---
title: "Built-up column with battens, 6 m, 1200 kN"
subject: "Design of Steel and Timber Structure"
chapter: 6
key: "semester_6/design_of_steel_and_timber_structure:builtup-battens-1200kn-6m"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/builtup-battens-1200kn-6m
site: BCT Engineering
---

# Built-up column with battens, 6 m, 1200 kN

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

- Asked in: 2076 Baisakh, 12 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-1200kn-6m
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-6

## Question

Design a built-up column to carry an axial load of 1200 kN and composed of two channels placed back to back. The effective length of the member is 6m. Design the column using battens system and Fe410 grade of steel.

## Short answer

Select 2 ISMC 250 back to back with 150 mm gap. Provide battens 260×8 mm, 310 mm long, spaced at 1000 mm c/c, attached by 4 M20 bolts at each connection.

## Answer

**Given/assumptions:** axial load $P_u=1200$ kN (taken as factored); effective length 6 m ($KL=6$ m); two channels back to back with battens, bolted (M20, grade 4.6); Fe410 / E250 ($f_y=250$, $f_u=410$).

#### 1. Section: 2 ISMC 250 back to back (flanges outward)
$A=7734$ mm², $I_{xx}=2\times3816.8\times10^4$ mm⁴, $r_{xx}=99.3$ mm, $C_{yy}=23.0$ mm, $b_f=80$ mm, $r_{yy,ch}=23.8$ mm. Built-up members are **buckling class c**.
- $\lambda_{xx}=\dfrac{KL}{r_{xx}}=\dfrac{6000}{99.3}=60.4$
- Battened column: $\lambda_e=1.1\lambda=66.4$ (cl. 7.6.1.5)
- $f_{cd}=158.0$ N/mm² (curve c, $f_y=250$, $\gamma_{m0}=1.10$) → $P_d=A f_{cd}=1222$ kN $\ge1200$ kN ✓

#### 2. Clear gap (so that $I_{yy}\ge I_{xx}$)
The centroid of each section is at $x_c=23.0+s/2$ from the column axis:
$$
x_c\ge\sqrt{\frac{I_{xx}-I_{yy,ch}}{A_{ch}}}=96.5\ \text{mm}\Rightarrow s\ge147\ \text{mm}
$$
Provide **$s=150$ mm**. $I_{yy}=7866\times10^4$ mm⁴, $r_{yy}=100.8$ mm, $1.1\lambda_{yy}=65.4$, $P_{dy}=1234$ kN $\ge P_u$ ✓

#### 3. Spacing of battens (cl. 7.7.1.4)
The slenderness of one section between battens must be $\le\min(50,\ 0.7\lambda_e)=46.5$:
$$
C\le 46.5\times23.8=1107\ \text{mm}
$$
For the length of 6000 mm provide 6 bays, i.e. 7 battens at **$C=1000$ mm c/c**.

#### 4. Size of battens (cl. 7.7.2)
- Distance between the connection lines $a=S=s+b_f=230$ mm
- End battens: depth $\ge a$ → **260 mm**; intermediate battens: depth $\ge\max(0.75a,\ 2b_f)$ → **260 mm**
- Thickness $\ge a/50=230/50$ → **8 mm**; length of batten $=a+80=310$ mm.

#### 5. Design forces on a batten
Transverse shear $V_t=2.5\%P_u=30.0$ kN, number of batten planes $N=2$:
$$
V_b=\frac{V_tC}{NS}=\frac{30.0\times1000}{2\times230}=65.22\ \text{kN},\qquad M=\frac{V_tC}{2N}=7.500\ \text{kN·m}
$$
- Plate in bending: $Z=\dfrac{t d^2}{6}=90133$ mm³ (intermediate batten); $\sigma=\dfrac{M}{Z}=83.2$ N/mm² $<f_y/\gamma_{m0}=227$ N/mm² ✓ ($M_d=20.48$ kN·m)
- Shear: $\tau=\dfrac{V_b}{dt}=31.4$ N/mm² $<\dfrac{f_y}{\sqrt3\gamma_{m0}}=131$ N/mm² ✓

#### 6. Bolts connecting battens to the sections (M20, 4.6)
4 bolts in a vertical line at each end (pitch 60 mm, end 40 mm): $V_{dsb}=45.3$ kN, bearing on 8 mm $=79.5$ kN → bolt value $45.3$ kN.
- Direct shear per bolt $=V_b/n=16.30$ kN; moment effect on the extreme bolt $=\dfrac{M y_{max}}{\Sigma y^2}=37.50$ kN
- Resultant $R=\sqrt{16.30^2+37.50^2}=40.89$ kN $<45.3$ kN ✓

**Answer: 2 ISMC 250 back to back (flanges outward), clear gap 150 mm; battens 260 × 8 mm (end and intermediate), 310 mm long, at 1000 mm c/c, with 4 M20 bolts at each end.**

## Check yourself

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

1. With lambda_xx = 60.4, what slenderness is used for f_cd of the battened column?
   A. 60.4
   B. 1.05 x 60.4 = 63.4
   C. 1.1 x 60.4 = 66.4
   D. 0.7 x 60.4 = 42.3
   Answer: C. Battened columns use 1.1 times the slenderness (cl. 7.6.1.5): 66.4. (Key point: Battened column effective slenderness)

2. How is C <= 1107 mm obtained, and what spacing is provided?
   A. C <= min(50, 0.7 lambda_e) x r_yy,ch = 46.5 x 23.8; provide 1000 mm c/c
   B. C <= 50 x column depth; provide 1107 mm
   C. C <= S; provide 230 mm
   D. C <= 6000/2; provide 3000 mm
   Answer: A. The slenderness of a single channel between battens is limited to 46.5, so C <= 1107 mm. 6000 mm in 6 bays gives 1000 mm. (Key point: Batten spacing)

3. With the distance between connection lines a = 230 mm, which size rule applies?
   A. Thickness >= a/50 (about 8 mm) and end batten depth >= a
   B. Thickness >= a/10 and depth <= a/2
   C. Thickness equals 2 mm
   D. Depth must be less than the flange width
   Answer: A. IS 800 cl. 7.7.2 requires t >= a/50 = 4.6, taken 8 mm, and end batten depth >= a; 260 mm was provided. (Key point: Batten thickness and depth)

4. The extreme bolt resultant is 40.89 kN with bolt value 45.3 kN. What does this mean?
   A. The bolts are safe because R is less than the bolt value
   B. The bolts fail
   C. The bolts are over-designed by 50%
   D. Bolt value is irrelevant here
   Answer: A. R = sqrt(16.30^2 + 37.50^2) = 40.89 kN < 45.3 kN, so 4 M20 bolts at each end are adequate. (Key point: Bolt group check)

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