---
title: "Weld length for ISA 90x90x12, 100 kN"
subject: "Design of Steel and Timber Structure"
chapter: 5
key: "semester_6/design_of_steel_and_timber_structure:tension-fillet-weld-isa90x90x12"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/tension-fillet-weld-isa90x90x12
site: BCT Engineering
---

# Weld length for ISA 90x90x12, 100 kN

Design of Steel and Timber Structure · Chapter 5: Tension Members

- Asked in: 2071 Bhadra, 10 marks
- 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/tension-fillet-weld-isa90x90x12
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-5

## Question

In truss ISA 90 × 90 × 12 mm is subjected to the factored tension load of 100 kN. It is to be connected to a gusset using fillet welds at the toe and back. Find the length of welds required so that the centre of gravity of the welds lies in plane of the centre of gravity of angle. Fe410

## Short answer

Use 6 mm fillet welds. Total weld length 126 mm; for balanced c.g., toe weld 38 mm and back weld 88 mm, each increased by 12 mm for craters to 50 mm and 105 mm.

## Answer

**Given:** ISA 90×90×12 in a truss, factored tension 100 kN, Fe410 ($f_u=410$ N/mm²), fillet welds at the toe and back. Centroid of the angle $c_x\approx27$ mm from the back of the connected leg (SP 6 value $\approx27$ mm).

#### Weld size
Thickness of the thicker part 12 mm → minimum size 5 mm (Table 21). Maximum size at the angle edge $=t-1.5=10.5$ mm. Use **$s=6$ mm** (shop weld).
- $f_{wd}=\dfrac{410}{\sqrt3\times1.25}=189.4$ N/mm²
- Strength per mm $=189.4\times0.7\times6=795.4$ N/mm

#### Length of weld
$$
L_w=\frac{100\times10^3}{795.4}=125.7\ \text{mm}
$$

#### Distribution so that the c.g. of the welds lies in the c.g. plane of the angle
Moments about the back weld ($L_2$): $L_1\times90=L_w\times c_x$, hence
- Toe weld: $L_1=\dfrac{125.7\times27}{90}=37.7$ mm (force $30.0$ kN)
- Back weld: $L_2=125.7-37.7=88.0$ mm (force $70.0$ kN)

Add $2s=12$ mm to each for the end craters, and keep every weld at least $4s=24$ mm long.

**Answer: 6 mm fillet weld; toe weld 50 mm, back weld 105 mm.**

## Check yourself

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

1. For a 12 mm angle edge, the maximum fillet size is
   A. t/2 = 6 mm
   B. t - 1.5 = 10.5 mm
   C. t + 1.5 = 13.5 mm
   D. 3 mm
   Answer: B. At the edge of a rounded plate/angle, maximum size is thickness minus 1.5 mm. (Key point: Weld size limits)

2. With 795.4 N/mm strength, what weld length carries 100 kN?
   A. 251.5 mm
   B. 125.7 mm
   C. 100 mm
   D. 62.8 mm
   Answer: B. L = 100000/795.4 = 125.7 mm. (Key point: Total length)

3. Toe weld L1 = Lw x cx/90 with cx = 27 mm. What is L1 and why?
   A. 88.0 mm; toe takes the larger share
   B. 37.7 mm; moment equilibrium about the back weld keeps the weld group centroid at the angle centroid
   C. 62.9 mm; equal split
   D. 27 mm; equals cx
   Answer: B. Taking moments about the back weld gives the short toe weld and longer back weld. (Key point: Distribution)

4. What is added to each computed weld length and why?
   A. 2s for end returns
   B. s only
   C. nothing
   D. 4s for tolerance
   Answer: A. 2s is added to account for the ineffective ends of each weld. (Key point: End returns)

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