Retainging Wall Sliding Factor Problem 1

[kenrbc]

I have 10' high concrete block 2.5'x2.5'x5' (W,H,L). DL= 3750#/ft
P= .5(.32)110(10)^2 = 1760#
When I check the sliding factor, then I never get the safety value over 1.5.
u = .5;
F = .5 x 3750 = 1875

Sliding factor = 1875 / 1760 = 1.1 Which is less than 1.5

Does anyone know why? or any other method to solve sliding problem?
TIA

 

[KBVT]

Don't have time to look into your calcs, but if you can't satisfy sliding without making your footing ridiculously big, I would suggest a shear key on the bottom of the footing.

 

[youngstructural]

Sliding by considering friction is a gross approximation, personally I only rarely use it...

You need to remember that the coefficient of friction is a dimensionless simplification used by Physicists; It does not really relate well to the physical world without empiracle data to back it up.

It would be much better to consider the full, formal, formulation:

Rs = 0.8·[(0.9·tandb) + (Kp·0.5·g·(k+d+s)^2) + (2·Cb(k+d+s)?Kp) + (Cb·B)]

where
Rs = Factored Sliding Resistance
db = active resultant angle (typically 1/3 to 2/3 of internal angle of friction for the BEARING soil)
Kp = Coulomb passive pressure coefficient
k = Depth of key
d = Depth of footing
s = Depth to top of footing from ground surface
Cb = Cohesion
B = Sum of Heel & Toe Lenghts (ie: Full fooling lenght in cross section)

Let me know if you have a hard time interpreting this...

And Good luck!
Cheers,

YS


B.Eng (Carleton)
Working in New Zealand, thinking of my snow covered home...

 

[dougantholz]

Are you looking at your passive pressure pushing back? Don't forget that takes away from your net horizontal load.

 

[kenrbc]

These are huge concrete block size 2.5'x2.5'x5'
They are stacked on top another. Crushed rock is located at the bottom. There is no concrete footing.

 

[JedClampett]

If you don't have a footing, I don't think you're ever going to get 1.5. Traditionally, I add to the heel of the footing to get more weight of soil to increase the friction. If the wall hasn't been built, maybe you can add some layers of geotextile to tie the wall beyond the soil wedge. And .5 is a pretty large friction factor. If you can't get 1.5 with that, you're in trouble.
To develop passive pressure, you need to have the wall move some, so be careful.

 

[RWF7437]

In my neighborhood ( Oregon ) these blocks go by several different names. "Ultra Block" is one of them.

The simple solution to your problem may be to simply rotate the lowest course of these blocks so that the 5 foot dimension is perpendicular to the exposed face of the wall. Then check for sliding based on this configuration. If that doesn't work you may need to rotate two courses to get a F.S. of 1.5 or you may need to embed the first course so that you can get some passive pressure at the toe of the wall.

There is some good design guidance usually available from the Block supplier.

If you can't get this let me know.

good luck

 

[RWF7437]

You can also improve the performance of such walls by designing them to be mechanically stabilized earth (MSE) walls employing geogrid fabrics.

Ignore cohesion to be conservative since "c" depends on soil moisture content remaining constant.

good luck

 

[civeng80]

I would agree with most of the comments here. I would rely on passive pressure either with backfill in front of wall or a shear key developing it.

Probably 1' of backfill in front would be ample to develop sliding resistance.

 

[asixth]

Sliding is something that I always have trouble with satisfying and have often thought to myself that maybe a friction co-efficient of 0.5 is a poor representation.

Generally passive resistance is ignored because there may be an excavtion infront of the wall (or block in your example) sometime in the future.

I would also like to hear of whether others are using a higher friction co-efficient than 0.5.

 

[youngstructural]

Friction is a poor representation? Where have I heard that before?

It's not like I said that two days ago and bothered to type in a full lenght cohesion, passive pressure, etc, formula...

Oh well, maybe if my name was YoungGeotech... lol!

I just hope that kenrbc can solve the problem. And I agree with you asixth, for the record, that passive pressure (ie: in the Kp·0.5·g·(k+d+s)^2 term) would solve this problem very well.

Cheers,

YS

B.Eng (Carleton)
Working in New Zealand, thinking of my snow covered home...

 

[dougantholz]

Looks like RWF had several good ideas - and you could also sink the first course 2 or 3 feet and develop a lot of passive pressure. If passive is usually 3-4 time active then it would only take 1/3 to 1/4 the embedment to balance active. Of course you then also have to worry about toe pressure and overturning - but one problem at a time please.

 

[RWF7437]

http://www.ultrablock.com/ultrablock_2_002.htm

 

[apsix]

YS
What you typed is a fairly radical departure from what is typically used in my experience. With all respect, I think it would be foolhardy to use a 'radical' formula someone posted on a forum without some independent verification.

Can you supply any references?

(Please don't let my post deter you from continuing your excellent contributions here.)

 

[youngstructural]

Aspix;

Despite my nasty combination of French, Irish & Scottish temper, I do have faily thick skin... And an unfortunate preclivity for sarcasm.

In all seriousness, I honestly think that many of the failures for retaining walls that I have heard of were for the most part caused by an over-simplification of the problem. Friction is NOT an appropriate solution for anything but the most routine of retaining walls, and this is NOT a routine retaining wall.

The formula is from "Principles of Foundation Engineering, Fifth Edition" by Braja M. Das. It can be found on page 347, and is used in the worked example on page 350. I believe the same formula (or one very similar) is found the other common foundation text by Bowles.

Regards,

YS

B.Eng (Carleton)
Working in New Zealand, thinking of my snow covered home...

 

[apsix]

Thanks YS

I think I have Bowles in storage, I'll try to dig it out over the weekend.

Weight of the wall doesn't seem to be included in the formula, is that correct?

Cheers

 

[youngstructural]

Ouch... Missed a term! What were you saying about not using a formula posted to a web forum?!?

The formula should have included 'G' for the total vertical load next to the first tan, AND also explained that the little g I have used is Gamma for the soil density involved. Thus the formula should be modified like so:

Rs = 0.8·[(0.9·G·tandb) + (Kp·0.5·g·(k+d+s)^2) + (2·Cb(k+d+s)?Kp) + (Cb·B)]

where
Rs = Factored Sliding Resistance
G = Total vertical load acting down
db = active resultant angle (typically 1/3 to 2/3 of internal angle of friction for the BEARING soil)
Kp = Coulomb passive pressure coefficient
g = unit weight of soil (kN/m^3)
k = Depth of key
d = Depth of footing
s = Depth to top of footing from ground surface
Cb = Cohesion
B = Sum of Heel & Toe Lenghts (ie: Full fooling lenght in cross section)

Try that one on for size; My profuse appologies for the oversight!

Oh, and a little further information: The first term can be thought of as something akin to friction (the effect of the vertical load), the second and third terms are the passive pressure resistance, and the final term is the force due to adhesion between the footing and the supporting ground.

Cheers,

YS

B.Eng (Carleton)
Working in New Zealand, thinking of my snow covered home...

 

[OldPaperMaker]

kenrbc,
I think what most of the responders are trying to tell you is that there is no way in hell that a 2.5' wide series of blocks is going to retain 10' of soil.
Sliding is generally the major problem with any retaing wall. But, with your wall, if you assume a lateral soil pressure of 35 pcf the overturning moment is 5833 ft-lbs and the righting moment is 4688 ft-lbs, which gives a factor of safety of 0.8. I think that this means that the wall will tip over.

 

[cap4000]

YS

I checked Das' 3rd edition book and on page 318 its has a simplified formula compared to yours. Assuming no cohesion taken, the formula works out to what most engineers use and the IES QuickR Wall and Ram Advanse Retaining Wall Program uses. It uses the tangent angle of the bearing soil. For 30 degrees its 0.577. Typically 0.5 to 0.6 is used. I gave you a star for your nice work.

 

[kenrbc]

Thank you for all responses. The wall was constructed.
I used equivalent active pressure 35psf/ft and 250psf/ft passive pressure.
Since there is lots of room remains in front of the wall, therefore I used 4ft soil to restraint the sliding.

How come there are many Ultra Block walls that taller than my wall still standing? I assumed they must have tie back huh?