Worked examples

Connection example 6: balance two weld lengths about the load line

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Design the side fillet welds for a $65 \times 50 \times 8$ angle carrying $60 kN$ characteristic dead tension and $70 kN$ characteristic imposed tension through its centroid. Use S355 steel and Class 42 electrode.

Original source: LectureNotes/Ch 2_Connection.pdf — p. 38. 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
Given loads	$60 kN$ dead, $70 kN$ imposed, arrows through dashed centroidal axis.
Given geometry	The left sketch labels $65 \times 50 \times 8$ L; connected-leg width $65 mm$ , centroid distance $21.1$ and $43.9 mm$ .
Design choice	Lecturer tries $6 mm$ fillet and final side lengths $140 mm$ (X), $75 mm$ (Y), shown at right.
Lookup	Table 9.2a, S355 row/Class 42 column: $p_{w} = 250 \frac{N}{{mm}^{2}}$ . Throat $a = 0.7 s$ .

Before calculating: recognition and strategy

A longer weld must be put on the side closer to the load line. Treat the two parallel weld lines as supports of the tensile force across the $65 mm$ separation: force equilibrium gives $R_{X} + R_{Y} = P$ ; moment equilibrium about X gives $R_{Y} \times 65 = P \times 21.1$ . Equal weld lengths would move the connection resultant away from the angle centroid.

1. Factored tensile force

P = 1.4 \times 60 + 1.6 \times 70 = 84 + 112 = 196 kN

Animation labFrom characteristic to design load1 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.

G is permanent load; Q is imposed load. A surface load and a line load also have different units.
This illustration uses the course gravity case $1.4 G + 1.6 Q$ . Other combinations in the original text retain their own factors.
For illustrative $G = 5 \frac{kN}{m ^{2}}$ , change Q and watch each separate contribution.
Do not carry this ULS total automatically into deflection. Follow the stated SLS load case.

2. Select a trial leg and find capacity per length

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

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

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

Required total effective length:

\frac{P}{q} = \frac{196}{1.05} = 186.667 mm

The angle edge thickness is $8 mm$ , so the source maximum edge-weld leg $8 - 2 = 6 mm$ permits this selection. The supporting plate thickness is not given; its effect on the minimum weld-size rule remains a detailing condition.

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.

3. Share force and effective length between the sides

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

R_{Y} = \frac{196 \times 21.1}{65} = 63.6246 kN

R_{X} = 196 - 63.6246 = 132.3754 kN

Alternatively calculate directly

\frac{196 \times 43.9}{65}

L_{X, eff} = \frac{132.3754}{1.05} = 126.0718 mm

L_{Y, eff} = \frac{63.6246}{1.05} = 60.5949 mm

Check: the forces sum $196 kN$ and the required lengths sum $186.6667 mm$ . The larger $43.9 mm$ opposite lever arm produces the larger SideX force.

Animation labBalance two weld forces2 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 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.

4. Add end allowances, round up and check the actual lengths

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

Side X actual length:

\geq 126.0718 + 2 \times 6 = 138.0718 mm \to 140 mm

Side Y actual length:

\geq 60.5949 + 12 = 72.5949 mm \to 75 mm

Provided effective length:

X = 140 - 12 = 128 mm; Y = 75 - 12 = 63 mm

Resistance:

X = 128 \times 1.05 = 134.4 kN > 132.3754

Y = 63 \times 1.05 = 66.15 kN > 63.6246

Each effective length exceeds $\max (4 \times 6, 40) = 40 mm$ . The physical side lengths $140$ and $75 mm$ also exceed the $65 mm$ transverse separation, matching the source detailing check. The extra rounding increases available resistance; the actual force split still follows equilibrium. End-return/support-plate details are not fully specified, so this is the completed requested side-weld strength/length design, not a complete fabrication detail.

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

$P = 196 kN$ . Use $6 mm$ fillet, $q = 1.05 \frac{kN}{mm}$ . Force split X/Y= $132.375$ / $63.625 kN$ . Required effective lengths $126.072$ / $60.595 mm$ . Add $12 mm$ to each and provide $140 mm$ (X), $75 mm$ (Y); effective capacities $134.4$ / $66.15 kN$ exceed the assigned forces.

Mistakes to avoid

Read the angle label: the source says $8 mm$ thick, not $6 mm$ .
Use the opposite centroid lever arm when allocating weld force.
Add $2 s$ separately to both weld runs.
Do not use total capacity alone if one side is too short.

Procedure for an unfamiliar variant

Locate the centroidal load line and both weld lines.
Factor loads and select a permitted trial weld leg.
Use force and moment equilibrium for the two side forces.
Divide each force by weld strength per length.
Add individual end deductions back, round up, and recheck each side.

Independent self-check

Try it yourself. Suppose the same $196 kN$ force acts midway between the $65 mm$ -spaced weld lines. What physical length is required on each side for $6 mm$ weld?

Reveal answer and reasoning

Each side takes $98 kN$ . Required effective length $= \frac{98}{1.05} = 93.333 mm$ . Add $12 mm \to 105.333 mm$ ; choose $110 mm$ each. Centred loading makes equal lengths appropriate.

Animation labBalance two weld forces2 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 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.