NCEES Civil PE Sample Exam #518 & 519

[ktulu]

NCEES Sample Exam Caisson Problems

I am having a bit of trouble figuring out these two problems. I do not deal with caisson at all...

Any help would be greatly appreciated.

ktulu

 

[jvg2010]

I am having the same problems with problems #518 and #519 of the 2000 NCEES Geotech Afternoon problems. I can not find anything on caissons. Can anyone help?

 

[oz oz]

Hi,

I am also preparing for April PE Exam (Geotech Module).

I do not have NCEES sample exam study material therefore I do not know the exact content of your question but as far as I know Bowles (5th edition) is a good source.

It gives some sample questions and explanations for drilled piers and caissons in his book which I am planning to take to exam as a reference guide just in case.

By the way, I am planning to take CERM, Navfac 702-703, Six-mins (geotech+structural) and Bowles to the exam. I hope these are enough.

What are you planning to take?

Regards,

Oz

 

[jamie]

I'm not sure how they solve it, and my approach my not be what they were thinking of. but I would look at the caisson as being a cantilevered beam fixed at one end subject to a uniform load (need to take the portion of the load normal to the surface). There is a standard equation in structural refereneces for max bending for these conditions. if they're 5 feet on center, then you'd want to look at the tributary area for each caisson as contributing to the beam force. Then you have to use superposition for the tie back tension to get the total bending. I'm not the strongest at these, so I'm not sure what you should end up with, but I think that's how you'd go about doing it.

as far as the anchoring, basically they're telling you you get 10 psi of frictional force max from the rock. so figuring you have a 6" diameter hole, you need the total surface area (the perimeter times the length) for the embedment into the rock to equal or exceed the 10 kips tension load. For this part I think you should get 5. What do they come up with?

 

[jamie]

actually, for the length they should get 7, since there's a factor of safety of 1.5.

 

[Badger]

Hi,

Good question, here are my solutions:

Problem 518: You need to compute the moment at Point A.

Given: tieback tension T= 10 kips; Tieback; Landslide pressure p = 1.0 ksf; the depth of the active landslide is 18ft; slide plane and ground slope are at 3:1; the tieback acts at a 2:1 angle from the top of the caisson. And from the diagram the caissons are on 5 ft centers and therefore each retain a 5 ft wide section of the active landslide.

Compute horiz. tension, Th x cos (angle) from the 2:1 triangle the hypotenuse is the square root of 5.

So Th = (10 kips)(2/(5^.5)=8.94 kips; the moment arm r= 18’ +2’+1’

Now find P = p(Area active landslide) P= (1.0ksf)(5’ x 18’) =90 kips. P horiz. = Ph

Ph = Ph cos(angle) = (90 kips)(3/(10^.5)) =85.38 kips. R= (1’+2’+18’/2)

Ma=0 so M(caisson) = Ph® - Th® = (85.38 kips)(12’) - (8.94 kips)(21’) = 836.8 ft-kips, answer C.

Problem 519: Find: Le (length extended into rock)

Given: fictional resistance, fr=10 psi; dia. of grouted hole = 6”; FS =1.5, and T = 10 kips.

T= Le(perimeter of the tieback)(fr)/FS The perimeter = (pi)(dia)

10 kips (1.5) = Le(pi)(6”)(10 psi) Le = 15,000 lbs/(188.5 lbs/in) =79.6 inches or 6.6 ft. Answer (B) 7

I had difficulty with all the extra information, and also forgot to add the 2' space between the active slide and the rock. I like this problem, you don't need any references, but there are plenty of places to miss something, so it is good practice.