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Calculating Friction Factor
2

Calculating Friction Factor

Calculating Friction Factor

(OP)
Has anyone ever used this before to calculate the friction factor for a spread footing?  I have attached an excerpt from the Civil Engineer's Handbook (Merritt).

http://files.engineering.com/getfile.aspx?folder=40fa290b-010f-47e9-88a6-a6e7319a6e94&file=Frictional_Resistance.jpg

http://files.engineering.com/getfile.aspx?folder=07aa01b9-55c1-46ea-bbb0-42c7f4f2fd49&file=Friction_Angles.jpg

I am looking at placing about 2 feet of crushed stone under a spread footing to reach a factor of safety of 2 for sliding and overturning.  I have seen other engineers use the tangent of ¾ of the friction angle.  I am thinking of using 38 degrees for the crushed stone, which is at the lowest end for gravel.  This would yield a friction factor of 0.78.  It just seems much higher than I have used before.  I have typically just used the friction factor as determined in the soils report which I have not seen higher than about 0.55 for compacted structural fill; I just do not have a report to use in this situation.
 

"Structural engineering is the art of modeling materials we do not wholly understand into shapes we cannot precisely analyse so as to withstand forces we cannot properly assess in such a way that the public at large has no reason to suspect the extent of our ignorance." -Dr. A. R. Dykes

RE: Calculating Friction Factor

Note that when using a layer between natural soil and the footing the shear force is then passed to the natural soil and hence its lower angle of inner friction will control the slippage check, or something like that.

In any case, effectively codes have become truly conservative respect friction, for as you say .55 is common for gravels, and angles of inner friction sometimes are not allowed to be counted for calculus above 35º.

Again, as the average behaviour, the more high values really attained for friction and inner friction are available as average, and so it must seem that the conservatism in using lower values must be a way of getting characteristical values standing 95% of the times.
 

RE: Calculating Friction Factor

I have very often used 2/3tan(phi) = f
2/3Tan(38) = 0.52

RE: Calculating Friction Factor

In Foundation Analysis and Design, 1968, Page 317, Bowles says to use f = tan(2phi/3)as the coefficient of friction between the base of the retaining wall and the soil.

RE: Calculating Friction Factor

(OP)
ishvaag:  What codes are you talking about that are conservative regarding friction?

"Structural engineering is the art of modeling materials we do not wholly understand into shapes we cannot precisely analyse so as to withstand forces we cannot properly assess in such a way that the public at large has no reason to suspect the extent of our ignorance." -Dr. A. R. Dykes

RE: Calculating Friction Factor

(OP)
PEinc: I have seen the 2/3 number before as well as using 3/4.  Your two calculations are not the same, though.  2/3Tan(38) is not the same as Tan(2/3(38)). I think your second equation from the design book you quoted is correct, right?  So the friction factor would be 0.47 and you would still use a safety factor for sliding and overturning using this value, right?

"Structural engineering is the art of modeling materials we do not wholly understand into shapes we cannot precisely analyse so as to withstand forces we cannot properly assess in such a way that the public at large has no reason to suspect the extent of our ignorance." -Dr. A. R. Dykes

RE: Calculating Friction Factor

I realize that the two equations do not give identical friction factors; but they are similar.

Check an AASHTO Bridge Design Specifications Manual for a table of ultimate friction factors.  For mass concrete poured on clean gravel, gravel-sand mixtures, and coarse sand, f = 0.55 to 0.60 and the friction angle is delta = 29 to 31 degrees.  The table is based on U.S. Department of the navy (1982), probably NAVFAC.
Tan(29) = 0.55 and Tan(31) = 0.60.

Yes, you still need a safewty factor.  The f values are ultimate values.  

RE: Calculating Friction Factor

MTEng1

the same AASHTO code is. 29 deg for say cemented gravels against concrete? Inner friction is over 40º and the gravel is attached to the concrete. The 2/3 factor is customary but scarcely more than that.

RE: Calculating Friction Factor

(OP)
Thanks to both of you.  You have been helpful.  I just talked to a local geotech and they use Tan(0.7*friction angle) once they determine the soil classification.

"Structural engineering is the art of modeling materials we do not wholly understand into shapes we cannot precisely analyse so as to withstand forces we cannot properly assess in such a way that the public at large has no reason to suspect the extent of our ignorance." -Dr. A. R. Dykes

RE: Calculating Friction Factor

i think u should consider between 2/3 and 3/4.but is usually taken 2/3.

RE: Calculating Friction Factor

I would really like to know what soil you have under your retaining wall.  If you have a soft or firm clay (say Su ranging up to 25 to 30 kPa), it is that which will control the sliding - to put 600 mm (2 ft) of gravel under the retaining wall that has a width say of 3 m (or even less) and to think that the "increase" in sliding resistance between the retaining wall and the gravel will keep your retaining wall from sliding and ignoring the soft to firm clay below is, in my opinion, fallacy.

RE: Calculating Friction Factor

I have wrestled with this same question- what is the depth of gravel required below a retaining wall footing to effectively increase its sliding resistance.  As stated above, if there is only a thin layer of gravel the sliding forces are simply transferred to the underlying soil.  I have not come up with an answer, but it must be some function of the footing width (i.e. 0.5B).   

RE: Calculating Friction Factor

Several items come to mind:

1)  At some point you can include passive resistance.
2)  My professor (JM Duncan) told us the biggest uncertainty in soil strength is friction angle.  Apply your safety factor to friction angle and then see what you get.  If you took a friction angle of 38 degrees and a safety factor of 1.5 you'd get tan(38/1.5)=0.47, which may be what is being accomplished when you use the "2/3" value.
3)  Another dimension to this whole problem is to what extent you fully mobilize the interface friction between the two media (i.e., concrete and aggregate/soil). I'd think the wet concrete would pretty much knit to the aggregate and you'd get pretty much full moblization; however, some folks use 2/3rds to figure this out too.

Hope this helps.

f-d
 

¡papá gordo ain't no madre flaca!

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