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Slab base for ISHB 350@710.2, 1000 kN

Short answer

Bearing strength of M20 concrete is 9 MPa. Base plate 450×350 mm; pressure 6.35 MPa. Thickness from bending and flange requirement is 12 mm. Use 4 M20 anchor bolts and6 mm fillet weld.

Design a slab base for a column ISHB 350@710.2 N/m subjected to a factored axial compressive load of 1000 kN. Concrete pedestal of grade M20.

Data: ISHB 350 @ 710.2 N/m (72.4 kg/m): h=350h=350, bf=250b_f=250, tf=11.6t_f=11.6, tw=10.1t_w=10.1 mm. Factored load P=1000P=1000 kN; pedestal M20; steel E250 (fy=250f_y=250).

1. Bearing strength of concrete (IS 800:2007 cl. 7.4.1)

0.45fck=0.45×20=9.00.45f_{ck}=0.45\times20=9.0 N/mm²

2. Plate size

Areq=1000×1039.0=111111 mm2A_{req}=\frac{1000\times10^3}{9.0}=111111\ \text{mm}^2

Provide equal projections of 50 mm: L=350+2×50=450L=350+2\times50=450 mm, B=250+2×50=350B=250+2\times50=350 mm. Area =157500=157500 mm² >Areq>A_{req} ✓

w=1000×103450×350=6.35 N/mm2<9.0w=\frac{1000\times10^3}{450\times350}=6.35\ \text{N/mm}^2<9.0

3. Thickness (cl. 7.4.3.1)

Projections a=b=50a=b=50 mm:

ts=2.5w(a2−0.3b2)γm0fy=2.5×6.35×(502−0.3×502)×1.10250=11.06 mmt_s=\sqrt{\frac{2.5w(a^2-0.3b^2)\gamma_{m0}}{f_y}}=\sqrt{\frac{2.5\times6.35\times(50^2-0.3\times50^2)\times1.10}{250}}=11.06\ \text{mm}

Also ts≥tf=11.6t_s\ge t_f=11.6 mm. Provide ts=12t_s=12 mm.

4. Connections

  • Column to plate: 6 mm fillet weld all round (nominal; column end machined for bearing).
  • Anchor bolts: 4 nos. M20 (minimum), embedded 450 mm, with washer plate, placed outside the flanges.

Answer: base plate 450 × 350 × 12 mm, 4 anchor bolts M20, 6 mm weld.

✓ Answer checked by Claude Opus 5.5 (high effort)

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