---
title: "Ultimate strength of ISA 100x100x10, M20 bolts"
subject: "Design of Steel and Timber Structure"
chapter: 5
key: "semester_6/design_of_steel_and_timber_structure:angle-ultimate-strength-100x100x10"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/angle-ultimate-strength-100x100x10
site: BCT Engineering
---

# Ultimate strength of ISA 100x100x10, M20 bolts

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

- Asked in: 2072 Magh, 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-ultimate-strength-100x100x10
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-5

## Question

Find the ultimate design strength of angle 100×100×10 mm in tension which is connected to a gusset 12 mm thick through 100 mm leg using M20 bolts of product Grade C and property class 4.6 in single line. Assume that the bolt threads are outside the shear plane. The yield and ultimate strength of the steel are 250 MPa and 410 MPa respectively.

## Short answer

The tension member design strength is the minimum of yielding, rupture, and block shear. Block shear governs, giving 286.3 kN for the angle.

## Answer

**Data:** ISA 100×100×10 ($A_g=1903$ mm²), gusset 12 mm, connected through one 100 mm leg, M20 (4.6) bolts in a single line, threads **outside** the shear plane. The number of bolts is not given: **assumed 4 bolts**, pitch 50 mm, end distance 35 mm, gauge 55 mm from the back; $d_0=22$ mm; $f_y=250$, $f_u=410$.

#### 1. Yielding
$T_{dg}=\dfrac{1903\times250}{1.10}\times10^{-3}=432.5$ kN

#### 2. Rupture with shear lag
- $A_{nc}=(100-5-22)\times10=730$ mm², $A_{go}=(100-5)\times10=950$ mm²
- $b_s=w+w_1-t=100+55-10=145$ mm ($w_1$ = gauge), $L_c=3\times50=150$ mm
$$
\beta=1.4-0.076\times\frac{100}{10}\times\frac{250}{410}\times\frac{145}{150}=0.952\ \ (\text{limits }0.7\ \text{to }1.443)
$$
$$
T_{dn}=\frac{0.9\times730\times410}{1.25}+\frac{0.952\times950\times250}{1.10}=421.0\ \text{kN}
$$

#### 3. Block shear
$T_{db1}=343.1$ kN, $T_{db2}=286.3$ kN.

#### 4. Bolts (threads outside the shear plane)
$V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{314}{1.25}\times10^{-3}=58.01$ kN; $V_{dpb}=83.2$ kN (10 mm angle). Four bolts: $232.0$ kN.

**Answer: member design strength $=\min(432.5,\,421.0,\,286.3)=286.3$ kN (block shear governs); with the assumed 4 bolts the joint strength is $\min(286.3,\,232.0)=232.0$ kN.**

## Check yourself

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

1. Threads are outside the shear plane. Which area is used for bolt shear?
   A. Net tensile area 245 mm2
   B. Nominal shank area Asb = 314 mm2
   C. Hole area 380 mm2
   D. Area of 10 mm plate
   Answer: B. With threads outside the shear plane, the full shank area pi/4 x 20^2 = 314 mm2 is used, giving 58.01 kN. (Key point: Bolt shear area)

2. The question gives no bolt count. How does the answer deal with this?
   A. Ignores bolts entirely
   B. Assumes 1 bolt
   C. Uses 10 bolts
   D. Assumes 4 bolts and reports member strength and joint strength
   Answer: D. A bolt number is assumed (4) to get Lc = 150 mm and the joint strength. (Key point: Assumption of bolts)

3. Tdg=432.5, Tdn=391.0, Tdb=286.3 kN. What governs member strength?
   A. Yielding 432.5 kN
   B. Rupture 391.0 kN
   C. Bolts 232.0 kN
   D. Block shear 286.3 kN
   Answer: D. Minimum of the three member checks is block shear. (Key point: Member strength)

4. With four bolts giving 232.0 kN, what is the joint strength?
   A. 286.3 kN
   B. 518.3 kN
   C. 232.0 kN
   D. 391.0 kN
   Answer: C. Joint strength is min(member 286.3, bolts 232.0) = 232.0 kN. (Key point: Joint strength)

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