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R.C. Beam question

R.C. Beam question

R.C. Beam question

(OP)
This problem came out of a PE study guide.  The question is asking what is the amount of reinforcing steel required for the R.C. beam.  If you look at the solution to the question it states "The depth of the neutral axis is ct = 0.375dt"  Where did they get the 0.375 from? according to my calcs, the answer is not even correct. I am coming up with the As,rqd = 2.67 sq. in.  however the book says 2.93 sq. in. is required.  Has anyone seen this before?
 

RE: R.C. Beam question

I solved it and came up with As req'd = 2.86 in^2.  

RE: R.C. Beam question

This is interesting.  I get 2.67 in^2 required.  This equates to 3.38 #8's.  I don't know how I would answer this question, though.  The answer is MOST NEARLY 3, but couldn't be less than 4.  How would you be expected to answer this?  My gut says 3, since that is MOST NEARLY the correct answer.

RE: R.C. Beam question

Those study guides are fluky. They'll have 30 pages of errata for the solutions.  

Here's some advice:

In your test, if you get hung up and can't find a match to one of the multiple choice questions just forget trying to solve for one answer. instead start plugging in their answers into the equation and see which one works.

I know this goes backwards against the real world design process, and I may catch flack on this forum for saying that, but when you've only got 6 minutes on average per question you don't have time to waste.

RE: R.C. Beam question

Toad- How did you get 2.86 in^2?

RE: R.C. Beam question

I also get 2.67 in^2.  I think they left the phi (0.9) factor out of the equation.  The allowable moment capacity needs to be Mu/0.9; or the formula for the allowable capacity is multiplied by 0.9.

0.9 times their As equals 2.64 ~ 2.67.

gjc
 

RE: R.C. Beam question

(OP)
Ok, I got it now.  The book is incorrect because they are only checking 1 of the conditions required.  They are checking C = T, but they are not checking Mn = (C or T)(d-a/2).  Which both conditions must met.

Well that only took me half a day to figure out.  I guess I am *#&$ on the P.E. exam.  

RE: R.C. Beam question

DWHA-

I haven't gone through the problem, but I'm not sure what you mean by Mn=(C or T)(d-a/2). For equilibrium to be satisfied, T must equal C.

RE: R.C. Beam question

what "phi" are you guys using?>  

RE: R.C. Beam question

huh pretty simple problem no?

RE: R.C. Beam question

if I use "phi" = 0.9, I get As req'd = 2.94 in^2

RE: R.C. Beam question

Toad,

Nice catch.  I just realized that the strain is only 0.004408, so phi would be 0.85.  This would bump up the required steel from 2.67 to the 2.86 that you referenced earlier.

RE: R.C. Beam question

a = As*Fy/(0.85*fc'*b)
Mu = phi(As*fy)*(d-a/2)

Fy= 60
fc'= 3
b= 12
phi= 0.9
d= 18
mu= 1.2 dead + 1.6 live = 185 kip ft = 2220 kip-in.
solve simultaneously,

I get As req'd = 2.94in^2
 

RE: R.C. Beam question

not following you EIT?  

RE: R.C. Beam question

If I use phi = 0.9, I get 2.67in^2.

Mu = 185 k-ft = 2220 k-in

a = (As)(fy)/(0.85f'c(b)) = 1.9608As

Mu = phiMn = 0.9(As)fy(d-a/2)
2220 = 0.9*As*60*(18-a/2) --->   a = 36-82.22/As

Set a=a, and solve As, I get As = 2.67.

I would do a design with Rn, not solving the quadratic, but I did that to be consistent with your method.

RE: R.C. Beam question

But the train isn't high enough to get you to a phi of 0.9, so phi ends up being 0.85, and the required steel ends up being 2.86.

RE: R.C. Beam question

Refresh me...I can't remember the parameters for determining "Phi"...for bending, per ACI 318-02, I started off with phi=0.85

RE: R.C. Beam question

oops!!
318-02
Phi for tension controlled is 0.90.
Damn!! I wish they'd quit screwing around with these factors!!!!!!

RE: R.C. Beam question

If the tensile strain at the furthest line of steel is 0.005 (or greater) you can use phi=0.9.  If you have less strain that 0.005, it slides from 0.9 down to 0.65 (for compression controlled sections) with a tensile strain of 0.00207 (epsilon sub y).

RE: R.C. Beam question

how are you determining the strain as 0.004408?

RE: R.C. Beam question

a=6.2", Beta1 = 0.85, therefore, c=6.2"/0.85 = 7.29"

By similar triangles, epsilon steel/(d-c) = 0.003/7.29

therefore epsilon steel = (d-c)*(0.003)/7.29 = 0.0044

RE: R.C. Beam question

a=6.2"?
 

RE: R.C. Beam question

how can you determine "a" prior to solving for the Area of Steel?
The statement of the problem was to find the Area of Steel. Isn't "a" a function of that?  

RE: R.C. Beam question

I'm not solving a first.  I'm using the a based on the required steel to get the strain.

RE: R.C. Beam question

(OP)
That is what I was getting at.  I was trying c = 0.375d & a = beta * c.  then a = 0.375*beta*c.

with this you now have a, not as a function of As.

But you still have 2 equations you must satisify. C = T, & Mn = (C or T)(d-a/2).   

RE: R.C. Beam question

T=C is satisfied by a=Asfy/(o.85(f'c)b).  That is by definition T=C.  

c does not = 0.375d.  You should never start with that assumption.  

Here's how I would do it from scratch.

Mu=2220K-in
Rn,req'd=Mu/(phibd^2)=0.634431--> assume phi=0.9 check later

rho req'd = 0.85f'c/fy *(1-[(1-2Rn/(0.85f'c))]^0.5)= 0.01238

rho min=0.0033<rho req'd, ok
As,req'd = 0.01238*12*18=2.67 --> try 3#8's =

a=2.67*60/(0.85*3*12) = 5.24"
c=a/0.85 = 6.2"
epsilon steel = (d-c)*0.003/c = 0.0057 > epsilon yield, ok
required steel = 2.67in^2.  

Once you pick 4 #8's, however, a goes up, c goes up epsilon steel goes down, phi goes down and As,req'd goes up.

RE: R.C. Beam question

Agree with EIT.  I would never begin with the requirement that the steel tensile strain be exactly 0.005.  I would begin with the requirement that the minimum steel strain be 0.005, to keep things tention controlled.  With a strain that must equal exactly 0.005, the required steel is 2.93 in², but who cares?  

RE: R.C. Beam question

Is this a Lindeburge book?  I got his for the FE and it was pretty good.

RE: R.C. Beam question

Also, how would you answer the question?  Would you say 3, since that is MOST NEARLY the req'd area of steel (but the beam would fail) or 4, since that is what you'll actually need (but isn't MOST NEARLY the req'd area of steel)?

RE: R.C. Beam question

Wow!  Learned something new already today.  That is why I keep coming back.

I was taught concrete design as we were changing from Working Stress to Ultimate Strength Design.  They weren't sure it was going to catch on and wanted us to be prepared for either to take precedence.  We solved for As by trial and error (slide rule accuracy), but one of the textbooks referred to an exact solution that could be achieved by solving a quadratic equation.

When I developed my concrete design spreadsheets, I sought out that equation and set up my sheets so that I would have the "exact solution".

Today I learned that phi is not always 0.9.  A search of the Code indicates that it changed in 2002.  Startling!.

And I learned that my answer is wrong.  It is little consolation that when you round up As to the next increment of bar sizes, we get the same answer.  That won't always be the case.


DWHA - this thread has revealed that you would not be the only one stumped by this problem.


The one thing I really liked about concrete design is that there was no one right answer and you couldn't just pull the most economical section out of a table.  There was some art melded with the science.

I usually make conservative assumptions about loads so that I don't have to worry if something calcs out 2% or 4% over.  But with the possibility that phi can vary so drastically, I need to rethink my fundamental approach to these calculations.  Thanks for enlightening me.

gjc
 

RE: R.C. Beam question

EIT-
I have to admit, I never have adjusted Phi, or ever had to.
What are the parameters for adjusting it?
Strain less that 0.005, then use ????

RE: R.C. Beam question

If the strain is less than 0.005, then it varies linearly from 0.9 (at 0.005) to 0.65 (for ties members, and 0.7 for spirally reinforced members) at a strain of 0.00207 (ACI let's you use 0.002).  Anything less than that is automatically "compression controlled" and phi is 0.65 (or 0.7).

RE: R.C. Beam question

The problem states the the steel is Grade 60.  To determine the stress at a strain of 0.005, you would need the stress-strain curve for the steel, but you would be beyond the straight line branch of the curve.  For the purpose of design, it is usual to assume that the stress remains constant at Fy.

BA

RE: R.C. Beam question

To further BA's very good point....
The strain of 0.005 simply ensures that the steel will yield.
Fy= 60 ksi is a minimum yield stress. In looking at a normal stress-strain diagram yield stress is achieved over a fairly short portion of the curve. That is to say, the steel is still capable of carrying higher loads for considerably higher strains until the ultimate stress is achieved, which will coincide with the ultimate strain. The ultimate strain is appreciably more than the yield strain.   

RE: R.C. Beam question

It is common to assume that steel never goes higher than yield.  Any strain over fy results in perfectly plastic behavior of the rebar.  

The reason for the 0.005 strain requirement is to allow for enough ductility to be built into the member to allow large cracks and excessive deflection prior to failure.  It should also be noted that the tensile strain under consideration here is the strain in the row of steel closest to the tension face - not the strain at the location of the average d (which would be at a different location if there are multiple layers).

RE: R.C. Beam question

I still haven't heard how anyone would actually anwer the question.  I am very interested in the expected answer.

RE: R.C. Beam question

The tensile strain in the steel doesn't actually reach 0.005.  In order to ensure ductile behavior, 0.005 is the minimum tensile strain the steel can have when the concrete reaches it's compressive failure strain of 0.003.  However, the concrete never reaches it'e compressive failure strain, so the steel never reaches a strain of 0.005.  In this example, at factored loads, the concrete strain is 0.00095, and the steel strain is 0.00182.

RE: R.C. Beam question

SEIT
Not familar with this exam, but although the correct answer according to the wording of the question may be 3 bars, this would not cater for the applied moment.
As engineers we have to be practical thus i would round up to 4.
In this exam can you add comments to the answers that you circle. If so i would go for 4 and breifly explain why.

RE: R.C. Beam question

miecz-
How did you determine those strain conditions?

patswfc-
I thought that for the multiple choice it's either right or wrong, no partial credit.  

RE: R.C. Beam question

It's important to note that the code does not allow a strain smaller than .004 (at the strength limit) for steel in non-prestressed flexural members.

RE: R.C. Beam question

I used 3 ksi for the concrete stress.  Maybe I should have assumed 0.85*3 ksi, in which case,  the concrete strain is 0.00081, and the steel strain is 0.00155.  I was trying to correct the error in my previous post, where I said the minimum steel strain was 0.005.  See the attached worksheet.   

RE: R.C. Beam question

That assumes a linear stress-strain relationship.  That might be more appropriate for service level loading analysis, but that's not valid for strength level analysis, is it?  That assumes a linear stress-strain relationship, which breaks down for concrete at failure.  That's the whole reason for using the rectangular stress block.  If it were triangular, that would be easy, the rectangle is to approximate the parabolic (or whatever shape it takes, - we studied the Hognestaad model in a grad class) stress-strain relationship at failure level loading.  

I agree that the concrete doesn't reach a strain of 0.003 at it's maximum stress, but it won't fail (in compression)until (theoretically) it reaches that 0.003 strain.  This what we're concerned with.

Am I missing something?

RE: R.C. Beam question

EIT-

I agree that my approach assumes a linear stress-strain relationship and that doesn't seem appropiate for strength level analysis.  I'll have to take another look at this, but I can't today.  Got work to get out.

RE: R.C. Beam question

To answer the "Most Nearly" question, I would say 4 since the steel required is for a beam with capacity to carry the moment. Three (3) does not carry the moment, so it does not matter if that is closest to the calculted number.  It just doesn't work.

But that most nearly part always throws me off with these tests.

RE: R.C. Beam question

EIT-

Do you have PCA's Notes on ACI 318? Section 6 explains this pretty well.  My worksheet is based on Example 6.1.

RE: R.C. Beam question

I have the '05 version.  It shows the concrete strain at 0.003, and the steel strain at 0.00523.  Is this the one you're referring to, or do you using a different version?

RE: R.C. Beam question

As I recall, the strain is deemed to be linear but the stress-strain curve is not linear.  The actual stress curve is curved but is approximated by the Whitney stress block which is rectangular.  I believe that a curved block is used in Europe.  The Canadian code is similar to ACI although there could be minor differences in the phi values.  

BA

RE: R.C. Beam question

BA-
Right, plane sections remain plane, so the strain diagram is linear, but the stress-strain relationship of the concrete isn't.  Either way, we always assume a concrete strain of 0.003 when determining steel strains right?

RE: R.C. Beam question

EIT-

I have the '02 version, but example 6.1 has [ε]s=.00523.  Interesting, there is a article in the section called "Tension-Controlled Sections and Transition" that shows the area of steel required at the transition point between tension controlled and compression controlled.  The area required at this point for the problem above is 2.93in²/ft.  I believe that [ø] is 0.9 for any area of steel less than that amount.

RE: R.C. Beam question

SEIT,

Our code used to assume a maximum concrete strain of 0.003 but the 1994 version bumped it up to 0.0035.  I'm not sure why and I don't know what the current code says about that.

What I normally do is guess at the depth of compression block, then calculate As required based on my guess.  I then calculate the depth of compression block required to match T = As*Fy.  If it is greater than my first guess, which it almost never is, I recalculate As.  The reason I do it that way is because I can't be bothered to solve the quadratic equation.

Ordinarily, I don't worry about the strain in the concrete, because I choose under-reinforced beams whenever I can.  The maximum concrete strain, whether 0.003 or 0.0035 combined with the strain at first yield of the steel i.e. Fy/E determines the location of the neutral axis based on a straight line strain diagram.

To consider a strain of 0.005 in the steel seems to be a strange thing to do.  I am wondering why the examiner put that in the exam question.   

BA

RE: R.C. Beam question

miecz-
Are you referencing the problem in the OP?  If so, I agree.  What I was noting is that you can't select an adequate number of #8's and be under 2.93, therefore phi will go down just by selecting the appropriate number of #8 bars.  If you keep it as a req'd area of steel then phi is 0.9, because it's less than 2.93.

RE: R.C. Beam question

Quote (StructuralEIT):

What I was noting is that you can't select an adequate number of #8's and be under 2.93, therefore phi will go down just by selecting the appropriate number of #8 bars

Somehow, I missed that. One would think that if 2.93 in² is adequate, then 3.14 ² would surely suffice, but you never know.  I once had a case with a bridge stringer that was composite in the negative moment region and worked with a certain amount of deck reinforcing.  But, if we added reinforcing to the deck, it moved the plastic neutral axis away from the compression flange, which made the section non-compact, and effectively dropped the strength of the section by 30%, according to the (AASHTO) code.

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