Worked examples

Connection example 10: welded splice and a minimum-size discrepancy

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Design a welded double-cover splice transferring $700 kN$ ultimate tension between $150 \times 20 mm$ S355 main plates. Cover thickness is $12 mm$ . Use Class 42 electrode. Explain the lecturer’s $6 mm$ -weld solution, then reconcile it with the chapter’s minimum weld-size table.

Original source: LectureNotes/Ch 2_Connection.pdf — p. 43, p. 44. 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.

Chapter 2 p.43: one cover plate on each face of the main plate. The selected cover-plate width is

100 mm

, within a main-plate width of

150 mm

, leaving

25 mm

on each side. Each half of the splice has four longitudinal welds: two edge welds on each of the two cover plates. The cover plate is labelled with thickness

t_{1} = 12 mm

; the main plate with thickness

t_{2} = 20 mm

. The outward arrows at the two ends indicate tension.Open full-size image

Lecturer's arrangement, Chapter 2 p.43: cover-plate width

100 mm

, actual longitudinal weld length

180 mm

; the end return is labelled

15 mm

. The corrected

8 mm

detail is explained in the calculation below and is not drawn into this original image. Some local dimension lettering overlaps; it must not be read as an additional dimension. This lesson uses the readable labels and the source written solution. The original equation above the drawing is

\frac{0.7 \times 6 \times 250}{1,000} = 1.05 \frac{kN}{mm}

\frac{700}{1.05} = 667 mm

; beneath the drawing the source also writes

\frac{667}{4} = 167 mm

. The last two source equalities involve rounding. The detailed solution here retains

666.6667

and

166.6667 mm

before adding the end allowance and rounding the selected length upwards. Figure 6.21 is the source figure identifier.Open full-size image

Read the diagram and collect the data

Input	Origin
Given demand	$700 kN$ ultimate; do not factor again.
Given plates	Main width $150 mm$ and thickness $20 mm$ ; two covers $12 mm$ thick.
Selected cover width	Lecturer chooses $100 mm$ to leave flat room for edge welds; not a given dimension.
Lookup	S355 $p_{y} = 345$ for $20 mm$ main,355 for $12 mm$ covers. $p_{w} = 250 \frac{N}{{mm}^{2}}$ for Class 42.
Detailing lookup	Chapter 2 p.33 Table 9.1: thicker part over $19 mm$ requires minimum leg $8 mm$ . Here the thicker connected main plate is $20 mm$ .

Before calculating: recognition and strategy

The force enters one half-splice from the main plate and divides equally between two covers. Each cover has two longitudinal weld runs, so four runs share the $700 kN$ on each side of the butt. Size runs using effective length, check laps/returns and check unperforated plate tension. The lecturer’s strength arithmetic is reproducible, but a final choice must also satisfy the supplied minimum-size rule.

1. Select and interpret the covers

Selected cover-plate width

= 100 mm

. Allowance on each side:

\frac{150 - 100}{2} = 25 mm

This is an explicit design choice providing space to place the weld on the main plate. Two $12 mm$ covers have combined thickness $24 mm$ but are checked individually at half force $350 kN$ . The source uses the full welded plate area: no bolt-hole deductions.

Animation labTension through an unperforated plate2 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.

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.

2. Reproduce the lecturer’s strength calculation

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 source trials

s = 6 mm

q = \frac{0.7 \times 6 \times 250}{1,000} = 1.05 \frac{kN}{mm}

Required total effective length per half:

\frac{700}{1.05} = 666.6667 mm

4

welds each require effective length:

\frac{666.6667}{4} = 166.6667 mm

Actual length per weld:

\geq 166.6667 + 2 \times 6 = 178.6667 mm \to 180 mm

Provided effective length per weld:

180 - 12 = 168 mm

Half-splice resistance:

4 \times 168 \times 1.05 = 705.6 kN > 700 kN

The four runs on the other side of the butt transfer the same $700 kN$ in series. Do not add both halves to claim $1,411.2 kN$ splice resistance.

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.

3. Reconcile the minimum-size table and give a corrected detail

The lecturer checks lap $\geq 5 \times 12 = 60 mm$ , physical run $180 \geq 100 mm$ transverse spacing, and returns $15 \geq 2 \times 6 = 12 mm$ . However the $20 mm$ main plate triggers the $8 mm$ minimum in Table 9.1 on p.33; the $6 mm$ selection is inconsistent with that supplied table.

Select

s = 8 mm

q = \frac{0.7 \times 8 \times 250}{1,000} = 1.40 \frac{kN}{mm}

Required effective length per weld:

\frac{700}{4 \times 1.40} = 125 mm

Actual length:

\geq 125 + 2 \times 8 = 141 mm \to select 145 mm

Provided effective length:

145 - 16 = 129 mm

Resistance per half:

4 \times 129 \times 1.40 = 722.4 kN > 700

Minimum effective length:

\max (4 \times 8, 40) = 40 mm < 129

Lap length:

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

Actual weld length

145 mm \geq 100 mm

transverse spacing. Provide end return

20 mm \geq 2 \times 8 = 16 mm

12 mm

cover-plate edge, maximum weld leg

= 12 - 2 = 10 mm \geq 8 mm

The corrected detail uses $8 mm$ fillets, $145 mm$ physical longitudinal runs on each half of each cover edge, and $20 mm$ returns. This is an authored correction, clearly distinguished from the lecturer’s dimensions.

Animation labFrom fillet leg to effective throat2 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.

4. Verify main and cover plate tension

Simple explanation: Why holes reduce tension resistance

The pull must squeeze through the steel left beside the holes.

Choose a possible fracture line across the member.
Subtract the holes crossed by that line using hole diameter.
Apply the course effective-area rule and its gross-area limit.

Remember: Do not subtract every hole anywhere in the connection.

Related concept and full method

Main plate:

A = 150 \times 20 = 3,000 {mm}^{2}

P_{t} = \frac{3,000 \times 345}{1,000} = 1,035 kN > 700

Each cover plate:

A = 100 \times 12 = 1,200 {mm}^{2}

P_{t} = \frac{1,200 \times 355}{1,000} = 426 kN > 350

Both plates together:

P_{t} = 2 \times 426 = 852 kN > 700

The corrected welded load path is governed by $722.4 kN$ weld resistance, above $700 kN$ . Main and cover tension pass. Every area is unperforated, so it equals the effective area for this concentric plate connection.

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.

Compact exam answer

Source arithmetic: $6 mm$ fillets, $180 mm$ physical runs give $705.6 kN$ per half; plates give $1,035 kN$ main and $852 kN$ cover pair. But Table 9.1 requires $8 mm$ for the $20 mm$ thicker part. Corrected detail: $8 mm$ fillets, $145 mm$ physical runs on each of four runs per half, $20 mm$ returns. Effective length $129 mm$ /run gives $722.4 kN$ ; lap, length and edge-leg limits pass.

Mistakes to avoid

Four parallel runs occur on each half-splice; opposite halves are in series.
The cover width $100 mm$ is selected, not the $150 mm$ main width.
Strength adequacy does not override the supplied minimum weld-size table.
Increasing leg size changes the $2 s$ length deduction.

Procedure for an unfamiliar variant

Select cover width with room for welds and check plate strength.
Count parallel weld runs on one transfer side only.
Find required effective length, add $2 s$ and round up.
Check minimum/maximum leg, lap, returns and transverse-spacing rules.
Resolve source discrepancies explicitly and verify the corrected detail.

Independent self-check

Try it yourself. For the corrected $8 mm$ welds, would $140 mm$ physical longitudinal runs be sufficient?

Hint

Recompute effective length after selecting the physical length.

Reveal answer and reasoning

Effective length $140 - 16 = 124 mm$ . Resistance $= 4 \times 124 \times 1.40 = 694.4 kN < 700 kN$ , so no. Rounding the $141 mm$ minimum down to $140 mm$ loses required capacity.

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.