Contents KBC 2009 Example 001
User Manual: KBC 2009 Example 001
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Software Verification PROGRAM NAME: REVISION NO.: ETABS 0 KBC 2009 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 6.5 kN/m (D) and 11 kN/m (L). The flexural moment capacity is checked for three unsupported lengths in the weak direction, Lb = 1.75 m, 4 m and 12 m. GEOMETRY, PROPERTIES AND LOADING Member Properties W460x74 E = 205,000 MPa Fy = 345 MPa Loading w = 6.5 kN/m (D) w = 11.0 kN/m (L) Geometry Span, L = 12m TECHNICAL FEATURES TESTED Section Compactness Check (Bending) Member Bending Capacities Unsupported length factors KBC 2009 Example 001 - 1 Software Verification PROGRAM NAME: REVISION NO.: ETABS 0 RESULTS COMPARISON Independent results are comparing with the results of ETABS. ETABS Independent Percent Difference Compact Compact 0.00% Cb ( Lb =1.75m) 1.004 1.002 0.20% φb M n ( Lb =1.75m) (kN-m) 515.43 515.43 0.00% Cb ( Lb =4m) 1.015 1.014 0.10% φb M n ( Lb =4m) (kN-m) 394.8 394.2 0.15% Cb ( Lb =12m) 1.136 1.136 0.00% φb M n ( Lb =12m) (kN-m) 113.48 113.45 0.03% Output Parameter Compactness COMPUTER FILE: KBC 2009 EX001 CONCLUSION The results show an acceptable comparison with the independent results. KBC 2009 Example 001 - 2 Software Verification PROGRAM NAME: REVISION NO.: ETABS 0 HAND CALCULATION Properties: Material: E = 205,000 MPa, Fy = 345 MPa Section: W460x74 bf = 191 mm, tf = 14.5 mm, d = 457 mm, tw = 9 mm h = d − 2t f = 457 − 2 • 14.5 = 428 mm h0 = d − t f = 457 − 14.5 = 442.5 mm S33 = 1457.3 cm3, Z33 = 1660 cm3 Iy =1670 cm4, ry = 42 mm, Cw = 824296.4 cm6, J = 51.6 cm4 = rts 1670 • 824296.4 = 50.45 mm 1457.3 I y Cw = S33 Rm = 1.0 for doubly-symmetric sections Other: c = 1.0 L = 12 m Loadings: wu = (1.2wd + 1.6wl) = 1.2(6.5) + 1.6(11) = 25.4 kN/m Mu = ∙ wu L2 = 25.4 122/8 = 457.2 kN-m 8 Section Compactness: Localized Buckling for Flange: = λ bf 191 = = 6.586 2t f 2 • 14.5 KBC 2009 Example 001 - 3 Software Verification PROGRAM NAME: REVISION NO.: ETABS 0 E 205, 000 = λ p 0.38 = 0.38 = 9.263 Fy 345 λ < λ p , No localized flange buckling Flange is Compact. Localized Buckling for Web: λ = h 428 = = 47.56 9 tw E 205, 000 = = 3.76 = 91.654 λ p 3.76 Fy 345 λ < λ p , No localized web buckling Web is Compact. Section is Compact. Section Bending Capacity: M p =Fy Z 33 =345 • 1660 =572.7 kN-m Lateral-Torsional Buckling Parameters: Critical Lengths: E 205, 000 Lp = 1.76 ry = 1.76 • 42 = 1801.9 mm = 1.8 m Fy 345 E = Lr 1.95rts 0.7 Fy 0.7 Fy S33 ho Jc 1 + 1 + 6.76 S33 ho Jc E 2 205, 000 51.6 • 1 0.7 • 345 1457.3 • 44.8 Lr = 1.95 • 50.45 1 + 1 + 6.76 0.7 • 345 1457.3 • 44.25 205, 000 51.6 • 1 2 KBC 2009 Example 001 - 4 Software Verification PROGRAM NAME: REVISION NO.: ETABS 0 Lr = 5.25 m 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 = 1.75 m: M= M = 1.00 max B 2 2 1 L 1 1.75 1− b = 1− 0.995 MA = MC = = 4 L 4 12 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 = 572.7 kN-m n p φb M= 0.9 • 572.7 n φb M n = 515.43 kN-m KBC 2009 Example 001 - 5 Software Verification PROGRAM NAME: REVISION NO.: ETABS 0 Member Bending Capacity for Lb = 4 m: M= M = 1.00 max B 2 2 1 L 1 4 MA = MC = 1− b = 1− = 0.972 4 L 4 12 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 ≤Mp M n = C b M p − (M p − 0.7 Fy S 33 ) L − L r p 4.00 − 1.80 = = M n 1.014 572.7 − ( 572.7 − 0.7 • 0.345 • 1457.3) 437.97 kN-m 5.25 − 1.80 φb M= 0.9 • 437.97 n φb M n = 394.2 kN-m Member Bending Capacity for Lb = 12 m: M= M = 1.00 max B 2 2 1 L 1 12 MA = MC = 1− b = 1− = 0.750 . 4 L 4 12 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: KBC 2009 Example 001 - 6 Software Verification PROGRAM NAME: REVISION NO.: Fcr = Cbπ 2 E Lb rts 2 Jc 1 + 0.078 S33 ho Lb rts ETABS 0 2 1.136 • π 2 • 205, 000 51.6 • 1 12000 1 + 0.078 86.5 MPa Fcr = = 2 1457.3 • 44.25 50.45 12000 50.45 2 M n = Fcr S 33 ≤ M p Mn = 86.5 • 1457.3 = 126.056kN-m φb M= 0.9 • 126.056 n φb M n = 113.45 kN-m KBC 2009 Example 001 - 7
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