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Unequal angle with fillet welds, 150 kN

Short answer

Use ISA 100×75×8 angle. 6 mm fillet welds, toe weld 105 mm, heel weld 210 mm, balansing the centrod.

Design a single angle (unequal angle) to carry a working tensile load of 150 KN, If the end connection is done using fillet welds. The yield and ultimate strength of the steel are 250 MPa and 410 MPa respectively.

Given: working tension 150 kN → factored Tu=1.5×150=225T_u=1.5\times150=225 kN; fillet welds; fy=250f_y=250, fu=410f_u=410 N/mm².

Section

Ag≥225×103×1.10250=990A_g\ge\dfrac{225\times10^3\times1.10}{250}=990 mm². Try ISA 100×75×8 (Ag=1336A_g=1336 mm²), long leg connected, centroid 31.5 mm from the heel.

  • Tdg=1336×2501.10×10−3=303.6T_{dg}=\dfrac{1336\times250}{1.10}\times10^{-3}=303.6 kN >225>225 ✓

Weld

  • s=6s=6 mm (angle 8 mm: min 3-5 mm, max 6.5 mm). Shop weld, fwd=4103×1.25=189.4f_{wd}=\dfrac{410}{\sqrt3\times1.25}=189.4 N/mm².
  • Strength per mm =189.4×0.7×6=795.4=189.4\times0.7\times6=795.4 N/mm.
  • Total length Lw=225×103795.4=282.9L_w=\dfrac{225\times10^3}{795.4}=282.9 mm.
  • For no eccentricity: toe weld L1=Lw×31.5100=89.1L_1=L_w\times\dfrac{31.5}{100}=89.1 mm; heel weld L2=193.8L_2=193.8 mm.
  • Adding 2s=122s=12 mm for end craters: toe 105 mm and heel 210 mm.

Rupture check (shear lag; welded, so bs=w=75b_s=w=75 mm; conservatively Lc=89L_c=89 mm)

β=1.4−0.076×758×250410×7589=1.034\beta=1.4-0.076\times\dfrac{75}{8}\times\dfrac{250}{410}\times\dfrac{75}{89}=1.034

Tdn=0.9×(100−4)×8×4101.25+β×(75−4)×8×2501.10=360.2 kN>225 ✓T_{dn}=\frac{0.9\times(100-4)\times8\times410}{1.25}+\frac{\beta\times(75-4)\times8\times250}{1.10}=360.2\ \text{kN}>225\ \checkmark

Answer: ISA 100×75×8, 6 mm fillet welds; toe weld 105 mm and heel weld 210 mm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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