Content text PSAD 27 - Steel Design - Allowable Bending Stresses.pdf
PSAD 27: Steel Design – Allowable Bending Stresses Video Transcript SITUATION 1: A simply supported W360×134 beam spans 9 m and has the following properties: A = 17,100 mm2 Ix = 416 × 106 mm4 d = 356 mm Iy = 151 x106 mm4 tw = 11 mm Sx = 2340 × 103 mm3 bf = 368 mm Sy = 818 × 103 mm3 tf = 18 mm Fy = 345 MPa rT = 102 mm Cb = 1.0 Consider bending about the x-axis. 1. If the compression flange of the beam is laterally supported every 3 meters, find the allowable flexural stress, Fb. 2. If the compression flange of the beam is laterally supported at the midspan, find the allowable flexural stress, Fb. 3. If the compression flange of the beam is only laterally supported at the simple supports, find the allowable flexural stress, Fb. Solution: Consider bending about the x-axis. Compare the unbraced length to Lc , Lc = 200bf √Fy = (200)(368) √345 = 3962.491 mm Lc = 137,900 ( d Af ) (Fy) = 137900 356(345) 368(18) = 7437.303 mm The smaller value of Lc governs, Lc = 3962.491 mm. Thus, Lb = 3000 mm < 3962.491 mm (Lc ) This means that the beam is laterally supported, there will be no lateral torsional buckling. If Lb < Lc , check the compactness of the section. For the flange, bf 2tf = 368 2(18) = 10.222 170 √Fy = 170 √345 = 9.152 250 √Fy = 250 √345 = 13.460 Therefore, 170 √Fy < bf 2tf < 250 √Fy The flange is non-compact.
For the web, d tw = 356 11 = 32.364 1680 √Fy = 1680 √345 = 90.4482 Therefore, d tw < 1680 √Fy The web is compact. For members where Lb < Lc with non-compact elements, Fb = Fy (0.79 − 0.000762 bf 2tf √Fy) Fb = 345[0.79 − 0.000762(10.2222)(√345)] Fb = 222. 635 MPa If the beam is supported at midspan, then the new unsupported length will be Lb = 9000 2 = 4500 mm Lb = 4500 mm > 3962.491 mm (Lc ) This means that the beam is laterally unsupported, there will be lateral torsional buckling. For laterally unsupported beams, compute the lateral slenderness ratio, Lb rT = 4500 102 = 44.118 √ 703270Cb Fy = √ (703270)(1.0) 345 = 45.149 √ 3516330Cb Fy = √ (3516330)(1.0) 345 = 100.957 Since Lb rT < √ 703,270Cb Fy , the case falls under (506-8) only: Fbc = 82740Cb (Lb d Af ) = 82740(1.0) 4500 [ 356 (368)(18) ] = 342.116 MPa ( allowable bending stress in compression ) Fbt = 0.60Fy = (0.60)(345) = 207 MPa ( allowable bending stress in tension ) The capacity is the smaller value, Fb = 207 MPa.
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