Design a suitable base and anchor bolts for a column subjected to an axial load 500 kN and a wind moment of 175 kNm.

Given data:

Axial load = 500KN = 500000N

Wind moment = 175 KNm

Assuming, grade of concrete = M20 i.e., fck = 20N/mm²

Bearing strength of concrete = 0.45 × fck = 0.45 × 20 = 9N/mm²

bearing strength, a reduced value of 0.45fck is used against maximum of 0.60fck ; as recommended by the IS code 

Now, area of slab, A = factored load / bearing strength of concrete

= (0.15 × 500000) / 9 = 8333.34 mm² = 8334 mm²

Square base plate is provided; therefore

Sides of the base plate = L = B = √A = √8334 = 91.29 mm = 92 mm

Thickness of the base plate = t = √{[ 2.5 w ( a² – 0.3 b² ) ¥m] / fy}

Where, w = intensity of pressure = Load / area provided

a = bigger projection of base plate beyond column

b = smaller projection of base plate beyond column

In this case , a = b because we are assuming, base plate and column both are square in shape.

For holding down , anchor bolts are used.

According to standards, when axial compressive load is on action then only two bolts will be enough for holding .