--- title: "Weld ISA 100x100x10 to 12 mm gusset" subject: "Design of Steel and Timber Structure" chapter: 4 key: "semester_6/design_of_steel_and_timber_structure:fillet-weld-angle-gusset-12mm" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/fillet-weld-angle-gusset-12mm site: BCT Engineering --- # Weld ISA 100x100x10 to 12 mm gusset Design of Steel and Timber Structure · Chapter 4: Connections in Steel Structures - Asked in: 2078 Chaitra, 6 marks - Video: [Design of fillet weld ISA toe_length-part-II](https://www.youtube.com/watch?v=31b2jMlZtog) — Dr. Shrikant Harle - 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/fillet-weld-angle-gusset-12mm - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-4 ## Question Design welded connection to connect ISA 100×100×10 with gusset plate of thickness 12mm. ## Short answer To develop the angle's full tension capacity (432.5 kN), use 6 mm fillet weld (strength 0.795 kN/mm). To balance centroid, provide toe weld 170 mm and heel weld 405 mm long. Total effective length 544 mm. ## Answer The load is not given, so the weld is designed to develop the **full design tensile strength of the angle** (ISA 100×100×10, $A_g=1903$ mm², centroid $c_x=28.4$ mm from the back), which is the usual approach. #### Load to be carried $$ T_{dg}=\frac{A_gf_y}{\gamma_{m0}}=\frac{1903\times250}{1.10}\times10^{-3}=432.5\ \text{kN} $$ #### Weld size (IS 800:2007 cl. 10.5) - Thicker part $=12$ mm (10 < t ≤ 20): minimum size 5 mm. - Maximum size along the rounded toe of the angle: $\tfrac34t=\tfrac34\times10=7.5$ mm. Use **$s=6$ mm** shop fillet weld. - Throat $t_t=0.7\times6=4.2$ mm; $f_{wd}=\dfrac{f_u}{\sqrt3\gamma_{mw}}=\dfrac{410}{\sqrt3\times1.25}=189.4$ N/mm² - Strength per mm: $q=189.4\times4.2=795.4$ N/mm (= 0.795 kN/mm) #### Weld length Total effective length $L=\dfrac{432.5\times1000}{795.4}=544$ mm. To avoid eccentricity, the welds are placed so that their resultant passes through the centroid of the angle. Let $L_1$ be the weld at the toe (far from the centroid) and $L_2$ the weld at the heel (back): - $F_1$ (toe) $=T\dfrac{c_x}{100}=122.8$ kN → $L_1=154.4$ mm - $F_2$ (heel) $=T\dfrac{100-c_x}{100}=309.7$ kN → $L_2=389.3$ mm The effective length of a fillet weld is its actual length minus $2s$ (end craters), so add $2s=12$ mm to each weld. **Provide:** 6 mm fillet weld, toe-side length $=170$ mm and heel-side length $=405$ mm (about 544 mm effective in total); the minimum length $4s=24$ mm is satisfied. The 12 mm gusset must be long enough to take the 405 mm weld; otherwise use an 8 mm weld at the heel or add an end (transverse) weld across the angle. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Since no load is given, what load is the weld designed for? A. The full design tensile strength of the angle, Ag fy/gamma_m0 = 432.5 kN B. A load of 100 kN C. fu Ag D. Zero eccentric load Answer: A. The weld develops the angle's design tensile strength T_dg = 1903 x 250/1.10 = 432.5 kN. (Key point: Design load) 2. How is total weld length found? A. Load divided by strength per mm, 432.5x1000/795.4 = 544 mm B. Equal to the angle length C. 4 times the size D. 100 mm each side Answer: A. L = T / (f_wd t_t) = 544 mm. (Key point: Weld length) 3. How is the load divided between the two weld sides? A. Equally B. So the resultant passes through the angle centroid: the heel (back) side carries more C. Toe side carries more D. Only one side welded Answer: B. Forces are inversely proportional to the distance from the centroid, so toe weld 122.8 kN and heel weld 309.7 kN. (Key point: Eccentricity avoided) 4. What is the minimum weld size for a 12 mm thick part and maximum at the toe of a 10 mm angle? A. 3 mm and 10 mm B. 5 mm and t - 1.5 = 8.5 mm C. 8 mm and 12 mm D. 6 mm and 6 mm Answer: B. For thicker part 10-20 mm minimum is 5 mm; the maximum at the angle edge is t - 1.5 = 8.5 mm. 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