---
title: "Unequal angle with fillet welds, 150 kN"
subject: "Design of Steel and Timber Structure"
chapter: 5
key: "semester_6/design_of_steel_and_timber_structure:tension-unequal-angle-fillet-150kn"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/tension-unequal-angle-fillet-150kn
site: BCT Engineering
---

# Unequal angle with fillet welds, 150 kN

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

- Asked in: 2070 Magh, 10 marks
- Video: [Length of welding for ISA section Problem 5](https://www.youtube.com/watch?v=QjILKodNnEA) — 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/tension-unequal-angle-fillet-150kn
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-5

## Question

Design a single angle (unequal angle) to carry a working tensile load of 150 KN, If the end connection is done using fillet welds. The yield and ultimate strength of the steel are 250 MPa and 410 MPa respectively.

## Short answer

Use ISA 100×75×8 angle. 6 mm fillet welds, toe weld 105 mm, heel weld 210 mm, balansing the centrod.

## Answer

**Given:** working tension 150 kN → factored $T_u=1.5\times150=225$ kN; fillet welds; $f_y=250$, $f_u=410$ N/mm².

#### Section
$A_g\ge\dfrac{225\times10^3\times1.10}{250}=990$ mm². Try **ISA 100×75×8** ($A_g=1336$ mm²), long leg connected, centroid 31.5 mm from the heel.
- $T_{dg}=\dfrac{1336\times250}{1.10}\times10^{-3}=303.6$ kN $>225$ ✓

#### Weld
- $s=6$ mm (angle 8 mm: min 3-5 mm, max 6.5 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.
- Total length $L_w=\dfrac{225\times10^3}{795.4}=282.9$ mm.
- For no eccentricity: toe weld $L_1=L_w\times\dfrac{31.5}{100}=89.1$ mm; heel weld $L_2=193.8$ mm.
- Adding $2s=12$ mm for end craters: **toe 105 mm and heel 210 mm**.

#### Rupture check (shear lag; welded, so $b_s=w=75$ mm; conservatively $L_c=89$ mm)
$\beta=1.4-0.076\times\dfrac{75}{8}\times\dfrac{250}{410}\times\dfrac{75}{89}=1.034$
$$
T_{dn}=\frac{0.9\times(100-4)\times8\times410}{1.25}+\frac{\beta\times(75-4)\times8\times250}{1.10}=360.2\ \text{kN}>225\ \checkmark
$$

**Answer: ISA 100×75×8, 6 mm fillet welds; toe weld 105 mm and heel weld 210 mm.**

## Check yourself

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

1. The load 150 kN is a working load. What is the design load?
   A. 150 kN
   B. 187.5 kN
   C. 300 kN
   D. 225 kN (x 1.5)
   Answer: D. Limit state design uses factored load = 1.5 x working load. (Key point: Working to factored load)

2. Required Ag for 225 kN is
   A. 990 mm2
   B. 900 mm2
   C. 1336 mm2
   D. 675 mm2
   Answer: A. 225 x 10^3 x 1.10 / 250 = 990 mm2. (Key point: Minimum area)

3. Total weld length at 795.4 N/mm for 225 kN is
   A. 282.9 mm
   B. 251.5 mm
   C. 314.3 mm
   D. 225 mm
   Answer: A. 225000/795.4 = 282.9 mm. (Key point: Weld length)

4. Toe/heel are distributed in the ratio of the centroid distances. Toe = 89.1 mm, heel = 193.8 mm. Why shorter toe?
   A. Toe weld is stronger
   B. Toe needs no returns
   C. Centroid is nearer the heel (31.5 mm), so heel weld takes more load
   D. Heel is thinner
   Answer: C. Moments about the heel put more weld length near the heel to avoid eccentricity. (Key point: Toe and heel)

## Related in Design of Steel and Timber Structure

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