---
title: "Angle tension member 250 kN with M20"
subject: "Design of Steel and Timber Structure"
chapter: 5
key: "semester_6/design_of_steel_and_timber_structure:tension-angle-250kn-m20"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/tension-angle-250kn-m20
site: BCT Engineering
---

# Angle tension member 250 kN with M20

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

- Asked in: 2077 Chaitra, 10 marks
- Video: [Tensile Strength Of Angle | Tension Members | Design Of Steel Structures](https://www.youtube.com/watch?v=iV5EdGhvzco) — Jawad Khan
- 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-m20
- 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 factored tensile force of 250KN assuming a single row of M20 bolts. The yeild strength and ultimate strength of the material is 250MPa and 410MPa respectively. The length of member is 3m.

## Short answer

ISA 100×75×8 (long leg connected) with 6 M20 bolts at 50 mm pitch. Design strength is 271.6 kN, which safely carries the 250 kN factored load.

## Answer

**Given:** factored tension 250 kN, $L=3$ m, single row of M20 bolts (grade 4.6 assumed), $f_y=250$, $f_u=410$ N/mm².

#### 1. Bolts
- $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26$ kN (single shear, thread in the shear plane)
- Bearing on the 8 mm angle ($e=40$, $p=50$): $k_b=\min(0.606,\ 0.508,\ 0.976,\ 1)=0.508$; $V_{dpb}=66.6$ kN → bolt value $45.26$ kN
- $n=\dfrac{250}{45.26}=5.5$ → **6 bolts**, pitch 50 mm, end distance 40 mm, bolt line 55 mm from the heel; capacity $271.6$ kN.

#### 2. Section
$A_g\ge\dfrac{250\times1.10}{250}\times10^3=1100$ mm². Because the angle is connected through one leg, choose **ISA 100×75×8** ($A_g=1336$ mm², $r_{vv}=15.9$ mm), long leg connected.
- Yielding: $T_{dg}=\dfrac{1336\times250}{1.10}\times10^{-3}=303.6$ kN $>250$ ✓
- Rupture (cl. 6.3.3): $A_{nc}=(100-4-22)\times8=592$ mm², $A_{go}=(75-4)\times8=568$ mm², $b_s=w+w_1-t=75+55-8=122$ mm, $L_c=250$ mm
$$
\beta=1.4-0.076\times\frac{75}{8}\times\frac{250}{410}\times\frac{122}{250}=1.188,\qquad T_{dn}=\frac{0.9\times592\times410}{1.25}+\frac{1.188\times568\times250}{1.10}=328.1\ \text{kN}
$$
- Block shear: $A_{vg}=2320$, $A_{vn}=1352$, $A_{tg}=360$, $A_{tn}=272$ mm² → $T_{db1}=384.7$, $T_{db2}=312.2$ kN
- Slenderness: $\dfrac{3000}{15.9}\approx189<400$ ✓

**Design strength $=\min(303.6,328.1,312.2,271.6)=271.6$ kN $>250$ kN.**
**Answer: ISA 100×75×8 (long leg connected) with 6 M20 bolts at 50 mm pitch.**

## Check yourself

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

1. In the bolted connection, which value governs the bolt value of an M20 (4.6) bolt in single shear, and what is it?
   A. Shear, 45.26 kN
   B. Bearing, 66.6 kN
   C. Bearing, 45.26 kN
   D. Shear, 66.6 kN
   Answer: A. Shear capacity 45.26 kN is less than bearing 66.6 kN on the 8 mm angle, so the lower value (shear) is the bolt value. (Key point: Bolt value)

2. How many bolts are needed for 250 kN with a bolt value of 45.26 kN?
   A. 6
   B. 5
   C. 4
   D. 8
   Answer: A. 250/45.26 = 5.5, rounded up to 6 bolts (capacity 271.6 kN). (Key point: Number of bolts)

3. Why is the section chosen larger than the minimum Ag = 1100 mm2 (ISA 100x75x8, 1336 mm2) and why is beta used?
   A. Because the angle is connected through one leg, the outstanding leg is less effective in rupture
   B. Because bolts reduce the yield strength of steel
   C. Because slenderness limit is 40
   D. Because beta increases the gross area
   Answer: A. With only one leg connected, the net-section strength is reduced by shear lag via beta, so extra area is provided. (Key point: Trial section and shear lag)

4. The strengths are Tdg=303.6, Tdn=318.0, Tdb=312.2 and bolts=271.6 kN. What is the design strength?
   A. 303.6 kN (yielding)
   B. 318.0 kN (rupture)
   C. 271.6 kN (bolts)
   D. 312.2 kN (block shear)
   Answer: C. The design strength is the minimum of all failure modes, which is the bolt group capacity 271.6 kN, greater than 250 kN. (Key point: Governing strength)

## Related in Design of Steel and Timber Structure

- [Angle tension member 250 kN, weld and bolt](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Atension-angle-250kn-weld-bolt) — 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
- [Tensile strength of ISA 100x75x8, six M20 bolts](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Aangle-strength-isa100x75x8-m20) — Design of Steel and Timber Structure, chapter 5
- [Ultimate strength of ISA 100x100x10, M20 bolts](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Aangle-ultimate-strength-100x100x10) — 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

## Related in other subjects

- [Splitting tensile strength of cylinder (400 kN)](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Asplitting-tensile-cylinder-400kn) — Concrete Technology and Masonry Structure, chapter 5, asked 2×
- [Splitting tensile strength of cylinder and cube (500 kN)](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Asplitting-tensile-cylinder-cube-500kn) — Concrete Technology and Masonry Structure, chapter 5, asked 2×