Contents AISC 360 10 Example 003
User Manual: AISC-360-10 Example 003
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Software Verification PROGRAM NAME: REVISION NO.: ETABS 0 AISC-360-10 Example 003 COMPOSITE COLUMN DESIGN EXAMPLE DESCRIPTION Determine if the 14-ft.-long filled composite member illustrated below is adequate for the indicated axial forces, shears, and moments. The composite member consists of an ASTM A500 Grade B HSS with normal weight (145 lb/ft3) concrete fill having a specified compressive strength, f c′ = 5 ksi. GEOMETRY, PROPERTIES AND LOADING Member Properties HSS10x6 x⅜ E = 29,000 ksi Fy = 46 ksi Loading Pr = 129.0 kips Mr = 120.0 kip-ft Vr = 17.1 kips Geometry Height, L = 14 ft AISC-360-10 Example 003 - 1 Software Verification PROGRAM NAME: REVISION NO.: ETABS 0 TECHNICAL FEATURE OF ETABS TESTED Tension capacity of composite column design. RESULTS COMPARISON Independent results are referenced from Example I.1 from the AISC Design Examples, Version 14.0. ETABS Independent Percent Difference 129 129 0.00% Available Strength, ΦPn (kip) 342.9 354.78 -3.35% Required Strength, Mu (k-ft) 120 120 0.00% 130.58 130.5 0.06% 1.19 1.18 0.85% Output Parameter Required Strength, Fu (k) Available Strength, ΦbMn (k-ft) Interaction Equation H1-1a COMPUTER FILE: AISC-360-10 EXAMPLE 003.EDB CONCLUSION The ETABS results show an acceptable comparison with the independent results. AISC-360-10 Example 003 - 2 Software Verification PROGRAM NAME: REVISION NO.: ETABS 0 HAND CALCULATION Properties: Materials: ASTM A500 Grade B Steel E = 29,000 ksi, Fy = 46 ksi, Fu = 58 ksi 5000 psi normal weight concrete Ec = 3,900 ksi, f c′ = 5 ksi, wconcrete = 145 pcf Section dimensions and properties: HSS10x6x⅜ H = 10.0 in, B= 6.00 in, t = 0.349 in As = 10.4 in2, Isx = 137 in4, Zsx=33.8 in3, Isy = 61.8 in4 Concrete area ht = 9.30 in., bt = 5.30 in., Ac = 49.2 in.2, Icx = 353 in4, Icy = 115 in4 Compression capacity: Nominal Compressive Strength: ΦcPn= 354.78 kips as computed in Example I.4 Bending capacity: Maximum Nominal Bending Strength: Zsx = 33.8 in3 bi • hi 2 − 0.192 • ri 3 where ri = t 4 5.30 • (9.30) 2 3 114.7 in.3 = − 0.192 • (0.349) = 4 Zc = 0.85 • f c′ • Zc 2 0.85 • 5 • 115 1, 798.5 kip-in. =46 • 33.8 + = =149.9 kip-ft 2 12 in./ft M D = Fy • Z sx + AISC-360-10 Example 003 - 3 Software Verification PROGRAM NAME: REVISION NO.: ETABS 0 Available Bending Strength: = hn 0.85 • f c′ • Ac h ≤ i 2(0.85 • f c′ • bi + 4 • t • Fy ) 2 0.85 • 5 • 49.2 9.30 ≤ 2(0.85 • 5 • 5.30 + 4 • 0.349 • 50) 2 = 1.205 ≤ 4.65 =1.205 in. = Z sn = 2 • t • hn2 = 2 • 0.349 • (1.205) 2 = 1.01 in.3 Z cn =bi • hn2 =5.30 • (1.205) 2 =7.70 in.3 0.85 • f c′ • Z cn 2 0.85 • 5 • 7.76 1, 740 kip-in. = 1,800 − 46 • 1.02 − = = 144.63 kip-ft 2 12 in./ft M nx = M D − Fy • Z sn − Φ b M nx =0.9 • 144.63 = 130.16 kip-ft Interaction Equation H1-1a: Pu 8 Mu + Φ c • Pn 9 Φ b • M n ≤ 1.0 129 8 120 + ≤ 1.0 354.78 9 130.16 1.18 > 1.0 n.g. AISC-360-10 Example 003 - 4
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