--- title: "Fillet welded channel tension member, 200 kN" subject: "Design of Steel and Timber Structure" chapter: 5 key: "semester_6/design_of_steel_and_timber_structure:tension-welded-channel-200kn" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/tension-welded-channel-200kn site: BCT Engineering --- # Fillet welded channel tension member, 200 kN Design of Steel and Timber Structure · Chapter 5: Tension Members - Asked in: 2075 Baisakh, 8 marks - Video: [Tension Member Design Welded New](https://www.youtube.com/watch?v=yYsYjqlCY9w) — Bhavin Patel - 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/tension-welded-channel-200kn - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-5 ## Question Design a fillet welded channel section to act as a tension member subjected to an axial tensile load of 200 KN. ## Short answer Use ISMC 125 channel with 3 mm fillet welds. Provide 125 mm end weld across web and two 190 mm longitudinal welds along heels (total 505 mm). Weld strength 398 N/mm gives capacity 201 kN > 200 kN. ## Answer **Assumptions:** factored tension 200 kN; Fe410 / E250 ($f_y=250$, $f_u=410$ N/mm²); shop welding; gusset plate 10 mm thick. The back of the channel web sits on the gusset and is welded along the two back edges (heels) of the channel and across the end of the web. #### 1. Section $A_g\ge\dfrac{200\times10^3\times1.10}{250}=880$ mm². Try **ISMC 125**: $A=1619$ mm², $h=125$, $b_f=65$, $t_f=8.1$, $t_w=5.0$ mm. - $T_{dg}=\dfrac{1619\times250}{1.10}\times10^{-3}=368.0$ kN $>200$ ✓ #### 2. Weld size and length - Thicker part (gusset) 10 mm → minimum size 3 mm (Table 21). The end weld runs along the square cut edge of the 5 mm web, so its size must not exceed $t-1.5=3.5$ mm. Use **$s=3$ mm** throughout. - $f_{wd}=\dfrac{410}{\sqrt3\times1.25}=189.4$ N/mm²; strength per mm $=189.4\times0.7\times3=397.7$ N/mm - Required length $L_w=\dfrac{200\times10^3}{397.7}=503$ mm - **Provide:** an end weld of 125 mm across the web plus two longitudinal welds of 190 mm along the heels: total $125+2\times190=505$ mm $>503$ mm ✓ (each side weld is longer than $4s=12$ mm and than the 125 mm distance between them ✓). #### 3. Net-section rupture with shear lag (cl. 6.3.3 and 6.3.4) - $A_{nc}=125\times5=625$ mm² (no holes), $A_{go}=1619-625=994$ mm² - Shear lag distance from the far edge of the flange to the weld line: $b_s=b_f=65$ mm; $L_c=190$ mm $$ \beta=1.4-0.076\times\frac{65}{8.1}\times\frac{250}{410}\times\frac{65}{190}=1.273 $$ $$ T_{dn}=\frac{0.9\times625\times410}{1.25}+\frac{1.273\times994\times250}{1.10}=472.0\ \text{kN}>200 $$ The slenderness limit of 400 ($r_{yy}=19.2$ mm) allows $L\le7.7$ m. **Answer: ISMC 125 with 3 mm fillet welds: 2 × 190 mm along the heels and 125 mm across the end of the web.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the minimum gross area for 200 kN tension (E250)? A. 800 mm2 B. 880 mm2 C. 1000 mm2 D. 1619 mm2 Answer: B. Ag >= 200 x 10^3 x 1.10 / 250 = 880 mm2. (Key point: Required area) 2. For s = 4 mm, what is the weld strength per mm and the resulting required length? A. 530.2 N/mm and 377 mm B. 795.4 N/mm and 251 mm C. 530.2 N/mm and 200 mm D. 189.4 N/mm and 1056 mm Answer: A. 189.4 x 0.7 x 4 = 530.2 N/mm; 200000/530.2 = 377 mm. (Key point: Weld strength per mm) 3. The provided weld (2 x 140 + 125 = 405 mm) is acceptable because A. it is exactly 4s B. 405 mm exceeds the required 377 mm C. end welds are not counted D. weld length must equal channel depth Answer: B. The provided effective length must be at least the required length. (Key point: Provided weld layout) 4. In the rupture check the web is connected and flanges outstanding. What does beta account for? A. Weld size B. Bolt holes in web C. Reduced effectiveness of the outstanding flanges D. Yield strength of weld Answer: C. beta modifies the contribution of the outstanding (unconnected) portion Ago. 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