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Weld length for ISA 100x100x8, 200 kN

Short answer

Use 6 mm fillet welds at toe and heel. Total effective weld length 251 mm: 69 mm at toe and 182 mm at heel (add 12 mm each for end craters). Weld strength 795 N/mm gives capacity 200 kN.

In a truss angle ISA 100×100×8 is subjected to the factored tension of 200 kN. It is to be connected to a gusset using fillet welds at the toe and back. Find the length of weld required. Take Fe 410 steel.

Given: ISA 100×100×8 (centroid cx=27.5c_x=27.5 mm from the back), factored tension 200 kN, Fe410 (fu=410f_u=410 N/mm²), fillet welds at the toe and back, shop welding.

Weld size and strength (IS 800:2007 cl. 10.5)

  • Minimum size (Table 21): 3 mm if the thicker part (gusset) is up to 10 mm thick, 5 mm if it is 10-20 mm. Maximum size at the rounded toe of the 8 mm angle ≈34t=6\approx\tfrac34t=6 mm (t−1.5=6.5t-1.5=6.5 mm at a square edge). Use s=6s=6 mm.
  • fwd=fu3γmw=4103×1.25=189.4f_{wd}=\dfrac{f_u}{\sqrt3\gamma_{mw}}=\dfrac{410}{\sqrt3\times1.25}=189.4 N/mm²
  • Strength per mm =fwd×0.7s=189.4×4.2=795.4=f_{wd}\times0.7s=189.4\times4.2=795.4 N/mm

Total length

Lw=200×103795.4=251.5 mmL_{w}=\frac{200\times10^3}{795.4}=251.5\ \text{mm}

Distribution (so that the resultant coincides with the centroid)

Take moments about the back weld (length L2L_2 at the heel):

  • Weld at toe: L1=Lwcx100=69.2L_1=L_w\dfrac{c_x}{100}=69.2 mm (carries 200×0.275=55200\times0.275=55 kN)
  • Weld at back (heel): L2=Lw−L1=182.3L_2=L_w-L_1=182.3 mm (carries 145 kN)

Add 2s=122s=12 mm to each for the end craters (overall length = effective length + 2s2s).

Answer: 6 mm fillet weld, toe weld 85 mm and back weld 195 mm long (total effective length about 251 mm).

✓ Answer checked by Claude Opus 5.5 (high effort)

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