RC Design using Eurocode in Robot Structural.doc

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/m2 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.

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.

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.

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.

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.

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

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: - members that will be calculated

- 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.

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:

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

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.

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.

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

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.

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

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. In the windows below we can see the parameters for our example:

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.

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.

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.

When the calculation is finished we can switch to the BeamDiagrams 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.

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 Exposure

   

Concrete creep coefficient cement class Concrete age (loading moment) Concrete age  Structure class  Fire resistance class

2

Beam: Beam36...37 2.1

Material properties:

: : --: 0,30 (mm) : X0 :  = 2,75 :N : 28 (days) : 50 (years) : S1 : no requirements

Number: 1

 Concrete

:

C25/30 fck = 25,00 (MPa) Bi-linear stress distribution [3.1.7(2)] 2501,36 (kG/m3) 20,0 (mm) B500C fyk = 500,00 (MPa) Horizontal branch of the stress-strain

Density : Aggregate size :  Longitudinal reinforcement: : diagram  Transversal reinforcement:

2.2

Ductility class : C B500A fyk = 500,00 (MPa)

:

Geometry: 2.2.1

Span

Position

L.supp. (m) 5,00

L (m) 0,40

R.supp. (m)

L.supp. (m) 5,65

L (m) 0,30

R.supp. (m)

P1 Span 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 2.2.2

Span

Position

P2 Span 0,40 Span length: Lo = 6,00 (m) Section from 0,00 to 5,65 (m) 30,0 x 60,0 (cm) without left slab without right slab

2.3     

Calculation options:

Regulation of combinations Calculations according to Seismic dispositions Precast beam Cover

 Cover deviations

: EN 1990:2002 : EN 1992-1-1:2004 AC:2008 : No requirements : no : bottom c = 2,5 (cm) : side c1= 2,5 (cm) : top c2= 2,5 (cm) : Cdev = 1,0(cm), Cdur = 0,0(cm)

 Coefficient 2 =0.50  Method of shear calculations

: long-term or cyclic load : strut inclination

2.4

Calculation results: The deflection L/500 (7.4.1(5)) is not verified No. Type 1. M [kN*m] 2. M [kN*m] 3. Areq [cm2]

State ULS ALS SLS

Span 2 2 2

x(m) Value 11.45 -50.76 11.45 -58.39 11.45 0.22

Capacity -39.55 -45.64 0.16

n* 0.78 0.78 0.74

n* - Safety factor

2.4.1

Internal forces in ULS

Span

Mt max. (kN*m) 48,55 53,61

P1 P2

Mt min. (kN*m) -2,26 -0,00

Ml (kN*m) -33,72 -62,67

Mr (kN*m) -61,10 -50,76

Ql (kN) 60,99 68,86

Qr (kN) -66,77 -46,33

-8 0 [k N * m ] -6 0 -4 0 -2 0 0 20 40 60 [m ] 80

0 B e n d in g M o m e n t U L S :

2

4

M

Mr

Mt

6

8

10

6

8

10

Mc

200 [k N ] 150 100 50 0 -5 0 -1 0 0 -1 5 0 [m ] -2 0 0

0 S h e a r F o rc e U L S :

2 V

Vr

2.4.2 Span P1 P2

V c (s tir r u p s )

4 V c ( to ta l )

Internal forces in SLS Mt max. (kN*m) 31,29 34,64

Mt min. (kN*m) 0,00 0,00

Ml (kN*m) -21,86 -40,62

Mr (kN*m) -39,58 -32,85

Ql (kN) 44,43 50,18

Qr (kN) -48,65 -33,80

-8 0 [k N * m ] -6 0 -4 0 -2 0 0 20 40 60 [m ] 80

0 B e n d in g M o m e n t S L S :

2 M _r

M r_ r

M c_r

4 M c_qp

6

8

10

4

6

8

10

4

6

8

10

4

6

8

10

M _qp

M r_ q p

60 50

[k N ]

40 30 20 10 0 -1 0 -2 0 -3 0 -4 0 -5 0

[m ] 0 S h e a r F o rc e S L S :

2 V_r

V r_ r

V_qp

V r_ q p

0 .2 [0 .1 % ] 0 .1 5 0 .1 0 .0 5 0 - 0 .0 5 - 0 .1 - 0 .1 5 - 0 .2 - 0 .2 5

[m ] 0 S tr a in s :

2 At

Ac

B

40 [M P a ] 30 20 10 0 -1 0 -2 0 -3 0 [m ] -4 0

0 S tr e s s e s :

2 A ts

A cs

2.4.3 Span P1 P2

Bs

Required reinforcement area Span (cm2) bottom top 2,08 0,00 2,31 0,00

Left support (cm2) bottom top 0,51 1,84 0,08 2,70

Right support (cm2) bottom top 0,00 2,64 0,10 2,17

4 [c m 2 ] 3 2 1 0 1 2 3 [m ] 4

0 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 :

2 Abt

Abr

4

6

8

10

4

6

8

10

A b m in

10 [c m 2 /m ] 8 6 4 2 0 2 4 6 8 10

[m ] 0 R e i n fo r c e m e n t A r e a fo r S h e a r :

2 A st

A s t_ s tr u t

A sr

AsH ang

2.4.4 Deflection and cracking fs_r fs_qp fl_qp f f_adm wk Span P1 P2

2.5

- short-term due to rare load combination - short-term deflection due to quasi-permanent load combination - long-term due to quasi-permanent load combination - total deflection - allowable deflection

- width of perpendicular cracks fs_r (cm) 0,0 0,1

fs_qp (cm) 0,1 0,2

fl_qp (cm) 0,1 0,2

f (cm) 0,1 0,2

Theoretical results - detailed results: 2.5.1

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

Abscissa (m) 0,40 0,74 1,28 1,82 2,36 2,90 3,44 3,98 4,52 5,06 5,40

ULS M max. (kN*m) 0,00 7,24 25,20 41,28 47,88 48,55 43,78 29,90 9,78 0,00 0,00

M min. (kN*m) -33,72 -27,60 -5,52 -0,00 -0,00 -0,00 -0,00 -2,26 -19,96 -54,57 -61,10

SLS M max. (kN*m) 0,00 0,00 10,34 23,42 30,52 31,29 25,70 14,00 0,00 0,00 0,00

Abscissa

ULS V max.

SLS V max.

afp

M min. (kN*m) -21,86 -7,93 0,00 0,00 0,00 0,00 0,00 0,00 -3,19 -24,66 -39,58

A bottom (cm2) 0,51 0,78 1,12 1,76 2,06 2,08 1,88 1,27 0,40 0,05 0,00

A top (cm2) 1,84 1,62 0,67 0,14 0,00 0,00 0,00 0,15 0,83 2,35 2,64

f_adm (cm) 2,2 2,4

wk (mm) 0,00 0,00

2.6

(m) 0,40 0,74 1,28 1,82 2,36 2,90 3,44 3,98 4,52 5,06 5,40

(kN) 60,99 55,04 41,96 26,58 9,98 -7,09 -23,83 -39,61 -53,46 -63,81 -66,77

(kN) 44,43 40,08 30,55 19,34 7,27 -5,15 -17,34 -28,82 -38,92 -46,47 -48,65

(mm) 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0,0

2.5.2

P2 : Span from 5,80 to 11,45 (m)

Abscissa (m) 5,80 6,20 6,80 7,40 8,00 8,60 9,20 9,80 10,40 11,00 11,45

ULS M max. (kN*m) 0,00 0,79 13,35 37,22 50,09 53,61 49,42 35,90 12,60 0,68 0,00

M min. (kN*m) -62,67 -51,84 -14,20 -0,00 -0,00 -0,00 -0,00 -0,00 -12,34 -41,42 -50,76

SLS M max. (kN*m) 0,00 0,00 1,20 19,44 30,80 34,64 30,13 18,61 1,22 0,00 0,00

Abscissa (m) 5,80 6,20 6,80 7,40 8,00 8,60 9,20 9,80 10,40 11,00 11,45

ULS V max. (kN) 68,86 65,22 54,10 39,57 23,26 -7,47 -23,88 -38,59 -50,01 -54,74 -46,33

SLS V max. (kN) 50,18 47,50 39,38 28,79 16,91 -5,41 -17,35 -28,06 -36,40 -39,87 -33,80

afp (mm) 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0,0

M min. (kN*m) -40,62 -22,63 0,00 0,00 0,00 0,00 0,00 0,00 0,00 -19,53 -32,85

A bottom (cm2) 0,08 0,33 0,75 1,58 2,15 2,31 2,12 1,51 0,93 0,43 0,10

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

support (B500C)

from 0,04

to

11,71

A top (cm2) 2,70 2,31 0,98 0,22 0,00 0,00 0,00 0,28 0,99 1,88 2,17

3

12

l = 12,01

from 0,04

to

11,71

Transversal reinforcement:  main (B500A)

3

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)

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 (mm) 12 12

Length (m) 11,89 12,01

Weight (kG) 10,56 10,67

NumberTotal weight (No.) (kG) 3 31,67 3 32,01

 Steel B500A  Total weight = 48,02 (kG)  Density = 22,70 (kG/m3)  Average diameter = 8,0 (mm)  Survey according to diameters: Diameter Length (mm) (m) 8 1,41

Weight (kG) 0,56

NumberTotal weight (No.) (kG) 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.

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.

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

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 Exposure

 Concrete creep coefficient  cement class

: : --: 0,30 (mm) : X0 :  = 2,75 :N

 Concrete age (loading moment)  Concrete age  Structure class  Fire resistance class

2

: 28 (days) : 50 (years) : S1 : no requirements

Beam: Beam36...37 2.1

Number: 1

Material properties:

 Concrete

:

C25/30 fck = 25,00 (MPa) Bi-linear stress distribution [3.1.7(2)] 2501,36 (kG/m3) 20,0 (mm) B500C fyk = 500,00 (MPa) Horizontal branch of the stress-strain

Density : Aggregate size :  Longitudinal reinforcement: : diagram  Transversal reinforcement:

2.2

Ductility class : C B500A fyk = 500,00 (MPa)

:

Geometry: 2.2.1

Span

Position

L.supp. (m) 5,00

L (m) 0,40

R.supp. (m)

L.supp. (m) 5,65

L (m) 0,30

R.supp. (m)

P1 Span 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 2.2.2

Span

Position

P2 Span 0,40 Span length: Lo = 6,00 (m) Section from 0,00 to 5,65 (m) 30,0 x 60,0 (cm) without left slab without right slab

2.3     

Calculation options:

Regulation of combinations Calculations according to Seismic dispositions Precast beam Cover

 Cover deviations

: EN 1990:2002 : EN 1992-1-1:2004 AC:2008 : No requirements : no : bottom c = 2,5 (cm) : side c1= 2,5 (cm) : top c2= 2,5 (cm) : Cdev = 1,0(cm), Cdur = 0,0(cm)

 Coefficient 2 =0.50  Method of shear calculations

2.4

: long-term or cyclic load : strut inclination

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 Span

Mt max. (kN*m) 48,55 53,61

P1 P2

Mt min. (kN*m) -2,26 -0,00

Ml (kN*m) -33,72 -62,67

Mr (kN*m) -61,10 -50,76

Ql (kN) 60,99 68,86

Qr (kN) -66,77 -46,33

-1 5 0 [k N * m ] -1 0 0

-5 0

0

50

100 [m ] 150

0 B e n d in g M o m e n t U L S :

2

4

M

Mr

Mt

6

8

10

6

8

10

Mc

400 [k N ] 300 200 100 0 -1 0 0 -2 0 0 -3 0 0 [m ] -4 0 0

0 S h e a r F o rc e U L S :

2 V

Vr

2.4.2 Span P1 P2

V c (s tir r u p s )

4 V c ( to ta l )

Internal forces in SLS Mt max. (kN*m) 31,29 34,64

Mt min. (kN*m) 0,00 0,00

Ml (kN*m) -21,86 -40,62

Mr (kN*m) -39,58 -32,85

Ql (kN) 44,43 50,18

Qr (kN) -48,65 -33,80

-1 5 0 [k N * m ] -1 0 0

-5 0

0

50

100 [m ] 150

0 B e n d in g M o m e n t S L S :

2 M _r

M r_ r

M c_r

4 M c_qp

6

8

10

4

6

8

10

4

6

8

10

4

6

8

10

M _qp

M r_ q p

60 50

[k N ]

40 30 20 10 0 -1 0 -2 0 -3 0 -4 0 -5 0

[m ] 0 S h e a r F o rc e S L S :

2 V_r

V r_ r

V_qp

V r_ q p

0 .2 [0 .1 % ] 0 .1 5 0 .1 0 .0 5 0 - 0 .0 5 - 0 .1 - 0 .1 5 - 0 .2 - 0 .2 5

[m ] 0 S tr a in s :

2 At

Ac

B

40 [M P a ] 30 20 10 0 -1 0 -2 0 -3 0 [m ] -4 0

0 S tr e s s e s :

2 A ts

A cs

2.4.3 Span P1 P2

Bs

Required reinforcement area Span (cm2) bottom top 2,08 0,00 2,31 0,00

Left support (cm2) bottom top 0,51 1,84 0,08 2,70

Right support (cm2) bottom top 0,00 2,64 0,10 2,17

8 [c m 2 ] 6 4 2 0 2 4 6 [m ] 8

0 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 :

2 Abt

Abr

4

6

8

10

4

6

8

10

A b m in

20 [c m 2 /m ] 15 10 5 0 5 10 15 [m ] 20

0 R e i n fo r c e m e n t A r e a fo r S h e a r :

2 A st

A sr

A sH ang

2.4.4 Deflection and cracking fs_r fs_qp fl_qp f f_adm wk Span P1 P2

2.5

- short-term due to rare load combination - short-term deflection due to quasi-permanent load combination - long-term due to quasi-permanent load combination - total deflection - allowable deflection

- width of perpendicular cracks fs_r (cm) 0,0 0,1

fs_qp (cm) 0,1 0,1

fl_qp (cm) 0,1 0,1

f (cm) 0,1 0,1

Theoretical results - detailed results: 2.5.1

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

Abscissa (m) 0,40 0,74 1,28 1,82 2,36 2,90 3,44 3,98 4,52 5,06 5,40

ULS M max. (kN*m) 0,00 7,24 25,20 41,28 47,88 48,55 43,78 29,90 9,78 0,00 0,00

M min. (kN*m) -33,72 -27,60 -5,52 -0,00 -0,00 -0,00 -0,00 -2,26 -19,96 -54,57 -61,10

SLS M max. (kN*m) 0,00 0,00 10,34 23,42 30,52 31,29 25,70 14,00 0,00 0,00 0,00

Abscissa

ULS V max.

SLS V max.

afp

M min. (kN*m) -21,86 -7,93 0,00 0,00 0,00 0,00 0,00 0,00 -3,19 -24,66 -39,58

A bottom (cm2) 0,51 0,78 1,12 1,76 2,06 2,08 1,88 1,27 0,40 0,05 0,00

A top (cm2) 1,84 1,62 0,67 0,14 0,00 0,00 0,00 0,15 0,83 2,35 2,64

f_adm (cm) 2,2 2,4

wk (mm) 0,00 0,00

2.6

(m) 0,40 0,74 1,28 1,82 2,36 2,90 3,44 3,98 4,52 5,06 5,40

(kN) 60,99 55,04 41,96 26,58 9,98 -7,09 -23,83 -39,61 -53,46 -63,81 -66,77

(kN) 44,43 40,08 30,55 19,34 7,27 -5,15 -17,34 -28,82 -38,92 -46,47 -48,65

(mm) 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0,0

2.5.2

P2 : Span from 5,80 to 11,45 (m)

Abscissa (m) 5,80 6,20 6,80 7,40 8,00 8,60 9,20 9,80 10,40 11,00 11,45

ULS M max. (kN*m) 0,00 0,79 13,35 37,22 50,09 53,61 49,42 35,90 12,60 0,68 0,00

M min. (kN*m) -62,67 -51,84 -14,20 -0,00 -0,00 -0,00 -0,00 -0,00 -12,34 -41,42 -50,76

SLS M max. (kN*m) 0,00 0,00 1,20 19,44 30,80 34,64 30,13 18,61 1,22 0,00 0,00

Abscissa (m) 5,80 6,20 6,80 7,40 8,00 8,60 9,20 9,80 10,40 11,00 11,45

ULS V max. (kN) 68,86 65,22 54,10 39,57 23,26 -7,47 -23,88 -38,59 -50,01 -54,74 -46,33

SLS V max. (kN) 50,18 47,50 39,38 28,79 16,91 -5,41 -17,35 -28,06 -36,40 -39,87 -33,80

afp (mm) 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0,0

M min. (kN*m) -40,62 -22,63 0,00 0,00 0,00 0,00 0,00 0,00 0,00 -19,53 -32,85

A bottom (cm2) 0,08 0,33 0,75 1,58 2,15 2,31 2,12 1,51 0,93 0,43 0,10

A top (cm2) 2,70 2,31 0,98 0,22 0,00 0,00 0,00 0,28 0,99 1,88 2,17

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

support (B500C)

from 11,70

to

0,03

3

16

l = 12,06

from 0,03

to

11,73

Transversal reinforcement:  main (B500A)

3

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)

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)  Survey according to diameters: Diameter (mm) 16 16

Length (m) 12,03 12,06

Weight (kG) 18,99 19,04

NumberTotal weight (No.) (kG) 3 56,97 3 57,12

 Steel B500A  Total weight = 91,36 (kG)  Density = 43,20 (kG/m3)  Average diameter = 10,0 (mm)  Survey according to diameters: Diameter Length (mm) (m) 10 1,70

Weight (kG) 1,05

NumberTotal weight (No.) (kG) 87 91,36

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

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 four tabs: - General - Concrete - 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.

In the windows below we can see the parameters for our example:

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.

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.

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.

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 ColumnInteraction N-M tab and see the graphic results.

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:      

2

Name Reference level Concrete creep coefficient cement class Environment class Structure class

: : 0,00 (m) : p = 2,77 :N : X0 : S1

Column: Column2 2.1

Number: 1

Material properties:

 Concrete Unit weight Aggregate size

: C25/30 : 2501,36 (kG/m3) : 20,0 (mm)

fck = 25,00 (MPa)

 Longitudinal reinforcement: Ductility class  Transversal reinforcement:

2.2

        

fyk = 500,00 (MPa) fyk = 500,00 (MPa)

Geometry: 2.2.1 2.2.2 2.2.3 2.2.4 2.2.5

2.3

: B500B :B : B500A

Rectangular Height: L Slab thickness Beam height Cover

40,0 x 40,0 (cm) = 3,80 (m) = 0,15 (m) = 0,60 (m) = 3,5 (cm)

Calculation options:

Calculations according to : EN 1992-1-1:2004 AC:2008 Seismic dispositions : No requirements 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

DL1 dead load DL2 dead load DL3 dead load LL1 live load SN1 snow SEI_X7 seismic SEI_Y8 seismic SEI_Z8 seismic SPE_NEW10 SPE_NEW11 SPE_NEW12 SPE_NEW13 SPE_NEW14 SPE_NEW15 SPE_NEW16 SPE_NEW17 SPE_NEW18 SPE_NEW19 SPE_NEW20 SPE_NEW21

f - load factor 2.5

Group

f

2 2 2 2 2 2 2 2 seismic seismic seismic seismic seismic seismic seismic seismic seismic seismic seismic seismic

1,35 1,35 1,35 1,50 1,50 1,00 1,00 1,00 2 2 2 2 2 2 2 2 2 2 2 2

N (kN) 505,05 110,64 177,63 132,97 23,28 -24,76 40,26 -83,67 1,00 1,00 1,00 1,00 1,00 1,00 1,00 1,00 1,00 1,00 1,00 1,00

My(s) (kN*m) 0,38 0,07 0,14 0,09 -0,00 -10,24 1,73 -0,89 -37,78 -61,94 -11,74 12,42 7,73 -72,79 -22,59 57,94 -79,02 -103,18 64,16 88,32

Calculation results: Seismic dispositions: No requirements! Safety factors Rd/Ed = 1,57 > 1.0

My(i) (kN*m) -0,32 -0,07 -0,11 -0,08 -0,01 -11,63 1,29 -0,45 -9,98 -10,80 -10,44 -9,63 -1,88 -4,61 -4,25 -1,52 -3,25 -4,07 -2,88 -2,06

Mz(s) (kN*m) -14,52 -4,99 -9,81 -6,60 -0,09 -0,67 -3,06 3,53 -11,35 -12,07 -11,89 -11,17 -2,38 -4,77 -4,59 -2,20 -3,43 -4,14 -3,55 -2,83

Mz(i) (kN*m) 7,05 2,44 4,83 3,23 0,02 1,55 6,14 -1,82 -0,53 1,31 -0,81 -2,64 -2,20 3,92 1,80 -4,32 2,41 4,25 -2,81 -4,65

2,85 -0,84 0,25 3,94 6,06 -6,22 -5,13 7,15 0,49 -3,20 0,44 4,13

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: static Not intended Initial Minimal total

ez (My/N) eEd: 0,1 (cm) ea: 0,0 (cm) e0: 0,1 (cm) emin: 2,0 (cm) etot: 2,0 (cm)

ey (Mz/N) -3,8 (cm) 0,9 (cm) 3,8 (cm) 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) 3,50

Lo (m) 3,50

 30,31

lim 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) 3,50

Lo (m) 3,50

 30,31

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

2.5.2

Reinforcement:

Real (provided) area Ratio:

2.6

Asr = 4,71 (cm2) = 0,29 %

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

3

28 8 28 8

l = 1,42 (m) l = 0,50 (m)

Material survey:  Concrete volume  Formwork = 5,12 (m2)

= 0,51 (m3)

 Steel B500B  Total weight = 13,93 (kG)  Density = 27,21 (kG/m3)  Average diameter = 10,0 (mm)  Reinforcement survey: Diameter 10

Length (m) 3,77

Weight (kG) 2,32

Number (No.) 6

Total weight (kG) 13,93

Number (No.) 28 28

Total weight (kG) 5,58 15,66

 Steel B500A  Total weight = 21,23 (kG)  Density = 41,47 (kG/m3)  Average diameter = 8,0 (mm)  Reinforcement survey: Diameter 8 8

Length (m) 0,50 1,42

Weight (kG) 0,20 0,56

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.

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.

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.

Below we can see the calculation note for the beam with the reinforcement proposed by us. 1

Level:      

2

Name Reference level Concrete creep coefficient cement class Environment class Structure class

: : 0,00 (m) : p = 2,77 :N : X0 : S1

Column: Column2 2.1

Number: 1

Material properties:

 Concrete Unit weight Aggregate size  Longitudinal reinforcement: Ductility class  Transversal reinforcement:

2.2

        

fck = 25,00 (MPa) fyk = 500,00 (MPa) fyk = 500,00 (MPa)

Geometry: 2.2.1 2.2.2 2.2.3 2.2.4 2.2.5

2.3

: C25/30 : 2501,36 (kG/m3) : 20,0 (mm) : B500B :B : B500A

Rectangular Height: L Slab thickness Beam height Cover

40,0 x 40,0 (cm) = 3,80 (m) = 0,15 (m) = 0,60 (m) = 3,5 (cm)

Calculation options:

Calculations according to : EN 1992-1-1:2004 AC:2008 Seismic dispositions : No requirements 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

DL1 DL2 DL3

dead load dead load dead load

2 2 2

f

N (kN) 1,35 505,05 1,35 110,64 1,35 177,63

My(s) (kN*m) 0,38 0,07 0,14

My(i) (kN*m) -0,32 -0,07 -0,11

Mz(s) (kN*m) -14,52 -4,99 -9,81

Mz(i) (kN*m) 7,05 2,44 4,83

LL1 live load SN1 snow SEI_X7 seismic SEI_Y8 seismic SEI_Z8 seismic SPE_NEW10 SPE_NEW11 SPE_NEW12 SPE_NEW13 SPE_NEW14 SPE_NEW15 SPE_NEW16 SPE_NEW17 SPE_NEW18 SPE_NEW19 SPE_NEW20 SPE_NEW21

2 2 2 2 2 seismic seismic seismic seismic seismic seismic seismic seismic seismic seismic seismic seismic

1,50 1,50 1,00 1,00 1,00 2 2 2 2 2 2 2 2 2 2 2 2

132,97 23,28 -24,76 40,26 -83,67 1,00 1,00 1,00 1,00 1,00 1,00 1,00 1,00 1,00 1,00 1,00 1,00

0,09 -0,00 -10,24 1,73 -0,89 -37,78 -61,94 -11,74 12,42 7,73 -72,79 -22,59 57,94 -79,02 -103,18 64,16 88,32

-0,08 -0,01 -11,63 1,29 -0,45 -9,98 -10,80 -10,44 -9,63 -1,88 -4,61 -4,25 -1,52 -3,25 -4,07 -2,88 -2,06

-6,60 -0,09 -0,67 -3,06 3,53 -11,35 -12,07 -11,89 -11,17 -2,38 -4,77 -4,59 -2,20 -3,43 -4,14 -3,55 -2,83

3,23 0,02 1,55 6,14 -1,82 -0,53 1,31 -0,81 -2,64 -2,20 3,92 1,80 -4,32 2,41 4,25 -2,81 -4,65

2,85 -0,84 0,25 3,94 6,06 -6,22 -5,13 7,15 0,49 -3,20 0,44 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: static Not intended Initial Minimal total

ez (My/N) eEd: 0,1 (cm) ea: 0,0 (cm) e0: 0,1 (cm) emin: 2,0 (cm) etot: 2,0 (cm)

ey (Mz/N) -3,8 (cm) 0,9 (cm) 3,8 (cm) 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) 3,50

Lo (m) 3,50

 30,31

lim 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)

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) 3,50

Lo (m) 3,50

 30,31

lim 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 Ratio:

2.6

Asr = 16,08 (cm2) = 1,01 %

Reinforcement: Main bars (B500B):  8 16 l = 4,44 (m) Transversal reinforcement: (B500A): stirrups: 30 10 l = 1,50 (m) pins

3

30 10

l = 1,50 (m)

Material survey:  Concrete volume  Formwork = 5,12 (m2)

= 0,51 (m3)

 Steel B500B  Total weight = 56,08 (kG)  Density = 109,53 (kG/m3)  Average diameter = 16,0 (mm)  Reinforcement survey: Diameter

Length

Weight

Number

Total weight

16

(m) 4,44

(kG) 7,01

(No.) 8

(kG) 56,08

Number (No.) 30

Total weight (kG) 27,80

 Steel B500A  Total weight = 27,80 (kG)  Density = 54,31 (kG/m3)  Average diameter = 10,0 (mm)  Reinforcement survey: Diameter 10

Length (m) 1,50

Weight (kG) 0,93

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.

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 - Reinforcing bars

- Wire fabrics - Reinf. for punching In these tabs we will provide information about material quality (concrete and steel. In the windows below we can see the parameters for our example:

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.

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.

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

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

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 Main reinforcement direction Main reinforcement grade  

 

Ductility class Bar diameters



Cover

 Cover deviations

: RC Floor1 : 0° : B500A; Characteristic strength = 500,00 MPa Horizontal branch of the stress-strain diagram :A bottom d1 = 1,0 (cm) d2 = 1,0 (cm) top d1 = 1,0 (cm) d2 = 1,0 (cm) bottom c1 = 1,5 (cm) top c2 = 1,5 (cm) Cdev = 1,0(cm), Cdur = 0,0(cm)

1.2. Concrete Class   

Density Concrete creep coefficient cement class

: C25/30; Characteristic strength = 25,00 MPa Rectangular stress distribution [3.1.7(3)] : 2501,36 (kG/m3) : 1,81 :N

1.3. Hypothesis   

  

 

Calculations according to Method of reinforcement area calculations Allowable cracking width - upper layer - lower layer Allowable deflection Verification of punching Exposure - upper layer - lower layer Calculation type Structure class

: EN 1992-1-1:2004 AC:2008 : analytical : 0,40 (mm) : 0,40 (mm) : 3,0 (cm) : yes : X0 : X0 : simple bending : S4

1.4. Slab geometry Thickness 0,15 (m) Contour: edge 1 2 3

beginning x1 y1 -0,00 -5,40 5,40 -5,40 5,40 -0,00

end x2 5,40 5,40 0,00

y2 -5,40 -0,00 0,00

length (m) 5,40 5,40 5,40

4

0,00

Support: n°

Name

10 10 10 10 12 12 12 14 16 16

point linear linear point point linear point linear point point

0,00

-0,00 dimensions (m) 0,40 / 0,40 5,40 / 0,30 0,30 / 5,40 0,40 / 0,40 0,40 / 0,40 0,30 / 5,40 0,40 / 0,40 5,40 / 0,30 0,40 / 0,40 0,40 / 0,40

-5,40

5,40

coordinates x y -0,00 -5,40 -0,00 -2,70 2,70 -5,40 -0,00 -5,40 0,00 0,00 2,70 -0,00 0,00 0,00 5,40 -2,70 5,40 -5,40 5,40 -5,40

edge — — — — — — — — — —

* - head present

1.5. Calculation results: 1.5.1. Maximum moments + reinforcement for bending Ax(+)

Ax(-)

Ay(+)

Ay(-)

0,00

0,00

0,00

0,00

3,14

0,00

0,00

0,00

0,00

10,80;-28,20

10,80;-28,20

10,80;-

Ax(-)

Ay(+)

Ay(-)

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

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

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

Provided reinforcement (cm2/m): 0,00 Modified required reinforcement (cm2/m): 3,14 Original required reinforcement (cm2/m): 0,00 Coordinates (m): 10,80;-28,20 28,20

1.5.2. Maximum moments + reinforcement for bending Ax(+) Symbol: required area/provided area Ax(+) (cm2/m) 3,14/0,00 Ax(-) (cm2/m) 0,00/0,00 Ay(+) (cm2/m) 3,14/0,00 Ay(-) (cm2/m) 0,00/0,00

SLS Mxx (kN*m/m) Myy (kN*m/m) Mxy (kN*m/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

Nxx (kN/m) Nyy (kN/m) Nxy (kN/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

ULS Mxx (kN*m/m) Myy (kN*m/m) Mxy (kN*m/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

Nxx (kN/m)

0,00

0,00

0,00

0,00

Nyy (kN/m) Nxy (kN/m)

0,00 0,00

0,00 0,00

0,00 0,00

0,00 0,00

Coordinates (m) 28,20 Coordinates* (m)

10,80;-28,20

10,80;-28,20

10,80;-28,20

10,80;-

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)