Design of reinforced concrete sections according to EN 1992-1-1 and EN 1992-2 Validation Examples Brno, 21.10.2010 IDEA RS s.r.o. Jihomoravské inovační centrum, U Vodárny 2a, 616 00 BRNO tel.: +420 - 541 142 063, fax: +420 - 541 143 011, www.idea-rs.cz 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 1992-1-1 and EN 1992-2, 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. Jihomoravské inovační centrum, U Vodárny 2a, 616 00 BRNO tel.: +420 - 541 142 063, fax: +420 - 541 143 011, www.idea-rs.cz Design of reinforced concrete sections according to EN 1992-1-1 and EN 1992-2 October 2010 Contents 1. Rib T1 .............................................................................................................................................................................. 4 1.1. Project details.......................................................................................................................................................... 4 Actions and analysis of Rib T1 ............................................................................................................................................. 5 1.1.1. 1.2. Cross section ........................................................................................................................................................... 9 1.3. Ultimate section resistance.................................................................................................................................... 10 1.4. Shear check ........................................................................................................................................................... 13 1.5. Torsional check ..................................................................................................................................................... 16 1.6. Interaction ............................................................................................................................................................. 17 1.7. Crack width calculation ........................................................................................................................................ 19 1.7.1. Crack witdh according to EN 1992-1-1 ....................................................................................................... 19 1.7.2. Example - Calculation of crack width according to EN 1992-1-1 .............................................................. 20 1.8. Calculating stiffness .............................................................................................................................................. 22 1.8.1. 2. Section forces ................................................................................................................................................ 7 Example - calculating the stiffness of the T-section according to EN 1992-1-1 ................................ 22 Column .......................................................................................................................................................................... 26 2.1. Project details........................................................................................................................................................ 26 2.2. Second order effects .............................................................................................................................................. 28 2.2.1. Simplified method based on nominal stiffness ............................................................................................ 29 2.2.1. Simplified method based on nominal curvature .......................................................................................... 29 2.2.2. Biaxial bending ........................................................................................................................................... 30 Page 3 Validation Examples Reinforced Concrete Section October 2010 1. Rib T1 1.1. Project details Example is taken from: Ing. Miloš Zich, Ph.D. and others, online publication "Konstrukční Eurokódy - Příklady posouzení betonových prvků dle Eurokódů", nakl. Verlag Dashöfer s. r. o., 2010, http://www.stavebniklub.cz/konstrukcni-eurokody-onbecd/ First floor slab Figure 1.1 - Schematic layout of structure Page 4 Validation Examples Reinforced Concrete Section October 2010 Section A Section B Figure 1.2 – Sections Actions and analysis of Rib T1 Figure 1.3 –Static schema of Rib T1 Page 5 Validation Examples Reinforced Concrete Section October 2010 Figure 1.4 - Floor composition for the calculation of loads Figure 1.5 - Permanent load calculation Variable load: qk = 10 kN/m2 * 2.0m = 20 kN/m quasi-permanent value: Factors defining the representative values of variable actions, A1.1 of EN1990 (also in attachment A4 in this document) 0, 1, 2 are shown in table. Page 6 Validation Examples Reinforced Concrete Section October 2010 Actions for Serviceability limit states (SLS) Loads for serviceability limit state are determined acc. to EN 1990 clause 6.5.3. There are 3 SLS-combinations: Characteristic combination of loads (Unacceptable cracking or deformation) Frequent load combination Quasi-permanent load combination Actions for Ultimate limit states (ULS) It is considered as a persistent design situation for ultimate limit state where partial factors are: G= 1,35, Q = 1,50. To determine the design load in Article 6.4.3.2 EN 1990 is prescribed the following equation marked as the equation (6.10) Substituting, we get the value of design load Alternatively, load can be further reduced according to equation (6.10) and (6.10b) and consider the less favorable value of both terms: 1.1.1. Section forces Page 7 Validation Examples Reinforced Concrete Section October 2010 Combination/Value Loads [kN/m] Vz(a) [kN] My (b) [kNm] SLS characteristic SLS frequent SLS quasi-permanent ULS 29.87 23.87 21.87 41.33 96.33 76.98 70.53 133.29 155.33 124.13 113.73 214.93 Table 1.1 - 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: VEd1 and MEd1 are calculated at distance lx = 0.225 + 0.458 = 0.683 m from the theoretical support.VEd1= 105,05 kN, MEd1 = 84,29 kNm. Page 8 Validation Examples Reinforced Concrete Section 1.2. October 2010 Cross section Figure 1.6 – Cross section Materials Concrete C25/30 fck = 25 MPa fcd = fck / c = 25 / 1,5 = 16,66 MPa fctm = 2,6 MPa fctd = 0,7 fctm / Steel B500B c = 0,7 2,6 / MPa fyk = 500 MPa fyd = fyk / s = 500 / 1,15 = 434,78 MPa Page 9 Validation Examples Reinforced Concrete Section 1.3. October 2010 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 1.7 - Response - given by program IDEA RCS Input data, Plane of strain: x = 0,0005876y =0,0 z= - 0,01034961 Figure 1.8 - Resulted plane of strain calculated by IDEA RCS Strain calculation in end fibers: Figure 1.9–Strain in ultimate compression fiber (picture from program IDEA RCS) Modulus of elasticity is calculated from stress-strain diagram Page 10 Validation Examples Reinforced Concrete Section October 2010 Defining the depth of compression zone (depth to neutral axis) in concrete, follow from: Concrete force in compression (as, the strain in concrete is outside the plastic branch, the stress along the section is linear in concrete) Concrete lever arm in the compression Concrete moment in compression Strain in reinforcing steel 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) Tensile force in reinforcement Page 11 Validation Examples Reinforced Concrete Section October 2010 Moment in tensile reinforcement Figure 1.11 – Comparing with results calculated by IDEA RCS program Equilibrium of forces 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 Validation Examples Reinforced Concrete Section 1.4. October 2010 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*0,083+ 1,85*0,08*0,0912+1/12*0,2*0,423+ 0,2*0,42*0,1592= 7,8933e-5+0,001225588+0,0012348+0,002123604= 0,004684m4 S= 1,85 * 0,08 * 0,091 + 0,2 * 0,051 *0,051/2 = 0,01373 m3 bw= 0,2 m 0,0 MPa cp= l= 1 Deriving from the above text, the concrete part does not carry all the shear force, hence shear reinforcement will be required. Figure 1.12 - Comparing with results calculated by RCS program Resistance without shear reinforcement in zones with cracks under bending loads CRd,c= 0,18 / c = 0,18 / 1,5 = 0,12 k1= 0,15 0,0 MPa cp= bw= 0,2 m d= 0,458 m min = 0.035 k3/2 fck1/2 = 0.035 1,6613/2 251/2 = 0,3745 MPa Page 13 Validation Examples Reinforced Concrete Section October 2010 Minimally 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 Asw= 2 * 0.0062 * PI /4 = 5,655e-5 m2 s= 0.24m z= 0.9*0.458 = 0.412 m exact value is 0,437 m fywd = fyd = 434,7MPa = 21,8 cw = 1,0 bw= 0,2 m z= 0.9*0.458 = 0.412 m exact value is 0,437m For calculation of strength reduction factor for concrete cracked in shear 1must be checked if the design stress of the shear reinforcement is over the 80% of the characteristic yield stress fywk, . Page 14 Validation Examples Reinforced Concrete Section October 2010 Figure 1.13 – Comparison of results calculated by RCS program Page 15 Validation Examples Reinforced Concrete Section 1.5. October 2010 Torsional check Section characteristics for torsional check u= A= tef = 2 * (1.85 + 0.5) = 4.7 m 2 * (1.85 + 0.5) = 0.232 m2 A / u = 0.232 / 4.7 = 0.049 m Torsional capacity without shear reinforcement Torsional capacity with shear reinforcement = 0.6 cw = 1,0 Figure 1.14 - Comparison of results calculated by RCS program Page 16 Validation Examples Reinforced Concrete Section 1.6. October 2010 Interaction Combined shear and torsion Shear reinforcement is not allowed to design according to detailing rules Compression strut check for combined shear and torsion Shear reinforcement check for combined shear and torsion Longitudinal reinforcement check for shear, torsion and bending = Page 17 Validation Examples Reinforced Concrete Section October 2010 Figure 1.15 - Comparing with results calculated by RCS program Page 18 Validation Examples Reinforced Concrete Section 1.7. October 2010 Crack width calculation 1.7.1. Crack width according to EN 1992-1-1 Check is introduced at midsection of beam My= 113,73 kNm Plane of strain calculated by program IDEA RCS: x = 0,0002092938 y= 0,0 z = -0,00282806 Defining the depth of compression zone (depth to neutral axis) in concrete, follow from: Figure 1.16 - Strain-stress diagram on fully cracked cross section Strain calculation in end concrete fibres: Stress calculation in end concrete fibres: Concrete force in compression: Concrete moment in compression: Strain in reinforcing steel: Page 19 Validation Examples Reinforced Concrete Section October 2010 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) Tensile force in the bar: Moment in tensile reinforcement: Equilibrium of forces: = 258,97– 258,97 = 0 Equilibrium of moments: 1.7.2. Example - Calculation of crack width according to EN 1992-1-1 Effective ratio of reinforcement: Maximal spacing of the cracks: Factors: k1= 0,8 in example is considered steel B500B k2= 0,5 Cross section loaded by bending moment, pure bending k3= 3,4 k4= 0,5 Effective height 7.3.2 (3) hc,ef: Page 20 Validation Examples Reinforced Concrete Section October 2010 Effective area: Mean strain in the reinforcement Mean value of the tensile strength of the concrete effective at the time when the cracks may first be expected to occur: factor: , long term action Crack width according to (EN 1992-1-1, clause 7.3.4) is : Figure 1.17 - Comparison of values with IDEA RCS results Page 21 Validation Examples Reinforced Concrete Section 1.8. October 2010 Calculating stiffness 1.8.1. Example - calculating the stiffness of the T-section according to EN 1992-1-1 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: (,t0) is the final value of creep coefficient Calculation will be carried out at mid-span section of quasi-permanent combination My = 113,73 kNm x = 0.0002092938, y = 0.0, z = -0.00282805826 Figure 1.18 - 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) Center of gravity of transformed cross section Moment of inertia of original cross section Page 22 Validation Examples Reinforced Concrete Section October 2010 Moment of inertia of transformed cross section Sectional characteristics of transformed concrete section with cracks Compression zone: Cross sectional area of transformed cross section (steel area is transformed to concrete) Page 23 Validation Examples Reinforced Concrete Section October 2010 Center of gravity of transformed cross section Moment of inertia of original cross section Moment of inertia of transformed cross section 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: 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. x = 0.00007225525, y = 0.0, z = -0.0009763408 Page 24 Validation Examples Reinforced Concrete Section October 2010 Strain in reinforcing steel: Stress in reinforcing steel Reduction factor/distribution coefficient bending stiffness of uncracked cross section: bending stiffness of fully cracked cross section: Stiffness is interpolated according to following expresion (Interpolation is done on level of stiffnesses) Page 25 Validation Examples Reinforced Concrete Section October 2010 2. Column 2.1. Project details Square cross section 0.4 x 0, 4 m2 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, oneboth-sidedly fixed in that is unbraced to the Z axis . Laterally fixed in the XY plane, and the plane XZ. It is stand-alone element perpendicular to the Y-axis and braced Figure 2.1 - Cross section and column geometry The internal forces obtained by calculating a linear structure in the investigated section: Combination for the ultimate limit state: , , . Quasi-permanent combination for the serviceability limit state: , , . First order end moments: At the beginning: At the end: , . , . Page 26 Validation Examples Reinforced Concrete Section October 2010 Calculating geometrical imperfections: Effective length l0 Reduction factor for length: , . Reduction factor for number of members . Inclination =0,00447. Eccentricity: , . Total eccentricity including effects of geometrical imperfections: , . Minimum eccentricity according to paragraph 6.1 (4): , , . The first order moment with geometrical imperfections: , . Page 27 Validation Examples Reinforced Concrete Section 2.2. October 2010 Second order effects Slenderness and limit slenderness: Slenderness ratio . Necessary values for calculating the limit slenderness: End moments ratio: , because member is unbraced perpendicularly to Y axis, , because end moments are equal ( ). Relative normal force . Mechanical reinforcement ratio . The effect of creep may be ignored, if the following free conditions are met . Conditions are not fulfilled, the effect of creep must not be ignored Effective creep ratio: , , 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. , , Limit slenderness: , , Slenderness criterion: slender column, non-slender column, 2nd order effects can be neglected. Page 28 Validation Examples Reinforced Concrete Section October 2010 2.2.1. Simplified method based on nominal stiffness Necessary factors: , method can be used. , , , , Nominal stiffness: Euler critical load: Second order moment: Total design moment including second order moment: 2.2.1. Simplified method based on nominal curvature Necessary factors: , , , Effective depth: , , , Page 29 Validation Examples Reinforced Concrete Section October 2010 Deflection: The nominal second order moment: Total design moment including second order moment: 2.2.2. Biaxial bending No further check is necessary if the slenderness ratios satisfy the following conditions first condition is not fulfilled, biaxial bending must be taken account according to paragraph 5.8.9 (4). Page 30