Question

Make any reasonable assumptions Q1 (50 pts): Design the T-beam shown below for bending: 4 in 24 in 4ft Span=24' Use: Driveway and car parking fc=4,500 psi f= 60 ksi

Answer 1

L hy b = 24 ft 4 in 12 in (assumed) 24 in 22 in (=h -2.5] = h d = GIVEN DATA Span of beam Slab thickness Breadth of web, Overall depth of beam, Effective depth, Center-center distance left side right side Strength of concrete, Yield Strength of steel, Service live load, fo= fy 4 ft 4 ft 4500 psi 60000 psi 250 psf (for driveways and parking) = 4 X 4 /12 + 12 x 20 (144) CALCULATIONS STEP-1: Computation of factored bending moment Self-weight of beam, 150 pcf xl 450 plf Uniformly distributed live load, WLL = 250 psf x 1000 plf W DL = 4 Therefore, factored design load, w = 1.2 W DL + 1.6 W LL 2140 plf 2.1 kplf Factored design moment, M, = w ,1/8 154.1 kip.ft STEP-2: Effective flange width, b Clear distance left side 6 ft 12 in /12 5 ft right side 6 ft 12 in /12 5 ft a) b shall not exceed one-fourth of the span length b = 1/4 24 x 12 /4 72 in b) The effective overhanging slab width on each side of the web shall not exceed eight times the slab thickness, that is, = 8 xht 32 in b 32 + 12 + 32 76 in c) half the clear distance to the next beam web on each side b 60 / 2 + 12 + 60 / 2 72 in Therefore, b = 72 in (the smallest of the three values)

STEP-3: Computation of area of reinforcement, As Assuming j = 0.95, jd = 20.425 As = M º fy jd 60000 х ### ) 154.1 x 12000 / (0.9 x 1.68 in Provide 2 -# 9 bars and these will fit into the web in single layer Area of steel provided, As = 1.99 in? STEP-4: Check if A, >= A s,min A 31 foi byd > fy 200bd fy s, min 0.87 > 0.86 As > As, min OK CAPACITY CHECK STEP-5: Computation of a a 0.85 fib 2 x 0.9 X 0.43 60000 4500 x 72 in < 4 in Since a is less than hf, the beam has rectangular beam action STEP-6: Check whether f = f, a 0.02 d Note: B. = 0.825 i) For fi' <= 4000 psi, K1=0.85 ii) For 4000<fc' < 8000 psi, B = 1.05 – 0.05f/1000 iii) For f.' >= 8000 psi, ß,=0.65 87,000 = 1 d 87,000+ fy 0.4883 > 0.02 [= a/d] Therefore, fs = fy ab Check if the section is tension-controlled or not Note: 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 = 21.5 in ald 0.02 acid, 0.375 Bi 0.3094 > 0.02 [= a/d] Therefore, the section is tension-controlled and 0.90 Note: i) ©= 0.9 for tension-controlled sections ii) 0 = 0.7 for compression-controlled sections iii) 0 = 0.356 + 0.204/(a /(B d) for transition sections STEP-7: Computation of nominal moment capcaity of beam, M, Lever arm, jd = (d-a/2) 21.285 in Nominal moment capacity, M, = 4f,(d – a/2) 1.99 60000 = х X 21.3 2541429 in.lb 211.8 ft-kips The design moment capacity of the beam is obtained as: OM, = 0.90 х 211.8 n 190.61 ft-kips 154.1 ft-kips Hence, it is SAFE 4 in 24 in 2 -# 9 12 in