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Slab base for SC 220, 1000 kN, M30

Short answer

Bearing strength13.5 MPa. Plate320×320 mm, pressure9.77 MPa. Thickness16 mm (≥ flange).4 M20 anchor bolts.,6 mm weld.

Design a slab base for a column SC220 to transfer an axial load of 1000 kN. Take Fe410 grade steel and M30 for concrete.

Assumption: SC 220 (IS 808 column section) is taken as h=220h=220 mm, bf=220b_f=220 mm, tf≈16t_f\approx16 mm. The load of 1000 kN is taken as the factored axial load. Concrete M30 (fck=30f_{ck}=30), steel Fe410/E250 (fy=250f_y=250).

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

0.45fck=0.45×30=13.500.45f_{ck}=0.45\times30=13.50 N/mm²

2. Plate size

Areq=P0.45fck=1000×10313.50=74074 mm2A_{req}=\frac{P}{0.45f_{ck}}=\frac{1000\times10^3}{13.50}=74074\ \text{mm}^2

Projection of about 50 mm all round the column (not less than 50 mm): L=220+2×50=320L=220+2\times50=320 mm, B=220+2×50=320B=220+2\times50=320 mm. Area provided =102400=102400 mm² >Areq>A_{req} ✓

w=PBL=1000×103320×320=9.77 N/mm2<13.50 N/mm2w=\frac{P}{BL}=\frac{1000\times10^3}{320\times320}=9.77\ \text{N/mm}^2<13.50\ \text{N/mm}^2

3. Thickness (cl. 7.4.3.1)

Projections a=50a=50 mm and b=50b=50 mm (larger and smaller):

ts=2.5w(a2−0.3b2)γm0fy=2.5×9.77×(502−0.3×502)×1.10250=13.71 mmt_s=\sqrt{\frac{2.5w(a^2-0.3b^2)\gamma_{m0}}{f_y}}=\sqrt{\frac{2.5\times9.77\times(50^2-0.3\times50^2)\times1.10}{250}}=13.71\ \text{mm}

It must also be not less than the flange thickness tf=16t_f=16 mm. Provide ts=16t_s=16 mm.

4. Connections

  • The column end is machined to bear on the plate; connect with a nominal 6 mm fillet weld all round (the load is transferred mainly by contact).
  • Anchor bolts: 4 nos. M20, embedded about 450 mm with a washer plate (for erection and to resist any uplift).

Answer: base plate 320 × 320 × 16 mm, 4 anchor bolts M20, 6 mm weld, pedestal of M30 concrete.

✓ Answer checked by Claude Opus 5.5 (high effort)

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