---
title: "Shop welded bracket, 5 mm fillet, safe load"
subject: "Design of Steel and Timber Structure"
chapter: 4
key: "semester_6/design_of_steel_and_timber_structure:bracket-weld-safe-load-5mm"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/bracket-weld-safe-load-5mm
site: BCT Engineering
---

# Shop welded bracket, 5 mm fillet, safe load

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

- Asked in: 2079 Chaitra, 10 marks
- Video: [#Design of steel structures#Eccentric model problem#Welding#Determine the load acting on the bracket](https://www.youtube.com/watch?v=jR2ZPEqpqLg) — CIVIL ENGINEERING COURSE Theory & Practical
- 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-safe-load-5mm
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-4

## Question

A bracket plate is shop welded to the flange of a column as shown in figure. If the size of the fillet weld is 5 mm, find the safe load P.
[Figure: bracket plate welded to column flange with a C-shaped weld, weld height 400 mm and flange-to-weld width 150 mm; load P acts at 120 mm from the weld]

## Short answer

For a 5 mm fillet weld (throat 3.5 mm) with total length 700 mm, weld strength f_wd = 189.37 N/mm². Direct shear and bending give resultant force per mm 0.001989P. Equating to strength per mm 662.80 N/mm gives factored load P_u = 333.2 kN; safe working load P ≈ 222.1 kN.

## Answer

**Assumptions:** the bracket is shop-welded to the column flange by a C-shaped weld lying on the face of the flange: two horizontal welds of 150 mm and one vertical weld of 400 mm. The load $P$ acts parallel to the flange at $e=120$ mm out from the weld plane, so the weld group carries **direct shear $P$ plus bending $M=Pe$**. Steel E250 ($f_u=410$ N/mm²), fillet size $s=5$ mm, throat $t_t=0.7\times5=3.5$ mm.

#### Weld strength (IS 800:2007 cl. 10.5.7)
$$
f_{wd}=\frac{f_u}{\sqrt3\gamma_{mw}}=\frac{410}{\sqrt3\times1.25}=189.37\ \text{N/mm}^2,\qquad q_{allow}=f_{wd}t_t=662.80\ \text{N/mm}
$$

#### Weld group properties (per mm of throat)
- Total length $L=2\times150+400=700$ mm
- Neutral axis at mid-depth (symmetrical about the horizontal axis).
- $I_{xx}=\dfrac{400^3}{12}+2\times150\times200^2=17333333$ mm³ (the weld thickness itself is neglected)

#### Forces per unit factored load $P$ at the extreme weld (y = 200 mm)
- Shear: $q_s=\dfrac{P}{L}=0.001429P$ N/mm
- Bending: $q_b=\dfrac{M y}{I_{xx}}=\dfrac{120P\times200}{17333333}=0.001385P$ N/mm
- Resultant: $q=\sqrt{q_s^2+q_b^2}=0.001989P$ N/mm

#### Safe load
$$
0.001989P\le662.80\Rightarrow P_u=333.2\ \text{kN}
$$
**Answer: design (factored) load $P_u\approx333.2$ kN, i.e. working load $P\approx222.1$ kN.** (If the load instead acted in the plane of the weld group, the weld would be checked for direct shear plus torsion rather than bending.)

## Check yourself

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

1. What is the throat thickness of a 5 mm fillet weld?
   A. 5 mm
   B. 3.5 mm
   C. 7 mm
   D. 2.5 mm
   Answer: B. t_t = 0.7 s = 0.7 x 5 = 3.5 mm. (Key point: Throat thickness)

2. How is the design strength of a shop fillet weld calculated?
   A. fu / (sqrt3 gamma_mw) = 410/(sqrt3 x 1.25) = 189.4 N/mm2
   B. fy / 1.10
   C. 0.6 fy
   D. fu / 1.25 only
   Answer: A. f_wd = fu/(sqrt3 gamma_mw) = 189.37 N/mm2, with gamma_mw = 1.25 for shop weld. (Key point: Weld design stress)

3. Which actions does the weld group carry?
   A. Torsion only
   B. Direct shear and bending due to the eccentric load
   C. Axial tension only
   D. Prying
   Answer: B. The load is parallel to the flange at e = 120 mm, giving shear P and bending M = P e. (Key point: Stresses on weld)

4. Where is the weld stress maximum and what is the factored load?
   A. At the neutral axis, P = 100 kN
   B. At the extreme weld y = 200 mm, Pu = 333.2 kN
   C. At the centre, P = 662.8 kN
   D. At the extreme weld, P = 222.1 kN factored
   Answer: B. Bending stress is greatest at y = 200 mm; 0.001989 P <= 662.8 gives Pu = 333.2 kN (working load 222.1 kN). (Key point: Critical point and result)

## Related in Design of Steel and Timber Structure

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