Worked examples

2024 BQ2(a): back-to-back bolted angles

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Chinese–English terminology

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Check bolt shear, bolt bearing, angle bearing and angle tension for two $125 \times 75 \times 8$ S355 unequal angles connected back-to-back to the web of $305 \times 305 \times 118$ UC by threeM20 Grade 8.8 bolts in $22 mm$ holes. Design concentric tension is $350 kN$ . The paper excludes checking the supporting UC.

Original source: Pastpaper/23ENGTY004.pdf — p. 3. 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	Source/type
Force	Given ultimatePu 350 kN, symmetric pair → $175 kN$ per angle.
Bolt material	M20 $A_{s} = 245 {mm}^{2}$ ; Grade 8.8 $p_{s} = 375$ , $p_{bb} = 1,000$ , $U_{b} = 800 \frac{N}{{mm}^{2}}$ .
Central ply	Data File p.11, $305 \times 305 \times 118$ UC web $t = 12.0 mm$ ; read the web column, not flange $T = 18.7 mm$ .
Angle material	$8 mm$ S355 →py 355; coursepbs 550,Us 510,Ke 1.1.
Area model	Use exact given $125$ / $75$ / $8$ rectangular leg split as in the lecture. The current unequal-angle table has a mislabelled $125 \times 75$ thickness row; do not import its inconsistent area $22.7 {cm}^{2}$ as an $8 mm$ angle.

Before calculating: recognition and strategy

The section view determines the number of shear planes. Follow $350 kN$ through three bolts into two angle legs, allocating $175 kN$ to each angle. For angle tension, deduct one hole on a transverse fracture path, apply effective-area caps and then the double-bolted-angle shear-lag reduction.

(i) Bolt shear —2 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

Each bolt has two shear interfaces: angle/UC web and UC web/angle. Per shear plane:

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

Double shear per bolt:

P_{s} = 2 \times 91.875 = 183.75 kN

Whole group:

3 \times 183.75 = 551.25 kN > 350

Passes. Equivalently, demand per interface:

\frac{175}{3} = 58.333333 kN < 91.875

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 —2 printed marks

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

Central UC web contact thickness

12 mm

(table lookup). Combined contact thickness of the two outer angles

8 + 8 = 16 mm

. Governing projected thickness:

\min (12, 16) = 12 mm

Bolt bearing per bolt:

\frac{20 \times 12 \times 1,000}{1,000} = 240 kN

Three-bolt group:

3 \times 240 = 720 kN > 350

Passes.

Each outer $8 mm$ contact individually provides $160 kN$ against $58.333333 kN$ ; the paired $320 kN$ exceeds the central $240 kN$ . This checks the bolt bearing limit. It does not perform the expressly excluded UC material-bearing or web-strength design.

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.

(iii) Bearing of the angles —6 printed marks

Evaluate one angle under $175 kN$ . Use d 20 for projected bearing, d₀22 for ligament deductions, pitch $80$ and end $50$ . The ordinary-hole factor $k_{bs} = 1$ .

Net clearance between holes:

l_{c} = 80 - 22 = 58 mm

B_{1} = k d t p_{bs} = \frac{1 \times 20 \times 8 \times 550}{1,000} = 88 kN

B_{2} = 0.5 k e t p_{bs} = \frac{0.5 \times 1 \times 50 \times 8 \times 550}{1,000} = 110 kN

Net-clearance term:

\frac{1.5 \times 58 \times 8 \times 510}{1,000} = 354.96 kN

Upper cap:

\frac{2 \times 20 \times 8 \times 800}{1,000} = 256 kN

B_{3} = \min (354.96, 256) = 256 kN

Bearing per bolt in each angle:

\min (88, 110, 256) = 88 kN

Three-bolt resistance for each angle:

3 \times 88 = 264 kN > 175

For the angle pair,

528 kN > 350

: passes.

Optional detailing observation, beyond the four requested strength checks: the closest transverse edge $41 mm$ exceeds either M20 minimum $34$ / $26$ . The opposite $84 mm$ free-edge distance exceeds the ordinary $11 t ε = 11 \times 8 \times \sqrt{\frac{275}{355}} = 77.452 mm$ maximum, so a complete detail would require review of that provision. It does not change the arithmetic bearing result above.

Animation labBearing and the remaining ligament1 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.

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) Tensile capacity of the angles —7 printed marks

Simple explanation: One connected leg does not load both legs equally

The connected leg receives the pull first; the other leg receives it through the angle.

Identify connected and outstanding legs from the drawing.
Use the relevant bolted or welded angle rule.
Keep the lecturer’s area convention consistent.

Remember: Bolted and welded reduction expressions are not the same rule.

Related concept and full method

The connected $125 mm$ leg contains the hole; the outstanding $75 mm$ leg does not. Split the shared $8 mm$ heel equally so no area is counted twice. As in the lecture, applyKe to the connected net leg and cap the unconnected leg at its gross area.

Connected-leg gross area:

a_{g 1} = (125 - \frac{8}{2}) \times 8 = 121 \times 8 = 968 {mm}^{2}

Outstanding leg:

a_{2} = (75 - \frac{8}{2}) \times 8 = 71 \times 8 = 568 {mm}^{2}

Angle gross area:

968 + 568 = 1,536 {mm}^{2}

Connected-leg net area:

a_{n 1} = (121 - 22) \times 8 = 99 \times 8 = 792 {mm}^{2}

Connected-leg effective area:

a_{e 1} = \min (1.1 \times 792, 968) = 871.2 {mm}^{2}

Outstanding-leg effective area:

a_{e 2} = \min (1.1 \times 568, 568) = 568 {mm}^{2}

Total effective area:

A_{e} = 871.2 + 568 = 1,439.2 {mm}^{2}

Reduction for the bolted double-angle connection:

0.25 a_{2} = 0.25 \times 568 = 142 {mm}^{2}

Resistance area per angle:

1,439.2 - 142 = 1,297.2 {mm}^{2}

P_{t, each} = \frac{355 \times 1,297.2}{1,000} = 460.506 kN > 175

P_{t, pair} = 2 \times 460.506 = 921.012 kN > 350

Passes.

The fracture path cuts one of the three longitudinally aligned holes. Deducting all three from the same transverse section is incorrect. Conversely, multiplying the outstanding $568$ by $1.1$ without its cap would invent extra gross area. Among the requested pair/group resistances, angle bearing $528 kN$ is the lowest.

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

Pu 350 kN; $175$ per angle. ThreeM20 in double shear: $551.25 kN$ . Bolt bearing with centralUCweb $12 mm$ : $720 kN$ . Angle bearing: $88 kN$ perbolt perangle, pair $528 kN$ . Angle effective area $1439.2 {mm}^{2}$ less $0.25 \times 568$ gives $1297.2$ ; pair tensile $921.012 kN$ . All four requested strength checks pass; supporting UC strength excluded. Full detailing still needs the $84 mm$ edge-distance provision reviewed.

Mistakes to avoid

Do not take the UC flange thickness when the bolt passes through its web.
Do not confuse two angles with six physical bolts.
Deduct one hole from the transverse angle net section.
Use the double-bolted coefficient $0.25$ , not the single-bolted $0.5$ .

Procedure for an unfamiliar variant

Read sectionA–A before counting interfaces.
Allocate the total force to bolts and angle legs.
Compare central and combined outer contact thicknesses.
Compute all angle-bearing bounds.
Build capped effective leg areas and apply the connection-specific reduction.

Independent self-check

Try it yourself. Invented variant: increasePu to $550 kN$ . Which of the four requested strength modes first fails?

Reveal answer and reasoning

Angle bearing is $528 kN < 550$ and fails. Bolt shear551.25 kN is a very narrow pass; bolt bearing720 and angle tension921.012 remain higher. The joint cannot be accepted just because the bolt shear still passes.

Animation labCount the bolt shear planes3 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.