# Circular Footings



## Blu1913 (Sep 25, 2006)

Heres another one...

SITUATION

A 10-foot diameter circular footing bears at a depth of 5 feet in clay. The clay has a total unit weight of 110 pcf and an unconfined compressive strength (qu) of 0.8 tsf. Assume vertical concentric loading.

REQUIREMENT

What is the ultimate bearing capacity?

A. 4.5 ksf

B. 5.5 ksf

C. 10.4 ksf

D. 1.1 ksf

SHOW WORK!!!


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## 3gorgesdam (Sep 25, 2006)

assume short ton is used.

c=Suc/2=0.8*2/2=0.8k/sf

qult=cNc+rDf=0.8*1.2+110/1000*5=1.510k/sf

It doesn't match any of the solution...Anyone???


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## EL Nica PE (Sep 26, 2006)

Using Vesic bearing (No typical in practice) factor.

C=Su/2= 0.8ksf since 1tsf=1/2ksf

5.1x1.2x0.8ksf+.11ksfx5 =5.45ksf

roundoff 5.5ksf

B


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## 3gorgesdam (Sep 26, 2006)

> Using Vesic bearing (No typical in practice) factor.
> C=Su/2= 0.8ksf since 1tsf=1/2ksf
> 
> 5.1x1.2x0.8ksf+.11ksfx5 =5.45ksf
> ...


By definition:

1 ton = 2 kips not 1/2 kips.


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## Road Guy (Sep 26, 2006)

i get something very similar

q[SIZE=8pt]u[/SIZE]= 800(5.7)+110(10)

=4560+1100 = 5600

=5.6K/SF


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## 3gorgesdam (Sep 26, 2006)

> i get something very similar
> q[SIZE=8pt]u[/SIZE]= 800(5.7)+110(10)
> 
> =4560+1100 = 5600
> ...


please explain where the How you got Nc (5.7) and Df (10). Thanks. I got that from CERM. Confusing.....


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## Blu1913 (Sep 27, 2006)

5.7 Comes from Terzaghi.

I dont know why the depth of footing is 10. THe problem states it to be 5 feet. Does that have to do with it being clay or something? Take the depth X 2?


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## Road Guy (Sep 27, 2006)

ok I'm dumb Df should have been 5'

thats what I get for trying to work a problem while my kids are at the table eating a late night bowel of cereal..


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## Blu1913 (Sep 27, 2006)

OK, so figure this. The answer is indeed 5.5.

If you use Df of 5, the answer is not correct. Any ideas...again they dont show how to do the problem, only the answer....


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## EL Nica PE (Sep 27, 2006)

> By definition:
> 1 ton = 2 kips not 1/2 kips.


Thanks

3gorgesdam

let me make it clearer what I meant to write.

I forgot the the 2

1 tsf (Qu)= (1qu/2)x2 (conversion factor) =1ksf (Su)


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## EL Nica PE (Sep 27, 2006)

Here is the equation for clay only.

SuNcSc+ gDf(Effective Overburden)Nq

Su=Shear strength of the soil or if unconfined strength is given then Su=qu/2

Nc= bearing factor (Terzaghi or Vesic) typical 5.7 or 5.1, respectively. Typically Terzaghi is use.

Sc= Shape factor

g= unit weight (effective if water is above )

Df= foundation depth

Nq=bearing factor 1 if phi =0

I hope this help.


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## 3gorgesdam (Sep 27, 2006)

> assume short ton is used.
> c=Suc/2=0.8*2/2=0.8k/sf
> 
> qult=cNc+rDf=0.8*1.2+110/1000*5=1.510k/sf
> ...


I found my own errors. The shape factor needs to multiply Nc which is 5.14. So the answer is:

c=Suc/2=0.8*2/2=0.8k/sf

qult=cNc+rDf=0.8*1.2*5.14+110/1000*5=5.48k/sf=5.5k/sf


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## Blu1913 (Sep 27, 2006)

well done...

It seems the Quick reference screws you by not including the multipliers in the equation.


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