# Assistance w/ Das: Bearing Capacity Problem 15.1

Started by
Dan06
, Jan 14 2012 08:46 PM

8 replies to this topic

### #1

Posted 14 January 2012 - 08:46 PM

Principles of Geotechnical Engineering (5th) Ed - Problem 15.1

Determine gross allowable bearing capacity. Use Terzaghi's bearing capacity factors.

Part A Given: soil unit weight = 120 lb/ft^3, c'=0, angle degrees = 40, Df = 3ft, B = 3.5ft

My approach:

Eq. 15.12: qu = (1.3 * c * Nc) + (q * Nq) + (1.4 * unit weight * B * Nunitweight)

Table 15.1 (Terzaghi's Bearing Capacity Factors) for given degrees:

Nc=95.66, Nq=81.27, Nunitweight=115.31

qall = qu / Fs, where Fs=4

qu = 48,629 lb/ft^2

qall = 12, 157 lb/ft^2

For part 15.1a I get 12,157 lb/ft^2; however the solutions indicate 13,368 lb/ft^2. I'm off on all subparts of this problem. Any thoughts? Are these type of problems representative of AM Breadth Geotech Problems?

Thanks in advance.

Determine gross allowable bearing capacity. Use Terzaghi's bearing capacity factors.

Part A Given: soil unit weight = 120 lb/ft^3, c'=0, angle degrees = 40, Df = 3ft, B = 3.5ft

My approach:

Eq. 15.12: qu = (1.3 * c * Nc) + (q * Nq) + (1.4 * unit weight * B * Nunitweight)

Table 15.1 (Terzaghi's Bearing Capacity Factors) for given degrees:

Nc=95.66, Nq=81.27, Nunitweight=115.31

qall = qu / Fs, where Fs=4

qu = 48,629 lb/ft^2

qall = 12, 157 lb/ft^2

For part 15.1a I get 12,157 lb/ft^2; however the solutions indicate 13,368 lb/ft^2. I'm off on all subparts of this problem. Any thoughts? Are these type of problems representative of AM Breadth Geotech Problems?

Thanks in advance.

### #2

Posted 15 January 2012 - 07:44 AM

Try using eq. 15.11 since this is a conrinuos footing; 15.12 is for square footing and has shape factors incorporated into it.

So q-ult= cNc+YDfNq +0.5YBNy = 0 + 29257.2 + 24215.1 = 53472.3 psf.

Good luck.

So q-ult= cNc+YDfNq +0.5YBNy = 0 + 29257.2 + 24215.1 = 53472.3 psf.

Good luck.

### #3

Posted 15 January 2012 - 01:19 PM

Success - Thanks Badger!

### #4

Posted 16 January 2012 - 06:34 AM

So, Badger is this mean that the solution in Das Book is wrong....13,368?

### #5

Posted 16 January 2012 - 09:58 PM

no, I think it just means that you need to be careful before using B.C. equation.. this is where you can be tricked!!

### #6

Posted 18 January 2012 - 03:48 AM

geo.....what you mean?.....where's the trick?

### #7

Posted 18 January 2012 - 04:22 PM

The trick is in the type of foundation, if you use the shape factors you get 12,157psf vs if you donot use the shape factors you get 13,368psf. So be careful and read the question again and watch for key words.... I hope it helps.geo.....what you mean?.....where's the trick?

### #8

Posted 19 January 2012 - 03:36 AM

No, but you made me look. I hadn't finished the problem in the reply, only q-ultimate, q-all = 53472/4=13368.So, Badger is this mean that the solution in Das Book is wrong....13,368?

I think in the test they will give you the shape factors or tell you whose to use. If you are in doubt I'd use the factors in the CERM. This book Principles of Geotechnical Engineering (5th) Ed doesn't go into shape factors much. I like the Principles of Foundation Engineering (5th) Ed by Das, it was real helpful to me on the exam. I recommend you getting it. I got used copy off B&N for $50, I still use it every so often.

### #9

Posted 20 January 2012 - 05:40 AM

Thanks Badger!

I got it.

I got it.

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