Worked examples

2023 BQ2(a): double welded unequal angles

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Check tensile capacity and design final $8 mm$ fillet-weld lengths for two $200 \times 100 \times 12$ S355 unequal angles connected through their long legs to an $18 mm$ gusset. Characteristic tension: dead $150 kN$ and imposed $160 kN$ . S355/Class 42; the question excludes gusset-plate checking.

Original source: Pastpaper/22ENGTY033.pdf — p. 3. 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
Materials/sizes	Given two $200 \times 100 \times 12$ angles, $18 mm$ gusset, $8 mm$ fillet, S355/Class 42.
Centroid lookup	Data File p.12, $200 \times 100$ row, $t = 12$ : $c_{x} = 7.04 cm = 70.4 mm$ , measured from the heel along the long leg; $c_{y} = 2.11 cm$ is the other coordinate.
Angle area method	Course rectangular leg split: share the $12 mm$ corner thickness equally. Do not silently mix this area with table root-fillet area $34.9 {cm}^{2}$ .
Weld detail choices	No longitudinal lengths are given; the final dimensions below are design selections, not measurements from the sketch.

Before calculating: recognition and strategy

Check the angles first, then split the pair force equally and balance each angle’s two weld forces about its centroid. Force divided by weld strength gives effective length, not final physical length. Finally apply all supplied length/leg rules; a strength-only length can be too short for detailing.

(i) Tensile capacity of the double angles —6 printed marks

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

Ultimate tension:

P = 1.4 \times 150 + 1.6 \times 160 = 210 + 256 = 466 kN

Demand per angle:

\frac{466}{2} = 233 kN

Connected leg:

a_{1} = (200 - \frac{12}{2}) \times 12 = 194 \times 12 = 2,328 {mm}^{2}

Outstanding leg:

a_{2} = (100 - \frac{12}{2}) \times 12 = 94 \times 12 = 1,128 {mm}^{2}

Gross/effective area without holes:

2,328 + 1,128 = 3,456 {mm}^{2}

Welded double-angle reduction

0.15 a_{2}

. Resistance area per angle:

3,456 - 0.15 \times 1,128 = 3,286.8 {mm}^{2}

T = 12 \leq 16

, giving

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

. Per angle:

P_{t} = \frac{3,286.8 \times 355}{1,000} = 1,166.814 kN

For the pair:

P_{t} = 2 \times 1,166.814 = 2,333.628 kN > 466

Passes.

A single welded angle would use $0.3$ rather than $0.15$ . Gusset capacity is deliberately not checked because the question expressly excludes it.

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.

(ii) Weld force balance, physical lengths and detailing —9 printed marks

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

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

The connected leg is $200 mm$ high. The angle centroid is $70.4 mm$ from SideX, so the opposite lever arm to SideY is $200 - 70.4 = 129.6 mm$ . Use the opposite lever arm in each force share.

Taking moments about X:

R_{Y} \times 200 = 233 \times 70.4

R_{Y} = 82.016 kN

R_{X} = 233 - 82.016 = 150.984 kN

Check

R_{X} \times 70.4 = R_{Y} \times 129.6

. S355／Class 42:

p_{w} = 250 \frac{N}{{mm}^{2}}

8 mm

fillet-weld throat:

0.7 \times 8 = 5.6 mm

q_{capacity} = \frac{5.6 \times 250}{1,000} = 1.4 \frac{kN}{mm}

Required effective length:

L_{X} = \frac{150.984}{1.4} = 107.845714 mm

L_{Y} = \frac{82.016}{1.4} = 58.582857 mm

Add to each segment

2 s = 16 mm

: X actual length

\geq 123.845714

; select for strength

125 mm

; Y actual length

\geq 74.582857

; select for strength

75 mm

Those are the arithmetic strength minima. However Ch 2 p.33 provision(6) says an end-connection weld length should not be less than the transverse spacing. Here the separation is $200 mm$ . Therefore do not present $125$ / $75 mm$ as a finished detail satisfying every supplied lecture rule.

Thicker part

18 mm

, minimum weld leg

6 mm

; given

8 \geq 6

. Angle edge thickness

12 mm

, maximum weld leg

12 - 2 = 10 mm

;

8 \leq 10

. Minimum effective weld length

\max (4 s, 40) = 40 mm

. Minimum lap length

\max (5 \times 12, 25) = 60 mm

. End return at least

2 s = 16 mm

; select

20 mm

end return.

One deliberately conservative final choice is SideX392 mm and SideY220 mm physical, on each angle. Both exceed $200 mm$ , and their effective lengths $376$ / $204$ nearly preserve the centroid-balancing ratio $129.6$ / $70.4$ . These are chosen design dimensions; the gusset outline must accommodate them. To avoid relying on an approximate balance after rounding, verify the actual two-line group:

L_{X, eff} = 392 - 16 = 376 mm

L_{Y, eff} = 220 - 16 = 204 mm

Total effective

L = 580 mm

. Side X resistance:

376 \times 1.4 = 526.4 kN > 150.984

Side Y resistance:

204 \times 1.4 = 285.6 kN > 82.016

Take a common effective starting point

x = 0

, Side X

y = 0

, Side Y

y = 200

\bar{x} = \frac{376 \times (\frac{376}{2}) + 204 \times (\frac{204}{2})}{580} = 157.751724 mm

\bar{y} = \frac{376 \times 0 + 204 \times 200}{580} = 70.344828 mm

Remaining eccentricity of the angle load line:

70.4 - 70.344828 = 0.055172 mm

J_{line} = \frac{376^{3}}{12} + 376 [{(188 - 157.751724)}^{2} + {70.344828}^{2}]

+ \frac{204^{3}}{12} + 204 [{(102 - 157.751724)}^{2} + {(200 - 70.344828)}^{2}]

= 11,405,292.616092 {mm}^{3}

Radius to the furthest endpoint:

\sqrt{{(376 - 157.751724)}^{2} + {70.344828}^{2}} = 229.304829 mm

Use the sum of direct- and torsional-force magnitudes as a conservative upper bound:

resultant force magnitude \leq direct force magnitude + torsional force magnitude

= \frac{233}{580} + \frac{(233 \times 0.055172) \times 229.304829}{11,405,292.616092}

= 0.401983 \frac{kN}{mm} < 1.4

Passes.

The bound adds magnitudes conservatively, so it is at least as severe as the actual vector resultant. Final $392$ / $220 mm$ runs with $20 mm$ returns satisfy the stated leg, length and lap rules. There are other valid overstrength choices; no official expected final length is supplied. The strength-only $125$ / $75$ calculation is shown so the exam method remains recognizable, while its detailing limitation is explicit.

Animation labBalance two weld forces2 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.

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.

Compact exam answer

$P = 466 kN$ ; the complete pair of angles $P_{t} = 2,333.628 kN$ . Each angle $233 kN$ , centroid lever arm $70.4$ ／ $129.6 mm$ ; $R_{X} = 150.984$ , $R_{Y} = 82.016 kN$ . $8 mm$ fillet weld $q = 1.4 \frac{kN}{mm}$ , required effective length $107.8457$ ／ $58.5829$ , minimum actual length selected for strength $125$ ／ $75 mm$ . The lecture’s end-connection spacing rule requires a longer weld. Here the fully checked selection exceeds $200 mm$ : each angle $X = 392$ ／ $Y = 220 mm$ ， $20 mm$ return welds; conservative actual-group result $q = 0.401983 < 1.4$ . The question excludes the gusset-plate check.

Mistakes to avoid

Do not put $466 kN$ on each angle.
Do not use the centroid coordinate perpendicular to the connected long leg.
Do not stop at effective lengths.
Do not overlook the lecture’s length-versus-transverse-spacing rule.

Procedure for an unfamiliar variant

Check tension with the correct double-welded reduction.
Balance side forces about the angle centroid.
Calculate throat strength and effective lengths.
Add end allowances and apply all stated detailing bounds.
When rounding changes the group centroid, verify the actual provided geometry.

Independent self-check

Try it yourself. Invented reasoning check: why are the strength-only SideX andSideY lengths unequal?

Reveal answer and reasoning

The angle centroid is closer to SideX than SideY. To place the resultant through that centroid, the nearer SideX carries $150.984 kN$ while SideY carries $82.016 kN$ . Equal weld stress/size therefore requires unequal effective lengths. It is not because the long leg carries the full pair’s load.