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Slab base for ISMB 350, 850 kN service

Short answer

Factored load1275 kN. Bearing strength11.25 MPa. Base plate460×250 mm, pressure11.09 MPa. Thickness18 mm.4 M20 anchor bolts.,6 mm weld.

Design a slab base for a column ISMB 350 @ 52.4 kg/m to carry a service load of 850 KN. Assume Fe410 grade steel and M25 concrete.

Data: ISMB 350 @ 52.4 kg/m: h=350h=350, bf=140b_f=140, tf=14.2t_f=14.2, tw=8.1t_w=8.1 mm. Service load 850850 kN → factored load P=1.5×850=1275P=1.5\times850=1275 kN. Steel Fe410 (fy=250f_y=250); concrete M25.

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

0.45fck=0.45×25=11.250.45f_{ck}=0.45\times25=11.25 N/mm²

2. Plate size

Areq=P0.45fck=1275×10311.25=113333 mm2A_{req}=\frac{P}{0.45f_{ck}}=\frac{1275\times10^3}{11.25}=113333\ \text{mm}^2

Projection of about 55 mm all round the column (not less than 50 mm): L=350+2×55=460L=350+2\times55=460 mm, B=140+2×55=250B=140+2\times55=250 mm. Area provided =115000=115000 mm² >Areq>A_{req} ✓

w=PBL=1275×103460×250=11.09 N/mm2<11.25 N/mm2w=\frac{P}{BL}=\frac{1275\times10^3}{460\times250}=11.09\ \text{N/mm}^2<11.25\ \text{N/mm}^2

3. Thickness (cl. 7.4.3.1)

Projections a=55a=55 mm and b=55b=55 mm (larger and smaller):

ts=2.5w(a2−0.3b2)γm0fy=2.5×11.09×(552−0.3×552)×1.10250=16.07 mmt_s=\sqrt{\frac{2.5w(a^2-0.3b^2)\gamma_{m0}}{f_y}}=\sqrt{\frac{2.5\times11.09\times(55^2-0.3\times55^2)\times1.10}{250}}=16.07\ \text{mm}

It must also be not less than the flange thickness tf=14.2t_f=14.2 mm. Provide ts=18t_s=18 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 460 × 250 × 18 mm, 4 anchor bolts M20, 6 mm weld, pedestal of M25 concrete.

✓ Answer checked by Claude Opus 5.5 (high effort)

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