Worked examples

Tutorial 2 Q7: separate flange and web weld design

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Design welds connecting a bracket cut from $533 \times 210 \times 82$ UB S355 to a column. Ultimate vertical load is $600 kN$ at eccentricity $250 mm$ . Use Class 42 electrode and the question’s assumption that flange welds resist moment while web welds resist shear.

Original source: LectureNotes/Ch 2_Connection.pdf — p. 45, p. 48. 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	Meaning
Moment force couple	Use the source example’s overall-depth lever-arm approximation $D = 528.3 mm$ between flange force lines.
Flange run	One $190 mm$ physical weld line at each flange, carrying the flange couple force.
Web runs	Two $340 mm$ physical runs share $600 kN$ direct shear.
Section thickness	Data File p.9 row $533 \times 210 \times 82$ : flange $13.2 mm$ ,web $9.6 mm$ .
Unspecified input	Column flange thickness is absent, affecting the minimum permitted weld size at the web/column joint. The strength design can be calculated; unconditional detailing depends on that thickness.

Before calculating: recognition and strategy

Use the stated simplified load split rather than a full elastic weld-group analysis. Moment $M = P e$ becomes an equal-and-opposite pair of flange forces $\frac{M}{D}$ ; the web takes the vertical shear. Size each family of welds using its own effective length. Because $L_{eff} = L - 2 s$ , changing the leg also changes the effective length; recheck the chosen trial.

1. Moment and flange welds

Simple explanation: How two flange forces make a moment

Two opposite forces form a turning pair, like two hands turning a wheel.

Identify the separation between their actual force lines.
Required force equals moment divided by that separation.
Design the relevant flange group for that force.

Remember: The force-line separation is not automatically the overall section depth.

Related concept and full method

M = 600 \times 250 = 150,000 kN \cdot mm = 150 kN \cdot m

Flange force:

F = \frac{M}{D} = \frac{150,000}{528.3} = 283.929585 kN

For trial weld leg

s

, resistance per flange weld:

(190 - 2 s) \times 0.175 s kN

Trial

8 mm

: effective length

174 mm

，

q = 1.4 \frac{kN}{mm}

, resistance

243.6 kN

, fails. Trial

9 mm

: effective length

172 mm

，

q = 1.575 \frac{kN}{mm}

, resistance

270.9 kN

, fails. Trial

10 mm

: effective length

170 mm

，

q = 1.75 \frac{kN}{mm}

, resistance

297.5 kN > 283.929585 kN

: passes.

Provide $10 mm$ fillet along each $190 mm$ flange run. For $13.2 mm$ flange edge, maximum leg $13.2 - 2 = 11.2 mm$ permits 10. The thicker-part minimum is at most $8 mm$ in the supplied table, so 10 meets it even if the column flange is thicker. Effective $170 \geq \max (4 \times 10, 40) = 40 mm$ . The force in each flange is $283.93 kN$ ; do not divide it by two merely because the couple has two flanges.

This $\frac{M}{D}$ force-couple idealization is explicitly tied to the lecturer’s corresponding bracket method. A refined connection analysis would establish the actual flange force resultants; the nominal $528.3 mm$ lever arm is not a newly measured weld-centroid separation.

Animation labSeparate the moment couple and shear3 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 simplified moment connection assigns different actions to different fastener groups.
Opposite flange forces separated by z resist moment: $F = \frac{M}{z}$ .
The web fasteners or welds carry the assigned vertical shear in this model.
Flange force, web shear, plate bearing and detailing each need their specified checks.

2. Web shear welds and detailing condition

Simple explanation: Why the weld throat is smaller than its leg

The shortest cut through the weld is thinner than the outside leg.

For the stated equal-leg 90° fillet, throat is approximately 0.7 × leg.
Multiply throat area by the matching weld design strength.
For force per length, use a one-millimetre weld strip.

Remember: Choose strength from both the steel grade and electrode class.

Related concept and full method

Two web welds share

V = 600 kN

, each taking

300 kN

. Trial

5 mm

: effective length per weld

340 - 10 = 330 mm

; combined resistance

2 \times 330 \times 0.175 \times 5 = 577.5 kN

, fails. Trial

6 mm

: effective length per weld

340 - 12 = 328 mm

; combined resistance

2 \times 328 \times 0.175 \times 6 = 688.8 kN > 600 kN

, passes. Resistance per weld

328 \times 1.05 = 344.4 kN > 300 kN

. Effective length

328 \geq \max (4 \times 6, 40) = 40 mm

. Maximum weld leg at the web edge

9.6 - 2 = 7.6 mm

, so

6 mm

satisfies this cap.

The $6 mm$ web weld meets the thicker-part minimum if the column flange thickness is at most $19 mm$ under the supplied table. If it exceeds $19 mm$ , the minimum would be $8 mm$ , incompatible with the $7.6 mm$ edge maximum in this edge-detail model; a different detail would be needed. Since column flange thickness is not stated, report $6 mm$ as the completed strength selection with this explicit detailing condition, not an unconditional fabrication approval. Provide any required end returns of at least $2 s$ where the actual joint permits; their geometry is not dimensioned here.

Animation labFrom fillet leg to effective throat2 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.

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.

Compact exam answer

M150 kNm; source lever arm $528.3 mm$ gives flange force $283.9296 kN$ . Use $10 mm$ flange fillets on $190 mm$ runs: effective $170 mm$ , capacity $297.5 kN$ each. For web shear use $6 mm$ fillets on two $340 mm$ runs: effective $328 mm$ each, pair capacity 688.8 kN>600. Web minimum-size compliance remains conditional on the omitted column flange thickness ( $6 mm$ selection permitted by the supplied minimum table only up to $19 mm$ thicker part).

Mistakes to avoid

Use the given $190 mm$ weld run rather than nominal $210 mm$ section width.
Each flange carries the full couple force.
Use both web runs for shear, but deduct $2 s$ from each.
Do not claim an omitted column thickness has been checked.

Procedure for an unfamiliar variant

Factor loads if required, then calculate $P e$ .
Convert moment into flange force using the stated force-couple model.
Try flange leg sizes and update each effective length.
Size the pair of web welds for the full shear force.
Check known maximum/minimum size and length conditions; state missing detailing inputs.

Independent self-check

Try it yourself. Invented variant: ultimate load remains $600 kN$ but its eccentricity rises to $300 mm$ . Can the selected flange weld remain $10 mm$ ?

Reveal answer and reasoning

$F = \frac{600 \times 300}{528.3} = 340.7155 kN$ , above the $10 mm$ weld resistance $297.5 kN$ . The web shear remains $600 kN$ , so its strength demand is unchanged. A larger flange weld is limited by the $13.2 - 2 = 11.2 mm$ edge maximum; even $11 mm$ gives $(190 - 22) \times 1.925 = 323.4 kN$ , still insufficient. A changed flange weld layout/detail is needed.

Animation labSeparate the moment couple and shear2 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 simplified moment connection assigns different actions to different fastener groups.
Opposite flange forces separated by z resist moment: $F = \frac{M}{z}$ .
The web fasteners or welds carry the assigned vertical shear in this model.
Flange force, web shear, plate bearing and detailing each need their specified checks.