Bending and Shear in Beams

Bending and Shear in Beams Lecture 3 5 th October 017 Contents Lecture 3 What reinforcement is needed to resist M Ed? Bending/ Flexure Section analysis, singly and doubly reinforced Tension reinforcement, A s neutral axis depth limit & K x = Neutral axis depth Compression reinforcement, A s Flexure Worked Example Doubly reinforced What shear links are needed to resist V Ed? Shear in Beams - Variable strut method Beam Examples Bending, Shear & High shear Exercise - Design a beam for flexure and shear EC Webinar Autumn 017 Lecture 3/1

Bending/ Flexure Section Design: Bending In principal flexural design is generally the same as BS8110 EC presents the principles only Design manuals will provide the standard solutions for basic design cases. There are modifications for high strength concrete ( f ck > 50 MPa ) Note: TCC How to guide equations and equations used on this course are based on a concrete f ck 50 MPa EC Webinar Autumn 017 Lecture 3/

Section Analysis to determine Tension & Compression Reinforcement EC contains information on: Information needed to derive equations Concrete stress blocks to find A s for a given M Ed. 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, f cd and reinforcement, f yd In EC there are no equations to determine A s, tension steel, and A s, 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 = bd f ck If K > K Compression steel required Value of K for maximum value of M with no compression steel and when x is at its maximum value. Rectangular Concrete Stress Block EC: Cl 3.1.7, Fig 3.5 Ac εcu3 η fcd x λx Fc Concise: Fig 6.1 d Parabolic stress block As Fs f ck 50 MPa εs 50 < f ck 90 MPa λ 0.8 = 0.8 (f ck 50)/400 η 1.0 = 1,0 (f ck 50)/00 f cd = α cc f ck /γ c = 0.85 f ck /1.5 f ck λ η 50 0.8 1 55 0.79 0.98 60 0.78 0.95 70 0.75 0.9 80 0.73 0.85 90 0.7 0.8 For f ck 50 MPa failure concrete strain, ε cu, = 0.0035 EC Webinar Autumn 017 Lecture 3/3

Reinforcement Design Stress/Strain Curve EC: Cl 3..7, Fig 3.8 σ Idealised kfyk In UK f yk = 500 MPa fyk kfyk/γs fyd = fyk/γs f yd = f yk /γ s = 500/1.15 = 435 MPa Design E s may be taken to be 00 GPa Steel yield strain = f yd /E s (ε s at yield point) = 435/00000 = 0.00 fyd/ Es ε ud ε uk ε At failure concrete strain is 0.0035 for f ck 50 MPa. If x/d is 0.6 steel strain is 0.003 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 EC: Cl 3.1.7 Design equations can be derived as follows: b M For grades of concrete up to C50/60, ε cu = 0.0035, η = 1 and λ = 0.8. f cd = 0.85f ck/1.5 f yd = f yk/1.15 = 0.87 f yk F c = (0.85 f ck / 1.5) b (0.8 x) = 0.453 f ck b x For no compression F st = 0.87A s f yk reinforcement F sc = 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 EC Webinar Autumn 017 Lecture 3/4

Analysis of a singly reinforced beam Determine A s Iterative method b M For horizontal equilibrium F c = F st 0.453 f ck b x = 0.87A s f yk Guess A s 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, F st : M = F c z = 0.453 f ck b x z (1) (1) Now z = d - 0.4 x x =.5(d - z) M = 0.453 f ck b.5(d - z) z = 1.1333 (f ck b z d - f ck b z ) M Let K = M / (f ck b d ) K M f bd f bdz = 1.1333 f bd ck ck = - ck ck ck (K may be considered as the normalised bending resistance) 0 = 1.1333 [(z/d) (z/d)] + K 0 = (z/d) (z/d) + 0.8835K f bz f bd EC Webinar Autumn 017 Lecture 3/5

0 = (z/d) (z/d) + 0.8835K Solving the quadratic equation: z/d = [1 + (1-3.53K) 0.5 ]/ z = d [ 1 + (1-3.53K) 0.5 ]/ M The lever arm for an applied moment is now known Quadratic formula Higher Concrete Strengths f ck 50MPa z = d[1+ (1 3.59K )]/ Normal strength f ck = 60MPa z = d[1+ (1 3.715K )]/ f ck = 70MPa z = d[1+ (1 3.9K )]/ f ck = 80MPa z = d[1+ (1 4.15K )]/ f ck = 90MPa z = d[1+ (1 4.41K )]/ EC Webinar Autumn 017 Lecture 3/6

Tension steel, A s Concise: 6..1 Take moments about the centre of the compression force, F c : M = F st z = 0.87A s f yk z Rearranging A s = M /(0.87 f yk 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) K = M / (f ck 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 EC Webinar Autumn 017 Lecture 3/7

Design aids for flexure- method (c) TCC Concrete Buildings Scheme Design Manual, Fig B.3 K = M / (f ck b d ) 1.4 Design chart for singly reinforced beam Maximum neutral axis depth EC: 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 k 1 + k x u /d where k 1 = 0.4 k = 0.6 + 0.0014/ ε cu = 0.6 + 0.0014/0.0035 = 1 x u = depth to NA after redistribution δ = Redistributed Bending Moment Elastic Bending Moment = Redistribution ratio x u d (δ - 0.4) Therefore there are limits on K and this limit is denoted K For K > K Compression steel needed EC Webinar Autumn 017 Lecture 3/8

K and Beams with Compression Reinforcement, A s The limiting value for K (denoted K ) can be calculated as follows: As before M = 0.453 f ck b x z (1) and K = M / (f ck b d ) & z = d 0.4 x & x u = d (δ 0.4) Substituting x u for x in eqn (1) and rearranging: M = b d f ck (0.6 δ 0.18 δ - 0.1) Concise: 6..1 K = M /(b d f ck ) = (0.6 δ 0.18 δ - 0.1) Min δ = 0.7 (30% redistribution). Steel to be either Class B or C for 0% 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 EC requirement and is not accepted by all engineers. Note: For plastic analysis x u /d must be 0.5 for normal strength concrete, EC cl 5.6. (). Compression steel, A s EC: Fig 3.5 Concise: 6..1 For K > K compression reinforcement A s is required. A s can be calculated by taking moments about the centre of the tension force: Rearranging M = K f ck b d + 0.87 f yk A s (d - d ) A s = (K - K ) f ck b d / (0.87 f yk (d - d )) EC Webinar Autumn 017 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: F sc = 0.87A s f yk 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: A s = K f ck b d /(0.87 f yk z) + A s 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: M K = & K' = 0.6δ 0.18δ 0. 1 bd f ck Note: δ =1.0 means no redistribution and δ = 0.8 means 0% moment redistribution. Yes No Is K K? δ K 1.00 0.08 0.95 0.195 0.90 0.18 0.85 0.168 0.80 0.153 0.75 0.137 No compression steel needed singly reinforced Compression steel needed - doubly reinforced 0.70 0.10 It is often recommended in the UK that K is limited to 0.168 to ensure ductile failure EC Webinar Autumn 017 Lecture 3/10

Flow Chart for Singly-reinforced Beam/Slab K K Calculate lever arm z from: d z = (Or look up z/d from table or from chart.) [ 1+ 1 3.53K ] 0.95d * * A limit of 0.95d is considered good practice, it is not a requirement of Eurocode. Calculate tension steel required from: A s = M f z yd Check minimum reinforcement requirements: 0.6f b d A ctm t s,min 0. 0013 bt d f yk (Cl.9..1.1) Exp. (9.1N) Check max reinforcement provided A s,max 0.04A c (Cl. 9..1.1) Check min spacing between bars > Ø bar > 0 > A gg + 5 Check max spacing between bars Minimum Reinforcement Area EC: Cl. 9..1.1, Exp. 9.1N The minimum area of reinforcement for beams and slabs is given by: 0.6f b d A ctm t s,min 0. 0013b f yk t d EC Webinar Autumn 017 Lecture 3/11

Flow Chart for Doubly- Reinforced Beam K > K Calculate lever arm z from: d z = [ 1+ 1 3.53K' ] Calculate excess moment from: M ' = bd f ( K K' ) ck Calculate compression steel required from: M' As = f d d yd ( ) Calculate tension steel required from: K' fck bd As = + A f z yd s Check max reinforcement provided A s,max 0.04A c (Cl. 9..1.1) Check min spacing between bars > Ø bar > 0 > A gg + 5 Flexure Worked Example (Doubly reinforced) EC Webinar Autumn 017 Lecture 3/1

Worked Example 1 Design the section below to resist a sagging moment of 370 knm assuming 15% moment redistribution (i.e. δ = 0.85). Take f ck = 30 MPa and f yk = 500 MPa. d Worked Example 1 Initially assume 3 mm φ for tension reinforcement with 30 mm nominal cover to the link all round (allow 10 mm for link) and assume 0mm φ for compression reinforcement. = 50 d = h c nom - Ø link - 0.5Ø = 500 30-10 16 = 444 mm = 444 d = c nom + Ø link + 0.5Ø = 30 + 10 + 10 = 50 mm EC Webinar Autumn 017 Lecture 3/13

Worked Example 1 K' = 0.168 δ K M K = bd f ck 6 370 10 = 300 444 30 = 0. 09 > K' provide compression steel [ 1+ 1 3.53 '] d z = K 444 = = 363 mm [ 1+ 1 3.53 0.168 ] 1.00 0.08 0.95 0.195 0.90 0.18 0.85 0.168 0.80 0.153 0.75 0.137 0.70 0.10 Worked Example 1 M ' = bd f ck ( K K' ) = 300 444 30 (0.09 0.168) 10 = 7.7 knm A s = f yd M ' ( d d ) 6 7.7 x 10 = 435 (444 50) = 44 mm 6 M M ' As = + A f z yd = 307 mm s ( 370 7. 7) 10 = 435 363 K ' bd f = f z 6 yd + 44 ck + A s EC Webinar Autumn 017 Lecture 3/14

Worked Example 1 Provide H0 for compression steel = 68mm (44 mm req d) and 3 H3 tension steel = 41mm (307 mm req d) By inspection does not exceed maximum area (0.04 A c ) or maximum spacing of reinforcement rules (cracking see week 6 notes) Check minimum spacing, assuming H10 links Space between bars = (300 30 x - 10 x - 3 x 3)/ = 6 mm > 3 mm* OK * EC Cl 8. () Spacing of bars for bond: Clear distance between bars > Ф bar > 0 mm > Agg + 5 mm Poll Q1: Design reinforcement strength, f yd For H type bar reinforcement what is f yd? a. 435 MPa b. 460 MPa c. 476 MPa d. 500 MPa EC Webinar Autumn 017 Lecture 3/15

Poll Q: Neutral axis depth, x A beam section has an effective depth of 500mm and the ultimate elastic bending moment has been reduced by 30%. What is the maximum depth of the neutral axis, x u? a. 150 mm b. 50 mm c. 300 mm d. 450 mm Shear in Beams Variable strut method EC Webinar Autumn 017 Lecture 3/16

Shear EC: Cl 6.., 6..3, 6.4 Concise: 7., 7.3, 8.0 There are three approaches to designing for shear: When shear reinforcement is not required e.g. slabs, week 5 Shear check uses V Rd,c When shear reinforcement is required e.g. Beams Variable strut method is used to check shear in beams Strut strength check using V Rd,max Links strength using V Rd,s Punching shear requirements e.g. flat slabs, week 5 The maximum shear strength in the UK should not exceed that of class C50/60 concrete. Cl 3.1. () P and NA. Shear in Beams EC: Cl 6..3 Concise: 7.3 Shear design is different from BS8110. EC uses the variable strut method to check a member with shear reinforcement. Definitions: V Ed V Rd,c V Rd,s - Applied shear force. For predominately UDL, shear may be checked at d from face of support Resistance of member without shear reinforcement - Resistance of member governed by the yielding of shear reinforcement V Rd,max - Resistance of member governed by the crushing of compression struts Whilst Eurocode deals in Resistances (capacities), V Rd,c,V Rd,s,V Rd,max and Effect of actions, V Ed in kn, in practice, it is often easier to consider shear strengths v Rd, v Rd,max and shear stresses, v Ed, in MPa. EC Webinar Autumn 017 Lecture 3/17

Members Requiring Shear Reinforcement EC: 6..3(1) Concise: Fig 7.3 compression chord compression chord strut F cd V(cot θ - cotα ) d α θ V ½ z ½ z z = 0.9d N V M s shear reinforcement tension chord F td α θ z angle between shear reinforcement and the beam axis Normally links are vertical. α = 90 o and cot α is zero angle between the concrete compression strut and the beam axis inner lever arm. In the shear analysis of reinforced concrete without axial force, the approximate value z = 0,9d may normally be used. Members Requiring Shear Reinforcement EC: 6..3(1) Concise: Fig 7.3 b w A sw is the minimum width Area of the shear reinforcement f ywd design yield strength = f yk /1.15 f cd design compressive strength = α cc f ck /1.5 = f ck /1.5 (α cc = 1.0 for shear) α cw = 1.0 Coefficient for stress in compression chord ν 1 strength reduction factor concrete cracked in shear ν 1 = ν = 0.6(1-f ck /50) Exp (6.6N) EC Webinar Autumn 017 Lecture 3/18

Strut Inclination Method EC: Equ. 6.8 & 6.9 for Vertical links Equ 6.9 αcw bw z ν 1 fcd VRd,max = cotθ + tanθ V Ed Strut angle limits 1.8 < θ < 45 Cot θ =.5 Cot θ = 1 V = Rd, s A sw s Equ 6.8 z fywd cotθ Eurocode vs BS8110: Shear Safer Shear reinforcement density A s f yd /s BS8110: V R = V C + V S Eurocode : V Rmax Fewer links (but more critical) Test results V R Minimum links Shear Strength, V R EC Webinar Autumn 017 Lecture 3/19

Shear Design: Links Variable strut method allows a shallower strut angle hence activating more links. As strut angle reduces concrete stress increases Min curtailment for 45 o strut V high s V low z d θ z d x Angle = 45 V carried on 3 links Max angle - max shear resistance x Angle = 1.8 V carried on 6 links Min strut angle - Minimum links Shear Resistance of Sections with Vertical Shear Reinforcement Concise: 7.3.3 V s V x Basic equations shear reinforcement control Exp (6.8) V Rd,s = A sw z f ywd cot θ /s 1 cotθ,5 concrete strut control Exp (6.9) z θ = 45 θ = 1.8 V Rd,max = α cw z b w ν 1 f cd /(cotθ + tanθ) d x θ z where: f ywd = f yk /1.15 = 0.5 z b w ν 1 f cd sin θ d ν 1 = ν = 0.6(1-f ck /50) α cw = 1.0 EC Webinar Autumn 017 Lecture 3/0

Shear links EC: Cl 6..3 Equation 6.9 is first used to determine the strut angle θ and then equation 6.8 is used to find the shear link area, A sw, and spacing s. Equation 6.9 gives V Rd,max values for a given strut angle θ e.g. when cot θ =.5 (θ = 1.8 ) Equ 6.9 becomes V Rd,max = 0.138 b w z f ck (1 - f ck /50) or in terms of stress: v Rd,max = 0.138 f ck (1 - f ck /50) Values are in the middle column of the table. Re-arranging equation 6.9 to find θ: θ = 0.5 sin -1 [v Rd /(0.0 f ck (1 - f ck /50))] Suitable shear links are found from equation 6.8: A sw /s = v Ed b w /( f ywd cot θ) v Rd, cot θ v Rd, cot θ f ck =.5 = 1.0 0.54 3.68 5 3.10 4.50 8 3.43 4.97 30 3.64 5.8 3 3.84 5.58 35 4.15 6.0 40 4.63 6.7 45 5.08 7.38 50 5.51 8.00 v Rd,max values, MPa, for cot θ = 1.0 and.5 EC Shear Flow Chart for vertical links Determine v Ed where: v Ed = design shear stress [v Ed = V Ed /(b w z) = V Ed /(b w 0.9d)] Determine the concrete strut capacity v Rd when cot θ =.5 v Rdcot θ =.5 = 0.138f ck (1-f ck /50) (or look up from table) Is v Rd,cot θ =.5 > v Ed? No Is v Rd,cot θ = 1.0 > v Ed? No Re-size Yes (cot θ =.5) Calculate area of shear reinforcement: A sw /s = v Ed b w /(f ywd cot θ) Check minimum area, cl 9..: A sw /s b w ρ w,min ρ w,min = (0.08 f ck )/f yk 0.001 Yes (cot θ > 1.0) Determine θ from: θ = 0.5 sin -1 [(v Ed /(0.0f ck (1-f ck /50))] Check maximum spacing of shear reinforcement, cl 9..: s l,max = s t,max = 0.75 d EC Webinar Autumn 017 Lecture 3/1

Design aids for shear Concise Fig 15.1 a) V Ed = 4.3 MPa A sw /s = 5.3 per m width Short Shear Spans with Direct Strut Action EC: 6..3 d d a v a v Where a v d the applied shear force, V Ed, for a point load (eg, corbel, pile cap etc) may be reduced by a factor a v /d where 0.5d a v d provided: The longitudinal reinforcement is fully anchored at the support. Only that shear reinforcement provided within the central 0.75a v is included in the resistance. Note: see PD6687-1:010 Cl.14 for more information EC Webinar Autumn 017 Lecture 3/

Beam examples Bending, Shear and High Shear Beam Example 1 G k = 75 kn/m, Q k = 50 kn/m, assume no redistribution and use EC0 equation 6.10 to calculate ULS loads. 8 m 450 1000 Nominal cover = 40mm to each face f ck = 30 Determine the flexural and shear reinforcement required (try 10mm links and 3mm main steel) EC Webinar Autumn 017 Lecture 3/3

Beam Example 1 Bending ULS load per m = (75 x 1.35 + 50 x 1.5) = 176.5 M ult = 176.5 x 8 /8 = 1410 knm How to Ch 4 - Beams d = 1000-40 - 10 3/ = 934 M K = bd 6 1410 10 = 450 934 30 f = ck 0.10 K = 0.08 K < K No compression reinforcement required 934 [ 1+ 1 3.53K ] = [ 1+ 1 3.53 x 0.10] = 8 0. d d z = 95 6 M 1410 x 10 A = = = 3943 mm s f z 435 x 8 yd Provide 5 H3 (401 mm ) Beam Example 1 Shear Shear force, V Ed = 176.5 x 8/ = 705 kn Shear stress: (We could take 505 kn @ d from face) v Ed = V Ed /(b w 0.9d) = 705 x 10 3 /(450 x 0.9 x 934) = 1.68 MPa v Rdcot θ =.5 = 3.64 MPa f ck v Rd, cot θ =.5 v Rd, cot θ = 1.0 v Rdcot θ =.5 > v Ed cot θ =.5 A sw /s = v Ed b w /(f ywd cot θ) A sw /s = 1.68 x 450 /(435 x.5) A sw /s = 0.70 mm Try H10 links with 4 legs. A sw = 314 mm s < 314 /0.70 = 448 mm provide H10 links at 450 mm spacing 0.54 3.68 5 3.10 4.50 8 3.43 4.97 30 3.64 5.8 3 3.84 5.58 35 4.15 6.0 40 4.63 6.7 45 5.08 7.38 50 5.51 8.00 EC Webinar Autumn 017 Lecture 3/4

Beam Example 1 Provide 5 H3 (401) mm ) with H10 links at 450 mm spacing Max spacing is 0.75d = 934 x 0.75 = 700 mm Beam Example High shear UDL not dominant Find the minimum area of shear reinforcement required to resist the design shear force using EC. Assume that: f ck = 30 MPa and f yd = 500/1.15 = 435 MPa EC Webinar Autumn 017 Lecture 3/5

Beam Example High shear Find the minimum area of shear reinforcement required to resist the design shear force using EC. fck fyd = 30 MPa and = 435 MPa Shear stress: v Ed = V Ed /(b w 0.9d) = 31.5 x 10 3 /(140 x 0.9 x 500) = 4.96 MPa v Rdcot θ =.5 = 3.64 MPa v Rdcot θ = 1.0 = 5.8 MPa v Rdcot θ =.5 < v Ed < v Rdcot θ = 1.0.5 > cot θ > 1.0 Calculate θ f ck v Rd, cot θ =.5 v Rd, cot θ = 1.0 0.54 3.68 5 3.10 4.50 8 3.43 4.97 30 3.64 5.8 3 3.84 5.58 35 4.15 6.0 40 4.63 6.7 45 5.08 7.38 50 5.51 8.00 Beam Example High shear Calculate θ θ = 0.5 sin θ = 0.5 sin = 35.0 cot θ = 1.43 1 1 0.0f ck ved (1 f ck 4.96 0.0 x 30 / 50) ( 1-30 / 50) A sw /s = v Ed b w /(f ywd cot θ ) A sw /s = 4.96 x 140 /(435 x 1.43) A sw /s = 1.1 mm Try H10 links with legs. A sw = 157 mm s < 157 /1.1 = 140 mm provide H10 links at 15 mm spacing EC Webinar Autumn 017 Lecture 3/6

Exercise Lecture 3 Design a beam for flexure and shear Beam Exercise Flexure & Shear G k = 10 kn/m, Q k = 6.5 kn/m (Use EC0 eq. 6.10) 8 m 450 Nominal cover = 35 mm to each face f ck = 30MPa Design the beam in flexure and shear 300 EC Webinar Autumn 017 Lecture 3/7

Aide memoire Exp (6.10) Remember this from the first week? Or Concise Table 15.5 Workings:- Load, M ult, d, K, K, (z/d,) z, A s, V Ed, A sw /s Φ mm Area, mm 8 50 10 78.5 1 113 16 01 0 314 5 491 3 804 EC Webinar Autumn 017 Lecture 3/8

Working space Working space EC Webinar Autumn 017 Lecture 3/9

Working space End of Lecture 3 Email: pgregory@concretecentre.com EC Webinar Autumn 017 Lecture 3/30