---
title: "Ultimate load of 2 ISA 100x100x8, M16 5.6"
subject: "Design of Steel and Timber Structure"
chapter: 5
key: "semester_6/design_of_steel_and_timber_structure:two-angles-ultimate-load-m16"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/two-angles-ultimate-load-m16
site: BCT Engineering
---

# Ultimate load of 2 ISA 100x100x8, M16 5.6

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

- Asked in: 2071 Bhadra, 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/two-angles-ultimate-load-m16
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-5

## Question

Find the ultimate load carrying capacity of 2 ISA 100 × 100 × 8 mm in tension which is connected to both sides of gusset plate 12 mm thick using M16 bolts of property class 5.6 in single line. One shear is in shaft and another is in thread. The yield and ultimate strength of the steel are 250 MPa and 410 MPa respectively.

## Short answer

The two angles themselves can carry 510.0 kN, but the assumed four M16 bolts limit the connection capacity. Bolt shear governs at 330.7 kN.

## Answer

**Data:** 2 ISA 100×100×8 ($A_g=1539$ mm² each) on both faces of a 12 mm gusset; M16 bolts class 5.6 ($f_{ub}=500$) in one line, one shear plane through the shank and the other through the threads (double shear). Number of bolts not given: **assumed 4**, pitch 50 mm, end distance 40 mm, gauge 55 mm; $d_0=18$ mm; $f_y=250$, $f_u=410$.

#### 1. Bolts
- $V_{dsb}=\dfrac{500}{\sqrt3\times1.25}(A_{nb}+A_{sb})=\dfrac{500}{\sqrt3\times1.25}(157+201)\times10^{-3}=82.7$ kN
- Bearing on gusset 12 mm ($e=40,\ p=50$): $k_b=0.676$; $V_{dpb}=106.4$ kN
- Bolt value $82.7$ kN; four bolts: $330.7$ kN

#### 2. One angle
- $T_{dg}=\dfrac{1539\times250}{1.10}\times10^{-3}=349.8$ kN
- Rupture: $A_{nc}=(100-4-18)\times8=624$ mm², $A_{go}=(100-4)\times8=768$ mm², $b_s=w+w_1-t=100+55-8=147$ mm, $L_c=150$ mm, $\beta=0.832$
$$
T_{dn}=\frac{0.9\times624\times410}{1.25}+\frac{0.832\times768\times250}{1.10}=329.5\ \text{kN}
$$
- Block shear: $\min(284.5,\ 255.0)=255.0$ kN
- One angle: $\min=255.0$ kN; two angles $=510.0$ kN

#### 3. Result
**Ultimate (design) load capacity $=\min(510.0\ \text{kN (angles)},\ 330.7\ \text{kN (bolts)})=330.7$ kN**, governed by shear of the bolts. (More bolts raise the capacity until the angle strength is reached.)

## Check yourself

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

1. One shear plane in shank and another in thread uses which areas?
   A. 2 x Anb
   B. 2 x Asb
   C. Asb + Anb = 201 + 157 mm2
   D. Only Asb
   Answer: C. Shank plane uses nominal area (201), threaded plane uses net tensile area (157). (Key point: Bolt shear area)

2. Shear 82.7 kN and bearing 106.4 kN. Four bolts give
   A. 425.6 kN
   B. 165.4 kN
   C. 510 kN
   D. 330.7 kN
   Answer: D. Bolt value is the lesser, 82.7 kN; 4 x 82.7 = 330.7 kN. (Key point: Bolt value)

3. Per angle, the minimum of Tdg, Tdn and Tdb is 255.0 kN. For two angles?
   A. 255.0 kN
   B. 765 kN
   C. 510.0 kN
   D. 127.5 kN
   Answer: C. Two angles carry twice the per angle strength. (Key point: Angle capacity)

4. What finally governs the ultimate load capacity?
   A. Angles: 510 kN
   B. Block shear: 255 kN
   C. Gross yielding: 699.6 kN
   D. Bolt shear: 330.7 kN
   Answer: D. Minimum of angle strength and bolt capacity is 330.7 kN. (Key point: Governing)

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