Tensile strength of ISA 100x75x8, M16 8.8

Data: ISA 100×75×8 ($A_g=1336$ mm²), long (100 mm) leg bolted to a 10 mm gusset by four M16 bolts (class 8.8), pitch 40 mm, end distance 40 mm, assumed gauge 55 mm; Fe410 ($f_y=250$, $f_u=410$); $d_0=18$ mm.

1. Yielding

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

2. Net-section rupture with shear lag

• $A_{nc}=(100-4-18)\times8=624$ mm², $A_{go}=(75-4)\times8=568$ mm²
• $b_s=w+w_1-t=75+55-8=122$ mm ($w_1$ = gauge), $L_c=3\times40=120$ mm

$$\beta=1.4-0.076\times\frac{75}{8}\times\frac{250}{410}\times\frac{122}{120}=0.958$$ $$T_{dn}=\frac{0.9\times624\times410}{1.25}+\frac{0.958\times568\times250}{1.10}=307.9\ \text{kN}$$

3. Block shear

$A_{vg}=1280$, $A_{vn}=776$, $A_{tg}=360$, $A_{tn}=288$ mm²

• $T_{db1}=253.0$ kN; $T_{db2}=214.1$ kN

4. Result

Design tensile strength of the angle $=\min(303.6,\ 307.9,\ 214.1)=214.1$ kN (block shear governs).

(Bolt check: $V_{dsb}=58.0$ kN, bearing on 8 mm leg $k_b=0.491$, $V_{dpb}=51.5$ kN; four bolts give $206.0$ kN, so the connection itself limits the joint to about $206$ kN.)