1. DEPARTMENT OF CIVIL ENGINEERING CE-2105 DESIGN OF CONCRETE STRUCTURE-I DESIGN OF BEAM (AS PER ACI CODE) Course Teacher- Presented By- Dr. Md. Rezaul Karim Md. Jahidur Rahman Associate Professor S.ID- 121041 Dhaka University of Engineering & Technology, Gazipur-17001 2. Content 2 USD(Ultimate Strength Design) Classification of beam with respect to design system Assumptions Evolution of design parameters Moment Factors Kn, Balanced Reinforcement Ratio b Calculating Strength Reduction Factor Calculating Design procedure for Singly Reinforced Beam Design procedure for Doubly Reinforced Beam Design procedure for T-Beam Appendix Jahidur Rahman 3. Ultimate Strength Design(USD) 3 Assuming tensile failure condition Additional strength of steel after yielding ACI code emphasizes this method Jahidur Rahman 4. Classification of beam with respect to design system Rectangular beam (reinforced at tension zone only) Doubly reinforced beam (reinforced at both tension and compression zone) T section beam (both beam and slab are designed together) 5. ASSUMPTIONS 5 Plane sections before bending remain plane and perpendicular to the N.A. after bending Strain distribution is linear both in concrete & steel and is directly proportional to the distance from N.A. Strain in the steel & surrounding concrete is the same prior to cracking of concrete or yielding of steel Concrete in the tension zone is neglected in the flexural analysis & design computation c=0.003 s = fy / Es h d c 0.85fc a a/2 d-a/2 b C T Jahidur Rahman 6. 6 Concrete stress of 0.85fc is uniformly distributed over an equivalent compressive zone. fc = Specified compressive strength of concrete in psi. Maximum allowable strain of 0.003 is adopted as safe limiting value in concrete. The tensile strain for the balanced section is fy/Es Moment redistribution is limited to tensile strain of at least 0.0075 s y fy fs Idealized Actual Es 1 Jahidur Rahman 7. EVALUATION OF DESIGN PARAMETERS Total compressive force - C = 0.85fc ba (Refer stress diagram) Total Tensile force- T = As fy C = T 0.85fc ba = As fy a = As fy / (0.85fc b) = d fy / (0.85 fc) [ = As / bd] Moment of Resistance/Nominal Moment- Mn = 0.85fc ba (d a/2) or, Mn = As fy (d a/2) = bd fy [ d (dfyb / 1.7fc) ] = fc [ 1 0.59 ] bd2 = fy / fc Mn = Kn bd2 Kn = fc [ 1 0.59 ] Ultimate Moment- Mu = Mn = Kn bd2 ( = Strength Reduction Factor) 7 Jahidur Rahman 8. Balaced Reinforcement Ratio ( b) From strain diagram, similar triangles cb / d = 0.003 / (0.003 + fy / Es); Es = 29x106 psi cb / d = 87,000 / (87,000+fy) b = Asb / bd = 0.85fc ab / (fy. d) = 1 ( 0.85 fc / fy) [ 87,000 / (87,000+fy)] Relationship b / n the depth `a of the equivalent rectangular stress block & depth `c of the N.A. is a = 1c 1= 0.85 ; fc 4000 psi 1= 0.85 - 0.05(fc 4000) / 1000 ; 4000 < fc 8000 1= 0.65 ; fc> 8000 psi 8 Jahidur Rahman 9. For beams the ACI code limits the max. amount of steel to 75% of that required for balanced section. 0.75 b Min. reinforcement is greater of the following: Asmin = 3fc x bwd / fy or 200 bwd / fy min = 3fc / fy or 200 / fy For statically determinate member, when the flange is in tension, the bw is replaced with 2bw or bf whichever is smaller The above min steel requirement need not be applied, if at every section, Ast provided is at least 1/3 greater than the analysis 9 Jahidur Rahman 10. Yes N o Yes N o Calculating 11. Calculating 12. Jahidur Rahman12 SINGLY REINFORCED BEAM Beam is reinforced near the tensile face Reinforcement resists the tension. Concrete resists the compression. 13. DESIGN PROCEDURE FOR SINGLY REINFORCED BEAM 13 1. Determine the service loads 3. Calculate d= h Effective cover 2. Assume `h` as per the support conditions [As per ACI code in table 9.5(a)] 4. Assume the value of `b` by the rule of thumb 5. Estimate self weight 6. Primary elastic analysis and derive B.M (M), Shear force (V) values 7. Compute min and b 8.Choose between min and b 14. Note Below: Design the steel reinforcement arrangement with appropriate cover and spacing stipulated in code. Bar size and corresponding no. of bars based on the bar size #n. Check crack width as per codal provisions. 14 Jahidur Rahman 9. Calculate , Kn 10. From Kn & M calculate `d required 11. Check the required `d with assumed `d 12. With the final values of , b, d determine the Total As required OK 15. DESIGN PROCEDURE FOR SINGLY REINFORCED BEAM BY FLOWCHART 15 Actual steel ratio Maximum steel ratio Over reinforced beam Yes Under reinforced beam N o Ultimate moment 16. Jahidur Rahman16 DOUBLY REINFORCED BEAM Beam is fixed for Architectural purposes. Reinforcement are provided both in tension and compression zone. Concrete has limitation to resist the total compression so extra reinforcement is required. 17. Actual steel ratio Maximum steel ratio rectangular beam Yes Doubly reinforced beam N o DESIGN PROCEDURE FOR DOUBLY REINFORCED BEAM Concrete section, Area of steel are known YesN o Compression and tension bar reach in yielding Compression bar does not reach in yielding 18. DESIGN PROCEDURE FOR DOUBLY REINFORCED BEAM Load or ultimate moment is given Maximum steel ratio Reinforcement required in comp. zone too rectangular beam YesDoubly reinforced beam N o 19. Jahidur Rahman19 T- REINFORCED BEAM A part of slab acts as the upper part of beam. Resulting cross section is T shaped. The slab portion of the beam is flange. The beam projecting bellow is web or stem. 20. DESIGN PROCEDURE FOR T-REINFORCED BEAM Are of concrete section, Area of steel are known Additional moment Nominal moment calculation Rectangula r beam analysis Yes T - beam analysis N o modification Total moment 21. DESIGN PROCEDURE FOR T- REINFORCED BEAM Moment or loading is given Total area of steel Nominal moment calculation Rectangula r beam analysis YesT - beam analysis N o Area of steel in flange Area of steel in web Additional moment T section: L section: Isolated beam: Check for 22. Jahidur Rahman22 APPENDIX 23. AS PER TABLE 9.5 (a) Simply Supported One End Continuous Both End Continuous Cantilever L / 16 L / 18.5 L / 21 L/8 Values given shall be used directly for members with normal weight concrete (Wc = 145 lb/ft3) and Grade 60 reinforcement For structural light weight concrete having unit wt. In range 90-120 lb/ft3 the values shall be multiplied by (1.65 0.005Wc) but not less than 1.09 For fy other than 60,000 psi the values shall be multiplied by (0.4 + fy/100,000) `h` should be rounded to the nearest whole number 23 Jahidur Rahman 24. RULE OF THUMB d/b = 1.5 to 2.0 for beam spans of 15 to 25 ft. d/b = 3.0 to 4.0 for beam spans > 25 ft. `b` is taken as an even number Larger the d/b, the more efficient is the section due to less deflection CLEAR COVER Not less than 1.5 in. when there is no exposure to weather or contact with the ground For exposure to aggressive weather 2 in. Clear distance between parallel bars in a layer must not be less than the bar diameter or 1 in. 24 Jahidur Rahman 25. BAR SIZE #n = n/8 in. diameter for n 8. Ex. #1 = 1/8 in. . #8 = 8/8 i.e., I in. Weight, Area and Perimeter of individual bars inch mm 3 0.376 0.375 9 0.11 1.178 4 0.668 0.500 13 0.20 1.571 5 1.043 0.625 16 0.31 1.963 6 1.502 0.750 19 0.44 2.356 7 2.044 0.875 22 0.60 2.749 8 2.670 1.000 25 0.79 3.142 9 3.400 1.128 28 1.00 3.544 10 4.303 1.270 31 1.27 3.990 11 5.313 1.410 33 1.56 4.430 14 7.650 1.693 43 2.25 5.319 18 13.600 2.257 56 4.00 7.091 Perimeter (in.) Stamdard Nominal Dimensions Bar No Wt.per Foot (lb) Diameter db C/S Area, Ab (in2 ) 25 Jahidur Rahman 26. 26 Jahidur Rahman