---
title: "Built-up column with battens, 10 m, 1080 kN"
subject: "Design of Steel and Timber Structure"
chapter: 6
key: "semester_6/design_of_steel_and_timber_structure:builtup-battens-1080kn-10m"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/builtup-battens-1080kn-10m
site: BCT Engineering
---

# Built-up column with battens, 10 m, 1080 kN

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

- Asked in: 2073 Magh, 15 marks
- Video: [DSS || Lecture : 07 || Lacing & Battens in Comp. Members || General Requirements ||](https://www.youtube.com/watch?v=uHkhor0mAg8) — VijayPath Civil Academy ( Amravati )
- 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-battens-1080kn-10m
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-6

## Question

Design a built up column 10 m long to carry a factored axial compressive load of 1080 kN. The column is restrained in position but not in direction at both ends. Design the column with connecting system as battens with bolted connection. Use two channel back to back assume steel of grade Fe 410, E250A and bolts grade 4.6.

## Short answer

Two ISMC 350 channels back-to-back with 220 mm web gap (overall width 420 mm). Batten plates 420×330×8 mm at 909 mm spacing (11 bays) using five M16 grade 4.6 bolts per connection. Compressive capacity 1463 kN > 1080 kN, safe.

## Answer

**Approach.** Battened column of two channels, designed to IS 800:2007 (cl. 7.1.2 for compressive strength and cl. 7.7 for battened columns). Fe 410 (E250): $f_y = 250$ N/mm², $f_u = 410$ N/mm²; bolts M16 grade 4.6 ($f_{ub} = 400$ N/mm²), $\gamma_{m0} = 1.10$, $\gamma_{mb} = 1.25$.

#### Data

- $P = 1080$ kN (factored). Ends restrained in position but not in direction, so $KL = 1.0 \times 10$ m $= 10000$ mm (Table 11).
- Buckling curve c, $\alpha = 0.49$ (built-up member, Table 10). Effective slenderness of a battened column $= 1.1 \times$ actual maximum slenderness (cl. 7.7.1.4).

#### Step 1: Trial section

Try 2 × **ISMC 350**: $A = 54.4$ cm² each, $I_{xx} = 10000$ cm⁴, $I_{yy} = 434$ cm⁴, $C_{yy} = 2.44$ cm, $B = 100$ mm, $t_w = 8.3$ mm, $t_f = 13.5$ mm, $r_{y} = 2.82$ cm.

$$
\begin{aligned} r_{zz} &= \sqrt{I_{xx}/A} = 13.56\ \text{cm} \\ \lambda_{zz} &= 10000/135.6 = 73.8 \\ \lambda_{eff} &= 1.1 \times 73.8 = 81.1 \end{aligned}
$$

Curve c at $\lambda_{eff} = 81.1$: $f_{cc} = 300$ N/mm², $\lambda_n = 0.913$, $\phi = 1.092$, $\chi = 0.592$.

$$
\begin{aligned} f_{cd} &= 0.592 \times 250/1.1 = 134.5\ \text{N/mm}^2 \\ P_d &= 2A f_{cd} = 10880 \times 134.5/10^3 = 1463\ \text{kN} > 1080\ \text{kN} \end{aligned}
$$

#### Step 2: Spacing of channels

For $I_{yy} \ge I_{zz}$: $h \ge 2\sqrt{(I_x - I_y)/A} = 2\sqrt{(10000 - 434) \times 10^4/5440} = 265$ mm.

Gap between webs $g = h - 2C_{yy} = 265 - 48.8 = 216$ mm; provide **g = 220 mm**, so $h = 269$ mm.

$I_{yy} = 2[434 + 54.4(13.44)^2] = 20521$ cm⁴, $r_{yy} = 13.73$ cm, $\lambda_{yy} = 72.8$. This is not more than $\lambda_{zz} = 73.8$, so the capacity $P_d = 1463$ kN is safe. Overall width $= 2B + g = 420$ mm.

#### Step 3: Spacing of battens (cl. 7.7.3)

The slenderness of one channel between batten connections must not exceed 50 or $0.7\lambda_{max} = 0.7 \times 73.8 = 51.6$. So $\lambda_1 \le 50.0$, and with $r_{min} = r_y = 28.2$ mm:

$$
\begin{aligned} C &\le 50.0 \times 28.2 = 1410\ \text{mm} \\ \text{Bays} &\ge 10000/1410 = 7.1 \to 8\ (\ge 3) \end{aligned}
$$

The main-member check of Step 7 (bending from the batten moment) needs a smaller spacing, so adopt **11 bays**: $C = 10000/11 = 909$ mm, $\lambda_1 = 909/28.2 = 32.2$.

#### Step 4: Forces on battens (cl. 7.7.2.1)

Transverse shear $V_t = 0.025 \times 1080 = 27.0$ kN, $N = 2$ planes. Bolt lines are at mid-flange, so $S = g + B = 220 + 100 = 320$ mm.

$$
\begin{aligned} V_b &= \frac{V_t C}{N S} = \frac{27.0 \times 909}{2 \times 320} = 38.4\ \text{kN} \\ M &= \frac{V_t C}{2N} = \frac{27.0 \times 909}{4} = 6136\ \text{kN·mm} \end{aligned}
$$

#### Step 5: Size of battens (cl. 7.7.2.3)

The effective depth of a batten is the longitudinal distance between its outermost bolts. End battens: effective depth $\ge h = 269$ mm; intermediate battens $\ge 0.75h = 202$ mm; in no case $< 2B = 200$ mm. Use the same size throughout: effective depth 270 mm (outermost bolts 270 mm apart), overall depth $= 270 + 2 \times 30 = 330$ mm.

Thickness $\ge S/50 = 320/50 = 6.4$ mm; provide **8 mm**. Batten plate: **420 mm × 330 mm × 8 mm**.

Plate check: $\sigma_b = M/Z = 6136364/145200 = 42.3$ N/mm² $< f_y/\gamma_{m0} = 227$; $\tau = V_b/(t\,d) = 14.5$ N/mm² $< f_y/(\sqrt3\gamma_{m0}) = 131$. OK.

#### Step 6: Bolts (M16, grade 4.6)

$V_{dsb} = 29.0$ kN (single shear, threaded), $V_{dpb} = 58.3$ kN ($t = 8$ mm, $k_b = 0.556$). Bolt value $= 29.0$ kN.

Try 5 bolts in one line on each channel at a pitch of 67.5 mm, so the outermost bolts are $4 \times 67.5 = 270$ mm apart (= the effective depth). Bolt distances from the centre of the group: 0, ±67.5, ±135 mm, $\Sigma r^2 = 2(67.5^2 + 135^2) = 45563$ mm². Force on the farthest bolt due to $M$ $= M r_{max}/\Sigma r^2 = 6136 \times 135/45563 = 18.2$ kN; due to shear $= V_b/n = 38.4/5 = 7.7$ kN.

$$
R = \sqrt{18.2^2 + 7.7^2} = 19.7\ \text{kN} \le 29.0\ \text{kN}
$$

Provide **5 bolts M16** per channel at each batten. Edge distance 30 mm, pitch 67.5 mm ($\ge 2.5d = 40$ mm).

#### Step 7: Check of the main member (cl. 7.7.2.1)

Each channel carries $P/2 = 540$ kN with a local moment $M = 6.14$ kN·m about its own $y$-axis. $\lambda_1 = 32.2$, $f_{cd} = 208.4$ N/mm², $P_{d1} = 5440 \times 208.4/10^3 = 1134$ kN. $M_d = Z_y f_y/\gamma_{m0} = 57.3 \times 10^3 \times 250/1.1 = 13.02$ kN·m ($Z_y = 57.3$ cm³).

$$
\frac{P/2}{P_{d1}} + \frac{M}{M_d} = \frac{540}{1134} + \frac{6.14}{13.02} = 0.95 < 1.0
$$

The main member is safe.

#### Final design

- 2 × ISMC 350 back to back, gap 220 mm, $P_d = 1463$ kN $> 1080$ kN.
- 11 equal bays of 909 mm; battens 420 × 330 × 8 mm at both ends and at every bay point, each connected by 5 M16 bolts (pitch 67.5 mm) to each channel.

```text
  |==|            |==|
  |  |------------|  |  <- end batten
  |  |            |  |
  |  |------------|  |  <- batten
  |  |            |  |  (bay C = 909 mm)
  |  |------------|  |  <- batten
  |==|            |==|
  channel     channel
```

## Check yourself

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

1. What factor multiplies the maximum actual slenderness of a battened column to get its effective slenderness?
   A. 1.0
   B. 1.05
   C. 1.1
   D. 1.5
   Answer: C. For battened columns the effective slenderness is 1.1 times the actual maximum slenderness (cl. 7.7.1.4); 1.05 is for laced columns. (Key point: Effective slenderness of battened column)

2. What limit governs the slenderness of one channel between batten connections?
   A. At most 145
   B. At most 180
   C. At most the smaller of 50 and 0.7 x maximum slenderness of the whole column
   D. At least 50
   Answer: C. Cl. 7.7.3 limits it to 50 or 0.7 lambda_max, whichever is less, which gives the maximum batten spacing C. (Key point: Spacing of battens)

3. The slenderness limit alone needs 8 bays, but 11 bays are adopted. Why?
   A. Because bolts need 11 rows
   B. Because the batten thickness must be 11 mm
   C. Because the main-member check for the batten moment needs a smaller spacing
   D. Because the load is 1080 kN
   Answer: C. The combined axial plus bending check of one channel (Step 7) needs a smaller C, so 11 bays of 909 mm are used. (Key point: Why more bays than the slenderness limit needs)

4. Which expressions give the batten shear and moment for transverse shear V_t, spacing C, N planes and S between bolt lines?
   A. V_b = V_t C/(N S), M = V_t C/(2N)
   B. V_b = V_t/N, M = V_t S
   C. V_b = V_t S/(N C), M = V_t C
   D. V_b = V_t C/S, M = V_t C/N
   Answer: A. IS 800 cl. 7.7.2.1 gives V_b = V_t C/(N S) = 38.4 kN and M = V_t C/(2N) = 6136 kN·mm. (Key point: Forces on battens)

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