Contents AISC LRFD 93 Example 002
User Manual: AISC LRFD-93 Example 002
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Software Verification PROGRAM NAME: REVISION NO.: SAP2000 0 AISC LRFD-93 Example 002 WIDE FLANGE MEMBER UNDER COMBINED COMPRESSION & BIAXIAL BENDING EXAMPLE DESCRIPTION A check of the column adequacy is checked for combined axial compression and flexural loads. The column is 14 feet tall and loaded with an axial load, Pu = 1400 kips and bending, M ux , M uy = 200k-ft and 70k-ft, respectively. It is assumed that there is reverse-curvature bending with equal end moments about both axes and no loads along the member. The column demand/capacity ratio is checked against the results of Example 6.2 in the 3rd Edition, LRFD Manual of Steel Construction, pages 6-6 to 6-8. GEOMETRY, PROPERTIES AND LOADING Member Properties W14X176 E = 29000 ksi Fy = 50 ksi Loading Pu = 1,400 kips Mux = 200 kip-ft Muy = 70 kip-ft Geometry H = 14.0 ft TECHNICAL FEATURES TESTED Section compactness check (compression) Member compression capacity Member bending capacity Demand/capacity ratio, D/C AISC LRFD-93 Example 002 - 1 Software Verification PROGRAM NAME: REVISION NO.: SAP2000 0 RESULTS COMPARISON Independent results are hand calculated and compared with the results from Example 6.2 in the 3rd Edition, LRFD Manual of Steel Construction, pages 6-6 to 6-8. SAP2000 Independent Percent Difference Compactness Compact Compact 0.00% φc Pn (kips) 1937.84 1937.84 0.00% φb M nx (k-ft) 1200 1200 0.00% φb M ny (k-ft) 600.478 600.478 0.00% 0.974 0.974 0.00% Output Parameter D/C COMPUTER FILE: AISC LRFD-93 EX002 CONCLUSION The results show an exact comparison with the independent results. AISC LRFD-93 Example 002 - 2 Software Verification PROGRAM NAME: REVISION NO.: SAP2000 0 HAND CALCULATION Properties: Material: ASTM A992 Grade 50 Steel Fy = 50 ksi, E = 29,000 ksi Section: W14x176 A = 51.8 in2, bf = 15.7 in, tf = 1.31 in, d = 15.2 in, tw = 0.83 in hc = d − 2t f = 15.2 − 2 • 1.31 = 12.58 in Ix = 2,140 in4, Iy = 838 in4, rx = 6.4275 in, ry = 4.0221 in Sx = 281.579 in3, Sy = 106.7516 in3, Zx = 320.0 in3, Zy = 163.0 in3. Member: Kx = Ky = 1.0 L = Lb = 14 ft Other φc =0.85 φb =0.9 Loadings: Pu = 1400 kips Mux = 200 k-ft Muy = 70 k-ft Section Compactness: Localized Buckling for Flange: / 2) ( b= (15.7 f / 2) = λ = 5.99 tf 1.31 λ= p 65 = Fy 65 = 9.19 50 λ < λ p , No localized flange buckling Flange is Compact. AISC LRFD-93 Example 002 - 3 Software Verification PROGRAM NAME: REVISION NO.: SAP2000 0 Localized Buckling for Web: h 12.58 λ= c= = 15.16 tw 0.83 φb Py = φb Ag Fy = 0.9 • 51.8 • 50 = 2331 kips Pu 1400 = = 0.601 φb Py 2331 Pu Since = 0.601 > 0.125 φb Py = λp 191 Fy P 2.33 − u φb Py 253 ≥ Fy 191 253 = ( 2.33 − 0.601 ) 46.714 ≥ = 35.780 50 50 λ < λ p , No localized web buckling Web is Compact. = λp Section is Compact. Member Compression Capacity: For braced frames, K = 1.0 and KxLx = KyLy = 14.0 ft, From AISC Table 4-2, φc Pn = 1940 kips Or by hand, = λc K y L Fy 1.0 • 14 • 12 50 = = 0.552 ry π E 4.022 • π 29000 Since λ c < 1.5, ( Fcr = Fy 0.658λc 2 50 • 0.658 )= 0.5522 = 44.012 ksi φc Pn = φc Fcr Ag = 0.85 • 44.012 • 51.8 φc Pn = 1937.84 kips AISC LRFD-93 Example 002 - 4 Software Verification PROGRAM NAME: REVISION NO.: SAP2000 0 From LRFD Specification Section H1.2, Pu 1400 = = 0.722 > 0.2 φc Pn 1937.84 Therefore, LRFD Specification Equation H1-1a governs. Section Bending Capacity 50 • 310 = = 1333.333 k-ft M F= px yZx 12 M py = Fy Z y Zy 163 However,= = 1.527 > 1.5, S y 106.7516 So Zy = 1.5 S y = 1.5 • 106.7516 = 160.1274in 3 = M py 50 • 160.1274 = 667.198 k-ft 12 Member Bending Capacity From LRFD Specification Equation F1-4, L p = 1.76ry E Fyf 1.76 • 4.02 L= p 29000 1 • = 14.2 ft > L= 14 ft b 12 50 φb M nx = φb M px φb M nx =0.9 • 1333.333 φb M nx = 1200 k-ft φb M ny = φb M py φb M ny = 0.9 • 667.198 φb M ny = 600.478 k-ft AISC LRFD-93 Example 002 - 5 Software Verification PROGRAM NAME: REVISION NO.: SAP2000 0 Interaction Capacity: Compression & Bending From LRFD Specification section C1.2, for a braced frame, Mlt = 0. M ux = B1x M ntx , where M ntx = 200 kip-ft; and M uy = B1 y M nty , where M nty = 70 kip-ft B1 = Cm P 1 − u Pe1 ≥1 For reverse curvature bending and equal end moments: M1 = +1.0 M2 M C m = 0.6 − 0.4 1 M2 C m = 0.6 − 0.4(1.0 ) = 0.2 pe1 = = pe1x π2 EI ( KL ) 2 π2 • 29000 • 2140 = 21, 702 kips 2 (14.0 • 12 ) π2 • 29000 • 838 = 8, 498 pe1 y = 2 (14.0 • 12 ) B1x = C mx ≥1 Pu 1 − Pe1x 0.2 = = 0.214 ≥ 1 B1x 1400 1 − 21702 AISC LRFD-93 Example 002 - 6 Software Verification PROGRAM NAME: REVISION NO.: SAP2000 0 B1x = 1 C my ≥1 P u 1 − P e1 y 0.2 B1 y = = 0.239 ≥ 1 1400 1 − 8498 B1 y = 1 B1 y = M ux = 1.0 • 200 = 200 kip-ft; and M uy = 1.0 • 70 = 70 kip-ft From LRFD Specification Equation H1-1a, 1400 8 200 70 + + 0.974 < 1.0 , OK = 1940 9 1200 600.478 D = 0.974 C AISC LRFD-93 Example 002 - 7
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