6 mins enough for PE Civil struc PM per Prob?

[STEEL MAN]

i tried solving problems in RC, steel from NCEEs sample questions, it seems it will took me more than 6 mins for Struc depth specially if it is code dependent, I dont think this isnt right. Let me know what your think guys for those who took it already?

 

[playboyman007]

Which problem in particular are you having the most trouble with?

 

[pmblair]

Just started my depth review, got a late start on it.

I do have the 6 min solutions, but for now i am going through the TAMU stuff on Statics and Strength pof materials first For a review of the basics.

 

[pmblair]

Just started looking through the 6 min solutions..... this might cover it. I am not sure how strong you are in statics/ mechanics of materials/ and structures. I think the CERM/ Six minutes/ and the codes will be adequate. But with out ever taking it, who knows.

 

[STEEL MAN]

I dont have a problem with statics and mechnics even with moving loads. But to give a quick example.

Calculate the Area of steel in RC beam given, Lspan=30 feet Wdl=4kip/ft Wll=3 Kip/ft the beam is fixed end, design midspan. B= 20inches H= 36" D=32". Honestly can you finish this in 6 mins?

In steel design axial + bending member of a W section with P dl=, Pll Pwind load..can this be done in 6 mins?

If you can do this in 6 mins even if you memorize the formulas isnt possible, the solns are quite lengthy. Like in RC calculating rho p involves quadratic eqn. I think I have to get me programmable calculator in less than 2 weeks.

 

[STEEL MAN]

add on to RC prob Fc'=5000psi and Fy=60000psi, theoretical this should be finish in less than 6mins in PE exam, but in reality it is tough to do that. Same with steel design say A36 steel. Fy=36ksi Fu 58 ksi.

 

[bootlegend]

Steel Man,

I think that problem you are describing can easily be done in six minutes. If you don't have the concrete design formulas memorized try downloading the FE reference manual and taking the concrete design section in with you. Work a couple problems like that one and you'll see that once you have the equations right in front of you it is pretty easy.

 

[STEEL MAN]

FE Supplied reference Handbook, yes, I have it too, I can even derive it really quickly and made my quick formulas not found from there, now to show you a quick step.

1. Compute Max moment, load factors considered.

2. Ru=Mu/(phi bd^2) then compute rho based on Max moment

3. Compare rho min=200/fy etc other code required

4. compare which is logical for rho to decide based on code

5. As=pbd

Can you do these in less than 6mins?

 

[STEEL MAN]

In steel design

1. Compute loads Mu based on gravity and wind loads daw shear and moment.

2. design for flexure, you can use look up tables from FE supplied reference handook or AISC

3. for axially loaded design members, compute slenderness ratio, determine section properties from AISC.

4. Compute eulers buckling eqn, biaxial eqn, and axially loaded compression member design

5. Use interation eqn.

Can you do these in 6min or less?

 

[pmblair]

Actually i can solve that in 6 min, but i design concrete all the time.

You either iterate ,and use some approximate method to get your As first. Or you can use the quadratic method by solving for #As in C=T first.

But i just started using the new steel code at work, and it two me an hour to check a column with 2 moments and axial load.

I think its relative. We all have certain strengths that we will call on during the test.

I don't think you will see all the parts you shown in the same problem. In the analysis part we do analysis. In the design part we design the members.

I think....

 

[STEEL MAN]

no, you cant do that C=T, that is wrong, you have to do it based on Mu, that approach is totally wrong.

your eqn will look like this. C=0.85Fc'Ba , T=Asfy, this eqn is only valid if you have balance failure, and not based on Mu. And in this case you have to assume an "a" value which is totally wrong and not based on Mu.

 

[STEEL MAN]

that is why I emphasized in checking rho p based on Mu, pmin, and pbal and there's another one 3/4of pbal.

These things Im really good at. Same with steel design.

To me logically NCEES should be clear enough what to solve for the PE Exam. Not like this.

 

[STEEL MAN]

here's a quick formula for rho p based on Mu

Mu/(phi b d^2) = pfy-0.59p^2 fy^2/fc'

you have two values for p but to make it quick use the negative value to solve for

then compare it with pmin, etc....etc.

then As can be solved.

anyways sorry to say that and good luck on your PE exam.

 

[STEEL MAN]

an example an attached file. notice soln 1 is direct approach while soln2 needs trial an error of "a" which is inaccurate and requires more than 6 minutes.

ACI_Rectangular_Beam_Concrete_Design.pdf

 

[pmblair]

Of course you have to check that it is tensioned controlled failure.

"C=0.85Fc'Ba , T=Asfy, this eqn is only valid if you have balance failure, and not based on Mu"

You solve a in terms of As then set that equation to Mu and solve quadratic for As. Then you have to round up to actual steel such as 2#5 or 3#6 or whatever and then check that beam, with that configuration for tension controlled failure. Only in extreme high moments with small sections will you get non tension controlled failure or will you require compression steel.

 

[Ble_PE]

One thing you have to realize Steelman is that the average length of a problem is 6 minutes. That does not mean that you can do every problem in 6 minutes. There will be some problems that you'll be able to answer after reading the question and others will take 10-15 minutes. To me, the problem you describe above is the best type of problem to have. It's very basic design and is simply plug-n-chug.

Good luck!

 

[STEEL MAN]

this discussion is going nowhere my concern was this type of problem cant be done in time as on less than 6mins. even if i use my short cut approach in soln1. the other soln which is Blairs soln2, Im sure cant be solve in 6mins you have to guess "a" and is very inaccurate and time consuming.

 

[pmblair]

There are some common rules of thumb on how to get As within like 5% just based on Mu. When i get in the office today i will edit and add them. It will give you an As to start that most likely will be right due to the need to round to whole bars. You have to also realize if it is a 8" single layer reinforced beam you cant get more then say about 3#6 in a layer and maintain proper cover and clearance. And you wouldn't put anything less then #5 in the beam.

maybe i am taking to many assumptions, i am used to being the one who controls the parameters and on the test that wont be the case.

 

[pmblair]

roe=(.85k1f'c)/(fy)[87/(87+fy)]

k1=.85-.05(fc-4) -------when f'c>4ksi

.75Asb=.75(roe)B(d)

a=(.75(Asb)fy)/((.85F'c)xB)

Mu=theta(.75Asb)fy[d-a/2]------------------Maximum moment with no compression steel.

As=Mu/(4d)

Mu in kip-ft

d=effective depth in inches

round to actual steel sizes.

Solve for thetaMn.

or

Solve for a in terms of As then quadratic for As. Check moment for tension controlled.

Steelman, the more you make me think about it, the harder it does seem to do in 6 min. But i dont think its a 15 min problem. Maybe 8 or 10 i dont know.......

 

[civilized_naah]

roe=(.85k1f'c)/(fy)[87/(87+fy)]k1=.85-.05(fc-4) -------when f'c>4ksi

.75Asb=.75(roe)B(d)

By the way, as of 2002, the ACI code does not use the language of 'balanced reinforcement' anymore. It now bases the reinforcement limit on the tensile strain in the reinforcement, which is not allowed to be below 0.004. This criterion ends up being very close to the criterion (rho_max = 0.75 x rho_balanced), so using either concept to come up with the answer is OK, but just saying that the criteria is stated differently now.