Question

w Design en rectangular section for the la Snow sketches of cross sectionn and E-4000 psi. s are included in the loads ment, and spacing e L-MR DIV TITUTI 16- Usein

Answer 1

GIVEN DATA Span of beam, 16 ft Reinforcement ratio, 0.5 P UDL (throughout) Dead load, kip/ft Concentrated Ioads (16 ft from support) 30 kips Live load, PL = Grades of materials 4000 psi Strength of concrete, Yield Strength of steel 60000 psi CALCULATIONS STEP-1: Computation of factored moment Factored dead load, 1.2 x 2. Wn = 2.4 kip/ft Factored live load (udl), PL= 1.6 х 30 48 kips - 162/2 Factored design moment, My 2.4 x 48 x 16 1075.2 kip.ft STEP-2: Computation of b and d 0.85Bf 87000 P = 87000 0.029 Pfylfc 0.18 knfc' a(1-0.59a 643.54 м, bd Therefore, 12000 Hence 12000 /(0.9 x bd 1075.2 x 643.54 22277 in Possible choices: i)b 12 in d 43 in and d+(1 layer of reinf) h 46.1 in 16 in d 37 in h d+(1 layer of reinf) i) b and 40.3 in 18 in d 35 in b and d+(1 layer of reinf) h =

38.2 in Minimum overall depth for limiting deflection 1/16 12 O.K Breadth of beam, in 18 = Overall depth of beam Effective depth of beam 40 in d 37.5 in STEP-3: Computation of area of reinforcement jd (d-a/2) Assumin 0.9 d 33.75 in A, M fjd 7.08 in 200b d - b,d fy As S,n 2.25 in 2.135 2.25 in2 ОК Select 6 H 10 This will fit in the width of the beam. Area of steel provided 7.36 in A STEP-4: Computation of a, the depth of equivalent rectangular compressive stress block Assuming that the stress in steel, f =fy the depth a of the equivalent stress block is obtained as: A,5, 0.85 fb 60000 7.36 x х 4000 х 0.85 18 7.22 in STEP-5: Check whether f, fy а - 0.193 Note 0.85 ) For f <= 4000 psi, B-0.85 i) For 4000<f B, = 1.05-0.05 f. /1000 ii) For f 8000 psi, B1-0.65 В - 87,000 а, В 87,000 f d 0.193 [= a/d 0.503 Therefore, fs fy Check if the section is tension-controlled or not No i) 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 dt= 37.5 in ald, 0.193 0.375 B 0.19 a/d 0.319 Therefore, the section is tension-controlled and 0.90 i) 0.9 for tension-controlled sections i 0.7 for compression-controlled sections i0.356+0.204/ (a/(p d) for transition sections Note: STEP-6: Computation of nominal moment capcaity of beam, M Lever arm jd =(d-a/2) 33.890 in A,f,(d-a/2) Nominal moment capacity, м, 3 60000 = 14965824.0 in.lb 33.890 7 x X 14965824.0 /12000 ft-kips 1247.2 The design moment capacity of the beam is obtained as: = 1122.5 ft-kips M= 0.90 1247.2 X n > 1075.2 ОК 6 10 18 in 2.5 in 40 in 37.5 in