---
title: "Tensile strength of ISA 100x75x8, six M20 bolts"
subject: "Design of Steel and Timber Structure"
chapter: 5
key: "semester_6/design_of_steel_and_timber_structure:angle-strength-isa100x75x8-m20"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/angle-strength-isa100x75x8-m20
site: BCT Engineering
---

# Tensile strength of ISA 100x75x8, six M20 bolts

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

- Asked in: 2073 Bhadra, 8 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/angle-strength-isa100x75x8-m20
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-5

## Question

A single angle ISA 100×75×8 mm is connected to 12 mm thick gusset plate at the ends with six bolts of M20 in one row to transfer tension. Determine the design tensile strength of the angle if gusset is connected to the 100 mm leg. Take $f_y$ = 250 MPa and $F_u$ = 410 MPa, pr.cl.4.6.

## Short answer

Design tensile strength of angle is 303.6 kN (gross section yielding governs). However, six M20 grade 4.6 bolts in single shear limit connection capacity to 271.6 kN.

## Answer

**Data:** ISA 100×75×8 ($A_g=1336$ mm²), gusset 12 mm connected to the 100 mm leg, six M20 bolts (4.6) in one row. Assumed pitch $p=50$ mm, end distance 40 mm, gauge 55 mm from the heel; $d_0=22$ mm; $f_y=250$, $f_u=410$.

#### 1. Yielding of gross section
$T_{dg}=\dfrac{A_gf_y}{\gamma_{m0}}=\dfrac{1336\times250}{1.10}\times10^{-3}=303.6$ kN

#### 2. Net-section rupture with shear lag
- $A_{nc}=(100-\tfrac82-22)\times8=592$ mm²; $A_{go}=(75-\tfrac82)\times8=568$ mm²
- $w=75$, $b_s=w+w_1-t=75+55-8=122$ mm ($w_1$ = gauge), $L_c=5\times50=250$ mm
$$
\beta=1.4-0.076\times\frac{75}{8}\times\frac{250}{410}\times\frac{122}{250}=1.188\quad(0.7\le\beta\le1.443)
$$
$$
T_{dn}=\frac{0.9A_{nc}f_u}{\gamma_{m1}}+\frac{\beta A_{go}f_y}{\gamma_{m0}}=328.1\ \text{kN}
$$

#### 3. Block shear
$A_{vg}=2320$, $A_{vn}=1352$, $A_{tg}=360$, $A_{tn}=272$ mm²: $T_{db1}=384.7$ kN, $T_{db2}=312.2$ kN.

#### 4. Bolt group (check)
$V_{dsb}=45.26$ kN (single shear); bearing ($t=8$): $V_{dpb}=66.6$ kN; six bolts: $6\times45.26=271.6$ kN. This is less than the strength of the angle, so the **bolts govern the strength of the joint**.

**Answer: design tensile strength of the angle $=\min(303.6,\ 328.1,\ 312.2)=303.6$ kN** (yielding of the gross section governs). With six M20 (4.6) bolts, the connection limits the load the member can transfer to **271.6 kN**.

## Check yourself

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

1. Which three failure modes are checked for the design tensile strength of the angle?
   A. Buckling, bending and shear
   B. Bearing, slip and fatigue
   C. Gross yielding, net-section rupture and block shear
   D. Gross yielding, buckling and torsion
   Answer: C. Design tensile strength is the minimum of Tdg, Tdn and Tdb. (Key point: Failure modes)

2. How is Lc obtained for six bolts at 50 mm pitch?
   A. Lc = 6 x 50 = 300 mm
   B. Lc = 5 x 50 = 250 mm
   C. Lc = 50 mm
   D. Lc = 40 mm end distance
   Answer: B. Lc is the length of the end connection between outermost bolts: 5 pitches x 50 mm. (Key point: Lc for shear lag)

3. What is Tdg for Ag = 1336 mm2, fy = 250 N/mm2?
   A. 303.6 kN
   B. 334 kN
   C. 273 kN
   D. 318 kN
   Answer: A. Tdg = 1336 x 250 / 1.10 = 303.6 kN. (Key point: Gross yielding)

4. Tdg=303.6, Tdn=318.0, Tdb=312.2 kN. What is the design strength?
   A. 312.2 kN
   B. 303.6 kN
   C. 318.0 kN
   D. 933.8 kN
   Answer: B. Take the minimum: gross yielding governs. (Key point: Governing value)

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