Two Layers of Tension Rebar

[ottles]

How often do you use two layers of tension rebars distance 1 inch apart (the 2 layers). The outer side would yield first before the inner ones. Usually how many percentage yielding of the outside side before the inner ones yield? and why is this allowed in structural books. I only want single layer in my design but I saw others doing 2 layers. What is your experience and say on this? Thank you.

 

[ottles]

You can make an upper bound estimate of your lower bar strain pretty easily. Draw a straight line from -0.0035 on your stran diagram at the compression edge to +0.002 at the inner layer of reinforcing. Then just read the outer layer strain from that same line. You'll find it much less than the rupture strain for most practical situations.

Algebraically, upper bound strain = d_max x (0.002 + 0.0035) / d_min - 0.0035

Kootk, I understand it's just for rough idea. But would like to confirm if the following computation is correct. See graphics.

I tested the common depth of 19.685". Here d_max = 18.11 and d_min 17.11.. computing for upper bound strain = d_max x (0.002 + 0.0035) / d_min - 0.0035 = 0.00231

Do you know the formula to get the distance to the neutral axis in your sample? I'm poor in trigo.. lol

So the outer strain is 0.00231 for inner bar strain of 0.002. I know this is just for rough idea and illustrating the idea plane remains plane.

(For the actual neutral axis of the beam at ultimate load. It's 6.87 and the neutral axis is higher so the difference is even lesser when the neutral axis gets higher up)

 

[rapt]

Ottles,

Come on, an engineer who cannot do middle school math!

But this calculation you are doing has nothing to do with the real neutral axis depth. You need to do these calculations for the reinforcing pattern you are analysing for, and equate the tension and compression forces to determine the actual neutral axis depth.

 

[KootK]

All that looks good ottles. It doesn't affect your calculations but why is your d greater than your d_max? I would expect d to be between the min and max values.

Neutral axis depth would be: d_min x 0.0035 / (0.0035 + 0.002)

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[rapt]

Kootk,

I agree with 2.

1 is the problem. The strain in the steel after cracking is completely unrelated to the cracking moment for a T section.

For tension on the flange side, Mcr is very dependant on the flange width, while steel strain after cracking is dependant on the web width. The flange is no longer an item as it is cracked and has no effect on the strain in the steel after cracking.

For tension on the web side, Mcr is very dependant on the web width, while steel strain after cracking is dependant on the flange width. The flange is the main influence on the strain in the steel after cracking.

So, while adding more steel will reduce the strain, basing the amount you add on Mcr will not give any idea of the strain in the steel and its ability to cause multiple cracking and a ductile section.

 

[ottles]

All that looks good ottles. It doesn't affect your calculations but why is your d greater than your d_max? I would expect d to be between the min and max values.

Neutral axis depth would be: d_min x 0.0035 / (0.0035 + 0.002)

Kootk, d is the actual depth of the beam which is 19.685" (or 500mm), although I know the d used for calculation is from compression edge to the bars (or the centroid which is what you may be talking about).

Anyway. What is most important is the inner rebars can contribute to strain at elastic service load and not just at yield.. because if you are waiting for inner bars to suddenly yield when the outer ones yield.. this is not good.. What is good is both all layers contribute at elastic strain.. right?

But this calculation you are doing has nothing to do with the real neutral axis depth. You need to do these calculations for the reinforcing pattern you are analysing for, and equate the tension and compression forces to determine the actual neutral axis depth.

Rapt. Our company mostly used software in design. They never computed each manually. But I want to verify manually. Anyway. I computed for the real neutral axis depth. For the following data:

As = 3.894848
fy = 60000
alpha = 0.72
fc' = 4000
b = 11.8"
d = 19.685"
neutral axis formula at ultimate strength = As*fy/(alpha*fc'*b) = 6.87

(this is the not the neutral axis of Kootk illustration but the actual beam (Kootk's is 10.88818)

Now I can compute for the strain of each layer of rebars and then stress given the real neutral axis at ultimate load.

 

[KootK]

Anyway. What is most important is the inner rebars can contribute to strain at elastic service load and not just at yield.. because if you are waiting for inner bars to suddenly yield when the outer ones yield.. this is not good.. What is good is both all layers contribute at elastic strain.. right?

I think that "what is important" with regard to rebar strain is best expressed by a paraphrasing of one of rapt's previous statements, if I have it right:

For ultimate moment resistance, what is important is an accurate estimate of the stress levels in the reinforcing at the ultimate limit state condition. An accurate understanding of rebar strain is required for this. No shortcuts.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[ottles]

I think that "what is important" with regard to rebar strain is best expressed by a paraphrasing of one of rapt's previous statements, if I have it right:

For ultimate moment resistance, what is important is an accurate estimate of the stress levels in the reinforcing at the ultimate limit state condition. An accurate understanding of rebar strain is required for this. No shortcuts.

Of course. But at elastic strain, if the outer and inner bars both contribute, then the beam can functions as designed (in books, they always take the centroid and never argue or give the illustration they both contribute). My initial worry is the outer bar can yield and rupture without the inner bars contributing then the inner bars will engage and rupture too after the main one rupture and failing the beam.. but if the full layers are engaged at elastic load.. then they would function as designed.

Last question. I can't find this in books too. In the following neutral axis for the actual beam given:
As = 3.894848
fy = 60000
alpha = 0.72
fc' = 4000
b = 11.8"
d = 19.685"
neutral axis formula at ultimate strength = As*fy/(alpha*fc'*b) = 6.87
See illustration in the following

How do you compute the strain at the 2 bars given a distance to them? What's the formula (I can't find this in structural textbooks). Sorry I'm really poor in trigo and most designers I know don't do this manual computation style. Thank you!

 

[KootK]

Of course. But at elastic strain,

I'm finding your focus on elastic strain confusing. Ultimate moment resistance is your concern, right? At that load level, we are dealing with plastic strain in some or all of our reinforcement. Or, at least, we'd prefer that.

My initial worry is the outer bar can yield and rupture without the inner bars contributing then the inner bars will engage and rupture too after the main one rupture and failing the beam.

Have you googled the rupture strain of rebar? Do that and compare it the post yield values that we've been discussing. To quote myself:

I developed it with a singular, important use in mind: helping OP out with this thread. With that diagram / equation in hand, OP can easily run a few test cases and convince himself that, while the outer layer strain may indeed exceed 0.002, it's not likely to be anywhere near the rupture strain.

in books, they always take the centroid and never argue or give the illustration they both contribute

You must be reading some exceptionally crappy books.

How do you compute the strain at the 2 bars given a distance to them?

The triangles above and below the neutral axis are similar in the geometric sense (all three angles identical between the two). As such, you can ratio the sides of the triangles to figure out their lengths. 6.87 is to 0.0035 as 10.24 is to ???.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[ottles]

My textbook is called Design of Concrete Structures 14th edition by Arthur H. Nilson, David Darwin, Charles Dolan.

I have practically mastered the chapter on flexure.

Elastic strain is strain before reaching yield
Plastic strain is the strain hardening after yield and I'm not mentioning this when I mentioned "elastic strain".

In the book the rupture strain of rebar is many times over after first reaching yield plateau then reaching tensile strength and then go down to rupture strain.

Thanks for the tips on computations. It makes everything clearer.

 

[ottles]

Kootk.. you seemed to ignore the elastic strain (the service load strain before yield.. see above where plastic range means the strain after yield) so I tried to compute it and got the following facts. For this neutral axis formula As*fy/(alpha*fc'*b), if you want to use elastic strain (service load), then you make the fc' = 2000 only (or corresponding to the actual service load) instead of 4000 psi (ultimate load), is this correct? So with the following data:

fc' = 2000 (instead of ultimate load 4000 psi)
As = 3.894848
fy = 60000
alpha = 0.72
b = 11.8"
d = 19.685"
neutral axis formula at elastic service load = As*fy/(alpha*fc'*b) = 13.75
now see the following illustration

the strain in the bars seemed to be only 0.000122 and 0.000158.. they are so small.

Can the strain diagram be used at all for small service load or only for ultimate strength? If it can be used for small service load.. the fact the rebar strains are so low means for elastic (service load range) the strains of the two can be taken for granted and you only need it more importantly for plastic (yielding strain hardening) ultimate load? is this why you keep ignoring the elastic service load strain and focusing on plastic yielding strain?

 

[jayrod12]

Technically speaking, a strain diagram is always applicable. However with concrete beams when choosing standard proportions the service load case is not required to be calculated. Or at least not at vigorously as you are attempting.

Are you even beyond the cracking moment at your service loads? That would be check number one.

I must reiterate Koot's question, why are yu so fixated on service level strain?

 

[Ingenuity]

As = 3.894848

...not to mention your fixation with INsignificant figures.

 

[EZBuilding]

Service level strain is used in deflection calculations, strength level strain is used in capacity/ductility design. This is an important distinction. Multiple layers of reinforcement in reinforced concrete beams is common in practice and to design to only have one layer of reinforcement would lead to constructability, architectural and financial constraints.

 

[rapt]

I cannot believe this discussion has continued so long. This is basic reinforced concrete ultimate section strength theory.

The strain in the concrete is assumed to be .003 or .0035 or whatever depending on your design code and possibly the concrete strength (Eurocode).

The compression force C in your concrete and the steel on the compression side of the neutral axis has to equal the tension force T in all of your steel on the tension side of the neutral axis.

You adjust the neutral axis depth which changes the strain in each layer of reinforcing. From this you calculate the stress in each layer and from that the tension or compression force in the layer.

This gives you C and T. When they are equal you have the right neutral axis depth and the final strains in each layer of reinforcement.

If you cannot calculate the strains in each layer of reinforcement or the associated stresses from an assumed compression face strain and assumed neutral axis depth, you need to go back to school, (and I do not mean university, I mean school).

PS This is all true only if you forget about Strain Localisation, a topic we will leave for the theoreticians.

 

[ottles]

A bit in page 77 of the book Design of Concrete Structure 14th Edition confuse me for a day already thinking of it every hour. So I'd like to ask for a little assistance.

Let's take C=T or alpha*fc'*bc= As*fs and at ultimate load
fc'=4000
fs=60,000
let's use the book example of b=10 and As=2.37
actual reinforcement ratio of the beam is As/bd = 2.37/(10x23) = 0.0103.
Balanced reinforcement ratio is alpha(fc'/fy)(strain(conc)/(strain(conc)+strain(steel))=0.0284.
The book says "Since the amount of steel in the beam is less than that which would cause failure by crushing of the concrete, the beam will fail in tension by yielding of the steel. It's nominal moment, from Eq. (3.206), is
Mn= 0.0103 x 60,000 x 10 x 23^2 (1-0.59 ((0.0103x60,000)/4000) = 2,970,000in-lb= 248 ft-kips
When the beam reaches Mn, the distance to the neutral axis, from Eq, (3.19b) is
c=0.0103 x 60,000 x 23 / 0.72 x 4000 = 4.94"

The c is the neutral axis which is derived from C=T or alpha*fc'*bc= As*fs
At balance point neutral axis corresponding to simultaneous crushing of the concrete and initiation of yielding in the steel, formula is c = strain(conc)/(strain(conc)+strain(steel)*d or c = 0.003/(0.003+0.002069)* 23 = 13.61.
But for ultimate strength and underreinforced for ductile failure,
neutral axis is only 4.94" instead of 13.61". If you will draw the strain diagram.. see following illustration:

Here's the problem. Which of the above is correct? For the one of the left, if the concrete fails at ultimate strain of 0.003, the steel strain is 4 times beyond yield already. For the one of the right. Just as steel yields at strain of fy/Es = 60,000/29,000,000 = 0.002069, the concrete is just tiny strain of 0.000566.. but the right one seems to be correct because remember it is underreinforced, so steel yields first, but how could the concrete strain be only 0.000566 when concrete reaches ultimate strength of 4000 psi corresponding to 0.003 strain. Anyone can give a clue? I've been thinking for it for a day already.

And in the spirit of Rapt message, I know how to solve for the stresses.. but still in the left illustration where the steel is beyond yield.. fs = strain(conc) * Es ((d-c/c)) = 318265 psi instead of 60,000 psi.. but if you set it to 60,000 psi, the concrete strength corresponding to strain of 0.000566 would be less than 1000 psi. Anyone? Many thanks.

 

[rapt]

Steel stress is not limited to yield strain. The steel stress is assumed to be constant at the yield stress for any strain above the yield strain. The stress does normally rise a few percent above this value (strain hardening) but that is normally ignored in design.

Depending on the type of steel you are using, the failure strain could be as high as 15 or more times the yield strain (earthquake class reinforcement).

Ultimate strength is determined at the concrete compression limit, so for ACI code a concrete compression strain of .003.

 

[ottles]

Steel stress is not limited to yield. The steel stress is assumed to be constant at the yield stress for any strain above the yield strain. The stress does normally rise a few percent above this value (strain hardening) but that is normally ignored in design.

Depending on the type of steel you are using, the failure strain could be as high as 15 or more times the yield strain (earthquake class reinforcement).

Ultimate strength is determined at the concrete compression limit, so for ACI code a concrete compression strain of .003.

But for sake of computation. If you want to get the neutral axis c just when the steel yield at 0.002069 strain (even thought concrete is below 0.003).. how do you compute for it? Most of our reinforcement are way below the reinforcement balance point.. in the case of the book example.. it's only 0.0103 instead of 0.0284 or only 36%. So for b=10, As=2.37

C=T
alpha*fc'*bc= As*fs
what value to be used for fc'..

 

[ottles]

After the above question (please answer). The following is my last question
In page 73 of the book Design of Concrete Structures 14th Edition, there is a section called "Stresses Elastic and Section Cracked". It uses transformed section to compute for fs and fc (elastic range below ultimate strength) given the moment. I have 2 questions.

1. Can't it be done without using transformed section.. why doesn't the book mention it? What books have you come across that computes for elastic stresses without using transformed section.

2. How do you solve for the quadratic equation for kd in the above illustration (I drew it so I won't have to scan the book and copyright problem)? I asked the physicsforums site and they thought it was homework and didn't wanna answer. Just give me steps how to solve for "kd" so I can enter it at excel and try different moments and corresponding fc and fs. The book just mentions the kd is 7.6" without giving the steps. Please show the steps. Thank you all!

 

[rapt]

Set the steel strain at the value you want and adjust the neutral axis depth and concrete strain until C = T.

But that will not be an ultimate limit condition. And you cannot do rectangular stress block for the concrete compression as the concrete stress block is only for an ultimate condition with concrete strain = .003. You would have to do a curvilinear concrete stress/stain relationship similar to the one defined in Eurocode or elsewhere.

For ultimate condition you set the concrete to .003 and adjust the steel strain to give C = T as discussed earlier.

If you want to have the steel strain as the controlling value at ultimate, that will set the neutral axis depth with compression strain at .003. You then have a value for C and strain in each layer of the steel.

You then have to add more and more steel to the steel layers until T = C.

 

[jayrod12]

To solve for kd you could guess and check. Or use a calculator to solve it, my calculator has a way to solve quadratics, and I'm fairly certain most people do.