Contents AISC 360 10 Example 001
User Manual: AISC-360-10 Example 001
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Software Verification PROGRAM NAME: REVISION NO.: ETABS 0 AISC-360-10 Example 001 COMPOSITE COLUMN DESIGN EXAMPLE DESCRIPTION Determine if the 14-ft.-long filled composite member illustrated below is adequate for the indicated dead and live loads. The composite member consists of an ASTM A500 Grade B HSS with normal weight (145 lb/ft3) concrete fill having a specified compressive strength, fc′ = 5 ksi. GEOMETRY, PROPERTIES AND LOADING Member Properties HSS10x6 x⅜ E = 29,000 ksi Fy = 46 ksi Loading PD = 32.0 kips PL = 84.0 kips Geometry Height, L = 14 ft AISC-360-10 Example 001 - 1 Software Verification PROGRAM NAME: REVISION NO.: ETABS 0 TECHNICAL FEATURE OF ETABS TESTED Compression capacity of composite column design. RESULTS COMPARISON Independent results are referenced from Example I.4 from the AISC Design Examples, Version 14.0. ETABS Independent Percent Difference Required Strength Pu (kip) 172.8 172.8 0.00% Available Strength ΦPn (kip) 342.93 354.78 3.34% Output Parameter COMPUTER FILE: AISC-360-10 EXAMPLE 001.EDB CONCLUSION The ETABS results show an acceptable comparison with the independent results. AISC-360-10 Example 001 - 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, Isy = 61.8 in4 Concrete area hi = H − 2 • t = 10 − 2 • 0.349 = 9.30 in. bi = B − 2 • t = 6 − 2 • 0.349 = 5.30 in. Ac= bi • hi − t 2 • (4 − π)= 5.30 • 9.30 − (0.349) 2 • (4 − π)= 49.2 in.2 Moment of inertia for bending about the y-y axis: ( H − 4 • t ) • bt3 t • ( B − 4 • t )3 (9π2 − 64) • t 4 B − 4•t 4•t = + + + π • t2 − I cy 12 6 36 • π 2 3• π 2 (10 − 4 • 0.349) • 5.303 0.349 • (6 − 4 • 0.349)3 (9π2 − 64) • 0.3494 + + + 12 6 36 • π 6 − 4 • 0.349 4 • 0.349 2 ) π • 0.3492 ( − 2 3• π = 114.3 in.4 = Design for Compression: Required Compressive Strength: Pu = 1.2 • PD + 1.6 • PL = 1.2 • 32.0 + 1.6 • 84.0 = 172.8 kips AISC-360-10 Example 001 - 3 Software Verification PROGRAM NAME: REVISION NO.: ETABS 0 Nominal Compressive Strength: E Pno = Pp = Fy • As + C2 • f c′ Ac + Asr s Ec where C2 = 0.85 for rectangular sections Asr = 0 when no reinforcing is present within the HSS Pno = 46 • 10.4 + 0.85 • 5 • (49.2 + 0.0) = 687.5 kips Weak-axis Elastic Buckling Force: As 0.6 + 2 C3 = ≤ 0.9 Ac + As 10.4 = 0.6 + 2 ≤ 0.9 49.2 + 10.4 0.9 controls = 0.949 > 0.9 EI eff = Es • I sy + Es • I sr + C3 • Ec • I cy = 29, 000 • 62.1 + 0 + 0.9 • 3,900 • 114.3 = 2, 201, 000 kip-in 2 Pe = π2 ( EI eff ) ( KL) 2 where K = 1.0 for a pin-ended member = Pe π2 • 2, 201, 000 = 769.7 kips 1.0 • (14.0 • 12) 2 Available Compressive Strength: Pno 688 = = 0.893 < 2.25 Pe 769.7 Therefore, use AISC Specification Equation I2-2: Pno ΦPn = ΦPno 0.658 Pe = 0.75 • 687.5 • (0.658)0.893 = 354.8 kips AISC-360-10 Example 001 - 4
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