （ $kN \cdot m$ ）	$M_{\max}$ （ $kN \cdot m$ ）	$m^{LT} M_{\max}$ （ $kN \cdot m$ ）
$AB$	$0.75, 1.5, 2.25$	$107.99375, 213.625, 316.89375$	$417.8$	$254.105625$
$BC$	$3.75, 4.5, 5.25$	$402.34375, 384.525, 364.34375$	$417.8$	$390.825625$
$CD$	$6.75, 7.5, 8.25$	$259.89375, 175.625, 88.99375$	$341.8$	$208.505625$

Segment BC:

M (3.75) = 145.566667 \times 3.75 - 2.1 \times {3.75}^{2} - 152 \times 0.75 = 402.34375

M (4.5) = 145.566667 \times 4.5 - 2.1 \times {4.5}^{2} - 152 \times 1.5 = 384.525

M (5.25) = 145.566667 \times 5.25 - 2.1 \times {5.25}^{2} - 152 \times 2.25 = 364.34375

m_{LT} = 0.2 + \frac{[0.15 \times 402.34375 + 0.5 \times 384.525 + 0.15 \times 364.34375]}{417.8}

= 0.935437 > 0.44

Equivalent moment demand:

0.2 \times 417.8 + 0.15 \times 402.34375 + 0.5 \times 384.525 + 0.15 \times 364.34375

= 390.825625 kN \cdot m < 427.177626

Passes.

For AB insert its three ordinates into the same expression: $0.2 \times 417.8 + 0.15 \times 107.99375 + 0.5 \times 213.625 + 0.15 \times 316.89375 = 254.105625$ . For CD: $0.2 \times 341.8 + 0.15 \times 259.89375 + 0.5 \times 175.625 + 0.15 \times 88.99375 = 208.505625$ . Their $m$ factors also exceed $0.44$ . BC governs and all three pass.

Animation labA beam bends sideways and twists1 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.

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.

8. The omitted local web checks: what can and cannot be concluded

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

Page 30 asks students to check bearing and web buckling, but neither the support bearing geometry nor the load-introduction details at B and C are supplied. A lateral-restraint cross does not specify the stiff bearing length or rotational restraint of the loaded flange relative to the web. Therefore the numerical local-web follow-up remains incomplete from the available source data.

Known section values:

t = 9 mm

p_{yw} = 355

k = T + r = 14.5 + 10.2 = 24.7 mm

. At A:

n = \min (2 + \frac{0.6 b_{e, A}}{24.7}, 5)

P_{bw, A} = \frac{(b_{1, A} + 24.7 n) \times 9 \times 355}{1,000}

= 3.195 (b_{1, A} + 24.7 n) kN

Require

\geq 145.5667 kN

. At D, use its actual bearing inputs and require

\geq 120.2333 kN

. If B/C are internal bearing locations,

n = 5

, giving

P_{bw} = 3.195 (b_{1} + 123.5) kN

; compare with the point force actually transferred at that contact.

Once $b_{1}$ , $b_{e}$ , $a_{e}$ and both local restraint conditions are known, evaluate $P_{x}$ using $d = 407.6 mm$ and the correct end/interior branch in the web-buckling procedure. If local restraint is absent, Lᴱ,web is also needed. Do not copy Example 1’s $160 mm$ , $80 mm$ and $0 mm$ values into this different diagram. The source’s overall “adequate” statement is supported for shear, section bending and LTB, but local-web adequacy is not established by the supplied information.

Animation labSpread a concentrated force into the web3 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.

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.

9. Complete the omitted off-centre deflection calculation

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

Use only the locations $x = 3 m$ and their imposed loads $60 kN$ , and $x = 6 m$ of $30 kN$ ; there is no imposed UDL. This calculation uses $N$ , $mm$ , constant $E = 205,000$ , and exact section-table value $I_{x} = 33,320 \times 10^{4} = 333,200,000 {mm}^{4}$ . Imposed-load reactions are $R_{A} = \frac{60 \times 6 + 30 \times 3}{9} = 50 kN$ , $R_{D} = 40 kN$ .

The general method starts from the curvature relationship. Taking downward deflection $v$ as positive, $E I v^{″} = - M$ . Integrate the bending moment twice. Define ${⟨ x - a ⟩}^{n}$ as follows: when $x < a$ , its value is zero; when $x \geq a$ , its value is ${(x - a)}^{n}$ . Use the support conditions $v (0) = v (9,000) = 0$ to determine the integration constants.

E I v (x) = C x - \frac{50,000 x^{3}}{6} + \frac{60,000 {⟨ x - 3,000 ⟩}^{3}}{6} + \frac{30,000 {⟨ x - 6,000 ⟩}^{3}}{6}

C = \frac{[50,000 \times {9,000}^{3} - 60,000 \times {6,000}^{3} - 30,000 \times {3,000}^{3}]}{6 \times 9,000}

= 420,000,000,000 N \cdot {mm}^{2}

E I v^{'} (x) = C - \frac{50,000 x^{2}}{2} + \frac{60,000 {⟨ x - 3,000 ⟩}^{2}}{2} + \frac{30,000 {⟨ x - 6,000 ⟩}^{2}}{2}

Maximum downward deflection occurs where slope $v^{'} = 0$ . It lies between B and C: slope numerator at $3,000$ is $+ 195 \times 10^{9}$ ; at $6,000$ it is $- 210 \times 10^{9}$ , and the positive bending moment makes slope decrease continuously.

In the interval

3,000 < x < 6,000

0 = 420 \times 10^{9} - 25,000 x^{2} + 30,000 {(x - 3,000)}^{2}

= 5,000 x^{2} - 180,000,000 x + 690 \times 10^{9}

x = \frac{[180,000,000 - \sqrt{{180,000,000}^{2} - 4 \times 5,000 \times 690 \times 10^{9}}]}{2 \times 5,000}

= 4,361.818 mm

The other root is outside this interval and is rejected.

v_{\max} = \frac{[420 \times 10^{9} \times 4,361.818 - \frac{50,000 \times {4,361.818}^{3}}{6} + \frac{60,000 \times {1,361.818}^{3}}{6}]}{205,000 \times 333,200,000}

= 17.06545 mm

At midspan the deflection is $17.04645 mm$ , close but not the exact maximum. The source does not specify the finish category for this example. If the brittle-finish $\frac{L}{360}$ criterion used in preceding examples is adopted explicitly, the limit is $\frac{9,000}{360} = 25 mm$ and $17.06545 < 25$ passes. The calculated deflection is determined; the acceptance statement is conditional on that chosen serviceability limit.

Animation labSee stiffness and deflection1 concept · 15 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.

Compact exam answer

$457 \times 191 \times 74$ UB S355 is Class 1. Rₐ/Rᴅ= $145.5667$ / $120.2333 kN$ ; Mᴮ/Mᶜ= $417.8$ / $341.8 kN \cdot m$ . $V_{c} = 842.998 kN$ ; $M_{c} = 586.815 kN \cdot m$ . All segments Lᴱ=3 m: λᴸᵀ=59.5807,pᵦ=258.426,Mᵦ=427.178 kNm. Exact equivalent demands AB/BC/CD $= 254.106$ / $390.826$ / $208.506 kN \cdot m$ ; all pass, BC controls. Imposed $δ_{\max} = 17.065 mm$ at $4.362 m$ ; passes if $\frac{L}{360}$ applies. Local bearing/web-buckling numerical checks are unresolved because bearing and local-restraint data are missing.

Mistakes to avoid

Do not turn B and C into vertical supports.
Do not use $3 m$ as the span for the whole-beam deflection.
Do not omit BC end moments when evaluating its quarter points.
Do not present assumed bearing dimensions as given.

Procedure for an unfamiliar variant

Identify the vertical supports, span, lateral restraints and individual load positions.
Keep dead and imposed loads separate; form the required ultimate and serviceability cases.
Find reactions and the maximum shear/moment by equilibrium, with units.
Choose a section-table row, confirm thickness-dependent strength, and classify both flange and web.
Check shear and bending; add segment LTB where restraint is discrete.
Complete the requested web and deflection checks; state the governing result and any missing data.

Independent self-check

Try it yourself. Invented variant: with all loads unchanged, the lateral restraint at C is removed. Which parts of this solution must be recalculated?

Reveal answer and reasoning

Vertical reactions, whole-beam SFD/BMD, section shear/bending and elastic deflection do not change. LTB segments become $AB = 3 m$ and $BD = 6 m$ . For BD use Lᴱ=6,000 mm (if the same effective-length rule is stipulated), recompute $λ$ , $v$ ,λᴸᵀ,pᵦ and Mᵦ, and use quarter positions $4.5, 6, 7.5 m$ with the true BD maximum atB. The point load atC remains even though its lateral restraint is removed.

Animation labSee stiffness and deflection4 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.

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.

Beam example 4: a tight LTB check with three point loads

Open this example on its own page

Select/check a S355 simply supported beam of $5 m$ span carrying three characteristic point loads, each $22 kN$ dead plus $12 kN$ imposed, and a $2 \frac{kN}{m}$ dead UDL. Loading is normal. Ends are restrained against torsion, with compression flange free to rotate in plan; there is no intermediate lateral restraint.

Original source: LectureNotes/Ch 3_Beam.pdf — p. 31, p. 32, p. 33. 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	Source/use
Span and load positions	Given dimension chain; central point is at $2.5 m$ .
Lateral model	The question specifies normal loading, end torsional restraint and free in-plane rotation, so $L^{E} = L = 5,000 mm$ .
Trial section	The source properties correspond to $457 \times 152 \times 52$ UB: $D = 449.8$ , $t = 7.6$ , $T = 10.9 mm$ , $\frac{b}{T} = 6.99$ , $\frac{d}{t} = 53.6$ , $r_{y} = 3.11 cm$ , $Z_{x} = 950$ , $S_{x} = 1,096 {cm}^{3}$ , $u = 0.859$ , $x = 43.9$ .
Table source	Data File pp.9–10 exact section row; p.5 Table 8.3a strength; p.6 moment factors.

Before calculating: recognition and strategy

Symmetry makes analysis simple but does not provide lateral restraint. Check the entire $5 m$ length for LTB. Because there are several point loads plus a UDL, the general quarter-point moment factor is appropriate. The result is close to the resistance, so use actual $u$ and $v$ and interpolate without early rounding.

1. Ultimate loading

Each point load:

P = 1.4 \times 22 + 1.6 \times 12 = 30.8 + 19.2 = 50 kN

w = 1.4 \times 2 = 2.8 \frac{kN}{m}

Total load:

3 \times 50 + 2.8 \times 5 = 164 kN

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.

2. Reactions, moment and quarter-point ordinates

Simple explanation: Why reactions balance the loads

Think of a seesaw that must neither fall nor turn.

Upward and downward forces must balance.
Take moments about a support to eliminate its reaction.
Use perpendicular distance from the point to the force line.

Remember: A lateral restraint is not automatically a vertical support.

Related concept and full method

R_{A} = R_{B} = \frac{164}{2} = 82 kN

Immediately left of midspan:

V = 82 - 2.8 \times 2.5 - 50 = 25 kN

Immediately right of midspan:

V = 25 - 50 = - 25 kN

Therefore the maximum moment occurs at midspan.

M_{\max} = 82 \times 2.5 - \frac{2.8 \times {2.5}^{2}}{2} - 50 (2.5 - 1)

= 205 - 8.75 - 75 = 121.25 kN \cdot m

In the interval

\frac{1}{4}

of the span, namely

x = 1.25 m

M_{2} = 82 \times 1.25 - \frac{2.8 \times {1.25}^{2}}{2} - 50 (1.25 - 1)

= 102.5 - 2.1875 - 12.5 = 87.8125 kN \cdot m

M_{3} = 121.25

; by symmetry,

M_{4} = 87.8125 kN \cdot m

The lower three-quarter point is at $3.75 m$ . The equality to $M_{2}$ follows from the symmetric loading; alternatively include both loads to its left in the free body. No moment contribution is assigned to a load lying to the right of the cut.

Animation labRead the moment shape within one segment2 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.

Moment ordinates must belong to the same effective unbraced segment.
Same-side and reverse-curvature diagrams have different signed end ratios.
The markers show ¼, ½ and ¾ of this segment, not of the entire beam.
LTB and column flexural factors are different. Preserve their individual bounds and coefficient sets.

3. Trial section and section moment capacity

Preliminary section-modulus guide:

S \geq \frac{121.25 \times 10^{6}}{355} = 341,549.3 {mm}^{3} = 341.549 {cm}^{3}

The lecturer selects

457 \times 152 \times 52

, whose

S_{x} = 1,096 {cm}^{3}

, because LTB is more restrictive than yielding.

T = 10.9 \leq 16 mm \to p_{y} = 355 \frac{N}{{mm}^{2}}

M_{c} = \min (\frac{355 \times 1,096}{1,000}, \frac{1.2 \times 355 \times 950}{1,000})

= \min (389.08, 404.7) = 389.08 kN \cdot m > 121.25

A small section satisfying M/pᵧ alone would not necessarily pass LTB. The chosen section is a verified trial, not a proof that every lighter section fails.

Animation labCompression and tension across a section2 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.

For sagging, the top flange is in compression and the bottom in tension; hogging reverses this.
Elastic bending stress varies with distance from the neutral axis: $σ = \frac{M y}{I}$ .
The section class governs whether elastic, plastic or effective properties may be used.
Use the shear at the section under examination; the largest shear elsewhere is not automatically coexistent.

4. Section class and shear-web threshold

ε = \sqrt{\frac{275}{355}} = 0.880141

\frac{b}{T} = 6.99 < 9 ε = 7.92127; \frac{d}{t} = 53.6 < 80 ε = 70.4113

is Class 1.

\frac{d}{t} = 53.6 < 70 ε = 61.6099

No separate web shear-buckling calculation is needed.

Animation labWhy thin elements buckle locally2 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 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.

5. Shear and low-shear bending applicability

A_{v} = 7.6 \times 449.8 = 3,418.48 {mm}^{2}

V_{c} = \frac{355 \times 3,418.48}{\sqrt{3} \times 1,000} = 700.649 kN

V_{\max} = 82 kN < 700.649

: shear passes. Shear at the same midspan section

25 kN < 0.6 \times 700.649 = 420.390 kN

, so the low-shear

M_{c}

above applies.

Animation labSee shear in the web3 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.

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.

6. Equivalent moment from Table 8.4b

Simple explanation: Why the shape of the moment diagram matters

The same peak moment is more demanding when a long region stays near that peak.

Use the diagram of this unrestrained segment.
Choose the applicable course sketch or quarter-point rule.
Apply its factor to the demand, following the stated check.

Remember: Column flexural-buckling factors and LTB factors are not interchangeable.

Related concept and full method

m_{LT} = 0.2 + \frac{0.15 \times 87.8125 + 0.5 \times 121.25 + 0.15 \times 87.8125}{121.25}

= 0.917268 > 0.44

Equivalent moment:

m_{LT} M_{\max}

= 0.2 \times 121.25 + 0.3 \times 87.8125 + 0.5 \times 121.25

= 111.21875 kN \cdot m

This factor applies to member LTB only. The cross-section still must carry the actual $121.25 kN \cdot m$ .

Animation labRead the moment shape within one segment2 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.

Moment ordinates must belong to the same effective unbraced segment.
Same-side and reverse-curvature diagrams have different signed end ratios.
The markers show ¼, ½ and ¾ of this segment, not of the entire beam.
LTB and column flexural factors are different. Preserve their individual bounds and coefficient sets.

7. Effective length and equivalent slenderness

Simple explanation: Which length belongs in the calculation?

A sideways tie can shorten the buckling region without shortening the beam’s vertical span.

Separate vertical-support spacing from lateral-restraint spacing.
Apply the course rule for the actual end restraint and loading.
Keep the original vertical load analysis unless its supports change.

Remember: Restraints in one direction may not restrain the other direction.

Related concept and full method

L_{E} = 5,000 mm; r_{y} = 3.11 cm = 31.1 mm

λ = \frac{5,000}{31.1} = 160.771704

\frac{λ}{x} = \frac{160.771704}{43.9} = 3.662226

v = {[1 + 0.05 {(3.662226)}^{2}]}^{- 0.25} = 0.879594

u = 0.859; β_{w} = 1

λ_{LT} = 0.859 \times 0.879594 \times 160.771704 \times \sqrt{1} = 121.474465

The source rounds $λ$ to $160.8$ and $v$ to $0.880$ , obtaining $121.6$ . Both approaches pass, but the exact values avoid losing the small margin in this example.

Animation labA beam bends sideways and twists2 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.

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.

8. Interpolate strength and compare

Simple explanation: A beam can escape sideways

The compressed flange can move sideways while the section twists.

Divide the beam at effective lateral restraints.
Use each segment’s effective length to obtain its buckling resistance.
Compare that resistance with the segment’s equivalent moment demand.

Remember: A section bending check alone does not check lateral-torsional buckling.

Related concept and full method

Table 8.3a: S355，

p_{y} = 355

column.

λ_{LT} = 120

p_{b} = 104

;

125

p_{b} = 97 \frac{N}{{mm}^{2}}

. Interpolation fraction:

f = \frac{121.474465 - 120}{125 - 120} = 0.294893

p_{b} = 104 + 0.294893 (97 - 104) = 101.935748 \frac{N}{{mm}^{2}}

M_{b} = p_{b} S_{x} = \frac{101.935748 \times 1,096}{1,000} = 111.721580 kN \cdot m

Demand

111.218750 <

resistance

111.721580

: LTB passes. Margin:

111.721580 - 111.218750 = 0.502830 kN \cdot m

Approximately the following fraction of resistance:

0.45 %

The lecturer’s rounded route gives pᵦ=101.8 and Mᵦ=111.6 kNm against $111.2$ , also a pass. Do not claim a large reserve. The illustrated solution checks shear, section bending and LTB; bearing dimensions are absent. The characteristic imposed load arrangement is known, but a finish limit is not specified in this example’s question. Those additional design conditions must be supplied before a blanket full-design conclusion.

Animation labA beam bends sideways and twists3 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.

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

Use S355 $457 \times 152 \times 52$ UB, Class 1. Each $P = 50 kN$ ， $w = 2.8 \frac{kN}{m}$ ， $R = 82 kN$ ， $M_{\max} = 121.25 kN \cdot m$ . $V_{c} = 700.649 kN$ ， $M_{c} = 389.08 kN \cdot m$ . $L^{E} = 5 m$ ; $m^{LT} = 0.917268$ , demand $111.21875 kN \cdot m$ . $λ = 160.7717$ , $v = 0.879594$ , $λ^{LT} = 121.4745$ ; interpolate $p_{b} = 101.93575 \frac{N}{{mm}^{2}}$ . $M_{b} = 111.72158 kN \cdot m > 111.21875$ ; LTB passes by only a small margin.

Mistakes to avoid

Point-load positions do not imply intermediate restraints.
Use quarter points $1.25, 2.5, 3.75 m$ , not the load positions $1, 2.5, 4 m$ .
Read pᵦ using λᴸᵀ, not $λ$ .
Do not round a narrow pass into an unsupported generous margin.

Procedure for an unfamiliar variant

Identify the vertical supports, span, lateral restraints and individual load positions.
Keep dead and imposed loads separate; form the required ultimate and serviceability cases.
Find reactions and the maximum shear/moment by equilibrium, with units.
Choose a section-table row, confirm thickness-dependent strength, and classify both flange and web.
Check shear and bending; add segment LTB where restraint is discrete.
Complete the requested web and deflection checks; state the governing result and any missing data.

Independent self-check

Try it yourself. Invented variant: all ultimate loads increase by $1 %$ , with geometry and section unchanged. Does the LTB check still pass?

Reveal answer and reasoning

The moment shape and mᴸᵀ stay unchanged; resistance stays $111.72158 kN \cdot m$ . Demand becomes $111.21875 \times 1.01 = 112.33094 kN \cdot m$ , exceeding resistance. The beam fails LTB even though section bending and shear have ample reserve.

Animation labRead the moment shape within one segment2 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.

Moment ordinates must belong to the same effective unbraced segment.
Same-side and reverse-curvature diagrams have different signed end ratios.
The markers show ¼, ½ and ¾ of this segment, not of the entire beam.
LTB and column flexural factors are different. Preserve their individual bounds and coefficient sets.

Animation labFollow the calculation sequence5 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.

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.

Practise this chapter: tutorials, assignments and past-paper answers →

Beams: loads, section checks and lateral stability

1. Identify what is restrained

2. Shear capacity and web slenderness

3. Bending capacity at low shear

4. What changes when shear is high?

5. A concentrated force can crush the web locally

6. The web can buckle as a short compression member

7. Select and check a fully restrained beam

Read the diagram and collect the data

Before calculating: recognition and strategy

1. Factor each load component

2. Reactions, shear and maximum moment

3. Select a trial section and read the correct columns

4. Establish the section class and web-shear branch

5. Resist the largest shear

6. Check coexistent shear, then bending

7. Support web bearing

8. Support web buckling

9. Deflection under the specified serviceability case

Compact exam answer

Mistakes to avoid

Procedure for an unfamiliar variant

Independent self-check

Read the diagram and collect the data

Before calculating: recognition and strategy

1. Slab to B1

2. B1 reactions to B2

3. Assemble B3 point load and direct slab UDL

4. Reactions, shear and moment of B3

5. Select 457×152×52457×152×52 UB

6. Classify the trial section

7. Check shear resistance

8. Check moment resistance

9. Check B3 serviceability

Compact exam answer

Mistakes to avoid

Procedure for an unfamiliar variant

Independent self-check

8. Normal versus destabilizing loads

9. Select the LTB effective length from restraints

10. Convert unbraced length to a resistance

11. Use the moment shape of each unbraced segment

12. Put the checks together

Read the diagram and collect the data

Before calculating: recognition and strategy

1. Factored loads

2. Reactions without premature rounding

3. Construct the SFD and BMD from the left-hand free body

4. Confirm strength and section class

5. Shear resistance

6. Actual section moment resistance

7. Check every unbraced segment

8. The omitted local web checks: what can and cannot be concluded

9. Complete the omitted off-centre deflection calculation

Compact exam answer

Mistakes to avoid

Procedure for an unfamiliar variant

Independent self-check

Read the diagram and collect the data

Before calculating: recognition and strategy

1. Ultimate loading

2. Reactions, moment and quarter-point ordinates

3. Trial section and section moment capacity

4. Section class and shear-web threshold

5. Shear and low-shear bending applicability

6. Equivalent moment from Table 8.4b

7. Effective length and equivalent slenderness

8. Interpolate strength and compare

Compact exam answer

Mistakes to avoid

Procedure for an unfamiliar variant

Independent self-check

5. Select $457 \times 152 \times 52$ UB