Worked examples

2024 BQ2(b): four-sided weld on a channel web

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Determine the required fillet-weld leg for the S355 plate welded on four sides to a parallel-flange-channel web, using Class 42 electrode. Rectangle $250 mm$ wide by $200 mm$ high. Characteristic dead $250 kN$ and imposed $150 kN$ act on the right edge line.

Original source: Pastpaper/23ENGTY004.pdf — p. 4. 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
Loads	Given dead $250$ and live $150 kN$ ; apply the course ULS combination once.
Geometry	GivenB250,H200 mm; symmetry gives corner coordinates $\pm 125$ / $\pm 100 mm$ .
Given formulas	The formula on the complete original exam page is $I_{x} = \frac{y^{3}}{6} + \frac{x y^{2}}{2}$ and $I_{y} = \frac{x^{3}}{6} + \frac{x^{2} y}{2}$ ; in that source expression, $x$ , $y$ denote full rectangle width and height respectively. The website working uses $B = x$ , $H = y$ to avoid confusion with corner coordinates, writing $I_{x} = \frac{H^{3}}{6} + \frac{B H^{2}}{2}$ and $I_{y} = \frac{B^{3}}{6} + \frac{B^{2} H}{2}$ . These are weld-line inertias, in ${mm}^{3}$ ; the two notations are equivalent after matching their definitions.
Weld strength	S355/Class 42 gives $p_{w} = 250 \frac{N}{{mm}^{2}}$ ; equal-leg weld throat $0.7 s$ .
Absent detail	Plate and channel-web thicknesses are not given; exposed-edge maximum leg requires their actual values.

Before calculating: recognition and strategy

Separate the direct vertical line force from the moment-induced tangential force. The critical corner is the side where the torsional vertical component adds to direct shear. Use the given line-inertia formulas with full rectangle dimensions, then resolve components before selecting a weld leg.

Complete 13-mark solution: load, line inertia, corner resultant and size

Simple explanation: Only the weld lines resist as weld

Imagine a wire rectangle: its empty middle is not more wire.

Count only the weld runs actually shown.
Use their total length and line second moments.
Combine direct and torsional forces at the critical location.

Remember: Line second moments have units ${mm}^{3}$ ; plate-area moments use ${mm}^{4}$ .

Related concept and full method

Simple explanation: Add arrows before taking the magnitude

Walking east and walking north do not point in the same direction.

Choose signed horizontal and vertical directions.
Add contributions along each direction separately.
For perpendicular components, use the right-triangle resultant.

Remember: Check the corner where direct and torsional components reinforce each other.

Related concept and full method

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

Ultimate load:

P = 1.4 \times 250 + 1.6 \times 150 = 350 + 240 = 590 kN

Effective closed-perimeter line length:

L = 2 (250 + 200) = 900 mm

e = \frac{250}{2} = 125 mm

M = P e = 590 \times 125 = 73,750 kN \cdot mm

Direct line shear:

q_{s} = \frac{P}{L} = \frac{590}{900} = 0.655555556 \frac{kN}{mm}

For the continuous closed-perimeter course model, use the rectangle’s full effective line. Separate terminated runs would require their actual end deductions and revised geometry; the question’s provided inertia formulas assume this continuous rectangle.

I_{x} = \frac{200^{3}}{6} + \frac{250 \times 200^{2}}{2}

= 1,333,333.333333 + 5,000,000 = 6,333,333.333333 {mm}^{3}

I_{y} = \frac{250^{3}}{6} + \frac{250^{2} \times 200}{2}

= 2,604,166.666667 + 6,250,000 = 8,854,166.666667 {mm}^{3}

J = I_{x} + I_{y} = 15,187,500 {mm}^{3}

Corner distances relative to the centroid

x = 125

y = 100 mm

. Vertical torsional component:

\frac{M \times 125}{J} = \frac{73,750 \times 125}{15,187,500} = 0.606995885 \frac{kN}{mm}

Total vertical line force:

q_{v} = 0.655555556 + 0.606995885 = 1.262551440 \frac{kN}{mm}

Horizontal component:

q_{h} = \frac{M \times 100}{J} = \frac{73,750 \times 100}{15,187,500} = 0.485596708 \frac{kN}{mm}

Resultant:

q = \sqrt{q_{v}^{2} + q_{h}^{2}}

= \sqrt{{1.262551440}^{2} + {0.485596708}^{2}} = 1.352715899 \frac{kN}{mm}

Both stresses are now expressed as force per effective weld length, so vector combination is dimensionally valid. The opposite side subtracts the torsional vertical component and does not govern.

Resistance per unit length:

q_{cap} = \frac{0.7 s p_{w}}{1,000} = \frac{0.7 s \times 250}{1,000} = 0.175 s \frac{kN}{mm}

Required weld leg:

s \geq \frac{1.352715899}{0.175} = 7.729805 mm

Trial for strength

8 mm

. Throat

0.7 \times 8 = 5.6 mm

;

q_{cap} = 1.4 \frac{kN}{mm}

. Utilisation:

\frac{1.352715899}{1.4} = 0.966226 < 1

Strength passes.

$8 mm$ meets every minimum-leg value listed in the source table, whose largest minimum is $8 mm$ . However the thinner exposed edge must allow that leg: under the source $t \geq 6 mm$ rule, $8 \leq t - 2$ requires $t \geq 10 mm$ . Neither thickness is supplied, so that maximum-size applicability remains unverified.

Animation labA weld group is a set of lines3 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.

The centre is empty: resistance comes from the weld lines, not a solid plate filling the group.
Use line lengths for centroid weighting. Line second moments have units $mm^{3}$ ; add the parallel-axis terms.
Direct force per length is $\frac{P}{L}$ . The eccentric moment adds tangential flow proportional to distance from the centroid.
Combine signed components at every candidate corner, then compare the maximum with throat resistance per length.

Compact exam answer

$P = 590 kN$ , $e = 125 mm$ , $M = 73,750 kN \cdot mm$ , $L = 900 mm$ . $I_{x} = 6.333333 \times 10^{6}$ , $I_{y} = 8.854167 \times 10^{6}$ , $J = 15.1875 \times 10^{6} {mm}^{3}$ . Governing $q_{v} = 1.26255144$ , $q_{h} = 0.485596708$ , resultant $1.352715899 \frac{kN}{mm}$ . $s \geq 7.729805 mm$ ; trial for strength $8 mm$ ， $q_{cap} = 1.4$ . Maximum weld leg at the edge depends on unprovided thickness information, so fabrication details cannot be confirmed unconditionally.

Mistakes to avoid

Do not use $250 mm$ as the centroid eccentricity.
Do not halve $590 kN$ : there is one loaded group.
Use ${mm}^{3}$ line inertia, not ${mm}^{4}$ plate-area inertia.
Do not ignore the maximum permitted edge weld size.

Procedure for an unfamiliar variant

Locate the load line relative to the weld centroid.
Factor characteristic loads and computePe.
Compute line length and both line inertias.
Resolve the corner torsional components and add direct shear with signs.
Convert resultant to leg and check the actual detail limits.

Independent self-check

Try it yourself. Invented variant: both characteristic loads increase $4 %$ proportionally. Does the $8 mm$ strength trial pass?

Reveal answer and reasoning

The resultant increases to $1.04 \times 1.352715899 = 1.406824535 \frac{kN}{mm}$ , above $1.4$ . The $8 mm$ trial fails strength even though the original utilisation was below $1$ . Geometry/materials unchanged means resistance does not scale with load.

Animation labA weld group is a set of lines1 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 centre is empty: resistance comes from the weld lines, not a solid plate filling the group.
Use line lengths for centroid weighting. Line second moments have units $mm^{3}$ ; add the parallel-axis terms.
Direct force per length is $\frac{P}{L}$ . The eccentric moment adds tangential flow proportional to distance from the centroid.
Combine signed components at every candidate corner, then compare the maximum with throat resistance per length.