DESIGN OF BASE PLATE AS PER BS PROVISIONS

1 Strength of materials

Grade of concrete C = N/sq.mm

Grade of steel = N/sq.mm

2 Design loads

Max moment M = kN-m

Max axial force N = kN

Min. axial force N' = kN

Horizontal shear = kN

3 Assumptions

Assume base plate size = 1300 x 1000 x 75

Provided column size = 625 x 1000 x

4 Check for tension in bolts

a) Max moment + Max axial force

If there is no tension in bolts

b = M/N = mm

Distance to edge of comp. stress

X/2 = bp/2 - b = mm

Conc. Compressive stress 0.6xfcu = 24 N/sq.mm

Compression (area of stress block) = kN

Actual design force compression = kN

Required design stress = N/sq.mm

Overhang e = L1 - 0.8xSw = mm

Moment in the cantilever portion = Nmm per mm width

b) Max moment + Min axial force

Calculating the bolt tension and concrete compression

M = 0.6 fcu bp x (h-x/2) - N' (h-hp/2)

0.6xfcuxbpxhx = x

0.6xfcuxbpx0.5X^2= x^2

N (h-hp/2) =

substituting the values and the resulting quadratic equation

x^2 - 1850x + 136217.948717949 = 0

A = 1

B =

c =

X = 76.8 mm

solving the quadratic equation

Compression C = 2397 kN

Tension T = 2097 kN

Moment per mm width mc = Nmm per mm width

1850

136217.9487

536684.8945

28860000

15600

2125000000

40

275

1680

5000

300

75

314

336

19593.6

19593.6

11.4469

118

329.5

621395.6616

5 Base plate thickness

a) Compression side bending

Required base plate thicknes tp = sqrt ((4*mc)/pyp*bp)

= mm

b) Tension side bending

Required base plate thicknes tp = sqrt(4*mT)/pyp

mT = T*m

m = L1 - k - 0.8*sw

= 255 mm

mT = N mm

tp = 77.3 mm

Provide a Base plate thickness of size 96 mm

6 Holding down bolts & Anchorage

Number of bolts provided = 8

Force per bolt = 262 kN

Selecting the bolt spec. = M30 mm

Grade of bolt = 8.8

Capacity of bolt = 314 kN

Spacing of bolt = 180 mm

Provide 8 no's of M30 dia bolts at a spacing of 180 mm c/c of bolts

7 Shear Transfer to concrete

Check for horizontal shear transferred by friction

Available shear resistance = 0.3*Cmin

= 90 kN

Horizontal Shear can be entirely resisted in friction

8 welds - Base plate to column

Tension flange weld Column Properties

Force in the tension flange weld to be lesser of B =

a) The tenion capacity of flange = Bc*Tc*Py T = 19

= kN D =

Ac =

b) Force in the tension flange = (M/Dc-Tc) - N*Af/Ac Tf = 12

= kN

Therefore weld force per mm = kN / mm

Thickness of weld required @215 Nsqmm = 11.8 mm

Provide 12 mm fillet weld alround the column in flanges and web

1520.35

306.8

533642112

15000

5574.39

2.52675

95.0709

314