---
title: "Weld size for bracket on ISMB 400"
subject: "Design of Steel and Timber Structure"
chapter: 4
key: "semester_6/design_of_steel_and_timber_structure:bracket-weld-ismb400"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/bracket-weld-ismb400
site: BCT Engineering
---

# Weld size for bracket on ISMB 400

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

- Asked in: 2077 Chaitra, 12 marks
- Video: [Problem 1 on Design of Welded connection | Design of Steel Structures](https://www.youtube.com/watch?v=zUgtcwk6VS8) — 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-weld-ismb400
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-4

## Question

A bracket plate is welded to the flanges of a column section ISMB 400 @ 61.5kg/m as shown in figure below. If the width of weld is 250mm, depth 300mm and eccentricity from the face of column is 100mm, determine the size of weld to support a factorized load of 200KN.
[Figure: bracket plate welded on three sides (width 250 mm, depth 300 mm) to the flange of column ISMB 400 @ 61.5 kg/m; 200 kN load acts 100 mm from the column face]

## Short answer

Factored load 200 kN at 100 mm eccentricity, weld group length 800 mm, Ixx=13.5e6 mm³. Shear 250 N/mm, bending 222.2 N/mm, resultant 334.5 N/mm. Throat required 1.77 mm, size 2.52 mm, but minimum size for 16 mm flange is 5 mm. Provide 5 mm fillet weld all round.

## Answer

**Assumptions:** the bracket plate is welded to the column flange by fillet welds on three sides: two horizontal welds of 250 mm (top and bottom) and one vertical weld of 300 mm. The factored load $P=200$ kN acts parallel to the flange at $e=100$ mm from the weld plane, so the weld carries **direct shear and bending** $M=Pe$. ISMB 400: $t_f=16$ mm. Shop weld, E250 ($f_u=410$).

#### Weld group (per mm of throat)
- $L=2\times250+300=800$ mm
- Neutral axis at mid-depth. $I_{xx}=\dfrac{300^3}{12}+2\times250\times150^2=13500000$ mm³
- $M=200\times10^3\times100=20\times10^6$ N·mm

#### Force per mm of weld at the extreme weld (y = 150 mm)
- Shear: $q_s=\dfrac{P}{L}=\dfrac{200\times10^3}{800}=250.0$ N/mm
- Bending: $q_b=\dfrac{My}{I_{xx}}=\dfrac{20\times10^6\times150}{13500000}=222.2$ N/mm
- Resultant: $q=\sqrt{q_s^2+q_b^2}=334.5$ N/mm

#### Size of weld
$$
f_{wd}=\frac{410}{\sqrt3\times1.25}=189.4\ \text{N/mm}^2,\quad t_t=\frac{q}{f_{wd}}=\frac{334.5}{189.4}=1.77\ \text{mm},\quad s=\frac{t_t}{0.7}=2.52\ \text{mm}
$$
The calculated size is small, so the **minimum size** of IS 800:2007 Table 21 governs: thicker part 16 mm (10 < t ≤ 20) gives 5 mm.

**Answer: provide 5 mm fillet weld all round the three sides.** Check: capacity of 5 mm weld $=662.8$ N/mm against demand $334.5$ N/mm (utilisation $0.50$).

## Check yourself

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

1. The 200 kN load acts parallel to the column flange at 100 mm from the weld plane. What does the weld group carry?
   A. Direct shear only
   B. Pure axial tension only
   C. Direct shear plus bending from M = P x e
   D. Pure torsion about the weld centroid only
   Answer: C. The load is eccentric to the weld plane, so it causes direct shear and a moment M = Pe that produces bending-type forces in the weld. (Key point: Type of loading on weld)

2. The three-sided weld has total length L = 2 x 250 + 300 = 800 mm. What is the direct shear per mm of weld?
   A. 400 N/mm
   B. 125 N/mm
   C. 250 N/mm
   D. 666.7 N/mm
   Answer: C. q_s = P/L = 200000/800 = 250 N/mm, the same along the whole weld. (Key point: Direct shear per mm)

3. How is the resultant force per mm at the extreme weld found from q_s = 250 and q_b = 222.2 N/mm?
   A. Add them directly: 472.2 N/mm
   B. Subtract them: 27.8 N/mm
   C. Take the larger one only
   D. Square root of the sum of squares: 334.5 N/mm
   Answer: D. Shear and bending forces act at right angles, so the resultant is sqrt(q_s^2 + q_b^2) = 334.5 N/mm. (Key point: Resultant force per mm)

4. The calculated weld size is only 2.52 mm, yet 5 mm is provided. Why?
   A. IS 800 minimum size for the thicker part (16 mm) is 5 mm and governs
   B. The weld must equal the flange thickness
   C. Bending always doubles the size
   D. Shop welds are always 5 mm
   Answer: A. When the calculated size is smaller than the code minimum (Table 21, 10 < t <= 20 mm gives 5 mm), the minimum size is provided. (Key point: Minimum size governs)

## Related in Design of Steel and Timber Structure

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