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RRufino wrote:
Hello,
I'm starting to use beams reinforcement design in Robot, instead of using external excel sheets. I'm following the EN1992-1-1:2004:Ac:2008 code.
By my first approach i´ve this questions:
- is there a way to define more then one diameter to reinforcement bars? Or even combine different diameters in the same section face?
Please mind that the result produced by this module is the required area of reinforcement. The amount of bars is just the secondary information calculated as required reinforcement area / specified reinforcement bar diameter area. This module is not capable of providing reinforcement bar distribution in a cross section or along a beam. This is done in the RC Beam design or RC Column design modules.
- when i´m designing a beam to a group of combinations which have seismic action, for the same point, per example over the columns, they have My+ and My-, but the design moment My that displays on the table is just the absolute biggest one...i think that by this way i'm always missing information for the face that have less moment.
Which table do you refer to? The RC Member design module calculates the necessary amount of reinforcement to withstand all loads from the list of load cases/ combinations you entered. You can see all the values of internal forces in the Forces table. The table displayed on the RC Members Required reinforcement layout shows only the forces from the combination for which the element's design ratio (force/strength) is the highest one. In addition mind to use the signed combinations while designing against seismic loads (Analysis type dialog > Combination sign tab).
- using this code the maximum longitudinal reinforcement should be 0,04xAc, where Ac is concrete area on the transversal section, but it displays 60,20 cm2 to 25x65 cm2 beam, where 0,04 Ac is 65cm2?
I'm not sure if I understand this point correctly. As 60.2 < 65 then I can't see a problem - am I missing something?
- to transversal reinforcement is there a way to see in table the value of reinforcement per meter or reinforcement per meter per leg?
No, such way of displaying results is not available. On the other hand the detailed bar schedule (including amount of steel) is available in the RC Beam design module not mentioning the fact that the distribution of stirrups along a beam is usually not uniform.
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Artur Kosakowski
Hi Artur, thanks by your answers.
Please mind that the result produced by this module is the required area of reinforcement. The amount of bars is just the secondary information calculated as required reinforcement area / specified reinforcement bar diameter area. This module is not capable of providing reinforcement bar distribution in a cross section or along a beam. This is done in the RC Beam design or RC Column design modules.
- ok, but when you say RC Beam design are you refering to RC design -> Provided Reinforcement or are you refering to a design module tha i don´t have? Please see my menu image in attachement.
Which table do you refer to? The RC Member design module calculates the necessary amount of reinforcement to withstand all loads from the list of load cases/ combinations you entered. You can see all the values of internal forces in the Forces table. The table displayed on the RC Members Required reinforcement layout shows only the forces from the combination for which the element's design ratio (force/strength) is the highest one. In addition mind to use the signed combinations while designing against seismic loads (Analysis type dialog > Combination sign tab).
- i'm refering to RC design -> RC members - Required Reinforcement -> Required Member reinforcement. I think this table should show design moment to the highest ratio to bottom reinforcement and to highest ratio to top reinforcement, not the design moment of the highest of highest ratio. Please see attachement.
I'm not sure if I understand this point correctly. As 60.2 < 65 then I can't see a problem - am I missing something?
- Maximum reinforcement column, in my thought, should display 65cm2, not 60,2.
No, such way of displaying results is not available. On the other hand the detailed bar schedule (including amount of steel) is available in the RC Beam design module not mentioning the fact that the distribution of stirrups along a beam is usually not uniform.
- I can see different distribution of stirrups along a beam by setting a number of reinforcement sections (@ Calculation Parameters -> Transversal reinforcement) equal to number of beam calculation points.
- ok, but when you say RC Beam design are you refering to RC design -> Provided Reinforcement or are you refering to a design module tha i don´t have? Please see my menu image in attachement.
What I mean is the Provided reinforcement. Selected beams and columns will be automatically exported from the model to the RC Beam and RC Column design modules respectively (Design > Provided reinforcement of RC Elements
- i'm refering to RC design -> RC members - Required Reinforcement -> Required Member reinforcement. I think this table should show design moment to the highest ratio to bottom reinforcement and to highest ratio to top reinforcement, not the design moment of the highest of highest ratio. Please see attachement.
I think this is not that easy as it looks like at first. Imagine a beam under 5 different set of loads:
1. Mtop = 100 kNm
2. M bottom = - 100kNm
3. Ntension = 100 kN Mtop=80kNm
4. Ntension = 100 kN Mbottom = -80kNm
4. Ncompression = 5000kN (compression reinforcement required)
The amount of reinforcement can be governed by each of these set of forces (values are of course 'some' and entered only to illustrate the situation) and even not directly governed by any of them (e.g. area of top and bottom reinforcement calculated in such a way that they secure beam's capacity for each of the applied set of loads but also in the way that the sum of the top and bottom reinforcement is the smallest possible). The solution is found by checking section capacity with assumed reinforcement rather than calculating theoretical values of top and bottom reinforcement for each of the cases separately and then assuming maximal from obtained values at the top and at the bottom. This is why the forces from the case that produced the highest design ratio are displayed as the design values. More detailed access to the results is of course possible in the required reinforcement modules.
- Maximum reinforcement column, in my thought, should display 65cm2, not 60,2.
Thank you for notifying this issue - it will be investigated. As far as I can see instead of the total area of the cross section its effective area is used instead (influence of a cover). Luckily for most of the cases this is not a big difference and usually it is not economically justified to have such large amount of steel in a beam to meet this limitation.
- I can see different distribution of stirrups along a beam by setting a number of reinforcement sections (@ Calculation Parameters -> Transversal reinforcement) equal to number of beam calculation points.
This is the only way the transversal steel is shown in this module. Again - more detailed way of accessing this information is available in the RC Beam design module.
If you find your post answered press the Accept as Solution button please. This will help other users to find solutions much faster. Thank you.
Artur Kosakowski
Theoretically I agree with you, but in practice usually axial forces are not enough to condition beams design. Following your logic i think you cannot label a combination as design combination and this table does it. It´s quite strange that regarding a simple supported beam with two load cases, one P1=5 kN/m and other P2= -2 kN/m, this table will always show P1 as the design case. But I understand that in the software perspective is not so simple.
As it is the simply supported beam one load will govern the area of top reinforcement whereas the other the bottom one. For a simple bending there is no problem to calculate such value of required area of reinforcement for which the beam safety ratio will be 1. What happens is that the area of top and bottom reinforcement vs. case 1 and case 2 respectively will give the same safety ratios. In such case the first load that gives the same ratio as other cases will be displayed. I can understand that you would expect the load with larger value to be shown but for this particular design situation both are 'equal'.
I agree, robot is considering effective concrete area. In this point there’s other thing, robot displays on extremes that maximum reinforcement is 0,02Ac(eff), but in fact I think that doesn’t explicitly exist nothing in En 1992 saying that.
I'm sorry but I don't understand this point. As Ac(eff) < Ac then 0.02 * Ac = 0.02*25*65 = 32.5 has to be larger than 0.02*Ac(eff). The value of maximal reinforcement displayed by Robot is as you mentioned earlier 60.2 which is larger than 32.5. Why do you assume that such formula is used here?
Artur Kosakowski
As far as I can see that happens only for negative moments (top reinforcement) of small values. We will check why.
Artur Kosakowski
As far as I can see this is due to calculations of minimal reinforcement for tension. Having small axial force you may ignore it. Try Robot 2015 
If you find your post answered press the Accept as Solution button please. This will help other users to find solutions much faster. Thank you.
Artur Kosakowski
I could repair on your picture about top reinforcement at ARSA2014 vs. 2015 -> 2,09 cm2 vs. to 2,31 cm2 ( about 10% more), any explanation to this?
Robot 2014 - 2.09 for bending
Robot 2015 - 2.31 for tension + bending (change in calculations of calculations of minimal reinforcement for tension done in Robot 2015)
Robot 2015 - 2.09 for bending (as in Robot 2014)
If you find your post answered press the Accept as Solution button please. This will help other users to find solutions much faster. Thank you.
Artur Kosakowski
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