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Double angle tension member, M18, same side

Short answer

Use 2 ISA 75×75×6 placed back-to-back on same side of gusset plate. Each angle connected with 5 M18 grade 4.6 bolts at 50 mm pitch (10 bolts total). Bolt capacity 354.7 kN governs and exceeds 300 kN.

Design a tensile member having length 2.5m, subjected to design axial load of 300 kN. Use double angle section provided at same side of gusset plate and M18 bolts to connect with gusset plate.

Assumptions: the two angles are placed back to back (one leg of each touching) with their other legs bolted to the same face of the gusset plate, each angle with its own line of bolts. The bolts are therefore in single shear, and each angle carries half the load, 150150 kN. M18 bolts of grade 4.6, d0=20d_0=20 mm; gusset plate 10 mm; E250.

1. Bolts

  • Single shear, thread in the shear plane: Vdsb=4003×1.25×192×10−3=35.47V_{dsb}=\dfrac{400}{\sqrt3\times1.25}\times192\times10^{-3}=35.47 kN
  • Bearing (e=35e=35, p=50p=50): kb=min⁡(35/60, 50/60−0.25, 0.976, 1)=0.583k_b=\min(35/60,\ 50/60-0.25,\ 0.976,\ 1)=0.583; on the 6 mm angle leg Vdpb=2.5×0.583×18×6×4101.25×10−3=51.7V_{dpb}=\dfrac{2.5\times0.583\times18\times6\times410}{1.25}\times10^{-3}=51.7 kN (on the 10 mm gusset 86.1 kN)
  • Bolt value =35.47=35.47 kN → per angle n=150/35.47=4.2n=150/35.47=4.2 → 5 bolts per angle (10 in all), pitch 50 mm (min 2.5d=452.5d=45 mm), end distance 35 mm (≥1.5d0=30\ge1.5d_0=30 mm); capacity 5×35.47=177.45\times35.47=177.4 kN per angle.

2. Section

Ag≥300×1.10250×103=1320A_g\ge\dfrac{300\times1.10}{250}\times10^3=1320 mm². Try 2 ISA 75×75×6 (A=866A=866 mm² each, total 1732 mm²), bolt line 40 mm from the heel.

  • Yielding per angle: Tdg=866×2501.10=196.8T_{dg}=\dfrac{866\times250}{1.10}=196.8 kN
  • Rupture per angle: Anc=(75−3−20)×6=312A_{nc}=(75-3-20)\times6=312 mm², Ago=(75−3)×6=432A_{go}=(75-3)\times6=432 mm², bs=75+40−6=109b_s=75+40-6=109 mm, Lc=4×50=200L_c=4\times50=200 mm, β=1.4−0.076×756×250410×109200=1.084\beta=1.4-0.076\times\dfrac{75}{6}\times\dfrac{250}{410}\times\dfrac{109}{200}=1.084
Tdn=0.9×312×4101.25+1.084×432×2501.10=198.6 kN per angleT_{dn}=\frac{0.9\times312\times410}{1.25}+\frac{1.084\times432\times250}{1.10}=198.6\ \text{kN per angle}
  • Block shear per angle (Avg=1410A_{vg}=1410, Avn=870A_{vn}=870, Atg=210A_{tg}=210, Atn=150A_{tn}=150 mm²): min⁡(229.3, 196.0)=196.0\min(229.3,\ 196.0)=196.0 kN
  • Strength of the pair =2×min⁡(196.8, 198.6, 196.0, 177.4)=354.7=2\times\min(196.8,\ 198.6,\ 196.0,\ 177.4)=354.7 kN >300>300 kN ✓ (bolts govern)

3. Slenderness and tacking

The minimum radius of gyration of the pair (about the axis parallel to the gusset) equals that of one angle, r≈23r\approx23 mm, so L/r=2500/23=109<400L/r=2500/23=109<400 ✓. Tack the two angles together with bolts at intervals of not more than 1000 mm.

Answer: 2 ISA 75×75×6 placed back to back on the same side of the gusset, each connected with 5 M18 bolts (4.6) at 50 mm pitch (10 bolts in all).

✓ Answer checked by Claude Opus 5.5 (high effort)

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