---
title: "Tension splice for 250x16 and 250x8 flats"
subject: "Design of Steel and Timber Structure"
chapter: 5
key: "semester_6/design_of_steel_and_timber_structure:tension-splice-250x16-flat"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/tension-splice-250x16-flat
site: BCT Engineering
---

# Tension splice for 250x16 and 250x8 flats

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

- Asked in: 2076 Bhadra, 10 marks
- 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-250x16-flat
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-5

## Question

Design and detail a tension splice to connect 250mm×16mm flat with 250mm×8mm flat using two cover plate of thickness 6mm each to carry a tension of 400KN. Use M20 high strength bolts of the property class 8.8 if,
(i) No slip is permitted at the ultimate load
(ii) Slip is permitted at the ultimate load.
Take $f_y$ = 250 MPa and $f_u$ = 410 MPa.

## Short answer

For no slip at ultimate load, use 4 M20 grade 8.8 bolts per side (8 total) with 8 mm packing on thinner flat. For slip permitted, use 6 M20 grade 8.8 bolts per side (12 total) with same packing. Bolt values: 105.4 kN (slip-critical) and 79.5 kN (bearing) per bolt.

## Answer

**Data:** factored tension 400 kN; main flats 250×16 and 250×8; two cover plates 6 mm each ($2\times6=12>8$ mm); M20 HSFG bolts class 8.8 ($f_{ub}=800$ N/mm²); $d_0=22$ mm; $f_y=250$, $f_u=410$ N/mm². Bolts are in **double shear** ($n_e=2$). The thinner flat (8 mm) governs the plate checks. The flats differ in thickness by $16-8=8$ mm, so an **8 mm packing plate** is placed on one face of the thinner flat to let the cover plates bear evenly.

#### Plate checks (common)
- Gross section of 8 mm flat: $T_{dg}=\dfrac{250\times8\times250}{1.10}=454.5$ kN $>400$ ✓
- Net section (2 bolts in a line): $T_{dn}=\dfrac{0.9(250-2\times22)\times8\times410}{1.25}=486.5$ kN $>400$ ✓
- Covers: $2\times6\times250$: $T_{dg}=681.8$ kN ✓

#### (i) No slip at ultimate load (slip-critical, $\gamma_{mf}=1.25$)
- $F_o=0.7f_{ub}A_{nb}=0.7\times800\times245\times10^{-3}=137.2$ kN
- $V_{nsf}=\dfrac{\mu_fn_eK_hF_o}{\gamma_{mf}}=\dfrac{0.48\times2\times1.0\times137.2}{1.25}=105.4$ kN per bolt ($\mu_f=0.48$, sand-blasted)
- $n=\dfrac{400}{105.4}=3.8$ → **4 bolts** on each side (2 rows of 2), capacity $421.5$ kN.

#### (ii) Slip permitted at ultimate load (bearing type)
- Shear (double shear, one plane in thread): $V_{dsb}=\dfrac{800}{\sqrt3\times1.25}(245+314)\times10^{-3}=206.6$ kN; packing thicker than 6 mm: $\beta_{pkg}=1-0.0125\times8=0.90$ → $185.9$ kN
- Bearing on 8 mm flat ($e=40$, $p=60$): $k_b=0.606$; $V_{dpb}=\dfrac{2.5k_bdtf_u}{1.25}=79.5$ kN → bolt value $79.5$ kN
- $n=\dfrac{400}{79.5}=5.03$ → **6 bolts** on each side (2 rows of 3), capacity $477.1$ kN.
- On the 16 mm flat side the bolts bear on the two cover plates (12 mm): $V_{dpb}=119.3$ kN, so $400/119.3=3.35$ → 4 bolts would be enough there; 6 bolts are used on both sides for symmetry.

#### Detailing
Pitch 60 mm ($\ge2.5d$), gauge 100 mm (edge distance 75 mm), end distance 40 mm; cover plates 250 mm wide, about 290 mm long in case (i) and 410 mm long in case (ii); 8 mm packing under one cover plate over the length of the thinner flat. Surface preparation (clean, sand-blasted faying surfaces) is required in case (i).

**Answer: (i) 4 bolts M20 (8.8) on each side; (ii) 6 bolts M20 (8.8) on each side; 8 mm packing on the thinner flat.**

## Check yourself

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

1. Which plate governs the plate checks and why?
   A. The 16 mm flat, because it is thicker
   B. The cover plates, because there are two
   C. None, plates are not checked
   D. The 8 mm flat, because it is the thinner main plate
   Answer: D. The splice strength is limited by the thinner 8 mm flat; the covers (12 mm total) are thicker than 8 mm. (Key point: Plate governs)

2. In case (i), no slip at ultimate load, which formula gives the bolt value?
   A. Vdsb = fub/(sqrt3 gamma_mb) x Anb
   B. Vdpb = 2.5 kb d t fu / gamma_mb
   C. Vdb = fy Ag / gamma_m0
   D. Vnsf = mu_f ne Kh Fo / gamma_mf
   Answer: D. Slip-critical (friction-grip) joints use slip resistance Vnsf with the pretension Fo = 0.7 fub Anb. (Key point: Slip-critical design)

3. With 105.4 kN per bolt, how many bolts are needed each side for 400 kN?
   A. 4
   B. 3
   C. 5
   D. 6
   Answer: A. 400/105.4 = 3.8, so 4 bolts (2 rows of 2) with capacity 421.5 kN. (Key point: Bolts for slip-critical case)

4. In case (ii), why are more bolts (6) required than in case (i)?
   A. Shear capacity is only 79.5 kN
   B. Bearing on the 8 mm flat gives a lower bolt value (79.5 kN) than shear (206.6 kN)
   C. Bolts are in single shear
   D. Bearing-type bolts have no pretension so need extra edge distance
   Answer: B. Bearing governs at 79.5 kN so n = 400/79.5 = 5.03, i.e. 6 bolts. (Key point: Bearing-type case)

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