Diseño Placa Base

Project

Job Ref.

Highland Engineering P.C.
700 Industrial Drive Suite A Cary, IL 60013 RSF Date Chk'd by Date Section Sheet no./rev.

1
App'd by Date

R

10/22/2010

COLUMN BASE PLATE DESIGN (AISC360-05) In accordance with the LRFD method
TEDDS calculation version 2.0.00;

1.5 " W 12x96 250.0 kips

Bolt diameter - 0.8 " Bolt embedment - 18.0 " Flange/base weld - 0.3 "Web/base weld - 0.2 "

19.0 "

20.0 kips

2800.0 kip_in

4.42 ksi 1.5 " 19 " 1.5 " Plan on baseplate 1.5 " 20.8 kips Elevation on baseplate

Design forces and moments Axial force; Bending moment; Shear force; Eccentricity; Anchor bolt to center of plate; Column details Column section; Depth; Breadth; Flange thickness; Web thickness; Baseplate details Depth; Breadth; Thickness; Design strength;Foundation geometry Member thickness; Dist center of baseplate to left edge foundation; Dist center of baseplate to right edge foundation; Dist center of baseplate to bot edge foundation; Dist center of baseplate to top edge foundation; Holding down bolt and anchor plate details Total number of bolts; Bolt diameter; Nbolt = 4 do = 0.750 in ha = 20.000 in xce1 = 50.000 in xce2 = 50.000 in yce1 =50.000 in yce2 = 50.000 in N = 22.000 in B = 22.000 in tp = 2.750 in Fy = 36.0 ksi Pu = 250.0 kips; (Compression) Mu = 2800.0 kip_in Fv = 20.0 kips e = ABS(Mu / Pu) = 11.200 in f = N/2 - e1 = 9.500 in

W 12x96 d = 12.700 in bf = 12.200 in tf = 0.900 in tw = 0.550 in

Project

Job Ref.

Highland Engineering P.C.
700 Industrial Drive Suite A Cary, IL 60013 RSF Date Chk'd by Date SectionSheet no./rev.

2
App'd by Date

R Bolt spacing; Edge distance; Minimum tensile strength of steel; Compressive strength of concrete; Strength reduction factors Compression; Flexure; Weld shear; Plate cantilever dimensions Area of base plate; Maximum area of supporting surface; Nominal strength of concrete under base plate; Plate bending coefficient X; Plate bending coefficient λ; Bending linecantilever distance m; Bending line cantilever distance n; Yield line theory cantilever distance n’; Maximum bending line cantilever; Check eccentricity Maximum bearing stress; Maximum bearing pressure; Critical eccentricity;

10/22/2010 sbolt = 19.000 in e1 = 1.500 in Fy = 36 ksi f’c = 4 ksi φc = 0.65 φb = 0.90 φv = 0.75 A1 = B × N = 484.000 in2 A2 = (N + 2 × lmin) × (B + 2 × lmin) = 10000.000 in2Pp = 0.85 × f'c × A1 × min(√(A2 / A1), 2) = 3291.2 kips X = (4 × d × bf) / (d + bf)2 × min(1.0, Pu / (Pp × φc)) = 0.12 λl = min(1, 2 × √(X) / (1 + √(1 - X))) = 0.35 m = (N - 0.95 × d) / 2 = 4.967 in n = (B - 0.8 × bf) / 2 = 6.120 in n’ = √(d × bf) / 4 = 3.112 in l = max(m, n, λl × n') = 6.120 in fp,max = 0.85 × f'c × φc × min(√(A2 / A1), 2) = 4.42 ksi qmax = fp,max × B = 97.24 kips/in ecrit = N /2 - Pu / (2 × qmax) = 9.715 in

e > ecrit so loads cannot be resisted by bearing alone. Therefore consider as a large moment Plate dimensions adequate as (f + N/2)2 >= (2 × Pu × (e + f))/qmax and a real solution for bearing length exists Bearing length - quadratic solution 1; Bearing length - quadratic solution 2; Bearing length; Tension force in bolts; Max tension in single bolt; Base plateyielding limit at bearing interface Required plate thickness; tp,req = √((4 × fp,max × Y × (l - Y/2))/(φb × Fy)) = 2.680 in PASS - Thickness of plate exceeds required thickness Base plate yielding limit at tension interface Distance from bolt CL to plate bending lines; Plate thickness required; x = abs(l - e1) = 4.620 in tp,req = 2.11 × √((Tu × x)/(B × Fy)) = 0.736 in PASS - Thickness of plateexceeds required thickness Frictional shear resistance Steel / concrete friction coefficient; Frictional shear resistance; μ = 0.4 φVn = min(φv × μ × qmax × Y, φv × 0.2 × f'c × Y × B, φv × 800psi × Y × B) = 36.77 kips PASS - Frictional shear resistance exceeds applied shear Y1 = (f + N/2) + √((f + N/2)2 - (2 × Pu × (e + f))/qmax) = 38.215 in Y2 = (f + N/2) - √((f + N/2)2 - (2 × Pu × (e + f))/qmax) =...