--- title: "Laced column with welds, 6 m, 1000 kN" subject: "Design of Steel and Timber Structure" chapter: 6 key: "semester_6/design_of_steel_and_timber_structure:builtup-laced-welded-1000kn-6m" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/builtup-laced-welded-1000kn-6m site: BCT Engineering --- # Laced column with welds, 6 m, 1000 kN Design of Steel and Timber Structure · Chapter 6: Compression Members - Asked in: 2069 Bhadra, 14 marks - Video: [Laced column design# Design of compression members#welded connection#](https://www.youtube.com/watch?v=2Hzd-YHwkxc) — CIVIL ENGINEERING COURSE Theory & Practical - Page: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/builtup-laced-welded-1000kn-6m - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-6 ## Question Design a built-up column of the effective length of 6m to carry an axial load of 1000KN using two channels and laces. Design the connections using welds. The grade of the steel is E250C. ## Short answer Two ISMC 225 channels back-to-back with 130 mm web gap (overall width 290 mm). Single lacing ISF 60×8 mm flats at 45° in two planes, connection spacing 420 mm, welded with 4 mm fillet welds. Capacity 1016 kN > 1000 kN, safe. ## Answer **Approach.** Laced column of two channels (back to back), designed to IS 800:2007 (cl. 7.1.2 for the compressive strength, cl. 7.6 for laced columns). Steel Fe 410 (E250): $f_y = 250$ N/mm², $f_u = 410$ N/mm², $\gamma_{m0} = 1.10$. #### Data - Factored axial load $P = 1000$ kN; effective length $KL = 6.0$ m (given, same for both axes). - Built-up member: buckling curve **c**, $\alpha = 0.49$ (Table 10 and Table 7). Effective slenderness of laced column $= 1.05 \times$ actual maximum slenderness (cl. 7.6.1.5). #### Step 1: Trial section Try 2 × **ISMC 225** (IS 808): $A = 33.3$ cm² each, $I_{xx} = 2710$ cm⁴, $I_{yy} = 188$ cm⁴, $C_{yy} = 2.31$ cm, $B = 80$ mm, $t_w = 6.5$ mm, $t_f = 12.4$ mm. Total area $A_e = 2 \times 3330 = 6660$ mm². $r_{zz} = \sqrt{I_{xx}/A} = 9.02$ cm. $$ \begin{aligned} \lambda_{zz} &= \frac{KL}{r_{zz}} = \frac{6000}{90.2} = 66.5 \\ \lambda_{eff} &= 1.05 \times 66.5 = 69.8 \end{aligned} $$ For $\lambda_{eff} = 69.8$, curve c: $f_{cc} = \pi^2E/\lambda^2 = 405$ N/mm², $\lambda_n = \sqrt{f_y/f_{cc}} = 0.786$, $\phi = 0.5[1 + 0.49(\lambda_n - 0.2) + \lambda_n^2] = 0.952$, $\chi = 0.671$. $$ \begin{aligned} f_{cd} &= \chi f_y/\gamma_{m0} = 0.671 \times 250/1.1 = 152.5\ \text{N/mm}^2 \\ P_d &= A_e f_{cd} = 6660 \times 152.5/10^3 = 1016\ \text{kN} > 1000\ \text{kN} \end{aligned} $$ #### Step 2: Spacing of channels To make the column equally strong about the other axis, $I_{yy} \ge I_{zz}$ (cl. 7.6.1.1). With $h$ the distance between the centroids of the two channels: $$ \begin{aligned} 2\left[I_{y} + A\left(\frac{h}{2}\right)^2\right] &\ge 2I_{x} \\ h &\ge 2\sqrt{\frac{I_x - I_y}{A}} = 2\sqrt{\frac{(2710 - 188) \times 10^4}{3330}} = 174\ \text{mm} \end{aligned} $$ Gap between webs $g = h - 2C_{yy} = 174 - 46.2 = 128$ mm. Provide **g = 130 mm** (centroid distance $h = 176$ mm). $$ \begin{aligned} I_{yy} &= 2[188 + 33.3(8.81)^2] = 5545\ \text{cm}^4 \\ r_{yy} &= \sqrt{I_{yy}/2A} = 9.12\ \text{cm}, \quad \lambda_{yy} = 6000/91.2 = 65.8 \end{aligned} $$ $\lambda_{yy} = 65.8 \le \lambda_{zz} = 66.5$, so the zz-axis governs and the capacity $P_d = 1016$ kN $\ge 1000$ kN. **Safe.** Overall width of column $= 2B + g = 290$ mm. ```text +-------+ gap g +-------+ | ISMC |<--------->| ISMC | +-------+ +-------+ webs face each other, flanges point outward lacing bolted to the flanges (top and bottom) ``` #### Step 3: Lacing system (single lacing, two planes) Transverse shear (cl. 7.6.6.1): $V_t = 2.5\% P = 0.025 \times 1000 = 25.0$ kN; shared by two lacing planes: $12.50$ kN each. Bolt lines are taken at mid-width of the flanges, so the transverse distance between lacing connections is $s = g + B = 130 + 80 = 210$ mm. Lacing angle $\theta = 45^\circ$ to the column axis (permitted 40° to 70°, cl. 7.6.4). $$ \begin{aligned} \text{Length of bar } l &= s/\sin\theta = 210/\sin 45^\circ = 297\ \text{mm} \\ \text{Force in bar } F &= \frac{V_t/2}{\sin\theta} = \frac{12.50}{\sin 45^\circ} = 17.68\ \text{kN} \end{aligned} $$ Spacing of connections along one channel $a_1 = 2s\cot\theta = 420$ mm. Check (cl. 7.6.5.1): $a_1/r_{min} = 420/23.8 = 17.7$, which is less than the smaller of 50 and $0.7\lambda_{max} = 46.6$. OK. **Flat size.** Width $\ge 3d = 48$ mm for M16 bolts; take 60 mm. Thickness $\ge l/40 = 7.4$ mm (cl. 7.6.3) and for $KL/r \le 145$, $t \ge l\sqrt{12}/145 = 7.1$ mm. Provide **ISF 60 × 8 mm**. $r = t/\sqrt{12} = 2.31$ mm, $l/r = 128.6 < 145$ OK. $f_{cd}$ (curve c) $= 75.5$ N/mm², $P_{d,lace} = 480 \times 75.5/10^3 = 36.2$ kN $> F = 17.68$ kN. OK. #### Step 4: Welded connection of lacing For a welded lacing the effective length is $0.7\times$ the distance between the inner ends of the welds (cl. 7.6.6.3); the check above was done on $l = 297$ mm, which is conservative. Fillet weld size $s = 4$ mm (min 3 mm for 6 to 10 mm plates, max $t - 1.5$). Design strength per mm: $$ \begin{aligned} f_w &= \frac{0.7 s f_u}{\sqrt3\,\gamma_{mw}} = \frac{0.7 \times 4 \times 410}{\sqrt3 \times 1.25} = 530\ \text{N/mm} \\ L_w &= \frac{F}{f_w} = \frac{17678}{530} = 33\ \text{mm} \end{aligned} $$ Provide a weld of length $35$ mm on each side of the bar at each end (total 70 mm > 33 mm). Lap on the channel flange $\ge 4t = 32$ mm (cl. 7.6.7). Weld along both sides for the full lap length. #### Step 5: End tie plates Tie plates are provided at the ends and designed like battens (cl. 7.7.2): effective depth $\ge h = 176$ mm; thickness $\ge s/50 = 4.2$ mm. Provide **290 mm wide × 180 mm deep × 6 mm thick**. The tie plate is fillet welded (size 5 mm) to the flange of each channel along two lines of length 180 mm. With $C = 420$ mm, $N=2$: $V_b = V_tC/(NS) = 25.0$ kN and $M = V_tC/2N = 2625$ kN·mm. Weld modulus per unit throat $Z_w = d^2/3 = 10800$ mm²: $q_m = M/Z_w = 243$ N/mm, $q_v = V_b/2d = 69$ N/mm, resultant $= 253$ N/mm $<$ weld strength $663$ N/mm. OK. #### Final design - Column: 2 × ISMC 225 back to back, gap 130 mm, $P_d = 1016$ kN $> 1000$ kN. - Lacing: single, flats 60 × 8 mm at 45°, 4 mm fillet welds. - End tie plates 290 × 180 × 6 mm. ```text +--------------------+ <- end tie plate | \ /\ /\ /| | \ / \ / \ / | single lacing | \/ \/ \/ | (same direction | /\ /\ /\ | on both faces) +--------------------+ <- end tie plate ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which formula gives the design strength per mm of a fillet weld of size s? A. f_w = 0.7 s f_y / gamma_m0 B. f_w = 0.7 s f_u / (sqrt(3) gamma_mw) C. f_w = s f_u / gamma_mb D. f_w = 0.6 s f_u / 2 Answer: B. Throat 0.7 s times f_u/(sqrt3 gamma_mw) gives 530 N/mm for a 4 mm weld. (Key point: Design weld strength) 2. F = 17.68 kN and f_w = 530 N/mm. What weld length is required, and what is provided? A. 33 mm required; 35 mm each side (70 mm total) provided B. 530 mm required; 600 mm provided C. 17.7 mm required; 18 mm provided D. 70 mm required; 35 mm total provided Answer: A. L_w = 17678/530 = 33 mm; 35 mm on each side of the bar gives 70 mm total, with lap at least 4t. (Key point: Weld length required) 3. The computed minimum gap is 128 mm and 130 mm is provided. Which condition gave 128 mm? A. Lacing angle of 45 degrees B. I_yy >= I_zz C. Weld length requirement D. Tie plate thickness Answer: B. Equal stiffness about both axes, I_yy >= I_zz, gives h >= 174 mm and hence g = 128 mm. 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