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

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

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

- Asked in: 2076 Bhadra, 10 marks
- Video: [Design of steel structures# Design of battened column#Channel section placed back to back#Bolted#](https://www.youtube.com/watch?v=58CGHPM9CCY) — 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-battens-1200kn-6m-restrained
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-6

## Question

Design a built up column using double channel section. Unsupported length of column is 6m, both end of column are restrained against lateral displacement but free in rotation. Column is subjected to design axial load of 1200 kN. Use battens.

## 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:** design axial load $P_u=1200$ kN; unsupported length 6 m; both ends restrained against lateral displacement but free to rotate (hinged) → $K=1.0$, $KL=6$ m. Two channels back to back with battens, bolted (M20, grade 4.6); 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. Ends are restrained against lateral displacement but free in rotation. What effective length factor is used?
   A. K = 0.5
   B. K = 0.8
   C. K = 1.0
   D. K = 2.0
   Answer: C. Such ends act as hinges, so K = 1.0 and KL = 6 m. (Key point: Effective length for hinged ends)

2. What is the purpose of the 150 mm clear gap?
   A. To make I_yy at least I_xx so the column buckles no earlier about y-y
   B. To provide space for painting only
   C. To reduce the number of battens
   D. To raise the steel grade
   Answer: A. x_c >= 96.5 mm gives s >= 147 mm, so 150 mm; then P_dy = 1234 kN >= P_u. (Key point: Gap from equal stiffness)

3. What transverse shear is assumed for designing the battens?
   A. 2.5% of the axial load (30 kN)
   B. 10% of the axial load
   C. 25% of the axial load
   D. Zero for axial members
   Answer: A. IS 800 takes V_t = 2.5% of P_u = 30 kN, shared by the batten planes. (Key point: Transverse shear)

4. The batten moment is 7.5 kN.m with Z = 90133 mm^3. What is checked?
   A. Bending stress 83.2 N/mm^2 is below 227 N/mm^2
   B. Bending stress 83.2 N/mm^2 exceeds 227 N/mm^2
   C. Stress is 7.5 N/mm^2
   D. Only the weld is checked
   Answer: A. sigma = M/Z = 7.5x10^6/90133 = 83.2 N/mm^2, below f_y/gamma_m0 = 227 N/mm^2. (Key point: Batten bending)

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