Question

4 (30 points). (a) Compute M, for the three beams shown in the figure below. In each case, c=4000 psi, f = 60 ksi, b = 12 in., d = 32.5 in., and h = 36 in. (b) From the results of part (a), comment on whether adding compression reinforcement is a cost-effective way of increasing the strength, pM,, of a beam. 12 in. 3.5 in. 2.5 in 2 No. 9 bars 4 No. 9 bars 32.5 in. 6 No. 9 bars The procedure for computing strength of a beam with compression reinforcement is on the next page for your convenience. Use the procedure for the 2nd and 3rd beams. For this problem, take d, as d. HT

Answer 1

GIVEN DATA b = h = da Breadth of beam, Overall depth of beam, Effective depth of beam, Reinforcement Strength of concrete, Yield Strength of steel, 12 in 36 in 32.5 in 6 No.9 4000 psi 60000 psi (double layer) fe's fy = CALCULATIONS STEP-1: Computation of a , the depth of equivalent rectangular compressive stress block Area of steel, A. = 5.96 in Assuming that the stress in steel, f. = f, the depth a of the equivalent stress blcok is obtained as: 0.85fb 5.96 X 60000 0.85 x 4000 x 8.76 in 12 STEP-2: Check whether f = f. a = B, = 0.27 Note: 0.85i) Forf,' <= 4000 psi, B,=0.85 ii) For 4000<fc' < 8000 psi, B = 1.05 -0.05%/1000 iii) Forf,' >= 8000 psi, B,=0.65 Š more on Bl 87,000 P87,000+f,) 0.503 > 0.270 (= a/d] = Therefore, fs = fy Check if the section is tension-controlled or not Note: i) For single layer of reinforcement, d, - a ii) For double layer, d, = d-clear cover-stirrup dia - half of bar dia Since the tension reinforcement is in double layer d = 32.5 in ald, = 0.27 adid, = 0.375 B. = 0.319 > 0.270 [= a/d] Therefore, the section is tension-controlled and % = 0.90 Note: i) = 0.9 for tension-controlled sections

ii) iii) = 0.7 for compression-controlled section: = 0.356 + 0.204/(a / B d.) for transition sections STEP-3: Computation of nominal moment capcaity of beam, M. Lever arm, jd = (d-a/2) = 28.120 in Nominal moment capacity, M, = 4,1, (d-a/2) = 5.96 X 60000 X 28.120 = 10055712.0 in.lb = 10055712.0 / 12000 = 838.0 ft-kips The design moment capacity of the beam is obtained as: OM, = 0.90 X 838.0 = 754.2 ft-kips

GIVEN DATA Breadth of beam, Overall depth of beam Effective depth of beam, Depth of compression reinf. Reinforcements-tension -compression Strength of concrete, Yield Strength of steel, b = 12 in h = 36 in d = 32.5 in d' = 3.5 in 6 No. 9 double layer 2 No. 9 single layer fc's 4000 psi 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 fo=fy Area of tension steel, A, = 5.97 in? Area of compression steel. As = 1.99 in? Area of steel in Beam-1 A = A,' = 1.99 in? Area of steel in Beam-2 A = A, A1 = 3.98 in? STEP-2: Compute a for beam-2 0.85 fb 4 x 60000 1 x 4000 x 5.86 in 12 = STEP-3: Check if compression steel yields or not that is whether f,'=fy d' = 3.5 in d = 0.6 Note: i) For f' <= 4000 psi, :-0.85 B = 0.85 ii) For 4000<fc' < 8000 psi, B. = 1.05-0.05 f/1000 iii) For f' >= 8000 psi, B1=0.65 87000) = 1 (1- 60000/ 87000) 0.85 = 0.366 < 0.6 [= d'/a] Therefore, compression steel does not yield STEP-4: Reevaluate value of a (0.85% ba? +(0.003E, 4 - 4,1)a -(0.003E, A, B,d') = 0 40800 a? -185070 a -515061.75 = 0 Solving the quadratic equation in a, we get a = 6.49 in

STEP-5: Check whether f = fy a = 0.2 ay B 87,000 87,000+ f, = 0.504 > 0.2 [= 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 double layer d = 32.5 in ald, = 0.2 ad/d, = 0.375 B. = 0.319 > 0.2 [= a/d] Therefore, the section is tension-controlled and = 0.90 Note: i) =0.9 for tension-controlled sections ii) Ø=0.7 for compression-controlled sections iii) 6=0.356+ 0.204/(a /(B d) for transition sections STEP-5: Check if minimum reinforcement is satisfied A3,00 = 3/1.b.dz 2005d 1.3 = 1.24 Since As>As,min O.K STEP-6: Computation of nominal moment capcaity of beam, M, Beam-1 Lever arm, jd = (d-d) = 29 in Design moment capacity. Mn3 = PEA (1-B, 1X0.003x (d - d') = 203.9 ft-kips Beam-2 Lever arm, id = (d-a/2) = 29.260 in MR2 = 6 x 0.85f.ba(d-a/2) = 581.1 ft-kips Design moment capacity. The total design moment capacity of the beam is obtained as: O M,= ØM 1 + M2 = 785.0 ft-kips

GIVEN DATA Breadth of beam, b = 12 in Overall depth of beam h = 36 in Effective depth of beam, d = 32.5 in Depth of compression reinf. d' = 3.5 in Reinforcements-tension 6 No. 9 double layer -compression 4 No. 9 double layer Strength of concrete, fi'= 4000 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 = 5.97 in? Area of compression steel, A, = 3.98 in? Area of steel in Beam-1 A = A' = 3.98 in Area of steel in Beam-2 A = A.-Ag = 1.99 in? STEP-2: Compute a for beam-2 o = = A , zf, 0.85 Sib 2 X 60000 1 x 4000 x 2.93 in 12 = Note: STEP-3: Check if compression steel yields or not that is whether f'af d' = 3.5 in d' = 1.2 a i) For f' <= 4000 psi, B1=0.85 B = 0.85 ii) For 4000<fc' < 8000 psi, P = 1.05 -0.05f/1000 iii) Forf' >= 8000 psi, B=0.65 a = 111 f, la B 87000 = 1 (1- 60000/ 87000) 0.85 = 0.366 < 1.2 [= 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 40800 a -11940 a -1030123.5 = 0 Solving the quadratic equation in a, we get a = 5.18 in

STEP-5: Check whether f = f, a 0.16 a a B 87,000 = Pil 37.000+ Jy ) = 0.504 > 0.16 (= 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 double layer di = 32.5 in ald, = 0.16 add, = 0.375 B, = 0.319 > 0.16 (= a/d] Therefore, the section is tension-controlled and Ø = 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 3 Asmin = 3f lab.d>2005, 1.3 = 1.24 Since As>As, min > O.K STEP-6: Computation of nominal moment capcaity of beam, M. Beam-1 Lever arm, jd = = (d-d') 29 in Design moment capacity, M = E x 0.003x (d - d) = 320.6 ft-kips Beam-2 Lever arm, jd = (d-a/2) = 29.910 in Design moment capacity. Mn2 = x 0.85 ba(d-a/2) = 474.1 ft-kips The total design moment capacity of the beam is obtained as: % M, - M + M - 794.7 ft-kips

% increase Singly reinforced Doubly reinforced (single layer) Doubly reinforced (double layer) Design moment capacity 754.2 ft-kips 785 ft-kips 794.7 ft-kips = 4.08 5.37 The percentage increase is very low. Hence, increasing compression reinforcement isnot cost effective.