---
title: "Double cover butt joint, M16 class 10.9"
subject: "Design of Steel and Timber Structure"
chapter: 4
key: "semester_6/design_of_steel_and_timber_structure:double-cover-butt-joint-hsfg-300kn"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/double-cover-butt-joint-hsfg-300kn
site: BCT Engineering
---

# Double cover butt joint, M16 class 10.9

Design of Steel and Timber Structure · Chapter 4: Connections in Steel Structures

- Asked in: 2069 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/double-cover-butt-joint-hsfg-300kn
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-4

## Question

Design a double cover butt joint to transmit a working load of 300KN to connect two flats 100mm wide and 20mm thick using M16 high strength bolts of property class 10.9 if slip is permitted at design load. The cover plates are 12 mm thick. Assume that one shear plane intercepts the threads of the bolts.

## Short answer

Provide 4 M16 HSFG bolts of class 10.9 on each side in a single line at 50 mm pitch, with 12 mm cover plates on both faces.

## Answer

**Data:** working load 300 kN → factored $P_u=1.5\times300=450$ kN. Flats 100×20 mm, covers 12 mm each, M16 HSFG bolts of class 10.9 ($f_{ub}=1000$ N/mm²), $d_0=18$ mm, E250. "Slip permitted at design load" means the bolts are designed as **bearing-type** (for shear and bearing, IS 800:2007 cl. 10.3/10.4). Double shear, with one plane through the threads.

#### Bolt value
- Shear: $V_{dsb}=\dfrac{f_{ub}}{\sqrt3\gamma_{mb}}(A_{nb}+A_{sb})=\dfrac{1000}{\sqrt3\times1.25}(157+201)\times10^{-3}=165.35$ kN
- Bearing on the 20 mm flat ($e=30$, $p=50$): $k_b=\min(0.556,\ 0.676,\ 2.44,\ 1)=0.556$
$$
V_{dpb}=\frac{2.5k_bdtf_u}{\gamma_{mb}}=\frac{2.5\times0.556\times16\times20\times410}{1.25}\times10^{-3}=145.8\ \text{kN}
$$
- Bolt value $=145.8$ kN (bearing governs)

#### Number of bolts
$n=\dfrac{450}{145.8}=3.08\Rightarrow$ **provide 4 bolts on each side of the joint**, in a single line along the centre of the flat at 50 mm pitch with 30 mm end distance, so the flat is weakened by only one hole.

#### Checks
- Flat gross yielding: $T_{dg}=\dfrac{100\times20\times250}{1.10}=454.5$ kN $\ge450$ ✓
- Flat net rupture (1 hole): $T_{dn}=\dfrac{0.9(100-18)\times20\times410}{1.25}=484.1$ kN $\ge450$ ✓
- Cover plates $2\times12=24>20$ mm: $T_{dg}=545.5$ kN ✓
- Bolts: $4\times145.8=583.1$ kN $\ge450$ ✓
- (For information, the slip resistance per bolt at service load is $V_{dsf}=\mu_f n_eK_hF_o/\gamma_{mf}=95.9$ kN with $\mu_f=0.48$, $n_e=2$, $K_h=1.0$, $F_o=0.7f_{ub}A_{nb}=109.9$ kN and $\gamma_{mf}=1.10$; the total of 384 kN exceeds 300 kN, so the joint does not slip under the working load either.)

**Answer: 4 M16 (10.9) HSFG bolts on each side in a single line at 50 mm pitch, with 12 mm cover plates on both faces.**

## Check yourself

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

1. What does 'slip permitted at design load' imply for design?
   A. Bolts are designed as bearing-type HSFG, checked in shear and bearing
   B. Bolts are designed against slip only
   C. Bolts carry tension only
   D. Welds replace the bolts
   Answer: A. If slip is allowed at the design load, the bolts act in shear and bearing (IS 800 cl. 10.4). (Key point: Meaning of slip permitted)

2. Shear capacity is 165.35 kN and bearing on the 20 mm flat is 145.8 kN. What is the bolt value?
   A. 165.35 kN
   B. 311.2 kN
   C. 145.8 kN, bearing governs
   D. 155.6 kN
   Answer: C. The smaller governs; here bearing at 145.8 kN. (Key point: Governing bolt value)

3. The factored load is 1.5 x 300 = 450 kN. How many bolts per side?
   A. 2
   B. 3 exactly
   C. 6
   D. 4, since 450/145.8 = 3.08
   Answer: D. n = 450/145.8 = 3.08, rounded up to 4 bolts. (Key point: Number of bolts)

4. Why is a single line of bolts used?
   A. Code requires one line
   B. Bearing is higher
   C. So the flat is weakened by only one hole
   D. To reduce pitch
   Answer: C. One hole in the cross-section keeps net rupture strength (484.1 kN) above the load. (Key point: Single line)

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