AASHTO Rho min help needed

[buening]

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

 

[buening]

Ah nevermind, i think i'm overcomplicating it.

 

[buening]

Equation 8-2 specifies Mcr which is the same as my textbook. I then used Equation 8-15, substituting Mn for 1.2Mcr and solving for rho. Thanks for letting me talk to myself LOL

 

[shin25]

The equation for cracking moment that you have described as fr*Ig/Yt- the Yt should be Yb, which is the distance from the cetroid of the section to the extreme tension fiber of the uncracked section at locations of positive moment.

 

[minorchord2000]

based on the requirement that Phi*Mn must exceed 1.2Mcr, I devised my own equation for rho min and it has been checked and double checked by independent bridge engineers.

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.

 

[buening]

Yeah i used the Mn equation and substituted the equation with rho for As (rho*b*d=As), but that leaves a quadratic equation. My basic thought was to calculate my phi Mn and then back out an As assuming jd=0.9d and then solving for a until the As approaches the same value. Then calculate Mcr (fr*S) and compare that value to phi Mn. If phi Mn is less than Mcr, then i increase my rho provided by 4/3 according to AASHTO.

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.

 

[minorchord2000]

I derived it using extensive algebraic manipulation. I welcome someone else to do it.

 

[buening]

minorchord, attached is the proof to your formula using the equation of 1.2Mcr=phiMn I'm having problems getting your 0.255 coefficient. I basically broke you equation down to a quadratic equation and manipulated it to be the same as the formula for 1.2Mcr=phiMn. Is your 0.255 coefficient rounded? Not that it makes a huge difference, but just curious. Also, did you factor that equation by hand or did you use a computer program? Impressive if it was done by hand LOL.

 

[buening]

Ok, so apparently it didn't show the picture. Here is the link

http://picshome.com/v2/download.php?id=ABC0381C1

 

[minorchord2000]

Buening:

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.

 

[buening]

It appears you are using a different equation for "a". The equation i am using for phiMn is 0.85*As*Fy*(d-a/2). For a=As*fy/(0.85*f'c). I then subsitute rho*b*d for As, which yields a=rho*d*fy/(0.85*f'c). The value for b drops out of that equation. I guess i wasn't using Mu, rather phiMn as AASHTO states. Using phiMn, the 0.85 cancels when you factor out 0.85*As*fy*a/2 using the above equation for a.

 

[buening]

Ah, it appears you used phi=0.9 whereas i used 0.85 After looking at AASHTO, 0.9 is the correct value for tension. My bad, this is probably the cause of the slight coefficient difference.

 

[buening]

The following link has my proof, and as you already know it does check out.

http://picshome.com/v2/download.php?id=0A05B4131

Hopefully others can open this, let me know if you cannot

 

[minorchord2000]

Buening:

That's it and you can see that the 0.255 is an exact number.