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COLUMN BASE PLATE DESIGN (AISC360)

COLUMN BASE PLATE DESIGN

In accordance with AISC 360-05 Tedds calculation version 2.0.10

7520

075

Design forces and moments Axial force Pu = -32.0 kN (Tension) Bending moment Mu = 33.0 kNm Shear force Fv = 19.0 kN Eccentricity e = ABS(Mu / Pu) = 1031 mm Anchor bolt to center of plate f = N/2 - e1 = 125 mm

Column details Column section W 150x29.8 Depth d = 157 mm Breadth bf = 153 mm Flange thickness tf = 9 mm Web thickness tw = 7 mm

Baseplate details Depth N = 350 mm Breadth B = 350 mm Thickness tp = 20 mm Design strength Fy = 460.0 N/mm2

Foundation geometry Member thickness ha = 1000 mm Dist center of baseplate to left edge foundation xce1 = 200 mm Dist center of baseplate to right edge foundation xce2 = 200 mm Dist center of baseplate to bot edge foundation yce1 = 200 mm Dist center of baseplate to top edge foundation yce2 = 200 mm

Holding down bolt and anchor plate details Total number of bolts Nbolt = 4 Bolt diameter do = 22.2 mm Bolt spacing sbolt = 200 mm Edge distance e1 = 50 mm Minimum tensile strength of steel Fy = 460 N/mm2

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Compressive strength of concrete fc = 30 N/mm2 Strength reduction factors Compression c = 0.60 Flexure b = 0.90 Weld shear v = 0.75 Plate cantilever dimensions Area of base plate A1 = B N = 122500 mm2 Maximum area of supporting surface A2 = (N + 2 lmin) (B + 2 lmin) = 160000 mm2 Nominal strength of concrete under base plate Pp = 0.85 f'c A1 min((A2 / A1), 2) = 3570.0 kN Plate bending coefficient X X = (4 d bf) / (d + bf)2 min(1.0, Pu / (Pp c)) = -0.01 Plate bending coefficient l = 1 = 1.00 Bending line cantilever distance m m = (N - 0.95 d) / 2 = 100 mm Bending line cantilever distance n n = (B - 0.8 bf) / 2 = 114 mm Yield line theory cantilever distance n n = (d bf) / 4 = 39 mm Maximum bending line cantilever l = max(m, n, l n') = 114 mm Maximum bearing stress fp,max = 0.85 f'c c min((A2 / A1), 2) = 17.5 N/mm2 Maximum bearing pressure qmax = fp,max B = 6120.0 N/mm Bearing length - quadratic solution 1 Y1 = (f + N/2) + ((f + N/2)2 - (2 Pu (f - e))/qmax) = 584 mm Bearing length - quadratic solution 2 Y2 = (f + N/2) - ((f + N/2)2 - (2 Pu (f - e))/qmax) = 16 mm Bearing length Y = min(Y1, Y2) = 16 mm Tension force in bolts Tu = Mu / (N - 2 e1) - Pu / 2 = 148.0 kN Max tensile force in single bolt Trod = Tu / Nbolty = 74.0 kN

Base plate yielding limit at bearing interface Required plate thickness tp,req = ((4 fp,max Y (l - Y/2))/(b Fy)) = 17 mm

PASS - Thickness of plate exceeds required thickness

Base plate yielding limit at tension interface Distance from bolt CL to plate bending lines x = abs(l - e1) = 64 mm Plate thickness required tp,req = 2.11 ((Tu x)/(B Fy)) = 15 mm

PASS - Thickness of plate exceeds required thickness

Shear lug Length of shear lug llug = 7.87 in Thickness of shear lug tlug = 0.79 in Height of shear lug hlug = 3.94 in Grout pad thickness G = 0.98 in Effective depth of shear lug dlug = hlug - G = 75 mm Distance from center of column base plate xlug = 0 mm Depth of shear lug required dlug,req = Fv / (c 0.85 f'c llug) = 6 mm

PASS - Depth of shear lug exceeds required depth of shear lug Dist - top edge of shear lug to top edge of conc a1,lug = yce2 - llug/2 = 100 mm Dist - bot edge of shear lug to bot edge of conc a2,lug = yce1 - llug/2 = 100 mm Dist - front edge of shear lug to front edge of conc b1,lug = xce2 - tlug/2 - xlug = 190 mm Length of projected shear area LAv = min(a1,lug, b1,lug) + min(a2,lug, b1,lug) + llug = 400 mm Depth of projected shear area DAv = min(b1,lug + dlug, ha) = 265 mm Projected shear area Av = (LAv DAv) - (llug dlug) = 91000 mm2 Shear strength of concrete in front of shear lug Vu = 4 v (f'c 1psi) Av = 124.16 kN

PASS - Shear strength of concrete exceeds applied shear force

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Moment in shear lug Mul = Fv (G + dlug / 2) = 1.19 kNm Thickness of shear lug required tlug,req = (4 Mul / (b Fy llug))0.5 = 8 mm

PASS - Thickness of shear lug exceeds required thickness of shear lug Lug weld leg length Dweld,lug = 6.0 mm Distance between centroids of welds s = tlug + 2 Dweld,lug 1/3 = 24 mm Axial force on weld per inch fc = Mul / (s llug) = 247.40 N/mm Shear force on weld per inch fv = Fv / (llug 2) = 47.50 N/mm Resultant weld load fr = ((fc)2 + (fv)2)0.5 = 251.91 N/mm Nominal weld stress Fw = v 0.60 FEXX (1.0 + 0.5 (sin(90deg))1.5) = 325.78 N/mm2 Weld resistance Rn = Fw Dweld,lug 1 / (2)0.5 = 1382.15 N/mm

PASS - Strength of shear lug exceeds required strength of shear lug

Flange weld Flange weld leg length tfw = 8.0 mm Tension capacity of flange Ptf = bf tf Fy = 652.4 kN Force in tension flange Ftf = Mu / (d - tf) - Pu (bf tf) / Acol = 235.4 kN Critical force in flange Ff = min(Ptf, max(Ftf, 0kips)) = 235.4 kN Flange weld force per in Rwf = Ff / (2 bf - tw) = 786.1 N/mm Electrode classification number FEXX = 482.6 N/mm2

Nominal weld stress Fw = v 0.60 FEXX (1.0 + 0.5 (sin(90deg))1.5) = 325.8 N/mm2 Design strength of weld per in Rnf = Fw twf / (2) = 1382.2 N/mm

PASS - Available strength of flange weld exceeds force in flange weld

Longitudinal web weld Web weld leg length tww = 6.0 mm Web weld force per in Rwl = Fv / (2 (d - 2 tf)) = 68.6 N/mm Electrode classification number FEXX = 482.6 N/mm2 Nominal weld stress Fw = v 0.60 FEXX (1.0 + 0.5 (sin(90deg))1.5) = 325.8 N/mm2 Design strength of weld per in Rnl = Fw tww / (2) = 1382.2 N/mm

PASS - Available strength of longitudinal web weld exceeds force in longitudinal web weld

Transverse web weld Web weld leg length tww = 6.0 mm Effective width for bending with 45deg distribution beff = 2 ((N - d)/2 - e1) = 93 mm Web weld force Ftw = abs(Pu) (Acol - 2 bf tf) / Acol = 8.0 kN Web weld force per in Rwt = Ftw / (2 (d - 2 tf)) = 29.1 N/mm Electrode classification number FEXX = 482.6 N/mm2 Nominal weld stress Fw = v 0.60 FEXX (1.0 + 0.5 (sin(90deg))1.5) = 325.8 N/mm2 Design strength of weld per in Rnt = Fw tww / (2) = 1382.2 N/mm

PASS - Available strength of transverse web weld exceeds force in transverse web weld

ANCHOR BOLT DESIGN

In accordance with ACI318-08 Tedds calculation version 2.0.12

Anchor bolt geometry Type of anchor bolt Cast-in headed end bolt anchor Diameter of anchor bolt da = 22 mm Number of bolts in x direction Nboltx = 2 Number of bolts in y direction Nbolty = 2

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Total number of bolts ntotal = (Nboltx 2) + (Nbolty - 2) 2 = 4 Total number of bolts in tension ntens = (NboltN 2) + (Nbolty - 2) = 2 Spacing of bolts in x direction sboltx = 250 mm Spacing of bolts in y direction sbolty = 200 mm Number of threads per inch nt = 9 Effective cross-sectional area of anchor Ase = / 4 (da - 0.9743 in / nt)2 = 297 mm2 Embedded depth of each anchor bolt hef = 300 mm

Material details Minimum yield strength of steel fya = 248.0 N/mm2 Nominal tensile strength of steel futa = 400.0 N/mm2 Compressive strength of concrete fc = 30 N/mm2 Concrete modification factor = 1.00 Strength reduction factors Tension of steel element t,s = 0.75 Shear of steel element v,s = 0.65 Concrete tension t,c = 0.65 Concrete shear v,c = 0.70 Concrete tension for pullout t,cB = 0.70 Concrete shear for pryout v,cB = 0.70 Steel strength of anchor in tension (D.5.1) Nominal strength of anchor in tension Nsa = Ase futa = 118.85 kN Steel strength of anchor in tension Nsa = t,s Nsa = 89.14 kN

PASS - Steel strength of anchor exceeds max tension in single bolt

Check concrete breakout strength of anchor bolt in tension (D.5.2)

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Plan on foundationConcrete breakout - tension

131.35 kN

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Section A-A

The anchors are located at less than 1.5hef from 4 edges. Therefore the effective embedded depth has to be limited to 3.28" in accordance with D.5.2.3 Limiting embedded depth hef,lim = 83 mm Coeff for basic breakout strength in tension kc = 10 Breakout strength for single anchor in tension Nb = kc (f'c 1 N/mm2) hef,lim1.5 1 mm0.5 = 41.67 kN Projected area for groups of anchors ANc = 80000 mm2 Projected area of a single anchor ANco = 9 hef,lim2 = 62500 mm2 Min dist center of anchor to edge of concrete ca,min = 75 mm Mod factor for groups loaded eccentrically ec,N = min(1 / (1 + ((2 e'N) / (3 hef,lim))), 1) = 1.000 Modification factor for edge effects ed,N = 0.7 + 0.3 (ca,min / (1.5 hef,lim)) = 0.880 Modification factor for no cracking at service loads c,N = 1.000 Modification factor for cracked concrete cp,N = 1.000 Nominal concrete breakout strength Ncbg = ANc / ANco ed,N c,N cp,N Nb = 46.93 kN Concrete breakout strength Ncbg = t,c Ncbg = 30.51 kN

FAIL - Tension in bolts exceed breakout strength

Pullout strength (D.5.3) Net bearing area of the head of anchor Abrg = 1000 mm2

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Mod factor for no cracking at service loads c,P = 1.000 Pullout strength for single anchor Np = 8 Abrg f'c = 240.00 kN Nominal pullout strength of single anchor Npn = c,P Np = 240.00 kN Pullout strength of single anchor Npn = t,cB Npn = 168.00 kN

PASS - Pullout strength of single anchor exceeds maximum axial force in single bolt

Side face blowout strength (D.5.4)

The sideface blowout will be checked in the x and y directions as the edge distances for the bolts in both directions are less than hef / 2.5

Check x direction Axial force in group of anchors Nsfb = (N1) = 131.35 kN Edge distance ca1 = 75 mm Side face blowout strength for single anchor Nsb = (160 ca1 (Abrg) (f'c 1psi)) = 172.58 kN Distance between outer anchors along the edge s = (Nbolty-1) sbolty = 200 mm Nom side face blowout strength multiple anchors Nsbg = (1 + s / (6 ca1)) Nsb = 249.29 kN Side face blowout strength for multiple anchors Nsbg = t,c Nsbg = 162.04 kN

PASS - Sideface blowout strength exceeds tension in bolts

Check y direction Axial force in single anchor Nsfb = N1 / Nbolty = 65.68 kN Nom side face blowout strength for single anchor Nsb = ((1 + max(1, min((ca2 / ca1), 3))) / 4) (160 ca1 (Abrg)

(f'c 1psi)) = 115.06 kN Side face blowout strength for multiple anchors Nsb = t,c Nsb = 74.79 kN

PASS - Sideface blowout strength exceeds tension in bolts