RC Design using Eurocode in Robot Structural.doc

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ROBOT STRUCTURAL ANALYSIS PROFESSIONAL 2011

DESIGN OF REINFORCED CONCRETE ELEMENTS

ACCORDING TO EN 1992-1-1:2004

The structure below has been modeled in order to present the design possibilities in Robot Structural Analysis Professional 2011 for reinforced concrete elements.

The structure was the following dimensions: 10.8mx28.2m and 15.5m height. There are 4 stories, 3m height each and the ground floor 3.5m height.

As loads acting on the structure we have modeled:

1. Self weight of elements: it is calculated automatically by the program.

2. Finishes: 1.5kN/m2

3. Non structural internal walls: 3.0kN/m2

4. Live loads: 2.0kN/m2_{for each floor and 1.0kN/m}2_{for roof.}

5. Snow: 1.6kN/m2

6. Seismic loads generated using the EC8 norm.

The loads were combined using the automatic combination options according to EC1 norm.

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The program offers the possibility to choose the design code from a large list. This can be done by accessing the Job Preferences option from the menu Tools. In the Job Preferences window presented below the user can choose the code for designing steel, aluminum, reinforced concrete, timber structures or geotechnical design.

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When choosing the code for design, sometimes, in the right side of the drop box there is a button that allows access to additional options. This is the case for eurocodes. The additional options available are presented in the window below and allows the user to personalize the safety coefficients according to the national annex of the user's country.

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If by default the required code is not present in the current code list, the user can search a particular code by checking the available list of codes. This can be done by pressing the More codes... button in the Jog Preferences window. Below you can see the window that appears by accessing the More codes.... button.

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Options for required reinforcement of Beams/Columns

In order to correctly calculate the required reinforcement the user has to provide some options for the program regarding the elements that has to be calculated. This will be done by accessing the Required

Reinforcement of Beams/Columns - options/Code parameters... command from the Design menu. The following window will appear.

In this window we will find two default set of elements defined: RC column and RC beam. These sets of options can't be modified. In order to personalize these elements press the new button in the upper part of the window. We have created two more types of elements named RC Beam 1 and RC Column 1. These elements were defined using the options

presented below.

The options that has to be selected refers to the supports of the elements and the way that these elements connects to other elements.

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Another set of options that has to be reviewed and personalize before calculation are the Calculation parameters... Here also there is a default set of options that can't be changed. We had to create a new set of options named standard1 in order to provide the correct options for

calculating the required reinforcement.

The calculation parameters refers to the material characteristics of the elements. The user has to provide information regarding the concrete and steel used for longitudinal and transversal reinforcement.

By pressing the new button the user has access to the window presented below, named Calculation Parameter Definition. This window has three tabs named:

General - options for concrete Longitudinal reinforcement Transversal reinforcement

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Required reinforcement for beams and columns

After providing the calculation parameters for the elements of the structure now we can proceed to calculate the required reinforcement for beams and columns.

For this select the Required reinforcement for Beams/Columns... command from the Design menu. By selecting this command the screen will change like in the picture below. We can see the following windows:

View

Calculations

Required Reinforcement Bars

In order to see the required reinforcement results the user has to introduce in the Calculation window the following information:

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- load cases that will be used to calculate the reinforcement - how many calculation points along the bar length will be used. After providing this information press the Calculate button.

After the calculation is finished the program will display the window presented below, where it will provide information regarding the calculation process and the results obtained. Here we can see if there were any errors or warnings during calculation. As we can see in our case all results were correct.

The results are presented in a table, in the lower part of the display. Here are the results for the provided reinforcement of the selected beam and column.

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Options for required reinforcement of slabs/walls

Same as for the beams and columns, before calculating the required reinforcement for slabs the user has to provide the calculation parameters for these type of elements. This can be done by accessing the Calculation parameters command form the Design menu/Required reinforcement for slabs/walls - option.

As default parameters sets we can find RC Floor and RC Wall. These sets can't be changed so we will have to create two new sets that can be personalized to our needs. We will call them RC Floor1 and RC Wall1.

The windows that allows the definition of the calculation parameters contain the following tabs:

- General - Materials

- SLS Parameters - Reinforcements

First we have defined the RC Floor1 options sets using the following parameters:

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For the RC Wall1 options set we have defined the following parameters:

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Required reinforcement for slab

After the definition of the calculation parameters we can proceed to calculate the required reinforcement for the slab. For this select the

Required reinforcement for slab/wall command from the Design menu. By selecting this command the display will change and we can see the

following windows: - View

- Plate and shell reinforcement - Reinforcement

In order to see the required reinforcement results the user has to introduce in the Plate and shell reinforcemet window the following

information:

- panels that will be calculated

- load cases that will be used to calculate the reinforcement After providing this information press the Calculate button.

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When the calculations are done we can use the Reinforcements window and select what results to be displayed in the view window. The following results are available:

- Reinforcement area - Reinforcement spacing - Number of bars - Minimum reinforcement - Displacement (SLS) - Cracking (SLS)

Important: In order to read results that are displayed in the SLS tab it is mandatory to have defined SLS Load Combination cases.

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Here are the results for provided reinforcement on all directions (X and Y) and layers (up and down).

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RC Beam design - Provided reinforcement

Beside the required reinforcement the program is capable to calculate provided reinforcement for all the elements. We will start by presenting the provided reinforcement for beams.

In order to calculate the provided reinforcement the user has to select the element that will be calculated and access the RC Beam Design command from the Design menu/Provided reinforcement for RC elements.

The first window that will appear will allow us to select the load cases that will be used for calculating the reinforcement.

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The display has changed and on screen we have some new windows where we can see the geometry of the element.

Before calculating the provided reinforcement we have to adjust two sets of calculation parameters.

- Analysis menu/Calculation options... - Analysis menu/Reinforcement pattern...

First we will start with Calculation options. This command will open a window with five tabs:

- General - Concrete

- Longitudinal reinforcement - Transversal reinforcement - Additional reinforcement

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In these tabs we will provide information about material quality (concrete and steel), cover, cracking and deflection options. Also we can provide information necessary to perform a fire resistance check.

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After the personalization of the calculation options we can save these sets in order to reuse them in other projects. We can do this by pressing the Save As... button.

The second set of options to be adjusted is the Reinforcement

pattern. Because the program will propose a real solution for reinforcement, we will have to provide some rules that the program will follow when it will draw the reinforcement.

Here are the parameters for reinforcement patterns used for calculating the provided reinforcement.

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Same as before, the user can save the reinforcement pattern options for later use with other projects by pressing the Save as button.

Next step is to indicate the options sets to be used for calculation. For this select the Options set... from the Analysis menu.

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For calculation the reinforcement select the Calculation command from the Analysis menu and press the Calculate button in the window that just opened.

When the calculation is done the program will display a window with information regarding events during calculation.

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When the calculation is finished we can switch to the

Beam-Diagrams tab and see the graphic results. Also below the diagrams there are displayed in a table the results in the characteristic points of the

element.

In the next tab, Beam-Reinforcement, we can see the reinforcement bars provided by the program. We can even change the reinforcement and recalculate the element.

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In the last tab called Beam-note we can see a full calculation note for the current calculated element. This calculation note can also be generated by selecting the Calculation note command from the Results menu. This calculation note can be save as a *.rtf file and can be attached to the project documentation.

Here it is the calculation note provided for our beam:

1 Level:

 Name :

 Reference level :

--- Maximum cracking : 0,30 (mm)

 Exposure : X0

 Concrete creep coefficient : _= 2,75

 cement class : N

 Concrete age (loading moment) : 28 (days)

 Concrete age : 50 (years)

 Structure class : S1

 Fire resistance class : no requirements

2 Beam: Beam36...37 Number: 1

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 Concrete : C25/30 fck = 25,00 (MPa)

Bi-linear stress distribution [3.1.7(2)]

Density : 2501,36 (kG/m3)

Aggregate size : 20,0 (mm)

 Longitudinal reinforcement: : B500C fyk = 500,00 (MPa)

Horizontal branch of the stress-strain diagram

Ductility class : C

 Transversal reinforcement: : B500A fyk = 500,00 (MPa)

2.2 Geometry:

2.2.1 Span Position L.supp. L R.supp.

(m) (m) (m)

P1 Span 0,40 5,00 0,40

Span length: Lo = 5,40 (m)

Section from 0,00 to 5,00 (m) 30,0 x 60,0 (cm) without left slab without right slab

(m) (m) (m)

P2 Span 0,40 5,65 0,30

2.3 Calculation options:

 Regulation of combinations : EN 1990:2002

 Calculations according to : EN 1992-1-1:2004 AC:2008

 Seismic dispositions : No requirements

 Precast beam : no

 Cover : bottom c = 2,5 (cm)

: side c1= 2,5 (cm)

: top c2= 2,5 (cm)

 Cover deviations : Cdev = 1,0(cm), Cdur = 0,0(cm)

 Coefficient 2 =0.50 : long-term or cyclic load

 Method of shear calculations : strut inclination

2.4 Calculation results:

The deflection L/500 (7.4.1(5)) is not verified

No. Type State Span x(m) Value Capacity n*

1. M [kN*m] ULS 2 11.45 -50.76 -39.55 0.78

2. M [kN*m] ALS 2 11.45 -58.39 -45.64 0.78

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n* - Safety factor

2.4.1 Internal forces in ULS

Span Mt max. Mt min. Ml Mr Ql Qr

(kN*m) (kN*m) (kN*m) (kN*m) (kN) (kN) P1 48,55 -2,26 -33,72 -61,10 60,99 -66,77 P2 53,61 -0,00 -62,67 -50,76 68,86 -46,33 0 2 4 6 8 1 0 8 0 6 0 4 0 2 0 0 - 2 0 - 4 0 - 6 0 - 8 0 [m ] [k N * m ] B e n d i n g M o m e n t U L S : M M r M t M c 0 2 4 6 8 1 0 - 2 0 0 - 1 5 0 - 1 0 0 - 5 0 0 5 0 1 0 0 1 5 0 2 0 0 [m ] [k N ] S h e a r F o r c e U L S : V V r V c ( s t i r r u p s ) V c ( to ta l ) 2.4.2 Internal forces in SLS

(kN*m) (kN*m) (kN*m) (kN*m) (kN) (kN)

P1 31,29 0,00 -21,86 -39,58 44,43 -48,65

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0 2 4 6 8 1 0 8 0 6 0 4 0 2 0 0 - 2 0 - 4 0 - 6 0 - 8 0 [m ] [k N * m ] B e n d i n g M o m e n t S L S : M _ r M r _ r M c _ r M c _ q p M _ q p M r _ q p 0 2 4 6 8 1 0 - 5 0 - 4 0 - 3 0 - 2 0 - 1 0 0 1 0 2 0 3 0 4 0 5 0 6 0 [m ] [k N ] S h e a r F o r c e S L S : V _ r V r _ r V _ q p V r _ q p 0 2 4 6 8 1 0 - 0 .2 5 - 0 .2 - 0 .1 5 - 0 .1 - 0 .0 5 0 0 .0 5 0 .1 0 .1 5 0 .2 [m ] [0 .1 % ] S tr a i n s : A t A c B 0 2 4 6 8 1 0 - 4 0 - 3 0 - 2 0 - 1 0 0 1 0 2 0 3 0 4 0 [m ] [M P a ] S tr e s s e s : A ts A c s B s

2.4.3 Required reinforcement area

Span Span (cm2) Left support (cm2) Right support (cm2) bottom top bottom top bottom top

P1 2,08 0,00 0,51 1,84 0,00 2,64

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0 2 4 6 8 1 0 4 3 2 1 0 1 2 3 4 [m ] [c m 2 ] R e i n fo r c e m e n t A r e a fo r B e n d i n g : A b t A b r A b m i n 0 2 4 6 8 1 0 1 0 8 6 4 2 0 2 4 6 8 1 0 [m ] [c m 2 /m ] R e i n fo r c e m e n t A r e a fo r S h e a r : A s t A s t_ s tr u t A s r A s H a n g

2.4.4 Deflection and cracking

fs_r - short-term due to rare load combination

fs_qp - short-term deflection due to quasi-permanent load combination fl_qp - long-term due to quasi-permanent load combination

f - total deflection f_adm - allowable deflection wk - width of perpendicular cracks

Span fs_r fs_qp fl_qp f f_adm wk

(cm) (cm) (cm) (cm) (cm) (mm)

P1 0,0 0,1 0,1 0,1 2,2 0,00

P2 0,1 0,2 0,2 0,2 2,4 0,00

2.5 Theoretical results - detailed results:

2.5.1 P1 : Span from 0,40 to 5,40 (m)

ULS SLS

Abscissa M max. M min. M max. M min. A bottom A top (m) (kN*m) (kN*m) (kN*m) (kN*m) (cm2) (cm2) 0,40 0,00 -33,72 0,00 -21,86 0,51 1,84 0,74 7,24 -27,60 0,00 -7,93 0,78 1,62 1,28 25,20 -5,52 10,34 0,00 1,12 0,67 1,82 41,28 -0,00 23,42 0,00 1,76 0,14 2,36 47,88 -0,00 30,52 0,00 2,06 0,00 2,90 48,55 -0,00 31,29 0,00 2,08 0,00 3,44 43,78 -0,00 25,70 0,00 1,88 0,00 3,98 29,90 -2,26 14,00 0,00 1,27 0,15 4,52 9,78 -19,96 0,00 -3,19 0,40 0,83 5,06 0,00 -54,57 0,00 -24,66 0,05 2,35 5,40 0,00 -61,10 0,00 -39,58 0,00 2,64 ULS SLS

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(m) (kN) (kN) (mm) 0,40 60,99 44,43 0,0 0,74 55,04 40,08 0,0 1,28 41,96 30,55 0,0 1,82 26,58 19,34 0,0 2,36 9,98 7,27 0,0 2,90 -7,09 -5,15 0,0 3,44 -23,83 -17,34 0,0 3,98 -39,61 -28,82 0,0 4,52 -53,46 -38,92 0,0 5,06 -63,81 -46,47 0,0 5,40 -66,77 -48,65 0,0 2.5.2 P2 : Span from 5,80 to 11,45 (m) ULS SLS

Abscissa M max. M min. M max. M min. A bottom A top (m) (kN*m) (kN*m) (kN*m) (kN*m) (cm2) (cm2) 5,80 0,00 -62,67 0,00 -40,62 0,08 2,70 6,20 0,79 -51,84 0,00 -22,63 0,33 2,31 6,80 13,35 -14,20 1,20 0,00 0,75 0,98 7,40 37,22 -0,00 19,44 0,00 1,58 0,22 8,00 50,09 -0,00 30,80 0,00 2,15 0,00 8,60 53,61 -0,00 34,64 0,00 2,31 0,00 9,20 49,42 -0,00 30,13 0,00 2,12 0,00 9,80 35,90 -0,00 18,61 0,00 1,51 0,28 10,40 12,60 -12,34 1,22 0,00 0,93 0,99 11,00 0,68 -41,42 0,00 -19,53 0,43 1,88 11,45 0,00 -50,76 0,00 -32,85 0,10 2,17 ULS SLS

Abscissa V max. V max. afp

(m) (kN) (kN) (mm) 5,80 68,86 50,18 0,0 6,20 65,22 47,50 0,0 6,80 54,10 39,38 0,0 7,40 39,57 28,79 0,0 8,00 23,26 16,91 0,0 8,60 -7,47 -5,41 0,0 9,20 -23,88 -17,35 0,0 9,80 -38,59 -28,06 0,0 10,40 -50,01 -36,40 0,0 11,00 -54,74 -39,87 0,0 11,45 -46,33 -33,80 0,0 2.6 Reinforcement: 2.6.1 P1 : Span from 0,40 to 5,40 (m) Longitudinal reinforcement: Transversal reinforcement:  main (B500A) stirrups 40 8 l = 1,41 e = 1*0,13 + 19*0,25 (m) pins 40 8 l = 1,41 e = 1*0,13 + 19*0,25 (m) 2.6.2 P2 : Span from 5,80 to 11,45 (m) Longitudinal reinforcement:  bottom (B500C) 3 12 l = 11,89 from 0,04 to 11,71  support (B500C)

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3 12 l = 12,01 from 0,04 to 11,71 Transversal reinforcement:  main (B500A) stirrups 46 8 l = 1,41 e = 1*0,07 + 22*0,25 (m) pins 46 8 l = 1,41 e = 1*0,07 + 22*0,25 (m) 3 Material survey:  Concrete volume = 2,11 (m3)  Formwork = 17,65 (m2)  Steel B500C  Total weight = 63,68 (kG)  Density = 30,11 (kG/m3)  Average diameter = 12,0 (mm)

 Survey according to diameters:

Diameter Length Weight NumberTotal weight

(mm) (m) (kG) (No.) (kG) 12 11,89 10,56 3 31,67 12 12,01 10,67 3 32,01  Steel B500A  Total weight = 48,02 (kG)  Density = 22,70 (kG/m3)  Average diameter = 8,0 (mm)

(mm) (m) (kG) (No.) (kG)

8 1,41 0,56 86 48,02

The user can erase the reinforcement provided by the program in the Beam-reinforcement tab, in order to define by himself a solution and see the capacity of the beam with that reinforcement.

After deleting the reinforcement the program will display the window below where we can see that the capacity of the element is zero.

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The user can define the reinforcement by selecting the Typical reinforcement command from the Reinforcement menu.

This way the user will have to go through six windows and provide information regarding stirrup diameter and distribution as well as main reinforcement parameters.

In the next windows we have indicated a possible reinforcement for the beam.

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In the Beam-reinforcement window we can see the reinforcement defined in the previous windows.

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When we switch to the Beam-diagrams tab the program will

automatically perform the calculation in order to provide results according to the new reinforcement. Every time we changed the reinforcement provided by the program, it will ask as before calculation if we wish to calculate the element with the modified reinforcement or the program will delete all the reinforcement and will propose again a solution. In our case we want to see the capacity of the beam with the reinforcement proposed by us, so we will choose YES.

When the calculation is finished we will see again the window with calculation status.

Below we can see the calculation note for the beam with the reinforcement proposed by us.

1 Level:

 Name :

 Reference level :

--- Maximum cracking : 0,30 (mm)

 Exposure : X0

 Concrete creep coefficient : _= 2,75

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 Concrete age (loading moment) : 28 (days)

 Concrete age : 50 (years)

 Fire resistance class : no requirements

2 Beam: Beam36...37 Number: 1

2.1 Material properties:

Bi-linear stress distribution [3.1.7(2)]

Density : 2501,36 (kG/m3)

 Longitudinal reinforcement: : B500C fyk = 500,00 (MPa)

Ductility class : C

2.2 Geometry:

(m) (m) (m)

P1 Span 0,40 5,00 0,40

(m) (m) (m)

P2 Span 0,40 5,65 0,30

2.3 Calculation options:

 Regulation of combinations : EN 1990:2002

 Precast beam : no

 Cover : bottom c = 2,5 (cm)

: side c1= 2,5 (cm)

: top c2= 2,5 (cm)

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 Coefficient 2 =0.50 : long-term or cyclic load

 Method of shear calculations : strut inclination

2.4 Calculation results:

The deflection L/500 (7.4.1(5)) is not verified

The "Freeze Reinforcement" option is switched on. The distribution of reinforcing

bars has not been modified. 2.4.1 Internal forces in ULS

(kN*m) (kN*m) (kN*m) (kN*m) (kN) (kN) P1 48,55 -2,26 -33,72 -61,10 60,99 -66,77 P2 53,61 -0,00 -62,67 -50,76 68,86 -46,33 0 2 4 6 8 1 0 1 5 0 1 0 0 5 0 0 - 5 0 - 1 0 0 - 1 5 0 [m ] [k N * m ] B e n d i n g M o m e n t U L S : M M r M t M c 0 2 4 6 8 1 0 - 4 0 0 - 3 0 0 - 2 0 0 - 1 0 0 0 1 0 0 2 0 0 3 0 0 4 0 0 [m ] [k N ] S h e a r F o r c e U L S : V V r V c ( s t i r r u p s ) V c ( to ta l ) 2.4.2 Internal forces in SLS

(kN*m) (kN*m) (kN*m) (kN*m) (kN) (kN)

P1 31,29 0,00 -21,86 -39,58 44,43 -48,65

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0 2 4 6 8 1 0 1 5 0 1 0 0 5 0 0 - 5 0 - 1 0 0 - 1 5 0 [m ] [k N * m ] B e n d i n g M o m e n t S L S : M _ r M r _ r M c _ r M c _ q p M _ q p M r _ q p 0 2 4 6 8 1 0 - 5 0 - 4 0 - 3 0 - 2 0 - 1 0 0 1 0 2 0 3 0 4 0 5 0 6 0 [m ] [k N ] S h e a r F o r c e S L S : V _ r V r _ r V _ q p V r _ q p 0 2 4 6 8 1 0 - 0 .2 5 - 0 .2 - 0 .1 5 - 0 .1 - 0 .0 5 0 0 .0 5 0 .1 0 .1 5 0 .2 [m ] [0 .1 % ] S tr a i n s : A t A c B 0 2 4 6 8 1 0 - 4 0 - 3 0 - 2 0 - 1 0 0 1 0 2 0 3 0 4 0 [m ] [M P a ] S tr e s s e s : A ts A c s B s

2.4.3 Required reinforcement area

Span Span (cm2) Left support (cm2) Right support (cm2) bottom top bottom top bottom top

P1 2,08 0,00 0,51 1,84 0,00 2,64

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0 2 4 6 8 1 0 8 6 4 2 0 2 4 6 8 [m ] [c m 2 ] R e i n fo r c e m e n t A r e a fo r B e n d i n g : A b t A b r A b m i n 0 2 4 6 8 1 0 2 0 1 5 1 0 5 0 5 1 0 1 5 2 0 [m ] [c m 2 /m ] R e i n fo r c e m e n t A r e a fo r S h e a r : A s t A s r A s H a n g

2.4.4 Deflection and cracking

fs_r - short-term due to rare load combination

fs_qp - short-term deflection due to quasi-permanent load combination fl_qp - long-term due to quasi-permanent load combination

f - total deflection f_adm - allowable deflection wk - width of perpendicular cracks

Span fs_r fs_qp fl_qp f f_adm wk

(cm) (cm) (cm) (cm) (cm) (mm)

P1 0,0 0,1 0,1 0,1 2,2 0,00

P2 0,1 0,1 0,1 0,1 2,4 0,00

2.5 Theoretical results - detailed results:

2.5.1 P1 : Span from 0,40 to 5,40 (m)

ULS SLS

Abscissa M max. M min. M max. M min. A bottom A top (m) (kN*m) (kN*m) (kN*m) (kN*m) (cm2) (cm2) 0,40 0,00 -33,72 0,00 -21,86 0,51 1,84 0,74 7,24 -27,60 0,00 -7,93 0,78 1,62 1,28 25,20 -5,52 10,34 0,00 1,12 0,67 1,82 41,28 -0,00 23,42 0,00 1,76 0,14 2,36 47,88 -0,00 30,52 0,00 2,06 0,00 2,90 48,55 -0,00 31,29 0,00 2,08 0,00 3,44 43,78 -0,00 25,70 0,00 1,88 0,00 3,98 29,90 -2,26 14,00 0,00 1,27 0,15 4,52 9,78 -19,96 0,00 -3,19 0,40 0,83 5,06 0,00 -54,57 0,00 -24,66 0,05 2,35 5,40 0,00 -61,10 0,00 -39,58 0,00 2,64 ULS SLS

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(m) (kN) (kN) (mm) 0,40 60,99 44,43 0,0 0,74 55,04 40,08 0,0 1,28 41,96 30,55 0,0 1,82 26,58 19,34 0,0 2,36 9,98 7,27 0,0 2,90 -7,09 -5,15 0,0 3,44 -23,83 -17,34 0,0 3,98 -39,61 -28,82 0,0 4,52 -53,46 -38,92 0,0 5,06 -63,81 -46,47 0,0 5,40 -66,77 -48,65 0,0 2.5.2 P2 : Span from 5,80 to 11,45 (m) ULS SLS

Abscissa M max. M min. M max. M min. A bottom A top (m) (kN*m) (kN*m) (kN*m) (kN*m) (cm2) (cm2) 5,80 0,00 -62,67 0,00 -40,62 0,08 2,70 6,20 0,79 -51,84 0,00 -22,63 0,33 2,31 6,80 13,35 -14,20 1,20 0,00 0,75 0,98 7,40 37,22 -0,00 19,44 0,00 1,58 0,22 8,00 50,09 -0,00 30,80 0,00 2,15 0,00 8,60 53,61 -0,00 34,64 0,00 2,31 0,00 9,20 49,42 -0,00 30,13 0,00 2,12 0,00 9,80 35,90 -0,00 18,61 0,00 1,51 0,28 10,40 12,60 -12,34 1,22 0,00 0,93 0,99 11,00 0,68 -41,42 0,00 -19,53 0,43 1,88 11,45 0,00 -50,76 0,00 -32,85 0,10 2,17 ULS SLS

Abscissa V max. V max. afp

(m) (kN) (kN) (mm) 5,80 68,86 50,18 0,0 6,20 65,22 47,50 0,0 6,80 54,10 39,38 0,0 7,40 39,57 28,79 0,0 8,00 23,26 16,91 0,0 8,60 -7,47 -5,41 0,0 9,20 -23,88 -17,35 0,0 9,80 -38,59 -28,06 0,0 10,40 -50,01 -36,40 0,0 11,00 -54,74 -39,87 0,0 11,45 -46,33 -33,80 0,0 2.6 Reinforcement: 2.6.1 P1 : Span from 0,40 to 5,40 (m) Longitudinal reinforcement: Transversal reinforcement:  main (B500A) stirrups 41 10 l = 1,70 e = 1*0,00 + 10*0,10 + 20*0,15 + 10*0,10 (m) pins 41 10 l = 1,70 e = 1*0,00 + 10*0,10 + 20*0,15 + 10*0,10 (m) 2.6.2 P2 : Span from 5,80 to 11,45 (m) Longitudinal reinforcement:  bottom (B500C) 3 16 l = 12,03 from 11,70 to 0,03  support (B500C)

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3 16 l = 12,06 from 0,03 to 11,73 Transversal reinforcement:  main (B500A) stirrups 46 10 l = 1,70 e = 1*0,00 + 10*0,10 + 25*0,15 + 10*0,10 (m) pins 46 10 l = 1,70 e = 1*0,00 + 10*0,10 + 25*0,15 + 10*0,10 (m) 3 Material survey:  Concrete volume = 2,11 (m3)  Formwork = 17,65 (m2)  Steel B500C  Total weight = 114,08 (kG)  Density = 53,94 (kG/m3)  Average diameter = 16,0 (mm)

(mm) (m) (kG) (No.) (kG) 16 12,03 18,99 3 56,97 16 12,06 19,04 3 57,12  Steel B500A  Total weight = 91,36 (kG)  Density = 43,20 (kG/m3)  Average diameter = 10,0 (mm)

(mm) (m) (kG) (No.) (kG)

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RC Column Design - Provided reinforcement

In order to calculate the provided reinforcement the user has to select the element that will be calculated and access the RC Column Design command from the Design menu/Provided reinforcement for RC elements.

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First we will start with Calculation options. This command will open a window with four tabs:

- Longitudinal reinforcement - Transversal reinforcement

In these tabs we will provide information about material quality (concrete and steel), cover and deflection options. Also we can provide information necessary to perform a fire resistance check.

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When the calculation is done the program will display a window with information regarding events during calculation.

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In the next tab, Column-Reinforcement, we can see the

reinforcement bars provided by the program. We can even change the reinforcement and recalculate the element.

When the calculation is finished we can switch to the Column-Interaction N-M tab and see the graphic results.

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In the last tab called Column-note we can see a full calculation note for the current calculated element. This calculation note can also be

generated by selecting the Calculation note command from the Results menu. This calculation note can be save as a *.rtf file and can be attached to the project documentation.

Here it is the calculation note provided for our column:

1 Level:

 Name :

 Reference level : 0,00 (m)

 Concrete creep coefficient : p = 2,77

 Environment class : X0

2 Column: Column2 Number: 1

Unit weight : 2501,36 (kG/m3)

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 Longitudinal reinforcement: : B500B fyk = 500,00 (MPa)

Ductility class : B

2.2 Geometry: 2.2.1 Rectangular 40,0 x 40,0 (cm) 2.2.2 Height: L = 3,80 (m) 2.2.3 Slab thickness = 0,15 (m) 2.2.4 Beam height = 0,60 (m) 2.2.5 Cover = 3,5 (cm) 2.3 Calculation options:

 Precast column : no

 Pre-design : no

 Slenderness taken into account : yes

 Compression : with bending

 Ties : to slab

 More than 50 % loads applied: after 90 day

 Fire resistance class : No requirements

2.4 Loads:

Case Nature Group _f N My(s) My(i) Mz(s) Mz(i) (kN) (kN*m) (kN*m) (kN*m) (kN*m) DL1 dead load 2 1,35 505,05 0,38 -0,32 -14,52 7,05 DL2 dead load 2 1,35 110,64 0,07 -0,07 -4,99 2,44 DL3 dead load 2 1,35 177,63 0,14 -0,11 -9,81 4,83 LL1 live load 2 1,50 132,97 0,09 -0,08 -6,60 3,23 SN1 snow 2 1,50 23,28 -0,00 -0,01 -0,09 0,02 SEI_X7 seismic 2 1,00 -24,76 -10,24 -11,63 -0,67 1,55 SEI_Y8 seismic 2 1,00 40,26 1,73 1,29 -3,06 6,14 SEI_Z8 seismic 2 1,00 -83,67 -0,89 -0,45 3,53 -1,82 SPE_NEW10 seismic 2 1,00 -37,78 -9,98 -11,35 -0,53 2,85 SPE_NEW11 seismic 2 1,00 -61,94 -10,80 -12,07 1,31 -0,84 SPE_NEW12 seismic 2 1,00 -11,74 -10,44 -11,89 -0,81 0,25 SPE_NEW13 seismic 2 1,00 12,42 -9,63 -11,17 -2,64 3,94 SPE_NEW14 seismic 2 1,00 7,73 -1,88 -2,38 -2,20 6,06 SPE_NEW15 seismic 2 1,00 -72,79 -4,61 -4,77 3,92 -6,22 SPE_NEW16 seismic 2 1,00 -22,59 -4,25 -4,59 1,80 -5,13 SPE_NEW17 seismic 2 1,00 57,94 -1,52 -2,20 -4,32 7,15 SPE_NEW18 seismic 2 1,00 -79,02 -3,25 -3,43 2,41 0,49 SPE_NEW19 seismic 2 1,00 -103,18 -4,07 -4,14 4,25 -3,20 SPE_NEW20 seismic 2 1,00 64,16 -2,88 -3,55 -2,81 0,44 SPE_NEW21 seismic 2 1,00 88,32 -2,06 -2,83 -4,65 4,13 f - load factor 2.5 Calculation results:

Seismic dispositions: No requirements!

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2.5.1 ULS Analysis

Design combination: 1.35DL1+1.35DL2+1.35DL3+1.50LL1+0.75SN1 (A) Internal forces: Nsd = 1287,90 (kN) Msdy = 0,92 (kN*m) Msdz = -49,53 (kN*m) Design forces: Upper node N = 1287,90 (kN) N*etotz = 25,76 (kN*m) N*etoty= -60,80 (kN*m) Eccentricity: ez (My/N) ey (Mz/N) static eEd: 0,1 (cm) -3,8 (cm)

Not intended ea: 0,0 (cm) 0,9 (cm)

Initial e0: 0,1 (cm) 3,8 (cm) Minimal emin: 2,0 (cm) 2,0 (cm) total etot: 2,0 (cm) -4,7 (cm) 2.5.1.1. Detailed analysis-Direction Y: 2.5.1.1.1 Slenderness analysis Non-sway structure L (m) Lo (m)  lim 3,50 3,50 30,31 57,97 Short column 2.5.1.1.2 Buckling analysis M2 = 0,92 (kN*m) M1 = -0,80 (kN*m)

Case: Cross-section at the column end (Upper node), Slenderness not taken into account M0 = 0,92 (kN*m)

ea = 0,0 (cm)

Ma = N*ea = 0,00 (kN*m) MEdmin = 25,76 (kN*m)

M0Ed = max(MEdmin,M0 + Ma) = 25,76 (kN*m) 2.5.1.2. Detailed analysis-Direction Z: 2.5.1.2.1 Slenderness analysis Non-sway structure L (m) Lo (m)  lim 3,50 3,50 30,31 49,36 Short column 2.5.1.2.2 Buckling analysis M2 = 24,18 (kN*m) M1 = -49,53 (kN*m)

Case: Cross-section at the column end (Upper node), Slenderness not taken into account M0 = -49,53 (kN*m) ea = *lo/2 = 0,9 (cm)  = h * m = 0,01  = 0,01 h = 1,00 m = (0,5(1+1/m))^0.5 = 1,00 m = 1,00 Ma = N*ea = 11,27 (kN*m) MEdmin = 25,76 (kN*m)

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2.5.2 Reinforcement:

Real (provided) area Asr = 4,71 (cm2)

Ratio: = 0,29 %

2.6 Reinforcement:

Main bars (B500B):

 6 10 l = 3,77 (m)

Transversal reinforcement: (B500A):

stirrups: 28 8 l = 1,42 (m) 28 8 l = 0,50 (m) pins 28 8 l = 1,42 (m) 28 8 l = 0,50 (m) 3 Material survey:  Concrete volume = 0,51 (m3)  Formwork = 5,12 (m2)  Steel B500B  Total weight = 13,93 (kG)  Density = 27,21 (kG/m3)  Average diameter = 10,0 (mm)  Reinforcement survey:

Diameter Length Weight Number Total weight

(m) (kG) (No.) (kG) 10 3,77 2,32 6 13,93  Steel B500A  Total weight = 21,23 (kG)  Density = 41,47 (kG/m3)  Average diameter = 8,0 (mm)  Reinforcement survey:

(m) (kG) (No.) (kG)

8 0,50 0,20 28 5,58

8 1,42 0,56 28 15,66

The user can erase the reinforcement provided by the program in the Column-reinforcement tab, in order to define by himself a solution and see the capacity of the column with that reinforcement.

After deleting the reinforcement the program will display the window below where we can see that the capacity of the element is zero.

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The user can define the reinforcement by selecting the Typical reinforcement command from the Reinforcement menu.

This way the user will have to go through three windows and provide information regarding stirrup diameter and distribution as well as main reinforcement parameters.

In the next windows we have indicated a possible reinforcement for the column.

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When we switch to the Column-Interaction N-M tab the program will automatically perform the calculation in order to provide results according to the new reinforcement. Every time we changed the reinforcement provided by the program, it will ask as before calculation if we wish to calculate the element with the modified reinforcement or the program will delete all the reinforcement and will propose again a solution. In our case we want to see the capacity of the beam with the reinforcement proposed by us, so we will choose YES.

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Below we can see the calculation note for the beam with the reinforcement proposed by us.

1 Level:

 Name :

 Reference level : 0,00 (m)

 Concrete creep coefficient : p = 2,77

 Environment class : X0

2 Column: Column2 Number: 1

Unit weight : 2501,36 (kG/m3)

 Longitudinal reinforcement: : B500B fyk = 500,00 (MPa)

Ductility class : B

2.2 Geometry: 2.2.1 Rectangular 40,0 x 40,0 (cm) 2.2.2 Height: L = 3,80 (m) 2.2.3 Slab thickness = 0,15 (m) 2.2.4 Beam height = 0,60 (m) 2.2.5 Cover = 3,5 (cm) 2.3 Calculation options:

 Precast column : no

 Pre-design : no

 Slenderness taken into account : yes

 Compression : with bending

 Ties : to slab

 More than 50 % loads applied: after 90 day

 Fire resistance class : No requirements

2.4 Loads:

Case Nature Group _f N My(s) My(i) Mz(s) Mz(i) (kN) (kN*m) (kN*m) (kN*m) (kN*m) DL1 dead load 2 1,35 505,05 0,38 -0,32 -14,52 7,05 DL2 dead load 2 1,35 110,64 0,07 -0,07 -4,99 2,44 DL3 dead load 2 1,35 177,63 0,14 -0,11 -9,81 4,83

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LL1 live load 2 1,50 132,97 0,09 -0,08 -6,60 3,23 SN1 snow 2 1,50 23,28 -0,00 -0,01 -0,09 0,02 SEI_X7 seismic 2 1,00 -24,76 -10,24 -11,63 -0,67 1,55 SEI_Y8 seismic 2 1,00 40,26 1,73 1,29 -3,06 6,14 SEI_Z8 seismic 2 1,00 -83,67 -0,89 -0,45 3,53 -1,82 SPE_NEW10 seismic 2 1,00 -37,78 -9,98 -11,35 -0,53 2,85 SPE_NEW11 seismic 2 1,00 -61,94 -10,80 -12,07 1,31 -0,84 SPE_NEW12 seismic 2 1,00 -11,74 -10,44 -11,89 -0,81 0,25 SPE_NEW13 seismic 2 1,00 12,42 -9,63 -11,17 -2,64 3,94 SPE_NEW14 seismic 2 1,00 7,73 -1,88 -2,38 -2,20 6,06 SPE_NEW15 seismic 2 1,00 -72,79 -4,61 -4,77 3,92 -6,22 SPE_NEW16 seismic 2 1,00 -22,59 -4,25 -4,59 1,80 -5,13 SPE_NEW17 seismic 2 1,00 57,94 -1,52 -2,20 -4,32 7,15 SPE_NEW18 seismic 2 1,00 -79,02 -3,25 -3,43 2,41 0,49 SPE_NEW19 seismic 2 1,00 -103,18 -4,07 -4,14 4,25 -3,20 SPE_NEW20 seismic 2 1,00 64,16 -2,88 -3,55 -2,81 0,44 SPE_NEW21 seismic 2 1,00 88,32 -2,06 -2,83 -4,65 4,13 f - load factor 2.5 Calculation results:

The "Freeze Reinforcement" option is switched on. The distribution of reinforcing bars has not been modified.

Seismic dispositions: No requirements!

The system of bars does not satisfy the cover requirements.

Safety factors Rd/Ed = 1,84 > 1.0

2.5.1 ULS Analysis

Design combination: 1.35DL1+1.35DL2+1.35DL3+1.50LL1+0.75SN1 (A) Internal forces: Nsd = 1287,90 (kN) Msdy = 0,92 (kN*m) Msdz = -49,53 (kN*m) Design forces: Upper node N = 1287,90 (kN) N*etotz = 25,76 (kN*m) N*etoty= -60,80 (kN*m) Eccentricity: ez (My/N) ey (Mz/N) static eEd: 0,1 (cm) -3,8 (cm)

Not intended ea: 0,0 (cm) 0,9 (cm)

Initial e0: 0,1 (cm) 3,8 (cm) Minimal emin: 2,0 (cm) 2,0 (cm) total etot: 2,0 (cm) -4,7 (cm) 2.5.1.1. Detailed analysis-Direction Y: 2.5.1.1.1 Slenderness analysis Non-sway structure L (m) Lo (m)  lim 3,50 3,50 30,31 66,64 Short column 2.5.1.1.2 Buckling analysis M2 = 0,92 (kN*m) M1 = -0,80 (kN*m)

Case: Cross-section at the column end (Upper node), Slenderness not taken into account M0 = 0,92 (kN*m)

ea = 0,0 (cm)

Ma = N*ea = 0,00 (kN*m) MEdmin = 25,76 (kN*m)

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M0Ed = max(MEdmin,M0 + Ma) = 25,76 (kN*m) 2.5.1.2. Detailed analysis-Direction Z: 2.5.1.2.1 Slenderness analysis Non-sway structure L (m) Lo (m)  lim 3,50 3,50 30,31 56,74 Short column 2.5.1.2.2 Buckling analysis M2 = 24,18 (kN*m) M1 = -49,53 (kN*m)

Case: Cross-section at the column end (Upper node), Slenderness not taken into account M0 = -49,53 (kN*m) ea = *lo/2 = 0,9 (cm)  = h * m = 0,01  = 0,01 h = 1,00 m = (0,5(1+1/m))^0.5 = 1,00 m = 1,00 Ma = N*ea = 11,27 (kN*m) MEdmin = 25,76 (kN*m)

M0Ed = max(MEdmin,M0 + Ma) = -60,80 (kN*m)

2.5.2 Reinforcement:

Real (provided) area Asr = 16,08 (cm2)

Ratio: = 1,01 %

2.6 Reinforcement:

Main bars (B500B):

 8 16 l = 4,44 (m)

Transversal reinforcement: (B500A):

stirrups: 30 10 l = 1,50 (m) pins 30 10 l = 1,50 (m) 3 Material survey:  Concrete volume = 0,51 (m3)  Formwork = 5,12 (m2)  Steel B500B  Total weight = 56,08 (kG)  Density = 109,53 (kG/m3)  Average diameter = 16,0 (mm)  Reinforcement survey:

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(m) (kG) (No.) (kG) 16 4,44 7,01 8 56,08  Steel B500A  Total weight = 27,80 (kG)  Density = 54,31 (kG/m3)  Average diameter = 10,0 (mm)  Reinforcement survey:

(m) (kG) (No.) (kG)

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RC Slab Design - Provided reinforcement

In order to calculate the provided reinforcement the user has to select the element that will be calculated and access the RC Slab Design command from the Design menu/Provided reinforcement for RC elements.

The display has changed and on screen we have some new

windows where we can see a map of the element with results that can be selected from the window on the right.

First we will start with Calculation options. This command will open a window with five tabs:

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- Wire fabrics

- Reinf. for punching

In these tabs we will provide information about material quality (concrete and steel.

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When the calculation is done we can switch to the window that will display the reinforcement proposed by the program. Below it is presented the path to that window.

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In the windows below are presented the upper and lower reinforcement for the slab that the program is providing.

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Same as for the other elements the user can generate a calculation note by selecting the Calculation note command from the Results menu:

Here it is the calculation note provided for our slab:

1. Slab: Plate51 - Panel no. 51

1.1. Reinforcement:

 Type : RC Floor1

 Main reinforcement direction : 0°

Main reinforcement grade : B500A; Characteristic strength = 500,00 MPa

 Ductility class : A

 Bar diameters bottom d1 = 1,0 (cm) d2 = 1,0 (cm)

top d1 = 1,0 (cm) d2 = 1,0 (cm)

 Cover bottom c1 = 1,5 (cm)

top c2 = 1,5 (cm)

 Cover deviations Cdev = 1,0(cm), Cdur = 0,0(cm)

1.2. Concrete

Class : C25/30; Characteristic strength = 25,00 MPa

Rectangular stress distribution [3.1.7(3)]

 Density : 2501,36 (kG/m3)

 Concrete creep coefficient : 1,81

1.3. Hypothesis

 Method of reinforcement area calculations : analytical

 Allowable cracking width

- upper layer : 0,40 (mm)

- lower layer : 0,40 (mm)

 Allowable deflection : 3,0 (cm)

 Verification of punching : yes

 Exposure

- upper layer : X0

- lower layer : X0

 Calculation type : simple bending

1.4. Slab geometry

Thickness 0,15 (m)

Contour:

edge beginning end length

x1 y1 x2 y2 (m)

1 -0,00 -5,40 5,40 -5,40 5,40

2 5,40 -5,40 5,40 -0,00 5,40

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4 0,00 0,00 -0,00 -5,40 5,40

Support:

n° Name dimensions coordinates edge

(m) x y 10 point 0,40 / 0,40 -0,00 -5,40 — 10 linear 5,40 / 0,30 -0,00 -2,70 — 10 linear 0,30 / 5,40 2,70 -5,40 — 10 point 0,40 / 0,40 -0,00 -5,40 — 12 point 0,40 / 0,40 0,00 0,00 — 12 linear 0,30 / 5,40 2,70 -0,00 — 12 point 0,40 / 0,40 0,00 0,00 — 14 linear 5,40 / 0,30 5,40 -2,70 — 16 point 0,40 / 0,40 5,40 -5,40 — 16 point 0,40 / 0,40 5,40 -5,40 — * - head present 1.5. Calculation results:

1.5.1. Maximum moments + reinforcement for bending

Ax(+) Ax(-) Ay(+) Ay(-)

Provided reinforcement (cm2/m):

0,00 0,00 0,00 0,00

Modified required reinforcement (cm2/m):

3,14 0,00 3,14 0,00

Original required reinforcement (cm2/m):

0,00 0,00 0,00 0,00

Coordinates (m):

10,80;-28,20 10,80;-28,20 10,80;-28,20 10,80;-28,20

1.5.2. Maximum moments + reinforcement for bending

Ax(+) Ax(-) Ay(+) Ay(-)

Symbol: required area/provided area

Ax(+) (cm2/m) 3,14/0,00 3,14/0,00 3,14/0,00 3,14/0,00 Ax(-) (cm2/m) 0,00/0,00 0,00/0,00 0,00/0,00 0,00/0,00 Ay(+) (cm2/m) 3,14/0,00 3,14/0,00 3,14/0,00 3,14/0,00 Ay(-) (cm2/m) 0,00/0,00 0,00/0,00 0,00/0,00 0,00/0,00 SLS Mxx (kN*m/m) 0,00 0,00 0,00 0,00 Myy (kN*m/m) 0,00 0,00 0,00 0,00 Mxy (kN*m/m) 0,00 0,00 0,00 0,00 Nxx (kN/m) 0,00 0,00 0,00 0,00 Nyy (kN/m) 0,00 0,00 0,00 0,00 Nxy (kN/m) 0,00 0,00 0,00 0,00 ULS Mxx (kN*m/m) 0,00 0,00 0,00 0,00 Myy (kN*m/m) 0,00 0,00 0,00 0,00 Mxy (kN*m/m) 0,00 0,00 0,00 0,00 Nxx (kN/m) 0,00 0,00 0,00 0,00

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Nyy (kN/m) 0,00 0,00 0,00 0,00 Nxy (kN/m) 0,00 0,00 0,00 0,00 Coordinates (m) 10,80;-28,20 10,80;-28,20 10,80;-28,20 10,80;-28,20 Coordinates* (m) 0,00;0,00;0,00 0,00;0,00;0,00 0,00;0,00;0,00 0,00;0,00;0,00

* - Coordinates in the structure global coordinate system

1.5.4. Deflection |f(+)| = 0,0 (cm) <= fdop(+) = 3,0 (cm) |f(-)| = 2,1 (cm) <= fdop(-) = 3,0 (cm) 1.5.5. Cracking upper layer ax = 0,24 (mm) <= adop = 0,40 (mm) ay = 0,26 (mm) <= adop = 0,40 (mm) lower layer ax = 0,00 (mm) <= adop = 0,40 (mm) ay = 0,00 (mm) <= adop = 0,40 (mm) 2. Loads:

Case Type List Value

1 self-weight 19to26 28 30to56 75to81 104to112 131to137

160to168 187to193 216to224 243to249 272to281 PZ Negative

2 (FE) uniform 48to56 104to112 160to168 216to224

PZ=-1,50(kN/m2)

2 (FE) uniform 272to281 PZ=-1,50(kN/m2)

PZ=-3,00(kN/m2)

PZ=-2,00(kN/m2)

Combination/Component Definition ULS/10 7*1.00+(8+9)*0.30 ULS/11 7*1.00+8*-0.30+9*0.30 ULS/12 7*1.00+(8+9)*-0.30 ULS/13 7*1.00+8*0.30+9*-0.30 ULS/14 (7+9)*0.30+8*1.00 ULS/15 (7+9)*0.30+8*-1.00 ULS/16 7*0.30+8*-1.00+9*-0.30 ULS/17 7*0.30+8*1.00+9*-0.30 ULS/18 (7+8)*0.30+9*1.00 ULS/19 7*0.30+8*-0.30+9*1.00 ULS/20 7*0.30+8*-0.30+9*-1.00 ULS/21 (7+8)*0.30+9*-1.00 3. Results - detailing List of solutions: Reinforcement: bars

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Diameter / Weight (kG) 1 - 301,03 2 - 305,33 3 - 315,33 4 - 317,42 5 - 327,41 6 - 331,71 7 - 360,65 8 - 380,64 9 - 466,16

Results for the solution no. 1 Reinforcement zones Bottom reinforcement

Name coordinates Provided reinforcement At

Ar x1 y1 x2 y2 (mm) / (cm) (cm2/m) (cm2/m) 1/1- Ax Main -0,00 -5,40 5,40 -0,00 8,0 / 15,0 3,14 < 3,35 1/2- Ay Perpendicular -0,00 -5,40 5,40 -0,00 8,0 / 15,0 3,14 < 3,35 Top reinforcement

Name coordinates Provided reinforcement At

Ar x1 y1 x2 y2  (mm) / (cm) (cm2/m) (cm2/m) 1/1+(1/4+) Ax Main 3,78 -4,59 5,40 -0,00 8,0 / 12,0 4,07 < 4,19 1/2+(1/4+) Ax Main -0,00 -5,40 1,08 -0,00 8,0 / 12,0 3,14 < 4,19 1/3+(1/4+) Ax Main 1,08 -5,40 5,40 -3,78 8,0 / 12,0 3,14 < 4,19 1/4+ Ax Main 1,08 -1,62 5,40 -0,00 8,0 / 12,0 3,14 < 4,19 1/5+(1/8+) Ay Perpendicular 1,08 -5,40 5,40 -3,78 8,0 / 13,0 3,75 < 3,87 1/6+(1/8+) Ay Perpendicular 1,08 -1,62 5,40 -0,00 8,0 / 13,0 3,14 < 3,87 1/7+(1/8+) Ay Perpendicular -0,00 -5,40 1,08 -0,00 8,0 / 13,0 3,33 < 3,87 1/8+ Ay Perpendicular 3,78 -3,78 5,40 -1,62 8,0 / 13,0 3,14 < 3,87 4. Material survey  Concrete volume = 4,37 (m3)  Formwork = 29,16 (m2)  Slab circumference = 21,60 (m)  Area of openings = 0,00 (m2)

(84)

 Steel B500A

 Total weight = 314,59 (kG)

 Density = 71,92 (kG/m3)

 Average diameter = 8,0 (mm)

Diameter Length Number:

(m)

8 1,53 18

8 2,07 58

(85)

RC Foundation Design - Provided reinforcement

In order to calculate the provided reinforcement the user has to select the element that will be calculated and access the RC Foundation Design command from the Design menu/Provided reinforcement for RC elements.

In the case of the foundations the program will provide additional calculation regarding ground under foundation: tensions, settlement, etc.

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In this window the user will provide the starting point of the

dimensioning process by indicating the dimensions of the foundation. Some of these dimensions can be fixed so that the program will not change them during calculation.

Before calculating the provided reinforcement we have to adjust four sets of calculation parameters:

(87)

- RC Element/Soil

- Analysis menu/Geotechnical options... - Analysis menu/Calculation options... - Analysis menu/Reinforcement pattern...

In the first set mentioned, Soil, the user will provide information regarding the soil under the foundation: layers, levels of the foundation relative to zero level, admissible stress for ground and water level.

The second set named Geotechnical options allows the user to choose what kind of verification to perform: settlement, rotation, sliding, allowable eccentricity.

(88)

The calculation option set contain four tabs: - General

- Concrete

- Longitudinal reinforcement - Transversal reinforcement

(89)

(90)

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The final set of options to be adjusted is the Reinforcement pattern. Because the program will propose a real solution for reinforcement, we will have to provide some rules that the program will follow when it will draw the reinforcement.

(92)

(93)

(94)

When the calculation is finished we can switch to the Foundation-Results tab and see the graphic results.

In the next tab, Foundation-Reinforcement, we can see the reinforcement bars provided by the program. We can even change the reinforcement and recalculate the element.

(95)

In the last tab called Foundation-note we can see a full calculation note for the current calculated element. This calculation note can also be generated by selecting the Calculation note command from the Results menu. This calculation note can be save as a *.rtf file and can be attached to the project documentation.

Here it is the calculation note provided for our foundation:

1 Spread footing: Foundation3 Number: 1

1.1 Basic data

1.1.1 Assumptions

 Geotechnic calculations according to : EN 1997-1:2008

 Concrete calculations according to : EN 1992-1-1:2004 AC:2008

 Shape selection : without limits

(96)

A = 1,90 (m) a = 0,60 (m) B = 1,90 (m) b = 0,60 (m) h1 = 0,40 (m) ex = 0,00 (m) h2 = 0,60 (m) ey = 0,00 (m) h4 = 0,05 (m) a' = 40,0 (cm) b' = 40,0 (cm) cnom1 = 6,0 (cm) cnom2 = 6,0 (cm)

Cover deviations: Cdev = 1,0(cm), Cdur = 0,0(cm)

1.1.3 Materials

 Concrete : C20/25; Characteristic strength = 20,00 MPa

Unit weight = 2501,36 (kG/m3) Bi-linear stress distribution [3.1.7(2)]

 Longitudinal reinforcement : type B500C Characteristic strength

= 500,00 MPa

Ductility class: C

 Transversal reinforcement : type B500C Characteristic strength

= 500,00 MPa

1.1.4 Loads:

Foundation loads:

Case Nature Group N Fx Fy Mx My

(kN) (kN) (kN) (kN*m) (kN*m) DL1 dead load 3 505,05 -0,20 6,16 -7,05 -0,32 DL2 dead load 3 110,64 -0,04 2,12 -2,44 -0,07 DL3 dead load 3 177,63 -0,07 4,18 -4,83 -0,11 LL1 live load 3 132,97 -0,05 2,81 -3,23 -0,08 SN1 snow 3 23,28 -0,00 0,03 -0,02 -0,01 SEI_X7 seismic 3 -24,76 6,26 0,62 -1,55 11,62 SEI_Y8 seismic 3 40,26 -0,73 2,61 -6,14 -1,19 SEI_Z8 seismic 3 -83,67 0,26 -1,53 1,82 0,30 SPE_NEW10 seismic 3 -37,78 6,12 0,95 -2,85 11,35 SPE_NEW11 seismic 3 -61,94 6,55 -0,62 0,84 12,07 SPE_NEW12 seismic 3 -11,74 6,40 0,30 -0,25 11,89 SPE_NEW13 seismic 3 12,42 5,96 1,86 -3,94 11,17 SPE_NEW14 seismic 3 7,73 1,22 2,34 -6,06 2,38 SPE_NEW15 seismic 3 -72,79 2,68 -2,88 6,22 4,77 SPE_NEW16 seismic 3 -22,59 2,53 -1,96 5,13 4,59 SPE_NEW17 seismic 3 57,94 1,07 3,25 -7,15 2,20

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SPE_NEW18 seismic 3 -79,02 1,91 -0,56 -0,49 3,43 SPE_NEW19 seismic 3 -103,18 2,35 -2,12 3,20 4,14 SPE_NEW20 seismic 3 64,16 1,84 0,93 -0,44 3,55 SPE_NEW21 seismic 3 88,32 1,40 2,50 -4,13 2,83 Backfill loads: Case Nature Q1 (kN/m2) 1.1.5 Combination list 1/ ULS A1 : 1.35DL1+1.35DL2+1.35DL3+1.50LL1+0.75SN1 2/ ULS A1 : 1.35DL1+1.35DL2+1.35DL3+1.50LL1 3/ ULS A1 : 1.35DL1+1.35DL2+1.35DL3 4/ ULS A1 : 1.00DL1+1.00DL2+1.00DL3+1.50LL1+0.75SN1 5/ ULS A1 : 1.00DL1+1.00DL2+1.00DL3+1.50LL1 6/ ULS A1 : 1.00DL1+1.00DL2+1.00DL3 7/ ULS A1 : 1.35DL1+1.35DL2+1.35DL3+1.05LL1+1.50SN1 8/ ULS A1 : 1.35DL1+1.35DL2+1.35DL3+1.50SN1 9/ ULS A1 : 1.00DL1+1.00DL2+1.00DL3+1.05LL1+1.50SN1 10/ ULS A1 : 1.00DL1+1.00DL2+1.00DL3+1.50SN1 11/ ULS A2 : 1.00DL1+1.00DL2+1.00DL3+1.30LL1+0.91SN1 12/ ULS A2 : 1.00DL1+1.00DL2+1.00DL3+1.30LL1 13/ ULS A2 : 1.00DL1+1.00DL2+1.00DL3 14/ ULS A2 : 1.00DL1+1.00DL2+1.00DL3+0.91LL1+1.30SN1 15/ ULS A2 : 1.00DL1+1.00DL2+1.00DL3+1.30SN1 16/ SLS : 1.00DL1+1.00DL2+1.00DL3+1.00LL1 17/ SLS : 1.00DL1+1.00DL2+1.00DL3 18/ SLS : 1.00DL1+1.00DL2+1.00DL3+1.00SN1 19/ SLS : 1.00DL1+1.00DL2+1.00DL3+1.00LL1+1.00SN1 20/* ULS : 1.35DL1+1.35DL2+1.35DL3+1.50LL1+0.75SN1 21/* ULS : 1.35DL1+1.35DL2+1.35DL3+1.50LL1 22/* ULS : 1.35DL1+1.35DL2+1.35DL3 23/* ULS : 1.00DL1+1.00DL2+1.00DL3+1.50LL1+0.75SN1 24/* ULS : 1.00DL1+1.00DL2+1.00DL3+1.50LL1 25/* ULS : 1.00DL1+1.00DL2+1.00DL3 26/* ULS : 1.35DL1+1.35DL2+1.35DL3+1.05LL1+1.50SN1 27/* ULS : 1.35DL1+1.35DL2+1.35DL3+1.50SN1 28/* ULS : 1.00DL1+1.00DL2+1.00DL3+1.05LL1+1.50SN1 29/* ULS : 1.00DL1+1.00DL2+1.00DL3+1.50SN1 30/* SLS : 1.00DL1+1.00DL2+1.00DL3+1.00LL1+0.50SN1 31/* SLS : 1.00DL1+1.00DL2+1.00DL3+1.00LL1 32/* SLS : 1.00DL1+1.00DL2+1.00DL3 33/* SLS : 1.00DL1+1.00DL2+1.00DL3+0.70LL1+1.00SN1 34/* SLS : 1.00DL1+1.00DL2+1.00DL3+1.00SN1 35/* SLS : 1.00DL1+1.00DL2+1.00DL3+0.50LL1 36/* SLS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1+0.20SN1 37/* SLS : 1.00DL1+1.00DL2+1.00DL3+0.20SN1 38/* SLS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1 39/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1+1.00SEI_X7 40/* ALS : 1.00DL1+1.00DL2+1.00DL3+1.00SEI_X7 41/* ALS : 1.00DL1+1.00DL2+1.00DL3 42/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1+1.00SEI_Y8 43/* ALS : 1.00DL1+1.00DL2+1.00DL3+1.00SEI_Y8 44/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1+1.00SEI_Z8 45/* ALS : 1.00DL1+1.00DL2+1.00DL3+1.00SEI_Z8 46/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1+1.00SPE_NEW10 47/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1+1.00SPE_NEW11 48/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1+1.00SPE_NEW12 49/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1+1.00SPE_NEW13 50/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1+1.00SPE_NEW14 51/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1+1.00SPE_NEW15 52/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1+1.00SPE_NEW16 53/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1+1.00SPE_NEW17 54/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1+1.00SPE_NEW18 55/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1+1.00SPE_NEW19 56/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1+1.00SPE_NEW20 57/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1+1.00SPE_NEW21 58/* ALS : 1.00DL1+1.00DL2+1.00DL3+1.00SPE_NEW10

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59/* ALS : 1.00DL1+1.00DL2+1.00DL3+1.00SPE_NEW11 60/* ALS : 1.00DL1+1.00DL2+1.00DL3+1.00SPE_NEW12 61/* ALS : 1.00DL1+1.00DL2+1.00DL3+1.00SPE_NEW13 62/* ALS : 1.00DL1+1.00DL2+1.00DL3+1.00SPE_NEW14 63/* ALS : 1.00DL1+1.00DL2+1.00DL3+1.00SPE_NEW15 64/* ALS : 1.00DL1+1.00DL2+1.00DL3+1.00SPE_NEW16 65/* ALS : 1.00DL1+1.00DL2+1.00DL3+1.00SPE_NEW17 66/* ALS : 1.00DL1+1.00DL2+1.00DL3+1.00SPE_NEW18 67/* ALS : 1.00DL1+1.00DL2+1.00DL3+1.00SPE_NEW19 68/* ALS : 1.00DL1+1.00DL2+1.00DL3+1.00SPE_NEW20 69/* ALS : 1.00DL1+1.00DL2+1.00DL3+1.00SPE_NEW21 70/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1-1.00SEI_X7 71/* ALS : 1.00DL1+1.00DL2+1.00DL3-1.00SEI_X7 72/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1-1.00SEI_Y8 73/* ALS : 1.00DL1+1.00DL2+1.00DL3-1.00SEI_Y8 74/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1-1.00SEI_Z8 75/* ALS : 1.00DL1+1.00DL2+1.00DL3-1.00SEI_Z8 76/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1-1.00SPE_NEW10 77/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1-1.00SPE_NEW11 78/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1-1.00SPE_NEW12 79/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1-1.00SPE_NEW13 80/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1-1.00SPE_NEW14 81/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1-1.00SPE_NEW15 82/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1-1.00SPE_NEW16 83/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1-1.00SPE_NEW17 84/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1-1.00SPE_NEW18 85/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1-1.00SPE_NEW19 86/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1-1.00SPE_NEW20 87/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1-1.00SPE_NEW21 88/* ALS : 1.00DL1+1.00DL2+1.00DL3-1.00SPE_NEW10 89/* ALS : 1.00DL1+1.00DL2+1.00DL3-1.00SPE_NEW11 90/* ALS : 1.00DL1+1.00DL2+1.00DL3-1.00SPE_NEW12 91/* ALS : 1.00DL1+1.00DL2+1.00DL3-1.00SPE_NEW13 92/* ALS : 1.00DL1+1.00DL2+1.00DL3-1.00SPE_NEW14 93/* ALS : 1.00DL1+1.00DL2+1.00DL3-1.00SPE_NEW15 94/* ALS : 1.00DL1+1.00DL2+1.00DL3-1.00SPE_NEW16 95/* ALS : 1.00DL1+1.00DL2+1.00DL3-1.00SPE_NEW17 96/* ALS : 1.00DL1+1.00DL2+1.00DL3-1.00SPE_NEW18 97/* ALS : 1.00DL1+1.00DL2+1.00DL3-1.00SPE_NEW19 98/* ALS : 1.00DL1+1.00DL2+1.00DL3-1.00SPE_NEW20 99/* ALS : 1.00DL1+1.00DL2+1.00DL3-1.00SPE_NEW21 100/* ALS : 1.00DL1+1.00DL2+1.00DL3+0.30LL1 1.2 Geotechnical design 1.2.1 Assumptions

 Cohesion reduction coefficient: 0,00

 Smooth precast foundation 6.5.3(10)

 Sliding with soil pressure considered: for X and Y directions

 Design approach: 1 A1 + M1 + R1 ' = 1,00 c' = 1,00 cu = 1,00 qu = 1,00  = 1,00 R,v = 1,00 R,h = 1,00 A2 + M2 + R1

(99)

' = 1,25 c' = 1,25 cu = 1,40 qu = 1,40  = 1,00 R,v = 1,00 R,h = 1,00 1.2.2 Soil: Soil level: N1 = 0,00 (m)

Column pier level: Na = 0,00 (m)

Minimum reference level: Nf = -0,50 (m)

Clay

• Soil level: 0.00 (m)

• Unit weight:2243.38 (kG/m3)

• Unit weight of solid: 2753.23 (kG/m3)

• Internal friction angle: 25.0 (Deg)

• Cohesion: 0.06 (MPa)

1.2.3 Limit states

Stress calculations

Soil type under foundation: not layered

Design combination ULS A1 :

1.35DL1+1.35DL2+1.35DL3+1.50LL1+0.75SN1

Load factors: 1.35 * Foundation weight

1.35 * Soil weight

Calculation results: On the foundation level

Weight of foundation and soil over it: Gr = 112,89 (kN) Design load:

Nr = 1400,79 (kN) Mx = -45,25 (kN*m) My = -1,30 (kN*m)

Load eccentricity:

eB = -0,00 (m) eL = 0,03 (m)

Equivalent foundation dimensions:

B' = B - 2|eB| = 1,90 (m)

L' = L - 2|eL| = 1,90 (m)

Foundation depth: Dmin = 1,00 (m)

Allowable stress calculation method: Semi-empirical - stress limit

qu = 0.50 (MPa)

ple* = 0,48 (MPa)

De = Dmin - d = 1,00 (m)

kp = 0,91

(100)

qu = kp * (ple*) + q'0 = 0,45 (MPa)

Stress in soil: qref = 0.43 (MPa)

Safety factor: qlim / qref = 1.061 > 1

Uplift

Uplift in ULS

1.35DL1+1.35DL2+1.35DL3+1.50LL1

Contact area: s = 0,02

slim = 0,33

Sliding

1.35DL1+1.35DL2+1.35DL3+1.50LL1

Weight of foundation and soil over it: Gr = 83,62 (kN)

Design load:

Nr = 1354,06 (kN) Mx = -45,21 (kN*m) My = -1,29 (kN*m)

Equivalent foundation dimensions: A_ = 1,90 (m)B_ = 1,90 (m)

Sliding area: 3,61 (m2)

Foundation/soil friction coefficient: tan(d = 0,30

Cohesion: cu = 0.06 (MPa)

Soil pressure considered:

Hx = -0,49 (kN) Hy = 21,04 (kN)

Ppx = 16,48 (kN) Ppy = -16,48 (kN)

Pax = -2,71 (kN) Pay = 2,71 (kN)

Sliding force value Hd = 7,27 (kN)

Value of force preventing foundation sliding:

- On the foundation level: Rd = 405,38 (kN)

Stabilility for sliding: 55.73 > 1

Average settlement

Soil type under foundation: not layered

Design combination SLS :

1.00DL1+1.00DL2+1.00DL3+1.00LL1+1.00SN1

Weight of foundation and soil over it: Gr = 83,62 (kN) Average stress caused by design load: q = 0,29 (MPa) Thickness of the actively settling soil: z = 4,75 (m) Stress on the level z:

- Additional: zd = 0,02 (MPa)

- Caused by soil weight: z = 0,13 (MPa)

Settlement:

- Original s' = 0,5 (cm)

- Secondary s'' = 0,0 (cm)

(101)

Safety factor: 10.25 > 1

Settlement difference

Design combination SLS :

1.00DL1+1.00DL2+1.00DL3+1.00LL1+1.00SN1

Settlement difference: S = 0,0 (cm) < Sadm = 5,0 (cm)

Safety factor: 1626 > 1

Rotation

About OX axis

1.35DL1+1.35DL2+1.35DL3+1.50LL1

Design load:

Nr = 1354,06 (kN) Mx = -45,21 (kN*m) My = -1,29 (kN*m)

Stability moment: _{Mstab = 1286,36 (kN*m)}

Rotation moment: _{Mrenv = 45,21 (kN*m)}

Stability for rotation: 28.46 > 1

About OY axis

Design combination: ULS A1 :

1.35DL1+1.35DL2+1.35DL3+1.50LL1

Design load:

Nr = 1354,06 (kN) Mx = -45,21 (kN*m) My = -1,29 (kN*m)

Stability moment: _{Mstab = 1286,36 (kN*m)}

Rotation moment: _{Mrenv = 1,29 (kN*m)}

Stability for rotation: 997.8 > 1

1.3 RC design

1.3.1 Assumptions

 Exposure : X0

1.3.2 Analysis of punching and shear Punching

Design combination ULS : 1.35DL1+1.35DL2+1.35DL3+1.50LL1+0.75SN1

(102)

Nr = 1400,79 (kN) Mx = -45,25 (kN*m) My = -1,30 (kN*m)

Length of critical circumference: 4,47 (m)

Punching force: 756,25 (kN)

Section effective height heff = 0,33 (m)

Reinforcement ratio:  = 0.22 %

Shear stress: 0,67 (MPa)

Admissible shear stress: 0,74 (MPa)

Safety factor: 1.104 > 1 1.3.3 Required reinforcement Spread footing: bottom: ULS : 1.35DL1+1.35DL2+1.35DL3+1.50LL1+0.75SN1 My = 186,21 (kN*m) Asx = 7,09 (cm2/m) ULS : 1.35DL1+1.35DL2+1.35DL3+1.50LL1+0.75SN1 Mx = 201,02 (kN*m) Asy = 7,68 (cm2/m) As min = 5,12 (cm2/m) top: A'sx= 0,00 (cm2/m) A'sy= 0,00 (cm2/m) As min = 0,00 (cm2/m) Column pier:

Longitudinal reinforcement A = 7,20 (cm2) A min. = 7,20 (cm2)

A = 2 * (Asx + Asy) Asx = 1,35 (cm2) Asy = 2,25 (cm2) 1.3.4 Provided reinforcement Spread footing: Bottom: Along X axis: 18 B500C 10 l = 2,00 (m) e = 1*-0,89 Along Y axis: 13 B500C 12 l = 2,13 (m) e = 0,14 Pier Longitudinal reinforcement Along X axis: 2 B500C 16 l = 2,70 (m) e = 1*-0,18 + 1*0,35 Along Y axis: 2 B500C 16 l = 2,76 (m) e = 1*-0,22 Transversal reinforcement 5 B500C 16 l = 2,07 (m) e = 1*0,24

(103)

2 Material survey:  Concrete volume = 1,66 (m3)  Formwork = 4,48 (m2)  Steel B500C  Total weight = 80,34 (kG)  Density = 48,40 (kG/m3)  Average diameter = 12,2 (mm)

Diameter Length Number:

(m) 10 2,00 18 12 2,13 13 16 2,07 5 16 2,70 2 16 2,76 2