Worked examples

2023 BQ2(b): ten bolts in an eccentric bracket

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Find shear and tension capacities, then check tenM20 Grade 8.8 bolts attaching the S355 bracket carrying $350 kN$ design load at $300 mm$ eccentricity. The end plate is $16 mm$ thick and the supporting section $305 \times 305 \times 137$ UC.

Original source: Pastpaper/22ENGTY033.pdf — p. 4. 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/model
Load	Given designed $350 kN$ ; no $1.4$ / $1.6$ factor.
Fasteners	M20 $A_{s} = 245 {mm}^{2}$ ; Grade 8.8 $p_{s} = 375$ , $p_{t} = 560 \frac{N}{{mm}^{2}}$ ; nominal tension $0.8 A_{s} p_{t}$ .
Approximate pivot	Use the course model rotating about the lowest bolt row: $y = 0, 70, 140, 210, 280 mm$ , with two bolts per row.
Prying applicability	Supporting UC flange widthB309.2 mm from Data File p.11. Bolt gaugeG is not dimensioned, so $G \leq 0.55 B$ cannot be independently verified.

Before calculating: recognition and strategy

The load tends to peel the upper plate away while every bolt also carries direct shear. Calculate nominal tension resistance for the course simplified no-prying treatment, then compare both individual actions and the combined ratio. The course combined limit is $1.4$ , not $1$ .

(i) Individual bolt capacities —5 printed marks

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

Single-shear resistance:

P_{s} = \frac{A_{s} p_{s}}{1,000} = \frac{245 \times 375}{1,000} = 91.875 kN

per bolt. Basic tension resistance:

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

Nominal tension resistance for the course reduced-resistance treatment:

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

State both $137.2$ and $109.76$ and identify their meanings. The following lecture-style interaction uses $109.76$ . The one end-plate/column interface is single shear.

Animation labOut-of-plane bolt tension and prying3 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.

(ii) Forces in the governing bolt and adequacy —10 printed marks

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

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

M = P e = 350 \times 300 = 105,000 kN \cdot mm

Σ y^{2} = 2 (0^{2} + 70^{2} + 140^{2} + 210^{2} + 280^{2})

= 2 (0 + 4,900 + 19,600 + 44,100 + 78,400) = 294,000 {mm}^{2}

F_{t, \max} = \frac{M y_{\max}}{Σ y^{2}} = \frac{105,000 \times 280}{294,000} = 100 kN

Direct shear:

F_{s} = \frac{P}{n} = \frac{350}{10} = 35 kN

Individual shear check:

35 < 91.875

: passes. Individual tension check:

100 < 109.76

: passes. Interaction:

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

= \frac{35}{91.875} + \frac{100}{109.76}

= 0.380952 + 0.911079 = 1.292031 < 1.4

Passes.

The bottom row has zero moment-induced tension in this approximation but still carries $35 kN$ shear. This is why the tension sum uses squared row distances, while direct shear simply divides by $10$ .

Flange transverse-spacing condition for no prying:

G \leq 0.55 B = 0.55 \times 309.2 = 170.06 mm

Animation labOut-of-plane bolt tension and prying4 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.

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

(i) Each M20: $P_{s} = 91.875$ , $P_{t} = 137.2$ , $P_{nom} = 109.76 kN$ . (ii) $P = 350$ , $e = 300$ , $M = 105,000 kN \cdot mm$ ; $Σ y^{2} = 294,000 {mm}^{2}$ ; $F_{t, \max} = 100$ , $F_{s} = 35 kN$ . Both individual limits pass; interaction $1.292031 < 1.4$ . This conclusion is conditional on the course no-prying model. Missing transverse spacing prevents verification of the required $G \leq 170.06 mm$ .

Mistakes to avoid

Count five rows, not six; top $70$ is a margin.
$Σ y^{2}$ Count two bolts per row.
Do not put all $350 kN$ on the top bolt.
Do not use $1$ as the combined-ratio limit.

Procedure for an unfamiliar variant

Identify bolt count, spacing and the stated approximate rotation line.
Compute shear and nominal tension resistances.
Calculate $P e$ and $Σ y^{2}$ , then the top-row tension.
Check both individual actions and the $1.4$ interaction.
State whether the no-prying prerequisites are actually specified.

Independent self-check

Try it yourself. Invented variant: eccentricity increases to $350 mm$ with the same $350 kN$ load. Does the nominal tension check pass?

Reveal answer and reasoning

$F_{t, \max}$ scales with $e$ : $\frac{100 \times 350}{300} = 116.666667 kN > 109.76$ , so individual tension fails. You must reject it even if considering only a combined sum might seem acceptable; all three bolt conditions are compulsory.

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.