---
title: "Built-up laced column 7.5 m, 1250 kN"
subject: "Design of Steel and Timber Structure"
chapter: 6
key: "semester_6/design_of_steel_and_timber_structure:builtup-laced-1250kn-7-5m"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/builtup-laced-1250kn-7-5m
site: BCT Engineering
---

# Built-up laced column 7.5 m, 1250 kN

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

- Asked in: 2073 Bhadra, 15 marks
- Video: [Design of double Laced column | Steel column connection details | 3d animation](https://www.youtube.com/watch?v=_omCPfxJGGM) — 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-laced-1250kn-7-5m
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-6

## Question

A 7.5m long built-up and laced column has to carry a factored axial load of 1250KN. The column is restrained in position but not in direction at each end. Design the column with single lacing system. Connection shall consist of two channels placed back to back at a suitable spacing.

## Short answer

With KL=7.5 m, use 2 ISMC 300 back to back, gap 190 mm (λe=66.7, fcd=157.5 N/mm², Pd=1438 kN). Single lacing: 60×10 mm flats at 45°, 1 M20 bolt per end. End tie plates 360×280×6 mm.

## Answer

**Given/assumptions:** factored load $P_u=1250$ kN; length 7.5 m, restrained in position but not in direction at each end → $K=1.0$, $KL=7.5$ m. Two channels back to back at a suitable spacing with single lacing; bolted connections (M20, grade 4.6); steel 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, $\alpha=0.49$).
- $\lambda_{xx}=\dfrac{KL}{r_{xx}}=\dfrac{7500}{118.1}=63.5$
- Laced column: $\lambda_e=1.05\lambda=66.7$ (cl. 7.6.1.4)
- $f_{cd}=\dfrac{f_y/\gamma_{m0}}{\phi+\sqrt{\phi^2-\lambda_n^2}}$ with $\phi=0.5[1+0.49(\lambda_n-0.2)+\lambda_n^2]$, $\lambda_n=\sqrt{f_y/f_{cc}}$, $f_{cc}=\pi^2E/\lambda_e^2$ → $f_{cd}=157.5$ N/mm²
- $P_d=A f_{cd}=1438$ kN $\ge P_u=1250$ 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, $\lambda_{yy}=61.8$, $1.05\lambda_{yy}=64.8$, $P_{dy}=1465$ 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 $V=2.5\%\,P_u=31.25$ kN; per lacing plane $V_p=15.62$ kN
- Force in lacing bar $F=\dfrac{V_p}{\sin\theta}=22.10$ 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. The column is restrained in position but not in direction at each end. What effective length is used?
   A. 0.65 x 7.5 m
   B. 1.0 x 7.5 m = 7.5 m
   C. 2.0 x 7.5 m
   D. 0.5 x 7.5 m
   Answer: B. Held in position but not restrained against rotation gives K = 1.0, so KL = 7.5 m. (Key point: Effective length)

2. How is the column's design strength P_d obtained and compared?
   A. P_d = A x f_y and compared with P_u
   B. P_d = A x f_cd with f_cd from the effective slenderness 1.05 lambda; P_d = 1438 kN >= 1250 kN
   C. P_d = 2.5% of P_u compared with the shear
   D. P_d = I_xx / r_xx
   Answer: B. f_cd depends on slenderness (curve c), so P_d = A f_cd = 1438 kN, which is at least P_u = 1250 kN. (Key point: Strength check of the section)

3. How is the transverse distance S between the lacing connection lines found?
   A. S = s + b_f = 190 + 90 = 280 mm
   B. S = s = 190 mm
   C. S = 2s = 380 mm
   D. S = b_f = 90 mm
   Answer: A. Bolt lines are at mid-flange, so S = clear gap plus one flange width = 280 mm. (Key point: Distance between connection lines)

4. Why is the lacing flat chosen as 60 x 10 mm?
   A. Thickness >= l/40 = 10 mm and width >= 3d = 60 mm for M20 bolts
   B. Thickness >= S/50 and width >= 2d
   C. Thickness is fixed at 6 mm by tie plates
   D. Width must equal the channel flange width
   Answer: A. Bar length is 396 mm so t >= l/40 = 10 mm, and width >= 3 x 20 = 60 mm. (Key point: Lacing bar size)

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