I was looking in Lindeburg's CERM and tried to find a connection between the Kh and Kv values that he gives in Appendix 37.A to the Rankine ka factor. In a problem he solves there is a note stating that the Rankine formula could be used and R broken into components. It is somewhat confusing as it is stated earlier that the Rankine formula disregards the angle of external friction, yet in the same book the previous statement is given in a problem where the delta angle of friction is not equal to 0 (example 37.2 vs formula 37.6). So is there actually a way once we have the Rankine ka value to derive mathematically the Ra,v and the Ra,h or do we need to use the Kh and Kv values from the graphs in the Appendix? .....which is quite a sloppy process!

are you asking about subgrade modulus or permeability? Can you define "Ra,v" and "Ra,h" also?

Not sure I'm going to be able to offer much assistance, but it'd be good to know what you are talking about. . .

f-d

¡papá gordo ain’t no madre flaca!

No it is not a modulus of permeability.
The Ra is the resultant inclined soil pressure force. The subscripts h and v correspond to horizontal and vertical components.
The example of CERM is number 37.2. The announcement of the problem follows:
The unkeyed retaining wall shown has been designed for a backfill of coarse-grained sand with silt having a weight of 125 lbf/ft^3. The angle of internal friction is (phi) 30 deg, and the angle of external friction (delta) is 17 deg. The backfill is sloped as shown (after calculation the beta angle is 18.4 deg). The maximum allowable soil pressure is 3000 lbf/ft^2. The adhesion is 950 lbf/ft^2. Calculate the factor of safety against sliding..

is this a homework assignment?

p.s., I knew it wasn't a modulus of permeability, 'cause no such thing exists.

f-d

¡papá gordo ain’t no madre flaca!

No actually it is not a homework assignment. In fact I don't recall using the CERM in college, but I do use it for PE prep if that is allowed within this forum. I thought that questions of clarification on material from the CERM, the ACI 318, the LRFD manual, the IBC etc are legitimate to ask here, and that the individuals that provide answers do that on a volunteer basis. Please correct me if I am wrong.

On the subject in question, I applied some formulae but the results are off by a factor of about 20% which makes me wonder if that has to do with the very loose (imprecise) input we get from graphs.
Thanks

The earth pressure thrust (Pa) as you know is 0.5 Ka Gamma H^2. On sloping backfill you will get vertical and horizontal component of this thrust. Say the sloping ground is at 26.6 Degrees, you can get say the horizontal component as Pa x Cosine 26.6 Deg. and the Vertical component as Pa x Sine 26.6 Deg. The units are lb/ft in horizontal or vertical direction.

The CERM most likely took the charts from '67 Terzaghi and Peck classic book. What they call Kh is simply = Ka x Gamma. Of course your Ka has to reflect the slopeing backfill per Rankine equations.

The Kv can be found as Kh x Sin 26.6 in this example. We discount the vertical component of the earth pressure (conservative) when performing stability analysis of sliding and overturning. In your exam if a Pv thrust is given, include it in your calcs. Alternatively, in your exam unless you are told discount the evrtical earth pressure component in sloping ground, include the Pv in your calcs.

Some books like that of Dr. Coduto, call the Kh as the equivalent horizontal fluid pressure. So if Kh is 45 psf/ft or 45 pcf, then it is similar to designing a wall that will retain a fluid having a density of 45 psf. Try to get Affi's 201 solved problems book and Basic Geotechnics textbook by McCarthy. It is plenty adequate for the Geotechnical PM depth of the PE exam.

Dear FixedEarth,
Thank you for the reply. I am actually not in Geotech, but I really want to clear this problem anyway. I would anticipate such problems in the AM to tell you the truth.
In that particular problem I calculate Ra=0.5*1ft*Ka*(gamma soil)*H^2 which returns an Ra value of 8364.2lbf (I multiplied by 1 ft considering the 1 ft strip standard - please correct me if my assumption is incorrect, but the units would not come right otherwise). http://books.google.com/books?id=_mfbUyqb7VQC&...
Thus my Rav and Rah are 2445.5 lbf and 7998.7 respectively versus those of the CERM which are 1680 & 6720 lbf. The latter raises the value of Kh that Lindeburg gets from Terzaghi's charts from 40 lbf/ft^2 to 47.6. That's seriously off and in the graph with the four different types of soil given, it is much closer to the Stiff residual silts than the Dirty sand and gravel. What is more confusing with Lindeburg is that he gives his formulas for Rah and Rav (formulas 37.27 &37.28) in a way that includes the lamda angle (90 degrees in this case) to the delta angle. That 90 degree phase difference should have either reversed the trigonometric functions …or I have spent too much time over this. Otherwise I am puzzled why the vertical component should be less than the horizontal anyway.

For now just forget about a slope. Pa = 0.5x(KaxGamma)xH^2 = 0.5xKhxH^2. So Kh = KaxGamma. When you have ascending slope backfill, if the angle is < 45 Deg., the horizontal component or Kh will be larger than Hv, the vertical component. In other words, Cos 30 Deg. > Sin 30 Deg.

When you have a slope, Ka is calculated with that nasty square root in both denominator and numerator. If you use Tan^2Phi equation, it will not work for sloped backfill. Then KaxGamma will give you the thrust at an inclined angle. To get the horizontal component of the earth pressure thrust = 0.5x(Kh)H^2xCos 30 Deg, to get vertical earth pressure thrust = 0.5x(Kh)H^2xSin 30 Deg. In the CERM it throws in wall friction angle, so that means he is using Coulomb because Rankine does not allow wall friction. The McCarthy book or Soils and Foundations by Lieu and Evett would help. May be you can get a older version from your library. Try not to get stuck at any one topic. Also if CERM explanation is not to your liking, the other books I mentioned may give you alternative solution. Good luck.

You're right. That is another cause of confusion with Lindeburg as he refers to his 37.6 formula (which is the Rankine formula) in this problem whilst we actually do need Coulomb's formula (37.5 in the CERM). The result between the two differs very slightly because the Lamda is 90 degrees and delta is too small in the problem but in such an important reference book you would expect the author to be more precise, or if in this case both formulas work, it should have been mentioned and explained why. But thank you, your note clears things quite a bit.