---
title: "Braced built-up column, 3 m, 1600 kN"
subject: "Design of Steel and Timber Structure"
chapter: 6
key: "semester_6/design_of_steel_and_timber_structure:braced-builtup-1600kn-3m"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/braced-builtup-1600kn-3m
site: BCT Engineering
---

# Braced built-up column, 3 m, 1600 kN

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

- Asked in: 2078 Chaitra, 14 marks
- Video: [Design of Built-up Column as per IS 800-2007 || Laced Columns || Steel Structure || Part-1](https://www.youtube.com/watch?v=J0ZoaF6ltCY) — Let's learn and design
- 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/braced-builtup-1600kn-3m
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-6

## Question

Design of braced built-up column using double channel-section, having unsupported length 3.0m, subjected to factored axial load of 1600 kN. Use single lacing as lattice member and bolted connection to connect lacing and column. The ends of the columns are effectively held in position at both ends but not restrained against rotation.

## Short answer

Use 2 ISMC 250 back to back with 150 mm clear gap. Provide single lacing with 60×10 mm flats at 45°, M20 bolts, and end tie plates 310×230×6 mm.

## Answer

**Given/assumptions:** factored load $P_u=1600$ kN; unsupported length 3.0 m; ends held in position but not restrained in rotation → $K=1.0$, $KL=3$ m. Two channels (back to back), single lacing, bolted (M20, grade 4.6); E250 ($f_y=250$, $f_u=410$).

#### 1. Section: 2 ISMC 250 back to back (webs facing, flanges outward)
$A=2\times3867=7734$ mm², $I_{xx}=2\times3816.8\times10^4$ mm⁴, $r_{xx}=99.3$ mm, $C_{yy}=23.0$ mm, $b_f=80$ mm. Built-up members use **buckling class c** (IS 800:2007 Table 10, $\alpha=0.49$).
- $\lambda_{xx}=\dfrac{KL}{r_{xx}}=\dfrac{3000}{99.3}=30.2$
- Laced column: $\lambda_e=1.05\lambda=31.7$ (cl. 7.6.1.4)
- $f_{cd}=\dfrac{f_y/\gamma_{m0}}{\phi+\sqrt{\phi^2-\lambda_n^2}}$ with $\phi=0.5[1+0.49(\lambda_n-0.2)+\lambda_n^2]$, $\lambda_n=\sqrt{f_y/f_{cc}}$, $f_{cc}=\pi^2E/\lambda_e^2$ → $f_{cd}=209.1$ N/mm²
- $P_d=A f_{cd}=1617$ kN $\ge P_u=1600$ kN ✓

#### 2. Clear gap for equal stiffness
The centroid of each section is at $x_c=23.0+s/2$ from the column axis. For $I_{yy}\ge I_{xx}$:
$$
x_c\ge\sqrt{\frac{I_{xx}-I_{yy,ch}}{A_{ch}}}=96.5\ \text{mm}\ \Rightarrow\ s\ge147\ \text{mm}
$$
Provide **$s=150$ mm**. Then $I_{yy}=2[I_{yy,ch}+A_{ch}x_c^2]=7866\times10^4$ mm⁴, $r_{yy}=100.8$ mm, $\lambda_{yy}=29.7$, $1.05\lambda_{yy}=31.2$, $P_{dy}=1622$ 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 $S=s+b_f=150+80=230$ mm
- Transverse shear $V=2.5\%\,P_u=40.0$ kN; per lacing plane $V_p=20.00$ kN
- Force in lacing bar $F=\dfrac{V_p}{\sin\theta}=28.28$ kN
- Length of bar $l=\dfrac{S}{\sin\theta}=325$ mm; effective length $=l$ (between the inner end bolts)
- $t\ge l/40=8.1$ mm → 10 mm; width $\ge3d=60$ mm → **flat 60 × 10 mm**
- $\lambda=\dfrac{l}{t/\sqrt{12}}=113<145$ ✓ (cl. 7.6.3); $f_{cd}=91$ N/mm², $P_c=54.9$ 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 $=\dfrac{2S}{\tan\theta}=460$ mm; slenderness of the section between them $=\dfrac{460}{23.8}=19.3\le\min(50,\ 0.7\lambda_{max})$ ✓ (cl. 7.6.3)

#### 4. Tie plates
Tie (end) plates at both ends of the lacing system: length $\ge S+$ edges $=310$ mm, depth $\ge S=230$ mm, thickness $\ge S/50=6$ mm → **310 × 230 × 6 mm** at each end, connected to each section by 2 bolts.

**Answer: 2 ISMC 250 back to back (webs facing, flanges outward), clear gap 150 mm, single lacing of 60 × 10 mm flats at 45°, tie plates 310 × 230 × 6 mm at the ends.**

## Check yourself

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

1. Ends are held in position but not restrained against rotation. What is KL for the 3 m column?
   A. KL = 3 m (K = 1.0)
   B. KL = 1.5 m
   C. KL = 6 m
   D. KL = 2.1 m
   Answer: A. Held in position but free to rotate is a hinged condition, so K = 1.0 and KL = 3000 mm. (Key point: Effective length factor)

2. The lacing bar for bolted connection is 60 x 10 mm. What decides this?
   A. t >= l/40 and width >= 3 times the bolt diameter (3 x 20 = 60 mm)
   B. t >= l/10 and width >= bolt diameter
   C. Thickness must equal the flange thickness
   D. Width must equal S
   Answer: A. l = 325 mm so t >= l/40 = 8.1, taken 10 mm; the width must allow bolt edge distances, at least 3d = 60 mm. (Key point: Lacing bar width and thickness)

3. How are the end tie plates sized?
   A. Depth at least S, length at least S plus edge, thickness at least S/50
   B. Depth equals the flange width and thickness 20 mm
   C. Depth at least 2S and thickness S/10
   D. No size rule applies
   Answer: A. With S = 230 mm: depth 230 mm, length 310 mm, thickness S/50 = 4.6, so 6 mm. They hold the lacing system together at the ends. (Key point: Tie plate size)

## Close variants set in other exams

- [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×

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