Worked examples

Tutorial 2 Q1: a bolted splice with a layout problem

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Check the $750 kN$ ultimate tension splice for bolt shear, bolt/plate bearing, plate tension and bolt layout. On each side of the splice are six Grade 8.8 M24 bolts in $26 mm$ holes, with threads in the shear planes. Plates are S355.

Original source: LectureNotes/Ch 2_Connection.pdf — p. 45, p. 46. Values tagged given are in the question or diagram; lookup values come from a named table; calculated values follow from the working; assumptions are stated explicitly.

Read the diagram and collect the data

Input	Origin and role
Force	Given $750 kN$ ultimate; each half-splice transfers all $750 kN$ .
Bolts	Given $6$ per half, M24, hole 26, Grade 8.8, threads in shear planes.
Plates	Main $150 \times 22 mm$ ; each cover $150 \times 12 mm$ . Two covers share the tension in parallel.
Geometry closure	Across the plate width, $40 + 70 + 40 = 150 mm$ ; between the three bolt columns in each half are two longitudinal spacings of $60 mm$ .
Lookups	M24: $A_{s} = 353 {mm}^{2}$ ; $p_{s} = 375$ , $p_{bb} = 1,000$ , $p_{bs} = 550$ , $U_{s} = 510$ , $U_{b} = 800 \frac{N}{{mm}^{2}}$ ; $k_{bs} = 1$ , $K_{e} = 1.1$ . Main plate $p_{y} = 345$ ; cover plate $p_{y} = 355 \frac{N}{{mm}^{2}}$ .

Before calculating: recognition and strategy

The two splice halves are in series, so do not add twelve bolts’ capacities against one $750 kN$ force. Within a half, six bolts share the concentric force equally. The two covers create two shear interfaces. Check resistance first, then check the actual spacing/edge rules: adequate strength does not waive an invalid layout.

1. Bolt shear

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

Force per bolt:

\frac{750}{6} = 125 kN

Double-shear resistance per bolt:

\frac{2 A_{s} p_{s}}{1,000} = \frac{2 \times 353 \times 375}{1,000} = 264.75 kN

Half-joint bolt-group resistance:

6 \times 264.75 = 1,588.5 kN > 750 kN

Passes.

Use $353 {mm}^{2}$ , not $\frac{π 24^{2}}{4}$ , because the question explicitly places threads in the shear planes.

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.

2. Bolt bearing

For this symmetric sandwich joint, effective thickness is the smaller of main-plate thickness and total thickness of both cover plates:

\min (22, 2 \times 12) = 22 mm

P_{bb, bolt} = \frac{d t p_{bb}}{1,000}

= \frac{24 \times 22 \times 1,000}{1,000} = 528 kN

Half-joint bolt group:

6 \times 528 = 3,168 kN > 750 kN

Passes.

Alternatively each cover takes $62.5 kN$ per bolt and has $\frac{24 \times 12 \times 1,000}{1,000} = 288 kN$ resistance. The main plate takes $125 kN$ per bolt; it governs the combined comparison.

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.

3. Connected-plate bearing and ligament bounds

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

Hole diameter

d_{0} = 26 mm

; longitudinal spacing

60 mm

l_{c} = 60 - 26 = 34 mm

End distance

e = 40 mm

;

k_{bs} = 1

B_{1} = \frac{1 \times 24 \times 22 \times 550}{1,000} = 290.4 kN

B_{2} = \frac{0.5 \times 1 \times 40 \times 22 \times 550}{1,000} = 242 kN

Net-clearance branch between holes:

\frac{1.5 \times 34 \times 22 \times 510}{1,000} = 572.22 kN

Upper cap:

\frac{2 \times 24 \times 22 \times 800}{1,000} = 844.8 kN

B_{3} = \min (572.22, 844.8) = 572.22 kN

P_{bs, bolt} = \min (290.4, 242, 572.22) = 242 kN

Half-joint bolt group:

6 \times 242 = 1,452 kN > 750 kN

Passes.

Using the minimum $40 mm$ end distance for the group is conservative for inner columns. For each $12 mm$ cover the corresponding minimum is $\frac{0.5 \times 40 \times 12 \times 550}{1,000} = 132 kN$ against $62.5 kN$ ; two covers give $264 kN$ per bolt, still above the $22 mm$ main plate’s 242.

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.

4. Main and cover net tension sections

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

A straight section across the width crosses two $26 mm$ holes, not six: the other bolt columns lie further along the load direction. For plate tension use $A_{e} = \min (A_{g}, K_{e} A_{n})$ , $K_{e} = 1.1$ for the supplied S355 method.

Main-plate gross section:

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

Main-plate net section:

A_{n} = (150 - 2 \times 26) \times 22 = 98 \times 22 = 2,156 {mm}^{2}

A_{e} = \min (3,300, 1.1 \times 2,156) = 2,371.6 {mm}^{2}

T = 22 mm

, so

p_{y} = 345 \frac{N}{{mm}^{2}}

P_{t, main} = \frac{2,371.6 \times 345}{1,000} = 818.202 kN > 750 kN

Each cover plate:

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

A_{n} = 98 \times 12 = 1,176 {mm}^{2}

A_{e} = 1.1 \times 1,176 = 1,293.6 {mm}^{2} < 1,800 {mm}^{2}

P_{t, cover} = \frac{1,293.6 \times 355}{1,000} = 459.228 kN > 375 kN

Combined resistance of both cover plates:

918.456 kN > 750 kN

Both tensile load paths pass.

The main net-section tension resistance $818.202 kN$ is the lowest of these requested strength checks.

Animation labSubtract holes on the failure path1 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.

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. Layout: a strength pass is not a complete pass

Minimum longitudinal spacing:

2.5 d = 2.5 \times 24 = 60 mm

Provided

60 mm

: just passes. Usual minimum transverse spacing:

3 d = 3 \times 24 = 72 mm

Provided

70 mm

, below the source's general requirement. Maximum longitudinal/transverse spacing:

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

60

and

70 mm

both pass. Maximum edge distance:

11 \times 12 \times \sqrt{\frac{275}{355}} = 116.1786 mm

40 < 116.1786

Table 9.3, M24 row gives minimum end/edge $42 mm$ for sheared/hand-flame-cut edges and $30 mm$ for rolled/gas-cut edges. The $40 mm$ dimensions pass the latter but fail the former. Edge preparation is not specified, so that conclusion is conditional. The $70 mm$ gauge is below the normal $72 mm$ rule regardless of this edge choice. The source says “normally” $3 d$ ; no justified exception is supplied, so do not approve it as meeting the ordinary course detailing requirement.

A clearly labelled redesign would use at least $72 mm$ gauge and compatible edges. To keep $42 mm$ sheared edges on both sides requires width $\geq 42 + 72 + 42 = 156 mm$ . Alternatively, a $150 mm$ width with $72 mm$ gauge leaves $39 mm$ edges and therefore requires a permitted edge type with minimum $\leq 39 mm$ . Recheck changed geometry; these are proposed modifications, not dimensions of the original.

Animation labBolt centres, holes and edges2 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.

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.

Compact exam answer

$750 kN$ per half-splice. Six M24 bolts in double shear: shear $1,588.5 kN$ ; bolt bearing $3,168 kN$ ; plate bearing $1,452 kN$ . Main effective tension $818.202 kN$ ; cover pair $918.456 kN$ . All requested strength checks pass. Layout: pitch $60 = 2.5 d$ passes, but gauge $70 < 3 d = 72$ . End/side $40 mm$ passes M24 rolled/gas-cut minimum 30 but fails sheared minimum 42. Given detail is not established as fully compliant; revise/justify layout.

Mistakes to avoid

Do not count the two halves as parallel resistance.
Do not remove six holes from one transverse net section.
Do not use $355$ for the $22 mm$ main plate.
Do not round $72 mm$ down to 70 for a minimum spacing.

Procedure for an unfamiliar variant

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

Independent self-check

Try it yourself. Invented variant: ultimate load increases to $825 kN$ with the same geometry. Which of the calculated strength checks fails first?

Reveal answer and reasoning

Main-plate effective tension is $818.202 kN$ , so it fails at $825 kN$ . The cover pair 918.456 and bolt/bearing group resistances still exceed 825. The original layout concerns also remain.