---
title: "Single angle tension member, 100 kN"
subject: "Design of Steel and Timber Structure"
chapter: 5
key: "semester_6/design_of_steel_and_timber_structure:tension-single-angle-100kn-m20"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/tension-single-angle-100kn-m20
site: BCT Engineering
---

# Single angle tension member, 100 kN

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

- Asked in: 2072 Asoj, 12 marks
- Video: [Tension Member Design (Single Angle Section )](https://www.youtube.com/watch?v=0ivqtH6Faew) — sagar paruthi
- 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-single-angle-100kn-m20
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-5

## Question

Design a single angle to carry a tension of 100 kN. The end connection is to done using M20 bolts of product Grade C and property class 4.6. The yield and ultimate strength of the steel are 250 MPa and 410 MPa respectively.

## Short answer

Use ISA 75×75×6 angle connected with 3 M20 grade 4.6 bolts at 50 mm pitch. Block shear capacity 129.5 kN governs and exceeds required 100 kN.

## Answer

**Given:** tension 100 kN (taken as the design/factored load; if it is a working load, multiply by 1.5), M20 bolts of product grade C, class 4.6 ($d_0=22$ mm), $f_y=250$, $f_u=410$ N/mm².

#### 1. Bolts
- $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26$ kN
- Bearing on 6 mm leg ($e=35,\ p=50$): $k_b=0.508$, $V_{dpb}=49.9$ kN → bolt value $45.26$ kN
- $n=100/45.26=2.2$ → **3 bolts**, pitch 50 mm, end distance 35 mm; capacity $135.8$ kN.

#### 2. Section
$A_g\ge\dfrac{100\times1.10}{250}\times10^3=440$ mm². The leg must hold an M20 bolt with edge distances $\ge1.5d_0=33$ mm on both sides, so the leg must be at least 70 mm: try **ISA 75×75×6** ($A=866$ mm²), bolt gauge 40 mm from the back.
- Yielding: $T_{dg}=\dfrac{866\times250}{1.10}\times10^{-3}=196.8$ kN
- Rupture with shear lag: $A_{nc}=(75-3-22)\times6=300$ mm², $A_{go}=(75-3)\times6=432$ mm², $b_s=w+w_1-t=75+40-6=109$ mm, $L_c=100$ mm, $\beta=0.769$
$$
T_{dn}=\frac{0.9\times300\times410}{1.25}+\frac{0.769\times432\times250}{1.10}=164.0\ \text{kN}
$$
- Block shear: $T_{db1}=148.8$ kN, $T_{db2}=129.5$ kN → $T_{db}=129.5$ kN

Design strength $=\min(196.8,\ 164.0,\ 129.5,\ 135.8)=129.5$ kN $>100$ kN ✓.

**Answer: ISA 75×75×6 connected by 3 M20 bolts (4.6) at 50 mm pitch.**

## Check yourself

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

1. With bolt value 45.26 kN, how many bolts for 100 kN?
   A. 2
   B. 4
   C. 3
   D. 5
   Answer: C. 100/45.26 = 2.2, rounded up to 3 bolts. (Key point: Bolt count)

2. beta came out at its lower limit 0.7 because
   A. Lc is short (100 mm), so shear lag is severe
   B. Lc is long
   C. fy is low
   D. the section is symmetric
   Answer: A. Short connection length makes beta computed below 0.7, so the minimum 0.7 is used. (Key point: Shear lag beta limit)

3. Tdg=196.8, Tdn=157.3, Tdb=129.5, bolts=135.8 kN. Design strength?
   A. 129.5 kN
   B. 135.8 kN
   C. 157.3 kN
   D. 196.8 kN
   Answer: A. Block shear is the least, 129.5 kN > 100 kN, so the section is safe. (Key point: Design strength)

## Related in Design of Steel and Timber Structure

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