Worked examples

Connection example 4: bracket bolts in shear and tension

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Check ten Grade 8.8 M16 bolts supporting a $250 kN$ ultimate bracket load at $200 mm$ eccentricity. Five bolts are arranged vertically on each side at $90 mm$ centres. Steel is S355.

Original source: LectureNotes/Ch 2_Connection.pdf — p. 27, p. 28. 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.27:

200 mm

is the external load lever arm beyond the column joint. The triangular bolt-force distribution rotates about the lowest bolt line; all distances are measured upwards from that line. The labels identify the joint and state that bolts on both sides have centre-to-centre spacing

90 mm

. “Bolt loads” identifies the bolt forces; “Maximum tension” identifies

F_{T}

. The other tensile forces shown, in order, are

\frac{F_{T} y_{2}}{y_{1}}

\frac{F_{T} y_{3}}{y_{1}}

\frac{F_{T} y_{4}}{y_{1}}

，

y_{1}

is the distance to the top row. The original conditions at the lower crop edge, cross-checked against the full p.27, are

F_{s} \leq P_{s}

F_{T} \leq P_{nom}

, and

\frac{F_{s}}{P_{s}} + \frac{F_{T}}{P_{nom}} \leq 1.4

. All three must hold simultaneously. Below, values are substituted separately into the shear, tension and interaction checks.Open full-size image

Read the diagram and collect the data

Input	Origin
Given load	$250 kN$ ultimate; the downward arrow is outside the connection plane.
Given bolt arrangement	Five bolts per side; two sides; $90 mm$ between consecutive rows; total $10$ .
Calculated row distances	On each side, measure upwards from the bottom rotation line: $0, 90, 180, 270, 360 mm$ .
Lookup	M16 $A_{s} = 157 {mm}^{2}$ , Grade 8.8 $p_{s} = 375$ and $p_{t} = 560 \frac{N}{{mm}^{2}}$ (Ch 2 pp.3,9–10).
Model assumption	Rigid bracket and linear tensile-force distribution about bottom bolt line, as stated in the lecturer approximation; use nominal tension $0.8 A_{s} p_{t}$ .

Before calculating: recognition and strategy

The load pulls the upper bolts along their axes while all bolts carry vertical shear. Use an out-of-plane rotation model, not the centroidal in-plane torsion formula. Check shear, tension and their interaction separately. The task’s bolt calculation does not provide the full plate/gauge dimensions needed for a complete joint/prying and bearing verification.

1. Find bolt shear and nominal tension resistance

P_{s} = \frac{375 \times 157}{1,000} = 58.875 kN

P_{t} = \frac{560 \times 157}{1,000} = 87.92 kN

P_{nom} = 0.8 \times 87.92 = 70.336 kN

The $0.8$ reduction is part of the adopted simplified model. Do not substitute $87.92$ in the nominal-tension interaction denominator.

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.

2. Sum squared distances about the bottom row

Each of the two vertical sides has the same distances. The bottom row contributes $0^{2}$ to moment resistance but remains in the count for direct shear.

J_{rows} = 2 (0^{2} + 90^{2} + 180^{2} + 270^{2} + 360^{2})

= 2 (0 + 8,100 + 32,400 + 72,900 + 129,600)

= 2 \times 243,000 = 486,000 {mm}^{2}

M = P e = 250 \times 200 = 50,000 kN \cdot mm

Animation labOut-of-plane bolt tension and prying1 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.

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.

3. Maximum tension is in the top row

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

Tension varies in proportion to distance $y$ . Let $F_{i} = C y_{i}$ ; moment equilibrium gives $M = C Σ y^{2}$ . Thus $C = \frac{M}{J}$ and the top bolt at $y = 360 mm$ has the largest tension.

F_{T, \max} = \frac{50,000 \times 360}{486,000} = 37.037 kN

37.037 < 70.336 kN

: individual tension check passes.

Animation labOut-of-plane bolt tension and prying1 concept · 5 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.

4. Uniform direct shear

F_{s} = \frac{P}{n} = \frac{250}{10} = 25 kN

25 < 58.875 kN

: individual shear check passes.

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. Combined action and conclusion

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

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

= \frac{25}{58.875} + \frac{37.037}{70.336}

= 0.424628 + 0.526573 = 0.951201 \leq 1.4

Passes.

The bolts satisfy all three calculated shear/tension conditions under the lecturer’s approximate model. This is the requested bolt result. Plate thickness, detailed bearing-edge geometry and flange gauge are not supplied sufficiently to certify every other joint limit state; do not claim those have been checked.

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.

Compact exam answer

$P_{s} = 58.875 kN$ ; $P_{nom} = 70.336 kN$ . $Σ y^{2} = 486,000 {mm}^{2}$ about bottom bolt line. $F_{T, \max} = \frac{250 \times 200 \times 360}{486,000} = 37.037 kN$ . $F_{s} = 25 kN$ . Individual limits pass and interaction $= 0.9512 < 1.4$ . Bolts are adequate under the specified approximate shear/tension model.

Mistakes to avoid

Do not factor $250 kN$ again.
Do not measure $y$ from the group centroid for this bottom-pivot model.
Do not omit the bottom bolts from direct shear.
Passing the $1.4$ interaction does not replace individual shear/tension limits.

Procedure for an unfamiliar variant

Identify out-of-plane rotation and the stated pivot.
Write all $y$ distances and include all physical bolts in $Σ y^{2}$ .
Calculate maximum tension and uniform shear.
Check each individual resistance and then the interaction.
State the model and any unprovided plate/prying information.

Independent self-check

Try it yourself. Move the same load to $400 mm$ eccentricity. Does the existing bolt group pass?

Reveal answer and reasoning

Tension doubles to $74.074 kN$ , exceeding $70.336$ . Interaction becomes $\frac{25}{58.875} + \frac{74.074}{70.336} \approx 1.478 > 1.4$ . Both tension and interaction fail; shear remains $25 kN$ .

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.