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
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.
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
2.1 Material properties:
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
2.2.2 Span Position L.supp. L R.supp.
(m) (m) (m)
P2 Span 0,40 5,65 0,30
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 : EN 1990:2002
Calculations according to : EN 1992-1-1:2004 AC:2008
Seismic dispositions : No requirements
Precast beam : no
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
3. Areq [cm2] SLS 2 11.45 0.22 0.16 0.74
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
2.4.2 Internal forces in SLS
Span Mt max. Mt min. Ml Mr Ql Qr
(kN*m) (kN*m) (kN*m) (kN*m) (kN) (kN)
P1 31,29 0,00 -21,86 -39,58 44,43 -48,65
P2 34,64 0,00 -40,62 -32,85 50,18 -33,80
0 2 4 6 8 1 0
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
P2 2,31 0,00 0,08 2,70 0,10 2,17
0 2 4 6 8 1 0
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
Abscissa V max. V max. afp
(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:
support (B500C)
3 12 l = 12,01 from 0,04 to 11,71
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)
Survey according to diameters:
Diameter Length Weight NumberTotal weight
(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.
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 : 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
2.1 Material properties:
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
2.2.2 Span Position L.supp. L R.supp.
(m) (m) (m)
P2 Span 0,40 5,65 0,30
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 : 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
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. 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
2.4.2 Internal forces in SLS
Span Mt max. Mt min. Ml Mr Ql Qr
(kN*m) (kN*m) (kN*m) (kN*m) (kN) (kN)
P1 31,29 0,00 -21,86 -39,58 44,43 -48,65
P2 34,64 0,00 -40,62 -32,85 50,18 -33,80
0 2 4 6 8 1 0
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
P2 2,31 0,00 0,08 2,70 0,10 2,17
0 2 4 6 8 1 0
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
Abscissa V max. V max. afp
(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:
support (B500C)
3 16 l = 12,06 from 0,03 to 11,73
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 Length Weight NumberTotal weight
(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)
Survey according to diameters:
Diameter Length Weight NumberTotal weight
(mm) (m) (kG) (No.) (kG)
10 1,70 1,05 87 91,36