---
title: "Strength of double cover butt joint, M20"
subject: "Design of Steel and Timber Structure"
chapter: 4
key: "semester_6/design_of_steel_and_timber_structure:double-cover-butt-joint-strength"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/double-cover-butt-joint-strength
site: BCT Engineering
---

# Strength of double cover butt joint, M20

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

- Asked in: 2078 Chaitra, 8 marks
- Video: [Calculation of strength of bolt in Double cover butt joint|| Design of steel structures|#Tutorial 03](https://www.youtube.com/watch?v=grZEIOkqTQo) — Infra Knowledge
- 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-strength
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-4

## Question

Two plates, 10 mm and 18 mm thick are connected by a double cover butt joint using 6 mm plates as shown in figure below. Find the strength of the joint. Given M20 bolts of grade 4.6 and E250 plates are used.
[Figure: double cover butt joint with 6 mm cover plates; two bolt groups on each side, each group 2 columns by 3 rows; bolt pitch 60 mm and 60 mm with 40 mm end distances vertically; horizontal spacing 40, 60, 40 mm; all dimensions in mm]

## Short answer

M20 grade 4.6 bolts in double shear with packing give bolt value 92.95 kN; 6 bolts yield 557.7 kN. Plate gross and net section strengths: net section rupture of the 10 mm main plate governs at 395.6 kN. Joint strength = 395.6 kN.

## Answer

**Assumptions from the figure:** main plates 10 mm and 18 mm, two cover plates 6 mm each (total 12 mm), plate width $=40+60+60+40=200$ mm; 6 bolts M20 (grade 4.6) on each side of the joint (3 across the width at 60 mm gauge, 2 along the load at 60 mm pitch), end distance 40 mm, $d_0=22$ mm. Steel E250: $f_y=250$, $f_u=410$ N/mm². The joint is in double shear; one shear plane is taken through the threads. An 8 mm packing is needed on the 10 mm plate side to make up the difference in thickness.

#### Strength of one bolt (IS 800:2007 cl. 10.3)
- Shear (double shear, one plane in thread): $V_{dsb}=\dfrac{f_{ub}}{\sqrt3\gamma_{mb}}(A_{nb}+A_{sb})=\dfrac{400}{\sqrt3\times1.25}(245+314)\times10^{-3}=103.28$ kN
- Packing of 8 mm (> 6 mm) reduces this: $\beta_{pkg}=1-0.0125\times8=0.90$, so $V_{dsb}=0.90\times103.28=92.95$ kN on the 10 mm plate side
- Bearing on the thinner main plate (10 mm < 12 mm of covers): $k_b=\min\left(\dfrac{40}{3\times22},\dfrac{60}{3\times22}-0.25,\dfrac{400}{410},1\right)=0.606$
$$
V_{dpb}=\frac{2.5k_bdtf_u}{\gamma_{mb}}=\frac{2.5\times0.606\times20\times10\times410}{1.25}\times10^{-3}=99.39\ \text{kN}
$$
- Bolt value $=\min(92.95,99.39)=92.95$ kN (shear with packing governs)

#### Strength of the joint
1. Bolts: $6\times92.95=557.7$ kN
2. Main plate 10 mm, gross section: $T_{dg}=\dfrac{200\times10\times250}{1.10}=454.5$ kN
3. Main plate net section (3 holes): $T_{dn}=\dfrac{0.9(200-3\times22)\times10\times410}{1.25}=395.6$ kN
4. Cover plates (2 × 6 = 12 mm): $T_{dg}=545.5$ kN, $T_{dn}=474.7$ kN
5. Block shear of the 10 mm plate (shear along the two outer bolt lines, $A_{vg}=2000$, $A_{vn}=1340$ mm²):
   - central block torn out ($A_{tg}=1200$, $A_{tn}=760$ mm²): $T_{db}=\min(486.8,\ 501.1)=486.8$ kN
   - two edge strips torn out ($A_{tg}=800$, $A_{tn}=580$ mm²): $T_{db}=\min(433.6,\ 410.2)=410.2$ kN

**Answer: strength of the joint $=\min(\ldots)=395.6$ kN**, governed by net-section rupture of the 10 mm plate (the 18 mm plate and the covers are stronger).

## Check yourself

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

1. How is the bolt shear strength calculated in this double cover joint?
   A. Single shear using Anb only
   B. Double shear using Anb + Asb (one plane in thread, one in shank)
   C. Using Asb twice
   D. Using bearing area
   Answer: B. With one shear plane in thread and one in shank, strength uses (Anb + Asb) = 103.28 kN. (Key point: Double shear strength)

2. Why is bearing checked on the 10 mm plate?
   A. It is the thinnest of 10 mm and the 12 mm total cover thickness
   B. It is the thickest
   C. It has the largest edge distance
   D. It is the cover plate
   Answer: A. Bearing is governed by the smaller of main plate (10 mm) and covers (12 mm). (Key point: Thinnest plate bearing)

3. What is the bolt value?
   A. 103.28 kN, shear
   B. 99.39 kN, bearing
   C. 99.39 kN, shear
   D. 596.4 kN
   Answer: B. min(103.28, 99.39) = 99.39 kN, bearing governs. (Key point: Governing bolt value)

4. What governs the strength of the joint?
   A. Bolt group, 596.4 kN
   B. Gross yielding of the 10 mm plate, 454.5 kN
   C. Net-section rupture of the 10 mm plate, 395.6 kN
   D. Block shear, 501.1 kN
   Answer: C. The minimum of all checks is net-section rupture of the 10 mm plate, 395.6 kN. (Key point: Joint strength)

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