---
title: "Eccentric bolted bracket, ISMB 300, 220 kN"
subject: "Design of Steel and Timber Structure"
chapter: 4
key: "semester_6/design_of_steel_and_timber_structure:bracket-bolted-ismb300-220kn"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/bracket-bolted-ismb300-220kn
site: BCT Engineering
---

# Eccentric bolted bracket, ISMB 300, 220 kN

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

- Asked in: 2076 Bhadra, 14 marks
- Video: [Problem 1 on Eccentric Bolted Connections | Eccentric Loading | Design of Steel Structures](https://www.youtube.com/watch?v=cNVysbieEm8) — 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/bracket-bolted-ismb300-220kn
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-4

## Question

Design an eccentrically loaded bolted connection to connect bracket plate of thickness 12 mm with flange of column ISMB 300. Bracket plate is subjected to design load 220 kN at 90 mm from inner face of column.

## Short answer

Factored load 220 kN at 90 mm eccentricity. M20 grade 4.6 bolts in two vertical lines of five, gauge 70 mm, pitch 60 mm, end distance 40 mm. Bolt shear capacity 45.26 kN, maximum bolt force 41.34 kN. Joint safe.

## Answer

**Assumptions:** bracket plate (12 mm) is lapped on the flange of ISMB 300 ($b_f=140$ mm, $t_f=12.4$ mm); the 220 kN factored load acts in the plane of the bolts at $e=90$ mm from the centroid of the bolt group. M20 bolts of grade 4.6 (shop bolting), $d_0=22$ mm, in two vertical lines.

#### Bolt value (IS 800:2007 cl. 10.3)
- Shear (single, thread in plane): $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26$ kN
- Bearing (12 mm plate, $e=40$, $p=60$): $k_b=\min(0.606,0.659,0.976,1)=0.606$; $V_{dpb}=119.3$ kN
- Bolt value $=45.26$ kN

#### Trial layout
Two vertical lines of 5 bolts: gauge $g=70$ mm (edge distance $(140-70)/2=35\ge1.5d_0=33$ mm), pitch $p=60$ mm (≥ $2.5d=50$), end distance 40 mm. Total 10 bolts.

```text
   e = 90      o   o   y=+120
   <------     o   o   y=+60
 P  |          o   o   y=0
 220 kN        o   o   y=-60
               o   o   y=-120
```

#### Forces
- $\Sigma(x^2+y^2)=10(35^2)+2\times2(60^2+120^2)=84250$ mm²
- Moment $T=220\times90=19800$ kN·mm
- Direct shear per bolt $=220/10=22.0$ kN (vertical)
- Critical corner bolt ($x=35,y=120$): torsional components $F_x=\dfrac{Ty}{\Sigma r^2}=28.20$ kN, $F_y=\dfrac{Tx}{\Sigma r^2}=8.23$ kN
- Resultant: $R=\sqrt{28.20^2+(22.0+8.23)^2}=41.34$ kN

$R=41.34\ \text{kN}<V_{db}=45.26$ kN. Safe.

**Answer: 10 bolts M20 (4.6), 2 lines of 5, pitch 60 mm, gauge 70 mm, end distance 40 mm.**

## Check yourself

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

1. For M20 grade 4.6 bolts, shear capacity is 45.26 kN and bearing capacity is 119.3 kN. What is the bolt value?
   A. 119.3 kN
   B. 164.6 kN
   C. 45.26 kN, the smaller
   D. 82.3 kN, the average
   Answer: C. Bolt value is the lesser of shear and bearing capacity, here the shear value 45.26 kN. (Key point: Bolt value)

2. Which bolt in the group is critical under an eccentric load in the plane of the bolts?
   A. The bolt nearest the centroid
   B. The extreme corner bolt farthest from the centroid
   C. Any bolt on the neutral axis
   D. The middle bolt of each line
   Answer: B. Torsional force on a bolt is proportional to its distance from the centroid, so the farthest corner bolt (x = 35, y = 120) is critical. (Key point: Critical bolt)

3. At the corner bolt, direct shear is 22.0 kN vertical and the torsional components are F_x = 28.20 kN and F_y = 8.23 kN. How is the resultant formed?
   A. sqrt(28.20^2 + (22.0 + 8.23)^2) = 41.34 kN
   B. 28.20 + 22.0 + 8.23 = 58.4 kN
   C. sqrt(28.20^2 + 8.23^2) = 29.4 kN
   D. sqrt(28.20^2 + (22.0 - 8.23)^2) = 31.4 kN
   Answer: A. Direct shear and the vertical torsional component act in the same direction at this bolt, so they add before combining with F_x. (Key point: Combining forces)

4. With a 70 mm gauge on a 140 mm flange and 22 mm holes, why is the 35 mm edge distance acceptable?
   A. It is at least 1.5 d0 = 33 mm
   B. It equals the pitch
   C. It is at least 2.5 d = 50 mm
   D. Edge distance has no limit
   Answer: A. The minimum edge distance is 1.5 times the hole diameter, 1.5 x 22 = 33 mm, and 35 mm satisfies it. (Key point: Edge distance)

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