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For this case 1/10(L) = 8’(12)/10 = 9.6 in. Therefore, please all your main bars within the beam web and bring to temperature/shrinkage bars from the slab as close to the phase of the beams and spaced not further apart that the spacing for cracking control. S =[ b w – 2(cover) –2(d v ) – 2(2d v )]/(n-1) S = [14in – 2(1.5 in) – 2 (3/8in) – 2(3/4)]/(3-1) = 4.75 in Assume f s = (2/3)(60) = 40 ksi, then c c = 2.25 in + 3/8in = 2.625 in (the value of 2.25in. is to provide for the slab bars ) ( ) ( ) 40000 40000 15 2.5 15 2.5(2.625) 8.4 2 60000 3 40000 40000 12 12 12 2 60000 3 c s req s c in f S or in f = = = = S = 4.75 in < S req = 8.4 in O.K. b e (1/10)L b e > (1/10)L (1/10)L Main bars extra bars extra bars Main bars

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Main bars 3-#6 extra 2 bars from T/S [email protected] 8” 7 in 18 in b w =14 in Exterior Beam: The tributary width is 7ft + half of the beam width = 7 ft + 7”/12 7.6 ft Since the tributary width is about half the value for the interior beam then the moments are approximately also half the value found before and the required area of steel will be also about half: A s = 2.37/2 = 1.18 in 2 . Using 3 bars to satisfy cracking control select 3-#6 bars ( 1.32 in 2 , b = 8.8). With this area and the dimensions already found check the capacity of the beam. D= (30 + 87.5)7.6 + 160.4 = 1,053 lb/ft L = 100(7.6) = 760 lb/ft U = 1.2D + 1.6L w u = 1.2(1053) + 1.6(760) = 2,480 lb/ft Design for w u = 2.48k/ft 2 2 2.48(8) 79.4 k-ft 2 2 u u w M = = = l Check capacity d = d t = 18” – 2.25” – 3/8”-1/2(6/8”) = 15” (2.25” has been used instead of the 1.5” to accommodate the slab bars) Check A smin : ' 2 min 2 3 3 4000 (14)(15) 0.66 in 60000 200 200 (14)(15) 0.70 in 60000 c w y s w y f b d f A b d f = = = Use A s min = 0.70 in 2 A s > A s min O.K.
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