Reinforced Concrete Design PDF

Reinforced Concrete Design CE 412 ENGR. DUSTIN GLENN CUEVAS, MSCE Intended Learning Outcomes Review the concepts of reinforced concrete design Solve practice problems regarding the following topics: Load Distribution ACI Moment Coefficients Cracking Mom...

Reinforced Concrete Design CE 412 ENGR. DUSTIN GLENN CUEVAS, MSCE Intended Learning Outcomes Review the concepts of reinforced concrete design Solve practice problems regarding the following topics: Load Distribution ACI Moment Coefficients Cracking Moment Stress Calculation Ultimate Strength Design of Beam Shear Design of Beam Column Design Footing Design One Way Slab Design Tributary Area The tributary area is the loaded area that directly contributes to a structural member It is the area bounded by the lines halfway to the next beam/column. Tributary Area for Beam Tributary Area for Column Tributary Area Tributary Area for Beam 5m 5m 3 @ 2m 3 @ 2m 3 @ 2m 6m 6m 6m Assume a uniform load of 2 kPa Dead Load (Section 204, NSCP) The structure’s own self weight Any permanent loads attached to the structure Common materials Concrete = 23.6 kN/m3 Steel = 77 kN/m3 Dead Load (Section 204, NSCP) Partition Loads Floors in office buildings where the partition are subject to change shall be designed to a uniformly distributed dead load of 1.0 kPa You can also check ASCE 7 -16 for other reference Dead Load Sample Calculation t h b Slab Weight = 𝛾𝑐𝑜𝑛𝑐 𝑡 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝐿𝑜𝑎𝑑 Uniformly Distributed Load to the Beam = = 𝛾𝑐𝑜𝑛𝑐 𝑡𝐴 𝑇 Remaining Beam Weight = 𝛾𝑐𝑜𝑛𝑐 𝑏(ℎ − 𝑡) Live Load (Section 205, NSCP) Live loads shall be the maximum loads expected by the intended use or occupancy Alternate Floor Live Load Reduction Section 205.6, NSCP 2015 LL reduction should not be applied on the following cases: The design live load can be reduced as follows: KLLAT ≤ 40 m2 L > 4.8 kPa 1 𝐿 = 𝐿𝑜 0.25 + 4.57 𝐴𝐼 Reduction is limited to: 0.5Lo (members supporting only one floor) where; 0.4Lo (members supporting multiple floor) 𝐿𝑜 = original live load 𝐴𝐼 = influence area = KLLAT Alternate Floor Live Load Reduction Element KLL Interior Columns 4 Exterior columns without cantilever slabs 4 Edge columns with cantilever slabs 3 Corner columns with cantilever slabs 2 Edge beams without cantilever slabs 2 Interior beams 2 All other members 1 Source: ASCE 7 -16 Source: NSCP 2015  Wind Load Seismic Load Depend on Depend on Wind speed Focus of earthquake Terrain Shaking intensity Topography of the location Ground conditions Force increase with height Mass and stiffness Geometry and exposed Area distribution 1 𝐹𝐷 = 𝜌𝐴𝐶𝐷 𝑣 2 𝑚𝑢ሷ + 2ξ𝜔𝑚𝑢ሶ + 𝑚𝜔2 u = −m𝑢ሷ 𝑔 2 Excitation is an applied pressure or force on the Excitation is an applied displacement at the base façade Force will be distributed along interior and Force will act mainly on exterior frames then exterior lateral load resisting elements transferred to floor diaphragms  Let’s consider a simply supported beam Stage 1 –Uncracked Stage 𝑓𝑐 𝑤𝑠 𝑓𝑐 (tension) less than 𝑓𝑟 2 Stage 2 – Concrete cracked but the stress in beam is linear 𝑓𝑐 𝑤𝑠 2 𝑤𝑠 8 2 𝑇 Stage 3 – Concrete cracked and the stress in the beam is nonlinear 𝑓′𝑐 compression 𝑇 tension Stages of Beam Failure ❑ Uncracked Stage – concrete resists compression and tension with concrete tensile stress below the modulus of rupture or when the moment is less than the cracking moment. ❑ Cracked Stage – member carries bending moment greater than cracking moment. In this stage, the steel carries all the tensile force, and the stresses are below elastic range ❑ Collapse Stage – member collapses either by crushing of concrete or yielding of steel bars. Cracking Moment ❑ The moment where the concrete crushes in tension and becomes ineffective. After this stage, the whole tension will be carried by the steel reinforcement. 𝑓𝑟 𝐼𝑔 𝑀𝑐𝑟 = 𝑦𝑡 where 𝑓𝑟 - modulus of rupture (stress at cracking) 𝒇𝒓 = 𝟎. 𝟔𝟐𝝀 𝒇′ 𝒄 𝐼𝑔 - gross moment of Inertia (steel bars neglected) 𝑦𝑡 - distance from the centroid to the extreme tension fiber General Notes on Beam eccf – extreme concrete compression fiber Concrete cover (cc) – NSCP 2015 420.6.1.3 under normal condition, if not exposed Column/beam, cc ≥ 40 mm Slabs/wall, cc ≥ 20 mm Footing, cc ≥ 75 mm Concrete spacing (cs) – NSCP 2015 425.2.1 𝑐𝑠 ≥ 𝑐𝑠𝑚𝑖𝑛 25 𝑚𝑚 𝑑𝑏 𝑤ℎ𝑒𝑟𝑒 𝑐𝑠𝑚𝑖𝑛 𝑖𝑠 𝑡ℎ𝑒 𝑙𝑎𝑟𝑔𝑒𝑠𝑡 𝑜𝑓 4 3 𝑑𝑎𝑔𝑔 (𝑚𝑜𝑠𝑡 𝑐𝑜𝑚𝑚𝑜𝑛 𝑖𝑠 𝑖𝑛) 3 4 Effective depth – distance from eccf to the centroid of bars d = h – cc – ds – db/2 Beam Section Detail 𝑏 − 2𝑐𝑐 − 2𝑑𝑠 − 𝑑𝑏 (𝑁 − 1) 𝑐𝑠 = where N = number of rebars 𝑁−1 𝐼𝑓 𝑐𝑠 < 𝑐𝑠𝑚𝑖𝑛 : ▪ Bundle ▪ Double Layer ▪ Change Bar Diameter ▪ Change Beam Width 25 𝑚𝑚 𝑑𝑒 = 𝑑 𝑏 𝑁 (𝑁𝑖𝑠 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑠𝑡 𝑏𝑢𝑛𝑑𝑙𝑒𝑑 𝑏𝑎𝑟𝑠) 𝑐𝑠 ≥ 𝑐𝑠𝑚𝑖𝑛 4 3 𝑑𝑎𝑔𝑔 (𝑚𝑜𝑠𝑡 𝑐𝑜𝑚𝑚𝑜𝑛 𝑖𝑠 𝑖𝑛) 3 4 Beam Section Detail Use Varignon’s theorem in determining the effective depth, d 𝑑1 = ℎ − 𝑐𝑐 − 𝑑𝑠 − 𝑑𝑏 /2 𝑑2 = 𝑑1 − 𝑑𝑏 − 𝑣𝑐𝑠 𝐴1 𝑁1 𝑑1 + 𝐴2 𝑁2 𝑑2 𝑑= 𝐴1 𝑁1 + 𝐴2 𝑁2 Beam Section Detail MINIMUM DEPTH, NSCP 2015 409.3.1.1 𝑙𝑛 where 𝑙𝑛 is the clear span ▪ Simply Supported 16 𝑙𝑛 ▪ One end continuous 18.5 Note that for beams reinforced 𝑙𝑛 ▪ Both end continuous with fy εy Under Reinforced Condition, Ductile Failure For relative high amount of tension reinforcement, failure may occur under conditions 1 & 2, causing brittle failure. It is for this reason that NSCP restricts maximum amount of reinforcement in member subjected to flexural load only. Tension Controlled Condition Balanced Condition Doubly Reinforced Concrete is not enough to provide the necessary compression force that can counteract the tension Mn1 – couple due to compression concrete and the part of tension steel As1 Mn2 – couple due to compression steel A’s and the other part of the tension steel area As2 Doubly Reinforced Design Solve for As1 from Tension controlled-condition Solve for Mu2 / φ = 𝑀𝑛2 𝑀𝑚𝑎𝑥 3 𝑀𝑛2 = − 𝑀𝑛1 𝑐5 = 𝑑 𝑎5 = 𝛽𝑐5 𝜑 8 Solve for 𝐴𝑠2 from Mn2 𝐶𝑐 = 𝑇1 0.85𝑓′𝑐 𝑎5 𝑏 = 𝐴𝑠1 𝑓𝑦 𝑀𝑛2 = 𝐴𝑠2 𝑓𝑦 (𝑑 − 𝑑 ′ ) 0.85𝑓 ′ 𝑐 𝑎5 𝑏 Total steel area in tension 𝐴𝑠1 = 𝑓𝑦 𝐴𝑠 = 𝐴𝑠1 + 𝐴𝑠2 Solve for 𝑀𝑛1 = Mu1 / φ 𝑎5 𝑀𝑛1 = 0.85𝑓 ′ 𝑎 𝑏(𝑑 − ) 𝑐 5 2 Doubly Reinforced Design Check for the strain of compression bars Check for the capacity after combing the reinforcement 𝑐 − 𝑑′ 𝑓𝑠′ = 600 𝐶𝑐 + 𝐶𝑠 = 𝑇 𝑐 0.85𝑓 ′ 𝑐 𝑎𝑏 + 𝐴′𝑠 (𝑓𝑠′ −0.85𝑓 ′ 𝑐 ) = 𝐴𝑠 𝑓𝑦 If yielding; a 𝑀𝑛2 = 𝐴′𝑠 (𝑓𝑦 − 0.85𝑓′𝑐 )(𝑑 − 𝑑′) 𝑀𝑛 = 0.85𝑓 ′ 𝑐 ab d − + 𝐴′𝑠 (𝑓𝑠′ −0.85𝑓 ′ 𝑐 )(𝑑 − 𝑑′) 2 If not yielding; 𝑀𝑛2 = 𝐴′𝑠 (𝑓𝑠 ′ − 0.85𝑓′𝑐 )(𝑑 − 𝑑 ′ ) Solve for 𝐴′𝑠 T-beams Analysis & Design Slabs are casted monolithically with beams Effective Flange Width 1. For T beams with flanges on both sides of the web, the overhanging slab width on either side of the beam web shall not exceed one-eighth of the beam clear span ℓn, 8 times the thickness of the slab h, or go beyond one-half the clear distance to the next beam sw. 2. For beams having a slab on one side only, the effective overhanging slab width shall not exceed one-twelfth the beam clear span ℓn, 6 times the thickness of the slab h, or go beyond one-half the clear distance to the next beam sw. 3. For isolated beams in which the flange is used only for the purpose of providing additional compressive area, the flange thickness shall not be less than one-half the width of the web bw, and the total flange width shall not be more than 4 times the web width bw. Effective Flange Width T-beam Analysis Case 1: The compression block is within the flange Re-computation of a: 𝑎 600 𝑑 − 𝛽 0.85𝑓′𝑐 𝑎𝑏𝑓 = 𝐴𝑠 𝑥 𝑎 𝛽 Step 3: Check for steel strain 𝑑−𝑐 𝐶=𝑇 𝜀𝑠 = 0.003 𝑐 Step 1: Assume steel is yielding: 𝑆𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 𝜑 0.85𝑓′𝑐 𝑎𝑏𝑓 = 𝐴𝑠 𝑓𝑦 𝐴𝑠 𝑓𝑦 Step 4: Solve for moment capacity 𝑎= 0.85𝑓′𝑐 𝑏𝑓 𝑎 𝜑𝑀𝑛 = 𝜑𝐴𝑠 𝑓𝑠 𝑑 − 𝑎 = 𝛽𝑐 2 Step 2: Check for stress in steel 𝑎 𝜑𝑀𝑛 = 𝜑0.85𝑓′𝑐 𝑎𝑏𝑓 𝑑 − 𝑖𝑓 𝑓𝑠 > 𝑓𝑦 → 𝑈𝑠𝑒 𝑓𝑦 2 𝑑−𝑐 𝑓𝑠 = 600 𝑐 𝑖𝑓 𝑓𝑠 < 𝑓𝑦 → 𝑅𝑒𝑐𝑜𝑚𝑝𝑢𝑡𝑒 𝑎 T-beam Analysis Case 2: The compression block is below the flange Re-computation of a: 𝑎 600 𝑑 − 𝛽 0.85𝑓′𝑐 𝑡𝑓 𝑏𝑓 + 0.85𝑓 ′ 𝑐 (𝑎 − 𝑡𝑓 )𝑏𝑤 = 𝐴𝑠 𝑥 𝑎 𝛽 Step 3: Check for steel strain 𝑑−𝑐 𝐶=𝑇 𝜀𝑠 = 0.003 𝑐 Step 1: Assume steel is yielding: 𝑆𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 𝜑 0.85𝑓′𝑐 𝑡𝑓 𝑏𝑓 + 0.85𝑓 ′ 𝑐 (𝑎 − 𝑡𝑓 )𝑏𝑤 = 𝐴𝑠 𝑓𝑦 Step 4: Solve for moment capacity Solve for a 𝑎 Solve for c 𝑎 = 𝛽𝑐 𝜑𝑀𝑛 = 𝜑𝐴𝑠 𝑓𝑠 𝑑 − 2 Step 2: Check for stress in steel ′ 𝑡𝑓 𝑎 𝜑𝑀𝑛 = 𝜑 0.85𝑓 𝑐 𝑡𝑓 𝑏𝑓 𝑑− + 0.85𝑓 ′ 𝑐 𝑎 − 𝑡𝑓 𝑏𝑤 d − 𝑡𝑓 − 2 2 𝑑−𝑐 𝑖𝑓 𝑓𝑠 > 𝑓𝑦 → 𝑈𝑠𝑒 𝑓𝑦 𝑓𝑠 = 600 𝑐 𝑖𝑓 𝑓𝑠 < 𝑓𝑦 → 𝑅𝑒𝑐𝑜𝑚𝑝𝑢𝑡𝑒 𝑎 Shear Failure of Concrete Beams Beams must have an adequate safety margin against other types of failure, some of which may be more dangerous than flexure failure. Shear failure of reinforced concrete, more properly called diagonal tension failure, is one example. Shear failure is a brittle failure, thus it will occur suddenly with little or no advance warning Truss Analogy of Concrete Beams Shear Design of Beams Shear Design of Beams Nominal Shear Strength Nominal Shear Strength, Vc Nominal Shear Strength, Vs Maximum Shear Reinforcement Steps in Vertical Stirrups Design Steps in Vertical Stirrups Design Shear reinforcement Spacing Shear reinforcement Spacing Shear reinforcement Spacing Shear reinforcement Spacing Axial Load Capacity of Columns Note: In actual practice, there are no perfect axially loaded columns. Axially loaded are columns such that the eccentricity is less than 0.10h for tied, and 0.05h for spiral. 𝑃𝑜 = 0.85𝑓 ′ 𝑐 𝐴𝑔 − 𝐴𝑠𝑡 + 𝐴𝑠𝑡 𝑓𝑦 𝑃𝑛 = 𝛼𝑃𝑜 Factor to account for the minimum eccentricity 𝛼 = 0.80 → 𝑓𝑜𝑟 𝑡𝑖𝑒𝑑 𝛼 = 0.85 → 𝑓𝑜𝑟 𝑠𝑝𝑖𝑟𝑎𝑙 𝑃𝑛 = 𝜑𝑃𝑛 Reduction factor to account for the uncertainties 𝜑 = 0.65 → 𝑓𝑜𝑟 𝑡𝑖𝑒𝑑 𝜑 = 0.75 → 𝑓𝑜𝑟 𝑠𝑝𝑖𝑟𝑎𝑙 Limits for Reinforcement of Compression Members 1. Ast shall not be less than 0.01Ag but not more than 0.08Ag. 2. The minimum number of longitudinal bars is 4 for bars within rectangular or circular ties, 3 for bars within triangular ties, and 6 for bars within spirals. Sizes and Spacing of Main Bars and Ties 1. Clear distance between longitudinal bars shall not be less than (4/3)dagg, 1.5db nor 40 mm. (Section 425.2.1) 2. Use 10mmø ties for 32mm bars or smaller and at least 12mmø for 36mm and bundled longitudinal bars. (Section 425.7.2.2) 3. Vertical spacing of ties shall be the following (Section 425.7.2.1): a. 16 x bar diameter b. 48 x tie diameter c. Least dimension of the column 4. Ties shall be arranged such that every corner and alternate longitudinal bar shall have lateral support provided by the corner of the tie with an included angle of not more than 135° and no bar shall be farther than 150 mm clear on each side along the tie from such a laterally supported bar. (Section 425.7.2.3) Sizes and Spacing of Main Bars and Ties 1.For cast-in-place construction, size of spirals shall not be less than 10 2.The clear spacing of spiral shall not exceed 75 mm, nor less than 25 3.Anchorage of spiral reinforcement shall be provided by 1 ½ extra turn of spiral bar. 4.Splices of spiral reinforcement shall be 1. Lap splices 48db but not less than 300 2. Welded splices in accordance with Sect 412.15.3 5.The percentage of spiral steel ρs is computed by where: as cross-sectional area of the spiral bar Dc diameter of the core out-to- out of the spiral db diameter of the spiral bar 6. Volumetric spiral ratio, ρs, shall not be less than the value given below Effect of Confinement for Column Design Effect of Confinement for Column Design