---
title: "Design strength of ISA 100x75x6, 6 bolts"
subject: "Design of Steel and Timber Structure"
chapter: 5
key: "semester_6/design_of_steel_and_timber_structure:angle-design-strength-100x75x6"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/angle-design-strength-100x75x6
site: BCT Engineering
---

# Design strength of ISA 100x75x6, 6 bolts

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

- Asked in: 2076 Baisakh, 10 marks; 2069 Bhadra, 10 marks (asked 2 times)
- 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-100x75x6
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-5

## Question

A single unequal angle 100x75x6 is connected to a 10mm thick gusseted plate at the ends with six 16mm diameter bolts to transfer tension as shown in figure below. Determine the design tensile strength of the angle assuming that the yield and the ultimate stress of steel used are 250 MPa and 410 MPa if the gusset plate is connected to the longer leg. Also design connection for its full capacity.
[Figure: ISA 100x75x6 (A = 1010 mm²) connected through its 100 mm leg to a 10 mm gusset plate with 16 mm bolts]

## Short answer

Design tensile strength is 229.5 kN (yielding of gross section). For full capacity, 8 M16 (grade 4.6) bolts are needed in the connection.

## Answer

**Data:** ISA 100×75×6 ($A_g=1010$ mm²), gusset 10 mm connected to the 100 mm leg by six 16 mm bolts ($d_0=18$ mm); $f_y=250$, $f_u=410$ N/mm². Assumed layout: single line of bolts, pitch $p=50$ mm, end distance 30 mm, bolt line 55 mm from the heel (edge distance 45 mm to the toe). Bolts assumed grade 4.6.

#### 1. Yielding of gross section (cl. 6.2)
$$
T_{dg}=\frac{A_gf_y}{\gamma_{m0}}=\frac{1010\times250}{1.10}\times10^{-3}=229.5\ \text{kN}
$$

#### 2. Rupture of net section with shear lag (cl. 6.3.3)
- Connected leg: $A_{nc}=(100-\tfrac{6}{2}-18)\times6=474$ mm²
- Outstanding leg: $A_{go}=(75-\tfrac{6}{2})\times6=432$ mm²
- $w=75$ mm, $b_s=w+w_1-t=75+55-6=124$ mm ($w_1$ = distance of the bolt line from the heel), $L_c=5\times50=250$ mm
$$
\beta=1.4-0.076\frac{w}{t}\frac{f_y}{f_u}\frac{b_s}{L_c}=1.4-0.076\times\frac{75}{6}\times\frac{250}{410}\times\frac{124}{250}=1.113\quad(0.7\le\beta\le1.443)
$$
$$
T_{dn}=\frac{0.9A_{nc}f_u}{\gamma_{m1}}+\frac{\beta A_{go}f_y}{\gamma_{m0}}=139.9+109.3=249.2\ \text{kN}
$$

#### 3. Block shear (cl. 6.4)
$A_{vg}=1680$, $A_{vn}=1086$, $A_{tg}=270$, $A_{tn}=216$ mm²
- $T_{db1}=\dfrac{A_{vg}f_y}{\sqrt3\gamma_{m0}}+\dfrac{0.9A_{tn}f_u}{\gamma_{m1}}=284.2$ kN
- $T_{db2}=\dfrac{0.9A_{vn}f_u}{\sqrt3\gamma_{m1}}+\dfrac{A_{tg}f_y}{\gamma_{m0}}=246.5$ kN

**Design tensile strength** $T_d=\min(229.5,\ 249.2,\ 246.5)=229.5$ kN (yielding of gross section governs).

#### Design of connection for full capacity (16 mm, 4.6 bolts)
- $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{157}{1.25}\times10^{-3}=29.01$ kN
- Bearing on the 6 mm angle ($e=30,\ p=50$): $k_b=0.556$, $V_{dpb}=43.7$ kN
- Number of bolts $=\dfrac{229.5}{29.01}=7.91$ → **8 bolts** (the six bolts are not enough for the full capacity).

**Answer: design tensile strength $\approx229.5$ kN; connect with 8 M16 bolts (4.6) at 50 mm pitch (a longer joint) or use higher-grade bolts.** With six bolts the connection strength would be only $6\times29.0=174.0$ kN.

## Check yourself

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

1. How is the design tensile strength determined?
   A. Highest of yielding, rupture and block shear
   B. Average
   C. Least of yielding, rupture and block shear
   D. Yielding only always
   Answer: C. The weakest mode controls; here 229.5 kN, yielding. (Key point: Design strength)

2. What is the yielding strength of the gross section?
   A. A_g f_y/gamma_m0 = 1010 x 250/1.10 = 229.5 kN
   B. 0.9 A_n f_u/1.25
   C. A_g f_u
   D. A_g f_y x 1.10
   Answer: A. T_dg = A_g f_y/γ_m0. (Key point: Yielding)

3. What does beta in the net rupture expression account for?
   A. Bolt hole size
   B. Slenderness
   C. Bolt grade
   D. Non-uniform stress because the outstanding leg is not connected
   Answer: D. β reduces the effectiveness of the unconnected outstanding leg. (Key point: Shear lag factor)

4. Each M16 bolt has value 29.01 kN. How many bolts for 229.5 kN?
   A. 6
   B. 8, since 229.5/29.01 = 7.91
   C. 4
   D. 10
   Answer: B. Six bolts give only 174 kN, so 8 are needed. (Key point: Bolts for full capacity)

## Related in Design of Steel and Timber Structure

- [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
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