Strain compatibility question
Strain compatibility question
(OP)
I'm trying to puzzle out something regarding strain compatibility and reinforced concrete (I'm making a spreadsheet to do concrete walls under axial compression and moment).
In the textbook Reinforced Concrete Mechanics and Design, 4th Edition (MacGregor & Wight) on page 203 and 204 they have an example of strain compatibility. I understand the whole process and how you assume a concrete strain of 0.003 and solve for the steel stresses, how you verify a tension-controlled section, and so on.
However, you end up with a Fs (steel stress) > Fy (yield stress with a concrete strain of 0.003 (unless you're compression controlled, obviously). Now, I assumed that you would proportionately reduce the strains in the concrete and steel until you brought the steel below yield. From there you could sum all the forces about the neutral axis and calculate the nominal moment capacity (moment to cause either yielding of steel or crushing).
However, the textbook example doesn't do this. It keeps the calculated Fs (beyond Fy) through-out. This obviously isn't right as your nominal moment capacity will result in steel being yielded at that moment.
So, I google a bit and look at this lecture notes PDF online: http:/ /anc-tbqui mby01.uaa. alaska.edu /courses/c e433/Quimb y/docs/CE4 33.L03.pdf In there it states that per ACI 10.2.4 "Stress in reinforcement below Fy shall be taken as Es times steel strain. For strains greater than that corresponding to Fy, stress in reinforcement shall be considered independent of strain and equal to Fy."
GAH! Which one is it?
I'm going to make the obvious assumption that the code is the correct way and my initial thought and my textbook are incorrect but I really need a second opinion here.
In the textbook Reinforced Concrete Mechanics and Design, 4th Edition (MacGregor & Wight) on page 203 and 204 they have an example of strain compatibility. I understand the whole process and how you assume a concrete strain of 0.003 and solve for the steel stresses, how you verify a tension-controlled section, and so on.
However, you end up with a Fs (steel stress) > Fy (yield stress with a concrete strain of 0.003 (unless you're compression controlled, obviously). Now, I assumed that you would proportionately reduce the strains in the concrete and steel until you brought the steel below yield. From there you could sum all the forces about the neutral axis and calculate the nominal moment capacity (moment to cause either yielding of steel or crushing).
However, the textbook example doesn't do this. It keeps the calculated Fs (beyond Fy) through-out. This obviously isn't right as your nominal moment capacity will result in steel being yielded at that moment.
So, I google a bit and look at this lecture notes PDF online: http:/
GAH! Which one is it?
I'm going to make the obvious assumption that the code is the correct way and my initial thought and my textbook are incorrect but I really need a second opinion here.
EIT with BS in Civil/Structural engineering.






RE: Strain compatibility question
That's the answer right there. The code assumes that the steel reinforcing behaves as an elastic element until it reaches its yield stress, and then behaves perfectly plastic. I.E. your Stress-Strain curve looks like...
_____________________
/
/
/
/
/
/
RE: Strain compatibility question
So, all I have to do is when I sum the forces about the neutral axis I cut any rebar stresses down to Fy but keep all the other stresses (both tensile and compressive) the same as they were in the strain compatibility check.
Correct?
EIT with BS in Civil/Structural engineering.
RE: Strain compatibility question
EIT with BS in Civil/Structural engineering.
RE: Strain compatibility question
You do limit Fs to Fy. You don't limit es to ey.
RE: Strain compatibility question
http
BA
RE: Strain compatibility question
Yes, I see now, I was confusing stress and strain (or, rather, I was realizing that you can of course not have a linear relationship between the two).
Okay, so what is the proper method?
Here is what my textbook example shows:
1) Solve for all the strains in concrete and rebar.
2) Find the resulting stress in the rebar and concrete. Do not limit it to Fy.
3) Based on the areas of rebar and the concrete compression block, solve for forces.
4) Sum the forces, verify they sum to zero. Check for tension controlled section.
5) Take the forces from 3) and multiply them times their respective distance from the centroid of the section to get a resultant moment.
6) Take that moment as the nominal moment capacity. Fs was never limited to Fy.
They never limit Fs to Fy. At the start of the example they have a stress-strain curve where, at 0.002 ksi strain they change the slope of the stress-strain curve but other than that they don't do anything.
So, my question now is where do I limit the Fs to Fy. In step 2 or step 6?
Maine EIT, Civil/Structural. Going to take the 1st part of the 16-hour SE test in October, wish me luck!
RE: Strain compatibility question
"Aspirin is prescribed for stress and strain...if you confuse the two on the final exam, you'll need the aspirin"
RE: Strain compatibility question
JDEngineer drew a stress/strain diagram for you, that should help. Strain is the x-axis, it can keep increasing but you see that the stress goes flat after Fy.
RE: Strain compatibility question
@Ron: Heh, that reminds me of my materials professor, he would make you do 5 push-ups in front of the class if you confused cement with concrete.
@bookowski: Okay, so the bar(s) basically gone plastic at that point and the extra stress is transferred to the other rebar with fs < fy until you develop a full plastic hinge (all fs >or= fy), correct?
So, to attempt to answer my own question. In step 2 of my list, I would set any stresses greater than Fy to be equal to Fy (which is not what the example does).
***To hopefully clarify this I've attached the example from my textbook as a PDF.***
Maine EIT, Civil/Structural. Going to take the 1st part of the 16-hour SE test in October, wish me luck!
RE: Strain compatibility question
You keep confusing stress and strain and it is critical you understand the difference.
RE: Strain compatibility question
So, if I understand it right the example has a mistake where, when summing the forces based on their trial depth "c" they didn't set the rebar with stress > Fy to Fy. Thus, they have more stress on their outer bars when really that stress should "shift" (I said "extra" but shift is the word I meant to be using) to the bars closer to the neutral axis. Correct?
I've never had a problem understanding stress vs strain. I've only been confusing what the stress-strain curve was supposed to be and how it's supposed to apply to the strain-compatibility.
@BARetired: I looked through that powerpoint but didn't find anything relevant. Did I miss something?
Maine EIT, Civil/Structural. Going to take the 1st part of the 16-hour SE test in October, wish me luck!
RE: Strain compatibility question
(0.002)(29000) + (0.00956-0.002)(1500) = 69.34
Now back to standard practice and building codes, ACI specifically says above yield strain you stop at fy. Read 10.2.4. The example (if worked in accordance with ACI) would limit the steel to fy. You are correct, this would shift the NA and change the moment capacity.
RE: Strain compatibility question
I still think your item 2) is bass-ackwords, it should read 'limit steel stress to Fy.' Your design should be limited Fs to Fy. Then the rest sounds about right. There is some FoS left in the steel, but the thought is that chunks of cracked conc. will be falling on your head before the steel fails in a catastrophic and sudden way.
RE: Strain compatibility question
@dhengr: Right, I thought the example had step too wrong according to ACI-318.
Just to make sure I'm clear:
1) Solve for all the strains in concrete and rebar by setting the concrete strain to 0.003 and assuming a depth to neutral axis "c".
2) Find the resulting stress in the rebar and concrete. ***If Fs (stress in the steel) is greater than Fy then reduce it to Fy.***
3) Based on the areas of rebar and the concrete compression block, solve for resultant forces.
4) Sum these forces, verify they sum to zero. Check for tension controlled section.
5) Take the forces from 3) and multiply them times their respective distance from the centroid of the section to get a resultant moment.
6) Take that moment as the nominal moment capacity.
Maine EIT, Civil/Structural. Going to take the 1st part of the 16-hour SE test in October, wish me luck!
RE: Strain compatibility question
Maine EIT, Civil/Structural. Going to take the 1st part of the 16-hour SE test in October, wish me luck!
RE: Strain compatibility question
Your textbook example assumes a stress-strain diagram as shown in Fig. 5-24b. This was stated at the bottom of page 20l. In other words, the author is considering strain hardening of the steel.
Usual practice is to ignore strain hardening and assume that the steel stress beyond yield strain remains Fy.
A spreadsheet would be a good way of developing interaction diagrams for walls of different thickness and reinforcement. I think there are some programs available on the Internet to do this but I haven't used any of them so I hesitate to recommend any particular one.
BA
RE: Strain compatibility question
RE: Strain compatibility question
Maine EIT, Civil/Structural. Going to take the 1st part of the 16-hour SE test in October, wish me luck!
RE: Strain compatibility question
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/