---
title: "Weld length for ISA 100x100x8, 200 kN"
subject: "Design of Steel and Timber Structure"
chapter: 5
key: "semester_6/design_of_steel_and_timber_structure:tension-angle-weld-isa100x100x8"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/tension-angle-weld-isa100x100x8
site: BCT Engineering
---

# Weld length for ISA 100x100x8, 200 kN

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

- Asked in: 2075 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-angle-weld-isa100x100x8
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-5

## Question

In a truss angle ISA 100×100×8 is subjected to the factored tension of 200 kN. It is to be connected to a gusset using fillet welds at the toe and back. Find the length of weld required. Take Fe 410 steel.

## Short answer

Use 6 mm fillet welds at toe and heel. Total effective weld length 251 mm: 69 mm at toe and 182 mm at heel (add 12 mm each for end craters). Weld strength 795 N/mm gives capacity 200 kN.

## Answer

**Given:** ISA 100×100×8 (centroid $c_x=27.5$ mm from the back), factored tension 200 kN, Fe410 ($f_u=410$ N/mm²), fillet welds at the toe and back, shop welding.

#### Weld size and strength (IS 800:2007 cl. 10.5)
- Minimum size (Table 21): 3 mm if the thicker part (gusset) is up to 10 mm thick, 5 mm if it is 10-20 mm. Maximum size at the rounded toe of the 8 mm angle $\approx\tfrac34t=6$ mm ($t-1.5=6.5$ mm at a square edge). Use **$s=6$ mm**.
- $f_{wd}=\dfrac{f_u}{\sqrt3\gamma_{mw}}=\dfrac{410}{\sqrt3\times1.25}=189.4$ N/mm²
- Strength per mm $=f_{wd}\times0.7s=189.4\times4.2=795.4$ N/mm

#### Total length
$$
L_{w}=\frac{200\times10^3}{795.4}=251.5\ \text{mm}
$$

#### Distribution (so that the resultant coincides with the centroid)
Take moments about the back weld (length $L_2$ at the heel):
- Weld at toe: $L_1=L_w\dfrac{c_x}{100}=69.2$ mm (carries $200\times0.275=55$ kN)
- Weld at back (heel): $L_2=L_w-L_1=182.3$ mm (carries 145 kN)

Add $2s=12$ mm to each for the end craters (overall length = effective length + $2s$).

**Answer: 6 mm fillet weld, toe weld 85 mm and back weld 195 mm long (total effective length about 251 mm).**

## Check yourself

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

1. What is the design shear strength of a fillet weld per mm length used here (s = 6 mm)?
   A. fwd x 0.7 s = 189.4 x 4.2 = 795.4 N/mm
   B. fwd x s = 1136 N/mm
   C. fu x 0.7 s = 1722 N/mm
   D. fwd x 0.7 s with fwd = fy/sqrt3 = 595 N/mm
   Answer: A. Throat thickness is 0.7s and fwd = fu/(sqrt3 gamma_mw) = 189.4 N/mm2. (Key point: Weld strength)

2. What total effective weld length is required for 200 kN?
   A. About 126 mm
   B. About 400 mm
   C. About 189 mm
   D. About 251 mm
   Answer: D. Lw = 200 x 10^3 / 795.4 = 251.5 mm. (Key point: Total weld length)

3. Why are the toe and back weld lengths unequal?
   A. So that the resultant of weld forces passes through the centroid of the angle
   B. Because toe welds are weaker
   C. To satisfy minimum size
   D. To avoid end returns
   Answer: A. Taking moments about the back weld: L1 = Lw x cx/100, so the weld nearer the centroid (back) is longer and no eccentricity arises. (Key point: Distribution of weld)

4. With cx = 27.5 mm and Lw = 251.5 mm, what is the toe weld length (before end returns)?
   A. 69.2 mm
   B. 182.3 mm
   C. 27.5 mm
   D. 125.7 mm
   Answer: A. L1 = 251.5 x 27.5/100 = 69.2 mm; back weld = 182.3 mm. (Key point: Toe weld length)

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