---
title: "Built-up column, single lacing, 8 m, 1100 kN"
subject: "Design of Steel and Timber Structure"
chapter: 6
key: "semester_6/design_of_steel_and_timber_structure:builtup-lacing-1100kn-8m-e250"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/builtup-lacing-1100kn-8m-e250
site: BCT Engineering
---

# Built-up column, single lacing, 8 m, 1100 kN

Design of Steel and Timber Structure · Chapter 6: Compression Members

- Asked in: 2074 Bhadra, 12 marks
- Video: [Design of single Laced column | Steel column connection details | 3d animation #steel](https://www.youtube.com/watch?v=2d_nhlkYXFY) — Comten Consulting Engineers
- 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/builtup-lacing-1100kn-8m-e250
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-6

## Question

Design a built up column to carry an axial load of 1100 KN. The length of column is 8m and is effectively held in position at both ends but not restrained against rotation. Use single lacing system with bolted connection. Grade of steel E250, M10 [?] Bolt, 4.6 grade. The built up column should be consists of double channel back to back.

## Short answer

With KL=8 m, use 2 ISMC 300 back to back, gap 190 mm (λe=71.1, fcd=150.4 N/mm², Pd=1373 kN). Single lacing: 60×10 mm flats at 45°, 1 M20 bolt per end. End tie plates 360×280×6 mm.

## Answer

**Reading of the doubtful value:** the bolt size is printed "M10 [?]". M10 is too small for the plate and flange used here, so **M20** is adopted (grade 4.6); with M10, two bolts would be needed at each end of a lacing bar.

**Given/assumptions:** factored load $P_u=1100$ kN; length 8 m, held in position at both ends but not restrained against rotation → $K=1.0$, $KL=8$ m. Two channels back to back, single bolted lacing; E250 ($f_y=250$, $f_u=410$).

#### 1. Section: 2 ISMC 300 back to back (webs facing, flanges outward)
$A=2\times4564=9128$ mm², $I_{xx}=2\times6362.6\times10^4$ mm⁴, $r_{xx}=118.1$ mm, $C_{yy}=23.6$ mm, $b_f=90$ mm. Built-up members use **buckling class c** (IS 800:2007 Table 10, $\alpha=0.49$).
- $\lambda_{xx}=\dfrac{KL}{r_{xx}}=\dfrac{8000}{118.1}=67.8$
- Laced column: $\lambda_e=1.05\lambda=71.1$ (cl. 7.6.1.4)
- $f_{cd}=\dfrac{f_y/\gamma_{m0}}{\phi+\sqrt{\phi^2-\lambda_n^2}}$ with $\phi=0.5[1+0.49(\lambda_n-0.2)+\lambda_n^2]$, $\lambda_n=\sqrt{f_y/f_{cc}}$, $f_{cc}=\pi^2E/\lambda_e^2$ → $f_{cd}=150.4$ N/mm²
- $P_d=A f_{cd}=1373$ kN $\ge P_u=1100$ kN ✓

#### 2. Clear gap for equal stiffness
The centroid of each section is at $x_c=23.6+s/2$ from the column axis. For $I_{yy}\ge I_{xx}$:
$$
x_c\ge\sqrt{\frac{I_{xx}-I_{yy,ch}}{A_{ch}}}=115.2\ \text{mm}\ \Rightarrow\ s\ge183\ \text{mm}
$$
Provide **$s=190$ mm**. Then $I_{yy}=2[I_{yy,ch}+A_{ch}x_c^2]=13461\times10^4$ mm⁴, $r_{yy}=121.4$ mm, $\lambda_{yy}=65.9$, $1.05\lambda_{yy}=69.2$, $P_{dy}=1402$ kN $\ge P_u$ ✓.

#### 3. Design of lacing (single lacing, two planes)
```text
   |\     /|
   | \   / |    theta = 45 deg to the axis
   |  \ /  |    S = distance between connection lines
   |  / \  |
   | /   \ |
```
- Distance between connection lines $S=s+b_f=190+90=280$ mm
- Transverse shear $V=2.5\%\,P_u=27.5$ kN; per lacing plane $V_p=13.75$ kN
- Force in lacing bar $F=\dfrac{V_p}{\sin\theta}=19.45$ kN
- Length of bar $l=\dfrac{S}{\sin\theta}=396$ mm; effective length $=l$ (between the inner end bolts)
- $t\ge l/40=10$ mm; width $\ge3d=60$ mm → **flat 60 × 10 mm**
- $\lambda=\dfrac{l}{t/\sqrt{12}}=137<145$ ✓ (cl. 7.6.3); $f_{cd}=68$ N/mm², $P_c=41.0$ kN $>F$ ✓
- Tension: $T_{dn}=\dfrac{0.9(w-d_0)tf_u}{\gamma_{m1}}=112.2$ kN $>F$ ✓
- Connection: M20 (4.6) $V_{dsb}=45.3$ kN, bearing on 10 mm $=99.4$ kN → **1 bolt at each end**
- Lacing angle $\theta=45°$ lies between 40° and 70° ✓
- Spacing of lacing points on one section $=\dfrac{2S}{\tan\theta}=560$ mm; slenderness of the section between them $=\dfrac{560}{26.1}=21.5\le\min(50,\ 0.7\lambda_{max})$ ✓ (cl. 7.6.3)

#### 4. Tie plates
Tie (end) plates at both ends of the lacing system: length $\ge S+$ edges $=360$ mm, depth $\ge S=280$ mm, thickness $\ge S/50=6$ mm → **360 × 280 × 6 mm** at each end, connected to each section by 2 bolts.

**Answer: 2 ISMC 300 back to back (webs facing, flanges outward), clear gap 190 mm, single lacing of 60 × 10 mm flats at 45°, tie plates 360 × 280 × 6 mm at the ends.**

## Check yourself

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

1. For the laced column, which slenderness is used to find f_cd after computing lambda_xx = KL/r_xx = 67.8?
   A. lambda_xx itself, 67.8
   B. 1.05 x lambda_xx = 71.1
   C. 1.1 x lambda_xx = 74.6
   D. 0.7 x lambda_xx = 47.5
   Answer: B. A laced column uses an effective slenderness of 1.05 times the actual maximum slenderness (cl. 7.6.1.4). 1.1 is for battened columns. (Key point: Effective slenderness of laced column)

2. Which buckling class (alpha) is used for the built-up double channel column?
   A. Class a, alpha = 0.21
   B. Class b, alpha = 0.34
   C. Class d, alpha = 0.76
   D. Class c, alpha = 0.49
   Answer: D. Built-up members are taken as buckling class c with imperfection factor alpha = 0.49 (IS 800 Table 10). (Key point: Buckling class for built-up members)

3. Why is the clear gap s between the channels chosen as 190 mm?
   A. To make I_yy at least equal to I_xx so both axes are equally strong
   B. To make the lacing angle exactly 45 degrees
   C. To keep the tie plate thickness at 6 mm
   D. To reduce the effective length of the column
   Answer: A. The gap is found from I_yy >= I_xx, giving x_c >= 115.2 mm and s >= 183 mm, so 190 mm is provided. (Key point: Clear gap for equal stiffness)

4. With V = 2.5% of P_u = 27.5 kN shared by two lacing planes, what is the force in a lacing bar at 45 degrees?
   A. 27.5 kN
   B. 13.75 kN
   C. 19.45 kN
   D. 38.9 kN
   Answer: C. Per plane V_p = 13.75 kN, so F = V_p / sin 45 = 19.45 kN. (Key point: Lacing bar force)

## Related in Design of Steel and Timber Structure

- [Built-up column, lacing, 8 m, 1100 kN](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abuiltup-lacing-1100kn-8m-fe410) — Design of Steel and Timber Structure, chapter 6, asked 4×
- [Braced built-up column, 3 m, 1600 kN](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abraced-builtup-1600kn-3m) — Design of Steel and Timber Structure, chapter 6, asked 2×
- [Built-up column, single lacing, 900 kN, 5 m](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abuiltup-single-lacing-900kn-5m) — Design of Steel and Timber Structure, chapter 6
- [Built-up column with battens, 4 m, 1000 kN](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abuiltup-battens-1000kn-4m) — Design of Steel and Timber Structure, chapter 6, asked 2×
- [Built-up laced column, 10 m, 1000 kN](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abuiltup-1000kn-10m-lacing) — Design of Steel and Timber Structure, chapter 6, asked 4×
- [Built-up column with battens, 10 m, 1080 kN](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abuiltup-battens-1080kn-10m) — Design of Steel and Timber Structure, chapter 6, asked 4×

## Related in other subjects

- [Column 35x40 cm, not restrained against rotation](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Acolumn-35x40-cm-not-restrained-against) — Design of RCC Structure, chapter 10
- [Brick column design for 80 kN axial load](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Abrick-column-80kn-roof) — Concrete Technology and Masonry Structure, chapter 8
- [Column 400x500, 1600 kN, 4.0 m](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Acolumn-400x500-1600-kn-4-0) — Design of RCC Structure, chapter 10