Reinforced Concrete (RC) Design PDF

REINFORCED CONCRETE (RC) DESIGN TEXTBOOK AND REFERENCES LECTURE OUTLINES Basis of Reinforced Concrete Design Singly Reinforced Beam  Eurocode 2 Eurocode 2, Part 1–1: General rules and rules for buildings Eurocode 2, Part 1–2: Structural fire design...

REINFORCED CONCRETE (RC) DESIGN TEXTBOOK AND REFERENCES LECTURE OUTLINES Basis of Reinforced Concrete Design Singly Reinforced Beam  Eurocode 2 Eurocode 2, Part 1–1: General rules and rules for buildings Eurocode 2, Part 1–2: Structural fire design Eurocode 2, Part 2: Bridges Eurocode 2, Part 3: Liquid-retaining and containment structures  EC2: PART 1–1 Part 1.1 is divided into twelve sections: Section 1: General Section 2: Basis of design Section 3: Materials Section 4: Durability and cover to reinforcement Section 5: Structural analysis Section 6: Ultimate limit states Section 7: Serviceability limit states Section 8: Detailing of reinforcement and prestressing tendons-General Section 9: Detailing of members and particular rules Section 10: Additional rules for precast concrete elements and structures Section 11: Lightweight aggregate concrete structures Section 12 Plain and lightly reinforced concrete structures  Principles of Ultimate Bending Strength Basic Principles/Assumptions of Ultimate Strength in Bending ❖ Simplified stress-strain relationship may be used ❖ The concrete has no tensile strength ❖ Plane sections normal to the axis remain plane after bending ❖ Perfect bonding at interface concrete-to-steel rebar: STRESS–STRAIN RELATIONS (STEEL) For steel reinforcement, the maximum stress is considered as steel yield stress (500 MPa) divided by partial safety factor of steel The modulus of elasticity is taken as 200 GPa The consideration of the partial safety factor is needed in order to obtain the design strength of concrete and steel STRESS–STRAIN RELATIONS (CONCRETE) ▪ When load is applied to a structure, the deformation occurred on the element will produce stress and strain ▪ The maximum stress for concrete is assumed to be 85% of its compressive strength divided by partial safety factor of concrete ▪ Whereas, the ultimate strain for concrete in compression is taken as 0.0035 The behaviour of structural concrete (figure) is represented by a parabolic stress–strain relationship, up to a strain 0.002, from which point the strain increases while the stress remains constant. The ultimate design stress is given by: This value 0.0035 is used in structural engineering codes and is based on experimental data. In practice, 0.0035 strain in concrete represents the extreme end of the concrete's behavior before it starts to exhibit cracks due to over- compression.  It is also assumed that plane sections of a structural member remain plane after straining, so that across the section there must be a linear distribution of strains. This assumption is used in simple bending theory to simplify However, for large deformations, highly plastic materials, or the mathematical analysis of beams subjected to bending when shear deformation becomes significant, the assumption loads. might not be accurate. This means that the cross-sections of the beam before bending (flat planes perpendicular to the beam's axis) will still remain flat and perpendicular to the neutral axis after the beam bends, even though the top part of the beam might compress and the bottom part might stretch.  STRESS DISTRIBUTION 1. The triangular stress distribution applies when the stresses are very nearly proportional to the strains, which generally occurs at the loading levels encountered under working conditions and is, therefore, used at the serviceability limit state. 2. The rectangular–parabolic stress block represents the distribution at failure when the compressive strains are within the plastic range, and it is associated with the design for the ultimate limit state. 3. The equivalent rectangular stress block is a simplified alternative to the rectangular– parabolic distribution.  BEAM BEHAVIOUR IN BENDING The theory of bending for reinforced concrete assumes that: The concrete will crack in the regions of tensile strains After cracking, all the tension is carried by the reinforcement. Tensile strength of concrete is neglected  FAILURE MODES There are 3 types of failure modes that could occur in beam design: 1. Under reinforced: Steel will reach its yield strength earlier than concrete 2.Balanced: Steel will reach its yield strength at the same time as concrete 3.Over reinforced: This is strictly not allowed Concrete will reach its maximum strength earlier than steel Failure occurs caused by early failure of concrete in compression Failure happens without warning (abrupt of sudden failure)  DESIGN OF RECTANGULAR SECTION There are two types of rectangular sections Singly reinforced: Consist only tension reinforcement, As. The top reinforcements are hanger bars (used to produce a cage like arrangement) Doubly reinforced: Consist of both tension, As and compression reinforcement, As’.  DESIGN OF SINGLY BEAM Bending of the section will induce a resultant tensile force Fst in the reinforcing steel, and a resultant compressive force in the concrete Fcc which acts through the centroid of the effective area of concrete in compression, as shown in figure. For equilibrium, the ultimate design moment, M, must be balanced by the moment of resistance of the section so that  THE BALANCE SECTION The concrete section with the depth of neutral axis at the specified maximum depth of 0.45d is often referred to as the balanced section because at the ultimate limit state the concrete and tension steel reach their ultimate strains at the same time. This occurs at the maximum moment of resistance for a singly reinforced section, that is a section with no compression steel. So for this section with: STEPS OF SINGLY BEAM DESIGN Page No: 182 Reinforced Concrete Design K is a unitless factor This should be less than or equal to 0.95d.  RECAP Basis of Reinforced Concrete Design Singly Reinforced Beam EXAMPLES Example 1 h=485 Example 1 h=485 Solution Note 2: If d is not given, calculate effective depth, d = ℎ − 𝑐 − 1 Note 1: If design action & bending 𝑠ℎ𝑒𝑎𝑟 𝑙𝑖𝑛𝑘𝑠 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 − moment is not given, calculate 𝑡𝑒𝑛𝑠𝑖𝑙𝑒 𝑏𝑎𝑟 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟/2 both; Assume, shear link diameter = 10mm and tensile bar diameter =20mm; c=25mm d= 485-25-10-20/2 =440mm; Minimum and maximum reinforcement area, Asmin= 0.26 (fctm/ fyk) bd = 0.26 (2.56/500) bd> 0.0013bd = 0.00133 bd> 0.0013bd use 0.00133bd = 0.00133*260*440 = 152 mm2 Asmax= 0.04Ac = 0.04 x b x h = 0.04 x 260 x 485 Use 4H20 (As = 1260mm2) = 5044 mm2 Example 2 Design tensile steel for a reinforced concrete beam with a length of 5 metres. A distributed permanent action of 20 kN/m and a variable action of 15 kN/m must be carried by the beam. Use b=250mm, h=600mm, C=25mm, fck = 30 N/mm2 and fyk = 500 N/mm2 Solution 2 Example 2 Design tensile steel for a reinforced concrete beam with a length of 5 metres. A distributed permanent action of 20 kN/m and a variable action of 15 kN/m must be carried by the beam. Use b=250mm, h=600mm, C=25mm, fck = 30 N/mm2 and fyk = 500 N/mm2 Solution 2 Permanent action gk= 20 kN/m Variable action qk= 15 kN/m Design Action, w= 1.35 gk+ 1.5qk = 1.35*20 + 1.5*15= 49.5kN/m Bending Moment, M = wL^2/8 = 49.5 x 5^2 /8= 155 kNm Effective depth, Assume 8mm shear link and 20mm tension bar d= h –C - steel bar diameter (link)-Main steel bar diameter/2= 600 –25 –8 –20/2 = 557 mm Design bending moment, M = 155 kNm K = M /bd2fck, Kbal= 0.167 = 155x106 / (250 x 5572x 30) = 0.07 < Kbal: No compression reinforcement required z= d[ 0.5 + (0.25 –K/1.134)1/2] = 0.93 d = 0.93 x 557 = 518 mm Example 2 Design tensile steel for a reinforced concrete beam with a length of 5 metres. A distributed permanent action of 20 kN/m and a variable action of 15 kN/m must be carried by the beam. Use b=250mm, h=600mm, C=25mm, fck = 30 N/mm2 and fyk = 500 N/mm2 Solution 2 Area of tension steel As= M/0.87fykz = 155 x106/(0.87 x 500 x 518) = 688 mm2 Use 3H20 (As = 943mm2) Minimum and maximum reinforcement area, Asmin= 0.26 (fctm/ fyk) bd = 0.26 (2.9/500) bd> 0.0013bd = 0.0015 bd> 0.0013bd use 0.0015bd = 0.0015*250*557 mm2 = 209 mm2 Asmax= 0.04Ac = 0.04 x b x h = 0.04 x 250 x 600 = 6000 mm2 EXTRA EXAMPLE  Example 3 Solution