---
title: "Tensile strength of ISA 100x75x8, M16 8.8"
subject: "Design of Steel and Timber Structure"
chapter: 5
key: "semester_6/design_of_steel_and_timber_structure:angle-design-strength-100x75x8-m16-88"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/angle-design-strength-100x75x8-m16-88
site: BCT Engineering
---

# Tensile strength of ISA 100x75x8, M16 8.8

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

- Asked in: 2071 Magh, 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/angle-design-strength-100x75x8-m16-88
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-5

## Question

Longer leg of a ISA 100 × 75 × 8 is connected to a gusset plate of thickness 10 mm by M16 bolts of property class 8.8 as shown. If Fe410 grade steel is used, determine the design tensile strength of the angle.
[Figure: four bolts in a line at 40 mm pitch (40, 40, 40, 40 mm), angle leg width 100 mm]

## Short answer

Design tensile strength is the minimum of yielding (303.6 kN), net-section rupture (307.9 kN), and block shear. Block shear governs at 214.1 kN.

## Answer

**Data:** ISA 100×75×8 ($A_g=1336$ mm²), long (100 mm) leg bolted to a 10 mm gusset by four M16 bolts (class 8.8), pitch 40 mm, end distance 40 mm, assumed gauge 55 mm; Fe410 ($f_y=250$, $f_u=410$); $d_0=18$ mm.

#### 1. Yielding
$T_{dg}=\dfrac{1336\times250}{1.10}\times10^{-3}=303.6$ kN

#### 2. Net-section rupture with shear lag
- $A_{nc}=(100-4-18)\times8=624$ mm², $A_{go}=(75-4)\times8=568$ mm²
- $b_s=w+w_1-t=75+55-8=122$ mm ($w_1$ = gauge), $L_c=3\times40=120$ mm
$$
\beta=1.4-0.076\times\frac{75}{8}\times\frac{250}{410}\times\frac{122}{120}=0.958
$$
$$
T_{dn}=\frac{0.9\times624\times410}{1.25}+\frac{0.958\times568\times250}{1.10}=307.9\ \text{kN}
$$

#### 3. Block shear
$A_{vg}=1280$, $A_{vn}=776$, $A_{tg}=360$, $A_{tn}=288$ mm²
- $T_{db1}=253.0$ kN; $T_{db2}=214.1$ kN

#### 4. Result
**Design tensile strength of the angle $=\min(303.6,\ 307.9,\ 214.1)=214.1$ kN** (block shear governs).

(Bolt check: $V_{dsb}=58.0$ kN, bearing on 8 mm leg $k_b=0.491$, $V_{dpb}=51.5$ kN; four bolts give $206.0$ kN, so the connection itself limits the joint to about $206$ kN.)

## Check yourself

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

1. Four bolts at 40 mm pitch: what is Lc?
   A. 160 mm
   B. 40 mm
   C. 80 mm
   D. 120 mm
   Answer: D. Lc = 3 pitches x 40 mm = 120 mm. (Key point: Lc)

2. Compared with a longer connection, the short Lc here makes beta
   A. larger, increasing strength
   B. irrelevant
   C. equal to 1.443
   D. smaller (0.795), reducing rupture strength
   Answer: D. beta has Lc in the denominator of the subtracted term, so short Lc lowers beta. (Key point: beta effect)

3. Tdg=303.6, Tdn=286.9, Tdb=214.1 kN. Which governs?
   A. Yielding 303.6 kN
   B. Rupture 286.9 kN
   C. Block shear 214.1 kN
   D. None
   Answer: C. Block shear gives the minimum. (Key point: Governing mode)

4. The bolt group gives about 206 kN. What does it imply?
   A. The angle strength becomes 206 kN
   B. Bolts are unnecessary
   C. The joint strength is limited to about 206 kN, below the angle's 214.1 kN
   D. Block shear is ignored
   Answer: C. Bolt capacity lower than the angle strength limits the joint. (Key point: Connection limit)

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