allowable vs ultimate coefficient of friction

[smvk3]

In my soils report, the geotechnical engineer has given me an ALLOWABLE coefficient of friction of 0.3. When using retaining wall software, I input this value. The software calculates all of the dead load of the wall, D, multiplies this value by the inputted coefficient of friction, and gives you a warning when the ratio of 0.3*D / F (where F is the total horizontal lateral force from the soil) is below 1.5. Is this correct? What would be the difference in calculation if I input the ULTIMATE coefficient of friction?

 

[jayrod12]

You would need to be using the factored loading. Allowable is checked against total service loads.

 

[mathcadboy]

There is no allowable and ultimate coefficient of friction.

There is only an allowable and ultimate dead load (your normal force to be multiplied by the coefficient to get the friction force).

 

[smvk3]

The geotech engineer has given me two values for the coefficient of friction; one that is ultimate (which is 0.45), and the another that is allowable (which is 0.3).

If I use the allowable coefficient of friction value (and use the ASD load combo 0.6D + H), I just need to satisfy the inequality 0.3*(0.6D) >= H.

If I use the ultimate coefficient of friction value (and use the LRFD load combo 0.9D + 1.6H), I just need to satisfy the inequality 0.45*(0.9D) >= 1.6H.

Am I seeing this correctly?

 

[msquared48]

Smvk3:

Sounds to me more like a factor of safety of 1.5 at the .3 level than a load factor. Depends how you look at it I guess.

Mike McCann, PE, SE (WA)

 

[mathcadboy]

Best to ask the geotech consultant on how to apply the allowable and ultimate coefficient of friction to avoid any misenterpretations. Its the first time I heard about this and would like to know his answer.

 

[smvk3]

Msquared,

Are you talking about 0.3*D/H >= 1.5? That's how I learned it in my geotech class.

 

[MacGruber22]

Seems reasonable there could be two levels of coefficients of friction, particularly if you approach it from a statistical/structural reliability point of view. Also, bearing pressures are reported as both allowable and ultimate, so why not friction? There is no single friction value - it is based on empirical testing, so there is some statistical distribution (hopefully bell) associated with the results. The farther (left) from the mean you proceed, the reliability of the resistance (in this case friction) is akin to allowable/working stresses, and is thus paired with load effects (combinations) which complement the resistance's reliability.

So I would be careful about believing this statement, without doing some research. Certainly, you are working with a Geotech who feels comfortable providing both. Maybe you can pick his/her brain?

There is no allowable and ultimate coefficient of friction.

mathcadboy - where does it say that there is not one, or that it is not good practice, etc. etc. ?

Smvk3: you might not have them, but don't forget any other load effects which act with the primary effect H. But, in general, I think your on the right page based on the info your geotech has provided,

I am curious now, because I have only ever been provided allowable coefficients. And that is super conservative when used with LRFD loads. :/



"It is imperative Cunth doesn't get his hands on those codes."

 

[smvk3]

I can't get a hold of the geotech engineer that's why I am hoping I can get some guidance here.

 

[MacGruber22]

smvk3 - I personally think you can run with it, and then before your plans go out for construction or permit, get a verification. Or is that day tomorrow? If you really have no time to waste and you don't feel good about it, just check your LRFD loads against the allowable friction. We haven't seen a sketch of your wall (that might be good now), but how much more heel mass do you need to get it to pass that way? If it isn't a lot extra, I wouldn't get worked up over the conservatism. Maybe you have some passive resistance you can use as well?

"It is imperative Cunth doesn't get his hands on those codes."

 

[MikeE55]

SMVK3, this is one reason why I don't like to get the coefficient of friction as provided by the geotech. If you design a retaining wall for 0.3, and then apply a factor of safety of 1.5 against sliding, you will end up with a very very conservative design. I asked this same question once of the engineer who prepared a soils report for us, and he said the 0.3 already contained a factor of safety. I think 0.45 is a lot more realistic number to use, and then apply the factor of safety.

 

[smvk3]

Mike55,

The 0.3 coefficient of friction factor does include a safety factor of 1.5.

So essentially instead of designing for a factory of safety of 1.5 for sliding, you are designing with a safety factor of 1.5*(1/0.6) = 2.5 for allowable stress design. This is what has confused me. In the past, I always thought that you were just given one coefficient of friction value (with no safety factor) and then make sure the ratio of the friction force to total lateral soil load is more than 1.5.

 

[PEinc]

Here is my take on this question:
If required safety factor for sliding = 1.5, N = normal force, 0.45 = ult. friction factor, 0.3 = allowable friction factor, SF = 0.45/0.3 = 1.5, then
Service Load Driving Force / (0.3 x N) > 1.0
or Service Load Driving Force / (0.45 x N) > 1.5

If required safety factor is > 1.5, then Service Load Driving Force / (0.45 x N) > Required SF

http://www.PeirceEngineering.com

 

[MikeE55]

PEinc, yes, the way I have always calculated in the past is to use 1.5 x sliding force < (0.45 x DL). Or to use another value for friction factor based on the type of soil actually used. smvk3, sorry I jumped in without fully understanding your question. I think applying 0.6 to the DL to reduce the amount available to resist sliding is being too conservative if you are also using a factor of safety of 1.5, as you have stated.

 

[smvk3]

PEinc,

Why would you want the driving force larger than your frictional resisting force? Shouldn't your equations be:

Service Load Driving Force / (0.3 x N) <= 1.0

(0.45 x N) / Service Load Driving Force >= 1.5

Am I seeing this correctly?

 

[msquared48]

Or .45/.3 = 1.5

Mike McCann, PE, SE (WA)