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Slab base for ISSC 200, 1200 kN

Short answer

Bearing strength9 MPa. Plate370×370 mm. Thickness from bending22.08 mm, use25 mm.4 M20 anchor bolts.,6 mm weld.

Design a slab base for ISSC 200 which has to carry a design axial load of 1200 kN resting on a concrete pedestal with M 20 grade of concrete.

Assumption: the section "ISSC 200" (SC 200) is taken with overall size h=200h=200 mm, bf=200b_f=200 mm and flange thickness tf≈15t_f\approx15 mm (only these three values are needed for the base plate). Factored axial load P=1200P=1200 kN, pedestal M20, E250 steel.

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

0.45fck=0.45×20=9.000.45f_{ck}=0.45\times20=9.00 N/mm²

2. Plate size

Areq=P0.45fck=1200×1039.00=133333 mm2A_{req}=\frac{P}{0.45f_{ck}}=\frac{1200\times10^3}{9.00}=133333\ \text{mm}^2

Projection of about 85 mm all round the column (not less than 50 mm): L=200+2×85=370L=200+2\times85=370 mm, B=200+2×85=370B=200+2\times85=370 mm. Area provided =136900=136900 mm² >Areq>A_{req} ✓

w=PBL=1200×103370×370=8.77 N/mm2<9.00 N/mm2w=\frac{P}{BL}=\frac{1200\times10^3}{370\times370}=8.77\ \text{N/mm}^2<9.00\ \text{N/mm}^2

3. Thickness (cl. 7.4.3.1)

Projections a=85a=85 mm and b=85b=85 mm (larger and smaller):

ts=2.5w(a2−0.3b2)γm0fy=2.5×8.77×(852−0.3×852)×1.10250=22.08 mmt_s=\sqrt{\frac{2.5w(a^2-0.3b^2)\gamma_{m0}}{f_y}}=\sqrt{\frac{2.5\times8.77\times(85^2-0.3\times85^2)\times1.10}{250}}=22.08\ \text{mm}

It must also be not less than the flange thickness tf=15t_f=15 mm. Provide ts=25t_s=25 mm.

4. Connections

  • The column end is machined to bear on the plate; connect with a nominal 6 mm fillet weld all round (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 370 × 370 × 25 mm, 4 anchor bolts M20, 6 mm weld, pedestal of M20 concrete.

✓ Answer checked by Claude Opus 5.5 (high effort)

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