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
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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
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October 2010
Section A
Section B
Figure 1.2 – Sections
Actions and analysis of Rib T1
Figure 1.3 –Static schema of Rib T1
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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.
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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
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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.
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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
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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 = 0,0005876y =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

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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
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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.
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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
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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,
.
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Validation Examples Reinforced Concrete Section
October 2010
Figure 1.13 – Comparison of results calculated by RCS program
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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
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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
=
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Validation Examples Reinforced Concrete Section
October 2010
Figure 1.15 - Comparing with results calculated by RCS program
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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:





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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:
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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
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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
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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)
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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
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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) 



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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:
,
.
,
.
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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:
,
.
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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.
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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:
,
,
,
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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).
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