Learn

Connections: trace the force through bolts, plates and welds

Open “Animation lab” beside a teaching step for a visual explanation or a walkthrough of its original expressions. Models are illustrative; source answers remain unchanged.

Chinese–English terminology

Need a simpler picture? Open “Simple explanation” beside a difficult step. These optional notes do not replace the full solution.

A connection transfers force between pieces of steel. Identify what crosses the joint before counting bolts or measuring weld. A force acting along a member can produce bolt shear, plate tension and local bearing at the same time; all relevant resistances must be satisfactory.

Lecture examples are included immediately after the relevant concepts. Expand an example to read its complete diagram, working, exam answer and self-check here.

Bolts, grades and weld types

Source: Chapter 2 pp.2–3 and 6–7. An ordinary clearance bolt transfers load through contact between its shank and the hole after any clearance movement. A friction-grip connection uses clamping and friction; ordinary bearing-bolt equations alone are not an HSFG slip check. For the ordinary Grade 8.8 bolts used in the worked examples, the course tables give $p_{s} = 375$ , $p_{bb} = 1,000$ and $p_{t} = 560 \frac{N}{{mm}^{2}}$ . Their nominal material ultimate tensile strength $U_{b} = 800 \frac{N}{{mm}^{2}}$ is a different quantity.

A grade label 8.8 indicates nominal ultimate strength $800 \frac{N}{{mm}^{2}}$ and yield ratio $0.8$ , hence nominal yield strength $640 \frac{N}{{mm}^{2}}$ in Chapter 2 Table 1. The code design strengths above are the values to use for capacity. Grade 4.6 has nominal yield/ultimate $240$ / $400$ ; Grade 10.9 has $900$ / $1,000 \frac{N}{{mm}^{2}}$ .

Chapter 2 p.7, Figure 3: weld leg and throat are different dimensions. The original labels identify Single V butt weld, Fillet weld, Single U/Double V/Double U preparations, Partial butt weld and Deep penetration fillet weld. Local labels identify Root, Angle of faces, Reinforcement, Gap and Throat;

Size = leg length

means the specified dimension equals the leg length; Fusion zone identifies the fused region. The middle detail in sketch (d) combines a partial-penetration butt weld with fillet welds.Open full-size image

Fillet welds have a triangular profile; the specified size $s$ is the leg, not the throat. Butt welds join prepared plate edges, such as single/double V or U preparations. Visual inspection, dye penetrant and magnetic-particle methods look for surface issues; ultrasonic and X-ray methods can detect internal defects as described in the notes.

Animation labCount the bolt shear planes3 concepts · 5 source expressions

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

Load must cross an interface between the connected plates.
A lap joint gives one shear plane through a bolt.
A symmetric double-cover joint may provide two shear planes. Count load-transfer interfaces, not just visible plates.
Use the source’s area and shear strength. Then check bearing, plate resistance and detailing separately.

Hole size, pitch, gauge and edge distances

Simple explanation: The bolt can crush or tear the plate

A strong bolt can still push through a weak hole edge.

Check the bolt and each connected plate’s bearing bounds.
Use nominal bolt diameter for bearing; hole size for removed material.
The smallest applicable resistance controls.

Remember: A short end distance may govern even when bolt shear passes.

Related concept and full method

Source: Chapter 2 pp.3–4. Nominal bolt diameter $d$ is the steel shank size. Hole diameter $d_{0}$ is larger: $d + 2 mm$ for $d \leq 24 mm$ , $d + 3 mm$ for larger ordinary bolts. Use $d$ for bearing area and spacing rules; $d_{0}$ for material removed by holes.

Chapter 2 p.4, Table 9.3 extract: choose the bolt size row and the correct edge-manufacture column. — Chapter 2 p.4, Table 9.3 extract: minimum end and edge distances for standard holes. Each pair gives, first, sheared/hand flame-cut edges and, second, rolled plate/section/bar edges or machine gas-cut edges, in $mm$ . Source rows: M12: $22$ ／ $18$ ; M16: $28$ ／ $22$ ; M18: $32$ ／ $24$ ; M20: $34$ ／ $26$ ; M22: $38$ ／ $28$ ; M24: $42$ ／ $30$ ; M27 and above: $1.75 d$ ／ $1.25 d$ . First select the bolt-size row, then the edge preparation.Open full-size image

Minimum pitch along the load direction:

2.5 d

.
General minimum gauge transverse to the load:

3 d

.
Maximum pitch/gauge:

\min (12 t, 150 mm)

, where

t

is the thinner connected-plate thickness.
Maximum edge distance:

11 t ε

;

ε = \sqrt{\frac{275}{p_{y}}}

End distance is from the hole centre to the plate end in the load direction. Edge distance is from hole centre to an adjacent side edge. Minimum values depend on edge preparation. For M20, the supplied Table 9.3 gives $34 mm$ at sheared/hand-flame-cut edges and $26 mm$ at rolled, sawn or machine-cut edges. You cannot call a layout satisfactory without identifying the applicable edge type. Example 2 explicitly checks a rolled edge.

Animation labBolt centres, holes and edges1 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.

Hole diameter d₀ differs from nominal bolt diameter d. Net-section deductions use the specified hole.
Pitch runs along the load direction; gauge measures spacing across rows.
End and edge distances start at the hole centre. The remaining ligament starts at the hole boundary.
Minimum spacing, edge distances, grip and plate thickness come from the specified rules, not this scaled illustration.

Recognise the competing failure modes

Source: Chapter 2 p.8. The bolt may shear across its cross section; bolt or plate may crush locally at the hole (bearing); the plate may tear across its net tensile section; a short plate end may tear out; or a block bounded by shear and tension paths may pull out. A larger bolt does not necessarily cure a weak net plate section, because its larger hole removes more material.

Use a load-path check: main plate → bolts or weld → cover/gusset → connected member. Resistances in series are compared using the minimum. Two identical cover plates sharing a load in parallel have capacities that can be added. The groups on opposite sides of a splice are in series: each side must transfer the whole splice force.

Animation labCount the bolt shear planes4 concepts

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

Load must cross an interface between the connected plates.
A lap joint gives one shear plane through a bolt.
A symmetric double-cover joint may provide two shear planes. Count load-transfer interfaces, not just visible plates.
Use the source’s area and shear strength. Then check bearing, plate resistance and detailing separately.

Shear planes and bolt shear capacity

Simple explanation: Count where the bolt can be sheared

The plate interfaces are the places trying to cut across the bolt.

Count actual loaded shear planes, not merely visible plates.
Choose shank or threaded area for the plane concerned.
Compare group resistance with the force that group transfers.

Remember: Bolts on opposite sides of a splice do not all act in parallel.

Related concept and full method

Chapter 2 p.3, Figure 2: the left lap joint has one shear interface; the central plate between two covers gives two interfaces. — Chapter 2 p.3, Figure 2: single shear at left, double shear at right. The upper side views show the plate layers; the lower plans show hole positions. Labels identify end distance, edge distance and centre spacing. $F$ and the load arrows show forces pulling the joint apart. The two shear interfaces come from sandwiching the main plate between two cover plates, not from having two rows of holes in plan.Open full-size image

Use tensile stress area $A_{s}$ where the threads may lie in the shear plane. Use full shank area only when it is certain threads cannot be there. For M16/M20/M24/M30 the tensile stress areas are $157$ / $245$ / $353$ / $561 {mm}^{2}$ in the supplied table.

One shear plane:

P_{s} = \frac{p_{s} A_{s}}{1,000} kN

Two identical shear planes:

P_{s, total} = \frac{2 p_{s} A_{s}}{1,000} kN

The “/ $1,000$ ” converts $N$ to $kN$ : stress $\frac{N}{{mm}^{2}}$ × area ${mm}^{2}$ = $N$ . For a concentric group of $n$ equally loaded bolts, applied force per bolt is $\frac{P}{n}$ . For an eccentric group, use the maximum resultant force found from direct force plus moment; $\frac{P}{n}$ alone misses torsion.

Try it yourself. A Grade 8.8 M20 bolt has threads in two shear planes. Find its shear capacity.

Reveal answer and reasoning

$\frac{2 \times 375 \times 245}{1,000} = 183.75 kN$ . One-plane capacity is $91.875 kN$ . Double shear is a geometry fact, not a property of Grade 8.8.

Animation labCount the bolt shear planes1 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.

Load must cross an interface between the connected plates.
A lap joint gives one shear plane through a bolt.
A symmetric double-cover joint may provide two shear planes. Count load-transfer interfaces, not just visible plates.
Use the source’s area and shear strength. Then check bearing, plate resistance and detailing separately.

Animation labCount the bolt shear planes1 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.

Load must cross an interface between the connected plates.
A lap joint gives one shear plane through a bolt.
A symmetric double-cover joint may provide two shear planes. Count load-transfer interfaces, not just visible plates.
Use the source’s area and shear strength. Then check bearing, plate resistance and detailing separately.

Bearing and end tear-out: calculate every bound

Simple explanation: The bolt can crush or tear the plate

A strong bolt can still push through a weak hole edge.

Check the bolt and each connected plate’s bearing bounds.
Use nominal bolt diameter for bearing; hole size for removed material.
The smallest applicable resistance controls.

Remember: A short end distance may govern even when bolt shear passes.

Related concept and full method

Source: Chapter 2 pp.9–10. Bearing is local crushing where a bolt pushes on its hole. The projected bearing area is $d \times t_{p}$ , not $\frac{π d^{2}}{4}$ . For symmetric double-cover splices, compare the main thickness with the combined cover thickness to find the governing load path; alternatively check the main plate at full load and each cover at half load.

Bolt bearing:

P_{bb} = \frac{d t_{p} p_{bb}}{1,000}

Connected-plate resistance corresponding to each bolt is

\min (B_{1}, B_{2}, B_{3})

, where:

B_{1} = \frac{k_{bs} d t_{p} p_{bs}}{1,000}

B_{2} = \frac{0.5 k_{bs} e t_{p} p_{bs}}{1,000}

B_{3} = \frac{\min (1.5 l_{c} t_{p} U_{s}, 2 d t_{p} U_{b})}{1,000}

$d$ , $t_{p}$ , $e$ and $l_{c}$ are in $mm$ . $e$ is end distance measured along load transfer. $l_{c}$ is the clear ligament between neighbouring hole edges in that direction: for aligned equal holes, pitch − $d_{0}$ . $k_{bs} = 1$ for standard holes, $0.7$ for oversize/short slots in this extract. For S355, $p_{bs} = 550$ and $U_{s} = 510 \frac{N}{{mm}^{2}}$ . For Grade 8.8, $U_{b} = 800 \frac{N}{{mm}^{2}}$ . This notation $l_{c}$ can look like $I_{c}$ in the source; it is a length, not a moment of inertia.

The first bound checks bearing strength, the second captures end-distance limitation, and the third includes a clear-ligament limit and an upper cap. Write all numbers before taking the minimum. Do not apply the largest result as capacity.

Animation labBearing and the remaining ligament1 concept · 5 source expressions

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

Force transfers through contact between bolt and connected plate.
The plate around the hole carries bearing stress. Bolt bearing and plate bearing are separate checks.
A short ligament can tear out towards the end. The direction of force determines the relevant edge.
Evaluate every specified bearing and ligament bound for each layer and retain the smallest applicable resistance.

Tension and the prying assumption

Simple explanation: Why the top bolt row is pulled hardest

The bracket tries to peel away from the support about the assumed contact line.

Use the rotation/contact line assumed by the course model.
Measure each bolt row’s distance from that line.
Check maximum tension, direct shear and their interaction.

Remember: A row at the pivot can still carry direct shear.

Related concept and full method

Source: Chapter 2 pp.10–11. Bolt tension acts along the bolt axis, trying to separate connected surfaces. The direct tensile capacity is $P_{t} = A_{s} p_{t}$ . For the course’s simplified treatment avoiding explicit prying-force design, use reduced nominal tension $P_{nom} = 0.8 A_{s} p_{t}$ .

M20, Grade 8.8:

P_{t} = \frac{245 \times 560}{1,000} = 137.2 kN

P_{nom} = 0.8 \times 137.2 = 109.76 kN

The source Clause 9.3.7.2 extract also restricts flange bolt gauge $G \leq 0.55 B$ for UB/UC/T flange details. Reducing capacity by $0.8$ is not blanket proof that every flexible plate has no prying. Where the exercise uses the simplified bracket model but does not supply sufficient gauge/plate details, state that model’s applicability and avoid claiming the whole joint has been comprehensively designed.

Animation labOut-of-plane bolt tension and prying1 concept · 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.

A bracket moment can create compression at the plate contact and tension in the bolt rows.
The original assumed pivot/compression line determines each row distance.
Under the elastic row model, a farther tension row attracts more tension. Direct shear may act simultaneously.
Flexible plates can add prying force. Use nominal tension and interaction rules exactly as specified by the course.

Gross, net and effective area

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

Source: Chapter 2 pp.11–12. Gross area includes the original steel. Net area subtracts hole area along a possible fracture line normal to force. Effective tension area applies the course net-area coefficient but cannot exceed the corresponding gross area.

Steel plate:

A_{g} = b t; A_{n} = (b - n d_{0}) t; A_{e} = \min (K_{e} A_{n}, A_{g})

K_{e} = 1.2

(S275) ,

1.1

(S355) ,

1.0

(S460) .

Here $n$ is the number of holes crossed by the fracture line, not the total bolts along the connection. In a two-row splice, a straight transverse section crosses two holes even when many bolts lie along the length.

Chapter 2 p.12, Figures 9.7–9.8: straight and zigzag fracture paths and the developed gauge for angle legs. — Chapter 2 p.12, Figures 9.7–9.8. Figure 9.7 identifies the direction of force; $D$ = hole diameter; $t$ = plate thickness; $a_{n}$ = net area; $b$ = plate width. Upper-case $S_{1}$ to $S_{4}$ marks longitudinal stagger; $g_{1}$ to $g_{3}$ marks transverse gauge. The three path expressions in source order are:
① $a_{n} = t (b - 2 D)$ ;
② $a_{n} = t (b - 3 D + \frac{0.25 S_{2}^{2}}{g_{1}})$ ;
③ $a_{n} = t (b - 4 D + \frac{0.25 S_{4}^{2}}{g_{1}} + \frac{0.25 S_{4}^{2}}{g_{2}} + \frac{0.25 S_{4}^{2}}{g_{3}})$ .
Figure 9.8: heel means the angle root. The developed gauge across the two legs is $g = g_{1} + g_{2} - t$ . All dimensions follow their source locations; do not estimate values from drawing scale.Open full-size image

For staggered holes, compare all credible straight and zigzag net areas. For each diagonal step, the deduction is reduced by $\frac{s^{2} t}{4 g}$ , where $s$ is longitudinal stagger and $g$ transverse gauge. Thus $A_{n} = A_{g} - Σ d_{0} t + Σ \frac{s^{2} t}{4 g}$ . Select the smallest net area, equivalent to the largest deduction. For bolts on both angle legs, use the developed gauge $g_{1} + g_{2} - t$ shown in the source. This course supplies the geometric rule, not a numbered staggered calculation.

For shear area, the notes permit ignoring holes only if $A_{v, net} \geq \frac{0.85 A_{v}}{K_{e}}$ . Otherwise the stated net shear capacity is $0.7 p_{y} K_{e} A_{v, net}$ . This is distinct from the ordinary gross rolled-beam shear formula.

Connection example 2: design a bolted tension splice

Open this example on its own page

Design a double-cover splice for an already ultimate tensile force of $700 kN$ . Main plates are $150 mm$ wide × $20 mm$ thick; two S355 cover plates are $12 mm$ thick. Use Grade 8.8 M20 bolts in $22 mm$ holes, $50 mm$ pitch and $40 mm$ end distance.

Original source: LectureNotes/Ch 2_Connection.pdf — p. 18, p. 19, p. 20. 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.18: two cover plates sandwich the main plate, giving each bolt two shear interfaces.

150 mm

is the plate width perpendicular to the tensile force. In the side view below,

t_{1}

labels cover-plate thickness, and

t_{2}

labels main-plate thickness. The opposing

700 kN

arrows indicate tension across the joint. The lecturer's chosen two-row bolt arrangement is shown separately.Open full-size image

Selected arrangement, Chapter 2 p.19: the

150 mm

plate width is divided into

30

–

90

–

30

, with four bolts on each side of the central butt joint. The transverse dimensions are the upper/lower edge distances

30 mm

, and the gauge between rows

90 mm

. The longitudinal dimensions identify end distance

40 mm

and hole-centre spacing

50 mm

. Above the original drawing, the lecturer first rounds the double-shear resistance per bolt to

183.8

, then writes

\frac{700}{183.8} = 3.8

and selects four bolts. That equality uses rounded values; the detailed solution below retains more significant figures before rounding the bolt count upwards.Open full-size image

Read the diagram and collect the data

Input	Origin
Given demand	$700 kN$ ultimate; no further factor.
Given thicknesses	Main $t_{2} = 20 mm$ ; each cover $t_{1} = 12 mm$ ; cross-section view labels each ply.
Given hole/pitch/end	M20 → $d = 20$ ; stated hole $d_{0} = 22$ ; longitudinal pitch $50$ ; end $40 mm$ .
Selected layout	p.19 drawing: two rows, gauge $90 mm$ , side edges $30 mm$ , two columns per half-splice.
Lookup	$A_{s} = 245 {mm}^{2}$ , $p_{s} = 375$ , $p_{bb} = 1,000$ ; S355 $p_{bs} = 550$ , $U_{s} = 510$ , $K_{e} = 1.1$ . Main $p_{y} = 345$ for $20 mm$ ; covers $p_{y} = 355$ for $12 mm$ . All stress units $\frac{N}{{mm}^{2}}$ .

Before calculating: recognition and strategy

Each half-splice must transfer the full $700 kN$ from a main plate into the covers. The covers each take half that force. Bolts have two shear planes; compare main thickness $20$ with total cover thickness $24$ for the bearing path. Select bolt count from the weakest per-bolt mode, round up, arrange the bolts, then check the plates cut by those holes.

1. Establish the capacity of one bolt and its bearing path

Simple explanation: Count where the bolt can be sheared

The plate interfaces are the places trying to cut across the bolt.

Count actual loaded shear planes, not merely visible plates.
Choose shank or threaded area for the plane concerned.
Compare group resistance with the force that group transfers.

Remember: Bolts on opposite sides of a splice do not all act in parallel.

Related concept and full method

Double shear:

P_{s} = \frac{2 \times 375 \times 245}{1,000} = 183.75 kN

t_{p} = \min (20, 2 \times 12) = 20 mm

Bolt bearing:

P_{bb} = \frac{20 \times 20 \times 1,000}{1,000} = 400 kN

l_{c} = pitch - hole diameter = 50 - 22 = 28 mm

B_{1} = \frac{1 \times 20 \times 20 \times 550}{1,000} = 220 kN

B_{2} = \frac{0.5 \times 1 \times 40 \times 20 \times 550}{1,000} = 220 kN

Net-clearance term between holes:

\frac{1.5 \times 28 \times 20 \times 510}{1,000} = 428.4 kN

Upper cap:

\frac{2 \times 20 \times 20 \times 800}{1,000} = 640 kN

Connected-plate resistance:

\min (220, 220, 428.4, 640) = 220 kN

Governing resistance per bolt:

\min (183.75, 400, 220) = 183.75 kN

Using total cover thickness $24$ is valid here because the two identical covers share the force symmetrically. An unequal cover arrangement would need separate load-sharing checks.

Animation labCount the bolt shear planes4 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.

Load must cross an interface between the connected plates.
A lap joint gives one shear plane through a bolt.
A symmetric double-cover joint may provide two shear planes. Count load-transfer interfaces, not just visible plates.
Use the source’s area and shear strength. Then check bearing, plate resistance and detailing separately.

2. Select bolts on each side of the splice

n \geq \frac{700}{183.75} = 3.80952

→ Select

4

bolts per half-joint. Resistance per half:

4 \times 183.75 = 735 kN > 700 kN

Demand per bolt:

\frac{700}{4} = 175 kN < 183.75 kN

Provide four on the left and four on the right: eight physical bolts in total. The eight do not jointly give “ $1,470 kN$ ” across the splice, because the left and right groups carry the same force in series.

Animation labCount the bolt shear planes1 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.

Load must cross an interface between the connected plates.
A lap joint gives one shear plane through a bolt.
A symmetric double-cover joint may provide two shear planes. Count load-transfer interfaces, not just visible plates.
Use the source’s area and shear strength. Then check bearing, plate resistance and detailing separately.

3. Verify the selected layout

The two transverse rows are separated by $90 mm$ . Width closure: $30 + 90 + 30 = 150 mm$ . Longitudinal pitch $50$ and end $40$ are the specified values. Use the source’s rolled-edge condition in Table 9.3, M20 row: minimum $26 mm$ .

Minimum pitch:

2.5 \times 20 = 50 mm \leq provided 50

Maximum pitch/gauge:

\min (12 \times 12, 150) = 144 mm

50

and

90 \leq 144

. General minimum gauge:

3 \times 20 = 60 mm < 90

The minimum end/edge distance requirement is

26 mm

. The provided end and edge distances respectively satisfy

26 mm < 40 mm

26 mm < 30 mm

Maximum cover-plate edge distance:

11 \times 12 \times \sqrt{\frac{275}{355}} = 116.18 mm > 30

Maximum main-plate edge distance:

11 \times 20 \times \sqrt{\frac{275}{345}} = 196.42 mm > 30

Animation labBolt centres, holes and edges1 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.

Hole diameter d₀ differs from nominal bolt diameter d. Net-section deductions use the specified hole.
Pitch runs along the load direction; gauge measures spacing across rows.
End and edge distances start at the hole centre. The remaining ligament starts at the hole boundary.
Minimum spacing, edge distances, grip and plate thickness come from the specified rules, not this scaled illustration.

4. Check main-plate tension across two holes

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

The critical straight transverse line crosses one hole in each row: two $22 mm$ holes. Do not deduct all four holes along a half-splice. Thickness $20 mm$ selects $p_{y} = 345 \frac{N}{{mm}^{2}}$ .

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

A_{n} = (150 - 2 \times 22) \times 20 = 106 \times 20 = 2,120 {mm}^{2}

A_{e} = \min (1.1 \times 2,120, 3,000) = 2,332 {mm}^{2}

P_{t} = \frac{2,332 \times 345}{1,000} = 804.54 kN > 700

Passes.

Animation labSubtract holes on the failure path1 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.

Gross area counts the complete plate width and thickness.
The highlighted transverse path passes through the bolt holes.
For the straight illustrative path, $A_{n} = (b - n d_{0}) t$ . Staggered paths require their specified correction.
Net area is not always effective area. Include the course’s strength ratio or shear-lag rule when applicable.

5. Check both covers and conclude

Each cover plate:

A_{g} = 150 \times 12 = 1,800 {mm}^{2}

A_{n} = (150 - 44) \times 12 = 1,272 {mm}^{2}

A_{e} = \min (1.1 \times 1,272, 1,800) = 1,399.2 {mm}^{2}

P_{t} = \frac{1,399.2 \times 355}{1,000} = 496.716 kN > 350 kN

Both plates together:

P_{t} = 2 \times 496.716 = 993.432 kN > 700 kN

All requested checks pass under the stated rolled-edge layout. The least load-path resistance is bolt shear $735 kN$ . The source gives approximately $993.2 kN$ for covers after intermediate rounding; carrying full precision gives $993.43 kN$ .

Animation labSubtract holes on the failure path2 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.

Gross area counts the complete plate width and thickness.
The highlighted transverse path passes through the bolt holes.
For the straight illustrative path, $A_{n} = (b - n d_{0}) t$ . Staggered paths require their specified correction.
Net area is not always effective area. Include the course’s strength ratio or shear-lag rule when applicable.

Compact exam answer

Four M20 Grade 8.8 bolts per side (eight total), two rows with $90 mm$ gauge, $50 mm$ pitch, $40 mm$ ends and $30 mm$ rolled edges. Double-shear resistance $735 kN$ per half; bearing exceeds this. Main tension $804.54 kN$ ; combined cover tension $993.43 kN$ . Each exceeds $700 kN$ . Layout passes for the source’s rolled-edge assumption.

Mistakes to avoid

Do not factor the already ultimate $700 kN$ .
Four bolts are required on each side of the plate butt.
Double shear doubles bolt shear, not the main-plate thickness.
A transverse net section crosses two holes, not four.

Procedure for an unfamiliar variant

Trace one half-splice at full force and each cover at its share.
Calculate per-bolt shear/bearing and take the minimum.
Round up bolt count and draw a feasible layout.
Check every pitch/edge rule using the thinner ply and correct edge finish.
Check main and cover effective tension areas independently.

Independent self-check

Try it yourself. If the required ultimate force becomes $760 kN$ but the splice is unchanged, which calculated resistance fails first?

Hint

Compare $760$ with the complete group resistance, not just one bolt.

Reveal answer and reasoning

Bolt shear $735 kN$ is inadequate. Main-plate tension $804.54 kN$ and cover-pair tension $993.43 kN$ still exceed $760$ . Adding bolts changes the layout and may change the fracture path, so repeat layout/net-area checks.

Animation labSubtract holes on the failure path1 concept · 7 source expressions

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

Gross area counts the complete plate width and thickness.
The highlighted transverse path passes through the bolt holes.
For the straight illustrative path, $A_{n} = (b - n d_{0}) t$ . Staggered paths require their specified correction.
Net area is not always effective area. Include the course’s strength ratio or shear-lag rule when applicable.

Block shear

Simple explanation: A whole block can pull out

Think of tearing a small tab out of a sheet along several connected edges.

Trace the complete block around the bolt group.
Separate edges sliding in shear from the edge pulled in tension.
Deduct holes only where the applicable path crosses them.

Remember: One straight net-section check does not cover block shear.

Related concept and full method

Chapter 2 p.13, Figure 9.9: a block can tear along the longitudinal bolt line and across its transverse end. — Chapter 2 p.13, Figure 9.9: Block Shear and Effective Shear Area. In the figure, $F_{t}$ is the applied tension; $L_{v}$ is measured along the candidate shear plane; $L_{t}$ is measured across the tensile plane. Cope identifies the beam-end notch. Arrows show the interaction between the block and the remaining member. Path directions differ with the detail; do not substitute the whole plate perimeter.Open full-size image

Read $L_{v}$ along the prospective shear face and $L_{t}$ across the tensile face, as labelled on the original detail. The source’s effective-area expression is $A_{v, eff} = t [L_{v} + K_{e} (L_{t} - k D_{t})]$ , followed by $P_{r} = 0.6 p_{y} A_{v, eff}$ . $D_{t}$ is the tensile-face hole diameter; for slots use the dimension normal to load. The extract gives $k = 0.5$ for a single bolt row or $2.5$ for two rows. Match the illustrated path; do not substitute the whole plate length or an arbitrary bolt-group perimeter. No numerical block-shear example with all path dimensions is supplied.

Animation labSeparate the block-shear paths1 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.

A block containing the connection can detach from the surrounding plate.
The paths parallel to the applied force carry shear.
The closing path across the end of the block carries tension.
Use the specified gross/net deductions and resistance expression. This is different from a single straight net-section fracture.

Single and double angles connected through one leg

Simple explanation: One connected leg does not load both legs equally

The connected leg receives the pull first; the other leg receives it through the angle.

Identify connected and outstanding legs from the drawing.
Use the relevant bolted or welded angle rule.
Keep the lecturer’s area convention consistent.

Remember: Bolted and welded reduction expressions are not the same rule.

Related concept and full method

Source: Chapter 2 pp.14–15. The unconnected leg does not receive the load directly, so the effective tension resistance is reduced. Let $a_{2}$ be the gross area of the unconnected leg and $A_{e}$ the sum of effective leg areas. Avoid counting the heel overlap twice: the lecture’s rectangular approximation allocates half the thickness to each leg.

Connection	Capacity
Single angle, bolted	$P_{t} = p_{y} (A_{e} - 0.5 a_{2})$
Single angle, welded	$P_{t} = p_{y} (A_{e} - 0.3 a_{2})$
Double angles on both sides of gusset, interconnected, bolted	$P_{t} = p_{y} (A_{e} - 0.25 a_{2})$
Same double-angle arrangement, welded	$P_{t} = p_{y} (A_{e} - 0.15 a_{2})$

Use either total areas for the pair or calculate one identical angle at half load and double its capacity. Do not use total area and then double the answer again. For tension members with separately applied moments the notes give $\frac{F_{t}}{P_{t}} + \frac{M_{x}}{M_{cx}} + \frac{M_{y}}{M_{cy}} \leq 1$ . Nominal connection eccentricity for the listed single-leg shapes is already allowed for by the specific reduction; do not invent an extra moment without a task requirement.

Connection example 1: single bolted unequal angle

Open this example on its own page

Check a single S355 $80 \times 60 \times 8 mm$ angle connected through its long leg to a $10 mm$ plate by the stated two Grade 8.8 M20 bolts. Dead tension is $75 kN$ ; imposed tension is $40 kN$ . Check bolt shear, bolt/plate bearing and angle tension.

Original source: LectureNotes/Ch 2_Connection.pdf — p. 16, p. 17. 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	Value and exact source
Given loads	$75 kN$ dead and $40 kN$ imposed, question text; the downward arrow is the resulting tension direction.
Given geometry	Long leg $80 mm$ ; short leg $60 mm$ ; $t = 8 mm$ . The split dimensions $76$ and $56$ allocate half the heel thickness to each leg.
Given fasteners	Text specifies TWO M20 Grade 8.8 bolts; standard $22 mm$ holes; $50 mm$ pitch; $45 mm$ end distance.
Given plate/model	$10 mm$ supporting plate; one shear plane explicitly stated.
Lookup	Ch 2 p.3 M20 tensile stress area $A_{s} = 245 {mm}^{2}$ . Ch 2 pp.9–10 Grade 8.8: $p_{s} = 375$ , $p_{bb} = 1,000$ , $U_{b} = 800 \frac{N}{{mm}^{2}}$ . S355: $p_{bs} = 550$ , $U_{s} = 510$ , $K_{e} = 1.1$ . Data File p.1, thickness $\leq 16 mm$ : $p_{y} = 355$ .

Before calculating: recognition and strategy

This is concentric tension through one connected angle leg. The load crosses the angle-to-plate interface as bolt shear, presses on holes as bearing, and remains tension in the angle. Start with the factored force. Check each fastener mode, then the reduced angle tension resistance. Prerequisites: stress×area gives force; “net area” removes holes; a single-leg connection needs the specific unconnected-leg reduction.

1. Convert characteristic loads to one design force

$P$ is the ULS tensile demand in $kN$ . Apply the course gravity factors once; the supplied loads are characteristic, not already ultimate.

P = 1.4 G + 1.6 Q = 1.4 \times 75 + 1.6 \times 40 = 105 + 64 = 169 kN

Load demand per bolt:

\frac{P}{n} = \frac{169}{2} = 84.5 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. Bolt shear across the single interface

Simple explanation: Count where the bolt can be sheared

The plate interfaces are the places trying to cut across the bolt.

Count actual loaded shear planes, not merely visible plates.
Choose shank or threaded area for the plane concerned.
Compare group resistance with the force that group transfers.

Remember: Bolts on opposite sides of a splice do not all act in parallel.

Related concept and full method

$A_{s}$ is the tensile stress area because no thread-free shear plane is guaranteed. There is one interface between the angle leg and the support plate, so no factor $2$ for double shear.

P_{s, one} = \frac{p_{s} A_{s}}{1,000} = \frac{375 \frac{N}{{mm}^{2}} \times 245 {mm}^{2}}{1,000} = 91.875 kN

P_{s, group} = 2 \times 91.875 = 183.750 kN > 169 kN

Passes.

Animation labCount the bolt shear planes1 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.

Load must cross an interface between the connected plates.
A lap joint gives one shear plane through a bolt.
A symmetric double-cover joint may provide two shear planes. Count load-transfer interfaces, not just visible plates.
Use the source’s area and shear strength. Then check bearing, plate resistance and detailing separately.

3. Bearing of the bolts

Simple explanation: The bolt can crush or tear the plate

A strong bolt can still push through a weak hole edge.

Check the bolt and each connected plate’s bearing bounds.
Use nominal bolt diameter for bearing; hole size for removed material.
The smallest applicable resistance controls.

Remember: A short end distance may govern even when bolt shear passes.

Related concept and full method

Use the thinner connected thickness $t_{p} = \min (8, 10) = 8 mm$ : the angle leg governs. $d = 20 mm$ is the bolt diameter, not the $22 mm$ hole.

P_{bb, one} = \frac{d t_{p} p_{bb}}{1,000} = \frac{20 \times 8 \times 1,000}{1,000} = 160 kN

Bolt-group resistance:

2 \times 160 = 320 kN > 169 kN

Passes.

Animation labBearing and the remaining ligament1 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.

Force transfers through contact between bolt and connected plate.
The plate around the hole carries bearing stress. Bolt bearing and plate bearing are separate checks.
A short ligament can tear out towards the end. The direction of force determines the relevant edge.
Evaluate every specified bearing and ligament bound for each layer and retain the smallest applicable resistance.

4. Bearing and ligament resistance of the angle leg

The $50 mm$ dimension is centre-to-centre pitch along tension. Subtract one full hole diameter to get the clear inter-hole ligament $l_{c}$ . The $45 mm$ dimension is end distance from end to nearest bolt centre.

l_{c} = 50 - 22 = 28 mm; e = 45 mm; k_{bs} = 1

B_{1} = \frac{1 \times 20 \times 8 \times 550}{1,000} = 88 kN

B_{2} = \frac{0.5 \times 1 \times 45 \times 8 \times 550}{1,000} = 99 kN

Net-clearance term between holes:

\frac{1.5 \times 28 \times 8 \times 510}{1,000} = 171.36 kN

Upper cap:

\frac{2 \times 20 \times 8 \times 800}{1,000} = 256 kN

B_{3} = \min (171.36, 256) = 171.36 kN

P_{bs, one} = \min (88, 99, 171.36) = 88 kN

Two bolts together:

2 \times 88 = 176 kN > 169 kN

Passes.

Plate bearing is closer to its limit than bolt shear: $\frac{169}{176} = 0.9602$ utilisation. This is why the larger bolt-bearing value $320 kN$ cannot be used as the joint’s governing resistance.

Animation labBearing and the remaining ligament1 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.

Force transfers through contact between bolt and connected plate.
The plate around the hole carries bearing stress. Bolt bearing and plate bearing are separate checks.
A short ligament can tear out towards the end. The direction of force determines the relevant edge.
Evaluate every specified bearing and ligament bound for each layer and retain the smallest applicable resistance.

5. Effective angle area and single-leg reduction

Simple explanation: One connected leg does not load both legs equally

The connected leg receives the pull first; the other leg receives it through the angle.

Identify connected and outstanding legs from the drawing.
Use the relevant bolted or welded angle rule.
Keep the lecturer’s area convention consistent.

Remember: Bolted and welded reduction expressions are not the same rule.

Related concept and full method

Use the lecturer’s rectangular leg-area approximation. The heel is shared: connected leg width $80 - \frac{8}{2} = 76 mm$ ; unconnected width $60 - \frac{8}{2} = 56 mm$ . A transverse fracture line crosses one hole, not both bolts arranged along the member. Apply $K_{e}$ separately to each leg and cap each at its own gross area.

a_{g 1} = 76 \times 8 = 608 {mm}^{2}; a_{2} = a_{g 2} = 56 \times 8 = 448 {mm}^{2}

A_{g} = 608 + 448 = 1,056 {mm}^{2}

a_{n 1} = (76 - 22) \times 8 = 54 \times 8 = 432 {mm}^{2}

a_{n 2} = 448 {mm}^{2}

The outstanding angle leg has no hole.

a_{e 1} = \min (1.1 \times 432, 608) = 475.2 {mm}^{2}

a_{e 2} = \min (1.1 \times 448, 448) = \min (492.8, 448) = 448 {mm}^{2}

A_{e} = 475.2 + 448 = 923.2 {mm}^{2}

P_{t} = \frac{p_{y} (A_{e} - 0.5 a_{2})}{1,000}

= \frac{355 \times (923.2 - 0.5 \times 448)}{1,000}

= \frac{355 \times 699.2}{1,000} = 248.216 kN > 169 kN

Passes.

The factor $0.5$ belongs specifically to a single bolted angle connected through one leg (Ch 2 p.14). It is not a generic reduction for all tension members.

Animation labWhy a connected angle leg matters1 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.

Locate the connected leg, outstanding leg and centroid before using the table.
Only the connected leg directly receives the fastener force.
The outstanding area may not become equally effective at the same section; this motivates the effective-area rule.
Bolted, welded, single-angle and double-angle details can have different rules. Preserve the formula attached to the original case.

6. State the governing result and the scope

For the requested resistance checks, the minimum is plate bearing $176 kN$ , above $169 kN$ . The angle tension check also passes. Units close correctly: stress× ${mm}^{2}$ gives $N$ before dividing by $1,000$ .

The pitch $50 mm$ equals $2.5 \times 20 = 50 mm$ and is less than $\min (12 \times 8, 150) = 96 mm$ . The end distance $45 mm$ exceeds both supplied M20 minimum values ( $34$ or $26 mm$ ). The transverse edge distances are not fully dimensioned in this example, so these observations do not prove that the entire arrangement has been verified.

Animation labFollow the calculation sequence5 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.

Locate the load, supports, connection geometry and any stated assumptions.
Keep given values, table lookups and calculated values distinct; reconcile their units.
The calculation player steps through the existing expressions in their original order.
Compare demand with resistance or the relevant limit. Keep missing inputs and conditional conclusions explicit.

Compact exam answer

$P = 169 kN$ . Two bolts in single shear: $P_{s} = 183.75 kN$ ; bolt bearing $= 320 kN$ ; angle bearing $= \min (176, 198, 342.72) = 176 kN$ . $A_{e} = 923.2 {mm}^{2}$ ; single angle $P_{t} = \frac{355 (923.2 - 0.5 \times 448)}{1,000} = 248.22 kN$ . All required resistances exceed the demand. Plate bearing governs, using the two-bolt model specified by the question.

Mistakes to avoid

Do not multiply by two for shear planes; the given joint is single shear.
Subtract $22 mm$ for a hole, use $20 mm$ for bearing.
Do not multiply the unconnected gross leg area by $1.1$ without its gross-area cap.
Do not silently replace the written two bolts with four based on the inconsistent sketch.

Procedure for an unfamiliar variant

Read the written bolt count and the interface geometry; reconcile any discrepancy.
Factor characteristic loads and distribute concentric load to the bolts.
Check shear, bolt bearing and all connected-part bounds.
Compute gross/net/effective leg areas and the correct single/double connection reduction.
Compare each complete load path with the same demand and name the governing one.

Independent self-check

Try it yourself. Keep the geometry but increase imposed load to $50 kN$ . Does the connection still pass the calculated strength checks?

Hint

Recalculate demand only; the capacities do not change when geometry and materials stay the same.

Reveal answer and reasoning

New $P = 1.4 \times 75 + 1.6 \times 50 = 185 kN$ . Plate bearing $176 kN$ and bolt shear $183.75 kN$ are both inadequate; angle tension still passes. A stronger angle alone would not fix the fastener path.

Animation labCount the bolt shear planes4 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.

Load must cross an interface between the connected plates.
A lap joint gives one shear plane through a bolt.
A symmetric double-cover joint may provide two shear planes. Count load-transfer interfaces, not just visible plates.
Use the source’s area and shear strength. Then check bearing, plate resistance and detailing separately.

Animation labWhy a connected angle leg matters2 concepts · 5 source expressions

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

Locate the connected leg, outstanding leg and centroid before using the table.
Only the connected leg directly receives the fastener force.
The outstanding area may not become equally effective at the same section; this motivates the effective-area rule.
Bolted, welded, single-angle and double-angle details can have different rules. Preserve the formula attached to the original case.

In-plane eccentric bolts: derive the force distribution

Simple explanation: An off-centre force also tries to turn the group

Pulling away from the centre causes a push plus a twist.

Find direct shear and moment about the bolt-group centroid.
Moment-induced forces act tangentially; farther bolts attract more.
Add horizontal and vertical components before finding the resultant.

Remember: Do not simply add two force magnitudes pointing in different directions.

Related concept and full method

Source: Chapter 2 pp.21–22. The eccentric force tries to rotate the bolt group in the plane of the plate. Assume a rigid plate, equal bolts, elastic force proportional to radius and uniform direct shear. Place the origin at the bolt-group centroid. Give every bolt coordinates $x_{i}$ , $y_{i}$ measured in $mm$ , with $r_{i}^{2} = x_{i}^{2} + y_{i}^{2}$ .

J_{bolts} = Σ r_{i}^{2} = Σ x_{i}^{2} + Σ y_{i}^{2}

Units:

{mm}^{2}

M = P e

Units:

kN \cdot mm

F_{T, i} = \frac{M r_{i}}{J_{bolts}}; F_{s} = \frac{P}{n}

Both have units

kN

Why this works: if torsional force is $C r_{i}$ , its resisting moment is $C r_{i}^{2}$ . Summing gives $M = C Σ r^{2}$ , so $C = \frac{M}{J}$ . The torsional force is tangent to a circle, perpendicular to the radius, not radial. For vertical applied load, its vertical component at a critical corner is $\frac{M x}{J}$ and horizontal component $\frac{M y}{J}$ . Choose the corner where the vertical component adds to direct shear.

F_{R} = \sqrt{{(\frac{P}{n} + \frac{M | x |}{J})}^{2} + {(\frac{M | y |}{J})}^{2}}

This is equivalent to the lecture vector formula $\sqrt{F_{s}^{2} + F_{T}^{2} + 2 F_{s} F_{T} \cos φ}$ , with $\cos φ = \frac{| x |}{r}$ for the shown vertical-load arrangement. Do not use $\frac{| y |}{r}$ merely because the load is vertical: torsional force is perpendicular to the radius.

Connection example 3: an eccentric bolt group in the plate plane

Open this example on its own page

Check the Grade 8.8 M24 bolt group carrying $100 kN$ dead plus $130 kN$ imposed load at a $525 mm$ eccentricity. Steel is S355. Use the lecturer’s one-side-plate load model.

Original source: LectureNotes/Ch 2_Connection.pdf — p. 25, p. 26. 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.25: the two bolt columns are separated by

500 mm

. Each column has six bolts at pitch

70 mm

, giving five spaces totalling

350 mm

. The centroid is midway between the columns and between the middle two rows. The original labels identify dead load, imposed load and the side plate. “All bolts 24 mm dia” gives a bolt diameter of

24 mm

. “Loads are on bolt shown on one plate” assigns the loads to the one side plate shown. The upper-right bolt is A; the diagonal

305.16

is the marked distance from the centroid to A, calculated to full precision below. Each horizontal outer edge distance is

57.9 mm

. These are neither the upper/lower

45 mm

end distances nor the load eccentricity. The plan below labels two

254 \times 254 \times 89

UC members; that member count does not change the explicitly stated one-side-plate load model.Open full-size image

Read the diagram and collect the data

Input	Origin
Given load/model	$100 kN$ dead+ $130 kN$ imposed. Drawing says bolt loads shown on one plate; do not halve the specified force again.
Given $e$	$525 mm$ horizontal distance from the dashed bolt-group centreline to vertical load line, upper dimension.
Coordinates	$x = \pm 250 mm$ from the two $250$ dimensions. $y = \pm 35, \pm 105, \pm 175 mm$ from $70 mm$ pitch and symmetric six-row arrangement.
Thickness and edge	Side plate $15 mm$ ; p.26 gives UC flange $17.3 mm$ , so $t_{p} = 15$ . Vertical end $45 mm$ ; pitch $70$ ; hole $26 mm$ .
Lookup	M24 $A_{s} = 353 {mm}^{2}$ ; $p_{s} = 375$ , $p_{bb} = 1,000$ , $p_{bs} = 550$ , $U_{s} = 510$ , $U_{b} = 800 \frac{N}{{mm}^{2}}$ .

Before calculating: recognition and strategy

The load rotates the plate in its own plane, so each bolt takes uniform direct shear plus tangential torsional shear. This is not a bolt-tension problem. Use centroidal coordinates, sum squared distances over all $12$ bolts, and resolve torsional force into components before adding direct shear. The far right corners have torsional vertical force in the same direction as direct shear.

1. Factored load and moment

P = 1.4 \times 100 + 1.6 \times 130 = 140 + 208 = 348 kN

M = P e = 348 kN \times 525 mm = 182,700 kN \cdot mm = 182.7 kN \cdot m

Animation labFrom characteristic to design load2 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.

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. Bolt-group polar sum and corner radius

Simple explanation: An off-centre force also tries to turn the group

Pulling away from the centre causes a push plus a twist.

Find direct shear and moment about the bolt-group centroid.
Moment-induced forces act tangentially; farther bolts attract more.
Add horizontal and vertical components before finding the resultant.

Remember: Do not simply add two force magnitudes pointing in different directions.

Related concept and full method

Each bolt contributes $x^{2} + y^{2}$ . There are $12$ identical $| x |$ values. At each absolute height $35$ , $105$ , $175 mm$ there are four bolts: two left/right positions multiplied by two positive/negative vertical positions.

Σ x^{2} = 12 \times 250^{2} = 12 \times 62,500 = 750,000 {mm}^{2}

Σ y^{2} = 4 (35^{2} + 105^{2} + 175^{2})

= 4 (1,225 + 11,025 + 30,625) = 171,500 {mm}^{2}

J = 750,000 + 171,500 = 921,500 {mm}^{2}

r_{A} = \sqrt{250^{2} + 175^{2}} = \sqrt{93,125} = 305.164 mm

\cos φ = \frac{250}{305.164} = 0.819232

Animation labAdd direct and torsional bolt forces1 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.

Assign signed coordinates to the bolts relative to the group centroid.
For the equal-bolt elastic model, the direct force is $\frac{P}{n}$ per bolt.
Moment magnitude is $P e$ ; its sign follows the load direction. For signed M: $F_{x} = \frac{- M y}{J}$ , $F_{y} = \frac{M x}{J}$ , $J = \sum r^{2}$ .
Add signed x and y components at each bolt, then take the resultant. The longest arrow shows the critical bolt in this illustration.

3. Direct force, torsion and maximum resultant

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 shear per bolt:

F_{s} = \frac{P}{12} = \frac{348}{12} = 29 kN

F_{T, A} = \frac{M r}{J} = \frac{182,700 \times 305.164}{921,500} = 60.503 kN

Vertical torsional-force component:

\frac{M \times 250}{J} = 49.566 kN

Horizontal torsional-force component:

\frac{M \times 175}{J} = 34.696 kN

F_{R, A} = \sqrt{{(29 + 49.566)}^{2} + {34.696}^{2}} = 85.886 kN

The units of $\frac{M r}{J}$ are $\frac{(kN \cdot mm) \times mm}{{mm}^{2}} = kN$ . The top/bottom right corners have equal resultant magnitude. Other rows have a smaller horizontal torsion component; the left column’s vertical torsion subtracts from direct shear, so it is less critical.

Animation labAdd direct and torsional bolt forces1 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.

Assign signed coordinates to the bolts relative to the group centroid.
For the equal-bolt elastic model, the direct force is $\frac{P}{n}$ per bolt.
Moment magnitude is $P e$ ; its sign follows the load direction. For signed M: $F_{x} = \frac{- M y}{J}$ , $F_{y} = \frac{M x}{J}$ , $J = \sum r^{2}$ .
Add signed x and y components at each bolt, then take the resultant. The longest arrow shows the critical bolt in this illustration.

4. Compare with single-shear bolt resistance

P_{s} = \frac{375 \times 353}{1,000} = 132.375 kN > 85.886 kN

Passes.

The source checks the side plate’s single shear interface. The plan view showing plates on both sides does not authorise doubling resistance while keeping only one side’s force model.

Animation labCount the bolt shear planes1 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.

Load must cross an interface between the connected plates.
A lap joint gives one shear plane through a bolt.
A symmetric double-cover joint may provide two shear planes. Count load-transfer interfaces, not just visible plates.
Use the source’s area and shear strength. Then check bearing, plate resistance and detailing separately.

5. Check bolt bearing against the same resultant

t_{p} = \min (15, 17.3) = 15 mm

P_{bb} = \frac{24 \times 15 \times 1,000}{1,000} = 360 kN > 85.886 kN

Passes.

Animation labBearing and the remaining ligament1 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.

Force transfers through contact between bolt and connected plate.
The plate around the hole carries bearing stress. Bolt bearing and plate bearing are separate checks.
A short ligament can tear out towards the end. The direction of force determines the relevant edge.
Evaluate every specified bearing and ligament bound for each layer and retain the smallest applicable resistance.

6. Connected-part bounds and conclusion

Simple explanation: The bolt can crush or tear the plate

A strong bolt can still push through a weak hole edge.

Check the bolt and each connected plate’s bearing bounds.
Use nominal bolt diameter for bearing; hole size for removed material.
The smallest applicable resistance controls.

Remember: A short end distance may govern even when bolt shear passes.

Related concept and full method

l_{c} = 70 - 26 = 44 mm

B_{1} = \frac{1 \times 24 \times 15 \times 550}{1,000} = 198 kN

B_{2} = \frac{0.5 \times 1 \times 45 \times 15 \times 550}{1,000} = 185.625 kN

Net-clearance term between holes:

\frac{1.5 \times 44 \times 15 \times 510}{1,000} = 504.9 kN

Upper cap:

\frac{2 \times 24 \times 15 \times 800}{1,000} = 576 kN

P_{bs} = \min (198, 185.625, 504.9, 576) = 185.625 kN > 85.886

Passes.

For the course’s specified connection checks, single-shear bolt capacity $132.375 kN$ is the governing resistance per critical bolt. The ratio $\frac{85.886}{132.375} \approx 0.649$ . This is not a total connection force capacity of $132.375 kN$ : it is compared with the critical bolt force.

Animation labBearing and the remaining ligament2 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.

Force transfers through contact between bolt and connected plate.
The plate around the hole carries bearing stress. Bolt bearing and plate bearing are separate checks.
A short ligament can tear out towards the end. The direction of force determines the relevant edge.
Evaluate every specified bearing and ligament bound for each layer and retain the smallest applicable resistance.

Compact exam answer

$P = 348 kN$ , $M = 182.7 kN \cdot m$ . $J = 921,500 {mm}^{2}$ ; $r = 305.164 mm$ . $F_{s} = 29 kN$ , $F_{T} = 60.503 kN$ ; $F_{R, \max} = 85.886 kN$ . Per-bolt resistances: shear $132.375 kN$ , bolt bearing $360 kN$ , plate bearing $185.625 kN$ . All exceed the resultant; requested checks pass.

Mistakes to avoid

Do not use $\frac{P}{12}$ as the complete bolt demand.
The sum for bolt-group geometry has units ${mm}^{2}$ , unlike weld line inertia ${mm}^{3}$ .
The $525 mm$ eccentricity starts at the group centroid, not the right bolt column.
Add force components; $F_{s} + F_{T}$ is not the actual resultant.

Procedure for an unfamiliar variant

Locate centroid and write every bolt coordinate.
Compute factored $P$ , centroidal eccentricity $e$ and $M$ .
Sum all $r^{2}$ and resolve $\frac{M r}{J}$ into components.
Find the corner where torsion adds to direct shear.
Compare that resultant with shear and all bearing limits.

Independent self-check

Try it yourself. If both characteristic loads increase by $20 %$ with unchanged geometry, how does the critical bolt force change?

Reveal answer and reasoning

$P$ and $M$ each multiply by $1.2$ ; every direct/torsional component does too. $F_{R} = 1.2 \times 85.886 = 103.063 kN$ , still below $132.375 kN$ . Capacity is unchanged.

Animation labAdd direct and torsional bolt forces1 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.

Assign signed coordinates to the bolts relative to the group centroid.
For the equal-bolt elastic model, the direct force is $\frac{P}{n}$ per bolt.
Moment magnitude is $P e$ ; its sign follows the load direction. For signed M: $F_{x} = \frac{- M y}{J}$ , $F_{y} = \frac{M x}{J}$ , $J = \sum r^{2}$ .
Add signed x and y components at each bolt, then take the resultant. The longest arrow shows the critical bolt in this illustration.

Animation labAdd direct and torsional bolt forces1 concept · 6 source expressions

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

Assign signed coordinates to the bolts relative to the group centroid.
For the equal-bolt elastic model, the direct force is $\frac{P}{n}$ per bolt.
Moment magnitude is $P e$ ; its sign follows the load direction. For signed M: $F_{x} = \frac{- M y}{J}$ , $F_{y} = \frac{M x}{J}$ , $J = \sum r^{2}$ .
Add signed x and y components at each bolt, then take the resultant. The longest arrow shows the critical bolt in this illustration.

Out-of-plane brackets: shear plus tension

Simple explanation: Two passing checks may still interact

One bolt is doing two jobs at the same time.

Check shear on its own and tension on its own.
Calculate the course’s combined-action expression.
Apply its own limit, not a limit borrowed from columns.

Remember: The supplied bolt interaction limit of 1.4 does not replace the individual checks.

Related concept and full method

Source: Chapter 2 pp.23–24. Here the bracket pulls the plate away from the support and rotates about a bottom contact/bolt line in the lecture approximation. Measure each bolt-row distance $y$ from that assumed rotation line. The bottom row has $y = 0$ and no moment-induced tension in this model, but still carries direct shear.

J_{rows} = Σ (distances of all bolts y^{2})

Units:

{mm}^{2}

F_{T, \max} = \frac{P e y_{\max}}{J_{rows}}

F_{s} = \frac{P}{n}

Check simultaneously:

F_{s} \leq P_{s}; F_{T, \max} \leq P_{nom}

\frac{F_{s}}{P_{s}} + \frac{F_{T, \max}}{P_{nom}} \leq 1.4

For two identical vertical rows of bolts, $J = 2 Σ y^{2}$ for one row. Do not multiply by $2$ again when the sum already includes both rows. The individual limits remain compulsory: a sum below $1.4$ alone does not excuse excessive tension or shear. Bearing also needs checking where the connection data permit it.

Connection example 4: bracket bolts in shear and tension

Open this example on its own page

Check ten Grade 8.8 M16 bolts supporting a $250 kN$ ultimate bracket load at $200 mm$ eccentricity. Five bolts are arranged vertically on each side at $90 mm$ centres. Steel is S355.

Original source: LectureNotes/Ch 2_Connection.pdf — p. 27, p. 28. 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.27:

200 mm

is the external load lever arm beyond the column joint. The triangular bolt-force distribution rotates about the lowest bolt line; all distances are measured upwards from that line. The labels identify the joint and state that bolts on both sides have centre-to-centre spacing

90 mm

. “Bolt loads” identifies the bolt forces; “Maximum tension” identifies

F_{T}

. The other tensile forces shown, in order, are

\frac{F_{T} y_{2}}{y_{1}}

\frac{F_{T} y_{3}}{y_{1}}

\frac{F_{T} y_{4}}{y_{1}}

，

y_{1}

is the distance to the top row. The original conditions at the lower crop edge, cross-checked against the full p.27, are

F_{s} \leq P_{s}

F_{T} \leq P_{nom}

, and

\frac{F_{s}}{P_{s}} + \frac{F_{T}}{P_{nom}} \leq 1.4

. All three must hold simultaneously. Below, values are substituted separately into the shear, tension and interaction checks.Open full-size image

Read the diagram and collect the data

Input	Origin
Given load	$250 kN$ ultimate; the downward arrow is outside the connection plane.
Given bolt arrangement	Five bolts per side; two sides; $90 mm$ between consecutive rows; total $10$ .
Calculated row distances	On each side, measure upwards from the bottom rotation line: $0, 90, 180, 270, 360 mm$ .
Lookup	M16 $A_{s} = 157 {mm}^{2}$ , Grade 8.8 $p_{s} = 375$ and $p_{t} = 560 \frac{N}{{mm}^{2}}$ (Ch 2 pp.3,9–10).
Model assumption	Rigid bracket and linear tensile-force distribution about bottom bolt line, as stated in the lecturer approximation; use nominal tension $0.8 A_{s} p_{t}$ .

Before calculating: recognition and strategy

The load pulls the upper bolts along their axes while all bolts carry vertical shear. Use an out-of-plane rotation model, not the centroidal in-plane torsion formula. Check shear, tension and their interaction separately. The task’s bolt calculation does not provide the full plate/gauge dimensions needed for a complete joint/prying and bearing verification.

1. Find bolt shear and nominal tension resistance

P_{s} = \frac{375 \times 157}{1,000} = 58.875 kN

P_{t} = \frac{560 \times 157}{1,000} = 87.92 kN

P_{nom} = 0.8 \times 87.92 = 70.336 kN

The $0.8$ reduction is part of the adopted simplified model. Do not substitute $87.92$ in the nominal-tension interaction denominator.

Animation labOut-of-plane bolt tension and prying2 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 bracket moment can create compression at the plate contact and tension in the bolt rows.
The original assumed pivot/compression line determines each row distance.
Under the elastic row model, a farther tension row attracts more tension. Direct shear may act simultaneously.
Flexible plates can add prying force. Use nominal tension and interaction rules exactly as specified by the course.

2. Sum squared distances about the bottom row

Each of the two vertical sides has the same distances. The bottom row contributes $0^{2}$ to moment resistance but remains in the count for direct shear.

J_{rows} = 2 (0^{2} + 90^{2} + 180^{2} + 270^{2} + 360^{2})

= 2 (0 + 8,100 + 32,400 + 72,900 + 129,600)

= 2 \times 243,000 = 486,000 {mm}^{2}

M = P e = 250 \times 200 = 50,000 kN \cdot mm

Animation labOut-of-plane bolt tension and prying1 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 bracket moment can create compression at the plate contact and tension in the bolt rows.
The original assumed pivot/compression line determines each row distance.
Under the elastic row model, a farther tension row attracts more tension. Direct shear may act simultaneously.
Flexible plates can add prying force. Use nominal tension and interaction rules exactly as specified by the course.

3. Maximum tension is in the top row

Simple explanation: Why the top bolt row is pulled hardest

The bracket tries to peel away from the support about the assumed contact line.

Use the rotation/contact line assumed by the course model.
Measure each bolt row’s distance from that line.
Check maximum tension, direct shear and their interaction.

Remember: A row at the pivot can still carry direct shear.

Related concept and full method

Tension varies in proportion to distance $y$ . Let $F_{i} = C y_{i}$ ; moment equilibrium gives $M = C Σ y^{2}$ . Thus $C = \frac{M}{J}$ and the top bolt at $y = 360 mm$ has the largest tension.

F_{T, \max} = \frac{50,000 \times 360}{486,000} = 37.037 kN

37.037 < 70.336 kN

: individual tension check passes.

Animation labOut-of-plane bolt tension and prying1 concept · 5 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 bracket moment can create compression at the plate contact and tension in the bolt rows.
The original assumed pivot/compression line determines each row distance.
Under the elastic row model, a farther tension row attracts more tension. Direct shear may act simultaneously.
Flexible plates can add prying force. Use nominal tension and interaction rules exactly as specified by the course.

4. Uniform direct shear

F_{s} = \frac{P}{n} = \frac{250}{10} = 25 kN

25 < 58.875 kN

: individual shear check passes.

Animation labCount the bolt shear planes1 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.

Load must cross an interface between the connected plates.
A lap joint gives one shear plane through a bolt.
A symmetric double-cover joint may provide two shear planes. Count load-transfer interfaces, not just visible plates.
Use the source’s area and shear strength. Then check bearing, plate resistance and detailing separately.

5. Combined action and conclusion

Simple explanation: Two passing checks may still interact

One bolt is doing two jobs at the same time.

Check shear on its own and tension on its own.
Calculate the course’s combined-action expression.
Apply its own limit, not a limit borrowed from columns.

Remember: The supplied bolt interaction limit of 1.4 does not replace the individual checks.

Related concept and full method

\frac{F_{s}}{P_{s}} + \frac{F_{T}}{P_{nom}}

= \frac{25}{58.875} + \frac{37.037}{70.336}

= 0.424628 + 0.526573 = 0.951201 \leq 1.4

Passes.

The bolts satisfy all three calculated shear/tension conditions under the lecturer’s approximate model. This is the requested bolt result. Plate thickness, detailed bearing-edge geometry and flange gauge are not supplied sufficiently to certify every other joint limit state; do not claim those have been checked.

Animation labOut-of-plane bolt tension and prying2 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 bracket moment can create compression at the plate contact and tension in the bolt rows.
The original assumed pivot/compression line determines each row distance.
Under the elastic row model, a farther tension row attracts more tension. Direct shear may act simultaneously.
Flexible plates can add prying force. Use nominal tension and interaction rules exactly as specified by the course.

Compact exam answer

$P_{s} = 58.875 kN$ ; $P_{nom} = 70.336 kN$ . $Σ y^{2} = 486,000 {mm}^{2}$ about bottom bolt line. $F_{T, \max} = \frac{250 \times 200 \times 360}{486,000} = 37.037 kN$ . $F_{s} = 25 kN$ . Individual limits pass and interaction $= 0.9512 < 1.4$ . Bolts are adequate under the specified approximate shear/tension model.

Mistakes to avoid

Do not factor $250 kN$ again.
Do not measure $y$ from the group centroid for this bottom-pivot model.
Do not omit the bottom bolts from direct shear.
Passing the $1.4$ interaction does not replace individual shear/tension limits.

Procedure for an unfamiliar variant

Identify out-of-plane rotation and the stated pivot.
Write all $y$ distances and include all physical bolts in $Σ y^{2}$ .
Calculate maximum tension and uniform shear.
Check each individual resistance and then the interaction.
State the model and any unprovided plate/prying information.

Independent self-check

Try it yourself. Move the same load to $400 mm$ eccentricity. Does the existing bolt group pass?

Reveal answer and reasoning

Tension doubles to $74.074 kN$ , exceeding $70.336$ . Interaction becomes $\frac{25}{58.875} + \frac{74.074}{70.336} \approx 1.478 > 1.4$ . Both tension and interaction fail; shear remains $25 kN$ .

Animation labOut-of-plane bolt tension and prying2 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 bracket moment can create compression at the plate contact and tension in the bolt rows.
The original assumed pivot/compression line determines each row distance.
Under the elastic row model, a farther tension row attracts more tension. Direct shear may act simultaneously.
Flexible plates can add prying force. Use nominal tension and interaction rules exactly as specified by the course.

Connection example 5: separate moment and shear bolt groups

Open this example on its own page

Design the bolt groups joining a $610 \times 229 \times 140$ UB floor beam to a $254 \times 254 \times 132$ UC. Steel is S355 and bolts are Grade 8.8 M24. Characteristic moments are $160 kN \cdot m$ dead and $90 kN \cdot m$ imposed; characteristic shears are $250 kN$ dead and $160 kN$ imposed.

Original source: LectureNotes/Ch 2_Connection.pdf — p. 29, p. 30. 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 actions	Separate moment and shear dead/imposed components in question table; factor each once.
Given force-couple arm	$595 mm$ between flange force lines in lecturer diagram/solution; not the nominal $610 mm$ beam depth.
Selected fasteners	Four upper flange bolts; four lower for opposite moment direction; eight web bolts, all M24.
Selected bearing detail	Lecturer p.30 selects $12 mm$ end plate, $55 mm$ end distance and $70 mm$ pitch. These are design choices, not task givens.
Lookup	M24 $A_{s} = 353 {mm}^{2}$ ; Grade 8.8 $p_{t} = 560$ , $p_{s} = 375$ , $p_{bb} = 1,000$ ; S355 $p_{bs} = 550$ , $U_{s} = 510$ ; $U_{b} = 800 \frac{N}{{mm}^{2}}$ .

Before calculating: recognition and strategy

Use the source idealisation: bending is a tension/compression force couple at the flanges; vertical shear is carried by a separate web-bolt group. Moment divided by the perpendicular couple arm gives flange force. Because groups have separate assigned actions, do not distribute the flange tensile force among the web bolts.

1. Design the tension flange group

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

M = 1.4 \times 160 + 1.6 \times 90 = 224 + 144 = 368 kN \cdot m

F_{flange} = \frac{M}{z} = \frac{368}{0.595} = 618.487 kN

P_{nom, one} = \frac{0.8 \times 560 \times 353}{1,000} = 158.144 kN

n \geq \frac{618.487}{158.144} = 3.911 \to select 4

Nominal tension resistance of the whole group:

P_{nom} = 4 \times 158.144 = 632.576 kN > 618.487

Passes.

For the moment direction shown, the lower flange region supplies compression and the upper group tension. The extra lower four bolts are not added to the upper group’s capacity; they provide the corresponding group for the opposite moment direction/detail.

Animation labOut-of-plane bolt tension and prying2 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 bracket moment can create compression at the plate contact and tension in the bolt rows.
The original assumed pivot/compression line determines each row distance.
Under the elastic row model, a farther tension row attracts more tension. Direct shear may act simultaneously.
Flexible plates can add prying force. Use nominal tension and interaction rules exactly as specified by the course.

2. Design the shear group

Simple explanation: Count where the bolt can be sheared

The plate interfaces are the places trying to cut across the bolt.

Count actual loaded shear planes, not merely visible plates.
Choose shank or threaded area for the plane concerned.
Compare group resistance with the force that group transfers.

Remember: Bolts on opposite sides of a splice do not all act in parallel.

Related concept and full method

V = 1.4 \times 250 + 1.6 \times 160 = 350 + 256 = 606 kN

P_{s, one} = \frac{375 \times 353}{1,000} = 132.375 kN

For

8

web bolts, demand per bolt:

\frac{606}{8} = 75.75 kN

P_{s, group} = 8 \times 132.375 = 1,059.0 kN > 606

Passes.

Animation labCount the bolt shear planes2 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.

Load must cross an interface between the connected plates.
A lap joint gives one shear plane through a bolt.
A symmetric double-cover joint may provide two shear planes. Count load-transfer interfaces, not just visible plates.
Use the source’s area and shear strength. Then check bearing, plate resistance and detailing separately.

3. Check selected end-plate bearing

Use $d = 24 mm$ and standard hole $d_{0} = 26 mm$ . The chosen $12 mm$ plate and $70 mm$ pitch give clear ligament $44 mm$ . The chosen end distance is $55 mm$ .

P_{bb} = \frac{24 \times 12 \times 1,000}{1,000} = 288 kN

l_{c} = 70 - 26 = 44 mm

B_{1} = \frac{24 \times 12 \times 550}{1,000} = 158.4 kN

B_{2} = \frac{0.5 \times 55 \times 12 \times 550}{1,000} = 181.5 kN

Net-clearance term between holes:

\frac{1.5 \times 44 \times 12 \times 510}{1,000} = 403.92 kN

Upper cap:

\frac{2 \times 24 \times 12 \times 800}{1,000} = 460.8 kN

P_{bs} = \min (158.4, 181.5, 403.92, 460.8) = 158.4 kN

Both bearing resistances exceed the $75.75 kN$ per-bolt demand and also exceed the $132.375 kN$ shear resistance, so shear governs the web group.

Animation labBearing and the remaining ligament1 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.

Force transfers through contact between bolt and connected plate.
The plate around the hole carries bearing stress. Bolt bearing and plate bearing are separate checks.
A short ligament can tear out towards the end. The direction of force determines the relevant edge.
Evaluate every specified bearing and ligament bound for each layer and retain the smallest applicable resistance.

4. State what has actually been designed

Simple explanation: What to do when one input is missing

A calculator cannot supply a dimension that the drawing never gave.

Separate given values, table values and calculated values.
Complete the checks whose required inputs are available.
State the missing input beside the remaining conditional result.

Remember: An illustrative assumption must not become an unstated exam given.

Related concept and full method

The flange tension and web shear/bearing bolt checks pass under the source’s force-couple idealisation. The source explicitly says only bolts have been designed; welds, end plates, stiffeners, column flange and column web require separate design. The $12 mm$ end-plate bearing calculation is not a complete end-plate bending/prying check.

Animation labSeparate the moment couple and shear5 concepts

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.

Compact exam answer

ULS $M = 368 kN \cdot m$ ; flange force $= 618.487 kN$ . Four M24 Grade 8.8 flange bolts give nominal tension $632.576 kN$ . ULS $V = 606 kN$ ; eight M24 web bolts give shear $1,059.0 kN$ . Selected $12 mm$ plate gives bolt bearing $288 kN$ and plate bearing $158.4 kN$ per bolt, both $> 75.75 kN$ . Bolts pass; remaining connection components are outside the source calculation.

Mistakes to avoid

Use $595 mm$ force-couple arm, not nominal beam depth $610 mm$ .
Do not add bottom compression-side bolts to the upper tensile group.
Do not call the end plate fully designed after checking only bearing.

Procedure for an unfamiliar variant

Factor moment and shear components separately.
Resolve bending moment into flange force using the stated couple arm.
Size the tension group using nominal bolt tension.
Assign shear to the web group and check its bearing path.
Document which other joint components are not included in the bolt-only exercise.

Independent self-check

Try it yourself. If the ultimate moment becomes $400 kN \cdot m$ , keeping $595 mm$ arm and four flange bolts, is the tension group adequate?

Reveal answer and reasoning

$F = \frac{400}{0.595} = 672.269 kN > 632.576 kN$ , so no. Required count $\geq \frac{672.269}{158.144} = 4.251$ ; at least $5$ by strength alone, but the final symmetric layout and plate/prying design must also be revised.

Animation labOut-of-plane bolt tension and prying1 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.

A bracket moment can create compression at the plate contact and tension in the bolt rows.
The original assumed pivot/compression line determines each row distance.
Under the elastic row model, a farther tension row attracts more tension. Direct shear may act simultaneously.
Flexible plates can add prying force. Use nominal tension and interaction rules exactly as specified by the course.

Animation labOut-of-plane bolt tension and prying2 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.

A bracket moment can create compression at the plate contact and tension in the bolt rows.
The original assumed pivot/compression line determines each row distance.
Under the elastic row model, a farther tension row attracts more tension. Direct shear may act simultaneously.
Flexible plates can add prying force. Use nominal tension and interaction rules exactly as specified by the course.

Long joints and large grip

Source: Chapter 2 p.24. Connections with more than two bolts and joint length over $500 mm$ require a shear-capacity reduction under the cited Clause 9.3.6.1.4. For ordinary bolts, total connected-ply thickness above five bolt diameters invokes the large-grip provision in Clause 9.3.6.1.5. The actual reduction equations are not reproduced in the supplied materials. Recognise the trigger, identify the missing clause information and avoid inventing a coefficient. None of the standard concentric splice examples here should be treated as an unrestricted formula for arbitrarily long connections.

Animation labLoad sharing along a long joint1 concept

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

Several bolts share a connection force, but a long joint need not distribute it perfectly evenly.
Compatibility between the connected plates influences force attracted by bolts near the ends.
Joint length is measured along the load path; grip is through the connected thickness.
Apply long-joint or large-grip provisions only when their conditions are met. Arrows here are qualitative.

Fillet weld throat and capacity

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

Source: Chapter 2 pp.31–33. A $90 °$ fillet with equal legs $s$ has effective throat $a \approx 0.7 s$ . The throat is the critical shortest section through the weld. The simplified method compares the resultant force per unit length $q$ with the throat resistance per unit length.

q_{capacity} = \frac{a p_{w}}{1,000} = \frac{0.7 s p_{w}}{1,000}

Units:

\frac{kN}{mm}

. Required weld leg:

s = \frac{1,000 q_{required}}{0.7 p_{w}}

Units:

mm

For S355 with Class 42 electrode the supplied Table 9.2a gives $p_{w} = 250 \frac{N}{{mm}^{2}}$ . Select both the parent-steel row and electrode-class column. “42” is not $p_{w}$ . The supplied S275 columns use $220 \frac{N}{{mm}^{2}}$ ; S460 with Class 50 uses $280 \frac{N}{{mm}^{2}}$ .

The directional method splits force per unit length into longitudinal $F_{L}$ and transverse resultant $F_{T}$ . $P_{L} = p_{w} a$ ， $P_{T} = K P_{L}$ , where $K = 1.25 \sqrt{\frac{1.5}{1 + \cos^{2} θ}}$ ， $θ$ is the force-to-throat angle specified in Figure 9.4. Check ${(\frac{F_{L}}{P_{L}})}^{2} + {(\frac{F_{T}}{P_{T}})}^{2} \leq 1$ . The numerical fillet-weld examples on this site use the lecturer’s simplified method, so do not add the directional-method enhancement.

Animation labFrom fillet leg to effective throat1 concept · 6 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.

Unnumbered example: a $6 mm$ S355/Class 42 fillet

Simple explanation: Keep the units on the same scale

Changing units is like changing the ruler, not the object.

$1 kN = 1,000 N$ ; $1 m = 1,000 mm$ .
Area needs two length conversions; inertia needs four.
Convert before substituting, then check the final unit.

Remember: A kN·m moment becomes 1,000,000 N·mm, not 1,000.

Related concept and full method

Source: Chapter 2 p.31 and L2 slide 6. Required result: strength per millimetre of weld. Given leg $s = 6 mm$ and material/electrode combination; lookup $p_{w} = 250 \frac{N}{{mm}^{2}}$ from Table 9.2a, S355 row/Class 42 column.

Throat area in a one-millimetre strip

A one-millimetre strip has throat area $a \times 1 mm$ . First calculate the throat; this is why the factor $0.7$ appears.

a = 0.7 \times 6 = 4.2 mm

;

A_{strip} = 4.2 mm \times 1 mm = 4.2 {mm}^{2}

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.

Multiply throat area by design strength

P_{strip} = 4.2 {mm}^{2} \times 250 \frac{N}{{mm}^{2}} = 1,050 N = 1.05 kN

q = \frac{1.05 kN}{1 mm} = 1.05 \frac{kN}{mm}

Exam script: $q = \frac{0.7 \times 6 \times 250}{1,000} = 1.05 \frac{kN}{mm}$ . This is capacity per effective length; multiply by effective weld length for total capacity.

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.

Try it yourself. What is the corresponding strength of an $8 mm$ weld with the same materials?

Reveal answer and reasoning

$a = 0.7 \times 8 = 5.6 mm$ . $q = \frac{5.6 \times 250}{1,000} = 1.40 \frac{kN}{mm}$ . Capacity scales with leg size when materials remain the same.

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.

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.

Effective length, minimum sizes, laps and returns

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

Chapter 2 p.34, Figure 9.3. End returns continue weld around corners; they are not the main longitudinal load-transfer lengths. — Chapter 2 p.34, Figure 9.3: an end return continues the weld around a corner; it is not the main longitudinal load-transfer weld. The source labels Return, $2 s$ , meaning return length $2 s$ ; $s$ is the weld leg. Other labels identify the Seat Bracket, Tension Member, End Connection Plate, Connection Angle and Beam End Connection.Open full-size image

Source: Chapter 2 pp.33–34. For the lecture straight welds, effective length $L_{eff} = L_{physical} - 2 s$ . The deduction represents ineffective starts/ends. If designing length, add $2 s$ to the required effective length, then round up. Apply it to each separately counted run, not once to the entire connection.

The minimum effective length is the greater of $4 s$ and $40 mm$ . The notes’ minimum leg table based on the thicker part is $3 mm$ through $6 mm$ thickness, $5 mm$ for $7$ – $13 mm$ , $6 mm$ for $14$ – $19 mm$ , and $8 mm$ over $19 mm$ . For a weld along a plate edge, maximum leg is plate thickness for thin plate below $6 mm$ ; for plate equal to or thicker than $6 mm$ , the maximum is thickness minus $2 mm$ . Do not increase a weld indefinitely on a thin angle edge.

Provide corner returns at least $2 s$ , consistent with the source diagram and worked detailing (the prose at p.33 omits “not” before “less than”). Lap length is at least $\max (5 t, 25 mm)$ , where $t$ is the thinner plate. The source also requires longitudinal weld length at least the transverse spacing between welds. Long lap joints over $100 s$ have further source-clause provisions; intermittent welds are unsuitable for fatigue conditions, and the notes limit unwelded gaps to $16 t$ / $300 mm$ in compression or $24 t$ / $300 mm$ in tension. Check the stated detail, not just total weld strength.

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

Open this example on its own page

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.

Connection example 9: a welded single unequal angle

Open this example on its own page

Check a $75 \times 50 \times 6$ S355 single angle connected by its long leg under $95 kN$ dead plus $40 kN$ imposed tension, and design the two side welds with Class 42 electrode.

Original source: LectureNotes/Ch 2_Connection.pdf — p. 42. 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	$95 kN$ dead and $40 kN$ imposed, task text.
Given dimensions	$75 \times 50 \times 6$ angle; source diagram shows $72$ and $47 mm$ half-heel widths and $12.1 mm$ centroid height.
Lookup centroid	$24.4 mm$ from SideX stated from section table in solution and labelled in diagram. Opposite arm $75 - 24.4 = 50.6 mm$ .
Lookup strengths	S355 thickness $6 mm$ → $p_{y} = 355$ ; S355/Class 42 → $p_{w} = 250 \frac{N}{{mm}^{2}}$ .
Design choice	$4 mm$ fillet, physical lengths $200 mm$ (X), $100 mm$ (Y), lecturer solution.

Before calculating: recognition and strategy

First check that the angle itself can carry the force: stronger welds cannot cure a weak angle. With no bolt holes, the gross area is effective, but a single welded connected leg still has the $0.3 a_{2}$ reduction for the unconnected leg. Then balance the two parallel weld forces about the centroid using their opposite lever arms.

1. Factored force and welded-angle resistance

Simple explanation: One connected leg does not load both legs equally

The connected leg receives the pull first; the other leg receives it through the angle.

Identify connected and outstanding legs from the drawing.
Use the relevant bolted or welded angle rule.
Keep the lecturer’s area convention consistent.

Remember: Bolted and welded reduction expressions are not the same rule.

Related concept and full method

P = 1.4 \times 95 + 1.6 \times 40 = 133 + 64 = 197 kN

a_{1} = (75 - \frac{6}{2}) \times 6 = 72 \times 6 = 432 {mm}^{2}

a_{2} = (50 - \frac{6}{2}) \times 6 = 47 \times 6 = 282 {mm}^{2}

A_{e} = A_{g} = 432 + 282 = 714 {mm}^{2}

No holes.

P_{t} = \frac{355 (714 - 0.3 \times 282)}{1,000}

= \frac{355 (714 - 84.6)}{1,000} = \frac{355 \times 629.4}{1,000} = 223.437 kN > 197

Passes.

The $0.3$ factor comes from the single welded angle rule on Ch 2 p.14. The bolted value $0.5$ would be a different connection model.

Animation labWhy a connected angle leg matters2 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.

Locate the connected leg, outstanding leg and centroid before using the table.
Only the connected leg directly receives the fastener force.
The outstanding area may not become equally effective at the same section; this motivates the effective-area rule.
Bolted, welded, single-angle and double-angle details can have different rules. Preserve the formula attached to the original case.

2. Trial $4 mm$ weld and balanced required lengths

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

q = \frac{0.7 \times 4 \times 250}{1,000} = 0.70 \frac{kN}{mm}

Total effective length:

L_{eff} = \frac{197}{0.70} = 281.4286 mm

R_{X} = \frac{197 \times 50.6}{75} = 132.9093 kN

R_{Y} = \frac{197 \times 24.4}{75} = 64.0907 kN

L_{X, eff} = \frac{132.9093}{0.70} = 189.8705 mm

L_{Y, eff} = \frac{64.0907}{0.70} = 91.5581 mm

The $12.1 mm$ centroid height is visible in the section but is not the transverse lever arm for sharing these two side-weld forces. The side-to-side $24.4$ /50.6 dimensions are the relevant ones.

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.

3. Final lengths, capacities and source 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

End allowance per weld

= 2 s = 8 mm

. Side X actual length:

\geq 189.8705 + 8 = 197.8705 mm \to 200 mm

Side Y actual length:

\geq 91.5581 + 8 = 99.5581 mm \to 100 mm

Provided effective length on side X

= 200 - 8 = 192 mm

, resistance

134.4 kN > 132.9093

. Side Y effective length

= 100 - 8 = 92 mm

, resistance

64.4 kN > 64.0907

Both effective lengths exceed $\max (16, 40) = 40 mm$ , and both physical lengths exceed $75 mm$ separation. For the $6 mm$ angle edge the maximum leg is $6 - 2 = 4 mm$ , matching the selection. The minimum-size rule depends on the unknown supporting plate thickness; do not assume it from the sketch.

Animation labFrom fillet leg to effective throat3 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.

Compact exam answer

$P = 197 kN$ . Welded single-angle capacity $= \frac{355 (714 - 0.3 \times 282)}{1,000} = 223.437 kN$ . Use $4 mm$ fillets with $q = 0.70 \frac{kN}{mm}$ . Required effective X/Y lengths $189.870$ / $91.558 mm$ . Provide $200$ / $100 mm$ physical lengths ( $192$ / $92 mm$ effective), giving $134.4$ / $64.4 kN$ against $132.909$ / $64.091 kN$ . Angle and weld strength pass; support-thickness detailing is unprovided.

Mistakes to avoid

Do not subtract holes from a welded angle with no holes.
The unconnected leg still needs the $0.3$ reduction.
For the $4 mm$ weld, the source writes “Add 12” (add $12$ ) is a textual error; it should be $2 s = 8$ .
Do not use $12.1 mm$ centroid height in the $75 mm$ transverse weld-force split.

Procedure for an unfamiliar variant

Check angle area and the appropriate welded single-leg resistance.
Read centroid location from the source/table.
Select a weld leg permitted by the angle edge.
Balance side forces, calculate effective lengths, then add $2 s$ to each.
Verify each rounded side and the detailing conditions.

Independent self-check

Try it yourself. If the imposed load rises to $45 kN$ , do the existing $200$ / $100 mm$ welds still pass?

Reveal answer and reasoning

$P = 133 + 72 = 205 kN$ . SideX demand $= \frac{205 \times 50.6}{75} = 138.307 kN > 134.4$ ; SideY $= 66.693 kN > 64.4$ . Both weld sides fail although angle capacity 223.437 still passes.

Animation labBalance two weld forces3 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.

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.

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

Open this example on its own page

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.

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.

Butt welds and inspection

Source: Chapter 2 p.34. A full-penetration butt weld can be taken as strong as its parent metal when the weld metal strength is not lower. Back-gouging/chipping before welding the reverse side or use of a backing plate can achieve penetration in the source discussion. A partial-penetration joint does not automatically have full parent-plate capacity. In an exam sketch, show the prepared edges and distinguish full from partial penetration.

Animation labFull and partial weld penetration1 concept

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

Two plates meet edge to edge. The open groove in this schematic exposes the thickness that must be joined.
A full-penetration butt weld joins through the complete thickness. The gold section shows the continuous weld metal.
A partial-penetration weld leaves part of the thickness unjoined. The dark root gap is not resisting weld metal.
The course permits parent-metal strength for full penetration when weld metal is not weaker. Confirm penetration and inspection; do not give the same capacity automatically to a partial joint.

Eccentric welds: use line properties, not plate properties

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

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

Source: Chapter 2 pp.35–37. For a closed rectangular weld of width $b$ and height $h$ , all four weld lines share direct shear. The plate interior is not weld. Uniform weld size allows the common throat to cancel while finding force per unit length, so the geometrical line second moments have units ${mm}^{3}$ , not ${mm}^{4}$ .

L = 2 (b + h)

I_{x, line} = 2 (\frac{h^{3}}{12}) + 2 b {(\frac{h}{2})}^{2} = \frac{h^{3}}{6} + \frac{b h^{2}}{2}

I_{y, line} = 2 (\frac{b^{3}}{12}) + 2 h {(\frac{b}{2})}^{2} = \frac{b^{3}}{6} + \frac{h b^{2}}{2}

I_{p, line} = I_{x, line} + I_{y, line}

The vertical lines each contribute $\frac{h^{3}}{12}$ about the centroidal $x$ axis. The two horizontal lines each have length $b$ at distance $\frac{h}{2}$ , giving $b {(\frac{h}{2})}^{2}$ by the parallel-axis rule. Reverse $b$ and $h$ for $I_{y}$ . At a corner, $x = \frac{b}{2}$ , $y = \frac{h}{2}$ and $r = \sqrt{x^{2} + y^{2}}$ . Measure eccentricity $e$ from the weld centroid to the applied load line, not automatically from the column face.

q_{s} = \frac{P}{L}

q_{T} = \frac{P e r}{I_{p, line}}

Both have units

\frac{kN}{mm}

q_{\max} = \sqrt{{(\frac{P}{L} + \frac{P e | x |}{I_{p, line}})}^{2} + {(\frac{P e | y |}{I_{p, line}})}^{2}}

The critical corner is where the vertical torsion component adds to direct shear. Calculate required leg from $q_{\max}$ and $p_{w}$ , then satisfy detailing. For a bracket in the perpendicular plane, use the flange/web force-couple model in Example 8, not these in-plane torsion equations. A three-sided weld has a different centroid, line length and inertia: recompute them from its actual weld lines.

Connection example 7: calculate a rectangular weld group from first principles

Open this example on its own page

Find the maximum force per unit weld length and select a fillet weld for one side plate carrying $60 kN$ dead plus $90 kN$ imposed load. The weld is on all four sides of a $200 mm \times 300 mm$ rectangle; the force eccentricity from its centre is $300 mm$ . Steel is S355 with Class 42 electrode.

Original source: LectureNotes/Ch 2_Connection.pdf — p. 39, p. 40. 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 geometry	$b = 200 mm$ , $h = 300 mm$ ; four weld sides. Centre is at half width/height: $x = 100$ , $y = 150 mm$ at a corner.
Given load arm	$e = 300 mm$ from group centroid, not from the right edge.
Given materials	S355/Class 42 → lookup $p_{w} = 250 \frac{N}{{mm}^{2}}$ , Table 9.2a.
Assumption	Uniform fillet size, rigid plate and line-weld elastic distribution, as in Ch 2 pp.35–37.

Before calculating: recognition and strategy

Separate direct shear $\frac{P}{L}$ from torsional force caused by $P e$ . Use the weld lines for centroid and inertia; the plate interior does not carry weld force. Derive line inertias with the parallel-axis rule. Then combine the horizontal and vertical force-per-length components at the right-hand corner, where vertical components add.

1. Geometry and factored force

L = 2 (b + h) = 2 (200 + 300) = 1,000 mm

P = 1.4 \times 60 + 1.6 \times 90 = 84 + 144 = 228 kN

M = P e = 228 \times 300 = 68,400 kN \cdot mm

r = \sqrt{100^{2} + 150^{2}} = \sqrt{32,500} = 180.278 mm

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 both line second moments

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

About the horizontal centroidal $x$ axis: two vertical welds each contribute $\frac{h^{3}}{12}$ ; two horizontal welds each contribute $b {(\frac{h}{2})}^{2}$ . About the vertical axis reverse the roles.

I_{x} = 2 (\frac{300^{3}}{12}) + 2 \times 200 \times 150^{2}

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

I_{y} = 2 (\frac{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}

Why is the unit ${mm}^{3}$ ? Weld-line length ( $mm$ ) multiplied by distance squared ( ${mm}^{2}$ ) gives that unit. Do not substitute the second moment of area of a solid rectangular plate, $\frac{b h^{3}}{12}$ , whose unit is ${mm}^{4}$ .

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.

3. Uniform direct shear per length

q_{s} = \frac{P}{L} = \frac{228 kN}{1,000 mm} = 0.228 \frac{kN}{mm}

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. Torsional components at the critical corner

q_{T} = \frac{M r}{I_{p}} = \frac{68,400 \times 180.278}{20,833,333.333} = 0.591888 \frac{kN}{mm}

Vertical component:

\frac{M x}{I_{p}} = \frac{68,400 \times 100}{20,833,333.333} = 0.32832 \frac{kN}{mm}

Horizontal component:

\frac{M y}{I_{p}} = \frac{68,400 \times 150}{20,833,333.333} = 0.49248 \frac{kN}{mm}

The source uses $\cos φ = \frac{100}{r} \approx 0.555$ for the angle between the vertical direct shear and the tangential torsion force. The corner radius itself is not the force direction.

Animation labA weld group is a set of lines1 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.

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.

5. Add components and calculate the maximum

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

q_{vertical} = 0.228 + 0.32832 = 0.55632 \frac{kN}{mm}

q_{\max} = \sqrt{{0.55632}^{2} + {0.49248}^{2}} = 0.742986 \frac{kN}{mm}

The opposite column of weld has vertical components subtracting, hence a smaller resultant. The upper and lower right corners have the same maximum 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.

6. Size the fillet and check its 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

s_{required} = \frac{1,000 q_{\max}}{0.7 p_{w}}

= \frac{1,000 \times 0.742986}{0.7 \times 250} = 4.24564 mm

The lecturer selects

s = 6 mm

q_{capacity} = \frac{0.7 \times 6 \times 250}{1,000} = 1.05 \frac{kN}{mm} > 0.742986

Weld strength passes.

The closed four-side weld is treated as the full line rectangle by the source method. This differs from separately terminated side welds with $2 s$ start/end deductions. The plate thickness is absent, so the numerical strength result is complete but minimum/maximum permitted leg size cannot be confirmed from Table 9.1/edge-thickness rules. The $6 mm$ choice is the lecturer’s selection, conditional on those missing detailing inputs.

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.

Compact exam answer

$P = 228 kN$ , $e = 300 mm$ , $L = 1,000 mm$ , $I_{p, line} = 20.83333 \times 10^{6} {mm}^{3}$ . $q_{s} = 0.228$ ; torsion components $0.32832$ vertical and 0.49248 horizontal kN/mm. $q_{\max} = 0.742986 \frac{kN}{mm}$ . $s \geq 4.246 mm$ ; $6 mm$ fillet supplies $1.05 \frac{kN}{mm}$ and passes weld strength. Plate-thickness detailing remains unverified because it is not given.

Mistakes to avoid

The two $300 mm$ dimensions mean different things: height and eccentricity.
Use line inertia in ${mm}^{3}$ , not solid-plate inertia in ${mm}^{4}$ .
The dashed right edge still belongs to the stated four-side weld.
Do not certify minimum/maximum weld size without the plate thickness.

Procedure for an unfamiliar variant

Identify every actual weld line and its load path.
Find centroid, length and line inertias.
Measure load eccentricity from the centroid.
Calculate direct and torsional components and the critical resultant.
Solve for leg size and check detailing separately.

Independent self-check

Try it yourself. Keep the same load and rectangle but move the load line through the weld centroid. Find required strength and leg by strength alone.

Reveal answer and reasoning

$e = 0$ so torsion vanishes. $q = \frac{228}{1,000} = 0.228 \frac{kN}{mm}$ . $s \geq \frac{1,000 \times 0.228}{175} = 1.303 mm$ by strength alone. The actual selected leg is still controlled by minimum/detailing rules; a $1.3 mm$ design cannot be adopted just from this strength result.

Animation labA weld group is a set of lines1 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.

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.

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

Open this example on its own page

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.

Animation labA weld group is a set of lines1 concept · 5 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.

Practise this chapter: tutorials, assignments and past-paper answers →

Connections: trace the force through bolts, plates and welds

Bolts, grades and weld types

Hole size, pitch, gauge and edge distances

Recognise the competing failure modes

Shear planes and bolt shear capacity

Bearing and end tear-out: calculate every bound

Tension and the prying assumption

Gross, net and effective area

Read the diagram and collect the data

Before calculating: recognition and strategy

1. Establish the capacity of one bolt and its bearing path

2. Select bolts on each side of the splice

3. Verify the selected layout

4. Check main-plate tension across two holes

5. Check both covers and conclude

Compact exam answer

Mistakes to avoid

Procedure for an unfamiliar variant

Independent self-check

Block shear

Single and double angles connected through one leg

Read the diagram and collect the data

Before calculating: recognition and strategy

1. Convert characteristic loads to one design force

2. Bolt shear across the single interface

3. Bearing of the bolts

4. Bearing and ligament resistance of the angle leg

5. Effective angle area and single-leg reduction

6. State the governing result and the scope

Compact exam answer

Mistakes to avoid

Procedure for an unfamiliar variant

Independent self-check

In-plane eccentric bolts: derive the force distribution

Read the diagram and collect the data

Before calculating: recognition and strategy

1. Factored load and moment

2. Bolt-group polar sum and corner radius

3. Direct force, torsion and maximum resultant

4. Compare with single-shear bolt resistance

5. Check bolt bearing against the same resultant

6. Connected-part bounds and conclusion

Compact exam answer

Mistakes to avoid

Procedure for an unfamiliar variant

Independent self-check

Out-of-plane brackets: shear plus tension

Read the diagram and collect the data

Before calculating: recognition and strategy

1. Find bolt shear and nominal tension resistance

2. Sum squared distances about the bottom row

3. Maximum tension is in the top row

4. Uniform direct shear

5. Combined action and conclusion

Compact exam answer

Mistakes to avoid

Procedure for an unfamiliar variant

Independent self-check

Read the diagram and collect the data

Before calculating: recognition and strategy

1. Design the tension flange group

2. Design the shear group

3. Check selected end-plate bearing

4. State what has actually been designed

Compact exam answer

Mistakes to avoid

Procedure for an unfamiliar variant

Independent self-check

Long joints and large grip

Fillet weld throat and capacity

Unnumbered example: a 6mm6mm S355/Class 42 fillet

Throat area in a one-millimetre strip

Multiply throat area by design strength

Effective length, minimum sizes, laps and returns

Read the diagram and collect the data

Before calculating: recognition and strategy

1. Factored tensile force

2. Select a trial leg and find capacity per length

3. Share force and effective length between the sides

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

Unnumbered example: a $6 mm$ S355/Class 42 fillet

2. Trial $4 mm$ weld and balanced required lengths

2. Trial $12 mm$ flange fillets