--- title: "Built-up column with battens, 4 m, 1000 kN" subject: "Design of Steel and Timber Structure" chapter: 6 key: "semester_6/design_of_steel_and_timber_structure:builtup-battens-1000kn-4m" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/builtup-battens-1000kn-4m site: BCT Engineering --- # Built-up column with battens, 4 m, 1000 kN Design of Steel and Timber Structure · Chapter 6: Compression Members - Asked in: 2080 Chaitra, 14 marks - Video: [Manual Design of Built-up Column Section (Battened) | Steel Structure | IS 800:2007 | Excel & ETABS](https://www.youtube.com/watch?v=1DVltntPBxs) — Parash Joshi - Civil Construction and Tutor - 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-1000kn-4m - 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 placed toe-to-toe, having unsupported length 4 m, subjected to factored axial compressive load of 1000 kN. Use batten as lattice member and bolted connection to connect batten and column. The ends of the columns are effectively held in position at both ends but not restrained against rotation. ## Short answer Use 2 ISMC 225 toe to toe with 70 mm clear gap; capacity about 1224 kN. Battens 260 x 8 mm, 230 mm long at 800 mm c/c, connected with 4 M20 bolts at each end. ## Answer **Given/assumptions:** factored load $P_u=1000$ kN; unsupported length 4 m; ends held in position but not restrained against rotation → $K=1.0$, $KL=4$ m. Two channels toe to toe, battens, bolted connections (M20, 4.6); E250 ($f_y=250$, $f_u=410$). #### 1. Section: 2 ISMC 225 toe to toe (flanges facing) $A=2\times3301=6602$ mm², $I_{xx}=2\times2694.6\times10^4$ mm⁴, $r_{xx}=90.3$ mm, $I_{yy,ch}=187.2\times10^4$ mm⁴, $C_{yy}=23.0$ mm, $b_f=80$ mm, $r_{yy,ch}=23.8$ mm. Built-up members are **buckling class c**. - $\lambda_{xx}=\dfrac{KL}{r_{xx}}=\dfrac{4000}{90.3}=44.3$ - Battened column: $\lambda_e=1.1\lambda=48.7$ (cl. 7.6.1.5) - $f_{cd}=185.4$ N/mm² (curve c, $f_y=250$, $\gamma_{m0}=1.10$) → $P_d=A f_{cd}=1224$ kN $\ge1000$ kN ✓ #### 2. Clear gap (so that $I_{yy}\ge I_{xx}$) The centroid of each section is at $x_c=57.0+s/2$ from the column axis: $$ x_c\ge\sqrt{\frac{I_{xx}-I_{yy,ch}}{A_{ch}}}=87.2\ \text{mm}\Rightarrow s\ge60.4\ \text{mm} $$ Provide **$s=70$ mm**. $I_{yy}=5962\times10^4$ mm⁴, $r_{yy}=95.0$ mm, $1.1\lambda_{yy}=46.3$, $P_{dy}=1247$ kN $\ge P_u$ ✓ #### 3. Spacing of battens (cl. 7.7.1.4) The slenderness of one section between battens must be $\le\min(50,\ 0.7\lambda_e)=34.1$: $$ C\le 34.1\times23.8=811\ \text{mm} $$ For the length of 4000 mm provide 5 bays, i.e. 6 battens at **$C=800$ mm c/c**. #### 4. Size of battens (cl. 7.7.2) - Distance between the connection lines (middle of the flanges) $S=s+b_f=150$ mm; distance between the centroids of the channels $=s+2(b_f-C_{yy})=184$ mm - End battens: depth $\ge184$ mm → **260 mm** (to fit 4 bolts); intermediate battens: depth $\ge\max(0.75\times184,\ 2b_f)=160$ mm → **260 mm** - Thickness $\ge S/50=3$ mm → **8 mm**; length of batten $=S+80=230$ mm. #### 5. Design forces on a batten Transverse shear $V_t=2.5\%P_u=25.0$ kN, number of batten planes $N=2$: $$ V_b=\frac{V_tC}{NS}=\frac{25.0\times800}{2\times150}=66.67\ \text{kN},\qquad M=\frac{V_tC}{2N}=5.000\ \text{kN·m} $$ - Plate in bending: $Z=\dfrac{t d^2}{6}=90133$ mm³; $\sigma=\dfrac{M}{Z}=55.5$ N/mm² $