---
title: "Built-up column, single lacing, 900 kN, 5 m"
subject: "Design of Steel and Timber Structure"
chapter: 6
key: "semester_6/design_of_steel_and_timber_structure:builtup-single-lacing-900kn-5m"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/builtup-single-lacing-900kn-5m
site: BCT Engineering
---

# Built-up column, single lacing, 900 kN, 5 m

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

- Asked in: 2071 Bhadra, 14 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-single-lacing-900kn-5m
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-6

## Question

Design a built-up column of the effective length of 5 m to carry an axial load of 900 kN using two channels and single lacing. Design the connections using bolt. The grade of the steel is Fe410.

## Short answer

Two ISMC 200 channels back-to-back with 110 mm web gap (overall width 260 mm). Single lacing ISF 60×8 mm flats at 45° in two planes, connection spacing 370 mm, with M16 bolts. Capacity 909 kN > 900 kN, safe.

## Answer

**Approach.** Laced column of two channels (back to back), designed to IS 800:2007 (cl. 7.1.2 for the compressive strength, cl. 7.6 for laced columns). Steel Fe 410 (E250): $f_y = 250$ N/mm², $f_u = 410$ N/mm², $\gamma_{m0} = 1.10$.

#### Data

- Factored axial load $P = 900$ kN; effective length $KL = 5.0$ m (given, same for both axes).
- Built-up member: buckling curve **c**, $\alpha = 0.49$ (Table 10 and Table 7). Effective slenderness of a laced column $= 1.05 \times$ actual maximum slenderness (cl. 7.6.1.5).

#### Step 1: Trial section

Try 2 × **ISMC 200** (IS 808): $A = 28.5$ cm² each, $I_{xx} = 1830$ cm⁴, $I_{yy} = 141$ cm⁴, $C_{yy} = 2.20$ cm, $B = 75$ mm, $t_w = 6.2$ mm, $t_f = 11.4$ mm.

Total area $A_e = 2 \times 2850 = 5700$ mm². $r_{zz} = \sqrt{I_{xx}/A} = 8.01$ cm.

$$
\begin{aligned} \lambda_{zz} &= \frac{KL}{r_{zz}} = \frac{5000}{80.1} = 62.4 \\ \lambda_{eff} &= 1.05 \times 62.4 = 65.5 \end{aligned}
$$

For $\lambda_{eff} = 65.5$, curve c: $f_{cc} = \pi^2E/\lambda^2 = 460$ N/mm², $\lambda_n = \sqrt{f_y/f_{cc}} = 0.737$, $\phi = 0.5[1 + 0.49(\lambda_n - 0.2) + \lambda_n^2] = 0.903$, $\chi = 0.701$.

$$
\begin{aligned} f_{cd} &= \chi f_y/\gamma_{m0} = 0.701 \times 250/1.1 = 159.4\ \text{N/mm}^2 \\ P_d &= A_e f_{cd} = 5700 \times 159.4/10^3 = 909\ \text{kN} > 900\ \text{kN} \end{aligned}
$$

#### Step 2: Spacing of channels

To make the column equally strong about the other axis, $I_{yy} \ge I_{zz}$ (cl. 7.6.1.1). With $h$ the distance between the centroids of the two channels:

$$
\begin{aligned} 2\left[I_{y} + A\left(\frac{h}{2}\right)^2\right] &\ge 2I_{x} \\ h &\ge 2\sqrt{\frac{I_x - I_y}{A}} = 2\sqrt{\frac{(1830 - 141) \times 10^4}{2850}} = 154\ \text{mm} \end{aligned}
$$

Gap between webs $g = h - 2C_{yy} = 154 - 44.0 = 110$ mm. Provide **g = 110 mm** (centroid distance $h = 154$ mm).

$$
\begin{aligned} I_{yy} &= 2[141 + 28.5(7.70)^2] = 3662\ \text{cm}^4 \\ r_{yy} &= \sqrt{I_{yy}/2A} = 8.01\ \text{cm}, \quad \lambda_{yy} = 5000/80.1 = 62.4 \end{aligned}
$$

$\lambda_{yy} = 62.4 \le \lambda_{zz} = 62.4$, so the zz-axis governs and the capacity $P_d = 909$ kN $\ge 900$ kN. **Safe.** Overall width of column $= 2B + g = 260$ mm.

```text
  +-------+   gap g   +-------+
  | ISMC  |<--------->| ISMC  |
  +-------+           +-------+
  webs face each other, flanges point outward
  lacing bolted to the flanges (top and bottom)
```

#### Step 3: Lacing system (single lacing, two planes)

Transverse shear (cl. 7.6.6.1): $V_t = 2.5\% P = 0.025 \times 900 = 22.5$ kN, shared by two lacing planes: $11.25$ kN each.

Bolt lines are taken at mid-width of the flanges, so the transverse distance between lacing connections is $s = g + B = 110 + 75 = 185$ mm. Lacing angle $\theta = 45^\circ$ to the column axis (permitted 40° to 70°, cl. 7.6.4).

$$
\begin{aligned} \text{Length of bar } l &= s/\sin\theta = 185/\sin 45^\circ = 262\ \text{mm} \\ \text{Force in bar } F &= \frac{V_t/2}{\sin\theta} = \frac{11.25}{\sin 45^\circ} = 15.91\ \text{kN} \end{aligned}
$$

Spacing of connections along one channel $a_1 = 2s\cot\theta = 370$ mm. Check (cl. 7.6.5.1): $a_1/r_{min} = 370/22.2 = 16.6$, which is less than the smaller of 50 and $0.7\lambda_{max} = 43.7$. OK.

**Flat size.** Width $\ge 3d = 48$ mm for M16 bolts; take 60 mm. Thickness $\ge l/40 = 6.5$ mm (cl. 7.6.3), and for $KL/r \le 145$, $t \ge l\sqrt{12}/145 = 6.3$ mm. Provide **ISF 60 × 8 mm**.

$r = t/\sqrt{12} = 2.31$ mm, $l/r = 113.3 < 145$ OK. $f_{cd}$ (curve c) $= 90.8$ N/mm², $P_{d,lace} = 480 \times 90.8/10^3 = 43.6$ kN $> F = 15.91$ kN. OK.

#### Step 4: Connection of lacing (M16, grade 4.6 bolts)

$$
\begin{aligned} V_{dsb} &= \frac{f_{ub}}{\sqrt3}\frac{0.78\,\pi d^2/4}{\gamma_{mb}} = \frac{400}{\sqrt3} \times \frac{157}{1.25 \times 10^3} = 29.0\ \text{kN} \\ V_{dpb} &= \frac{2.5\,k_b\, d\, t\, f_u}{\gamma_{mb}} = \frac{2.5 \times 0.556 \times 16 \times 8 \times 410}{1.25 \times 10^3} = 58.3\ \text{kN} \end{aligned}
$$

($k_b = e/3d_0 = 30/54 = 0.556$ with edge distance 30 mm $\ge 1.5d_0$.) Bolt value $= 29.0$ kN. Bolts needed $= 15.91/29.0 = 0.55$, so provide **1 bolt M16** at each end of every bar.

Net section of flat in tension: $A_n = (60 - 18) \times 8 = 336$ mm², $T_{dn} = 0.9 f_u A_n/\gamma_{m1} = 99.2$ kN $> 15.91$ kN. OK.

#### Step 5: End tie plates

Tie plates are provided at the ends and designed like battens (cl. 7.7.2): effective depth (distance between outermost bolts) $\ge h = 154$ mm; thickness $\ge s/50 = 3.7$ mm. Use 3 M16 bolts on each channel at 80 mm pitch (outermost bolts 160 mm apart) with 30 mm end distance, so the plate is **260 mm wide × 220 mm deep × 6 mm thick**.

With $C = 370$ mm, $N=2$: $V_b = V_tC/(NS) = 22.5$ kN and $M = V_tC/2N = 2081$ kN·mm. Bolt forces: from $M$ $= 2081 \times 80/(2 \times 80^2) = 13.0$ kN, from shear $= 22.5/3 = 7.5$ kN, resultant $= 15.0$ kN $\le 29.0$ kN. OK.

#### Final design

- Column: 2 × ISMC 200 back to back, gap 110 mm, $P_d = 909$ kN $> 900$ kN.
- Lacing: single, flats 60 × 8 mm at 45°, 1 M16 bolt at each end.
- End tie plates 260 × 220 × 6 mm, 3 M16 bolts per channel.

```text
  +--------------------+   <- end tie plate
  | \    /\    /\    /|
  |  \  /  \  /  \  / |   single lacing
  |   \/    \/    \/  |   (same direction
  |   /\    /\    /\  |    on both faces)
  +--------------------+   <- end tie plate
```

## Check yourself

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

1. For 2 ISMC 200, what P_d is found and what does it show?
   A. P_d = 909 kN, slightly above 900 kN, so safe
   B. P_d = 900 kN exactly from f_y A
   C. P_d = 1100 kN, much over-designed
   D. P_d = 800 kN, so unsafe
   Answer: A. lambda_eff = 65.5 gives f_cd = 159.4 N/mm^2, so P_d = 5700 x 159.4 = 909 kN > 900 kN. (Key point: Strength of trial section)

2. After providing a 110 mm gap, lambda_yy = 62.4 equals lambda_zz = 62.4. What does this mean?
   A. The column is unsafe about the yy axis
   B. The column buckles equally about both axes, so capacity is unchanged
   C. The lacing must be doubled
   D. The gap must be increased
   Answer: B. The gap makes I_yy >= I_zz, so slenderness about both axes is equal and the zz-axis value governs P_d. (Key point: Which axis governs)

3. Bar length is 262 mm. Which checks fix the 60 x 8 mm flat?
   A. t >= S/50 only
   B. t >= l/40 = 6.5 mm and slenderness l/r <= 145
   C. t >= 2.5% of load
   D. t equals the tie plate thickness
   Answer: B. Thickness must be at least l/40 (6.5 mm) and slenderness l/r <= 145, so 8 mm is used with l/r = 113. (Key point: Lacing flat thickness)

4. How are the end tie plates designed?
   A. Like lacing bars, by axial force only
   B. Like battens, with effective depth >= h and thickness >= s/50
   C. They need no design
   D. Thickness at least l/40
   Answer: B. Tie plates are designed like battens (cl. 7.7.2): depth >= h = 154 mm and t >= s/50 = 3.7 mm; 160 x 6 mm is provided. (Key point: Tie plate design)

## Related in Design of Steel and Timber Structure

- [Built-up column, single lacing, 8 m, 1100 kN](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abuiltup-lacing-1100kn-8m-e250) — Design of Steel and Timber Structure, chapter 6
- [Built-up column, two channels toe-to-toe, 1500 kN](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abuiltup-1500kn-10m-toe-to-toe) — Design of Steel and Timber Structure, chapter 6
- [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×
- [Built-up column with battens, 6 m, 1200 kN](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abuiltup-battens-1200kn-6m) — Design of Steel and Timber Structure, chapter 6
- [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 laced column 7.5 m, 1250 kN](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abuiltup-laced-1250kn-7-5m) — Design of Steel and Timber Structure, chapter 6

## Related in other subjects

- [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
- [Safe axial load on a timber column](https://www.bctengineer.com/question.md?key=semester_5%2Fdesign_of_timber_and_masonry_structures%3Aaxial-capacity-timber-column) — Design of Timber and Masonry Structures, chapter 3
- [Corner stresses of 500 x 900 column (300 kN)](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Acolumn-stress-500x900-300kn) — Concrete Technology and Masonry Structure, chapter 8, asked 7×