Worked examples

Assignment 1 Q2: a $1000 kN$ bolted splice

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Check bolt shear, bolt bearing, plate bearing and plate tension for the source S355 splice: $26 mm$ main plate, two $10 mm$ covers, Grade 8.8 M30 bolts in $33 mm$ holes, ultimate tension $1,000 kN$ .

Original source: Assignment/AY2627s 1-CON4334-Assignment 1.pdf — p. 2. 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

Value	Origin/type
Tension	Given ultimate $1,000 kN$ ; do not multiply by $1.4$ or $1.6$ . Each half-splice transfers this full force.
Fasteners	Given M30, Grade 8.8 and hole diameter $33 mm$ . Ch 2 table: $A_{s} = 561 {mm}^{2}$ , $p_{s} = 375 \frac{N}{{mm}^{2}}$ , $p_{bb} = 1,000 \frac{N}{{mm}^{2}}$ , $U_{b} = 800 \frac{N}{{mm}^{2}}$ .
Plates	Given S355; $p_{bs} = 550 \frac{N}{{mm}^{2}}$ , $U_{s} = 510 \frac{N}{{mm}^{2}}$ . Main plate $T = 26$ , giving $p_{y} = 345$ ; cover plate $T = 10$ , giving $p_{y} = 355$ .
End distance	Cover end $55 mm$ is explicitly dimensioned. For a numerical main-plate end check assume the butt lies centrally within the $110 mm$ gap, e_m= $55 mm$ ; this split is a stated symmetry assumption, not a measured drawing length.
Distribution	Six bolts per half; main force/bolt $\frac{1,000}{6} = 166.666667 kN$ ; each of two covers takes $\frac{500}{6} = 83.333333 kN$ .

Before calculating: recognition and strategy

Cut the joint at one half and follow force from the main plate through six double-shear bolts into the two covers. The other half performs the same transfer in reverse; twelve bolts are not one parallel group carrying $1,000 kN$ . Check each material layer’s actual force and thickness, then the two-hole net section.

(i) 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

Threads are assumed in every shear plane, so

A_{s} = 561 {mm}^{2}

. Per shear plane:

P_{s} = \frac{A_{s} p_{s}}{1,000} = \frac{561 \times 375}{1,000} = 210.375 kN

Double-shear resistance per bolt:

2 \times 210.375 = 420.75 kN

Six-bolt half-joint resistance:

6 \times 420.75 = 2,524.5 kN > 1,000

Passes. Equivalently, demand per bolt:

\frac{1,000}{6} = 166.666667 kN < 420.75

Demand per shear plane:

83.333333 kN < 210.375

Using the threaded stress area is the conservative course model when thread location is not specified. Do not use the $33 mm$ hole area as the bolt shear area.

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.

(ii) Bolt bearing against all connected layers

Bolt bearing uses nominal bolt diameter $30 mm$ and projected contact thickness. The bolt’s two cover contacts share the load; the thinner effective thickness is $\min (26, 10 + 10) = 20 mm$ .

Bolt bearing against the main plate:

\frac{30 \times 26 \times 1,000}{1,000} = 780 kN

per bolt. Contact at each cover plate:

\frac{30 \times 10 \times 1,000}{1,000} = 300 kN

Both plates together give, per bolt,

600 kN

, below

780

, so this governs. Six-bolt half-joint resistance:

6 \times 600 = 3,600 kN > 1,000

Passes.

The $600 kN$ is a bolt-material bearing limit. It is not the resistance of the weaker S355 plate material, which needs the next check.

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.

(iii) Plate bearing and ligament limits

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 Ch 2 pp.9–10: the ordinary, non-countersunk hole model uses $k_{bs} = 1$ ; nominal $d = 30 mm$ , longitudinal spacing $90 mm$ , hole diameter $33 mm$ , hence clear ligament between holes $l_{c} = 90 - 33 = 57 mm$ . For each plate, take the minimum of the three course bearing bounds.

P_{bs} = \min [k d t p_{bs}, 0.5 k e t p_{bs}, \min (1.5 l_{c} t U_{s}, 2 d t U_{b})]

Bound per bolt	Main $26 mm$ (e_m 55 assumed)	One $10 mm$ cover (e 55 given)
$k d t p_{bs}$	$\frac{1 \times 30 \times 26 \times 550}{1,000} = 429 kN$	$\frac{1 \times 30 \times 10 \times 550}{1,000} = 165 kN$
$0.5 k e t p_{bs}$	$\frac{0.5 \times 55 \times 26 \times 550}{1,000} = 393.25 kN$	$\frac{0.5 \times 55 \times 10 \times 550}{1,000} = 151.25 kN$
$1.5 l_{c} t U_{s}$	$\frac{1.5 \times 57 \times 26 \times 510}{1,000} = 1,133.73 kN$	$\frac{1.5 \times 57 \times 10 \times 510}{1,000} = 436.05 kN$
$2 d t U_{b}$ ceiling	$\frac{2 \times 30 \times 26 \times 800}{1,000} = 1,248 kN$	$\frac{2 \times 30 \times 10 \times 800}{1,000} = 480 kN$
Minimum	$393.25 kN$	$151.25 kN$

Main-plate half-joint group:

6 \times 393.25 = 2,359.5 kN > 1,000

Each cover-plate half-joint group:

6 \times 151.25 = 907.5 kN > 500

Both cover plates together:

1,815 kN > 1,000

; cover plates govern plate bearing.

If the butt is not centered, retain the actual main end distancee_m. For the end-distance bearing bound alone:

\frac{0.5 e_{m} \times 26 \times 550}{1,000} \geq \frac{1,000}{6}

e_{m} \geq \frac{2 \times 166.666667 \times 1,000}{26 \times 550} = 23.310023 mm

This is only the strength-derived minimum; detailing may require substantially more. For a standard M30 bolt, the minimum end/edge distance is $1.75 d = 52.5 mm$ for a sheared or hand flame-cut edge, and $1.25 d = 37.5 mm$ for a rolled or machine gas-cut edge. The known transverse $40 mm$ therefore satisfies the latter but not the former. The longitudinal spacing is $90 \geq 2.5 d = 75$ ; transverse spacing $120 \geq 3 d = 90$ ; maximum spacing $\min (12 \times 10, 150) = 120 mm$ , so the transverse spacing is exactly at the maximum. Confirm the edge preparation and the main-plate end distances on both sides of the butt joint before declaring the detail unconditionally satisfactory.

Animation labBearing and the remaining ligament1 concept · 19 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.

(iv) Main and cover tensile capacities

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 transverse net-section cut through an aligned bolt column crosses two $33 mm$ holes, not all six holes along a half-splice. Course effective areaAe=min(Ag,KeAn), with Ke 1.1 for the stated S355 course model.

Plate width:

40 + 120 + 40 = 200 mm

Main-plate gross area:

A_{g} = 200 \times 26 = 5,200 {mm}^{2}

Main-plate net area:

A_{n} = (200 - 2 \times 33) \times 26 = 134 \times 26 = 3,484 {mm}^{2}

Main-plate effective area:

A_{e} = \min (5,200, 1.1 \times 3,484) = 3,832.4 {mm}^{2}

26 mm

thickness uses

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

P_{t, main} = \frac{3,832.4 \times 345}{1,000} = 1,322.178 kN > 1,000

Passes. Gross area per cover plate:

A_{g} = 200 \times 10 = 2,000 {mm}^{2}

Net area per cover plate:

A_{n} = (200 - 66) \times 10 = 1,340 {mm}^{2}

Effective area per cover plate:

A_{e} = \min (2,000, 1.1 \times 1,340) = 1,474 {mm}^{2}

10 mm

thickness uses

p_{y} = 355 \frac{N}{{mm}^{2}}

P_{t, cover} = \frac{1,474 \times 355}{1,000} = 523.27 kN > 500

Both cover plates together:

2 \times 523.27 = 1,046.54 kN > 1,000

Passes.

The cover pair’s tension resistance gives the smallest of the requested numerical group capacities. Strength passes under the stated end-bearing assumption, with only $4.654 %$ capacity above demand in cover tension. The transverse edge preparation caveat remains a separate detailing issue.

Animation labSubtract holes on the failure path1 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.

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

Per half: sixM30 in double shear. Shear $2,524.5 kN$ ; bolt bearing $3,600 kN$ ; plate bearing main $2,359.5$ and cover pair $1,815 kN$ (centered-butt e_m 55 assumed). Tensile main $1,322.178$ and cover pair $1,046.54 kN$ . All exceed $1,000$ . Cover tension controls. $40 mm$ side edges require an edge category permitting $37.5 mm$ ; they fail the $52.5 mm$ sheared-edge minimum. Confirm the main end-distance split before final approval.

Mistakes to avoid

Do not divide $1,000 kN$ by twelve bolts for a half-splice check.
Do not add the full $26 + 10 + 10$ thickness for the controlling bolt bearing.
Use $33 mm$ for hole deductions and $30 mm$ for projected bearing.
Do not use $355 \frac{N}{{mm}^{2}}$ in the $26 mm$ main tensile calculation.

Procedure for an unfamiliar variant

Draw the half-splice force path.
Count shear planes and forces per plate.
Check bolt shear and bearing.
Evaluate every plate-bearing bound and detailing condition.
Cut the correct net section, cap effective area at gross, and check main and covers separately.

Independent self-check

Try it yourself. Invented variant: ultimate tension increases to $1,100 kN$ with no geometry change. Which demonstrated requested check first fails?

Reveal answer and reasoning

The cover-pair tensile resistance remains $1,046.54 kN$ , below $1,100$ . The main tensile, shear and the stated bearing capacities still exceed $1,100$ , but any failed required check makes the detail inadequate. Cover tension cannot be rescued by surplus bolt shear resistance.

Assignment 1 Q2: a 1000kN1000kN bolted splice

Read the diagram and collect the data

Before calculating: recognition and strategy

(i) Bolt shear capacity

(ii) Bolt bearing against all connected layers

(iii) Plate bearing and ligament limits

(iv) Main and cover tensile capacities

Compact exam answer

Mistakes to avoid

Procedure for an unfamiliar variant

Independent self-check

Assignment 1 Q2: a $1000 kN$ bolted splice