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Base plate for ISHB 450, 1400 kN

Short answer

Bearing strength 9 N/mm2; required area 155556 mm2; use 550 x 350 mm plate, pressure 7.27 N/mm2. Thickness 11.83 mm but not less than flange 13.7 mm, so provide 14 mm plate with 4 M20 anchor bolts and 6 mm fillet weld.

Design the base plate for the column ISHB 450 to carry a factored load of 1400 kN. Take E250 grade of steel and M20 grade of concrete.

Data: ISHB 450 @ 87.2 kg/m: h=450h=450, bf=250b_f=250, tf=13.7t_f=13.7, tw=9.8t_w=9.8 mm. Factored load P=1400P=1400 kN; steel E250; concrete M20.

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=1400×1039.00=155556 mm2A_{req}=\frac{P}{0.45f_{ck}}=\frac{1400\times10^3}{9.00}=155556\ \text{mm}^2

Projection of about 50 mm all round the column (not less than 50 mm): L=450+2×50=550L=450+2\times50=550 mm, B=250+2×50=350B=250+2\times50=350 mm. Area provided =192500=192500 mm² >Areq>A_{req} ✓

w=PBL=1400×103550×350=7.27 N/mm2<9.00 N/mm2w=\frac{P}{BL}=\frac{1400\times10^3}{550\times350}=7.27\ \text{N/mm}^2<9.00\ \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=2.5×7.27×(502−0.3×502)×1.10250=11.83 mmt_s=\sqrt{\frac{2.5w(a^2-0.3b^2)\gamma_{m0}}{f_y}}=\sqrt{\frac{2.5\times7.27\times(50^2-0.3\times50^2)\times1.10}{250}}=11.83\ \text{mm}

It must also be not less than the flange thickness tf=13.7t_f=13.7 mm. Provide ts=14t_s=14 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 550 × 350 × 14 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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