Question

Problem 2: The doubly reinforced section has 3 #9 bars as tension reinforcement and 2 #7 bars as compression reinforcement. Calculate the nominal moment capacity M n. Assume f'c= 4.5 ksi and fy = 60 ksi. The cover distances indicated on the figure are clear covers. 12" 1.5"I 16" As 1.5"I

Answer 1

GIVEN DATA Breadth of beam, b = 12 in Overall depth of beam h = 16 in Effective depth of beam, d = 13.94 in (=h-1.5-0.5(9/8)] Depth of compression reinf. d' = 1.94 in (= 1.5-0.5(7/8)] Reinforcements-tension 3 No. 9 single layer -compression 2 No. 7 single layer Strength of concrete, fi'= 4500 psi Yield Strength of steel, fy = 60000 psi CALCULATIONS STEP-1: The beam is divided into beam-1 and beam-2. Both compression and tension steel are assumed to yield, i.e., f;' = f, and f,f, Area of tension steel, As = 2.99 in? Area of compression steel, A, = 1.21 in? Area of steel in Beam-1 A = A' = 1.21 in? Area of steel in Beam-2 A = A.-Ag = 1.78 in? STEP-2: Compute a for beam-2 O= 4:21 = 0.85 Sub 2 X 60000 1 x 4500 x 2.33 in 12 = STEP-3: Check if compression steel yields or not that is whether f'af 1.94 in d' = 0.84 Note: i) For f' <= 4000 psi, B1=0.85 B = 0.825 ii) For 4000<fc' < 8000 psi, B = 1.05 -0.05 /1000 iii) Forf' >= 8000 psi, B=0.65 - 16 f La) B. 1 87000 1 (1- 60000/ 87000) 0.825 = 0.377 < 0.84 [= d'/a) Therefore, compression steel does not yield STEP-4: Reevaluate value of a (0.85f ba? +(0.003E A, - Afya-(0.003E, A, B,d) = 0 45900 a? -74130 a -168484.64 = 0 Solving the quadratic equation in a, we get a = 2.89 in

STEP-5: Check whether f = f, a = 0.208 a B 87,000 ) d 87,000+ S = 0.489 > 0.21 (= a/d] Therefore, fs = fy that is tension steel yields 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 le layer di = 33.5 in ald, = 0.087 a d , = 0.375 B = 0.31 > 0.09 (= a/d ] Therefore, the section is tension-controlled and 0 = 0.90 Note: i) Ø=0.9 for tension-controlled sections ii) =0.7 for compression-controlled sections iii) =0.356+ 0.204/(a / B d) for transition sections STEP-5: Check if minimum reinforcement is satisfied As min 31 feb 2005, d = 0.57 Since As>As,min 0.56 O.K STEP-6: Computation of nominal moment capcaity of beam, M. Beam-1 Lever arm, jd = = (d-d') 12 in Design moment capacity, M2 = 5 x 0.003x (d - d) = 42.3 ft-kips Beam-2 Lever arm, jd = (d-a/2) = 12.500 in Design moment capacity. 0M = 0x0.85fba(d -a/2) = 124.4 ft-kips The total design moment capacity of the beam is obtained as: % M, - M + M - 166.7 ft-kips