---
title: "Built-up column, two channels toe-to-toe, 1500 kN"
subject: "Design of Steel and Timber Structure"
chapter: 6
key: "semester_6/design_of_steel_and_timber_structure:builtup-1500kn-10m-toe-to-toe"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/builtup-1500kn-10m-toe-to-toe
site: BCT Engineering
---

# Built-up column, two channels toe-to-toe, 1500 kN

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

- Asked in: 2081 Chaitra, 14 marks
- Video: [Design of laced column problem# Channel section placed back to back#Bolted connection#](https://www.youtube.com/watch?v=W1lPuvbOmB4) — CIVIL ENGINEERING COURSE Theory & Practical
- 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-1500kn-10m-toe-to-toe
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-6

## Question

Design a built-up column consisting of two channels placed toe-to-toe. The column carries an axial factored load of 1500 kN. The effective length of column is 10 m. Also, design the lacing system using welded connection. Take Fe 415 grade steel.

## Short answer

Use 2 ISMC 350 toe to toe with 120 mm clear gap; laced column capacity about 1515 kN. Single lacing 50 x 8 mm flats at 45 degrees with 5 mm fillet welds, and end tie plates 300 x 220 x 6 mm.

## Answer

**Given/assumptions:** factored load $P_u=1500$ kN; effective length $KL=10$ m (given); two channels toe to toe with **single lacing and welded connection**. "Fe 415" is read as ordinary structural steel with $f_y=250$ N/mm², $f_u=410$ N/mm² (IS 2062 E250); if $f_y=415$ were intended, redo the design with that value. Shop fillet welds.

#### 1. Section: 2 ISMC 350 toe to toe (flanges facing, webs outward)
$A=2\times5366=10732$ mm², $I_{xx}=2\times10008.0\times10^4$ mm⁴, $r_{xx}=136.6$ mm, $C_{yy}=24.4$ mm, $b_f=100$ mm. Built-up members use **buckling class c** (IS 800:2007 Table 10, $\alpha=0.49$).
- $\lambda_{xx}=\dfrac{KL}{r_{xx}}=\dfrac{10000}{136.6}=73.2$
- Laced column: $\lambda_e=1.05\lambda=76.9$ (cl. 7.6.1.4)
- $f_{cd}=\dfrac{f_y/\gamma_{m0}}{\phi+\sqrt{\phi^2-\lambda_n^2}}$ with $\phi=0.5[1+0.49(\lambda_n-0.2)+\lambda_n^2]$, $\lambda_n=\sqrt{f_y/f_{cc}}$, $f_{cc}=\pi^2E/\lambda_e^2$ → $f_{cd}=141.2$ N/mm²
- $P_d=A f_{cd}=1515$ kN $\ge P_u=1500$ kN ✓

#### 2. Clear gap for equal stiffness
The centroid of each section is at $x_c=75.6+s/2$ from the column axis. For $I_{yy}\ge I_{xx}$:
$$
x_c\ge\sqrt{\frac{I_{xx}-I_{yy,ch}}{A_{ch}}}=133.6\ \text{mm}\ \Rightarrow\ s\ge116\ \text{mm}
$$
Provide **$s=120$ mm**. Then $I_{yy}=2[I_{yy,ch}+A_{ch}x_c^2]=20595\times10^4$ mm⁴, $r_{yy}=138.5$ mm, $\lambda_{yy}=72.2$, $1.05\lambda_{yy}=75.8$, $P_{dy}=1534$ 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 $S=s+b_f=120+100=220$ mm
- Transverse shear $V=2.5\%\,P_u=37.5$ kN; per lacing plane $V_p=18.75$ kN
- Force in lacing bar $F=\dfrac{V_p}{\sin\theta}=26.52$ kN
- Length of bar $l=\dfrac{S}{\sin\theta}=311$ mm; effective length for welded lacing $=0.7l=218$ mm
- $t\ge l_e/40=5.5$ mm; a 6 mm flat fails in compression (23.4 kN $<F$), so adopt 8 mm; width 50 mm → **flat 50 × 8 mm**
- $\lambda=\dfrac{l_e}{t/\sqrt{12}}=94<145$ ✓; $f_{cd}=115$ N/mm², $P_c=45.9$ kN $>F$ ✓
- Weld: 5 mm shop fillet; strength $=\dfrac{410}{\sqrt3\times1.25}\times0.7\times5=662.8$ N/mm → length required $=\dfrac{F}{q}=40$ mm; provide **55 mm of weld at each end** (not less than the bar width and $4s$), returned round the ends
- Lacing angle $\theta=45°$ lies between 40° and 70° ✓
- Spacing of lacing points on one section $=\dfrac{2S}{\tan\theta}=440$ mm; slenderness of the section between them $=\dfrac{440}{28.3}=15.5\le\min(50,\ 0.7\lambda_{max})$ ✓ (cl. 7.6.3)

#### 4. Tie plates
Tie (end) plates at both ends of the lacing system: length $\ge S+$ edges $=300$ mm, depth $\ge S=220$ mm, thickness $\ge S/50=6$ mm → **300 × 220 × 6 mm** at each end, fillet-welded to each section.

**Answer: 2 ISMC 350 toe to toe (flanges facing, webs outward), clear gap 120 mm, single lacing of 50 × 8 mm flats at 45° (welded), tie plates 300 × 220 × 6 mm at the ends.**

## Check yourself

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

1. For a laced column the slenderness about the x-x axis is 73.2. What value is used to find f_cd, and why?
   A. 73.2, unchanged
   B. 1.05 x 73.2 = 76.9, to allow for shear deformation in lacing
   C. 0.7 x 73.2 = 51.2, since lacing stiffens the column
   D. 1.1 x 73.2 = 80.5, as for battens
   Answer: B. IS 800 cl. 7.6.1.4 increases laced column slenderness by 5 percent: lambda_e = 1.05 lambda = 76.9. (Key point: Effective slenderness for laced column)

2. What condition fixes the clear gap s between the channels?
   A. I_yy must be at least I_xx so the column is not weaker about the y-axis
   B. The gap must equal the flange width
   C. I_yy must be exactly half of I_xx
   D. The gap must be 50 mm as a minimum
   Answer: A. The gap is chosen so that I_yy >= I_xx. This gave s >= 116 mm and s = 120 mm was provided. (Key point: Clear gap for equal stiffness)

3. How is the force in the lacing bar found?
   A. Transverse shear 2.5% of P_u, shared by the two lacing planes, divided by sin(theta)
   B. Full axial load divided by cos(theta)
   C. 2.5% of P_u divided by tan(theta) with one plane only
   D. Total shear multiplied by sin(theta)
   Answer: A. V = 2.5% x 1500 = 37.5 kN, V_p = 18.75 kN per plane, and F = V_p/sin 45 = 26.52 kN. (Key point: Lacing force and bar size)

4. Which statement about lacing geometry and end tie plates is correct?
   A. Lacing angle must be between 40 and 70 degrees; tie plates are provided at both ends of the lacing system
   B. Lacing angle must be below 30 degrees; tie plates are not needed
   C. Lacing angle must be 90 degrees; tie plates go only mid-height
   D. Lacing angle may be anything; tie plates thickness is 20 mm
   Answer: A. The 45 degree lacing lies between 40 and 70 degrees. Tie plates 300 x 220 x 6 mm are provided at each end, with thickness at least S/50. (Key point: Lacing angle and tie plates)

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