---
title: "Laced two-channel column, 800 kN, 6 m"
subject: "Design of Steel and Timber Structure"
chapter: 6
key: "semester_6/design_of_steel_and_timber_structure:column-two-channels-800kn-laced"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/column-two-channels-800kn-laced
site: BCT Engineering
---

# Laced two-channel column, 800 kN, 6 m

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

- Asked in: 2068 Bhadra (old course), 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/column-two-channels-800kn-laced
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-6

## Question

Design a column to carry an axial load of 800KN using two channels laced together. The length of the column is 6m and is effectively held in position at both ends but not restrained against rotation.

## Short answer

Two ISMC 225 channels back to back with 130 mm gap between webs; overall width 290 mm. The laced column provides a factored axial capacity of 1016 kN, exceeding the 800 kN load.

## 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) is assumed: $f_y = 250$ N/mm², $f_u = 410$ N/mm², $\gamma_{m0} = 1.10$. The 800 kN load is taken as the factored load.

#### Data

- Factored axial load $P = 800$ kN; effective length $KL = 6.0$ m (both axes; ends held in position but not restrained against rotation, $K = 1.0$, Table 11).
- 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 225** (IS 808): $A = 33.3$ cm² each, $I_{xx} = 2710$ cm⁴, $I_{yy} = 188$ cm⁴, $C_{yy} = 2.31$ cm, $B = 80$ mm, $t_w = 6.5$ mm, $t_f = 12.4$ mm. (2 × ISMC 200 gives only about 795 kN, so it is not enough.)

Total area $A_e = 2 \times 3330 = 6660$ mm². $r_{zz} = \sqrt{I_{xx}/A} = 9.02$ cm.

$$
\begin{aligned} \lambda_{zz} &= \frac{KL}{r_{zz}} = \frac{6000}{90.2} = 66.5 \\ \lambda_{eff} &= 1.05 \times 66.5 = 69.8 \end{aligned}
$$

For $\lambda_{eff} = 69.8$, curve c: $f_{cc} = \pi^2E/\lambda^2 = 405$ N/mm², $\lambda_n = \sqrt{f_y/f_{cc}} = 0.786$, $\phi = 0.5[1 + 0.49(\lambda_n - 0.2) + \lambda_n^2] = 0.952$, $\chi = 0.671$.

$$
\begin{aligned} f_{cd} &= \chi f_y/\gamma_{m0} = 0.671 \times 250/1.1 = 152.5\ \text{N/mm}^2 \\ P_d &= A_e f_{cd} = 6660 \times 152.5/10^3 = 1016\ \text{kN} > 800\ \text{kN} \end{aligned}
$$

#### 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:

$$
\begin{aligned} 2\left[I_{y} + A\left(\frac{h}{2}\right)^2\right] &\ge 2I_{x} \\ h &\ge 2\sqrt{\frac{I_x - I_y}{A}} = 2\sqrt{\frac{(2710 - 188) \times 10^4}{3330}} = 174\ \text{mm} \end{aligned}
$$

Gap between webs $g = h - 2C_{yy} = 174 - 46.2 = 128$ mm. Provide **g = 130 mm** (centroid distance $h = 176$ mm).

$$
\begin{aligned} I_{yy} &= 2[188 + 33.3(8.81)^2] = 5545\ \text{cm}^4 \\ r_{yy} &= \sqrt{I_{yy}/2A} = 9.12\ \text{cm}, \quad \lambda_{yy} = 6000/91.2 = 65.8 \end{aligned}
$$

$\lambda_{yy} = 65.8 \le \lambda_{zz} = 66.5$, so the zz-axis governs and the capacity $P_d = 1016$ kN $\ge 800$ kN. **Safe.** Overall width of column $= 2B + g = 290$ 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): $V_t = 2.5\% P = 0.025 \times 800 = 20.0$ kN, shared by two lacing planes: $10.00$ kN each.

Bolt lines are taken at mid-width of the flanges, so the transverse distance between lacing connections is $s = g + B = 130 + 80 = 210$ mm. Lacing angle $\theta = 45^\circ$ to the column axis (permitted 40° to 70°, cl. 7.6.4).

$$
\begin{aligned} \text{Length of bar } l &= s/\sin\theta = 210/\sin 45^\circ = 297\ \text{mm} \\ \text{Force in bar } F &= \frac{V_t/2}{\sin\theta} = \frac{10.00}{\sin 45^\circ} = 14.14\ \text{kN} \end{aligned}
$$

Spacing of connections along one channel $a_1 = 2s\cot\theta = 420$ mm. Check (cl. 7.6.5.1): $a_1/r_{min} = 420/23.8 = 17.7$, which is less than the smaller of 50 and $0.7\lambda_{max} = 46.6$. OK.

**Flat size.** Width $\ge 3d = 48$ mm for M16 bolts; take 60 mm. Thickness $\ge l/40 = 7.4$ mm (cl. 7.6.3), and for $KL/r \le 145$, $t \ge l\sqrt{12}/145 = 7.1$ mm. Provide **ISF 60 × 8 mm**.

$r = t/\sqrt{12} = 2.31$ mm, $l/r = 128.6 < 145$ OK. $f_{cd}$ (curve c) $= 75.5$ N/mm², $P_{d,lace} = 480 \times 75.5/10^3 = 36.2$ kN $> F = 14.14$ 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 $= 14.14/29.0 = 0.49$, 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 $> 14.14$ 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) $\ge h = 176$ mm; thickness $\ge s/50 = 4.2$ mm. Use 3 M16 bolts on each channel at 90 mm pitch (outermost bolts 180 mm apart) with 30 mm end distance, so the plate is **290 mm wide × 240 mm deep × 6 mm thick**.

With $C = 420$ mm, $N=2$: $V_b = V_tC/(NS) = 20.0$ kN and $M = V_tC/2N = 2100$ kN·mm. Bolt forces: from $M$ $= 2100 \times 90/(2 \times 90^2) = 11.7$ kN, from shear $= 20.0/3 = 6.7$ kN, resultant $= 13.4$ kN $\le 29.0$ kN. OK.

#### Final design

- Column: 2 × ISMC 225 back to back, gap 130 mm, $P_d = 1016$ kN $> 800$ kN.
- Lacing: single, flats 60 × 8 mm at 45°, 1 M16 bolt at each end.
- End tie plates 290 × 240 × 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. The load is 800 kN, but the section capacity is P_d = 1016 kN. What is the correct conclusion?
   A. The section is unsafe
   B. The section is safe, though conservative
   C. The lacing must be removed
   D. The effective length must be doubled
   Answer: B. P_d = 1016 kN exceeds 800 kN, so the 2 ISMC 225 section is safe. (Key point: Capacity versus load)

2. The ends are held in position but not restrained against rotation. What are K and KL?
   A. K = 0.5, KL = 3 m
   B. K = 2.0, KL = 12 m
   C. K = 1.0, KL = 6 m
   D. K = 0.8, KL = 4.8 m
   Answer: C. For this end condition K = 1.0 (Table 11), so KL = 6.0 m. (Key point: Effective length)

3. What is checked for the slenderness a_1/r_min of a channel between lacing connections?
   A. It must exceed 50
   B. It must not exceed the smaller of 50 and 0.7 lambda_max
   C. It must equal lambda_zz
   D. It must be below 10
   Answer: B. Cl. 7.6.5.1 limits a_1/r_min to the lesser of 50 and 0.7 lambda_max; here 17.7 < 46.6. (Key point: Check between lacing points)

4. For the end tie plate, bolt force from moment is 21.0 kN and from shear 6.7 kN. How is the bolt checked?
   A. Add them arithmetically and compare with 14 kN
   B. Take the resultant sqrt(21.0^2 + 6.7^2) = 22.0 kN against the bolt value 29.0 kN
   C. Use only the moment force
   D. Use only the shear force
   Answer: B. The two forces act at an angle and combine as a vector resultant, 22.0 kN, which is within the 29.0 kN bolt value. (Key point: Tie plate bolt check)

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