Worked examples

Connection example 2: design a bolted tension splice

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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.