Single channel tension member, 280 kN

Given: $T=280$ kN (design), $L=3.2$ m, gusset 10 mm, one line of M20 bolts (4.6), E250 ($f_y=250$, $f_u=410$ N/mm²), channel connected through its web.

1. Trial section

$A_g\ge\dfrac{T\gamma_{m0}}{f_y}=\dfrac{280\times10^3\times1.10}{250}=1232$ mm². Try ISMC 175: $A=2440$ mm², $h=175$, $b_f=75$, $t_f=10.2$, $t_w=5.7$ mm, $r_{yy}=22.3$ mm.

• Yielding: $T_{dg}=\dfrac{2440\times250}{1.10}=554.5$ kN $>280$ ✓
• Slenderness: $\dfrac{L}{r_{min}}=\dfrac{3200}{22.3}=143<400$ ✓ (IS 800 Table 3)

2. Bolts (M20, 4.6, single shear, $d_0=22$ mm)

• Shear: $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26$ kN
• Bearing on 5.7 mm web ($e=40$, $p=50$): $k_b=0.508$, $V_{dpb}=47.4$ kN
• Bolt value $=45.26$ kN → $n=\dfrac{280}{45.26}=6.2$ → provide 7 bolts at 50 mm pitch, end distance 40 mm (joint length $380$ mm); capacity $316.8$ kN.

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

• $A_{nc}=(175-22)\times5.7=872$ mm² (connected web), $A_{go}=A-175\times5.7=1442$ mm² (outstanding flanges)
• $w=75$, $b_s=w+w_1-t_w=75+87.5-5.7=156.8$ mm, $L_c=6\times50=300$ mm

$$\beta=1.4-0.076\frac{w}{t_f}\frac{f_y}{f_u}\frac{b_s}{L_c}=1.222\ (\le1.443)$$ $$T_{dn}=\frac{0.9A_{nc}f_u}{\gamma_{m1}}+\frac{\beta A_{go}f_y}{\gamma_{m0}}=658.0\ \text{kN}>280$$

4. Block shear (cl. 6.4)

$A_{vg}=1938$, $A_{vn}=1123$, $A_{tg}=499$, $A_{tn}=436$ mm² (failure along the bolt line and across to the web edge)

• $T_{db1}=\dfrac{A_{vg}f_y}{\sqrt3\gamma_{m0}}+\dfrac{0.9A_{tn}f_u}{\gamma_{m1}}=383.0$ kN
• $T_{db2}=\dfrac{0.9A_{vn}f_u}{\sqrt3\gamma_{m1}}+\dfrac{A_{tg}f_y}{\gamma_{m0}}=304.7$ kN → $T_{db}=304.7$ kN $>280$ ✓

Answer: ISMC 175 with 7 M20 (4.6) bolts at 50 mm pitch; design strength $=\min(554.5,\,658.0,\,304.7,\,316.8)=304.7$ kN $\ge280$ kN. Safe.