Reference

Companion slides and quick-reference notes

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

These are teaching notes for every slide in the four companion presentations. Each original slide remains readable locally. The linked chapter supplies the full derivation and worked applications; corrections to compressed slide statements are explicit.

PPT/Structural Steelwork Design L1.pptx · slide 1: Introduction: what you are learning

A design answer compares a demand caused by loads with a resistance supplied by steel. First calculate the loads and forces; then check the chosen component. A failure of any required check means the proposal needs revision.

Study the full explanation and applications.

Original slide 1: Introduction: what you are learning. No dimensions are inferred from slide scale. — Original slide 1, introduction: what you will learn. The title is Structural Steelwork Design; Chapter 1 covers foundations, steel metallurgy and HK Code 2011. The module is CON4334 Civil Engineering and the lecturer is credited as King Lai. Do not infer dimensions from slide scale.Open full-size image

Animation labFollow the calculation sequence1 concept

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.

PPT/Structural Steelwork Design L1.pptx · slide 2: The course roadmap

Learn foundations before connections, beams and columns. The same equilibrium equations recur: vertical forces balance, horizontal forces balance, and moments balance. The shape and restraints decide which resistance checks are needed.

Study the full explanation and applications.

Original slide 2: The course roadmap. No dimensions are inferred from slide scale. — Original slide 2, Learning Roadmap. The five topics are: materials and metallurgy (composition, alloys, fatigue, corrosion and fire protection); steel sections (hot-rolled UB/UC, built-up and cold-formed); limit state design (ULS/SLS and partial safety factors); analysis and initial imperfections ( $P - Δ$ effects and $0.5 %$ notional horizontal forces, NHF); and classification/deflection limits (plastic through slender classes, SLS Table 5.1). Do not infer dimensions from slide scale.Open full-size image

Animation labFollow the calculation sequence12 concepts

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.

PPT/Structural Steelwork Design L1.pptx · slide 3: Steel behaviour and protection

The slide describes steel as mostly iron, with controlled carbon and manganese. Carbon raises strength but can reduce ductility and weldability. Fatigue is repeated-load crack growth; corrosion removes material; heat reduces stiffness and strength. For an exam explanation, link each protection method to the damage mechanism it prevents.

Study the full explanation and applications.

Original slide 3: Steel behaviour and protection. No dimensions are inferred from slide scale. — Original slide 3, Material & Metallurgy. The left column lists composition: approximately $98 %$ iron, up to $0.25 %$ carbon, up to $1.6 %$ manganese, plus silicon, phosphorus and sulphur. Higher carbon increases strength but reduces ductility and weldability. The right column says to avoid stress concentrations and abrupt section changes for fatigue; corrosion protection includes galvanising, weathering steel and cathodic protection; fire protection uses solid, hollow or profile casing. The percentages are the slide's statements, not complete material specifications for every grade. Do not infer dimensions from slide scale.Open full-size image

Animation labElasticity, yielding and ductility4 concepts

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

Stress is force divided by area. Strain measures change in length relative to original length.
Before yielding, stress is approximately $E \times strain$ . E controls the elastic slope.
Further strain includes permanent deformation. Higher yield strength does not by itself increase E.
Strength, stiffness and ductility answer different questions. This schematic is not a measured stress–strain curve.

PPT/Structural Steelwork Design L1.pptx · slide 4: Recognise the section

An I or H section has two flanges and a web; angles and channels are open, unsymmetrical shapes. Rolled means formed hot at a mill. Built-up means plates or sections joined into a member. Identify the section family before choosing dimensions, axes or buckling curves.

Study the full explanation and applications.

Original slide 4: Recognise the section. No dimensions are inferred from slide scale. — Original slide 4, Steel Section Types. Hot-rolled UBs mainly resist major-axis bending; UCs accommodate larger axial compression. CHS/SHS/RHS hollow sections are efficient in compression and torsion. Built-up and compound sections include plate girders, crane girders, and battened/laced members. Cold-formed thin-walled Z, sigma and lipped channel sections can serve as purlins. Do not infer dimensions from slide scale.Open full-size image

Animation labExplore section geometry and axes1 concept

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

The flanges are the wide plates; the web connects them. Rotate the I-section to see both.
The same section has different stiffness and resistance about its two principal axes.
I controls elastic curvature; $Z = \frac{I}{y _{max}}$ is elastic section modulus. Plastic modulus S comes from plastic stress blocks.
Nominal section labels are not every actual dimension. Keep the row, axis and units together.

PPT/Structural Steelwork Design L1.pptx · slide 5: ULS versus SLS

ULS checks strength and stability using the prescribed factored load combination. SLS checks usable behaviour such as deflection. Do not carry the ULS load into an imposed-load-only deflection check. $G$ is permanent load, $Q$ imposed load and $W$ wind action.

Study the full explanation and applications.

Original slide 5: ULS versus SLS. No dimensions are inferred from slide scale. — Original slide 5, Limit State Design Philosophy. The left column describes ULS protection against collapse, yielding and fracture; $γ_{f} > 1.0$ denotes the partial load factor, and $γ_{m}$ the material factor reducing nominal strength. The right column describes everyday serviceability, controlling excessive deflection, floor vibration and dynamic sway. It lists unfactored service loads $γ_{f} = 1.0$ . Factors must still be selected for the actual combination and course rules; the left-column summary is not a universal rule for all favourable/unfavourable loads. Do not infer dimensions from slide scale.Open full-size image

Animation labTwo different design questions1 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.

ULS compares factored actions with the applicable resistance to yielding, rupture or instability.
SLS checks movement, vibration or another specified use requirement.
A beam can be strong enough but deflect too much. The two checks need their own loads and denominators.
A resistance pass cannot stand in for a serviceability pass. Complete all requested checks.

PPT/Structural Steelwork Design L1.pptx · slide 6: Choose a load combination

The slide lists $1.4 G + 1.6 Q$ , $1.4 G + 1.4 W$ and $1.2 G + 1.2 Q + 1.2 W$ . These are different cases, not extra multipliers applied one after another. Choose the relevant adverse combination and consider whether an action is beneficial. For example, $G = 4$ and $Q = 3 \frac{kN}{m^{2}}$ gives $1.4 \times 4 + 1.6 \times 3 = 10.4 \frac{kN}{m^{2}}$ .

Study the full explanation and applications.

Original slide 6: Choose a load combination. No dimensions are inferred from slide scale. — Original slide 6, Factored Load Combinations (HK Code 2011). The three source expressions are: $1.4 dead load + 1.6 imposed load$ ; $1.4 dead load + 1.4 wind load$ ; $1.2 dead load + 1.2 imposed load + 1.2 wind load$ . Dead, imposed and wind loads correspond respectively to $G$ ／ $Q$ ／ $W$ above. Do not infer dimensions from slide scale.Open full-size image

Animation labFrom characteristic to design load1 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.

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.

PPT/Structural Steelwork Design L1.pptx · slide 7: Choose the analysis model

Simple construction treats joints as nominally pinned for global analysis; continuous construction transfers joint moments. First-order analysis uses the undeformed shape; second-order analysis includes extra moments caused by displacement under axial force. Do not infer moment fixity from a visually thick line in a sketch.

Study the full explanation and applications.

Original slide 7: Choose the analysis model. No dimensions are inferred from slide scale. — Original slide 7, Methods of Structural Analysis. Simple design uses pinned connections with cores/bracing resisting lateral loads; continuous design uses rigid moment-resisting connections. First-order linear-elastic analysis ignores geometric deformation; second-order analysis includes geometric nonlinearity from overall sway $P - Δ$ and member bending $P - δ$ . The slide identifies HK Code 2011. Do not infer dimensions from slide scale.Open full-size image

Animation labRestraints, sway and imperfections2 concepts

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

A frame needs a defined path for horizontal force as well as gravity.
Pinned joints alone do not provide frame moment resistance; sway restraint needs a real structural system.
Diagonal bracing carries horizontal action through axial forces. This sketch does not assign a numerical frame classification.
Use the specified imperfection/notional-force model and critical-load criteria. Avoid counting alternative imperfection models twice.

PPT/Structural Steelwork Design L1.pptx · slide 8: Imperfections and notional forces

Real members are not perfectly straight or vertical. The slide uses a nominal frame imperfection $\frac{1}{200}$ and notional horizontal force $0.005$ times vertical load. For a $2000 kN$ vertical action, that force is $0.005 \times 2000 = 10 kN$ . Apply it in the stated analysis case; it does not replace actual wind automatically. Member initial bow and frame lean are different imperfections.

Study the full explanation and applications.

Original slide 8: Imperfections and notional forces. No dimensions are inferred from slide scale. — Original slide 8, Initial Imperfections & NHF. The slide says a structure is not $100 %$ perfectly vertical. The reference angle for overall inclination is $φ = \frac{1}{200} rad$ ; NHF acts horizontally and is at least the following fraction of total factored vertical load: $0.5 %$ ( $0.005$ ). Initial member bow is represented by $\frac{e_{0}}{L}$ . Do not infer dimensions from slide scale.Open full-size image

Animation labRestraints, sway and imperfections2 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 frame needs a defined path for horizontal force as well as gravity.
Pinned joints alone do not provide frame moment resistance; sway restraint needs a real structural system.
Diagonal bracing carries horizontal action through axial forces. This sketch does not assign a numerical frame classification.
Use the specified imperfection/notional-force model and critical-load criteria. Avoid counting alternative imperfection models twice.

PPT/Structural Steelwork Design L1.pptx · slide 9: Serviceability limits

For imposed-load vertical deflection, use the applicable row: brittle finishes $\frac{L}{360}$ , other beams $\frac{L}{200}$ , cantilevers $\frac{L}{180}$ . The slide also mentions crane $\frac{L}{600}$ and frame $\frac{H}{500}$ ; the full course table states the conditions, so a limit must be selected by function and displacement type. A span of $7200 mm$ at $\frac{L}{360}$ permits $20 mm$ .

Study the full explanation and applications.

Original slide 9: Serviceability limits. No dimensions are inferred from slide scale. — Original slide 9, SLS Deflection Limits (Table 5.1). The four rows are: beams supporting plaster/finishes, $deflection \leq \frac{L}{360}$ ; general beams, $deflection \leq \frac{L}{200}$ ; crane girders, $deflection \leq \frac{L}{600}$ ; building-frame sway, $sway \leq \frac{H}{500}$ . This reproduces the source summary table; actual uses and load conditions must follow the full table. Do not infer dimensions from slide scale.Open full-size image

Animation labSee stiffness and deflection1 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.

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.

PPT/Structural Steelwork Design L1.pptx · slide 10: Strength depends on thickness

S355 is a grade name, not a promise that every thickness has design strength $355 \frac{N}{{mm}^{2}}$ . Read the maximum-thickness row: a $22 mm$ flange falls in the $\leq 40 mm$ row, giving $345 \frac{N}{{mm}^{2}}$ . Then $ε = \sqrt{\frac{275}{345}} = 0.892804$ . Use the controlling plate thickness for the relevant property check.

Study the full explanation and applications.

Original slide 10: Strength depends on thickness. No dimensions are inferred from slide scale. — Original slide 10, Steel Grades & Design Strength. The left column lists BS EN grades S275, S355 and S460, with the shorthand $p_{y} = 275, 355, 460 \frac{N}{{mm}^{2}}$ ; it also lists GB grades Q235, Q345, Q390 and Q420. The right column says increasing $t$ reduces $p_{y}$ , and gives the width/thickness parameter $ε = \sqrt{\frac{275}{p_{y}}}$ . The left-column unit has a question-mark-like superscript. Here it is transcribed as $\frac{N}{{mm}^{2}}$ using the strength table from the same course, while retaining the source image. The left-column shorthand is not a universal lookup value across all thicknesses. Do not infer dimensions from slide scale.Open full-size image

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.

PPT/Structural Steelwork Design L1.pptx · slide 11: Section classes

Class 1 permits a plastic hinge with rotation; Class 2 reaches plastic moment with less rotation; Class 3 uses elastic capacity; Class 4 requires effective properties. Check flange and web separately under their actual stress distributions, then take the worse class. A stocky flange does not make a slender web Class 1.

Study the full explanation and applications.

Original slide 11: Section classes. No dimensions are inferred from slide scale. — Original slide 11, Cross-Section Classification. Class 1, Plastic: a plastic hinge has sufficient rotation capacity. Class 2, Compact: the section can reach plastic moment $M_{p}$ , but local buckling limits rotation. Class 3, Semi-Compact: the extreme fibre reaches $p_{y}$ , but local buckling prevents full plastic moment. Class 4, Slender: local buckling occurs before yielding. Do not infer dimensions from slide scale.Open full-size image

Animation labWhy thin elements buckle locally1 concept

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.

PPT/Structural Steelwork Design L1.pptx · slide 12: Turn discussion into a self-test

Without notes, explain why a beam can pass bending yet fail LTB, why a column has two slenderness ratios, and why bolt diameter differs from hole diameter. Answers: instability depends on unrestrained length; radii/restraints differ by axis; holes need clearance. Return to the linked chapters if you cannot explain the mechanism.

Study the full explanation and applications.

Original slide 12: Turn discussion into a self-test. No dimensions are inferred from slide scale. — Original slide 12, Questions & Discussion: no calculation question or dimension is given. The discussion topics are turned into self-checks above and are not presented as original exam questions. Do not infer dimensions from slide scale.Open full-size image

Animation labA beam bends sideways and twists3 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.

PPT/Joint Connections L2.pptx · slide 1: Joint connections: purpose

A connection transfers forces between members. Sketch the complete force path before calculating an individual bolt or weld; a strong fastener cannot rescue a weak connected plate.

Study the full explanation and applications.

Original slide 1: Joint connections: purpose. No dimensions are inferred from slide scale. — Original slide 1: Steelwork Design: Joint Connections; Module CON4334, Chapter 2; Bolted & Welded Connections. Do not infer dimensions from slide scale.Open full-size image

Animation labCount the bolt shear planes3 concepts

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

Load must cross an interface between the connected plates.
A lap joint gives one shear plane through a bolt.
A symmetric double-cover joint may provide two shear planes. Count load-transfer interfaces, not just visible plates.
Use the source’s area and shear strength. Then check bearing, plate resistance and detailing separately.

PPT/Joint Connections L2.pptx · slide 2: Bolts and welds

Bolts transfer force through shear, bearing, tension, or specified friction-grip action. Welds transfer force through fused metal along an effective throat. This course’s ordinary-bolt equations do not establish slip resistance for a friction-grip joint.

Study the full explanation and applications.

Original slide 2: Bolts and welds. No dimensions are inferred from slide scale. — Original slide 2, connection fundamentals: bolts and welds. Their structural purpose is to transfer tension, shear and moment safely between members. Ordinary clearance bolts are listed as Grades 4.6/8.8; high-strength friction-grip (HSFG) bolts are also listed. Bolts permit quick, straightforward site assembly without special equipment. Fillet welds join corners; butt welds join prepared grooves. Welding offers a neat appearance, compact details and structural continuity. Do not infer dimensions from slide scale.Open full-size image

Animation labCount the bolt shear planes5 concepts

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

Load must cross an interface between the connected plates.
A lap joint gives one shear plane through a bolt.
A symmetric double-cover joint may provide two shear planes. Count load-transfer interfaces, not just visible plates.
Use the source’s area and shear strength. Then check bearing, plate resistance and detailing separately.

PPT/Joint Connections L2.pptx · slide 3: Read the bolt layout

Pitch runs along the force transfer direction; gauge separates lines; end distance is from a hole centre to the end in that direction. Hole diameter is used in net area, nominal bolt diameter in projected bearing. Check standard-hole clearance and minimum/maximum spacing before claiming a layout is valid.

Study the full explanation and applications.

Original slide 3: Read the bolt layout. No dimensions are inferred from slide scale. — Original slide 3, hole clearance and layout rules. Standard holes: if $d \leq 24 mm$ ， $hole diameter = d + 2 mm$ ; if $d > 24 mm$ ， $hole diameter = d + 3 mm$ . Minimum pitch along the load direction $2.5 d$ ; minimum transverse spacing perpendicular to the load $3.0 d$ ; maximum spacing $\min (12 t, 150 mm)$ ， $t$ is the thinner plate thickness. End and edge distances prevent end tear-out and provide sufficient bearing area. Do not infer dimensions from slide scale.Open full-size image

Animation labBolt centres, holes and edges1 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.

Hole diameter d₀ differs from nominal bolt diameter d. Net-section deductions use the specified hole.
Pitch runs along the load direction; gauge measures spacing across rows.
End and edge distances start at the hole centre. The remaining ligament starts at the hole boundary.
Minimum spacing, edge distances, grip and plate thickness come from the specified rules, not this scaled illustration.

PPT/Joint Connections L2.pptx · slide 4: Five failure modes

Inspect bolt shear, bolt tension where present, bearing of the bolt/connected part, tensile rupture of the net section and block shear. For each mode draw what separates or deforms. Capacity is the smallest applicable resistance along the force path, not the sum of unrelated failure resistances.

Study the full explanation and applications.

Original slide 4: Five failure modes. No dimensions are inferred from slide scale. — Original slide 4, five core failure modes of bolted connections. (1) Bolt-shank shear: per plane in single/double shear, $P_{s} = p_{s} A_{s}$ ; (2) local bearing at the plate/bolt hole: the slide gives $P_{bb} = d t_{p} p_{bb}$ ; (3) net-section tensile rupture of the connected plate: the slide gives $A_{e} = K_{e} a_{n}$ , still subject to rules including the gross-area cap; (4) end shear tear-out when end distance is insufficient; (5) block shear along the bolt rows with tension across the end line. This five-mode source list uses a different grouping from the comprehensive checklist above. Bolt tension must still be checked separately whenever tension exists. Do not infer dimensions from slide scale.Open full-size image

Animation labSeparate the block-shear paths5 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.

A block containing the connection can detach from the surrounding plate.
The paths parallel to the applied force carry shear.
The closing path across the end of the block carries tension.
Use the specified gross/net deductions and resistance expression. This is different from a single straight net-section fracture.

PPT/Joint Connections L2.pptx · slide 5: Weld types and inspection

A fillet weld joins surfaces at an angle and is sized by leg length. A butt weld joins prepared edges; penetration matters. Visual inspection checks shape and visible defects; penetrant and magnetic-particle methods seek surface defects; ultrasonic and radiographic methods seek internal defects. Strong arithmetic assumes acceptable fabrication.

Study the full explanation and applications.

Original slide 5: Weld types and inspection. No dimensions are inferred from slide scale. — Original slide 5, weld geometry and types. A fillet weld has a triangular section, effective throat $a = 0.7 s$ and effective length $L_{effective} = overall length - 2 s$ . Butt welds may be full- or partial-penetration. The slide equates full-penetration resistance to parent-metal strength, subject to the course welding and inspection conditions. NDT means non-destructive testing; listed methods include visual, penetrant, magnetic-particle, ultrasonic and X-ray inspection. Do not infer dimensions from slide scale.Open full-size image

Animation labFrom fillet leg to effective throat2 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.

An equal-leg fillet between perpendicular plates has an approximately right-triangular section.
For this geometry, throat $a \approx 0.7 s$ . It is shorter than the leg.
Effective resisting area = $a \times L_{eff}$ . With $p_{w} = 220 \frac{N}{mm ^{2}}$ here, capacity per length is $0.7 s \times 220 \frac{N}{mm}$ .
Use the course end allowances, minimum size and length rules; increasing the geometric length alone does not resolve every detailing check.

PPT/Joint Connections L2.pptx · slide 6: The unnumbered 6 mm weld calculation

Given S355 steel and Class 42 electrode, Table 9.2a gives $p_{w} = 250 \frac{N}{{mm}^{2}}$ . A $90 °$ fillet of leg $s = 6 mm$ has throat $a = 0.7 s = 4.2 mm$ . Per millimetre of effective weld, area= $4.2 \times 1 = 4.2 {mm}^{2}$ ; resistance= $4.2 \times 250 = 1050 N = 1.05 kN$ . Thus $q = 1.05 \frac{kN}{mm}$ . Required length for $105 kN$ direct force is $\frac{105}{1.05} = 100 mm$ effective; physical straight length is $100 + 2 \times 6 = 112 mm$ before other detailing constraints.

Study the full explanation and applications.

Original slide 6: The unnumbered 6 mm weld calculation. No dimensions are inferred from slide scale. — Original slide 6, weld design resistance. In the simplified method, resistance per millimetre of weld length is $0.7 s p_{w}$ . The original unnumbered example uses a $6 mm$ weld leg, S355, $p_{w} = 250 \frac{N}{{mm}^{2}}$ , and resistance per unit length $\frac{0.7 \times 6 \times 250}{1000} = 1.05 \frac{kN}{mm}$ . Detailing rules: minimum effective weld length $\max (4 s, 40 mm)$ ; end return at least $2 s$ ; listed lap lengths $5 t$ or $25 mm$ , taking the larger under the full course rule. The $105 kN$ extension calculation above is this website's teaching addition, not another load given on the slide. Do not infer dimensions from slide scale.Open full-size image

Animation labFrom fillet leg to effective throat3 concepts · 11 source expressions

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

An equal-leg fillet between perpendicular plates has an approximately right-triangular section.
For this geometry, throat $a \approx 0.7 s$ . It is shorter than the leg.
Effective resisting area = $a \times L_{eff}$ . With $p_{w} = 220 \frac{N}{mm ^{2}}$ here, capacity per length is $0.7 s \times 220 \frac{N}{mm}$ .
Use the course end allowances, minimum size and length rules; increasing the geometric length alone does not resolve every detailing check.

PPT/Joint Connections L2.pptx · slide 7: Recognise eccentricity in three dimensions

An in-plane force rotates the bolt/weld group about its centroid and gives direct plus torsional shear. An out-of-plane bracket load opens the joint and creates fastener tension. For bolts, check individual shear and tension and $\frac{F_{s}}{P_{s}} + \frac{F_{t}}{P_{nom}} \leq 1.4$ ; $P_{nom} = 0.8 A_{s} p_{t}$ . The interaction limit is $1.4$ , not $1.0$ .

Study the full explanation and applications.

Original slide 7: Recognise eccentricity in three dimensions. No dimensions are inferred from slide scale. — Original slide 7, combined stresses in eccentric connections. Upper half, in-plane: direct shear $F_{s}$ is shared equally; torsional force $F_{T}$ is proportional to radius from the centroid; resultant $F_{R} = \sqrt{F_{s}^{2} + F_{T}^{2} + 2 F_{s} F_{T} \cos (φ)}$ ， $φ$ is the angle between the two forces, written as “phi” in the source. Lower half, out-of-plane: direct shear is shared equally; tension uses the rotation model about the lowest bolt row. The source still uses upper-case $F_{T}$ for tension and gives the interaction expression $\frac{F_{s}}{P_{s}} + \frac{F_{T}}{P_{nom}} \leq 1.4$ . The two uses of $F_{T}$ have different physical meanings and cannot be interchanged. The explanation above writes tension as $F_{t}$ ; the distinction is made explicit here. Do not infer dimensions from slide scale.Open full-size image

Animation labAdd direct and torsional bolt forces4 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.

Assign signed coordinates to the bolts relative to the group centroid.
For the equal-bolt elastic model, the direct force is $\frac{P}{n}$ per bolt.
Moment magnitude is $P e$ ; its sign follows the load direction. For signed M: $F_{x} = \frac{- M y}{J}$ , $F_{y} = \frac{M x}{J}$ , $J = \sum r^{2}$ .
Add signed x and y components at each bolt, then take the resultant. The longest arrow shows the critical bolt in this illustration.

PPT/Joint Connections L2.pptx · slide 8: A reusable connection workflow

Resolve factored actions; classify force direction; label geometry; calculate individual fastener demand; calculate all applicable fastener and plate resistances; check layout; state the governing result and assumptions. See the worked example before attempting the corresponding tutorial with the answer hidden.

Study the full explanation and applications.

Original slide 8: A reusable connection workflow. No dimensions are inferred from slide scale. — Original slide 8, student solution procedure. Step 1: calculate $P_{u} = 1.4 dead load + 1.6 imposed load$ . Step 2: check single/double shear, bearing, bolt tension or weld-throat resistance. Step 3: check plate tension using gross and effective net areas $A_{e}$ . Step 4: check code requirements for pitch, edge distance, transverse spacing and weld end returns. Do not infer dimensions from slide scale.Open full-size image

Animation labFollow the calculation sequence8 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.

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.

PPT/Steel Beam Design L3.pptx · slide 1: Beam design

A beam mainly resists transverse load by bending and shear. Its vertical supports and sideways restraints serve different purposes: a lateral restraint need not be a vertical support.

Study the full explanation and applications.

Original slide 1: Beam design. No dimensions are inferred from slide scale. — Original slide 1: Steel Beam Design, Chapter 3. Do not infer dimensions from slide scale.Open full-size image

Animation labBalance reactions and moments4 concepts

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.

PPT/Steel Beam Design L3.pptx · slide 2: Beam learning objectives

A complete beam answer progresses through load analysis, section classification, shear, bending, local web checks where required, LTB where unrestrained, and deflection. Section selection is iterative because a heavier trial section also changes self-weight.

Study the full explanation and applications.

Original slide 2: Beam learning objectives. No dimensions are inferred from slide scale. — Original slide 2, Learning Objectives: understand effective length, lateral/torsional restraint and destabilizing loads; calculate loads; and check simply supported beams under the Hong Kong steel code for classification, shear, bending, LTB, deflection, web bearing and web buckling. Do not infer dimensions from slide scale.Open full-size image

Animation labFollow the calculation sequence12 concepts

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.

PPT/Steel Beam Design L3.pptx · slide 3: Compression flange restraint

The flange in compression can move sideways and twist the beam. The slide’s $2.5 %$ restraint-force rule is a strength requirement for the restraint system, not evidence that any slab is automatically an effective restraint. Confirm which flange is in compression and whether the attachment prevents lateral movement/twist.

Study the full explanation and applications.

Original slide 3: Compression flange restraint. No dimensions are inferred from slide scale. — Original slide 3: full lateral restraint follows clause 8.2. The compression flange is restrained over its entire length, and the restraint system must resist at least the following fraction of the maximum design compressive force: $2.5 %$ . Partial restraint provides restraint only at selected positions. Under clause 8.3, check each segment for LTB, effective length $L_{E}$ and moment resistance $M_{b}$ above. Do not infer dimensions from slide scale.Open full-size image

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.

PPT/Steel Beam Design L3.pptx · slide 4: Classify the beam

Under major-axis bending the flanges and web have different stress distributions. Use the flange outstand $\frac{b}{T}$ and web $\frac{d}{t}$ from the section table, compare with the correct multiples of $ε$ , and select the worse class. Do not substitute overall depth $D$ for clear web depth $d$ .

Study the full explanation and applications.

Original slide 4: Classify the beam. No dimensions are inferred from slide scale. — Original slide 4: section classification depends on width/thickness ratios and steel strength. Class 1, plastic, can form a plastic hinge with adequate rotation. Class 2, compact, can reach plastic moment resistance with limited rotation. Class 3, semi-compact, can yield at the extreme fibre, but local buckling prevents full plastic moment resistance. Class 4, slender, buckles locally before yielding and requires an effective section. In the source expression $ε = \sqrt{\frac{275}{p_{y}}}$ , $p_{y}$ is measured in $\frac{N}{{mm}^{2}}$ is substituted. Do not infer dimensions from slide scale.Open full-size image

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.

PPT/Steel Beam Design L3.pptx · slide 5: Shear resistance

For a rolled I, H or channel under the course major-axis shear model, $A_{v} = t D$ . Then $V_{c} = \frac{p_{y} A_{v}}{\sqrt{3}}$ . Convert $N$ to $kN$ only after multiplying stress by area. If $V \leq 0.6 V_{c}$ the low-shear bending rule applies; passing $V \leq V_{c}$ alone does not establish low shear.

Study the full explanation and applications.

Original slide 5: Shear resistance. No dimensions are inferred from slide scale. — Original slide 5, clause 8.2.1: shear resistance $V_{c} = \frac{p_{y} A_{v}}{\sqrt{3}} \geq V$ . Hot-rolled I/H/channel sections use $A_{v} = t D$ ; the slide gives the following for RHS: $A_{v} = 2 t d$ . Low shear means $V \leq 0.6 V_{c}$ ; high shear means $V > 0.6 V_{c}$ and requires reduced moment resistance under clause 8.2.2.2. Do not infer dimensions from slide scale.Open full-size image

Animation labSee shear in the web2 concepts · 9 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.

PPT/Steel Beam Design L3.pptx · slide 6: Low and high shear bending

For Class 1/2 at low shear, $M_{c} = \min (p_{y} S, 1.2 p_{y} Z)$ . Above $0.6 V_{c}$ , reduce bending resistance using the course high-shear formula, with $ρ = {(\frac{2 V}{V_{c}} - 1)}^{2}$ . At $V = 0.8 V_{c}$ , $ρ = {(1.6 - 1)}^{2} = 0.36$ . Use the relevant shear-area term from the full formula, not $36 %$ of the entire bending resistance by guesswork.

Study the full explanation and applications.

Original slide 6: Low and high shear bending. No dimensions are inferred from slide scale. — Original slide 6: for low shear $V \leq 0.6 V_{c}$ . Class 1／2: $M_{c} = p_{y} S \leq 1.2 p_{y} Z$ ; Class 3: $M_{c} = p_{y} Z$ or, under the applicable effective-modulus method, use $p_{y} S_{eff}$ ; Class 4: $M_{c} = p_{y} Z_{eff}$ or, under the applicable reduced-strength method, use $p_{yr} Z$ . These alternatives are distinct code-permitted methods; do not simply choose the larger result. For high shear $V > 0.6 V_{c}$ in Class 1/2 sections: $M_{c} = p_{y} (S - ρ S_{v}) \leq 1.2 p_{y} (Z - \frac{ρ S_{v}}{1.5})$ ， $ρ = {(\frac{2 V}{V_{c}} - 1)}^{2}$ above. Do not infer dimensions from slide scale.Open full-size image

Animation labCompression and tension across a section3 concepts · 11 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.

PPT/Steel Beam Design L3.pptx · slide 7: Local web bearing and buckling

A support reaction or concentrated load presses through a finite bearing length into the web. Bearing and web buckling are separate checks. Identify the stiff bearing length, load position near an end, and lateral restraint. This slide only displays the $a_{e} \geq 0.7 d$ branch; the full lecture includes the other case. Missing contact dimensions prevent a definite local capacity.

Study the full explanation and applications.

Original slide 7: Local web bearing and buckling. No dimensions are inferred from slide scale. — Original slide 7, clause 8.4.10: web bearing $P_{bw} = (b_{1} + n k) t p_{yw}$ ; load spreads at $1 : 2.5$ to the root fillet. For hot-rolled sections, $k = T + r$ . The web-buckling resistance $P_{x}$ expression shown requires rotational and lateral restraint to the flange. When $a_{e} \geq 0.7 d$ , $P_{x} = \frac{25 ε t}{\sqrt{(b_{1} + n k) d}} P_{bw}$ . $b_{1}$ is bearing length; $n k$ is the additional dispersion length selected for the end location. Dimensions and the applicable branch must still be identified from the actual question. Do not infer dimensions from slide scale.Open full-size image

Animation labSpread a concentrated force into the web2 concepts · 5 source expressions

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

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.

PPT/Steel Beam Design L3.pptx · slide 8: LTB factors and effective length

Compute $λ = \frac{L_{E}}{r_{y}}$ , then $v$ and $λ_{LT} = u v λ \sqrt{β_{W}}$ , and read $p_{b}$ . The slide’s suggestion to multiply destabilizing length by $1.2$ is not a universal rule: use the applicable normal/destabilizing column and actual end-restraint row in the full lecture effective-length rules. Load height and freedom to twist determine applicability.

Study the full explanation and applications.

Original slide 8: LTB factors and effective length. No dimensions are inferred from slide scale. — Original slide 8: LTB must be checked when the compression flange is not restrained along its full length. Original notation is retained as $m_{Lt} M_{x} \leq M_{b}$ and $M_{x} \leq M_{cx}$ ; $λ_{lt} = u v λ \sqrt{β_{w}}$ ; for Class 1/2, $M_{b} = p_{b} S_{x}$ . The diagram symbol $m_{Lt}$ , $λ_{lt}$ , $β_{w}$ corresponds to this lesson's $m_{LT}$ , $λ_{LT}$ , $β_{W}$ ; source letter case is retained. The slide says destabilizing loading multiplies $L_{E}$ by $1.2$ . Applicability limits and the full lookup method are explained above; this is not a fixed multiplier for every restraint condition. Do not infer dimensions from slide scale.Open full-size image

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.

PPT/Steel Beam Design L3.pptx · slide 9: Fully restrained workflow

Resolve loads and reactions; draw the critical moment/shear; choose a trial section; check its class, $V$ and $M_{c}$ ; check local web behaviour; calculate service deflection. State the physical reason LTB is restrained rather than silently omitting it.

Study the full explanation and applications.

Original slide 9: Fully restrained workflow. No dimensions are inferred from slide scale. — Original slide 9, six steps for a fully restrained steel beam. (1) Determine dead and imposed loads. (2) Calculate design moment $M_{x}$ and shear $V$ . (3) Make a trial section selection using $S_{x} = \frac{M_{x}}{p_{y}}$ . (4) Classify using width/thickness ratios. (5) Check bending and shear resistances. (6) Check local web bearing/buckling and deflection. The trial-selection formula does not replace these checks. Do not infer dimensions from slide scale.Open full-size image

Animation labFollow the calculation sequence12 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.

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.

PPT/Steel Beam Design L3.pptx · slide 10: Unrestrained workflow

Divide the beam at effective lateral restraints. For each relevant segment find its maximum moment and moment shape, obtain $m_{LT}$ , calculate effective slenderness and $M_{b}$ , and compare $m_{LT} M_{\max} \leq M_{b}$ . The segment with the longest length is not necessarily the governing one.

Study the full explanation and applications.

Original slide 10: Unrestrained workflow. No dimensions are inferred from slide scale. — Original slide 10, six steps for a beam without full lateral restraint. (1) Determine design actions. (2) The trial may need to be heavier because $p_{b} < p_{y}$ . (3) Classify and identify normal/destabilizing loading. (4) Calculate slenderness using $λ = \frac{L_{E}}{r_{y}}$ . (5) Determine $λ_{lt}$ and look up bending strength in Table 8.3a. (6) Check $m_{Lt} M_{x} \leq M_{b}$ , where $M_{b} = p_{b} S_{x}$ . For the source symbol $λ_{lt}$ , $m_{Lt}$ and its correspondence to this lesson's upper-case subscript, see slide 8. Do not infer dimensions from slide scale.Open full-size image

Animation labA beam bends sideways and twists3 concepts · 5 source expressions

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

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

PPT/Steel Beam Design L3.pptx · slide 11: Finish with serviceability and a conclusion

Use the stated service load and unamplified elastic stiffness $E I$ . For central $P$ , $δ = \frac{P L^{3}}{48 E I}$ ; for UDL $w$ , $δ = \frac{5 w L^{4}}{384 E I}$ . Convert ${cm}^{4}$ to ${mm}^{4}$ by × $10^{4}$ . State the largest utilization and any incomplete local checks.

Study the full explanation and applications.

Original slide 11: Finish with serviceability and a conclusion. No dimensions are inferred from slide scale. — Original slide 11, design and exam tips: establish restraints before selecting the calculation method; include classification; keep serviceability deflection separate from strength checks. The example limit for brittle finishes is $\frac{L}{360}$ . State the applicable Hong Kong code clauses and the conclusion. Do not infer dimensions from slide scale.Open full-size image

Animation labSee stiffness and deflection1 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.

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.

PPT/Steel Column Design L4.pptx · slide 1: Column design

A column resists compression, often with bending. A short section can crush; a long member can buckle before reaching material strength. These mechanisms require separate checks.

Study the full explanation and applications.

Original slide 1: Column design. No dimensions are inferred from slide scale. — Original slide 1: Steel Column Design (CON4334), Chapter 4; understanding the content, effective length and member resistance. Do not infer dimensions from slide scale.Open full-size image

Animation labA strong slice can belong to an unstable member2 concepts

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

Combine axial compression and the two bending demands using the specified section resistances.
The whole member adds effective-length and buckling-curve effects.
This uses its own moment factor and bending resistance; it is not a copy of the section check.
Elastic, plastic and buckling resistances are not interchangeable. Read the three original expressions and their first-order/amplified moments.

PPT/Steel Column Design L4.pptx · slide 2: Column learning objectives

Learn restraint interpretation, two-axis slenderness, curve selection, compression resistance and combined axial/bending checks. A single axial capacity is insufficient when the load acts eccentrically.

Study the full explanation and applications.

Original slide 2: Column learning objectives. No dimensions are inferred from slide scale. — Original slide 2, Learning Objectives: understand effective length and section classification; correctly design columns carrying axial force alone; evaluate columns under combined axial force and bending in simple and continuous construction. Do not infer dimensions from slide scale.Open full-size image

Animation labWhy thin elements buckle locally3 concepts

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.

PPT/Steel Column Design L4.pptx · slide 3: Effective length and slenderness

For each axis, $L_{E} = K L$ and $λ = \frac{L_{E}}{r}$ . The dimensionless factor $K$ represents the stated end restraint/sway condition. Millimetres must be used in both numerator and denominator. Tying the weak direction reduces $L_{Ey}$ only if the tie actually restrains that buckling direction.

Study the full explanation and applications.

Original slide 3: Effective length and slenderness. No dimensions are inferred from slide scale. — Original slide 3, compression members and effective length. Buckling resistance depends on slenderness $λ = \frac{L_{E}}{r}$ ; effective length $L_{E} = K L$ is the product of factor $K$ and actual length $L$ . Use rotational and positional restraint conditions to select from Table 8.6. Compression resistance for plastic, compact and semi-compact sections: $P_{c} = A_{g} p_{c}$ above. Do not infer dimensions from slide scale.Open full-size image

Animation labEffective length and buckling axes1 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 column can bow sideways before its section reaches the crushing resistance.
Each axis has its own radius of gyration and restraint spacing.
$λ = \frac{L _{E}}{r}$ , with compatible length units. A tie affects only the directions it actually restrains.
Select each buckling curve and compressive strength before forming Pc. The governing axis is determined by resistance, not slenderness alone.

PPT/Steel Column Design L4.pptx · slide 4: Select a strut curve before a number

Read Table 8.7 by section type, maximum thickness and buckling axis. A rolled H section with thickness $\leq 40 mm$ uses curve b about $x$ and c about $y$ . Then read Table 8.8 using that curve, $p_{y}$ and slenderness; interpolate between bounding rows.

Study the full explanation and applications.

Original slide 4: Select a strut curve before a number. No dimensions are inferred from slide scale. — Original slide 4, strut-strength tables. Compressive strength $p_{c}$ depends on section type, design strength $p_{y}$ , slenderness $λ$ and buckling curve. Table 8.7 selects curves a, b, c and d using section geometry and residual stress. Tables 8.8(a) to (d) on the slide give design strength in $\frac{N}{{mm}^{2}}$ for steel grades including S275, S355 and S460. Use the applicable curve and table heading; steel grade alone does not select the table. Do not infer dimensions from slide scale.Open full-size image

Animation labEffective length and buckling axes3 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.

A column can bow sideways before its section reaches the crushing resistance.
Each axis has its own radius of gyration and restraint spacing.
$λ = \frac{L _{E}}{r}$ , with compatible length units. A tie affects only the directions it actually restrains.
Select each buckling curve and compressive strength before forming Pc. The governing axis is determined by resistance, not slenderness alone.

PPT/Steel Column Design L4.pptx · slide 5: Section and member interactions

Cross-section interaction combines axial utilization and bending utilizations at the same section. Member equations include reduced compression/LTB resistance and moment factors. Follow the full lecture’s elastic denominators and overbars: compact slide notation can hide whether a moment is first-order or amplified.

Study the full explanation and applications.

Original slide 5: Section and member interactions. No dimensions are inferred from slide scale. — Original slide 5, combined axial force and bending (clause 8.9). A column must satisfy both section resistance and member buckling resistance. Clause 8.9.1 gives the section expression $\frac{F_{c}}{A_{g} p_{y}} + \frac{M_{x}}{M_{cx}} + \frac{M_{y}}{M_{cy}} \leq 1$ . The member interaction expressions in clause 8.9.2 account for second-order overall sway and member bending ( $P - Δ$ and $P - δ$ ). Do not infer dimensions from slide scale.Open full-size image

Animation labA strong slice can belong to an unstable member2 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.

Combine axial compression and the two bending demands using the specified section resistances.
The whole member adds effective-length and buckling-curve effects.
This uses its own moment factor and bending resistance; it is not a copy of the section check.
Elastic, plastic and buckling resistances are not interchangeable. Read the three original expressions and their first-order/amplified moments.

PPT/Steel Column Design L4.pptx · slide 6: Frame classification and amplification

Elastic critical load factor $λ_{cr}$ measures proximity to global instability. Use the lecture’s non-sway/sway/ultra-sensitive boundaries and applicable amplification method. If the problem already supplies amplification factors, apply them once to the specified first-order moments; an already amplified $M_{LT}$ must not be amplified again.

Study the full explanation and applications.

Original slide 6: Frame classification and amplification. No dimensions are inferred from slide scale. — Original slide 6, sections 2.3 and 2.4: frame classification and amplification. Non-sway frames: $λ_{cr} \geq 10$ , allowing overall $P - Δ$ effects to be ignored. Sway frames: $5 \leq λ_{cr} < 10$ ; highly sway-sensitive frames: $λ_{cr} < 5$ , requiring a higher-order second-order analysis. The slide gives moment amplification factor $\frac{λ_{cr}}{λ_{cr} - 1}$ to account for second-order sway, subject to the applicability conditions of the course method. Do not infer dimensions from slide scale.Open full-size image

Animation labRestraints, sway and imperfections2 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.

A frame needs a defined path for horizontal force as well as gravity.
Pinned joints alone do not provide frame moment resistance; sway restraint needs a real structural system.
Diagonal bracing carries horizontal action through axial forces. This sketch does not assign a numerical frame classification.
Use the specified imperfection/notional-force model and critical-load criteria. Avoid counting alternative imperfection models twice.

PPT/Steel Column Design L4.pptx · slide 7: Simple construction

Nominally pinned beams can still deliver eccentric reactions to a column. Calculate reaction×eccentricity and share joint moment according to the prescribed column stiffnesses. For the course simple-construction LTB shortcut, $λ_{LT} = \frac{0.5 L}{r_{y}}$ uses actual column length $L$ , not the flexural effective length $L_{E}$ .

Study the full explanation and applications.

Original slide 7: Simple construction. No dimensions are inferred from slide scale. — Original slide 7, columns in simple construction. Beams are assumed fully loaded and simply supported, with all equivalent-moment factors $m = 1.0$ . Equivalent slenderness for lateral-torsional buckling $λ_{LT} = 0.5 (\frac{L}{r_{y}})$ ， $L$ is storey height, not $L_{E}$ . Under multi-storey clause 8.7.8, pattern loading need not be considered. Nominal moments come from reaction eccentricity, taking the larger eccentricity from a load line $100 mm$ from the column face or at the centre of stiff bearing. Do not infer dimensions from slide scale.Open full-size image

Animation labEccentric reactions and stiffness sharing3 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.

A beam reaction can act away from the column centre even at a nominally pinned beam connection.
$M = R e$ . Opposing reactions can cancel part of the signed moment, while both still add compression.
The course simple model distributes the joint moment in proportion to $\frac{I}{L}$ of the columns above and below.
Equal relevant stiffness gives half each. A roof joint with no upper column is a different case.

PPT/Steel Column Design L4.pptx · slide 8: Write both buckling checks accurately

Keep flexural-buckling and LTB interaction equations separate. The full lecture Eq. 8.80 uses amplified bending moments; Eq. 8.81 uses the specified amplified $M_{LT}$ and first-order minor-axis $M_{y}$ . The slide is a memory aid; use the full derivation to restore subscripts, bars and elastic moment denominators.

Study the full explanation and applications.

Original slide 8: Write both buckling checks accurately. No dimensions are inferred from slide scale. — Original slide 8, summary of column checks. The source section expression is $[\frac{F_{c}}{A_{g} p_{y}}] + [\frac{M_{x}}{M_{cx}}] + [\frac{M_{y}}{M_{cy}}] \leq 1$ , the same for simple and continuous construction. The source member expression for simple construction is $[\frac{F_{c}}{P_{cy}}] + [\frac{m_{LT} M_{LT}}{M_{b}}] + [\frac{m_{y} M_{y}}{M_{cy}}] \leq 1$ ， $m = 1.0$ ; continuous construction selects specific $m$ values from Tables 8.9 and 8.4. This reproduces the source diagram without silently replacing its $M_{cy}$ with the elastic denominator in the full lecture. Actual solutions must use the complete formulas and amplified-moment definitions linked above; this summary, which omits overbars, cannot replace them.Open full-size image

Animation labA strong slice can belong to an unstable member1 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.

Combine axial compression and the two bending demands using the specified section resistances.
The whole member adds effective-length and buckling-curve effects.
This uses its own moment factor and bending resistance; it is not a copy of the section check.
Elastic, plastic and buckling resistances are not interchangeable. Read the three original expressions and their first-order/amplified moments.

PPT/Steel Column Design L4.pptx · slide 9: The worked-example sequence

Example 1 teaches axial compression; Example 2 adds a weak-axis tie; Example 3 transfers eccentric beam reactions; Example 4 derives floor loads and reduced imposed load. The full notes also contain Example 5 for combined end moments, which remains required even though this summary omits it.

Study the full explanation and applications.

Original slide 9: The worked-example sequence. No dimensions are inferred from slide scale. — Original slide 9, design examples and learning resources. Example 1: pin-ended column under axial force. Example 2: pin-ended column tied against minor-axis buckling. Example 3: column under axial force and bending in simple construction. Example 4: axially loaded column using tributary area and load reduction. Do not infer dimensions from slide scale.Open full-size image

Animation labEffective length and buckling axes5 concepts

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

A column can bow sideways before its section reaches the crushing resistance.
Each axis has its own radius of gyration and restraint spacing.
$λ = \frac{L _{E}}{r}$ , with compatible length units. A tie affects only the directions it actually restrains.
Select each buckling curve and compressive strength before forming Pc. The governing axis is determined by resistance, not slenderness alone.

PPT/Steel Column Design L4.pptx · slide 10: References and further study

The slide names references and suggests external video searches. This offline course contains the needed teaching locally. When reading outside material, compare code edition, axis labels, yield strength rules and partial factors before using any equation; do not combine Eurocode and supplied HK design expressions.

Study the full explanation and applications.

Original slide 10: References and further study. No dimensions are inferred from slide scale. — Original slide 10, References & Useful Links: Hong Kong Code of Practice for the Structural Use of Steel 2011; and Steelwork Design Guide to BS 5950 / HK Code (CON4334 course notes). The original suggested YouTube search strings are retained for recognition: Steel Column Design EC3 / BS5950 Workflow (column design workflow) and Effective Length Buckling Civil Engineering (effective length and buckling). These are source suggestions for further study. This site does not depend on external videos or combine different codes. Do not infer dimensions from slide scale.Open full-size image

Animation labRead a table without losing the keys2 concepts

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.

NOTE.docx · rendered page 1: material and capacities

$E = 205000 \frac{N}{{mm}^{2}}$ is elastic stiffness. $ε = \sqrt{\frac{275}{p_{y}}}$ scales classification limits. Read $p_{y}$ from thickness before calculating $ε$ . The note lists $M_{c} = \min (p_{y} S, 1.2 p_{y} Z)$ , but this is the Class 1/2 low-shear rule, not the Class 3/4 rule. $V_{c} = \frac{p_{y} A_{v}}{\sqrt{3}}$ uses $A_{v} = t D$ for the specified rolled sections and shear direction. $P_{t} = A_{e} p_{y}$ needs net-area/shear-lag adjustments for connected angles. $P_{c} = A_{g} p_{c}$ needs separate curve and slenderness lookups about both axes. Despite the heading “Member Checks”, $M_{c}$ and $V_{c}$ alone are cross-section resistances; instability needs additional member checks.

Original Word page 1; the equations are embedded objects and were inspected as an image. — Original Word document p.1, Student Quick-Reference Guide: Steelwork Design. Equations are embedded objects inspected as images. (1) Materials: Young's modulus $E = 205,000 \frac{N}{{mm}^{2}}$ ; steel-grade factor $ε = {(\frac{275}{p_{y}})}^{0.5}$ ; design strength varies with grade and plate thickness, for example S355 at thickness $\leq 16 mm$ ， $p_{y} = 355 \frac{N}{{mm}^{2}}$ . (2) Bending: $M_{cx} = p_{y} S_{x} \leq 1.2 p_{y} Z_{x}$ ， $M_{cy} = p_{y} S_{y} \leq 1.2 p_{y} Z_{y}$ , subject to the classification and low-shear conditions above. Shear: $V_{c} = \frac{p_{y} A_{v}}{\sqrt{3}}$ ; for hot-rolled I, H and channel sections, use $A_{v} = t D$ . Tension: $P_{t} = A_{e} p_{y}$ , adjusting net area for the bolted/welded connection type. Compression: $P_{c} = A_{g} p_{c}$ , with slenderness $λ = \frac{L_{e}}{r}$ ; lower-case subscript $L_{e}$ is retained from the source.Open full-size image

Try it yourself. Why can $P_{c}$ be less than $A_{g} p_{y}$ ?

Reveal answer and reasoning

$p_{c}$ accounts for member buckling and is normally below $p_{y}$ ; $p_{y}$ only describes the material/section strength.

Animation labElasticity, yielding and ductility9 concepts · 18 source expressions

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

Stress is force divided by area. Strain measures change in length relative to original length.
Before yielding, stress is approximately $E \times strain$ . E controls the elastic slope.
Further strain includes permanent deformation. Higher yield strength does not by itself increase E.
Strength, stiffness and ductility answer different questions. This schematic is not a measured stress–strain curve.

NOTE.docx · rendered page 2: deflection and fasteners

For a simply supported prismatic beam with constant $E I$ , central-point-load deflection is $\frac{P L^{3}}{48 E I}$ , and full-span UDL deflection is $\frac{5 w L^{4}}{384 E I}$ . The Word note correctly shows $L^{3}$ for the point load; the separate Data File has a conflicting exponent. Select the service load and limit row, then use consistent $N$ and $mm$ . Bolt $P_{s} = p_{s} A_{s}$ is per shear plane; choose shank or thread area from where the plane crosses. $P_{t} = A_{s} p_{t}$ is individual tension resistance; $P_{nom} = 0.8 A_{s} p_{t}$ is the nominal value used in the specified interaction expression, so the “or” does not permit arbitrary selection. Bolt bearing $P_{bb} = d t_{p} p_{bb}$ still requires separate connected-part bearing checks. Weld $q = \frac{0.7 s p_{w}}{1000}$ is strength per millimetre of effective length, not total force.

Original Word page 2, including the correctly printed point-load exponent L³. — Original Word document p.2: deflection limits, connections and bolts. The source uses lower-case span $l$ and upper-case deflection $Δ$ : full-span UDL $Δ = \frac{5}{384} \frac{w l^{4}}{E I}$ ; midspan point load $Δ = \frac{P l^{3}}{48 E I}$ . The cubic power in the point-load formula is correct; this lesson writes it as $L^{3}$ . Common Table 5.1 limits: cantilevers $\frac{length}{180}$ ; beams carrying plaster/brittle finishes $\frac{span}{360}$ ; other beams $\frac{span}{200}$ . Bolt shear resistance $P_{s} = p_{s} A_{s}$ ; tension resistance $P_{t} = A_{s} p_{t}$ , or use the following in the specified interaction expression: $P_{nom} = 0.8 A_{s} p_{t}$ ; bolt bearing $P_{bb} = d t_{p} p_{bb}$ . Original fillet-weld expression $Strength = 0.7 \times (leg length) \times p_{w} \times 10^{- 3} \frac{kN}{mm}$ ; Strength means resistance per millimetre of effective length, leg length is the weld leg entered in millimetres, and $p_{w}$ is measured in $\frac{N}{{mm}^{2}}$ is substituted.Open full-size image

Try it yourself. A $6 mm$ S355/Class 42 fillet has $120 mm$ effective length. What direct force can it carry?

Reveal answer and reasoning

$q = \frac{0.7 \times 6 \times 250}{1000} = 1.05 \frac{kN}{mm}$ ; resistance $= q L = 1.05 \times 120 = 126 kN$ , subject to detailing and parent-metal checks.

Animation labFrom fillet leg to effective throat1 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.

An equal-leg fillet between perpendicular plates has an approximately right-triangular section.
For this geometry, throat $a \approx 0.7 s$ . It is shorter than the leg.
Effective resisting area = $a \times L_{eff}$ . With $p_{w} = 220 \frac{N}{mm ^{2}}$ here, capacity per length is $0.7 s \times 220 \frac{N}{mm}$ .
Use the course end allowances, minimum size and length rules; increasing the geometric length alone does not resolve every detailing check.

Animation labSee stiffness and deflection6 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.

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.