---
title: "Column with battens, 1500 kN, 8 m, fixed ends"
subject: "Design of Steel and Timber Structure"
chapter: 6
key: "semester_6/design_of_steel_and_timber_structure:column-rolled-i-1500kn-8m-fixed"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/column-rolled-i-1500kn-8m-fixed
site: BCT Engineering
---

# Column with battens, 1500 kN, 8 m, fixed ends

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

- Asked in: 2075 Baisakh, 20 marks
- 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-rolled-i-1500kn-8m-fixed
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-6

## Question

Design a column section subjected to axial load of 1500 KN using rolled steel 'I' sections. Height of the column is 8 m. Both ends of column are fixed and exits sway condition. Also design lateral bracing of column using battening system.

## Short answer

Effective length KL = 1.2×8 = 9.6 m. Use 2 ISMB 350 side by side, gap 150 mm (class c, λe=73.9, fcd=146 N/mm², Pd=1948 kN). Battens 320×8 mm at 1333 mm c/c, each end connected with 8 M20 bolts (two lines of 4).

## Answer

**Given/assumptions:** factored load $P_u=1500$ kN; height 8 m; both ends fixed against rotation but the frame is free to sway → effective length factor $K=1.2$ (IS 800:2007 Table 11, recommended value), so $KL=1.2\times8000=9600$ mm. The column is built up of **two rolled I-sections (ISMB)** placed side by side and tied by battens (bolted, M20 4.6); E250 ($f_y=250$, $f_u=410$). The battens are on the flange faces; the bolt group on each I-section is centred on its web line, so $S=s+b_f$.

#### 1. Section: 2 ISMB 350 placed side by side with a gap (webs parallel)
$A=13342$ mm², $I_{xx}=2\times13630.3\times10^4$ mm⁴, $r_{xx}=142.9$ mm, $C_{yy}=70$ mm, $b_f=140$ mm, $r_{yy,ch}=28.4$ mm. Built-up members are **buckling class c**.
- $\lambda_{xx}=\dfrac{KL}{r_{xx}}=\dfrac{9600}{142.9}=67.2$
- Battened column: $\lambda_e=1.1\lambda=73.9$ (cl. 7.6.1.5)
- $f_{cd}=146.0$ N/mm² (curve c, $f_y=250$, $\gamma_{m0}=1.10$) → $P_d=A f_{cd}=1948$ kN $\ge1500$ kN ✓

#### 2. Clear gap (so that $I_{yy}\ge I_{xx}$)
The centroid of each section is at $x_c=70.0+s/2$ from the column axis:
$$
x_c\ge\sqrt{\frac{I_{xx}-I_{yy,ch}}{A_{ch}}}=140.1\ \text{mm}\Rightarrow s\ge140\ \text{mm}
$$
Provide **$s=150$ mm**. $I_{yy}=29127\times10^4$ mm⁴, $r_{yy}=147.8$ mm, $1.1\lambda_{yy}=71.5$, $P_{dy}=2000$ kN $\ge P_u$ ✓

#### 3. Spacing of battens (cl. 7.7.1.4)
The slenderness of one section between battens must be $\le\min(50,\ 0.7\lambda_e)=50.0$:
$$
C\le 50.0\times28.4=1420\ \text{mm}
$$
For the length of 8000 mm provide 6 bays, i.e. 7 battens at **$C=1333$ mm c/c**.

#### 4. Size of battens (cl. 7.7.2)
- Distance between the connection lines $a=S=s+b_f=290$ mm
- End battens: depth $\ge a$ → **320 mm**; intermediate battens: depth $\ge\max(0.75a,\ 2b_f)=280$ mm → **320 mm**
- Thickness $\ge a/50=290/50$ → **8 mm**; length of batten $=290+70+2\times35=430$ mm (to take the two bolt lines on each flange).

#### 5. Design forces on a batten
Transverse shear $V_t=2.5\%P_u=37.5$ kN, number of batten planes $N=2$:
$$
V_b=\frac{V_tC}{NS}=\frac{37.5\times1333}{2\times290}=86.21\ \text{kN},\qquad M=\frac{V_tC}{2N}=12.500\ \text{kN·m}
$$
- Plate in bending: $Z=\dfrac{t d^2}{6}=136533$ mm³ (intermediate batten); $\sigma=\dfrac{M}{Z}=91.6$ N/mm² $<f_y/\gamma_{m0}=227$ N/mm² ✓ ($M_d=31.03$ kN·m)
- Shear: $\tau=\dfrac{V_b}{dt}=33.7$ N/mm² $<\dfrac{f_y}{\sqrt3\gamma_{m0}}=131$ N/mm² ✓

#### 6. Bolts connecting battens to the sections (M20, 4.6)
A bolt cannot be placed on the web line of an I-section, so at each end use **8 bolts in two vertical lines of 4**, one line on each side of the web (35 mm from the web centre line, pitch 80 mm, end 40 mm). $V_{dsb}=45.3$ kN, bearing on 8 mm $=79.5$ kN → bolt value $45.3$ kN.
- $\Sigma r^2=8\times35^2+4\times(40^2+120^2)=73800$ mm²
- Direct shear per bolt $=V_b/8=10.78$ kN (along the column); moment on the extreme bolt (35, 120): $\dfrac{M y}{\Sigma r^2}=20.33$ kN across and $\dfrac{M x}{\Sigma r^2}=5.93$ kN along the column
- Resultant $R=\sqrt{20.33^2+(10.78+5.93)^2}=26.3$ kN $<45.3$ kN ✓

**Answer: 2 ISMB 350 placed side by side with a gap (webs parallel), clear gap 150 mm; battens 320 × 8 mm (end and intermediate), 430 mm long, at 1333 mm c/c, with 8 M20 bolts (two lines of 4) at each end.**

## Check yourself

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

1. Both ends fixed but the frame is free to sway. What K and KL are used?
   A. K = 0.65, KL = 5.2 m
   B. K = 1.2, KL = 9.6 m
   C. K = 2.0, KL = 16 m
   D. K = 1.0, KL = 8 m
   Answer: B. IS 800 Table 11 recommends K = 1.2 for fixed ends with sway, so KL = 1.2 x 8000 = 9600 mm. (Key point: Effective length factor for sway)

2. Why is 2 ISMB 350 acceptable for 1500 kN?
   A. P_d = A f_cd = 13342 x 146.0 = 1948 kN, which is above 1500 kN
   B. A f_y is above 1500 kN so no check is needed
   C. lambda is below 20
   D. P_d is 1000 kN
   Answer: A. lambda_xx = 67.2, lambda_e = 1.1 x 67.2 = 73.9 and f_cd = 146.0 N/mm^2, giving 1948 kN. (Key point: Section capacity)

3. How is the batten spacing C = 1333 mm chosen?
   A. C <= min(50, 0.7 lambda_e) x r = 1420 mm, and 8000 mm is divided into 6 equal bays
   B. C is always 1000 mm
   C. C equals the batten depth
   D. C is half the column height
   Answer: A. Slenderness of one section is limited to 50, so C <= 50 x 28.4 = 1420 mm; 8000/6 = 1333 mm. (Key point: Batten spacing)

4. The extreme bolt resultant is 45.09 kN while the bolt value is 45.3 kN. What follows?
   A. The bolts are just adequate
   B. The bolts fail
   C. More than 10 bolts are needed
   D. The check is not needed
   Answer: A. R = sqrt(17.24^2 + 41.67^2) = 45.09 kN < 45.3 kN, so 5 M20 bolts at each end are just safe. (Key point: Batten bolts)

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