Worked examples

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

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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.