---
title: "Double cover butt joint for boiler plates"
subject: "Design of Steel and Timber Structure"
chapter: 4
key: "semester_6/design_of_steel_and_timber_structure:boiler-plate-butt-joint-efficiency"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/boiler-plate-butt-joint-efficiency
site: BCT Engineering
---

# Double cover butt joint for boiler plates

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

- Asked in: 2073 Bhadra, 10 marks
- Video: [Design of butt joint](https://www.youtube.com/watch?v=ktuEzGU1aKU) — Simpleciviltamil
- 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/boiler-plate-butt-joint-efficiency
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-4

## Question

Design a single bolted double cover butt joint to connect boiler plates of thickness 12mm for maximum efficiency. Use M16 bolts of grade 4.6. Boiler plates are of Fe410. Find the efficiency of the joint.

## Short answer

Use M16 4.6 bolts in a single row at 40 mm pitch, end distance 30 mm, with 8 mm cover plates on both sides; the joint efficiency is about 61%.

## Answer

**Joint type:** single-bolted (one row) double cover butt joint; bolts in double shear. Plates 12 mm, Fe410; M16 bolts grade 4.6, $d_0=18$ mm. Covers: two plates of 8 mm ($2\times8=16\ge12$ mm).

#### Strength of one bolt (IS 800:2007 cl. 10.3)
- Shear (double shear, one plane in thread): $V_{dsb}=\dfrac{400}{\sqrt3\times1.25}(157+201)\times10^{-3}=66.14$ kN
- Bearing on the 12 mm main plate: with a single row there is no pitch in the direction of load, so only the end distance $e=30$ mm ($\ge1.5d_0=27$ mm) controls: $k_b=\min(30/54,\ 0.976,\ 1)=0.556$; $V_{dpb}=\dfrac{2.5k_bdtf_u}{\gamma_{mb}}=87.5$ kN
- Bolt value $=66.14$ kN

#### Pitch for maximum efficiency
Efficiency is highest when the net strength of the plate between bolt holes equals the bolt value, so that no material is wasted:
$$
\frac{0.9(p-d_0)tf_u}{\gamma_{m1}}=V_b\Rightarrow p=18+\frac{66.14\times10^3}{0.9\times12\times410/1.25}=36.7\ \text{mm}
$$
This is below the minimum pitch $2.5d=40$ mm, so **take $p=40$ mm**. Check:
- Net strength of plate in a pitch strip: $T_{dn}=\dfrac{0.9(40-18)\times12\times410}{1.25}=77.9$ kN $>V_b$
- Gross strength of plate in a pitch strip: $T_{dg}=\dfrac{40\times12\times250}{1.10}=109.1$ kN

#### Efficiency
$$
\eta=\frac{\text{lesser of bolt value and }T_{dn}}{T_{dg}}=\frac{66.14}{109.1}\times100=60.6\%
$$
**Answer: M16 bolts at 40 mm pitch in one row, end distance 30 mm, with 8 mm cover plates on both sides; efficiency $\approx61\%$.** (A single row cannot do better than this; efficiency can be raised by using more rows of bolts in a zig-zag pattern.)

## Check yourself

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

1. In a double cover butt joint, why does the bolt shear use (157 + 201)?
   A. Bolts are in double shear: one plane through threads, one through the shank
   B. Two bolts act together
   C. It is a bearing area
   D. It is the gross plus net plate area
   Answer: A. Two shear planes act; one cuts the threaded area (157 mm^2) and one the shank (201 mm^2). (Key point: Shear planes)

2. How is the ideal pitch found?
   A. Pitch = 2.5 d
   B. Pitch = 16 t
   C. Equate plate net strength between holes to the bolt value
   D. Equate gross strength to the bolt value
   Answer: C. Efficiency is highest when net plate strength per pitch equals the bolt value, no material wasted; it gave 36.7 mm. (Key point: Pitch for maximum efficiency)

3. The calculated pitch is 36.7 mm. What pitch is adopted?
   A. 36.7 mm
   B. 30 mm
   C. 50 mm
   D. 40 mm, the minimum 2.5d
   Answer: D. Pitch cannot be below 2.5d = 40 mm, so 40 mm is used. (Key point: Adopted pitch)

4. How is efficiency obtained?
   A. Lesser of bolt value and net strength divided by gross strength of plate per pitch = 60.6%
   B. Bolt value divided by net strength
   C. Net divided by bolt value
   D. Cover plate area over main plate area
   Answer: A. η = 66.14/109.1 = 60.6%. (Key point: Efficiency)

## Related in Design of Steel and Timber Structure

- [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
- [Strength of double cover butt joint, M20](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Adouble-cover-butt-joint-strength) — Design of Steel and Timber Structure, chapter 4
- [Triple staggered lap joint, M16 8.6, efficiency](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Atriple-staggered-lap-joint) — Design of Steel and Timber Structure, chapter 4