Question

For Problems 1 and 2, design rectangular sections for the beams. Loads and Density values are shown. Beam weights are not included in the loads shown. Show sketches of cross sections, including bar sizes, arrangement, and spacing. Assume concrete weighs 150 lb/ft³, fy = 60,000 psi, and fc′ =4000 psi, unless given otherwise.

2. Design a cantilever beam. The loading cases are shown in Figure 2: Wp = 3 k/ft, w, = 1 kft MITIMIZ 12f- Use p= 0.1852 Figure 2

Answer 1

1 = 12 ft GIVEN DATA Span of beam, UDL Dead load, Live load, Grades of materials Strength of concrete, Yield Strength of steel, W W = = 3 1 kip/ft kip/ft f.' = fy = 4000 psi 60000 psi CALCULATIONS STEP-1: Computation of factored moment Assume self-weight = W, = = 0.4 kip/ft 1.2 x ( 3.4 5.68 kip/ft ) + 1.6 X 1 M,= = 5.68 x 122 12 408.96 kip.ft STEP-2: Computation of b and d = pfylfe' = 0.18 k, = fc' (1-0.590) = 643.54 Therefore, M - Øk, bd? = = bda 12000 408.96 8473 Hence, 12000 /(0.9 X 643.54 ) X in Possible choices: i) b and = h = = 12 in d = 27 in d+ (1 layer of reinf) 29.6 in ii) and b = h = = 16 in d = 23 in d+ (1 layer of reinf) 26 in iii) and b = h = = 18 in d = 22 in d+ (1 layer of reinf) 24.7 in Minimum overall depth for limiting deflection = 1/16 = 9 O.K Breadth of beam, Overall depth of beam Effective depth of beam, b h d = = = 16 in 26 in 23.5 in

16 x 26 / 144 STEP-3: Checking of dead load and revised Mu Self-weight = 0.15 kip/ft x = 0.43 kip/ft W, = 1.2 x 3 + = 5.716 kip/ft M, = 411.6 ft. kips 0.43 )+ 1.6 x 1 0.9 d = in STEP-4: Computation of area of reinforcement Assumin jd = (d-a/2) = 21.15 A, EM jd = 4.297 in? 3V1 30 Jcbd2f 2005 d = = 4# 1.189 < 1.253 in? 10 1.25 in? OK Select This will fit in the width of the beam. Area of steel provided, A, = 4.91 in? STEP-5: Computation of a, the depth of equivalent rectangular compressive stress block Assuming that the stress in steel, fs = fy the depth a of the equivalent stress block is obtained as: = 0.85fb 4.91 X 60000 0.85 X 4000 x 5.42 in 16 = STEP-6: Check whether f = fy a = 0.231 Note: B = 0.85 ) Forf' <= 4000 psi, B ,-0.85 ii) For 4000<f B. = 1.05-0.05f/1000 iii) For fe' >= 8000 psi, p=0.65 87,000 P87,000 + 1, = 0.503 > 0.231 (= a/d] Therefore, fs = fy Check if the section is tension-controlled or not Noi) For single layer of reinforcement, d, - d ii) For double layer, d = d-clear cover-stirrup dia - half of bar dia

Since the tension reinforcement is in single layer di = 23.5 in ald, = 0.231 al/d, = 0.375 BI = 0.319 > 0.23 [= a/d+] Therefore, the section is tension-controlled and Ø = 0.90 Note: i) = 0.9 for tension-controlled sections ii) = 0.7 for compression-controlled sections iii) = 0.356 +0.204/0/18 d.) for transition sections STEP-7: Computation of nominal moment capcaity of beam, M, Lever arm, id = (d-a/2) = 20.790 in Nominal moment capacity, M, = 4.f, (d-a/2) = 5 X 60000 x 20.790 = 6124734.0 in.lb = 6124734.0 / 12000 = 510.4 ft-kips The design moment capacity of the beam is obtained as: OM, = 0.90 x 510.4 = > 459.4 411.6 ft-kips OK k 16 in 4 # 10 23.5 in 26 in