AASHTO Rho min help needed
AASHTO Rho min help needed
(OP)
Hey all, I'm having troubles finding the minimum reinforcement required for AASHTO. I'm more familiar with ACI, and those requirements are 200/fy or 3f'c^0.5/fy. I've found in section 8.17.1.1 that minimum reinforcement should provide enough strength to develop at least 1.2Mcr. The only equations i've found for Mcr is in 8.16.5.2.7 and that is for compression members, whereas i'm designing for flexure. In the past i've used a spreadsheet that my "elders" have developed and the equations all have the value "m", and they claim they had tables based on the values of m and it made life easy. I learned LRFD design in school and my textbook covers only ACI code, so it's useless as well. Does anyone care to point me in the right direction or provide a formula? I'm aware of the 4/3 rule for rho, but i need to determine min rho as well. I need to know how to do this by hand when i eventually take the SE exam. Thanks!
BTW, the Mcr formula in my textbook is fr*Ig/yt I'd like to back out a rho min to make life a bit easier in the future. There is surely an easier way than calculating the yt for each section when checking for minimum reinfocement, but maybe not. Sorry if this seems trivial, but ACI has always been my friend :)
BTW, the Mcr formula in my textbook is fr*Ig/yt I'd like to back out a rho min to make life a bit easier in the future. There is surely an easier way than calculating the yt for each section when checking for minimum reinfocement, but maybe not. Sorry if this seems trivial, but ACI has always been my friend :)






RE: AASHTO Rho min help needed
RE: AASHTO Rho min help needed
RE: AASHTO Rho min help needed
RE: AASHTO Rho min help needed
rho min = .85f'c/fy*[1-sqrt(1-(h/d)^2*1/(.255 sqrt f'c))]
h = total height or depth of member being checked
d = depth from compression face to centroid of reinforcing steel
additonally 1.2Mcr = 1.2*7.5*sqrt(f'c)bh^2/(6*12000) =
bh^2sqrt(f'c)/8000
answer is in foot kips
works every time
be sure that phi*Mn exceeds 1.2Mcr
if this is the case, there is no need to check rho min but I have added that calc in any event.
RE: AASHTO Rho min help needed
Thanks for that equation. I am curious how it was derived and will look into if further, considering all of the equations i've separated rho from ended up being quadratic. I prefer not to blindly use equations without knowing how they are derived, so i'll see if i can reproduce that exact equation.
RE: AASHTO Rho min help needed
RE: AASHTO Rho min help needed
RE: AASHTO Rho min help needed
http://picshome.com/v2/download.php?id=ABC0381C1
RE: AASHTO Rho min help needed
It was all done by hand. In fact, I kept the proof and I am looking at it now.
My first line in the proof says:
Mu = phi*rho*fy*bd^2/12 - 0.5*rho*fy/(.85*f'c)*(phi*rho*fy*bd^2)/12
I did not factor the expression, I just used the quadratic formula after I arranged the equation into a(rho)^2 +b(rho) + c = 0
I manipulated it such that a = 1
b - -1.7f'c/fy
and c = 24Mu(.85f'c)/(phi*(fy)^2bd^2)
and solved for rho. where phi = 0.90
after solving for rho, I substitued Mu - 1.2Mcr where Mcr = h^2*7.5(root f'c)b/6
and 1.2Mcr = 1.2h^2*7.5*(root f'c)b/(6*12)
You will find that the 0.255 value is not rounded and is exact.
Maybe you used that famous 0.59 factor instead of its actual derivation of 0.50/0.85. I kept this division througouht the proof until it resolved and fell out during the algebra.
I will try to make a pdf of my proof and post it here in this forum.
RE: AASHTO Rho min help needed
RE: AASHTO Rho min help needed
RE: AASHTO Rho min help needed
http://picshome.com/v2/download.php?id=0A05B4131
Hopefully others can open this, let me know if you cannot
RE: AASHTO Rho min help needed
That's it and you can see that the 0.255 is an exact number.