Single angle tension member, 100 kN

Given: tension 100 kN (taken as the design/factored load; if it is a working load, multiply by 1.5), M20 bolts of product grade C, class 4.6 ($d_0=22$ mm), $f_y=250$, $f_u=410$ N/mm².

1. Bolts

• $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26$ kN
• Bearing on 6 mm leg ($e=35,\ p=50$): $k_b=0.508$, $V_{dpb}=49.9$ kN → bolt value $45.26$ kN
• $n=100/45.26=2.2$ → 3 bolts, pitch 50 mm, end distance 35 mm; capacity $135.8$ kN.

2. Section

$A_g\ge\dfrac{100\times1.10}{250}\times10^3=440$ mm². The leg must hold an M20 bolt with edge distances $\ge1.5d_0=33$ mm on both sides, so the leg must be at least 70 mm: try ISA 75×75×6 ($A=866$ mm²), bolt gauge 40 mm from the back.

• Yielding: $T_{dg}=\dfrac{866\times250}{1.10}\times10^{-3}=196.8$ kN
• Rupture with shear lag: $A_{nc}=(75-3-22)\times6=300$ mm², $A_{go}=(75-3)\times6=432$ mm², $b_s=w+w_1-t=75+40-6=109$ mm, $L_c=100$ mm, $\beta=0.769$

$$T_{dn}=\frac{0.9\times300\times410}{1.25}+\frac{0.769\times432\times250}{1.10}=164.0\ \text{kN}$$

• Block shear: $T_{db1}=148.8$ kN, $T_{db2}=129.5$ kN → $T_{db}=129.5$ kN

Design strength $=\min(196.8,\ 164.0,\ 129.5,\ 135.8)=129.5$ kN $>100$ kN ✓.

Answer: ISA 75×75×6 connected by 3 M20 bolts (4.6) at 50 mm pitch.