Worked examples

Connection example 9: a welded single unequal angle

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Check a $75 \times 50 \times 6$ S355 single angle connected by its long leg under $95 kN$ dead plus $40 kN$ imposed tension, and design the two side welds with Class 42 electrode.

Original source: LectureNotes/Ch 2_Connection.pdf — p. 42. 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	Origin
Given loads	$95 kN$ dead and $40 kN$ imposed, task text.
Given dimensions	$75 \times 50 \times 6$ angle; source diagram shows $72$ and $47 mm$ half-heel widths and $12.1 mm$ centroid height.
Lookup centroid	$24.4 mm$ from SideX stated from section table in solution and labelled in diagram. Opposite arm $75 - 24.4 = 50.6 mm$ .
Lookup strengths	S355 thickness $6 mm$ → $p_{y} = 355$ ; S355/Class 42 → $p_{w} = 250 \frac{N}{{mm}^{2}}$ .
Design choice	$4 mm$ fillet, physical lengths $200 mm$ (X), $100 mm$ (Y), lecturer solution.

Before calculating: recognition and strategy

First check that the angle itself can carry the force: stronger welds cannot cure a weak angle. With no bolt holes, the gross area is effective, but a single welded connected leg still has the $0.3 a_{2}$ reduction for the unconnected leg. Then balance the two parallel weld forces about the centroid using their opposite lever arms.

1. Factored force and welded-angle resistance

Simple explanation: One connected leg does not load both legs equally

The connected leg receives the pull first; the other leg receives it through the angle.

Identify connected and outstanding legs from the drawing.
Use the relevant bolted or welded angle rule.
Keep the lecturer’s area convention consistent.

Remember: Bolted and welded reduction expressions are not the same rule.

Related concept and full method

P = 1.4 \times 95 + 1.6 \times 40 = 133 + 64 = 197 kN

a_{1} = (75 - \frac{6}{2}) \times 6 = 72 \times 6 = 432 {mm}^{2}

a_{2} = (50 - \frac{6}{2}) \times 6 = 47 \times 6 = 282 {mm}^{2}

A_{e} = A_{g} = 432 + 282 = 714 {mm}^{2}

No holes.

P_{t} = \frac{355 (714 - 0.3 \times 282)}{1,000}

= \frac{355 (714 - 84.6)}{1,000} = \frac{355 \times 629.4}{1,000} = 223.437 kN > 197

Passes.

The $0.3$ factor comes from the single welded angle rule on Ch 2 p.14. The bolted value $0.5$ would be a different connection model.

Animation labWhy a connected angle leg matters2 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 connected leg, outstanding leg and centroid before using the table.
Only the connected leg directly receives the fastener force.
The outstanding area may not become equally effective at the same section; this motivates the effective-area rule.
Bolted, welded, single-angle and double-angle details can have different rules. Preserve the formula attached to the original case.

2. Trial $4 mm$ weld and balanced required lengths

Simple explanation: Why two weld lengths can be unequal

Two side welds must balance the load about its actual line of action.

Find the load line and the lever arm to each weld.
Balance moments as well as the total force.
Convert each weld’s force into its own required length.

Remember: Equal-looking legs do not justify equal weld forces without equilibrium.

Related concept and full method

q = \frac{0.7 \times 4 \times 250}{1,000} = 0.70 \frac{kN}{mm}

Total effective length:

L_{eff} = \frac{197}{0.70} = 281.4286 mm

R_{X} = \frac{197 \times 50.6}{75} = 132.9093 kN

R_{Y} = \frac{197 \times 24.4}{75} = 64.0907 kN

L_{X, eff} = \frac{132.9093}{0.70} = 189.8705 mm

L_{Y, eff} = \frac{64.0907}{0.70} = 91.5581 mm

The $12.1 mm$ centroid height is visible in the section but is not the transverse lever arm for sharing these two side-weld forces. The side-to-side $24.4$ /50.6 dimensions are the relevant ones.

Animation labBalance two weld forces2 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 angle centroid/load line is generally not halfway between the two weld runs.
$F_{1} + F_{2} = P$ . Both weld runs contribute to the applied force.
$F_{1} \times a = F_{2} \times b$ . The run nearer the load line carries more force.
For equal throat resistance per length, the required effective lengths follow the same ratio. Add detailing allowances afterwards.

3. Final lengths, capacities and source correction

Simple explanation: Drawn length and useful length differ

The start and end of a weld are not credited as fully effective in this course model.

Find the effective length required by strength.
Add the specified end allowance to each separate run.
Round up, then check minimum size, spacing and returns.

Remember: Strength alone does not prove a weld detail is acceptable.

Related concept and full method

End allowance per weld

= 2 s = 8 mm

. Side X actual length:

\geq 189.8705 + 8 = 197.8705 mm \to 200 mm

Side Y actual length:

\geq 91.5581 + 8 = 99.5581 mm \to 100 mm

Provided effective length on side X

= 200 - 8 = 192 mm

, resistance

134.4 kN > 132.9093

. Side Y effective length

= 100 - 8 = 92 mm

, resistance

64.4 kN > 64.0907

Both effective lengths exceed $\max (16, 40) = 40 mm$ , and both physical lengths exceed $75 mm$ separation. For the $6 mm$ angle edge the maximum leg is $6 - 2 = 4 mm$ , matching the selection. The minimum-size rule depends on the unknown supporting plate thickness; do not assume it from the sketch.

Animation labFrom fillet leg to effective throat3 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.

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

$P = 197 kN$ . Welded single-angle capacity $= \frac{355 (714 - 0.3 \times 282)}{1,000} = 223.437 kN$ . Use $4 mm$ fillets with $q = 0.70 \frac{kN}{mm}$ . Required effective X/Y lengths $189.870$ / $91.558 mm$ . Provide $200$ / $100 mm$ physical lengths ( $192$ / $92 mm$ effective), giving $134.4$ / $64.4 kN$ against $132.909$ / $64.091 kN$ . Angle and weld strength pass; support-thickness detailing is unprovided.

Mistakes to avoid

Do not subtract holes from a welded angle with no holes.
The unconnected leg still needs the $0.3$ reduction.
For the $4 mm$ weld, the source writes “Add 12” (add $12$ ) is a textual error; it should be $2 s = 8$ .
Do not use $12.1 mm$ centroid height in the $75 mm$ transverse weld-force split.

Procedure for an unfamiliar variant

Check angle area and the appropriate welded single-leg resistance.
Read centroid location from the source/table.
Select a weld leg permitted by the angle edge.
Balance side forces, calculate effective lengths, then add $2 s$ to each.
Verify each rounded side and the detailing conditions.

Independent self-check

Try it yourself. If the imposed load rises to $45 kN$ , do the existing $200$ / $100 mm$ welds still pass?

Reveal answer and reasoning

$P = 133 + 72 = 205 kN$ . SideX demand $= \frac{205 \times 50.6}{75} = 138.307 kN > 134.4$ ; SideY $= 66.693 kN > 64.4$ . Both weld sides fail although angle capacity 223.437 still passes.

Animation labBalance two weld forces3 concepts · 3 source expressions

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

The angle centroid/load line is generally not halfway between the two weld runs.
$F_{1} + F_{2} = P$ . Both weld runs contribute to the applied force.
$F_{1} \times a = F_{2} \times b$ . The run nearer the load line carries more force.
For equal throat resistance per length, the required effective lengths follow the same ratio. Add detailing allowances afterwards.

Connection example 9: a welded single unequal angle

Read the diagram and collect the data

Before calculating: recognition and strategy

1. Factored force and welded-angle resistance

2. Trial 4mm4mm weld and balanced required lengths

3. Final lengths, capacities and source correction

Compact exam answer

Mistakes to avoid

Procedure for an unfamiliar variant

Independent self-check

2. Trial $4 mm$ weld and balanced required lengths