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# 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-tbquimby01.uaa.alaska.edu/courses/ce433/Quimby/docs/CE433.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.

EIT with BS in Civil/Structural engineering.

### RE: Strain compatibility question

"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"

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...

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### RE: Strain compatibility question

(OP)
Alright. Now, I assume that for the strain compatibility check we don't limit Fs to Fy as I've never seen any example do that and it just doesn't make sense as you're trying to find the depth of compression.

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

(OP)
That is when I sum the forces to get my nominal moment capacity. Sorry if that wasn't clear.

EIT with BS in Civil/Structural engineering.

### RE: Strain compatibility question

It sounds like you're confusing the strain and stress.  The steel strain, call it es, can exceed the yield strain, ey.  But the stress, Fs, maxes out at Fy, even for strains greater than ey.

You do limit Fs to Fy.  You don't limit es to ey.

### RE: Strain compatibility question

(OP)
Just realized that a second ago and came to post here.

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

nutte...you reminded me of something my structures professor put in his syllabus...

"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

I think you're getting confused when you keep saying to 'limit Fy'. You are not limiting it, it's that the actual stress in the bar can not exceed Fy. The strain is limitless (until rupture) but after the yield point the stress in the bar does not increase (at least in the idealized way we design beams, it does harden a bit). If you're doing a spreadsheet you need something like =if(es > ey, FY,es*Es) to get the stress in your steel.

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

(OP)
@BARetired: Thanks, I posted my earlier comment before I saw yours. I'm looking over those lecture notes right now.

@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

No. There is no "extra" stress. The stress in the bars is what it is. If the bars haven't yielded, then the stress is equal to E*epsilon (epsilon being the strain); if it has yielded (for 60 ksi steel this is at about a strain of .002), then the stress in the bars is the yield stress (typically 60 ksi) regardless of strain. The strain can be an order of magnitude beyond yield (and quite frequently is) and your stress is still Fy.

You keep confusing stress and strain and it is critical you understand the difference.

### RE: Strain compatibility question

(OP)
@frv: I think all my confusion is stemming from that example I posted. In there they DO NOT have Fs > Fy being changed into Fy. They solve for their strains and then turn those strains into stresses. From there they turn those stresses into forces.

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

The textbook example does indeed have two stress-strain relationships: 58/0.002 = 29000 (the one we all use) and 3/.002 = 1500 (instead of a flat plateau of zero).  Even the example uses the 1500 relationship after reaching 0.002.

(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

Look at the stress/strain curves for various steels, a few have a well defined 'yield point,' but most do not.  So, the 'yield strength' is defined as the stress, on the curve, at a .2% strain offset, we had to pick something, fairly safe and not near Fu.  But, the curve does continue to show increased stress with increased strain, just at a much lower stress increase rate for a given strain increase.  The inherent material safety factor is that in working with Fy at the .002 strain offset, we can still get some fairly small stress increase without rupture of the steel; but that the strains and cracking, etc. will scare the hell out of us before for the steel actually ruptures in a sudden failure.

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

(OP)
@PMR06: Okay, so my confusion was because they didn't flatten out their stress-strain curve once they reached a strain of 0.002 as they should have. Yes, I referenced 10.2.4 in my initial post. So, all I have to do is follow that example but change their stress-strain curve and limit Fs to Fy.

@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

(OP)
"Step two" I mean. (It must be Friday)

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

TMP,

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

@TehMightyPirate, yes your last procedure basically looks correct.  Step 4 is easier said than done though.  You won't get neutral axis correct the first time, so there will be multiple iterations.  This is where computers earn their keep.  If you're using Excel, Two Words: Goal Seek

### RE: Strain compatibility question

(OP)
Yes, I've programed an excel sheet to do all this for me. Never tried out goal seek, though. Thanks for the tip.

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

Although there is nothing wrong with using goal-seek to find the depth of the NA, with a rectangular concrete stress block and a rectangular section it much quicker and almost as easy to solve a simple equation relating the nett force on the section to the depth of the NA.

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

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