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Tension splice for 250x16 and 250x8 flats

Short answer

For no slip at ultimate load, use 4 M20 grade 8.8 bolts per side (8 total) with 8 mm packing on thinner flat. For slip permitted, use 6 M20 grade 8.8 bolts per side (12 total) with same packing. Bolt values: 105.4 kN (slip-critical) and 79.5 kN (bearing) per bolt.

Design and detail a tension splice to connect 250mm×16mm flat with 250mm×8mm flat using two cover plate of thickness 6mm each to carry a tension of 400KN. Use M20 high strength bolts of the property class 8.8 if, (i) No slip is permitted at the ultimate load (ii) Slip is permitted at the ultimate load. Take fyf_y = 250 MPa and fuf_u = 410 MPa.

Data: factored tension 400 kN; main flats 250×16 and 250×8; two cover plates 6 mm each (2×6=12>82\times6=12>8 mm); M20 HSFG bolts class 8.8 (fub=800f_{ub}=800 N/mm²); d0=22d_0=22 mm; fy=250f_y=250, fu=410f_u=410 N/mm². Bolts are in double shear (ne=2n_e=2). The thinner flat (8 mm) governs the plate checks. The flats differ in thickness by 16−8=816-8=8 mm, so an 8 mm packing plate is placed on one face of the thinner flat to let the cover plates bear evenly.

Plate checks (common)

  • Gross section of 8 mm flat: Tdg=250×8×2501.10=454.5T_{dg}=\dfrac{250\times8\times250}{1.10}=454.5 kN >400>400 ✓
  • Net section (2 bolts in a line): Tdn=0.9(250−2×22)×8×4101.25=486.5T_{dn}=\dfrac{0.9(250-2\times22)\times8\times410}{1.25}=486.5 kN >400>400 ✓
  • Covers: 2×6×2502\times6\times250: Tdg=681.8T_{dg}=681.8 kN ✓

(i) No slip at ultimate load (slip-critical, γmf=1.25\gamma_{mf}=1.25)

  • Fo=0.7fubAnb=0.7×800×245×10−3=137.2F_o=0.7f_{ub}A_{nb}=0.7\times800\times245\times10^{-3}=137.2 kN
  • Vnsf=μfneKhFoγmf=0.48×2×1.0×137.21.25=105.4V_{nsf}=\dfrac{\mu_fn_eK_hF_o}{\gamma_{mf}}=\dfrac{0.48\times2\times1.0\times137.2}{1.25}=105.4 kN per bolt (μf=0.48\mu_f=0.48, sand-blasted)
  • n=400105.4=3.8n=\dfrac{400}{105.4}=3.8 → 4 bolts on each side (2 rows of 2), capacity 421.5421.5 kN.

(ii) Slip permitted at ultimate load (bearing type)

  • Shear (double shear, one plane in thread): Vdsb=8003×1.25(245+314)×10−3=206.6V_{dsb}=\dfrac{800}{\sqrt3\times1.25}(245+314)\times10^{-3}=206.6 kN; packing thicker than 6 mm: βpkg=1−0.0125×8=0.90\beta_{pkg}=1-0.0125\times8=0.90 → 185.9185.9 kN
  • Bearing on 8 mm flat (e=40e=40, p=60p=60): kb=0.606k_b=0.606; Vdpb=2.5kbdtfu1.25=79.5V_{dpb}=\dfrac{2.5k_bdtf_u}{1.25}=79.5 kN → bolt value 79.579.5 kN
  • n=40079.5=5.03n=\dfrac{400}{79.5}=5.03 → 6 bolts on each side (2 rows of 3), capacity 477.1477.1 kN.
  • On the 16 mm flat side the bolts bear on the two cover plates (12 mm): Vdpb=119.3V_{dpb}=119.3 kN, so 400/119.3=3.35400/119.3=3.35 → 4 bolts would be enough there; 6 bolts are used on both sides for symmetry.

Detailing

Pitch 60 mm (≥2.5d\ge2.5d), gauge 100 mm (edge distance 75 mm), end distance 40 mm; cover plates 250 mm wide, about 290 mm long in case (i) and 410 mm long in case (ii); 8 mm packing under one cover plate over the length of the thinner flat. Surface preparation (clean, sand-blasted faying surfaces) is required in case (i).

Answer: (i) 4 bolts M20 (8.8) on each side; (ii) 6 bolts M20 (8.8) on each side; 8 mm packing on the thinner flat.

✓ Answer checked by Claude Opus 5.5 (high effort)

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