---|------|------| e2=40 o o o | | 300 mm

e2=40 o o o |------|------|------| 80 70 70 80

| Dimension   | Value  | Ref                |
| ----------- | ------ | ------------------ |
| Plate width | 300 mm |                    |
| Plate thick | 12 mm  |                    |
| e1 (end)    | 40 mm  | Min 1.2d0 = 26.4 ✓ |
| e2 (edge)   | 40 mm  | Min 1.2d0 = 26.4 ✓ |
| p1 (gauge)  | 70 mm  | Min 2.2d0 = 48.4 ✓ |
| p2 (pitch)  | 80 mm  | Min 2.4d0 = 52.8 ✓ |

### Section Properties

| Property          | Value     |
| ----------------- | --------- |
| Plate grade       | S355JR    |
| Plate thickness t | 12 mm     |
| fy (t ≤ 16 mm)    | 355 N/mm² |
| fu (3 < t ≤ 100)  | 470 N/mm² |
| Bolt grade        | 8.8       |
| Bolt diameter d   | 20 mm     |
| Hole diameter d0  | 22 mm     |
| Tensile area As   | 245 mm²   |
| fub               | 800 MPa   |
| fyb               | 640 MPa   |

### Partial Factors (UK NA to EN 1993-1-8)

| Factor | Value | Application                 |
| ------ | ----- | --------------------------- |
| γ_M0   | 1.00  | Plate yielding              |
| γ_M2   | 1.25  | Bolts, bearing, net section |

---

## Step 1 — Bolt Shear Resistance per Shear Plane (Table 3.4)

For M20 Grade 8.8 bolts in shear (threads included in the shear plane, which is the usual case for structural connections):

Fv,Rd = αv × fub × As / γ_M2

For Grade 8.8: αv = 0.6 (Table 3.4, shear through threaded portion)

Fv,Rd = 0.6 × 800 × 245 / 1.25 = 94,080 N = 94.1 kN

The connection has 6 bolts in single shear:

Total V_Rd = 6 × 94.1 = 564.6 kN

**Utilization:** N_Ed / total V_Rd = 500 / 564.6 = **0.89 ✓** (89%)

Bolt shear is the governing check for this connection, utilising 89% of capacity.

---

## Step 2 — Bolt Bearing Resistance (Table 3.4)

Bearing resistance of the plate per bolt:

Fb,Rd = k1 × αb × fu × d × t / γ_M2

### Edge Bolts (end bolts in the direction of load transfer)

k1 = min(2.8 × e2 / d0 - 1.7, 2.5) = min(2.8 × 40 / 22 - 1.7, 2.5) = min(3.39, 2.5) = 2.5

αb = min(e1 / (3 × d0), fub / fu, 1.0) = min(40 / (3 × 22), 800 / 470, 1.0) = min(0.606, 1.70, 1.0) = 0.606

Fb,Rd,edge = 2.5 × 0.606 × 470 × 20 × 12 / 1.25 = 136,634 N = 136.6 kN

### Inner Bolts

k1 = min(1.4 × p2 / d0 - 1.7, 2.5) = min(1.4 × 80 / 22 - 1.7, 2.5) = min(3.39, 2.5) = 2.5

αb = min(p1 / (3 × d0) - 1/4, fub / fu, 1.0) = min(70 / (3 × 22) - 0.25, 1.70, 1.0) = min(0.811, 1.70, 1.0) = 0.811

Fb,Rd,inner = 2.5 × 0.811 × 470 × 20 × 12 / 1.25 = 182,885 N = 182.9 kN

### Total Bearing Capacity

ΣFb,Rd = 2 × 136.6 + 4 × 182.9 = 273.2 + 731.6 = 1,004.8 kN

**Utilization:** 500 / 1,004.8 = **0.50 ✓** (50%)

Bearing is not critical. If edge and end distances were reduced to minimum values (1.2d0 = 26.4 mm), bearing would approach bolt shear capacity.

---

## Step 3 — Net Cross-Section Tension Resistance (Cl. 6.2.3(2))

The net area of the plate at the bolt holes:

Anet = (b - n × d0) × t = (300 - 2 × 22) × 12 = 256 × 12 = 3,072 mm²

**Note:** For staggered holes, the net section would account for alternate load paths per Cl. 6.2.3(3), but with a simple double-row arrangement, the straight line across all holes governs.

Nu,Rd = Anet × fu / γ_M2 = 3,072 × 470 / 1.25 = 1,155 × 10³ N = 1,155 kN

**Utilization:** 500 / 1,155 = **0.43 ✓** (43%)

---

## Step 4 — Gross Section Yielding (Cl. 6.2.3(1))

The gross cross-section must also be checked for yielding:

Npl,Rd = A × fy / γ_M0 = (300 × 12) × 355 / 1.00 = 3,600 × 355 = 1,278 × 10³ N = 1,278 kN

**Utilization:** 500 / 1,278 = **0.39 ✓** (39%)

---

## Step 5 — Block Tearing (Cl. 3.10.2)

Block tearing considers a segment of the plate tearing out around a group of bolts.

For the symmetric bolt group under concentric tension:

Ant = (e2 - d0 / 2) × t = (40 - 11) × 12 = 348 mm² (net tension area per side) Anv = (e1 + (n1 - 1) × p1 - (n1 - 0.5) × d0) × t × 2 sides = (40 + 2 × 70 - 2.5 × 22) × 12 × 2 = (40 + 140 - 55) × 24 = 125 × 24 = 3,000 mm²

Veff,1,Rd = (fu × Ant) / γ_M2 + (fy × Anv) / (√3 × γ_M0) = (470 × 348) / 1.25 + (355 × 3,000) / (1.732 × 1.00) = 130,848 + 614,780 = 745,628 N = 745.6 kN

**Utilization:** 500 / 745.6 = **0.67 ✓** (67%)

---

## Step 6 — Summary

| Check                    | Clause       | Utilisation | Status |
| ------------------------ | ------------ | ----------- | ------ |
| Bolt shear resistance    | Table 3.4    | 0.89 (89%)  | ✓      |
| Plate bearing resistance | Table 3.4    | 0.50 (50%)  | ✓      |
| Net section tension      | Cl. 6.2.3(2) | 0.43 (43%)  | ✓      |
| Gross section yielding   | Cl. 6.2.3(1) | 0.39 (39%)  | ✓      |
| Block tearing            | Cl. 3.10.2   | 0.67 (67%)  | ✓      |

**Conclusion:** 6 × M20 Grade 8.8 bolts in a 300×12 mm S355JR plate are adequate for a factored tension load of 500 kN, governed by bolt shear at 89% utilisation. The bolt shear check is typically the governing failure mode for bearing-type connections when edge distances and spacing meet the standard minimums. Consider M24 bolts or Grade 10.9 if higher capacity is required.

---

_This worked example is for educational purposes. All designs must be verified by a qualified engineer. Use the [EN 1993 bolted connection calculator](/tools/bolted-connections/ 'EN 1993 bolted connection free calculator') to check other bolt sizes or configurations._

## See Also

- [Bolted Connections Calculator](/tools/bolted-connections/)
- [Welded Connections Calculator](/tools/welded-connections/)
- [Bolt Torque Calculator](/tools/bolt-torque/)
- [Bolt Hole Sizes Reference](/reference/bolt-holes/)
- [Connection Design Workflow](/reference/connection-design-workflow/)