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
RE: R.C. Beam question
RE: R.C. Beam question
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
RE: R.C. Beam question
0.9 times their As equals 2.64 ~ 2.67.
gjc
RE: R.C. Beam question
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
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
RE: R.C. Beam question
RE: R.C. Beam question
RE: R.C. Beam question
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
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
RE: R.C. Beam question
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
RE: R.C. Beam question
RE: R.C. Beam question
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
RE: R.C. Beam question
RE: R.C. Beam question
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
RE: R.C. Beam question
The statement of the problem was to find the Area of Steel. Isn't "a" a function of that?
RE: R.C. Beam question
RE: R.C. Beam question
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
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
RE: R.C. Beam question
RE: R.C. Beam question
RE: R.C. Beam question
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
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
RE: R.C. Beam question
RE: R.C. Beam question
BA
RE: R.C. Beam question
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
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
RE: R.C. Beam question
RE: R.C. Beam question
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
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
RE: R.C. Beam question
RE: R.C. Beam question
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
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
But that most nearly part always throws me off with these tests.
RE: R.C. Beam question
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
RE: R.C. Beam question
BA
RE: R.C. Beam question
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
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
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
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
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.
RE: R.C. Beam question
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RE: R.C. Beam question
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BA