Angle tension member 250 kN, weld and bolt

Given: factored tension 250 kN, $f_y=250$, $f_u=410$ N/mm². Trial section for both cases: ISA 100×75×8 ($A_g=1336$ mm²), connected through the 100 mm leg, centroid 31.5 mm from the heel (short-leg side).

$T_{dg}=\dfrac{1336\times250}{1.10}\times10^{-3}=303.6$ kN $>250$ ✓

(i) Welded connection

• Size: 8 mm angle → use $s=6$ mm (min 3-5 mm, max $t-1.5=6.5$ mm). $f_{wd}=189.4$ N/mm²; strength $=189.4\times0.7\times6=795.4$ N/mm.
• Total length $L_w=\dfrac{250\times10^3}{795.4}=314.3$ mm.
• Distribution about the centroid: toe weld $L_1=L_w\times\dfrac{31.5}{100}=99.0$ mm; heel weld $L_2=215.3$ mm. Add $2s=12$ mm for end craters: toe 115 mm, heel 230 mm.
• Rupture check (welded, so $b_s=w=75$ mm; conservatively $L_c\approx99$ mm, the shorter weld): $\beta=1.4-0.076\times\dfrac{75}{8}\times\dfrac{250}{410}\times\dfrac{75}{99}=1.071$; $T_{dn}=\dfrac{0.9\times(100-4)\times8\times410}{1.25}+\dfrac{\beta\times(75-4)\times8\times250}{1.10}=365.0$ kN $>250$ ✓

(ii) Bolted connection (M20, grade 4.6, $d_0=22$ mm)

• $V_{dsb}=45.26$ kN; bearing on 8 mm leg ($e=40,\ p=50$) $V_{dpb}=66.6$ kN → bolt value $45.26$ kN
• $n=\dfrac{250}{45.26}=5.5$ → 6 bolts, pitch 50 mm, end distance 40 mm, gauge 55 mm; capacity $6\times45.26=271.6$ kN.
• Rupture: $A_{nc}=(100-4-22)\times8=592$ mm², $A_{go}=(75-4)\times8=568$ mm², $b_s=75+55-8=122$ mm, $\beta=1.188$ → $T_{dn}=328.1$ kN
• Block shear: $\min(384.7,\ 312.2)$ kN $>250$ ✓

Answer: ISA 100×75×8 with (i) 6 mm fillet welds (toe 115 mm, heel 230 mm), or (ii) 6 bolts M20 at 50 mm pitch.