Bending and Shear in Beams Lecture 3 5th October 2016 Contents – Lecture 3 • Bending/ Flexure – Section analysis, singly and doubly reinforced – Tension reinforcement, A s – neutral axis depth limit & K’ – Compression reinforcement, A s2 • Flexure Worked Example – Doubly reinforced • Shear in Beams - Variable strut method • Beam Examples – Bending, Shear & High shear • Exercise - Design a beam for flexure and shear EC2 Webinar – Autumn 2016 Lecture 3/1 Bending/ Flexure Section Design: Bending • In principal flexural design is generally the same as BS8110 • EC2 presents the principles only • Design manuals will provide the standard solutions for basic design cases. • There are modifications for high strength concrete ( fck > 50 MPa ) Note: TCC How to guide equations and equations used on this course are based on a concrete fck ≤ 50 MPa EC2 Webinar – Autumn 2016 Lecture 3/2 Section Analysis to determine Tension & Compression Reinforcement EC2 contains information on: • Concrete stress blocks • Reinforcement stress/strain curves • The maximum depth of the neutral axis, x. This depends on the moment redistribution ratio used, δ. • The design stress for concrete, fcd and reinforcement, fyd In EC2 there are no equations to determine As, tension steel, and A s2 , compression steel, for a given ultimate moment, M, on a section. Equations, similar to those in BS 8110, are derived in the following slides. As in BS8110 the terms K and K’ are used: M K = Value of K for maximum value of M 2 with no compression steel and bd f ck when x is at its maximum value. If K > K’ Compression steel required Rectangular Concrete Stress Block EC2: Cl 3.1.7, Fig 3.5 εcu3 η fcd Concise: Fig 6.1 Fc Ac x λx d Remember this from last week? As Fs εs fck λ η fck ≤≤≤ 50 MPa 50 < fck ≤≤≤ 90 MPa 50 0.8 1 55 0.79 0.98 λλλ 0.8 = 0.8 –(fck – 50)/400 60 0.78 0.95 70 0.75 0.9 ηηη 1.0 = 1,0 –(fck – 50)/200 80 0.73 0.85 90 0.7 0.8 fcd = αcc fck /γc = 0.85 fck /1.5 For fck ≤ 50 MPa failure concrete strain, εcu , = 0.0035 EC2 Webinar – Autumn 2016 Lecture 3/3 Reinforcement Design Stress/Strain Curve EC2: Cl 3.2.7, Fig 3.8 σ Idealised kf yk kf yk /γs fyk In UK fyk = 500 MPa fyd = fyk /γs fyd = fyk /γs = 500/ 1.15 = 435 MPa Design Es may be taken to be 200 GPa Steel yield strain = f /E yd s ε ε ε fyd / Es ud uk (εεεs at yield point) = 435/200000 = 0.0022 At failure concrete strain is 0.0035 for fck ≤ 50 MPa. If x/d is 0.6 steel strain is 0.0023 and this is past the yield point. Design steel stress is 435 MPa if neutral axis, x, is less than 0.6d. Analysis of a singly reinforced beam EC2: Cl 3.1.7 Design equations can be derived as follows: b M For grades of concrete up to C50/60, εcu = 0.0035, ηηη = 1 and λλλ = 0.8. fcd = 0.85 fck /1.5 fyd = fyk /1.15 = 0.87 fyk Fc = (0.85 fck / 1.5) b (0.8 x) = 0.453 fck b x For no compression Fst = 0.87 As fyk reinforcement Fsc = 0 Methods to find As: • Iterative, trial and error method – simple but not practical • Direct method of calculating z, the lever arm, and then A s EC2 Webinar – Autumn 2016 Lecture 3/4 Analysis of a singly reinforced beam Determine A s – Iterative method b M For horizontal equilibrium Fc= Fst 0.453 fck b x = 0.87 As fyk Guess As Solve for x z = d - 0.4 x M = F c z Stop when design applied BM, M Ed ≃ M Analysis of a singly reinforced beam Determine A s – Direct method Take moments about the centre of the tension force, Fst : M = Fc z = 0.453 fck b x z (1) Now z = d - 0.4 x M ∴ x = 2.5( d - z) ∴ M = 0.453 fck b 2.5( d - z) z 2 = 1.1333 ( fck b z d - fck b z ) 2 Let K = M / ( fck b d ) 2 M fck bdz fck bz K = 2 = 1.1333 2 - 2 fck bd fck bd fck bd (K may be considered as the normalised bending resistance) ∴ 0 = 1.1333 [( z/d )2 –(z/d )] + K 0 = ( z/d )2 –(z/d ) + 0.88235 K EC2 Webinar – Autumn 2016 Lecture 3/5 0 = ( z/d )2 –(z/d ) + 0.88235 K Solving the quadratic equation: M z/d = [1 + (1 - 3.53 K)0.5 ]/2 z = d [ 1 + (1 - 3.53 K)0.5 ]/2 The lever arm for an applied moment is now known Quadratic formula Higher Concrete Strengths fck ≤ 50MPa z = d[1 + (1 − 3.529K )]/2 Normal strength fck = 60MPa z = d[1 + (1 − 3.715K )]/2 fck = 70MPa z = d[1 + (1 − 3.922K )]/2 fck = 80MPa z = d[1 + (1 − 4.152K )]/2 fck = 90MPa z = d[1 + (1 − 4.412K )]/2 EC2 Webinar – Autumn 2016 Lecture 3/6 Tension steel, A s Concise: 6.2.1 Take moments about the centre of the compression force, F c: M = Fst z = 0.87 As fyk z Rearranging As = M /(0.87 fyk z) The required area of reinforcement can now be found using three methods: a) calculated using these expressions b) obtained from Tables of z/d (eg Table 5 of How to beams or Concise Table 15.5, see next slide ) c) obtained from graphs (eg from the ‘ Green Book’ or Fig B.3 in Concrete Buildings Scheme Design Manual, next slide but one ) Design aids for flexure - method (b) 2 K = M / ( fck b d ) Concise: Table 15.5 . Traditionally z/d was limited to 0.95 max to avoid issues with the quality of ‘covercrete’. ‘Normal’ tables and charts are only valid up to C50/60 EC2 Webinar – Autumn 2016 Lecture 3/7 Design aids for flexure- method (c) TCC Concrete Buildings Scheme Design Manual, Fig B.3 ) 2 b d b ck f =/ M( K Design chart for singly reinforced beam Maximum neutral axis depth EC2: Cl 5.5 Linear elastic analysis with limited redistribution Concise: Table 6.1 According to Cl 5.5(4) the depth of the neutral axis is limited, viz: δ ≥ k1 + k2 xu/d where k1 = 0.4 k2 = 0.6 + 0.0014 / εcu2 = 0.6 + 0.0014/0.0035 = 1 xu = depth to NA after redistribution Redistribu ted Bending Moment δ = Elastic Bending Moment = Redistribution ratio ∴ xu ≤ d ( δ - 0.4) Therefore there are limits on K and this limit is denoted K’ For K > K’ Compression steel needed EC2 Webinar – Autumn 2016 Lecture 3/8 K’ and Beams with Compression Reinforcement, A s2 Concise: 6.2.1 The limiting value for K (denoted K’) can be calculated as follows: As before M = 0.453 fck b x z … (1) 2 and K = M / ( fck b d ) & z = d – 0.4 x & xu = d ( δ – 0.4) Substituting xu for x in eqn (1) and rearranging: 2 2 M’ = b d fck (0.6 δ – 0.18 δ - 0.21) 2 2 ∴ K’ = M’ /( b d fck ) = (0.6 δδδ – 0.18 δδδ - 0.21) Min δ = 0.7 (30% redistribution). Steel to be either Class B or C for 20% to 30% redistribution. Some engineers advocate taking x/d < 0.45, and ∴K’ < 0.168. It is often considered good practice to limit the depth of the neutral axis to avoid ‘over-reinforcement’ to ensure a ductile failure. This is not an EC2 requirement and is not accepted by all engineers. Note: For plastic analysis xu/d must be ≤ 0.25 for normal strength concrete, EC2 cl 5.6.2 (2). Compression steel, A s2 EC2: Fig 3.5 Concise: 6.2.1 For K > K’ compression reinforcement A s2 is required. As2 can be calculated by taking moments about the centre of the tension force: 2 M = K’ fck b d + 0.87 fyk As2 (d - d2) Rearranging 2 As2 = ( K - K’) fck b d / (0.87 fyk (d - d2)) EC2 Webinar – Autumn 2016 Lecture 3/9 Tension steel, A s for beams with Compression Reinforcement, The concrete in compression is at its design capacity and is reinforced with compression reinforcement. So now there is an extra force: Fsc = 0.87 As2 fyk The area of tension reinforcement can now be considered in two parts. The first part balances the compressive force in the concrete (with the neutral axis at x u). The second part balances the force in the compression steel. The area of reinforcement required is therefore: 2 As = K’ fck b d /(0.87 fyk z) + As2 where z is calculated using K’ instead of K Design Flowchart The following flowchart outlines the design procedure for rectangular beams with concrete classes up to C50/60 and grade 500 reinforcement Carry out analysis to determine design moments ( M) Determine K and K’ from: δδδ K’ M 2 K === & K' === 6.0 δδδ −−− 18.0 δδδ −−− .0 21 1.00 0.208 bd 2 f ck 0.95 0.195 Note: δδδ =1.0 means no redistribution and δδδ = 0.8 means 20% moment redistribution. 0.90 0.182 0.85 0.168 Yes No 0.80 0.153 Is K ≤ K’ ? 0.75 0.137 0.70 0.120 No compression steel Compression steel needed - needed – singly reinforced doubly reinforced It is often recommended in the UK that K’ is limited to 0.168 to ensure ductile failure EC2 Webinar – Autumn 2016 Lecture 3/10 Flow Chart for Singly-reinforced Beam/Slab K ≤ K’ Calculate lever arm z from: d z === [[[1+++ 1−−− 53.3 K ]]] ≤≤≤ 95.0 d * 2 (Or look up z/d from table or from chart.) * A limit of 0.95 d is considered good practice, it is not a requirement of Eurocode 2.