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?

 

[canwesteng]

The only thing I can think of is if the soil is saturated, I wouldn't apply Ka, as the surcharge pressure is being applied to water with no internal friction.

 

[MacGruber22]

I wouldn't provide a qz/qx factor on your surcharge unless your geotech recommends so.

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

 

[PEinc]

The factor you refer to is the active or at-rest earth pressure coefficient. It is used to calculate the lateral pressure on a wall due to a vertical surcharge pressure. If the wall is able to rotate or deflect a small amount, then the active earth pressure coefficient, Ka, is used and usually multiplied to the vertical surcharge pressure. If the wall is rigidly supported to prevent any rotation or deflection, then the at-rest earth pressure coefficient, Ko, is used and multiplied to the vertical surcharge pressure. Every soils and foundation book describes the lateral earth pressure coefficients in detail. For granular soils, Ka is often approximately 0.33 while Ko can often be about 0.45 to 0.5. These values depend on the soil type, density, percentage of fines, angularity of the soil particles, etc. For clays, the lateral earth pressure coefficients can be as high as 1.0. I don't know whether you are designing a wall or writing a geotech report. In either case, you have a lot of reading to do.

http://www.PeirceEngineering.com

 

[MacGruber22]

Don't forget that the type, intensity, and location of surcharge plays heavily into an appropriate K value. Generally, you can't apply the same K factor from the retained soil and fluid to any ole' surcharge unless the surcharge is very uniform and relatively small in scale. Depending on your surcharge conditions, your geotech my recommend a surcharge K which is higher than your typical K.

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

 

[monchie]

Hello,

Thank you very much for all the replies.

Pls. see attached files, I have a huge surcharge loading of 75 kN/m2 as required. That's why,ie, multiplied it by a certain factor of 1/3 is of great help in terms of reducing rebar as explained by "PE..", it depends on certain cases. In the past, I normally design retaining walls with a surcharge of, between 10 & 20 kN/m2 and applied it directly to the retained structure. I have no problems understanding in applying a factor to soil pressure(ie, because it depends on the basic principles of "soil mechanics", angle of friction ..etc.). What it is "new" to me is applying a factor to surcharge loading, as in my previous query, others using a factor and some are not. The retaining wall I am designing is propped at the top and it is 2.4 m high.
Again, any additional explanation or reference to the code will be greatly appreciated.

 

[IDS]

I'm not sure why some are suggesting that loads above top of wall level will have a much larger horizontal effect than loads below the top of the wall. Even if the soil is saturated, unless the water is confined or the water level will rise due to the surcharge the water will not have any greater pressure. The vertical load has to go through the soil, and the horizontal component will be controlled by the soil properties and the stiffness of the wall, just as it is for the backfill.

If you have a low propped wall you shouldn't be using the active pressure for the backfill or the surcharge, but the same at rest pressure coefficient would be appropriate for both.

Is it possible to post an extract of the I StructE document saying you should apply the full surcharge load?

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[monchie]

@IDS, pls. see attached pdf file.

 

[IDS]

Thanks for that monchie.

It says:

For stress due to active pressure,
Siga(Z) = Ka (... + q - ...) + ...

where q is the surcharge pressure, so it is multiplied by Ka.

It seems a confusing way to show it, but their intention in the highlighted note seems to be to show why it is constant with depth.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[IDS]

I see the worked example shows the surcharge pressure as not being multiplied by K. That is inconsistent with the first equation, and is wrong.

For a propped wall the horizontal pressure coefficient should be greater than Ka, but there is no reason why it should not be applied to the surcharge pressure.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[monchie]

Thank you very much Mr. Jenkins.

 

[PEinc]

You really should consult one or more different reference books on wall pressures and surcharge pressures. Q = q (lateral) x H, not q x Q. H = 3 and q = 10. Therefore, Q = 30. If the retained soil were a pure clay with phi = 0, then Ka would be 1.0 for level retained soils. Then, the uniform lateral Rankine surcharge pressure would be q (vertical) x 1.0 and Q would = q (vertical) x 1.0 x 10 x 3 = 30. If phi > 0, Ka < 1.0 for level retained soil.
It would be much more clear for you to read reference books with correct examples than to read ET postings.

http://www.PeirceEngineering.com

 

[IDS]

PEInc - I would be interested to see your references for Ka = 1.0 for a soil with zero friction angle.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[monchie]

@PEinc
"You really should consult one or more different reference books on wall pressures and surcharge pressures. Q = q (lateral) x H, not q x Q. H = 3 and q = 10. Therefore, Q = 30"

-Can you cite a reference/documents regarding this(even just one..)? I'm only interested with the "surcharge" loading,which is actually "loads" on top of the soil. Previously, when I have a surcharge loading of 10 to 20 kN/m2,I've just directly applied it directly to the retained structure(without any factor). But when 50 & 75kN/m2 are required for a surcharge loading, others(engineers) suggested that a certain factor(similar to the factor applied to the soil pressure) can be multiplied to the surcharge loading. I've done some reading, and I found out, as in my previous comments, others are doing it and others are not.

I've shared the views of Mr. Jenkins, How come that a loads on top of the soil(surcharge), can be directly applied vertically to the retained structure,when in fact,there's a soil in between?That's why multiplying a factor(less than 1.0) makes sense to me.It is just like a boxer hitting you in the face but you partially deflect it with your hands.

 

[PEinc]

Rankine Ka for phi = 45 degrees = tan^2(45+phi/2) = 1.0
You really should read books that discuss lateral earth pressures for wall design. This forum is not an easy place for us to explain earth pressure theory in detail.

http://www.PeirceEngineering.com

 

[PEinc]

If a retaining wall were retaining water, and if you could somehow surcharge the water, the vertical surcharge load would be applied laterally to the back of the wall with the same magnitude because the water has no shear strength. Soil usually has shear strength which gives it the ability to partially internally support some of the vertical surcharge. Therefore, the lateral pressure resulting from a vertical surcharge on top of the retained soil will be less than the magnitude of the vertical load. The vertical surcharge load would be multiplied by either Ka, the active earth pressure coefficient, or by Ko, the at-rest coefficient, depending on the rigidity or flexibility of the wall.

http://www.PeirceEngineering.com

 

[Signious]

Just to add in, a great couple of books for lateral earth pressures are:
-Principles of Foundation Engineering by Das
-Soil Mechanics for Foundations by Budhu

They both work their way up from basic principles to shorthand equations.

 

[CarlB]

That formula PEinc gave for "active" pressure is actually the "passive" pressure formula. But the math was incorrect for either.
Replace the "+" with a "-" for the active coefficient, which would be 0.17 for phi=45 degrees.

 

[PEinc]

CarlB is somewhat correct. I rushed my response and wrote plus instead of minus. Rankine Ka = tan^2(45-phi/2). Kp = tan^2(45+phi/2). At phi = 0 degrees, Rankine Ka = Rankine Kp = 1.0, not 0.17 as indicated by CarlB.
tan^2(45 degrees) = 1.0. 45 - (0/2) = 45. tan^2(45) = 1.0. CarlB incorrectly said (45-0/2) = 22.5 degrees. Only phi gets divided by 2, not the 45 degrees.

http://www.PeirceEngineering.com

 

[CarlB]

OK PEinc, to defend myself, I used phi=45 degrees, not 0 degrees, which is also what you wrote previously.
So I stand by my calcs.
I think you were thinking phi=0 degrees, so (45-phi/2)=45 degrees.