Worked examples

Tutorial 2 Q6: eccentric welds on two side plates

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Determine the fillet weld leg for two $20 mm$ side plates carrying characteristic dead $350 kN$ plus imposed $300 kN$ . The load is $200 mm$ from the column face. Steel is S355 and electrode Class 42.

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

Quantity	Source and derivation
Per-plate load	Two identical side plates in the plan/front arrangement share the total equally.
Group size	Use the labelled $200 mm$ width and $300 mm$ height for each four-sided weld group.
Load eccentricity	From group centre to load: half the $200 mm$ width plus $200 mm$ beyond the face → $e = 300 mm$ .
Weld model	Uniform all-round fillet represented as a closed line rectangle, as in Connection Example 7. The plate area is not the weld-line area.
Size bounds	$20 mm$ plate sets minimum $8 mm$ under Table 9.1 and edge maximum $18 mm$ under $t - 2$ .

Before calculating: recognition and strategy

Two different $200 mm$ labels play different roles. One defines the rectangular group width; the other begins at the column face. The centroid is halfway across the group, so the load arm is $300 mm$ . Calculate forces per length for one plate using half the load, then select a weld leg that meets resistance and detailing.

1. Factored force, centroid and moment

Total ultimate load:

P = 1.4 \times 350 + 1.6 \times 300 = 490 + 480 = 970 kN

Each side plate:

P_{1} = \frac{970}{2} = 485 kN

Weld group

b = 200 mm

h = 300 mm

. Total length:

L = 2 (b + h) = 1,000 mm

Coordinate distances from centroid to corner:

x = 100

y = 150 mm

e = 200 + \frac{b}{2} = 200 + 100 = 300 mm

M = P_{1} e = 485 \times 300 = 145,500 kN \cdot mm

Equal sharing assumes the displayed symmetric plate pair and centred attachment of the applied load. Using 485 with two weld groups already accounts for both plates; do not double the capacity again within a single group.

Animation labA weld group is a set of lines2 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 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.

2. Derive line-weld polar inertia

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

I_{x} = 2 (\frac{h^{3}}{12}) + 2 b {(\frac{h}{2})}^{2}

= \frac{2 \times 300^{3}}{12} + 2 \times 200 \times 150^{2}

= 4,500,000 + 9,000,000 = 13,500,000 {mm}^{3}

I_{y} = 2 (\frac{b^{3}}{12}) + 2 h {(\frac{b}{2})}^{2}

= \frac{2 \times 200^{3}}{12} + 2 \times 300 \times 100^{2}

= 1,333,333.333 + 6,000,000 = 7,333,333.333 {mm}^{3}

I_{p} = I_{x} + I_{y} = 20,833,333.333 {mm}^{3}

Treat each side as a line. A parallel-axis term is length times distance squared, giving units ${mm}^{3}$ . This calculation finds force per unit weld length; do not use solid-plate inertia with units ${mm}^{4}$ .

Animation labA weld group is a set of lines2 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 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.

3. Find the critical corner resultant per length

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

Direct vertical force:

q_{s} = \frac{P_{1}}{L} = \frac{485}{1,000} = 0.485 \frac{kN}{mm}

Vertical torsional component:

q_{Ty} = \frac{M x}{I_{p}}

= \frac{145,500 \times 100}{20,833,333.333} = 0.6984 \frac{kN}{mm}

Horizontal torsional component:

q_{Tx} = \frac{M y}{I_{p}}

= \frac{145,500 \times 150}{20,833,333.333} = 1.0476 \frac{kN}{mm}

At a corner where vertical components add:

q_{vertical} = 0.485 + 0.6984 = 1.1834 \frac{kN}{mm}

q_{\max} = \sqrt{{1.1834}^{2} + {1.0476}^{2}} = 1.580475 \frac{kN}{mm}

At the opposite edge the vertical components subtract, giving a smaller resultant. Top and bottom corners on the adding side have the same magnitude.

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.

4. Select and verify the fillet leg

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

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

s_{required} = \frac{1.580475}{0.175} = 9.031286 mm

For each plate, select an all-round

10 mm

fillet weld. Resistance per unit length:

0.175 \times 10 = 1.75 \frac{kN}{mm} > 1.580475 \frac{kN}{mm}

Minimum weld leg, selected size and plate-edge maximum:

8 \leq 10 \leq 20 - 2 = 18 mm

The closed all-round group follows the source’s continuous line-weld model; do not treat its four connected sides as four separately terminated $180$ / $280 mm$ runs without revising the model. Its $200$ and $300 mm$ sides exceed the minimum effective-length threshold $\max (4 s, 40) = 40 mm$ . The selected size completes the requested weld-group resistance under the displayed four-sided, symmetric model. Column local resistance is a separate check not requested or fully specified.

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.

Compact exam answer

Total ULS load $970 kN$ , per plate $485 kN$ . Rectangle $200 \times 300 mm$ ， $L = 1,000 mm$ , $e = 300 mm$ , $M = 145,500 kN \cdot mm$ . $I_{p, line} = 20.833333 \times 10^{6} {mm}^{3}$ . Direct force $0.485$ , torsional vertical $0.6984$ / horizontal $1.0476 \frac{kN}{mm}$ ; $q_{\max} = 1.580475 \frac{kN}{mm}$ . $s \geq 9.0313 mm$ ; select an all-around $10 mm$ fillet weld on each plate. Resistance $1.75 \frac{kN}{mm}$ , satisfying the $8 - 18 mm$ size range.

Mistakes to avoid

The load arm starts at the weld centroid, not the column face.
Split the force between the two plates once.
Do not use solid-rectangle inertia.
Vector-add direct and torsional components at the critical corner.

Procedure for an unfamiliar variant

Establish the number of parallel weld groups and force per group.
Locate the weld-line centroid and actual load eccentricity.
Calculate total line length and polar line inertia.
Resolve direct/torsional force-per-length components and find their maximum resultant.
Round weld leg up and verify its capacity and allowed size range.

Independent self-check

Try it yourself. Invented variant: the load is $200 mm$ from the group centroid instead of the column face. Find the required leg, keeping other data unchanged.

Reveal answer and reasoning

$M = 485 \times 200 = 97,000 kN \cdot mm$ Torsional vertical component $0.4656$ and horizontal component $0.6984 \frac{kN}{mm}$ . $q_{\max} = \sqrt{{(0.485 + 0.4656)}^{2} + {0.6984}^{2}} \approx 1.1795 \frac{kN}{mm}$ The strength requirement gives a weld leg of approximately $6.74 mm$ . However, Table 9.1 for a $20 mm$ plate still requires at least $8 mm$ , so select $8 mm$ , resistance $1.4 \frac{kN}{mm}$ , conditional on the same all-around weld model.

Animation labA weld group is a set of lines2 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.

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.