DESIGN OF BASE PLATE FOR PIPE Inputs

Column Pipe Section P406.00mm (pipe dia)Height of Stanchion h = 6.20m

Ph1 = 10.00 KN (Friction force)Ph2 = 10.00 KN (Wind force against Friction and structure)N = 56 KN

Dimension of Base Plate D = 0.650 m b = 0.650 mDia. Of anchor Bolt = 0.030 m No of Anchor bolt in either dir. = 3

Anchor Bolt capacity Tension = 65 KN Shear = 30 KNGrade of Concrete = 21 N/mm2Steel yield Strength = 235 N/mm2

dt = 0.060 mdt /3 = 0.020 md = 0.590 mM = 62.00 KN-me = 1.107 m

Tension Force Effects :-

D/6 = 0.108 m(D/6 + dt/3) = 0.128 m

3- Tension force affects (Anchor Bolts) бc = 2N(e+D/2-dt) / (b.Xn(d-dt-Xn/3)), Z = N(e-D/2+Xn/3)

x/d = (e-D/2) / dx/d = 1.326

r =r = 0.006Xn/d = 0.4 (From Graph) Xn = 0.24 m

So σc1 = 1.82 N/mm2Z1 = 87.43 KN (Per 3 Anchor Bolts)

For Ph2 = 10.00 KNM = 62.00 KN-me = 1.107 m

3- Tension force affects (Anchor Bolts) бc = 2N(e+D/2-dt) / (b.Xn(d-dt-Xn/3)), Z = N(e-D/2+Xn/3)

x/d = 1.326Xn/d = 0.45 (From Graph) Xn = 0.27 m

σc2 = 1.62 N/mm2

88.43 KN (Per 3 Anchor Bolts)

CHECKS

1- Compression Check of Concrete

(σc1 + σc2) / fc' = 0.163 OK

2- Anchor Bolt Check

at / (b.d)

Z2 =

i) Tension(Z1+Z2) /no. of bolt = 58.62 KN(Z1+Z2) /no. of bolt /(Anchor Bolt Tension Capacity) = 0.90 OK

ii) Shear

0.167 OKCombination of Anchor Bolt Shear force shall be checked

1.069 OK

3- Check Of Base Plate Thickness

i) Compression

l = 172 mmb = 294 mm

therefore l / b = 0.584

Mx = 0.06 σc b2My = -0.27285019 σc l2

Mx = 8784.63 N-mm/mmMy = -13003.16 N-mm/mm

t = (6*Mmax / fb)^0.5t = 20 mm

ii) Tension

b = 159 mmZ = 27 t^2

σ = Me / Z < fbM = (Z1+Z2) /3 fb = 0.66 fy 1.25

t > 51 mm TEFLON SHEET ARE USED

((Ph12 +Ph22) ^0.5 /8) /Anchor Bolt Shear Capacity

Z = bt2 / 6