---
title: "Angle tension member 250 kN, weld and bolt"
subject: "Design of Steel and Timber Structure"
chapter: 5
key: "semester_6/design_of_steel_and_timber_structure:tension-angle-250kn-weld-bolt"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/tension-angle-250kn-weld-bolt
site: BCT Engineering
---

# Angle tension member 250 kN, weld and bolt

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

- Asked in: 2070 Bhadra, 12 marks
- Video: [Tension carrying capacity in steel structure by limit state method of single angle Bolted](https://www.youtube.com/watch?v=pJhitlaMeDM) — Parag Pal
- 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-250kn-weld-bolt
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-5

## Question

Design a suitable angle section to carry a tensile force of 250 kN (Factored). The end connection is to be done by using (i) fillet welds (ii) bolts. The yield and ultimate strengths of the steel are 250 MPa and 410 MPa, respectively.

## Short answer

ISA 100×75×8 works. Welded: 6 mm fillet, toe 115 mm, heel 230 mm. Bolted: six M20 bolts at 50 mm pitch.

## Answer

**Given:** factored tension 250 kN, $f_y=250$, $f_u=410$ N/mm². Trial section for both cases: **ISA 100×75×8** ($A_g=1336$ mm²), connected through the 100 mm leg, centroid 31.5 mm from the heel (short-leg side).

$T_{dg}=\dfrac{1336\times250}{1.10}\times10^{-3}=303.6$ kN $>250$ ✓

#### (i) Welded connection
- Size: 8 mm angle → use $s=6$ mm (min 3-5 mm, max $t-1.5=6.5$ mm). $f_{wd}=189.4$ N/mm²; strength $=189.4\times0.7\times6=795.4$ N/mm.
- Total length $L_w=\dfrac{250\times10^3}{795.4}=314.3$ mm.
- Distribution about the centroid: toe weld $L_1=L_w\times\dfrac{31.5}{100}=99.0$ mm; heel weld $L_2=215.3$ mm. Add $2s=12$ mm for end craters: **toe 115 mm, heel 230 mm**.
- Rupture check (welded, so $b_s=w=75$ mm; conservatively $L_c\approx99$ mm, the shorter weld): $\beta=1.4-0.076\times\dfrac{75}{8}\times\dfrac{250}{410}\times\dfrac{75}{99}=1.071$; $T_{dn}=\dfrac{0.9\times(100-4)\times8\times410}{1.25}+\dfrac{\beta\times(75-4)\times8\times250}{1.10}=365.0$ kN $>250$ ✓

#### (ii) Bolted connection (M20, grade 4.6, $d_0=22$ mm)
- $V_{dsb}=45.26$ kN; bearing on 8 mm leg ($e=40,\ p=50$) $V_{dpb}=66.6$ kN → bolt value $45.26$ kN
- $n=\dfrac{250}{45.26}=5.5$ → **6 bolts**, pitch 50 mm, end distance 40 mm, gauge 55 mm; capacity $6\times45.26=271.6$ kN.
- Rupture: $A_{nc}=(100-4-22)\times8=592$ mm², $A_{go}=(75-4)\times8=568$ mm², $b_s=75+55-8=122$ mm, $\beta=1.188$ → $T_{dn}=328.1$ kN
- Block shear: $\min(384.7,\ 312.2)$ kN $>250$ ✓

**Answer: ISA 100×75×8 with (i) 6 mm fillet welds (toe 115 mm, heel 230 mm), or (ii) 6 bolts M20 at 50 mm pitch.**

## Check yourself

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

1. For an 8 mm angle, the weld size 6 mm is within
   A. min 8 mm and max 10 mm
   B. min 3-5 mm and max t - 1.5 = 6.5 mm
   C. min 2 mm and max 3 mm
   D. only the minimum matters
   Answer: B. Size must lie between the minimum from Table 21 and the maximum t - 1.5. (Key point: Weld size)

2. Toe weld = Lw x 31.5/100 = 99.0 mm. What is the heel weld?
   A. 314.3 mm
   B. 99.0 mm
   C. 215.3 mm
   D. 157 mm
   Answer: C. Heel = 314.3 - 99.0 = 215.3 mm. (Key point: Weld distribution)

3. Number of M20 bolts needed for 250 kN is
   A. 5
   B. 6
   C. 4
   D. 8
   Answer: B. 250/45.26 = 5.5 so 6 bolts. (Key point: Bolt count)

4. In the welded case, why does beta fall to its minimum 0.7?
   A. Welds weaken the steel
   B. Weld size is large
   C. Angle is symmetric
   D. The weld length Lc (99 mm) is short
   Answer: D. Short Lc gives a lower computed beta, which is limited to 0.7. (Key point: Shear lag with welds)

## Related in Design of Steel and Timber Structure

- [Unequal angle with fillet welds, 150 kN](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Atension-unequal-angle-fillet-150kn) — Design of Steel and Timber Structure, chapter 5
- [Angle tension member 250 kN with M20](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Atension-angle-250kn-m20) — Design of Steel and Timber Structure, chapter 5
- [Single angle tension member, 100 kN](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Atension-single-angle-100kn-m20) — Design of Steel and Timber Structure, chapter 5
- [Double angle tension member 400 kN, M20](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Atension-double-angle-400kn-class-c) — Design of Steel and Timber Structure, chapter 5
- [Unequal angle tension member with HSFG bolts](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Atension-unequal-angle-hsfg-200kn) — Design of Steel and Timber Structure, chapter 5
- [Fillet welded channel tension member, 200 kN](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Atension-welded-channel-200kn) — Design of Steel and Timber Structure, chapter 5

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- [Tensile test data of a steel bar](https://www.bctengineer.com/question.md?key=semester_3%2Fmaterial_science%3Atensile-test-numerical-steel-bar) — Material Science, chapter 3