Design of reinforced concrete sections according to EN and EN

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1 Design of reinforced concrete sections according to EN and EN Validation Examples Brno, IDEA RS s.r.o. South Moravian Innovation Centre, U Vodarny 2a, BRNO tel.: , fax: ,

2 Foreword The introduction of European standards is a significant event as, for the first time, all design and construction codes within the EU will be harmonized. These Eurocodes will affect all design and construction activities. The aim of this publication, Design of reinforced concrete sections according to EN and EN , is to illustrate how the Code is treated on practical examples. In order to explain the use of all relevant clauses of Eurocode 2, an example of a simply supported oneway rib-shaped slab and an example of column with high axial load and bi-axial bending is introduced. IDEA RS s.r.o. South Moravian Innovation Centre, U Vodarny 2a, BRNO tel.: , fax: ,

3 Design of reinforced concrete sections according to EN and EN October 2010 Contents 1. Rib T Project details... 4 Actions and analysis of Rib T Section forces Cross section Ultimate section resistance Shear check Torsional check Interaction Crack width calculation Crack width according to EN Example - Calculation of crack width according to EN Calculating stiffness Example - calculating the stiffness of the T-section according to EN Column Project details Second order effects Simplified method based on nominal stiffness Simplified method based on nominal curvature Biaxial bending Page 3

4 Validation Examples Reinforced Concrete Section October Rib T Project details Example is taken from: Ing. Miloš Zich, Ph.D. and others, others online publication "Konstrukční ční Eurokódy - Příklady posouzení betonových prvkůů dle Eurokódů", Eurokódů nakl. Verlag Dashöfer s. r. o., 2010, First floor slab Figure Schematic layout of structure Page 4

5 Section A Section B Figure 1.2 Sections Actions and analysis of Rib T1 Figure 1.3 Static schema of Rib T1 Page 5

Figure 1.4 - Floor composition for the calculation of loads Figure 1.5 - Permanent load calculation Variable load: quasi-permanent value: q k = 10 kn/m 2 * 2.

6 Figure Floor composition for the calculation of loads Figure Permanent load calculation Variable load: quasi-permanent value: q k = 10 kn/m 2 * 2.0m = 20 kn/m ψ = 0,6 20 = 12 / Factors defining the representative values of variable actions, ψ 0, ψ 1, ψ 2 are shown in table. A1.1 of EN1990 (also in attachment A4 in this document) Page 6

7 Actions for Serviceability limit states (SLS) Loads for serviceability limit state are determined acc. to EN 1990 clause There are 3 SLS-combinations: Characteristic combination of loads (Unacceptable cracking or deformation), "+"" + ", " + " ψ,,,ψ = 9, = 29,87 Frequent load combination, "+"" + "ψ,, " + " ψ,,,ψ = 9, , = 23,87 Quasi-permanent load combination, "+"γ " + ψ,,,ψ = 9, ,6 20 = 21,87 Actions for Ultimate limit states (ULS) It is considered as a persistent design situation for ultimate limit state where partial factors are: γ G = 1,35, γ Q = 1,50. To determine the design load in Article EN 1990 is prescribed the following equation marked as the equation (6.10) γ,, "+"γ " + "γ,, " + " γ, ψ,, Substituting, we get the value of design load = 1,35 9, , = 43,32 Alternatively, load can be further reduced according to equation (6.10) and (6.10b) and consider the less favorable value of both terms: γ,,"+"γ " + "γ, ψ, " + " γ, ψ,,, ξ,, "+"γ " + "γ,, " + " γ, ψ,, = 1 2 = 1,35 9, ,5 0, = 34,32 = 1,35 0,85 9, , = 41, Section forces = 1 8 Page 7

8 Combination/Value Loads [kn/m] Vz(a) [kn] My (b) [knm] SLS characteristic SLS frequent SLS quasi-permanent ULS Table Internal forces for individual SLS and ULS load combinations The shear force is calculated at distance d from the face of the support. Estimated value of d is based on the assumption that the moment near the support will be positive. Value d = 458 mm. Values: V Ed1 and M Ed1 are calculated at distance l x = = m from the theoretical support.v Ed1 = 105,05 kn, M Ed1 = 84,29 knm. Page 8

9 1.2. Cross section Figure 1.6 Cross section Materials Concrete C25/30 Steel B500B f ck = 25 MPa f cd = f ck / γ c = 25 / 1,5 = 16,66 MPa f ctm = 2,6 MPa f ctd = 0,7 f ctm / γ c = 0,7 2,6 / 1,5 = 1,213 MPa f yk = 500 MPa f yd = f yk / γ s = 500 / 1,15 = 434,78 MPa Page 9

10 1.3. Ultimate section resistance The cross section resistance (capacity) is the calculation of stress, strain and internal forces status on the calculated cross section for its limit state. For concrete the stress-strain relation is assumed to bi-linear. For reinforcing steel the stress-strain relation is assumed to be bi-linear without strain hardening. Bending moment at middle section from basic combination of loads. Figure Response - given by program IDEA RCS Input data, Plane of strain: ε x = 0, ε y =0,0 ε z = - 0, Figure Resulted plane of strain calculated by IDEA RCS Strain calculation in end fibers: ε = ε + ε + ε = 0, , = 0, Figure 1.9 Strain in ultimate compression fiber (picture from program IDEA RCS) Modulus of elasticity is calculated from stress-strain diagram = /ε = 16,7/ 0,00175 = 9,52 Page 10

11 σ = ε = 0, ,52 10 = 7,27 Defining the depth of compression zone (depth to neutral axis) in concrete, follow from: = + ε = 0, , = 0, ε 0, Concrete force in compression (as, the strain in concrete is outside the plastic branch, the stress along the section is linear in concrete) = 2 = 0, ,85 16,7 2 = 495,83 Concrete lever arm in the compression = 3 = 0,1305 0, = 0, Concrete moment in compression = = 495,83 0, = 52,52 Strain in reinforcing steel ε = ε + ε + ε = 0, , = 0, Figure 1.10 Stress in reinforcing steel ( Diagram is taken from program IDEA RCS) Calculating of stress in reinforcing steel (whereas, the section is loaded in the plane of symmetry and reinforcement is not in one layer, these layers can be replaced by one layer with an area equal to the sum of all areas of reinforcement) σ = ε = 0, = 559,34 > 434,78 434,78 MPa Tensile force in reinforcement Page 11

= σ = 0,0011404 434,783 = 495,83 KN Moment in tensile reinforcement M = F z = 495,83 0.327 = 162,134 knm Figure 1.

12 = σ = 0, ,783 = 495,83 KN Moment in tensile reinforcement M = F z = 495, = 162,134 knm Figure 1.11 Comparing with results calculated by IDEA RCS program Equilibrium of forces + = = 495,83 495,83 = 0 Equilibrium of moments + = = -162,134-52,52 = -214,654 Note: Due to coordination system that is used inside the program, the design moment My has opposite sign. Page 12

1.4. Shear check Resistance without shear reinforcement in zones without cracks under bending loads Calculated in center of gravity of concrete section I = 1/12*1,85*0,08 3 + 1,85*0,08*0,091 2

13 1.4. Shear check Resistance without shear reinforcement in zones without cracks under bending loads Calculated in center of gravity of concrete section I = 1/12*1,85*0, ,85*0,08*0, /12*0,2*0, ,2*0,42*0,159 2 = 7,8933e-5+0, , , = 0,004684m 4 S = 1,85 * 0,08 * 0, ,2 * 0,051 *0,051/2 = 0,01373 m 3 b w = 0,2 m σ cp = 0,0 MPa α l = 1, = + = 0, ,2 1, , = 82,76, = 82,76 <, = 105,05 Deriving from the above text, the concrete part does not carry all the shear force, hence shear reinforcement will be required. Figure Comparing with results calculated by RCS program Resistance without shear reinforcement in zones with cracks under bending loads C Rd,c = 0,18 / γ c = 0,18 / 1,5 = 0,12 = = = 1,661 = = 760 = 0, , k 1 = 0,15 σ cp = 0,0 MPa b w = 0,2 m d = 0,458 m ν min = k 3/2 fck 1/2 = ,661 3/2 25 1/2 = 0,3745 MPa Page 13

14 , =. (100 ) Minimally + = 0,12 1,661 (100 0, ) + 0 0,2 0,458 = 50,15, = + = 0, ,2 0,458 = 34,30, = 50,15 <, = 105,05 Deriving from the above text, the concrete part does not carry all the shear force, hence shear reinforcement will be required. Resistance without shear reinforcement = min (,,, ) A sw = 2 * * PI /4 = 5,655e-5 m 2 s = 0.24m z = 0.9*0.458 = m exact value is 0,437 m f ywd = f yd = 434,7MPa θ = 21,8, = cot = 2 2,827e 5 0, , ,8 0,24 = 105,70 >, = 105,05 h α cw = 1,0 b w = 0,2 m z = 0.9*0.458 = m exact value is 0,437m For calculation of strength reduction factor for concrete cracked in shear ν 1 must be checked if the design stress of the shear reinforcement is over the 80% of the characteristic yield stress f ywk, = =,,, = = = 0,54.,, 434,74 > 0.8 = 400, = 1,0 0,2 0,412 0,54 16,66 10 = cot + tan cot 21,8 + tan 21,8 = 255,76 >, = 133,27 h Page 14

15 Figure 1.13 Comparison of results calculated by RCS program Page 15

1.5. Torsional check Section characteristics for torsional check u = 2 * (1.85 + 0.5) = 4.7 m A = 2 * (1.85 + 0.5) = 0.232 m 2 t ef = A / u = 0.232 / 4.7 = 0.049 m = ± 16 = 4.7 ± 4.7 16 0.

16 1.5. Torsional check Section characteristics for torsional check u = 2 * ( ) = 4.7 m A = 2 * ( ) = m 2 t ef = A / u = / 4.7 = m = ± 16 = 4.7 ± = 4 4 2b. h = =.m ± = = m = = 4.504m. =.m Torsional capacity without shear reinforcement, = 2 = = Torsional capacity with shear reinforcement ν = 0.6 α cw = 1,0, = 2 = ,0 16, , sin 21.8 cos 21.8 = 36,09, = 2 = 2 0,1187 2,827e 5 0,24 7,603e 4 434, , ,507 = 14,59 Figure Comparison of results calculated by RCS program Page 16

17 1.6. Interaction Combined shear and torsion, +, =,, +. Shear reinforcement is not allowed to design according to detailing rules = 2,09 > 1,0 => Compression strut check for combined shear and torsion + = 105,05,, 255, = 0,41 < 1,0 => 36,09 Shear reinforcement check for combined shear and torsion, = Σ = 2,827e 5 0,24 434,78 10 = 51,21 = + tan = 105, , tan 21,8 = 50,99 2 0,1187 = 50,99 = 0,99 < 1,0 =>, 51,21 Longitudinal reinforcement check for shear, torsion and bending, = Σ =7,603e 4 (434,78 253,9) 10 = 137,52 = tan + 2 tan = 105,05 tan 21, ,1187 tan 21,8 = 262,64 = 262,64 = 1,91 > 1,0 =>, 137,52 Page 17

18 Figure Comparing with results calculated by RCS program Page 18

19 1.7. Crack width calculation Crack width according to EN Check is introduced at midsection of beam M y = 113,73 knm Plane of strain calculated by program IDEA RCS: ε x = 0, ε y = 0,0 ε z = -0, Defining the depth of compression zone (depth to neutral axis) in concrete, follow from: = + ε = 0, , = 0,0565 ε 0, Figure Strain-stress diagram on fully cracked cross section Strain calculation in end concrete fibres: ε = ε + ε + ε = 0, , ,1305 = 0, Stress calculation in end concrete fibres: σ = ε = 0, = 4,954 Concrete force in compression: = σ 2 = σ 0,0565 1,85 4,954 = 2 2 = 0,259 kn Concrete moment in compression: M = F z = F z x 3 = ,1305 0, = 28,92 knm Strain in reinforcing steel: ε = ε + ε + ε = 0, , ,32748 = 0, Page 19

20 Calculating of stress in reinforcing steel (whereas, the section is loaded in the plane of symmetry and reinforcement is in one layer, this layer can be replaced by one bar with an area equal to the sum of all areas of reinforcement) σ = ε = 0, = 227,1 Tensile force in the bar: = σ = 0, = 258,97 kn Moment in tensile reinforcement: M = F z = 258, = 84,81 knm Equilibrium of forces: + = = 258,97 258,97 = 0 Equilibrium of moments: + = = 84,81 28,92 = 113,725 knm Example - Calculation of crack width according to EN Effective ratio of reinforcement: ρ, = = 11,40 10 = 0,05429, 0,021 Maximal spacing of the cracks:, = + φ/ρ,, = 3,4 0, ,8 0,5 0,425 0,022/0,05649 = 0,1743 Factors: k 1 = k 2 = k 3 = 3,4 k 4 = 0,5 0,8 in example is considered steel B500B 0,5 Cross section loaded by bending moment, pure bending Effective height (3) h c,ef : h, = 2,5h ; h 3 ; h 2 = 2,50,5 0,458; 0,5 0,05649 ; 0,5 3 2 = 0,105; 0,1478; 0,25 = 0,105 Page 20

Effective area: Mean strain in the reinforcement ε ε = ε ε =, = h, = 0,105 0,2 = 0,021 σ, ρ, α ρ, 227,17 0,4, 1 + 6,45 0,05429, 0,6 227,17 200 200 ε ε = 0,0010065 > 0,0006815 ε ε = 0,0010065 0,6 σ

21 Effective area: Mean strain in the reinforcement ε ε = ε ε =, = h, = 0,105 0,2 = 0,021 σ, ρ, α ρ, 227,17 0,4, 1 + 6,45 0,05429, 0,6 227, ε ε = 0, > 0, ε ε = 0, ,6 σ Mean value of the tensile strength of the concrete effective at the time when the cracks may first be expected to occur:, = = 2,6 factor: = 0,4, long term action Crack width according to (EN , clause 7.3.4) is : =, ε ε = 174,3 0, = 0,175 Figure Comparison of values with IDEA RCS results Page 21

22 1.8. Calculating stiffness Example - calculating the stiffness of the T-section according to EN Considering the strain, stress and internal forces in the previous examples are already calculated, the plane of strain is computed for a cracked section loaded by internal forces at the time when the cracks may first be expected to occur from the quasi-permanent combination. To calculate, for short-term stiffness, the difference in the calculation of shortand long-term stiffness is only taking into account the effective modulus of elasticity:, = where: ϕ, ϕ(,t 0 ) is the final value of creep coefficient Calculation will be carried out at mid-span section of quasi-permanent combination My = 113,73 knm ε x = , ε y = 0.0, ε z = Figure Strain stress diagram on cracked concrete cross section Sectional characteristics of transformed concrete section without cracks Cross sectional area of transformed cross section (steel area is transformed to concrete) = + = h h h + α = 210 1,85 0,5 1,85 0,20,5 0, ,40 10 = 0, Center of gravity of transformed cross section = α = 11, ,4516 0,327 = 0, , Page 22

23 Moment of inertia of original cross section = 1 12 h + h h h h h h + h h 2 = ,85 0,5 + 1,85 0,5 0,1305 0, ,85 0,20,5 0, ,85 0,20,5 0,08 0, ,5 0,08 = 0, Moment of inertia of transformed cross section = + + α = 0, ,85 0,20,5 0,08 + 0, , ,4516 0,3274 = 0, Sectional characteristics of transformed concrete section with cracks Compression zone: = + ε = 0, , = 0,0565 ε Cross sectional area of transformed cross section (steel area is transformed to concrete) = + α = 1,85 0, , ,4516 = 0, Center of gravity of transformed cross section = ( 2) α = 1,85 0,0565 0,1305 0, , ,4516 0,327 = 0, , Moment of inertia of original cross section = = 1 1,85 0, ,85 0,5650,1305 0, = I = 0, m Moment of inertia of transformed cross section = + + α = 0,0565 1,85 0, , ,4516 0,327 = 0, Page 23

Rematk: Current IDEA RCS version calculates cross sectional characteristics related to original center of gravity of cross section Since the same assumptions for calculating the limit state and

24 Rematk: Current IDEA RCS version calculates cross sectional characteristics related to original center of gravity of cross section Since the same assumptions for calculating the limit state and stiffness and width of cracks were used, we assume the stress in the reinforcement from the example of the calculation of crack width: σ = 227,1 Now we calculate the tensile force from ultimate load on the cracked section immediately prior to cracking. This plane is taken over from program IDEA RCS. ε x = , ε y = 0.0, ε z = Strain in reinforcing steel: ε = ε + ε + ε = 0, , ,32748 = 0, Stress in reinforcing steel σ = ε = 0, = 78,3 Reduction factor/distribution coefficientξ = 1 β σ σ = 1 1,, = 0,8808 bending stiffness of uncracked cross section:, = = 0, = 169 Page 24

bending stiffness of fully cracked cross section:, =, = 0,00129725 31 = 40,215 Stiffness is interpolated according to

25 bending stiffness of fully cracked cross section:, =, = 0, = 40,215 Stiffness is interpolated according to following expresion (Interpolation is done on level of stiffnesses) α = ξα + 1 ξα = ξ + 1 ξ α = 0, , = 55,48 Page 25

26 2. Column 2.1. Project details Square cross section 0.4 x 0, 4 m 2 reinforced in four corners by bars of 25 mm, stirrup with diameter 10 mm. Material C35/45, Reinforcements B 500B, concrete cover 25 mm, creep coefficient in infinity φ (, t0) = 1,68. Column 5 m, one- Laterally fixed in the XY plane, and both-sidedly fixed in the plane XZ. It is stand-alone element that is unbraced perpendicular to the Y-axis and braced to the Z axis. Figure Cross section and column geometry The internal forces obtained by calculating a linear structure in the investigated section: Combination for the ultimate limit state: = 800,, = 50,, = 0. Quasi-permanent combination for the serviceability limit state:, = 700,,, = 45,,, = 0. First order end moments: At the beginning: At the end:, = 60,, = 0,, = 0., = 0. Page 26

27 Calculating geometrical imperfections: Effective length l 0, = 2 = 2 5 = 10,, = 0,5 = 0,5 5 = 2,5. Reduction factor for length:, =, = 2/ = 2/ 5 = 0,894, 2/3 0, Reduction factor for number of members,, = 0,51 + 1/ = 0,51 + 1/1 = 1. Inclination,, = = ,894 1=0, Eccentricity:, =, /2 = 0, /2 = 0,02235,, =, /2 = 0, ,5/2 = 0, Total eccentricity including effects of geometrical imperfections:, +, =, / +, = 50/ ,02235 = 0, ,02235 = 0,08485,, +, =, / +, = 0/ , = 0, Minimum eccentricity according to paragraph 6.1 (4):,, = maxh/30; 0,02 = max0,4/30; 0,02 = max0,0133 ; 0,02 = 0,02,, = max, +, ;, = max0,08485; 0,02 = 0,08485,, = max, +, ;, = max0, ; 0,02 = 0,02. The first order moment with geometrical imperfections:, =, = 800 0,08485 = 67,88,, =, = 800 0,02 = 16. Page 27

28 2.2. Second order effects Slenderness and limit slenderness: Slenderness ratio =, =, = 86,6, =, =,, = 21,65,, = h = 0,4 = 0,1155. Necessary values for calculating the limit slenderness: End moments ratio:, = 1, because member is unbraced perpendicularly to Y axis,, = 1, because end moments are equal (,, =,, ). Relative normal force = =,, Mechanical reinforcement ratio = = 0,214. =,,,, = 0,229. The effect of creep may be ignored, if the following free conditions are met (,) = 1,68 2 = 86,6 75, = 21,65 < 75, =, = 0,08485 h = 0,4,, = = 0,02 h = 0,4. Conditions are not fulfilled, the effect of creep must not be ignored Effective creep ratio:, =,,,,, =,,,, = 1,68,, = 1,501, = 1,68, = 0,411, the moment from the quasi-permanent combination, including the effects of the first order we received from the same calculation as for the design moment, only difference is we are not taking account the condition for minimum eccentricity. =,, =,, = 0,769, = = = ,229 = 1,207,, = 1,7,, = 1,7 1 = 0,7 Limit slenderness:,, = 0,924,, = 20 / = 20 0,769 1,207 0,7/ 0,214 = 28,09,, = 20 / = 20 0,924 1,207 0,7/ 0,214 = 33,75, Slenderness criterion: = 86,6 >, = 28,09 = 21,65 <, = 33,75 slender column, non-slender column, 2nd order effects can be neglected. Page 28

29 Simplified method based on nominal stiffness Necessary factors: = / = 0,00196/0,16 = 0,01225 > 0,002, method can be used. = /20 = 35/20 = 1,323,, = /170 = 0,214 86,6/170 = 0,109, = 1,, =, /(1 +, ) = 1,323 0,109/(1 + 1,501) = 0,0577, = =, = 28,397,, = / = /9,6 = 1,028 Nominal stiffness: =,, +, = 0, , , , = 12,631. Euler critical load:, =, Second order moment:, =, = 12, = 1246,63, / 1 = 67,88 1,028 = 124,99, 1246,63/800 1 Total design moment including second order moment:, =, +, = 67, ,99 = 192,87,, =, = Simplified method based on nominal curvature Necessary factors: = 1 + = 1 + 0,229 = 1,229, = 0,4, = / = 1,229 0,214/1,229 0,4 = 1,31 1, = 0,35 + = 0,35 +, = 0,052,, = 1 +, = 1 + 0,052 1,501 = 0, Effective depth: = (h/2) + = 0,3525 m, = =, = 0,00217, 1/ = /(0,45 ) = 0,00217/(0,45 0,3525) = 0,0137, Page 29

30 1/ =, 1/ = 1 1 0,0137 = 0,0137. Deflection:, = 1, / = 0, /10 = 0,137. The nominal second order moment:, =, = 800 0,137 = 109,6. Total design moment including second order moment:, =, +, = 67, ,6 = 177,48,, =, = Biaxial bending No further check is necessary if the slenderness ratios satisfy the following conditions =,, = 4 2, first condition is not fulfilled, biaxial bending must be taken account according to paragraph (4). Page 30

Eurocode Training EN : Reinforced Concrete

Eurocode Training EN : Reinforced Concrete Eurocode Training EN 1992-1-1: Reinforced Concrete Eurocode Training EN 1992-1-1 All information in this document is subject to modification without prior notice. No part of this manual may be reproduced,