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Bending and Shear in Beams - concretecentre.com

EC2 Webinar Autumn 2016 Lecture 3/1 Bending and Shear in BeamsLecture 35thOctober 2016 Contents Lecture 3 Bending / Flexure Section analysis , singly and doubly reinforced Tension reinforcement, As neutral axis depth limit & K Compression reinforcement, As2 Flexure Worked Example Doubly reinforced Shear in Beams -Variable strut method beam Examples Bending , Shear & High Shear Exercise -Design a beam for flexure and shearEC2 Webinar Autumn 2016 Lecture 3/2 Bending / FlexureSection 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 MPaEC2 Webinar Autumn 2016 Lecture 3/3 Section analysis to determineTension & Compression ReinforcementEC2 contains information on: concrete stress blocks Reinforcement stress/strain curves The maximum depth of the neut

Note: For plastic analysis xu/d must be ≤ 0.25 for normal strength concrete, EC2 cl 5.6.2 (2). Concise: 6.2.1 K’ and Beams with Compression Reinforcement, A s2 For K > K’ compression reinforcement A s2 is required. As2 can be calculated by taking moments about the centre of the tension force: M= K’ fck b d 2 + 0.87 f yk As2 (d - d2 ...

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Transcription of Bending and Shear in Beams - concretecentre.com

1 EC2 Webinar Autumn 2016 Lecture 3/1 Bending and Shear in BeamsLecture 35thOctober 2016 Contents Lecture 3 Bending / Flexure Section analysis , singly and doubly reinforced Tension reinforcement, As neutral axis depth limit & K Compression reinforcement, As2 Flexure Worked Example Doubly reinforced Shear in Beams -Variable strut method beam Examples Bending , Shear & High Shear Exercise -Design a beam for flexure and shearEC2 Webinar Autumn 2016 Lecture 3/2 Bending / FlexureSection 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 MPaEC2 Webinar Autumn 2016 Lecture 3/3 Section analysis to determineTension & Compression ReinforcementEC2 contains information on: concrete stress blocks Reinforcement stress/strain curves The maximum depth of the neutral axis, x.

2 This depends on the moment redistribution ratio used, . The design stress for concrete , fcdand reinforcement, fydIn EC2 there are no equations to determine As, tension steel, and As2, 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:ck2fbdM K =Value of K for maximum value of M with no compression steel and when x is at its maximum K > K Compression steel requiredAsd fcdFs x sx cu3Fc Acfck 50 MPa50 < fck 90 MPa (fck 50)/400 1,0 (fck 50)/200fcd= ccfck/ c= concrete Stress BlockForfck 50 MPa failure concrete strain, cu, = : Cl , Fig : Fig this from last week?

3 EC2 Webinar Autumn 2016 Lecture 3/4 ud fyd/ Esfykkfykfyd = fyk/ skfyk/ s Idealised Design ukReinforcementDesign Stress/Strain CurveEC2: Cl , Fig UK fyk= 500 MPafyd= fyk/ s= 500 435 MPaEsmay be taken to be 200 GPaSteel yield strain = fyd/Es( s at yield point) = 435/200000= At failure concrete strain is for fck 50 x/d is steel strain is and this is past the yield steel stress is 435 MPa if neutral axis, x, is less than analysis of a singly reinforced beamEC2: Cl Design equations can be derived as follows:For grades of concrete up to C50/60, cu= , = 1 and = fcd= fykFc= ( fck/ ) b( x) = fckb xFst= fykMbMethods to find As: Iterative, trial and error method simple but not practical Direct method of calculating z, the lever arm, and then AsFor no compression reinforcement Fsc= 0EC2 Webinar Autumn 2016 Lecture 3/5 analysis of a singly reinforced beamDetermine As Iterative methodFor horizontal equilibrium Fc= fckb x= fykGuess As Solve for x z= d- x M = Fc zMbStop when design applied BM, MEd MTake moments about the centre of the tension force, Fst.

4 M = Fcz = fckb x z(1)Now z= d- x x= (d-z) M= fckb (d-z)z= (fckb z d - fckb z2)LetK= M / (fckb d 2) (K may be considered as the normalised Bending resistance) 0= [(z/d)2 (z/d)] + K0= (z/d)2 (z/d) + ==2222 - bdfbzfbdfbdzfbdfMKckckckckckMAnalysis of a singly reinforced beamDetermine As Direct methodEC2 Webinar Autumn 2016 Lecture 3/60 = (z/d)2 (z/d) + the quadratic equation:z/d = [1 + (1 - ) ]/2z= d[ 1 + (1 - ) ]/2 The lever arm for an applied moment is now knownMQuadratic formulaHigher concrete Strengthsfck 50 MPa)] (1d[1z +=)] (1d[1z +=fck= 60 MPafck= 70 MPafck= 80 MPafck= 90 MPa)] (1d[1z +=)] (1d[1z +=)] (1d[1z +=Normal strengthEC2 Webinar Autumn 2016 Lecture 3/7 Take moments about the centre of the compression force, Fc:M= Fstz = fykzRearrangingAs = M/( )The required area of reinforcement can now be found using three methods.

5 A)calculated using these expressionsb)obtained from Tables of z/d (eg Table 5 of How to Beams or Concise Table , see next slide)c)obtained from graphs (eg from the Green Book or Fig in concrete Buildings Scheme Design Manual, next slide but one)Tension steel, AsConcise: aids for flexure -method (b)Concise: Table M / (fckb d 2) Normal tables and charts are only valid up to C50 z/d was limited to max to avoid issues with the quality of covercrete .EC2 Webinar Autumn 2016 Lecture 3/8 Design aids for flexure-method (c)TCC concrete Buildings Scheme Design Manual, Fig chart for singly reinforced beam K= M / (fckb d 2)Maximum neutral axis depthAccording to Cl (4) the depth of the neutral axis is limited, viz: k1+ k2 xu/dwherek1= + cu2= + = 1xu= depth to NA after redistribution= Redistribution ratio xu d ( - )Therefore there are limits on Kand this limit is denoted K For K > K Compression steel neededMoment Bending ElasticMoment Bending tedRedistribu = Concise: Table : Cl Linear elastic analysis with limited redistributionEC2 Webinar Autumn 2016 Lecture 3/9 The limiting value for K(denotedK ) can be calculated as follows.

6 As beforeM= fckb x (1)and K= M / (fckb d 2) & z = d x & xu= d ( )Substituting xufor x in eqn (1) and rearranging:M = b d2fck( 2- ) K = M /(b d2fck) = ( 2- )Min = (30% redistribution). Steel to be either Class B or C for 20% to 30% engineers advocate taking x/d< , and K < 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 : For plastic analysis xu/d must be for normal strength concrete , EC2 cl (2). Concise: and Beams with Compression Reinforcement, As2 For K > K compression reinforcement As2is be calculated by taking moments about the centre of the tension force:M= K fckb d 2+ (d -d2)Rearranging As2= (K-K )fckb d 2/ ( (d -d2))Compression steel, As2 Concise: : Fig Webinar Autumn 2016 Lecture 3/10 Tension steel, Asfor Beams with Compression Reinforcement, The concrete in compression is at its design capacity and is reinforced with compression reinforcement.

7 So now there is an extra force:Fsc= fykThe area of tensionreinforcement can now be considered in two parts. The first part balances the compressive force in the concrete (with the neutral axis at xu). The second part balances the force in the compression steel. The area of reinforcement required is therefore:As = K fckb d 2/( ) +As2where z is calculated using K instead of KThe following flowchart outlines the design procedure for rectangularbeams with concrete classes up to C50/60 and grade 500 reinforcementDetermine K and K from: Note: = means no redistribution and = means 20% moment steel needed -doubly reinforcedIs K K ?No compression steelneeded singly reinforcedYesNock2fdbMK==== '&2 ==== KCarry out analysis to determine design moments (M)It is often recommended in the UK that K is limited to to ensure ductile failure K FlowchartEC2 Webinar Autumn 2016 Lecture 3/11 Calculate lever arm zfrom:(Or look up z/d from table or from chart.)

8 *A limit of considered good practice, it is not a requirement of Eurocode 2. [[[[]]]]* ++++====Check minimum reinforcement requirements:dbfdbfAtyktctmmin, Check max reinforcement provided As,max (Cl. )Check min spacing between bars > bar> 20 > Agg+ 5 Check max spacing between barsCalculate tension steel required from:zfMAyds====Flow Chart for Singly-reinforced beam /Slab K K ( )Exp. ( )Minimum Reinforcement AreaThe minimum area of reinforcement for Beams and slabs is given by:EC2: Cl. , Exp. , EC2 Webinar Autumn 2016 Lecture 3/12 Flow Chart for Doubly-Reinforced beam K > K Calculate lever arm zfrom:[[[[]]]]' ++++====Calculate excess moment from:(((())))''2 KKfbdMck ====Calculate compression steel required from:(((())))2yd2s'ddfMA ====Calculate tension steel required from:Check max reinforcement provided As,max (Cl.)

9 Check min spacing between bars > bar> 20 > Agg+ 52syd2s'AzfbdfKAck++++====Flexure Worked Example(Doubly reinforced)EC2 Webinar Autumn 2016 Lecture 3/13 Worked Example 1 Design the section below to resist a sagging moment of 370 kNmassuming 15% moment redistribution ( = ).Takefck= 30 MPa andfyk= 500 assume 32 mm for tension reinforcement with 30 mmnominal cover to the link all round (allow 10 mm for link) and assume20mm for compression cnom- link- = 500 30 - 10 16= 444 mmd2=cnom+ link+ = 30 + 10 + 10= 50 mm= 50= 444 Worked Example 1EC2 Webinar Autumn 2016 Lecture 3/14 provide compression steel[][] ' += +=Kdz'.KfbdMK>= ==2090304443001037026ck21680.'=K K Example 1() ) (30444300''622= = = KKfbdMck()262yd2smm 424 50) (44443510 x = =ddfMA'262sydsmm230742436343510772370=+ =+ =).

10 ('AzfMMAW orked Example 1EC2 Webinar Autumn 2016 Lecture 3/15 Provide 2 H20 for compression steel = 628mm2 (424 mm2req d)and3 H32 tension steel = 2412mm2 (2307 mm2req d)By inspection does not exceed maximum area ( Ac) or maximum spacing of reinforcement rules (cracking see week 6 notes)Check minimum spacing, assuming H10 linksSpace between bars = (300 30 x 2 - 10 x 2 - 32 x 3)/2= 62 mm > 32 mm*..OK* EC2 Cl (2) Spacing of bars for bond:Clear distance between bars > Фbar> 20 mm > Agg + 5 mmWorked Example 1 For H type bar reinforcement what is fyd?Poll Q1:Design reinforcement strength, fyda. 435 MPab. 460 MPac. 476 MPad. 500 MPaEC2 Webinar Autumn 2016 Lecture 3/16A beam section has an effective depth of 500mm and the ultimate elastic Bending moment has been reduced by 30%.)


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