Rc Design Beam Shear design module doubts

[Anonymous]

Greetings, my problem is the following: I am doing a very detailed revision on the calculations in robot's RC design module (provided reinforcement) and I have found the following questions I would like someone could answer me, I really want to keep learning this software because I think it's better than ETABS, hence I am comparing both plus I am doing manual calculations on my own. The things I have encountered so far are the following: 1) I am trying to design a Special Moment Frame Beam in ACI 318 code, that means I have to select the option High risk or category D,E,F. When I do this I am also considering the plastic moments at the sides of the beam that generate a shear named Ve. I read in the manual that Robot needs a load combination case named SEISMIC SHEAR to calculate this force and the type of the load must be ACC. When I choose the code ACI 318-08 with simples cases option when I am tranfering the beam to the module, the software considers this load because I can see it in the results when I click on ALS layout and see the different combinations. The problem is that the value of shear V or Vc that the program is calculating is way too high from the one that ETABS or my own calculations based on the code. It should be around 10.000 Kg and Robot it's showing 25.000 Kg. When I read the manual about the nature of this force, it says "The formula from ve has been extended". Does this mean Robot is not using the exact formula of the ACI code? Can I see which formula is the program using to bring such a high result? 2) Another problem that is bothering me is that when this SEISMIC SHEAR combination is present (consider plastic moment option) the software automatically puts an addiotional stirrup column on the arregement when I use more than 2 longitudinal bars on either bottom or top fibers. When I try to delete this extra transversal reinforcement the program sends a warning saying that the cross section dimensions are too small. I don't understand this because if I decide to not consider the plastic moment such reinforcement is not needed. I am guessing that the program is doing this because Ve is way too high. I realy would like someone could help me with this, it's the only thing that I am missing so I can finally close the chapter in the beam module and conclude to my collegues that this software is bringing results very similar to the ones we do doing manual shear design calculations (Normally I wouldn't bother writing all this but the program does not show how it does these calculations, I managed to calculate the longitudinal reinforcement ones and everything was fine, but in this case, the Shear is way too high). Thanks for reading this, I wait your response!

[Artur.Kosakowski]

Could you attach Robot file and your hand calculations and indicate which elements of the model to look at please? Thank you.

[Anonymous]

Of course! 

I will explain my hand calculations right here (That way someone else can benefit from this topic)

When I choose to consider the plastic moments at the ends of the beam in the joint faces, the ACI 318 code says that we have to calculate a seismic shear named Ve.

The formula that you can find in the code is Ve= (Mpr1 + Mpr2)/Ln + Wu*Ln/2 

The code also indicates that Ve is the sum of the resulting shear from the plastic moments in counter-clockwise and clockwise directions plus the static shear that generates from the factored tributary gravity load along the beam's span.

So if we try to calculate this on robot, we get the following:

On the 1st span of the beam robot is designing we have a different value of provided reinforcement area depending on the fiber we are evaluating. For the left and right supports sections we have 8.55 cm2 for top reinforcement and 5.94 cm2 for the bottom one.

Mpr formula is M=1.25*Fy*As*(d-a/2) where a=(0.85*Fy*As) / (0.85fc*b)

So, this case our data is:

Fy=4218 Kg/cm2

d=35 cm

fc=249.83 Kg/cm2

Ln (free span lenght)= 450 cm

Factored Shear Force (The greater shear force value at the left joint face in the envelope)= 9314 Kg (Taken from the shear force diagram for ACC envelope combination)

In this case Counter-clockwise and clockwise Mprs are the same because both sections at the joint faces have the same area of reinforcement, so theres only the need to calculate one of those cases.

So Mpr1 (top reinforcement) would be Mpr1= 1.25*4218*8.55*(35-((0.85*4218*8.55) / (0.85*249*30) )/2) = 1.418.541 Kg.cm

For Mpr2 (Bottom reinforcement) it would be Mpr2= 1.25*4218*5.94*(35-((0.85*4218*5.94) / (0.85*249*30) )/2) = 1.019.351 Kg.cm

So the shear that generates from plastic moments is V= (1.418.541 + 1.019.351) / 450 = 5417 Kg

In order to obtain Ve we need to calculate the shear that comes from the factored tributary load too. In the ACI code its stated that the formula to calculate this one is wu*ln/2 and robot needs the SEISMIC SHEAR combination that its generated when the option "consider plastic moments" its on. 

In ETABS and SAP this shear is obtained directly from the shear force diagram that generates due to the SEISMIC SHEAR combination, for example, if the seicmic combination is 1.2DL+L+S, the SEISMIC SHEAR combination would be 1.2DL+L corresponing only to the gravity loads that the ACI code mentions. So, if I go to the respective shear force diagram, the value of V would be 8818 Kg at the left joint face. 

So, to conclude the result would be Ve= 5417 + 8818 =14235 Kg. thefore, because Ve>Factored shear force, 14953>9314, the shear force for design would be Ve=14953 Kg.

I do this exact procedure on ETABS and SAP2000 and I am obtaining the same value, plus the programs indicate that the V for design its 14253 Kg for the respective load combination. The software gives me the same exact value I am calculating with the previous hand calculation.

Robot is showing a value of 24.419 Kg for the Design Shear.

This is the thing I am not understanding. The manual says something about the Ve formula being extended, I would like more information on the matter. I dont know if I did something wrong in the design module and thats why I am getting such a high value or maybe I did something wrong creating the load combinations in the software or the software its doing something else i am not aware of.

I await your response and thanks for your time. I am looking forward to find the reason behind this.

Regarding my second question, i prefer to solve this one first.

[Anonymous]

Bump, did someone managed to look at the things I mentioned?

[Artur.Kosakowski]

Some investigations take more time 

In Robot the shear force is incorrectly multiplied by 2 which should be done for precast beams only. 

[Anonymous]

Oh, is there a way to configure it correctly? I have Robot 2015. I could divide it by 2 and see the real shear but the problem would be that the one robot uses for designing the steel reinforcement would be the one that its multiplied by 2. 

Its there a way to configure it? Or I must wait for a patch?

[Artur.Kosakowski]

You may try to edit load factors for shear combinations in the regulation file reducing them by half. 

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.

[Anonymous]

Alright, I just tried that and the new value of V is 16.000 Kg, it's not exactly the same but it's closer now. The only thing I am still noticing it's that the shear reinforcement it's still the same, the value of Vc it's 32.000 Kg. The software it's forcing the reinforcement to have a third transversal bar, if I decide to delete this one to only leave the normal stirrup, a message appears saying that "Cross sections dimensions are too small".

I don't understand that because with the Ve reduction, it shouldn't be neccesary to have that quantity of transversal reinforcement area.

What maybe be the cause of this? Thanks in advance for your attention, I appreciate it.

[Artur.Kosakowski]

Could you attach the model with reinforcement you have generated in the beam please? 

[Anonymous]

As you can see the picture above, the design shear has been reduced because I halved the load factors of the SESMIC SHEAR combination., but the shear capacity it´s still the same.

The reinforcement I mentioned in this one:

If I decide to delete it, then the warning appears. This kind of beam shoulnd´t need that extra transversal reinforcement. It should be enough with just the stirrup I have chosen.

This file is attached below. Once again, I wait your answer. Thanks in advance. I am really trying to figure all this out so I can tell my collegues to start using ROBOT.

[Artur.Kosakowski]

If I decide to delete it, then the warning appears. This kind of beam shoulnd´t need that extra transversal reinforcement. It should be enough with just the stirrup I have chosen.

I'm not very familiar with ACI however I think that seismic provisions usually require each of main rebars being "tied" wit transversal reinforcement. If I delete it I can't see any negative influence on beam's capacity for shear when I calculate the beam with frozen reinforcement. 

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.

[Anonymous]

You are right, when I try to delete the reinforcement there's no warning anymore. It's weird because there was before,so I deciced to check other model's saves and I found out that It happens on the one I just attached. The main difference with this one is that I generated the design combination using the simple cases option instead of the manual combinations one. (I did it this way because it was easier and faster but at the time I didn't know that I needed the manual combination to route to fix the problem with load factors affecting seismic shear).

The other thing I am noticing is that when I delete in the file you have, apart form the fact that it does not shows the warning, when I check the Vc shear it's still the same. Isn't Vc the shear capacity? Shouldn't the value of vc be less than it was when the beam had that extra transversal bar? That's the thing I don't understand. If I remove transversal reinfocerment, the shear provided capacity should be lower.

I will investigate more deeply the ACI code in search of that seismic provision that says that all main bars must be tied. Meanwhile I would appreciate a lot if you could check why the program shows the warning when I used simple cases as design combinations and not when I decide to use manual combinations option. Also I would like your opinion on why the software it's not reducting Vc value when we delete the 3rd tranversal bar. 

Thanks in advance, I appreciate your service.

[Anonymous]

I have been investigating the transverse detailing in the ACI code more deeply. I didn´t find anything about all main bars being forced to be tied al the time, I only found out an article that states that only the corner bars must be tied and it´s not a seismic provision chapter. In the seismic provisions chapter you can chek the section 21.5.3, it´s short and it does not say anything about all bars being tied.

Also I found out this useful guide to following correctly the ACI seismic provisions steel detailing. I will attach the document too. If you check the beam transverse reinforcement chapter you will see that there´s no condition that states that all bars must be tied.

I hope you find this information useful and I will be glad to hear your opinion on the matter. Thanks in advance again.

[Anonymous]

Arthur, were you able to verify the things I mentioned? I think we might be closer to finally end the topic

[Alireza.Kord]

for better understanding what's going on, could you attach etabs file please?

[Alireza.Kord]

Vc is nominal shear strength provided by concrete so it not depend on bars 

[Artur.Kosakowski]

I'm sorry for not answering earlier but I wanted to run test on SP4 for RSA 2016 before posting the answer on the forum:

1. Too small cross section error displayed on frozen reinforcement when you delete pins is no longer displayed on 2016SP4 which corresponds to me assumption of it being excessive as there is no such error displayed when you run calculations with such reinforcement after freezing it.

2. No difference in the shear capacity diagram after deleting pins is due to the limit from point 11.4.7.9 of ACI

3. For the requirement of having all main rebars tied - could you attach the content of point 7.10.5.3 of ACI 318 please? Does it say that every main rebar should be tied or every second main rebar should be tied? Thank you.

[Anonymous]

Alright Arthur, I will comment on the 3 points you made:

1- I will download Robots 2016 Educational version then, now that you have told me that this was fixed in SP4.

2- The thing about Vc I am not getting is the following:

In ACI 11.4.7.2 it is stated that:

Where Av corresponds to the area of transversal reinforcement. In this case, the diameter of stirrups I am using is #3 and the spacing in the confinement zone is 5 cm. 

Also, the number of vertical bars on the section must be taken in account for the calculation of Av. 

Evaluating all this variables, we can obtain the following result:

Vs= (#vertical bars * Av * fy * d)/ s

If we put the third pin in the section as robot is doing automatically, the Vs shall be:

Vs= (3* 0.71 * 4282 * (40-5) ) / 5 = 63844 Kg

If we remove the thrid pin:

Vs= (2 * 0.71 * 4282 * (40-5) ) / 5= 42.563 Kg

Also applying the 11.4.7.9 we have

Vsmax= 2.2 * (250)^0.5 * 30 * (40 -5 )= 36.524 Kg

So this means that the 32.000 robot is showing in the results is the 36.000 I am calculating?

Also, the software calculates a Vs inferior to 36.524 Kg in the center of the span.

3- The code 10.5.3 is refering to compresion members, we are currently evaluating the beam shear design. This beam is not presenting any axial force.

This article means that the lateral support must be there in the corner bars and along the internal bars in an alternate way. Also, this article its stating that this consideration must be applied on columns, not beams.

I await your responde and thanks again for answering me like you always do. I will be testing robot 2016 meanwhile.

[Anonymous]

Also, to clarify even more what the article 7.10.5.3 is saying I took the time to draw this:

(Warning: I am not an artist)

The considerations of 7.10.5.3 are stated for compression members. 

For flexural members (beams) this are the ones that ACI states: