Tensile strength of ISA 100x75x8, six M20 bolts

Data: ISA 100×75×8 ($A_g=1336$ mm²), gusset 12 mm connected to the 100 mm leg, six M20 bolts (4.6) in one row. Assumed pitch $p=50$ mm, end distance 40 mm, gauge 55 mm from the heel; $d_0=22$ mm; $f_y=250$, $f_u=410$.

1. Yielding of gross section

$T_{dg}=\dfrac{A_gf_y}{\gamma_{m0}}=\dfrac{1336\times250}{1.10}\times10^{-3}=303.6$ kN

2. Net-section rupture with shear lag

• $A_{nc}=(100-\tfrac82-22)\times8=592$ mm²; $A_{go}=(75-\tfrac82)\times8=568$ mm²
• $w=75$, $b_s=w+w_1-t=75+55-8=122$ mm ($w_1$ = gauge), $L_c=5\times50=250$ mm

$$\beta=1.4-0.076\times\frac{75}{8}\times\frac{250}{410}\times\frac{122}{250}=1.188\quad(0.7\le\beta\le1.443)$$ $$T_{dn}=\frac{0.9A_{nc}f_u}{\gamma_{m1}}+\frac{\beta A_{go}f_y}{\gamma_{m0}}=328.1\ \text{kN}$$

3. Block shear

$A_{vg}=2320$, $A_{vn}=1352$, $A_{tg}=360$, $A_{tn}=272$ mm²: $T_{db1}=384.7$ kN, $T_{db2}=312.2$ kN.

4. Bolt group (check)

$V_{dsb}=45.26$ kN (single shear); bearing ($t=8$): $V_{dpb}=66.6$ kN; six bolts: $6\times45.26=271.6$ kN. This is less than the strength of the angle, so the bolts govern the strength of the joint.

Answer: design tensile strength of the angle $=\min(303.6,\ 328.1,\ 312.2)=303.6$ kN (yielding of the gross section governs). With six M20 (4.6) bolts, the connection limits the load the member can transfer to 271.6 kN.