Worked examples

Connection example 8: flange welds carry moment, web welds carry shear

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Design the bracket welds for $100 kN$ dead plus $140 kN$ imposed load at $250 mm$ from the support. The bracket is cut from a $356 \times 171 \times 67$ UB. The specified web-weld leg is half the flange-weld leg. Use S355 and Class 42 electrode.

Original source: LectureNotes/Ch 2_Connection.pdf — p. 41. 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 load	$100 kN$ dead, $140 kN$ imposed; downward arrow at $e = 250 mm$ from connection face.
Given weld geometry	$B = 150 mm$ across flange run; $D = 364.6 mm$ vertical arm; $a = 280 mm$ web run; all explicitly dimensioned.
Given size relationship	$s_{web} = \frac{s_{flange}}{2}$ 。
Lookup/model	$p_{w} = 250 \frac{N}{{mm}^{2}}$ . Rotate about $X_{1} - X_{1}$ at bottom flange; flange weld takes moment, two web welds take direct shear. This is the source’s simplified force allocation.

Before calculating: recognition and strategy

Do not treat this as a rectangular in-plane torsion weld. The vertical load outside the supporting face produces flange tension and compression. Moment equilibrium gives tensile flange force $F = \frac{M}{D}$ . Divide that force by the effective flange weld length to get demand per millimetre. Because effective length depends on chosen leg, select a trial leg and verify it self-consistently.

1. Design force and force couple

P = 1.4 \times 100 + 1.6 \times 140 = 140 + 224 = 364 kN

M = P e = 364 \times 250 = 91,000 kN \cdot mm

Flange force:

F = \frac{M}{D} = \frac{91,000}{364.6} = 249.589 kN

Animation labSeparate the moment couple and shear1 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.

A simplified moment connection assigns different actions to different fastener groups.
Opposite flange forces separated by z resist moment: $F = \frac{M}{z}$ .
The web fasteners or welds carry the assigned vertical shear in this model.
Flange force, web shear, plate bearing and detailing each need their specified checks.

2. Trial $12 mm$ flange fillets

Simple explanation: How two flange forces make a moment

Two opposite forces form a turning pair, like two hands turning a wheel.

Identify the separation between their actual force lines.
Required force equals moment divided by that separation.
Design the relevant flange group for that force.

Remember: The force-line separation is not automatically the overall section depth.

Related concept and full method

The source tries $12 mm$ . The available $150 mm$ run loses $2 s = 24 mm$ at its ends, leaving $126 mm$ effective. Use the printed weld run $B$ , not the nominal $171 mm$ section width.

L_{flange, eff} = 150 - 2 \times 12 = 126 mm

q_{flange} = \frac{F}{L_{eff}} = \frac{249.589}{126} = 1.980862 \frac{kN}{mm}

s_{needed} = \frac{1,000 \times 1.980862}{0.7 \times 250} = 11.3192 mm

12 mm \geq 11.3192 mm; q_{cap} = \frac{0.7 \times 12 \times 250}{1,000} = 2.10 \frac{kN}{mm}

Flange-weld resistance:

2.10 \times 126 = 264.6 kN > 249.589

Passes.

This also solves the implicit inequality $\frac{0.7 p_{w} s (B - 2 s)}{1,000} \geq \frac{M}{D}$ by a verified trial. Increasing $s$ improves throat but shortens effective length, so do not omit the recheck.

Animation labFrom fillet leg to effective throat2 concepts · 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.

3. Apply the required half-size rule and check shear

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_{web} = \frac{12}{2} = 6 mm

Effective length per weld:

L_{web, eff} = 280 - 2 \times 6 = 268 mm

Total effective web-weld length:

2 \times 268 = 536 mm

q_{web} = \frac{364}{536} = 0.679104 \frac{kN}{mm}

s_{needed, web} = \frac{1,000 \times 0.679104}{175} = 3.88060 mm

The provided

6 mm

gives

q_{cap} = 1.05 \frac{kN}{mm}

; web-weld resistance:

536 \times 1.05 = 562.8 kN > 364

Passes.

Even though strength alone needs only about $3.9 mm$ , use $6 mm$ to satisfy the given relationship to the $12 mm$ flange weld.

Animation labFrom fillet leg to effective throat2 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.

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. Check lengths and identify the remaining thickness condition

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

Minimum effective flange-weld length:

\max (4 \times 12, 40) = 48 mm < 126

Minimum effective web-weld length:

\max (4 \times 6, 40) = 40 mm < 268

The source dimensions describe weld runs on the cut bracket, not a licence to infer other fabrication details from scale. The supporting column thickness is not stated. The result therefore establishes flange/web weld strength and the size ratio; applicability of the thicker-part minimum-size rule to the support must be confirmed separately.

Animation labFrom fillet leg to effective throat2 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.

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

$P = 364 kN$ , $M = 91 kN \cdot m$ . About $X_{1} - X_{1}$ , flange force $= 249.589 kN$ . $12 mm$ flange weld has $126 mm$ effective length and $264.6 kN$ resistance. $6 mm$ web welds each have $268 mm$ effective length, combined resistance $562.8 kN$ . Both pass and web leg is half flange leg. Support thickness for minimum weld-size verification is not supplied.

Mistakes to avoid

The moment arm $D$ is $364.6 mm$ ; the web run $a$ is $280 mm$ . They are not interchangeable.
Use two web welds, but the source’s tension-flange force is resisted by the specified flange run.
Subtract $2 s$ using that run’s own leg size.

Procedure for an unfamiliar variant

Identify the source pivot and the flange-force arm.
Factor load and calculate moment/force couple.
Try a flange leg, calculate its effective length and verify capacity.
Apply the specified web/flange size relationship and check web shear.
Check effective-length/detailing rules and state unprovided support data.

Independent self-check

Try it yourself. Keep the selected welds and increase the eccentricity to $300 mm$ . What happens to the flange strength check?

Reveal answer and reasoning

Flange force becomes $\frac{364 \times 300}{364.6} = 299.506 kN$ , greater than $264.6 kN$ . The flange weld fails while the unchanged direct web shear remains $364 kN$ and still passes.

Animation labSeparate the moment couple and shear2 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.

A simplified moment connection assigns different actions to different fastener groups.
Opposite flange forces separated by z resist moment: $F = \frac{M}{z}$ .
The web fasteners or welds carry the assigned vertical shear in this model.
Flange force, web shear, plate bearing and detailing each need their specified checks.

Connection example 8: flange welds carry moment, web welds carry shear

Read the diagram and collect the data

Before calculating: recognition and strategy

1. Design force and force couple

2. Trial 12mm12mm flange fillets

3. Apply the required half-size rule and check shear

4. Check lengths and identify the remaining thickness condition

Compact exam answer

Mistakes to avoid

Procedure for an unfamiliar variant

Independent self-check

2. Trial $12 mm$ flange fillets