Question

Concrete ..  

Q use only ultimate - strenght design (strenght design) and calculate the design strenght (ultimate strenght *) capacity 0mn) of the beam section given belowi total of steel bar & 11 is equal fi= 4700 psi, fy = 75000 psi 14 in 6.25 in? -1 2o in 32 I 4717 3 in 12 IN .

Answer 1

Given data: 20 in uin 4#11 bars As 6.25 in 32 in - 4700 psi = 4.7 ksi 4#11 fy = 75000 psi = 75 ksi 0 Tsin b: 20 in K 12" 에 Assuming neutral axis lies with in frange: he= 4in bw = 12 in d=32-3=29 in 0.003 LT 111 Es 75 x 6.25 : 0.85 x 4.4 x 5.866" as web take Efx no Fs=6 fs. As 0.85 feite (ax 20) y lehe). own asumption is wrong and N.A lies in frango Since Neutral axis lies in frange, divide the T section into two parts cis Flange overhangs il stem di web and corresponding steel :. web and corresponding steel, Arfi + As-Asf. Aff 0

for case (i) : Fsf = (CC), Ay fy: 0.85 te he (b-bw) Ay * 75 = 0.85* 4.7 x 4 ( 20 -12) 1.7045 in Aff for coue 2 (As-Af) fy Fsw = (Gen. = 0.85.fl. bwa. 0.85 x 4.7 x 12 xa → (6.25 – 1.7045) x 75 = (hence OK!) » a: 7.111 in. > ne CNA a BE ßia 0.85 for fé's yksi 0.65 for f = 8 ksi 0.85-0.05(n) for 4 < fe' < 8 where n=fc-4 fel 4 0.85 ? 4.7 5 interpolating the valve to we obtain 14-4.5) B; = 0.85-(085-0,8) (4-5) 0.8 B; - 0,815. . CNA 7.111 0.815 8.725 in. 11

steel: verifying yield of tc ES d-CNA (from law of similar Mangles applies to strain profile CNA Es = Ec d- CNA CNA 0.0069 >0.005 Es : 0.003 29-8.725 8.725 [tension controls failure). from ACI : 9.3.2. 0.9 0.65 tension Control, compression control 'E. 0.002 0.005 My = :: for ts=0.0069; corresponding value of 0=0.9. Nominal moment capacity: Ay ty (d-he) + fy (Ag - Age) (d- ) = (1.7845)x 75 (29-4) + 75 (6.25 – 1.40us) x (24 – 7") & Mn = 12125.96 Kip-in + * Mn = 1010.5 kip.ft. Ø.Mn design strength of the beam = 0.9 x 1010.5 = 909.447 kip-ft.