Base Plate DesignMS601a
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DESIGN OF BASE PLATE FOR PIPE Inputs Column Pipe Section P406.00mm (pipe dia) Height of Stanchio 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 m Dia. Of anchor Bolt 0.030 m No of Anchor bolt in either 3 Anchor Bolt capaci Tension = 65 KN Shear = 30 KN Grade of Concrete 21 N/mm2 Steel yield Strength 235 N/mm2 dt = 0.060 m dt /3 = 0.020 m d = 0.590 m M = 62.00 KN-m e = 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)) x/d = (e-D/2) / d x/d = 1.326 r = r = 0.006 Xn/d = 0.4 (From Graph) Xn = 0.24 m So σc1 = 1.82 N/mm2 Z1 = 87.43 KN (Per 3 Anchor Bolts) For Ph2 = 10.00 KN M = 62.00 KN-m e = 1.107 m 3- Tension force affects (Anchor Bolts) бc = 2N(e+D/2-dt) / (b.Xn(d-dt-Xn/3)) x/d = 1.326 Xn/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 =
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Base Plate DesignMS601a
Transcript of Base Plate DesignMS601a
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
