Question

5- Determine the ACI design moment capacity, $Mn, for the beam shown. Assume no axial force on the member. (150 mm 3 (75 mm) (175 mm (75 mm) (125 mm . 1200 mm 1100 mm) -22 (550 mm) (400 mm 18 (450 mm - 12 - (300 mm 4.B (4 x 25 mm 5.8 15 x 25 mm 2012" (50 mm 300 mm (50 mm) 12" (300 mm

Answer 1

EU = 0.003 GIVEN DATA Breadth of beam, Overall depth of beam Effective depth of beam, Reinforcement Strength of concrete, Yield Strength of steel, 12 in 22 in 20 in 4 No. 8 single layer) 5000 psi (assumed as they were 60000 psi not mentioned) fo's STEP-1: Computation of tension force Area of steel, A, = 3.14 in ? Assuming that the stress in steel, f. = f, the tension force in steel reinforcement can be calculated as: T = f,As = 60000 X 3.14 = 188400 lb STEP-2: Ccomputation of area of compression stress block From equilibrium, we have CET = 188400 lb Let a be the depth of equivalent rectangular compression stress block. Therefore, compressive force can be written as: C = 0.85 X 5000 x 3 x 4 x 2 + (a - 4 ) x 12 a = 5.69 in > 4 in Hence, STEP-3: Check whether fs =fy This check is done using strain compatibility. The ultimate compressive strain in concrete at its extreme fiber is: E = 0.003 B = 0.8 Note: i) Forf.' <= 4000 psi, B1=0.85 Therefore, C = a/B ii) For 4000<fc' < 8000 psi, B = 1.05 -0.05f/1000 = 7.11 in ii) For f.' >= 8000 psi, p1=0.65

From similar triangles of the strain diagram, we can write, Therefore, &, = 0.003 [( 20 - 7.11 ) / 7.11 ] = 0.00544 The yield strain in reinforcement for the given grade of steel is: E = 60000 / 2.9E+07 = 0.00207 Since E, > Ey, fs = fy and thus the assumption in step-1 is justified Since fs = fy, the section is tension-controlled and hence, = 0.90 Note: i) = 0.9 for tension-controlled sections ii) 0 = 0.7 for compression-controlled sections iii) = 0.356 +0.204/(a / d) for transition sections STEP-4: Computation of nominal moment capcaity of beam, M, The compressive stress block can be divided into two areas Compressive area, Aa= 4 x 3 x 2 24 in? Lever arm, L.A, = 20 .4 12 18 in Compressive area, ( 5.69 -4 ) x 12 20.28 in? Lever arm, 20 - 4 - 5.69 . 4)/2 15.16 in Nominal moment capacity, M = 0.855.[(, L.A)+(42 XL.A)] = = 3142640.4 in.lb 3142640.4 / 12000 261.9 ft-kips The design moment capacity of the beam is obtained as: OM, = 0.90 x 261.9 = 235.7 ft-kips

Ecu = 0.003 GIVEN DATA Breadth of beam, b = 12 in Effective depth of beam, Reinforcement Strength of concrete, Yield Strength of steel, 16 in 5 No. 8 (single layer) 5000 psi (assumed as they were 60000 psi not mentioned) fe = fy = NA -C STEP-1: Computation of tension force Area of steel, A,= 3.93 in ? Assuming that the stress in steel, f, Efy the tension force in steel reinforcement can be calculated as: T = f,As = 60000 X 3.93 = 235800 lb STEP-2: Ccomputation of area of compression stress block From equilibrium,we have C = T 235800 lb Let a be the depth of equivalent rectangular compression stress block. Therefore, compressive force can be written as: C = 0.85 x 5000 x 5 x 4 + (a - 4 ) x 12 Hence, a = 6.96 in > 4 in STEP-3: Check whether fs = fy This check is done using strain compatibility. The ultimate compressive strain in concrete at its extreme fiber is: S = 0.003 Bi= 0.8 Note: i) Forf' <= 4000 psi, P1=0.85 Therefore, C = a/P. ii) For 4000<fc' < 8000 psi, B = 1.05 -0.05f/1000 = 8.70 in ) For f' >= 8000 psi, =0.65

From similar triangles of the strain diagram, we can write, Therefore, &= 0.003 || 16 - 8.70 ) / 8.70 1 = 0.00252 The yield strain in reinforcement for the given grade of steel is: y = 60000 / 2.9E+07 = 0.00207 Since &,> fs = fy and thus the assumption in step-1 is justified Since fs = fy, the section is tension-controlled and hence, = 0.90 Note: i) = 0.9 for tension-controlled sections ii) p= 0.7 for compression-controlled sections iii) = 0.356 + 0.204/(a / B d.) for transition sections STEP-4: Computation of nominal moment capcaity of beam, M, The compressive stress block can be divided into two areas Compressive area, A = 4 x 5 20 in 2 Lever arm, 16 -4 / 2 14 in Compressive area, 16.96 4 ) x 12 35.52 in? Lever arm, LA2 = 16 - 4.1 6.96 - 4)/2 = 10.52 in Nominal moment capacity, M = 0.85 f. [(A x L.A)+(A, XL.A)] = = 2778099.2 in.lb 2778099.2 / 12000 231.5 ft-kips The design moment capacity of the beam is obtained as: OM, = 0.90 x 231.5 = 208.4 ft-kips