Angle tension member 250 kN with M20

Given: factored tension 250 kN, $L=3$ m, single row of M20 bolts (grade 4.6 assumed), $f_y=250$, $f_u=410$ N/mm².

1. Bolts

• $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26$ kN (single shear, thread in the shear plane)
• Bearing on the 8 mm angle ($e=40$, $p=50$): $k_b=\min(0.606,\ 0.508,\ 0.976,\ 1)=0.508$; $V_{dpb}=66.6$ kN → bolt value $45.26$ kN
• $n=\dfrac{250}{45.26}=5.5$ → 6 bolts, pitch 50 mm, end distance 40 mm, bolt line 55 mm from the heel; capacity $271.6$ kN.

2. Section

$A_g\ge\dfrac{250\times1.10}{250}\times10^3=1100$ mm². Because the angle is connected through one leg, choose ISA 100×75×8 ($A_g=1336$ mm², $r_{vv}=15.9$ mm), long leg connected.

• Yielding: $T_{dg}=\dfrac{1336\times250}{1.10}\times10^{-3}=303.6$ kN $>250$ ✓
• Rupture (cl. 6.3.3): $A_{nc}=(100-4-22)\times8=592$ mm², $A_{go}=(75-4)\times8=568$ mm², $b_s=w+w_1-t=75+55-8=122$ mm, $L_c=250$ mm

$$\beta=1.4-0.076\times\frac{75}{8}\times\frac{250}{410}\times\frac{122}{250}=1.188,\qquad T_{dn}=\frac{0.9\times592\times410}{1.25}+\frac{1.188\times568\times250}{1.10}=328.1\ \text{kN}$$

• Block shear: $A_{vg}=2320$, $A_{vn}=1352$, $A_{tg}=360$, $A_{tn}=272$ mm² → $T_{db1}=384.7$, $T_{db2}=312.2$ kN
• Slenderness: $\dfrac{3000}{15.9}\approx189<400$ ✓

Design strength $=\min(303.6,328.1,312.2,271.6)=271.6$ kN $>250$ kN. Answer: ISA 100×75×8 (long leg connected) with 6 M20 bolts at 50 mm pitch.