Worked examples

Tutorial 2 Q2: welded splice strength and minimum-leg correction

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Check the $850 kN$ ultimate tension splice in FigureQ2. Steel is S355, electrode Class 42, and the drawing specifies $6 mm$ fillet welds, $12 mm$ cover plates and $22 mm$ main plates.

Original source: LectureNotes/Ch 2_Connection.pdf — p. 45, p. 46. 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	How it is used
Parallel runs	Each cover has two longitudinal side welds on each half; with two covers there are four load-transfer runs per half.
Length	Each run has physical length $220 mm$ . Do not use $440 + 10$ as one effective run.
Material	The original website summary writes weld strength as $p_{v}$ , while the calculation below uses $p_{w}$ . This explanation follows the calculation notation. S355/Class 42 weld strength $p_{w} = 250 \frac{N}{{mm}^{2}}$ . Main-plate thickness $22 mm$ , giving $p_{y} = 345$ ; cover-plate thickness $12 mm$ , giving $p_{y} = 355 \frac{N}{{mm}^{2}}$ .
Detailing	Thicker joined part $22 mm$ selects minimum $8 mm$ from Ch 2 Table 9.1. Cover edge $12 mm$ permits at most $10 mm$ by $t - 2$ .

Before calculating: recognition and strategy

The two splice halves each transfer $850 kN$ . Sum the four parallel longitudinal runs within one half. Start/end deductions apply to each run. Check the main and cover plate tensile capacities, then examine weld-size and return detailing independently. Short corner returns are not needed as extra strength in this conservative longitudinal-run model.

1. Calculate the drawn $6 mm$ weld resistance

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

Throat:

a = 0.7 s = 0.7 \times 6 = 4.2 mm

Resistance per unit length:

q = \frac{a p_{w}}{1,000} = \frac{4.2 \times 250}{1,000} = 1.05 \frac{kN}{mm}

Effective length per weld:

220 - 2 \times 6 = 208 mm

Total effective length per half:

4 \times 208 = 832 mm

Resistance:

832 \times 1.05 = 873.6 kN > 850 kN

The original weld strength passes. Force per weld:

\frac{850}{4} = 212.5 kN

Resistance per weld:

208 \times 1.05 = 218.4 kN

Adding both left and right halves would double-count resistance in series.

Animation labFrom fillet leg to effective throat1 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.

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.

2. Plate tensile capacities

The main plate has no bolt holes:

A = 150 \times 22 = 3,300 {mm}^{2}

P_{t, main} = \frac{3,300 \times 345}{1,000} = 1,138.5 kN > 850 kN

Each cover plate:

A = 130 \times 12 = 1,560 {mm}^{2}

P_{t, cover} = \frac{1,560 \times 355}{1,000} = 553.8 kN > 425 kN

Both cover plates together:

1,107.6 kN > 850 kN

Plate strength passes.

Use $130 mm$ for each cover, not $150 mm$ . The width is fixed by the $10 + 130 + 10$ dimension chain.

Animation labTension through an unperforated plate1 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.

For this concentric, unperforated plate, the full width and thickness form the resisting section. There are no bolt holes to deduct.
The illustrative width is 100 mm. Adjust thickness t: $A = b \times t$ , in $mm^{2}$ .
Using illustrative $p_{y} = 275 \frac{N}{mm ^{2}}$ , $P_{t} = \frac{A p _{y}}{1000}$ in kN. Use the actual material and thickness band for the original question.
The covers are moved apart for visibility. A symmetric pair each carries half the force. Check the main plate, each cover and the welds separately.

3. Check the drawn detail and make a labelled 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

Thicker part

22 mm > 19 mm

, minimum weld leg

8 mm

. The original

s = 6 mm < 8 mm

does not meet the supplied minimum-size rule.

12 mm

cover-plate edge, maximum weld leg:

12 - 2 = 10 mm

Original

6 mm

weld effective length:

208 > \max (4 \times 6, 40) = 40 mm

Lap length:

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

Longitudinal length exceeds transverse spacing:

longitudinal length 220 mm > transverse spacing 130 mm

Original end return:

15 \geq 2 \times 6 = 12 mm

Thus the given detail has enough calculated strength but fails the course minimum fillet size. A supported correction is $8 mm$ leg with the same $220 mm$ longitudinal runs, plus enlarged returns of at least $16 mm$ ; choose $20 mm$ returns as a design choice.

Revise to

8 mm

, giving:

q = \frac{0.7 \times 8 \times 250}{1,000} = 1.4 \frac{kN}{mm}

Per weld:

L_{eff} = 220 - 16 = 204 mm

Half-joint resistance:

4 \times 204 \times 1.4 = 1,142.4 kN > 850 kN

8 mm

lies between minimum

8 mm

and maximum

10 mm

204 \geq \max (32, 40) = 40 mm

End return

20 mm \geq 16 mm

. Main-plate and combined cover-plate tension resistances remain respectively

1,138.5

and

1,107.6 kN

The corrected joint’s requested strength checks pass; cover tension is now the lowest resistance. The image remains the original $6 mm$ / $15 mm$ detail so the change is auditable.

Animation labFrom fillet leg to effective throat2 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.

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

Given $6 mm$ weld: four effective $208 mm$ runs per half, capacity 873.6 kN>850. Main/cover pair tension $1,138.5$ / $1,107.6 kN$ pass. However $22 mm$ thicker part requires $8 mm$ minimum fillet, so the drawn detail fails that condition. Proposed correction: $8 mm$ fillet, $220 mm$ runs, $20 mm$ returns; weld capacity $1,142.4 kN$ , acceptable $8 - 10 mm$ size range and plate capacities unchanged.

Mistakes to avoid

Deduct $2 s$ on each run.
Do not count corner returns as extra needed resistance without a consistent model.
Check minimum size even when strength passes.
After increasing weld size, update effective lengths and returns.

Procedure for an unfamiliar variant

Identify the force path and whether the joint is concentric, in-plane eccentric or out-of-plane eccentric.
Read all dimensions from the relevant view; do not measure drawing scale.
Distinguish characteristic from ultimate loads and factor only when needed.
Calculate demand per fastener/weld with the appropriate equilibrium model.
Check all requested resistances and detailing conditions separately.
State whether the given detail passes, fails, or remains conditional on missing data.

Independent self-check

Try it yourself. Invented variant: keep the corrected $8 mm$ weld but shorten each physical run to $165 mm$ . Does weld strength pass $850 kN$ ?

Reveal answer and reasoning

Effective length: $165 - 16 = 149 mm$ Resistance: $4 \times 149 \times 1.4 = 834.4 kN < 850 kN$ Fails. Required actual length: $\geq \frac{850}{4 \times 1.4} + 16 = 167.786 mm$ Adopt $170 mm$ or greater, then recheck all detailing requirements.

Animation labFrom fillet leg to effective throat1 concept · 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.

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.

Tutorial 2 Q2: welded splice strength and minimum-leg correction

Read the diagram and collect the data

Before calculating: recognition and strategy

1. Calculate the drawn 6mm6mm weld resistance

2. Plate tensile capacities

3. Check the drawn detail and make a labelled correction

Compact exam answer

Mistakes to avoid

Procedure for an unfamiliar variant

Independent self-check

1. Calculate the drawn $6 mm$ weld resistance