---
title: "Splice of 18 mm flats with M18 10.9 bolts"
subject: "Design of Steel and Timber Structure"
chapter: 5
key: "semester_6/design_of_steel_and_timber_structure:tension-splice-m18-hsfg-109"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/tension-splice-m18-hsfg-109
site: BCT Engineering
---

# Splice of 18 mm flats with M18 10.9 bolts

Design of Steel and Timber Structure · Chapter 5: Tension Members

- Asked in: 2070 Bhadra, 14 marks
- Video: [Problem 1 Design of Tension Members | Net Sectional Area of Plate in Chain Bolting](https://www.youtube.com/watch?v=vGm9kiE4btk) — Ganesh Shegar
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/tension-splice-m18-hsfg-109
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-5

## Question

Two 18 mm thick steel flats are spliced by two 8 mm thick plates with four M18 high strength bolts of property class 10.9. Determine the ultimate design load carrying capacity of the connection (i) if slip is permitted at the ultimate load and (ii) if slip is not permitted at ultimate load. Assume that one shear plane intercepts the threads of the bolts. $f_y$ = 250 MPa and $f_u$ = 410 MPa.
[Figure: flats 220 mm wide, 4 bolts in two rows (staggered), edge distances 55 mm, gauge 110 mm, 55 mm; longitudinal spacings 45 mm, 85 mm, 45 mm; cover plates 8 mm, main flats 18 mm]

## Short answer

If slip is permitted at ultimate load, bearing-type design gives 708.5 kN. If slip is not permitted, the slip-critical resistance governs at 412.9 kN.

## Answer

**Reading of the figure:** 4 bolts M18 (class 10.9, $f_{ub}=1000$ N/mm²) on each side of the splice, in two rows of two (gauge 110 mm, edge distances 55 mm, pitch 85 mm, end 45 mm); flats 220×18 mm; two cover plates 8 mm (total 16 mm); double shear; $d_0=20$ mm; $f_y=250$, $f_u=410$.

#### Plate checks
- Flat (18 mm): $T_{dg}=\dfrac{220\times18\times250}{1.10}=900.0$ kN; $T_{dn}=\dfrac{0.9(220-2\times20)\times18\times410}{1.25}=956.4$ kN
- Covers (2 × 8 = 16 mm): $T_{dg}=800.0$ kN; $T_{dn}=850.2$ kN

#### (i) Slip permitted at ultimate load (bearing-type)
- Shear (double shear, one plane in thread): $V_{dsb}=\dfrac{1000}{\sqrt3\times1.25}(192+254)\times10^{-3}=206.0$ kN
- Bearing on cover plates (16 mm, thinner than the flats): $k_b=\min(\tfrac{45}{60},\tfrac{85}{60}-0.25,2.44,1)=0.750$; $V_{dpb}=\dfrac{2.5k_bdtf_u}{1.25}=177.1$ kN
- Bolt value $177.1$ kN; four bolts $=708.5$ kN
- **Design load $=\min(708.5,\ 900.0,\ 956.4,\ 800.0)=708.5$ kN**

#### (ii) Slip not permitted at ultimate load (slip-critical)
- $F_o=0.7\times1000\times192\times10^{-3}=134.4$ kN
- $V_{nsf}=\dfrac{\mu_fn_eK_hF_o}{\gamma_{mf}}=\dfrac{0.48\times2\times1.0\times134.4}{1.25}=103.2$ kN per bolt ($\mu_f=0.48$, $\gamma_{mf}=1.25$)
- Four bolts: $412.9$ kN
- **Design load $=412.9$ kN** (slip resistance governs).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is bearing checked on the 16 mm cover plates?
   A. Covers are stronger
   B. Flats are ignored
   C. Total cover thickness 16 mm is less than 18 mm main flat so it governs bearing
   D. Covers have more bolts
   Answer: C. In double shear, bearing is checked on the thinner of the main plate and combined covers. (Key point: Thinner plate in bearing)

2. Bolt value 177.1 kN (bearing) with four bolts gives
   A. 708.5 kN
   B. 824 kN
   C. 177.1 kN
   D. 412.9 kN
   Answer: A. 4 x 177.1 = 708.5 kN, which is less than plate capacities 800 kN. (Key point: Bearing-type capacity)

3. In case (ii) the bolt value is Vnsf. What is Fo?
   A. 0.7 fub Anb = 134.4 kN
   B. fub Anb = 192 kN
   C. 0.7 fub Asb = 177.8 kN
   D. 0.9 fub Anb
   Answer: A. Proof load Fo = 0.7 x 1000 x 192 N = 134.4 kN. (Key point: Slip-critical formula)

4. Why is the capacity in (ii) (412.9 kN) much lower than in (i)?
   A. Because bolts are weaker
   B. Because plates are thinner
   C. Because there is single shear
   D. It is limited by friction (slip resistance), not shear or bearing
   Answer: D. No slip is permitted at ultimate load, so friction resistance 103.2 kN per bolt governs. (Key point: Result of slip-critical)

## Related in Design of Steel and Timber Structure

- [Ultimate strength of ISA 100x100x10, M20 bolts](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Aangle-ultimate-strength-100x100x10) — Design of Steel and Timber Structure, chapter 5
- [Tension splice for 250x16 and 250x8 flats](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Atension-splice-250x16-flat) — Design of Steel and Timber Structure, chapter 5
- [Design strength of ISA 100x75x6, 6 bolts](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Aangle-design-strength-100x75x6) — Design of Steel and Timber Structure, chapter 5, asked 2×
- [Double cover butt joint, M16 class 10.9](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Adouble-cover-butt-joint-hsfg-300kn) — Design of Steel and Timber Structure, chapter 4
- [Ultimate load of 2 ISA 100x100x8, M16 5.6](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Atwo-angles-ultimate-load-m16) — Design of Steel and Timber Structure, chapter 5
- [Splice connection for ISHB 250, 900 kN](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Acolumn-splice-ishb250-900kn) — Design of Steel and Timber Structure, chapter 6

## Related in other subjects

- [Ferrite and pearlite in high strength steel](https://www.bctengineer.com/question.md?key=semester_3%2Fcivil_engineering_materials%3Aferrite-pearlite-high-strength-steel) — Civil Engineering Materials, chapter 8
- [Ultimate moment of T-beam (2250 mm flange)](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Aultimate-moment-of-t-beam-2250) — Design of RCC Structure, chapter 4
- [Fusing current and current carrying capacity](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Afusing-current-carrying-capacity) — Switchgear and Protection, chapter 3
- [Current carrying capacity of cable](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Acable-current-capacity-factors) — Industrial Power Distribution and Illumination, chapter 3