R.C. Beam question

[DWHA]

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?

 

[ToadJones]

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

 

[Lion06]

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.

 

[vandede427]

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.

 

[Lion06]

Toad- How did you get 2.86 in^2?

 

[mtu1972]

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

 

[DWHA]

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.

 

[frv]

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.

 

[ToadJones]

what "phi" are you guys using?>

 

[iponom]

huh pretty simple problem no?

 

[ToadJones]

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

 

[Lion06]

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.

 

[ToadJones]

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

 

[ToadJones]

not following you EIT?

 

[Lion06]

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.

 

[Lion06]

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.

 

[ToadJones]

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

 

[ToadJones]

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

 

[Lion06]

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).

 

[ToadJones]

how are you determining the strain as 0.004408?