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Splice plate for ISHB 350 column

Short answer

Design splice for 50% of factored load, 500 kN. Flange splice plates 250 x 10 mm with 4 M20 bolts each side per flange; web splice plates 2 x (200 x 8 mm) with 2 M20 bolts each side.

Design splice plate and the connection for a column section ISHB 350 to support a factored axial load of 1000 kN. Use M20 bolts of grade 4.6 and steel of grade E250. Assume the column ends are machined. The connection is to be spliced at a height of 3 meters.

Data: ISHB 350 @ 72.4 kg/m (h=350h=350, bf=250b_f=250, tf=11.6t_f=11.6, tw=10.1t_w=10.1 mm, A=9221A=9221 mm²), factored axial load 10001000 kN at the splice, ends machined for bearing. M20 bolts grade 4.6 (d0=22d_0=22 mm); E250.

Design basis

Because the column ends are machined and in full contact, most of the load passes by direct bearing. The splice plates and bolts are designed to carry at least 50 % of the factored load (PsP_s) and to hold the two lengths in line (against bending from accidental eccentricity and erection loads).

Ps=0.5×1000=500 kNP_s=0.5\times1000=500\ \text{kN}

The load is shared in proportion to area: flanges Af=2×250×11.6=5800A_f=2\times250\times11.6=5800 mm², web Aw=(350−2×11.6)×10.1=3301A_w=(350-2\times11.6)\times10.1=3301 mm².

  • Flanges: Pf=PsAfA=314.5P_f=P_s\dfrac{A_f}{A}=314.5 kN → 157.2 kN per flange
  • Web: Pw=PsAwA=179.0P_w=P_s\dfrac{A_w}{A}=179.0 kN

Bolt values (M20, 4.6)

  • Single shear: Vdsb=4003×2451.25×10−3=45.26V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26 kN; double shear (threads conservatively in both shear planes): 2×45.26=90.52\times45.26=90.5 kN
  • Bearing (e=40e=40, p=60p=60): kb=0.606k_b=0.606; on flange 11.6 mm =115.3=115.3 kN, on web 10.1 mm =100.4=100.4 kN
  • Flange bolt value 45.2645.26 kN (single shear, outer splice plate); web bolt value 90.590.5 kN (double shear, two web plates).

Flange splice

n=157.245.26=3.5n=\dfrac{157.2}{45.26}=3.5 → 4 bolts on each side of the joint per flange (2 rows of 2, pitch 60, gauge 120, end 40 mm). Flange splice plate 250×10250\times10 mm, length =2×(40+60+40)+=2\times(40+60+40)+ gap ≈300\approx300 mm. Stress in the plate =157.2×103250×10=62.9=\dfrac{157.2\times10^3}{250\times10}=62.9 N/mm² <fy/γm0=227<f_y/\gamma_{m0}=227 N/mm² ✓

Web splice

n=179.090.5=1.98n=\dfrac{179.0}{90.5}=1.98 → 2 bolts per side in double shear (2×90.5=1812\times90.5=181 kN >179>179 kN ✓); web plates (two, one on each face) 200×8200\times8 mm, 250 mm long, with 2 rows of bolts.

Notes

  • The splice is at 3 m height, so bending from lateral loads is small; the plates and bolts also hold the two lengths in line and resist erection loads.
  • Packing plates are provided if the two lengths differ in size.

Answer: flange splice plates 250 × 10 mm with 4 M20 bolts each side per flange; web splice plates 2 × (200 × 8 mm) with 2 M20 bolts each side.

✓ Answer checked by Claude Opus 5.5 (high effort)

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