Question

Problem #2 (33%) Design the T-beam for the floor system shown below. The floor has a 4-in. slab suspoted by 22-ft-span-length beams cast monolithically with the slab. Beams are 8 ft- 0 in. on center and have a web width of 12 22 in.; fe' 3000 psi in. and a total depth and fy 60,000 psi. Service loads are 0.125- ksf live load and 0.200-ksf dead load. The given dead load does not include the weight of the floor system. h 22 b, 12 8-0 8-0-

Answer 1

GIVEN DATA Span of beam 22 ft L Slab thickness hf= 4 in Breadth of web, bw= 12 in Overall depth, 22 in h= 19.5 in Ed-2.5] Effective depth, d 7 ft 8-12" 7 ft 8'-12" Clear distance, left side right side Strength of concrete, psi 3000 60000 psi Yield Strength of steel, f, = Superimposed dead load, Service live load, 125 psf 200 psf CALCULATIONS STEP-1: Computation of factored bending moment Self-weight of beam 18 /144 150 pcf x 725 plf 10 x 4 /12 12 x + Uniformly distributed superimposed dead load, 125 psf x 10 = 1250 plf Total dead load, W 1975 plf 200 psf x Uniformly distributed live load, 10 W LL= 2000 plf 1.2 w 1.6 w Therefore, design load, = 5570 plf 5.57 kplf w2/8 336.985 kip.ft Factored design moment, M= W u STEP-2: Effective flange width, b a) b shall not exceed one-fourth of the span length =L/4 22 x 12/4 66 in b) The effective overhanging slab width on each side of the web shall not exceed eight = 8x h times the slab thickness, that is, 32 in b = 32 12 32 76 in c) half the clear distance to the next beam web on each side 84 2 12 84 /2 b = 96 in

Therefore (the smallest of the three values) b = 66 in STEP-3: Computation of area of reinforcement, As Assuming j 0.95, jd= 18.525 м, As= /(0.9 x 336.985 х 12000 60000 18.53 X n2 4.04 3-# 11 bars and these will fit into the web in single layer Provide in2 Area of steel provided, As 4.45 STEP-4: Check if A, Ain 200b,d -b d2 A smin= w 0.64 0.78 As Asmn ОК САРАСITY CHЕСК STEP-5: Computation of a a = 0.85fb 60000 4.45 x 0.85 x 3000 x 66 in < in 1.59 4 Since a is less than hf, the beam has rectangular beam action STEP-6: Check whether f - f a 0.08154 Note: i) For f 4000 psi, B-0.85 В 3 0.85 ii) For 4000<fc' < 8000 psi, B,1.05 0.05/1000 Hi) For f 8000 psi, B0.65 а, 87,000 d 87,000f = 0.50306 > Therefore, fs fy 0.082 a/d] Check if the section is tension-controlled or not Note: i) For single layer of reinforcement, dd i) For double layer, d d-clear cover-stirrup dia half of bar dia Since the tension reinforcement is in single layer dt = 25.5 in

ald 0.06235 0.375 B ad, 0.062 a/d 0.31875 Therefore, the section is tension-controlled and Note: 0.9 for tension-controlled sections ii) 0.7 for compress ion-controlled sections 0.90 0.356 0.204/(a/(B d,) for transition sections i STEP-7: Computation of nominal moment capcaity of beam, M jd = (d-a/2) Lever arm, 18.705 in M A,f,(d-a/ 2) Nominal moment capacity, 4.99E+06 i 60000 in.lb 4.45 x 18.71 х ft-kips 416.2 The design moment capacity of the beam is obtained as: 374.57 ft-kips Фм, - 0.90 416.2 X > 336.985 ft-kips Hence, it is SAFE 4 in 22 in 3 - 11 12 in