Unequal angle tension member with HSFG bolts

Given: design tension 200 kN; single unequal angle; HSFG bolts of property class 6.8 ($f_{ub}=600$ N/mm²); E250. The bolts are designed as a friction-grip connection (no slip at the design load): M20, $d_0=22$ mm, $\mu_f=0.48$ (sand-blasted surface), $K_h=1.0$, $n_e=1$ (single shear), $\gamma_{mf}=1.25$ (IS 800:2007 cl. 10.4.3).

1. Section

Required $A_g\ge\dfrac{200\times10^3\times1.10}{250}=880$ mm². Try ISA 100×75×8, $A_g=1336$ mm², connected through the 100 mm leg.

2. HSFG bolts

• Proof load $F_o=0.7f_{ub}A_{nb}=0.7\times600\times245\times10^{-3}=102.9$ kN
• Slip resistance $V_{dsf}=\dfrac{\mu_fn_eK_hF_o}{\gamma_{mf}}=\dfrac{0.48\times1\times1.0\times102.9}{1.25}=39.51$ kN
• Number of bolts $=\dfrac{200}{39.51}=5.1$ → 6 bolts in one line, pitch 50 mm ($\ge2.5d$), end distance 40 mm ($\ge1.5d_0=33$ mm), gauge 55 mm from the heel. Capacity $237.1$ kN.

3. Member checks

• Yielding: $T_{dg}=\dfrac{1336\times250}{1.10}=303.6$ kN
• Net rupture with shear lag: $A_{nc}=(100-4-22)\times8=592$ mm², $A_{go}=(75-4)\times8=568$ mm², $b_s=w+w_1-t=75+55-8=122$ mm, $L_c=250$ mm, $\beta=1.188$: $T_{dn}=174.8+153.4=328.1$ kN
• Block shear: $T_{db}=\min(384.7,312.2)=312.2$ kN

All exceed 200 kN. The slenderness limit (400) is checked with the member length (not given): $r_{min}=15.9$ mm allows $L\le6.3$ m.

Answer: ISA 100×75×8 connected by 6 M20 HSFG bolts (class 6.8) in a line at 50 mm pitch.