Design strength of ISA 100x75x6, 6 bolts

Data: ISA 100×75×6 ($A_g=1010$ mm²), gusset 10 mm connected to the 100 mm leg by six 16 mm bolts ($d_0=18$ mm); $f_y=250$, $f_u=410$ N/mm². Assumed layout: single line of bolts, pitch $p=50$ mm, end distance 30 mm, bolt line 55 mm from the heel (edge distance 45 mm to the toe). Bolts assumed grade 4.6.

1. Yielding of gross section (cl. 6.2)

$$T_{dg}=\frac{A_gf_y}{\gamma_{m0}}=\frac{1010\times250}{1.10}\times10^{-3}=229.5\ \text{kN}$$

2. Rupture of net section with shear lag (cl. 6.3.3)

• Connected leg: $A_{nc}=(100-\tfrac{6}{2}-18)\times6=474$ mm²
• Outstanding leg: $A_{go}=(75-\tfrac{6}{2})\times6=432$ mm²
• $w=75$ mm, $b_s=w+w_1-t=75+55-6=124$ mm ($w_1$ = distance of the bolt line from the heel), $L_c=5\times50=250$ mm

$$\beta=1.4-0.076\frac{w}{t}\frac{f_y}{f_u}\frac{b_s}{L_c}=1.4-0.076\times\frac{75}{6}\times\frac{250}{410}\times\frac{124}{250}=1.113\quad(0.7\le\beta\le1.443)$$ $$T_{dn}=\frac{0.9A_{nc}f_u}{\gamma_{m1}}+\frac{\beta A_{go}f_y}{\gamma_{m0}}=139.9+109.3=249.2\ \text{kN}$$

3. Block shear (cl. 6.4)

$A_{vg}=1680$, $A_{vn}=1086$, $A_{tg}=270$, $A_{tn}=216$ mm²

• $T_{db1}=\dfrac{A_{vg}f_y}{\sqrt3\gamma_{m0}}+\dfrac{0.9A_{tn}f_u}{\gamma_{m1}}=284.2$ kN
• $T_{db2}=\dfrac{0.9A_{vn}f_u}{\sqrt3\gamma_{m1}}+\dfrac{A_{tg}f_y}{\gamma_{m0}}=246.5$ kN

Design tensile strength $T_d=\min(229.5,\ 249.2,\ 246.5)=229.5$ kN (yielding of gross section governs).

Design of connection for full capacity (16 mm, 4.6 bolts)

• $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{157}{1.25}\times10^{-3}=29.01$ kN
• Bearing on the 6 mm angle ($e=30,\ p=50$): $k_b=0.556$, $V_{dpb}=43.7$ kN
• Number of bolts $=\dfrac{229.5}{29.01}=7.91$ → 8 bolts (the six bolts are not enough for the full capacity).

Answer: design tensile strength $\approx229.5$ kN; connect with 8 M16 bolts (4.6) at 50 mm pitch (a longer joint) or use higher-grade bolts. With six bolts the connection strength would be only $6\times29.0=174.0$ kN.