Contents AISC 360 05 Example 001
User Manual: AISC 360-05 Example 001
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Software Verification PROGRAM NAME: REVISION NO.: ETABS 0 AISC 360-05 Example 001 WIDE FLANGE MEMBER UNDER BENDING EXAMPLE DESCRIPTION The design flexural strengths are checked for the beam shown below. The beam is loaded with a uniform load of 0.45 klf (D) and 0.75 klf (L). The flexural moment capacity is checked for three unsupported lengths in the weak direction, Lb = 5 ft, 11.667 ft and 35 ft. GEOMETRY, PROPERTIES AND LOADING Member Properties W18X50 E = 29000 ksi Fy = 50 ksi Loading w = 0.45 klf (D) w = 0.75 klf (L) Geometry Span, L = 35 ft TECHNICAL FEATURES TESTED Section Compactness Check (Bending) Member Bending Capacities Unsupported length factors AISC 360-05 Example 001 - 1 Software Verification ETABS 0 PROGRAM NAME: REVISION NO.: RESULTS COMPARISON Independent results are comparing with the results of Example F.1-2a from the AISC Design Examples, Volume 13 on the application of the 2005 AISC Specification for Structural Steel Buildings (ANSI/AISC 360-05). ETABS Independent Percent Difference Compactness Compact Compact 0.00% Cb ( Lb =5ft) 1.004 1.002 0.20% 378.750 378.750 0.00% 1.015 1.014 0.10% 307.124 306.657 0.15% Cb ( Lb =35ft) 1.138 1.136 0.18% φb M n ( Lb =35ft) (k-ft) 94.377 94.218 0.17% Output Parameter φb M n ( Lb =5ft) (k-ft) Cb ( Lb =11.67ft) φb M n ( Lb =11.67ft) (k-ft) COMPUTER FILE: AISC 360-05 EX001 CONCLUSION The results show an acceptable comparison with the independent results. AISC 360-05 Example 001 - 2 Software Verification PROGRAM NAME: REVISION NO.: ETABS 0 HAND CALCULATION Properties: Material: ASTM A572 Grade 50 Steel E = 29,000 ksi, Fy = 50 ksi Section: W18x50 bf = 7.5 in, tf = 0.57 in, d = 18 in, tw = 0.355 in h = d − 2t f = 18 − 2 • 0.57 = 16.86 in h0 = d − t f =18 − 0.57 =17.43 in S33 = 88.9 in3, Z33 = 101 in3 Iy =40.1 in4, ry = 1.652 in, Cw = 3045.644 in6, J = 1.240 in4 rts = 40.1 • 3045.644 = 1.98 in 88.889 I y Cw = S33 Rm = 1.0 for doubly-symmetric sections Other: c = 1.0 L = 35 ft Loadings: wu = (1.2wd + 1.6wl) = 1.2(0.45) + 1.6(0.75) = 1.74 k/ft Mu = ∙ wu L2 = 1.74 352/8 = 266.4375 k-ft 8 Section Compactness: Localized Buckling for Flange: λ= bf 2t f = 7.50 = 6.579 2 • 0.57 AISC 360-05 Example 001 - 3 Software Verification PROGRAM NAME: REVISION NO.: λ p = 0.38 ETABS 0 E 29000 = 0.38 = 9.152 Fy 50 λ < λ p , No localized flange buckling Flange is Compact. Localized Buckling for Web: λ = h 16.86 = = 47.49 tw 0.355 λ p = 3.76 29000 E = 3.76 = 90.553 50 Fy λ < λ p , No localized web buckling Web is Compact. Section is Compact. Section Bending Capacity: M p =Fy Z 33 =50 • 101 =5050 k − in Lateral-Torsional Buckling Parameters: Critical Lengths: E 29000 Lp = 1.76 ry = 1.76 • 1.652 = 70.022 in = 5.835 ft Fy 50 E = Lr 1.95rts 0.7 Fy Lr = 1.95 • 1.98 0.7 Fy S33 ho Jc 1 + 1 + 6.76 S33 ho Jc E 2 29000 1.240 • 1.0 0.7 • 50 88.9 • 17.43 1 + 1 + 6.76 0.7 • 50 88.9 • 17.43 29000 1.240 • 1.0 2 AISC 360-05 Example 001 - 4 Software Verification PROGRAM NAME: REVISION NO.: ETABS 0 Lr = 16.966 ft Non-Uniform Moment Magnification Factor: For the lateral-torsional buckling limit state, the non-uniform moment magnification factor is calculated using the following equation: Cb = 2.5M max 12.5M max Rm ≤ 3.0 + 3M A + 4 M B + 3M C Eqn. 1 Where MA = first quarter-span moment, MB = mid-span moment, MC = second quarter-span moment. The required moments for Eqn. 1 can be calculated as a percentage of the maximum mid-span moment. Since the loading is uniform and the resulting moment is symmetric: 1 L M A = MC = 1− b 4 L 2 Member Bending Capacity for Lb = 5 ft: M= M = 1.00 max B 2 2 1 L 1 5 1− b = 1− = 0.995 MA = MC = 4 L 4 35 Cb = 12.5 (1.00 ) 2.5 (1.00 ) + 3 ( 0.995 ) + 4 (1.00 ) + 3 ( 0.995 ) Cb = 1.002 Lb < L p , Lateral-Torsional buckling capacity is as follows: M = M = 5050 k − in n p ϕb M= 0.9 • 5050 /12 n = ϕb M n 378.75 k − ft AISC 360-05 Example 001 - 5 Software Verification PROGRAM NAME: REVISION NO.: ETABS 0 Member Bending Capacity for Lb = 11.667 ft: M= M = 1.00 max B 2 2 1 L 1 11.667 1− b = 1− 0.972 MA = MC = = 4 L 4 35 Cb = 12.5 (1.00 ) 2.5 (1.00 ) + 3 ( 0.972 ) + 4 (1.00 ) + 3 ( 0.972 ) Cb = 1.014 L p < Lb < Lr , Lateral-Torsional buckling capacity is as follows: Lb − L p ≤ M p M n = C b M p − (M p − 0.7 Fy S 33 ) − L L p r 11.667 − 5.835 = = M n 1.014 5050 − ( 5050 − 0.7 • 50 • 88.889 ) 4088.733 k − in 16.966 − 5.835 ϕb M= 0.9 • 4088.733 /12 n = ϕb M n 306.657 k − ft Member Bending Capacity for Lb = 35 ft: M= M = 1.00 max B 2 2 1 L 1 35 1− b = 1− = 0.750 . MA = MC = 4 L 4 35 Cb = 12.5 (1.00 ) 2.5 (1.00 ) + 3 ( 0.750 ) + 4 (1.00 ) + 3 ( 0.750 ) (1.00 ) Cb = 1.136 Lb > Lr , Lateral-Torsional buckling capacity is as follows: AISC 360-05 Example 001 - 6 Software Verification ETABS 0 PROGRAM NAME: REVISION NO.: Fcr = Cbπ 2 E Lb r ts 2 Jc 1 + 0.078 S 33 ho L b rts 2 1.136 • π 2 • 29000 1.24 • 1 420 1 + 0.078 14.133 ksi Fcr = = 2 88.889 • 17.4 1.983 420 1.983 2 M n = Fcr S 33 ≤ M p M n= 14.133 • 88.9= 1256.245 k − in ϕb M= 0.9 • 1256.245 /12 n = ϕb M n 94.218 k − ft AISC 360-05 Example 001 - 7
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