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Splice of 18 mm flats with M18 10.9 bolts

Short answer

If slip is permitted at ultimate load, bearing-type design gives 708.5 kN. If slip is not permitted, the slip-critical resistance governs at 412.9 kN.

Two 18 mm thick steel flats are spliced by two 8 mm thick plates with four M18 high strength bolts of property class 10.9. Determine the ultimate design load carrying capacity of the connection (i) if slip is permitted at the ultimate load and (ii) if slip is not permitted at ultimate load. Assume that one shear plane intercepts the threads of the bolts. fyf_y = 250 MPa and fuf_u = 410 MPa. [Figure: flats 220 mm wide, 4 bolts in two rows (staggered), edge distances 55 mm, gauge 110 mm, 55 mm; longitudinal spacings 45 mm, 85 mm, 45 mm; cover plates 8 mm, main flats 18 mm]

Reading of the figure: 4 bolts M18 (class 10.9, fub=1000f_{ub}=1000 N/mm²) on each side of the splice, in two rows of two (gauge 110 mm, edge distances 55 mm, pitch 85 mm, end 45 mm); flats 220×18 mm; two cover plates 8 mm (total 16 mm); double shear; d0=20d_0=20 mm; fy=250f_y=250, fu=410f_u=410.

Plate checks

  • Flat (18 mm): Tdg=220×18×2501.10=900.0T_{dg}=\dfrac{220\times18\times250}{1.10}=900.0 kN; Tdn=0.9(220−2×20)×18×4101.25=956.4T_{dn}=\dfrac{0.9(220-2\times20)\times18\times410}{1.25}=956.4 kN
  • Covers (2 × 8 = 16 mm): Tdg=800.0T_{dg}=800.0 kN; Tdn=850.2T_{dn}=850.2 kN

(i) Slip permitted at ultimate load (bearing-type)

  • Shear (double shear, one plane in thread): Vdsb=10003×1.25(192+254)×10−3=206.0V_{dsb}=\dfrac{1000}{\sqrt3\times1.25}(192+254)\times10^{-3}=206.0 kN
  • Bearing on cover plates (16 mm, thinner than the flats): kb=min⁡(4560,8560−0.25,2.44,1)=0.750k_b=\min(\tfrac{45}{60},\tfrac{85}{60}-0.25,2.44,1)=0.750; Vdpb=2.5kbdtfu1.25=177.1V_{dpb}=\dfrac{2.5k_bdtf_u}{1.25}=177.1 kN
  • Bolt value 177.1177.1 kN; four bolts =708.5=708.5 kN
  • Design load =min⁡(708.5, 900.0, 956.4, 800.0)=708.5=\min(708.5,\ 900.0,\ 956.4,\ 800.0)=708.5 kN

(ii) Slip not permitted at ultimate load (slip-critical)

  • Fo=0.7×1000×192×10−3=134.4F_o=0.7\times1000\times192\times10^{-3}=134.4 kN
  • Vnsf=μfneKhFoγmf=0.48×2×1.0×134.41.25=103.2V_{nsf}=\dfrac{\mu_fn_eK_hF_o}{\gamma_{mf}}=\dfrac{0.48\times2\times1.0\times134.4}{1.25}=103.2 kN per bolt (μf=0.48\mu_f=0.48, γmf=1.25\gamma_{mf}=1.25)
  • Four bolts: 412.9412.9 kN
  • Design load =412.9=412.9 kN (slip resistance governs).

✓ Answer checked by Claude Opus 5.5 (high effort)

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