surcharge loading on retaining wall 3

[monchie]

Hello,

Regarding surcharge loads on retaining wall, I've notice that, in some instances, a factor of,ie, Ka =1/3 was multiplied to surcharge loading, while others do not have any. I've checked some references ISTructE example do not have any "factor" while "Structural Foundation Designer's Manual(Curtins)" have one(wKa, pp 307). Any ideas why others doing it differently?

 

[IDS]

PEInc - if you are going to adopt a condescending tone in your responses you should at least get your facts right. The formulas we use for calculating pressures on retaining walls are based on failure mechanisms. To get a Ka of 1.0 both the friction angle and cohesion would have to be zero; in other words a liquid. The Rankine formula is not applicable to cohesive soils.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[monchie]

" I admire people that despite their vast amount of knowledge, can explained things better and clearer even a child can understand it"

Unknown

 

[PEinc]

First, I meant my previous posts to be brief and to the point, not condescending. If I came across as such, I am sorry.
The point I was trying to make is that active lateral surcharge pressure cannot equal the vertical surcharge unless Ka = 1. That cannot happen unless both phi and c are 0. In my opinion, design examples showing the lateral and vertical pressures being the same are wrong or overly conservative.
Additionally, the only way that tan^2(45-phi/2) could equal 0.17, as CarlB indicated above, is if you incorrectly use Ka = tan^2((45-phi)/2) instead of tan^2(45-(phi/2)).
The Rankine formula is often used for cohesive soils in the drained condition where c is assumed to be 0 and a phi angle is used. Refer to various publications for graphs of phi angle vs. plasticity index.

http://www.PeirceEngineering.com

 

[IDS]

PEInc - thanks for the clarification. I totally missed your point (which is the same as my point), so apologies for jumping to incorrect conclusions.

I agree with everything in your last post, except that the Ka of 0.17 for phi = 45 is wrong. Tan^2(45 - 45/2) is equal to 0.17.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[PEinc]

Tan^2(45 + 0/2) = 1
Tan^2((45+0)/2) = tan^2(22.5) = 0.17

http://www.PeirceEngineering.com

 

[IDS]

Tan^2(45 + 0/2) = 1
Tan^2((45+0)/2) = tan^2(22.5) = 0.17

CarlB's numbers were for phi = 45, not phi = 0.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[PEinc]

Yes. CarlB did use phi = 45 degrees in his above example for Ka = 0.17, but I thought this thread was about having equal vertical and lateral surcharge pressures and how Rankne Ka could = 1 only if phi is 0 degrees. I don't think phi = 45 degrees has anything to do with this thread.

http://www.PeirceEngineering.com

 

[CarlB]

Good gawd man, own up and give it a rest!

If you read my posts you'll see I weighed in only to try to clear up mis-information/errors/typos put forth by you.

Rankine Ka for phi = 45 degrees = tan^2(45+phi/2) = 1.0

Both me and IDS apparently felt it was worthwhile to point out inconsistenies for any future thread readers, so we both attempted- twice each. Thanks IDS for following through, we should now better spend our time elsewhere

 

[PEinc]

I admitted and corrected the single typo after CarlB pointed it out. After that, it is you who are beating an off-topic, "dead horse." And that's my final response.

http://www.PeirceEngineering.com