rcs validation examples part 1 - idea-rs.nl · 2018-03-09 · validation examples reinforced...

IDEA RS s.r.o. South Moravian Innovation Centre, U Vodarny 2a, 616 00 BRNO tel.: +420 - 511 205 263, fax: +420 - 541 143 011, www.idea-rs.cz, www.idea-rs.com

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. South Moravian Innovation Centre, U Vodarny 2a, 616 00 BRNO tel.: +420 - 511 205 263, fax: +420 - 541 143 011, www.idea-rs.cz, www.idea-rs.com

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 one-way rib-shaped slab and an example of column with high axial load and bi-axial bending is introduced.

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. Section forces ....................................................................................................... 7

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 width 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. Example - calculating the stiffness of the T-section according to EN 1992-1-1 22

2. 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

Validation Examples Reinforced Concrete Section

1. Rib T1

1.1. Project details

Example is taken from:

Ing. Miloš Zich, Ph.D. and othersposouzení betonových prvků dle Eurokódůhttp://www.stavebniklub.cz/konstrukcni

First floor slab

Validation Examples Reinforced Concrete Section

Project details

others, online publication "Konstrukční Eurokódy ů dle Eurokódů", nakl. Verlag Dashöfer s. r. o., 2010,

http://www.stavebniklub.cz/konstrukcni-eurokody-onbecd/

Figure 1.1 - Schematic layout of structure

October 2010

ční Eurokódy - Příklady o., 2010,

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


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: ψ� ∗ �� = 0,6 ∗ 20 = 12 ��/��

Factors defining the representative values of variable actions, ψ0, ψ1, ψ2 are shown in table. A1.1 of EN1990 (also in attachment A4 in this document)


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) ∑ ��,�"+"�" + "��,�" + "∑ ψ�,��,��

�,ψ� = 9,87 + 0 + 20 + 0 = 29,87 �� ⁄

Frequent load combination

��,�"+"�" + "ψ�,��,�" + "�ψ�,��,��

�,ψ� = 9,87 + 0 + 0,7⋅ 20 + 0 = 23,87 �� ⁄

Quasi-permanent load combination

��,�"+"γ�" + �ψ�,��,��

�,ψ� = 9,87 + 0 + 0,6⋅ 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: γG= 1,35, γQ = 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)∑ γ,��,�"+"γ�" + "γ�,��,�" + "∑ γ�,�ψ�,��,��

Substituting, we get the value of design load

� = 1,35⋅9,87 + 0 + 1,5⋅ 20 + 0 = 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�,��,�� ,

� = 1,35⋅9,87 + 0 + 1,5⋅ 0,7⋅ 20 + 0 = 34,32 �� ⁄

� = 1,35⋅ 0,85⋅9,87 + 0 + 1,5⋅ 20 + 0 = 41,33 �� ⁄

1.1.1. Section forces

� �� =1

2� �

�� =1

8� ��


Combination/Value Loads [kN/m] Vz(a) [kN] My (b) [kNm]

SLS characteristic 29.87 96.33 155.33

SLS frequent 23.87 76.98 124.13

SLS quasi-permanent 21.87 70.53 113.73

ULS 41.33 133.29 214.93

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.


1.2. Cross section

Figure 1.6 – Cross section

Materials

Concrete C25/30 fck = 25 MPa

fcd = fck / γc = 25 / 1,5 = 16,66 MPa

fctm = 2,6 MPa

fctd = 0,7 fctm / γc = 0,7 2,6 / 1,5 = 1,213 MPa

Steel B500B fyk = 500 MPa

fyd = fyk / γs = 500 / 1,15 = 434,78 MPa


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 1.7 - Response - given by program IDEA RCS

Input data, Plane of strain:

εx = 0,0005876εy =0,0 εz= - 0,01034961

Figure 1.8 - Resulted plane of strain calculated by IDEA RCS

Strain calculation in end fibers:

ε� = ε� + ε�� + ε �� = 0,00058758 + 0 ∗ 0 − 0,01034961 ∗ 0.131 = − 0,00076304

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 ��


σ� = ε�� = −0,00076304 9,52 10� = −7,27 ��

Defining the depth of compression zone (depth to neutral axis) in concrete, follow from:

� = �� +ε�ε

= 0,1305 +0,00058758

−0,01034961= 0,0737268 �

Concrete force in compression (as, the strain in concrete is outside the plastic branch, the stress along the section is linear in concrete) �� = �� 2 = 0,0737268 ∗ 1,85 ∗ 16,7 2 = −495,83 ��⁄⁄

Concrete lever arm in the compression �� = �� − � 3⁄ = 0,1305 − 0,0737268 3⁄ = 0,1059244 �

Concrete moment in compression �� = �� = −495,83 ∗ 0,1059244 = −52,52 ��

Strain in reinforcing steel

ε� = ε� + ε�� + ε �� = 0,00058758 + 0 ∗ 0 − 0,01034961 ∗ −0.327 = 0,0039769

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�⋅ �� = 0,0039769 ∗ 200 = 559,34 > 434,78 ⇒ 434,78 MPa

Tensile force in reinforcement


�� = ��σ� = 0,0011404⋅434,783 = 495,83 KN

Moment in tensile reinforcement

M� = F�z� = 495,83⋅ − 0.327 = −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.


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*0,083+ 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 σcp= 0,0 MPa αl= 1

��,� = � �� + �� =

0,004684⋅0,2

0,01373!�1,213⋅10�� + 0 = 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 1.12 - Comparing with results calculated by RCS program

Resistance without shear reinforcement in zones with cracks under bending loads

CRd,c= 0,18 / γc = 0,18 / 1,5 = 0,12

� = 1 + "200# = 1 + "200

458= 1,661

$� =��# =

760

200 458= 0,008297 ≤ 0,02

k1= 0,15 σcp= 0,0 MPa bw= 0,2 m d = 0,458 m νmin = 0.035 k3/2 fck1/2 = 0.035 1,6613/2 251/2 = 0,3745 MPa


��,� = %&��.�� (100 $��)� � + �� ' ��#

= %0,12 1,661 (100⋅ 0,008297⋅ 25(3)� � + 0' 0,2⋅ 0,458 = 50,15��

Minimally

��,� = )*��! + �� +��# = �0,3745(3 + 0� ∗ 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 (��,� ,��,��) 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 θ = 21,8°

��,� = ��, � �� cot- =

2⋅2,827e − 5

0,24∗ 0,412 ∗ 434,78⋅10� ∗ ./021,8

= 105,70�� > ��,� = 105,05�� → &ℎ(.� 1, 23 αcw = 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 ν1must be checked if the design stress of the shear reinforcement is over the 80% of the characteristic yield stress

fywk, �� ="��

��

�#�$=

��%,%�⋅��,��

�,� �,&�� #��,'

434,74 �� > 0.8 ∗ �� = 400��

4� = 0.6⋅ 51 −(��%�6 = 0.6⋅ 51 −

�%�%�6 = 0,54.

��,�� = �� 4� ��cot- + tan-� =

1,0⋅ 0,2⋅ 0,412 ⋅ 0,54⋅ 16,66⋅10�

�cot 21,8 + tan 21,8�

= 255,76�� > ��,�� = 133,27�� → &ℎ(.� 1, 23


Figure 1.13 – Comparison of results calculated by RCS program


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 m2

tef = A / u = 0.232 / 4.7 = 0.049 m

� =−7 ± √7� − 16�

−4=

−4.7 ± √4.7� − 16 ∗ 0.232

−4=

−4.7 ± 4.287

−4= m�.��

�.�&)

ℎ =�7 − 2b�

2= m�.�&)

�.��

�� = �0.103 − 0.049��2.247 − 0.049� = 0.1187 m�

7� = 2)�0.103 − 0.049� + �2.247 − 0.049�+ = 4.504m

Torsional capacity without shear reinforcement

9��,� = 2��0*(�� = 2⋅0.1187⋅0.049⋅ 1.213 10� = 14.11��

Torsional capacity with shear reinforcement ν = 0.6 αcw = 1,0 9��,�� = 2 4 ��0*( �+,$ �-�$

= 2 ∗ 0.6 ∗ 1,0 ∗ 16,66⋅10� ∗ 0,1187

∗ 0.049 ∗ sin 21.8 ∗ cos 21.8 = 36,09��

9��,� = 2��"��, ��

��7� ��= 2⋅ 0,1187"2,827e − 5

0,24434,78⋅10� 7,603e − 4

4,507434,78⋅10� = 14,59��

Figure 1.14 - Comparison of results calculated by RCS program


1.6. Interaction

Combined shear and torsion "��

"��,�+

.��.��,�

=��%,�%%�,�% +

��&.�� = 2,09 > 1,0 => �/0 23

Shear reinforcement is not allowed to design according to detailing rules

Compression strut check for combined shear and torsion ��,��+

9��9��,��=

105,05

255,76+

0

36,09= 0,41 < 1,0 => 23

Shear reinforcement check for combined shear and torsion

��,�� = Σ�� =2,827e − 5

0,24434,78⋅10� = 51,21��

�� = :��;� � +9��

2 ��< tan- = :105,05

2 0,412+

0

2⋅ 0,1187< tan 21,8 = 50,99��

��,��=

50,99

51,21= 0,99 < 1,0 => �/0 23

Longitudinal reinforcement check for shear, torsion and bending

��,�� = Σ�� =7,603e − 4⋅(434,78 − 253,9)⋅10� = 137,52��

�� =��

tan - +9��7�

2 �� tan - =105,05

tan 21,8+

0⋅4.504

2⋅ 0,1187 tan 21,8= 262,64��

��,��=

262,64

137,52= 1,91 > 1,0 => �/0 23


Figure 1.15 - Comparing with results calculated by RCS program


1.7. 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:

εx = 0,0002092938

εy= 0,0

εz = -0,00282806

Defining the depth of compression zone (depth to neutral axis) in concrete, follow from:

� = �� +ε�ε

= 0,1305 +0,000209293

−0,002828058= 0,0565 �

Figure 1.16 - Strain-stress diagram on fully cracked cross section

Strain calculation in end concrete fibres:

ε�� = ε� + ε�� + ε �� = 0,000209293 − 0,002828058⋅ 0,1305 = − 0,0001598

Stress calculation in end concrete fibres:

σ�� = ε��⋅�� = −0,0001598⋅ 31 = −4,954 ��

Concrete force in compression:

�� =��σ�

2=

��⋅��σ�2

=�0,0565⋅1,85�⋅ − 4,954

2 = −0,259 kN

Concrete moment in compression:

M� = F�z� = F�⋅�z/01 − x 3⁄ � = −258937 − �0,1305 − 0,0565 3⁄ � = −28,92 kNm

Strain in reinforcing steel:

ε�� = ε� + ε�� + ε �� = 0,000209293 − 0,002828058⋅ − 0,32748 = 0,001135425


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�⋅�� = 0,001135425⋅ 200⋅10� = 227,1 ��

Tensile force in the bar: �� = ��σ� = 0,0011404⋅227085000 = 258,97 kN

Moment in tensile reinforcement:

M� = F�z� = 258,97 ⋅ − 0.32745 = −84,81 kNm

Equilibrium of forces: �� + �� = �� = 258,97– 258,97 = 0

Equilibrium of moments: �� + �� = �� = −84,81 − 28,92 = −113,725 kNm

1.7.2. Example - Calculation of crack width according to EN 1992-1-1

Effective ratio of reinforcement:

ρ�,*(( =��,*((

=11,40⋅102&

0,021= 0,05429

Maximal spacing of the cracks:

,3,�� = ��. + ��φ/ρ�,*((

,3,�� = 3,4⋅0,031 + 0,8⋅0,5⋅0,425⋅0,022/0,05649 = 0,1743 �

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:

ℎ�,*( = �1; =2,5�ℎ − #�;ℎ − �

3;ℎ

2> =

�1; =2,5�0,5 − 0,458�;0,5 − 0,05649

3;

0,5

2> =

�1;?0,105; 0,1478; 0,25@ = 0,105 �


Effective area:

��,*(( = ℎ�,*(�� = 0,105 ∗ 0,2 = 0,021��

Mean strain in the reinforcement ε�� − ε�� =σ�2��

��,��

ρ�,��4�5α�ρ�,��6

��≥ 0,6

σ�

��

ε�� − ε�� =227,17 − 0,4

�,7�,�%&�� 1 + 6,45⋅0,05429�

200≥ 0,6

227,17

200

ε�� − ε�� = 0,0010065 > 0,0006815 →

ε�� − ε�� = 0,0010065

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 1992-1-1, clause 7.3.4) is :

A� = ,3,��ε�� − ε�� = 174,3 0,0010065 = 0,175 ��

Figure 1.17 - Comparison of values with IDEA RCS results


1.8. 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 short-and long-term stiffness is only taking into account the effective modulus of elasticity:

��,*(( =��

�5ϕ89,��:

where: ϕ(∞,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

εx = 0.0002092938,

εy = 0.0,

εz = -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)

�� = �� + �� = �ℎ − �� − ��)ℎ − ℎ(+ + ��α* =

1,85⋅0,5 − �1,85 − 0,2��0,5 − 0,08� + 11,40⋅102& 210

31= 0,239357 ��

Center of gravity of transformed cross section

�� =��α*�� =

11,40⋅102&⋅6,4516⋅ − 0,327

0,239357= −0,01005 �


Moment of inertia of original cross section

�� = 1

12�ℎ� + �ℎ �� − ℎ 2⁄ �� −

1

12�� − ��)ℎ − ℎ(+�

− �� − ��)ℎ − ℎ(+)��! + )ℎ − ℎ(+ 2⁄ +� =

1

121,85⋅0,5� + 1,85⋅0,5⋅�0,1305 − 0,5 2⁄ �� −

1

12�1,85 − 0,2��0,5 − 0,08��

− �1,85 − 0,2��0,5 − 0,08�)−0,3274 + �0,5 − 0,08�+� = 0,0046628 �&

Moment of inertia of transformed cross section

�� = �� + �� + ��α*�� =

0,00468631 + �1,85 − 0,2��0,5 − 0,08� + �−0,01005�� + 11,40⋅102&⋅6,4516⋅ − 0,3274�

= 0,00542541 �&

Sectional characteristics of transformed concrete section with cracks

Compression zone:

� = �� +ε�ε

= 0,1305 +0,000209293

−0.002828058= 0,0565 �

Cross sectional area of transformed cross section (steel area is transformed to concrete)�� =� � + ��α* = 1,85 0,0565 + 11,40 102&6,4516 = 0,111901 ��

Center of gravity of transformed cross section

�� =�⋅�(�� − � 2⁄ )��α*�� =

1,85⋅0,0565⋅�0,1305 − 0,0565 2⁄ � + 11,40⋅102&⋅6,4516⋅ − 0,327

0,239357= 0,07401�

Moment of inertia of original cross section

�� =1

12�� + �� − � 2⁄ �� =

1

121,85 0,565 + 1,85 0,565�0,1305 − 0,565 2⁄ �� =

I; = 0,001909 m&

Moment of inertia of transformed cross section

�� = �� + �� + ��α*�� =

0,0565⋅1,85⋅0,07401� + 11,40⋅102&⋅6,4516⋅−0,327� = 0,00129725 �&


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.

εx = 0.00007225525,

εy = 0.0,

εz = -0.0009763408

Strain in reinforcing steel:

ε�� = ε� + ε�� + ε �� = 0,000072255 − 0,00097634⋅ − 0,32748 = 0,000392

Stress in reinforcing steel σ�3 = ε⋅�� = 0,000392⋅ 200 = 78,3 ��

Reduction factor/distribution coefficientξ = 1 − β5σ��σ�6� = 1 − 1 5 )',�

��),�6� = 0,8808

bending stiffness of uncracked cross section:

��,< = ��<�� = 0,00542541⋅31⋅10� = 169 ��


bending stiffness of fully cracked cross section:

��,<< = ��,<<�� = 0,00129725⋅31 = 40,215 ��

Stiffness is interpolated according to following expresion (Interpolation is done on level of stiffnesses) α = ξα<< + �1 − ξ�α< = ξ��<< + �1 − ξ��<

α = 0,88 ⋅ 40 + �1 − 0,88�⋅169 = 55,48 ��


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, 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 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:

�� = −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 ��.


Calculating geometrical imperfections:

Effective length l0 ��,� = 2 ∙ � = 2 ∙ 5 = 10 �,

��, = 0,5 ∙ � = 0,5 ∙ 5 = 2,5 �.

Reduction factor for length:

�>,� = �>, = 2/√� = 2/√5 = 0,894, 2/3 ≤ 0,894 ≤ 1.

Reduction factor for number of members

��,�, = !0,5�1 + 1/�� = !0,5�1 + 1/1� = 1.

Inclination-�,�, = -� ∙ �> ∙ �� = 1 200 ∙ 0,894 ∙ 1⁄ =0,00447.

Eccentricity:

(�,� = -��,�/2 = 0,00447 ∙ 10/2 = 0,02235 �,

(�, = -��, /2 = 0,00447 ∙ 2,5/2 = 0,0055875 �.

Total eccentricity including effects of geometrical imperfections:

(��!,� + (�,� = ��,�/�� + (�,� = 50/800 + 0,02235 = 0,0625 + 0,02235 = 0,08485 �,

(��!, + (�, = ��, /�� + (�, = 0/800 + 0,0055875 = 0,0055875 �.

Minimum eccentricity according to paragraph 6.1 (4):

(�,�, = max�ℎ/30; 0,02� = max�0,4/30; 0,02� = max�0,0133BBBB; 0,02� = 0,02 �,

(��,� = max)(��!,� + (�,�; (�,�+ = max�0,08485; 0,02� = 0,08485 �,

(��, = max)(��!, + (�, ; (�, + = max�0,0055875; 0,02� = 0,02 �.

The first order moment with geometrical imperfections:

��,� = �� ⋅ (��,� = 800 ⋅ 0,08485 = 67,88 ��,

��, = �� ⋅ (��, = 800 ⋅ 0,02 = 16 ��.


2.2. Second order effects

Slenderness and limit slenderness:

Slenderness ratioC� =��,�

��=

��,��%% = 86,6, C =

��,�

��=

�,%�,��%% = 21,65,

1�, = � �� ℎ� = � �

�� 0,4� = 0,1155 �.

Necessary values for calculating the limit slenderness:

End moments ratio:

D�,� = 1, because member is unbraced perpendicularly to Y axis,

D�, = 1, because end moments are equal (��, ,� = ��, ,�).

Relative normal force; =?��

@�(��=

'��∙��,�7×�%∙�� ,%⁄ = 0,214.

Mechanical reinforcement ratioE =@�(��@�(��

=�,��7×%��∙�� ,�%⁄

�,�7×�%∙�� ,%⁄ = 0,229.

• The effect of creep may be ignored, if the following free conditions are metF(9,��) =

1,68 ≤ 2

• C� = 86,6 ≰ 75, C = 21,65 < 75

• B��,�

?��=

7),'''�� = 0,08485 ≱ ℎ = 0,4,

B��,�

?��=

�7'�� = 0,02 ≱ ℎ = 0,4.

Conditions are not fulfilled, the effect of creep must not be ignored

Effective creep ratio:

F*(,� = F89,��:B��,��,�

B��,�= 1,68

7�,7%7),'' = 1,501,

F*(, = F89,��:B��,��,�

B��,�= 1,68

�,��7 = 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.

�� =�

C�5�,�D��,�E=

�8�5�,�⋅�,%��: = 0,769, � =

�8�5�,�⋅�,&��: = 0,924,

G = √1 + 2E = √1 + 2 ⋅ 0,229 = 1,207,

&�, = 1,7 − D�,�, = 1,7 − 1 = 0,7

Limit slenderness:

C��,� = 20 ⋅ �� ⋅ G ⋅ &�/√; = 20 ⋅ 0,769 ⋅ 1,207 ⋅ 0,7/√0,214 = 28,09,

C��, = 20 ⋅ � ⋅ G ⋅ & /√; = 20 ⋅ 0,924 ⋅ 1,207 ⋅ 0,7/√0,214 = 33,75,

Slenderness criterion:

C� = 86,6 > C��,� = 28,09 slender column,

C = 21,65 < C��, = 33,75 non-slender column, 2nd order effects can be neglected.


2.2.1. Simplified method based on nominal stiffness

Necessary factors:

H = ��/�� = 0,00196/0,16 = 0,01225 > 0,002, method can be used.

�� = !��/20 = !35/20 = 1,323,

��,� = ; ⋅ C�/170 = 0,214 ⋅ 86,6/170 = 0,109,

3� = 1, 3�,� = ��,�/(1 + F*(,�) = 1,323 ⋅ 0,109/(1 + 1,501) = 0,0577,

�� =��F��

=�&,�))�,� = 28,397 ��,

I = J�/.� = J�/9,6 = 1,028

Nominal stiffness:

�� = 3�,��,� + 3��,�= 0,0577 × 28,397 ⋅ 10� × 2,133 ⋅ 102� + 1 × 200 ⋅ 10� × 4,566 ⋅ 102%

= 12,631 ��.

Euler critical load:

�G,� =J��,�

� =J� × 12,631 ⋅ 10�

10� = 1246,63 ��

Second order moment:

��,� = ��,�I�)�G,�/��+ − 1

= 67,881,028�1246,63/800� − 1

= 124,99 ��,

Total design moment including second order moment:

��,� = ��,� + ��,� = 67,88 + 124,99 = 192,87 ��,

��, = ��, = 16 ��.

2.2.1. Simplified method based on nominal curvature

Necessary factors:

;� = 1 + E = 1 + 0,229 = 1,229,

;�� = 0,4,

33 = �;� − ;�/�;� − ;�� = �1,229 − 0,214�/�1,229 − 0,4� = 1,31 ≰ 1,

I� = 0,35 +(�� −

H��%� = 0,35 +

�%�� −

'7,7�%� = −0,052,

3D,� = 1 + I�F*(,� = 1 + −0,052 ⋅ 1,501 = 0,921 ≱ 1.

Effective depth:

# = (ℎ/2) + 1� = 0,3525 m,

K�� =(��

=

��

�,��

�� = 0,00217,

1/D� = K��/(0,45 #) = 0,00217/(0,45 ⋅ 0,3525) = 0,0137,


1/D� = 33 ⋅ 3D,� ⋅ 1/D� = 1 ⋅ 1 ⋅ 0,0137 = 0,0137.

Deflection:

(�,� = L1D�M ��,�� /. = 0,0137 ⋅ 10�/10 = 0,137 �.

The nominal second order moment:

��,� = ��(�,� = 800 ⋅ 0,137 = 109,6 ��.

Total design moment including second order moment:

��,� = ��,� + ��,� = 67,88 + 109,6 = 177,48 ��,

��, = ��, = 16 ��.

2.2.2. Biaxial bending

No further check is necessary if the slenderness ratios satisfy the following conditions

H�H�

='7,7��,7% = 4 ≰ 2, first condition is not fulfilled, biaxial bending must be taken account

according to paragraph 5.8.9 (4).

rcs validation examples part 1 - idea-rs.nl · 2018-03-09 · validation examples reinforced...

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