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Base Plate Report Sample Aisc

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Base Plate Sample calculation
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  Base Plate Analysis Analysis Classification (AISC Design Guide #1) Load Set: Load Set 1 Load Combination: 1.4DN direction:= 9.5603 in vs. eN = -10. inB direction:= 9.5603 in vs. eB = 0 in Classification:Axial Compression with a Large Moment in the strong direction. Pu = 140 K (Compression is positive)Mun = 0 K-ft     0 inMub = 116.67 K-ft     -10. inEdge Lengths:B = 22 inN = 22 in^2(Equations 3.3.7)Support Strength, f'c = 4 KsiLoaded Area, A1 = 484 in^2Support Area, A2 =The largest area contained on the support that is geometrically similar to and concentric with the loaded area. A2 = 484 in^2(ACI 10.17.1) = 2.21 Ksi Design Guide #1 sign conventionLoads and eccentricities shown in positive directionPage 1 of 3VAConnect 1.00.0000www.iesweb.com   Project: Base Plate 5 Billing Reference: 2200.025 IES Employee, IES File: Example Report.vcbp Monday, January 07, 2013  Load Set: Load Set 1Load Combination: 1.4D Base Plate Parameters: B = 22 in, N = 22 in, f = 9.5 in, Pu = 140 K, eN = 10. inFor large moments, fpu = f fpn (See Support Bearing Capacity)fpu = 2.21 Ksi Check Base Plate Dimensions: Base plate with large moment(Fig. 3.4.1 AISC Design Guide #1) 420.25 in^2 > 112.3 in^2Base Plate area is adequate= 2.9515 in= 2.21 Ksi * 22 in * 2.9515 in - 140 K = 3.501 K Base Plate Bending Demand: Tension Interface = 0.71959 ft-K/ftBase Plate Bending Demand: Bearing Interface = 4.9675 inMu(n) calculation is a crude attempt at including some 2-way bending in the analysis. The approach is based on the note at the end of section 3.4.2.= 6.12 in= 2.21 Ksi * 6.12 in^2.0 / 2.0 = 41.387 ft-K/ftSince Y < mMu(Bearing) = 41.387 ft-K/ft Figure shows bending lines of column and illustrates base plate under strong axis bending. Effective width shown by dashed lines. Column Depth, d = 12.7 inColumn Width, b = 12.2 inDepth Reduction Factor, = 0.95 (sect 3.1.3)Width Reduction Factor, g  = 0.95 (sect 3.1.3)= 2.21 Ksi * 2.9515 in * (4.9675 in - 2.9515 in / 2.0) = 22.776 ft-K/ftControlling Mu = max( 0.71959 ft-K/ft , 41.387 ft-K/ft) = 41.387 ft-K/ft Base Plate w/ Large Strong-Axis Moment (AISC Design Guide #1, Section 3.4) Moment arm for tension anchor group, x = 3.4675 inw = Effective width for tension anchor group, where the load distributes at 45 degree anglesw = 16.87 in Calculate Bearing Length, Y: Page 2 of 3VAConnect 1.00.0000www.iesweb.com   Project: Base Plate 5 Billing Reference: 2200.025 IES Employee, IES File: Example Report.vcbp Monday, January 07, 2013  Base Plate Design Support Bearing Capacity (ACI 318-08 10.14.1) Load Set: Load Set 1Load Combination: 1.4D fpu = 2.21 Ksi f  = 0.65 Support Strength, f'c = 4 KsiLoaded Area, A1 = 484 in^2Support Area, A2 = 484.01 in^2(The largest area contained on the support that is geometrically similar to and concentric with the loaded area) Unity = fpu / (   fpn) = 1.000 = 2.21 Ksi Plate Bending (AISC 360-10 F11) Load Set: Load Set 1Load Combination: 1.4D Mu = 3.4489 K-ft f= 0.90 , Fy = 36 Ksi, t = 2.5 in, w = width = 1 in   Mn = 4.2188 K-ftUnity = Mu / (   * Mn) = 0.818 (Eqn F11-1) For rectangular sections, 1.6 * My will not control Base Plate Detailing (AISC 360-10 J3) Messages:Bolt spacing is adequate.Bolt edge distances are adequate. Page 3 of 3VAConnect 1.00.0000www.iesweb.com   Project: Base Plate 5 Billing Reference: 2200.025 IES Employee, IES File: Example Report.vcbp Monday, January 07, 2013
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