Rankine Passive Condition (Coduto vs Bowles) 1

[CWEngineer]

I am trying to compare how Coduto and Bowles calculate the Passive Pressure using Rankine. I am doing this for my personal knowledge - I tend to learn something when I have a difficult time figuering something out. I have spent some time and I can't get Coduto's and Bowles answers to agree.

I would really appreciate if someone can help me out in determing if I am doing something wrong with my interpretations.

 

[oldestguy]

I'd try a more basic reference, possibly even look up where Rankine was first published.

If you can't find that, go back to early writings of Terzaghi or perhaps Taylor.

Most modern writers are using earlier work of others when they write books these days, especially when it comes to something like Rankine.

 

[BigH]

They are both basically telling you the same thing. Presume that Bowles is the one you are following "first". Without the cos(beta) in front, the force is acting parallel to the backfill slope (beta). Bowles notes that cos(beta) is a common entry - so he puts it in front of the fancy equation (11.8) by combining the two terms - the kp term and the "slope" effect. So, 11.8, analogous to 11.7a, becomes 11.8a (with the cos(beta) in front).

Conduto does it somewhat differently in that he doesn't "change" kp to the horizontal by hiding cos(beta) into the fancy equation - he forces you to use kp and then multiply by cos(beta) (see equation 23.11 - note he has kp*cos(beta) but Bowles just has kp' (where the ' is the combo effect).

In another "way" Conduto x cos(beta) should equal Bowles. From your post, 3.654 x cos(beta) =(?) 3.598.
cos(beta) = 0.9848 xo 3.654 x .9848 = 3.598. I'd say they are equal. The "cos(beta)" term really doesn't become that important until beta approaches 25deg when the cos(beta) = 0.9.

Anyway - that's my take on it. You just forgot to multiply Conduto's term by cos(beta).

 

[CWEngineer]

Really appreciate your feedback BigH. I really want to clarify this one out. Here is what I am seeing.

1. Coduto Eq. 23.13 in his book is not multiplied by cos(beta)
2. Coduto Eq. 23.11 (Normal force between soil and wall) is multiplied by cos(beta).
3. If I multiply Codutos Kp by cos(beta) as you shown in your calculation to obtain Kp=3.598, wouldn't that mean I would be multiplying by cos(beta) twice. Once in the Kp calculation and another one in the Pp/b.

Thanks For Your Help.

 

[BigH]

That's the rub - you don't multiply it twice. He shows Kp without it. (Where Bowles shows a "modified" one with it). When you calc 23.11, you simply plug in Conduto's Kp as shown - the cos(beta) term is in 23.11. I was showing that if you plug in Condutos Kp into 23.11 and then rearrange so that cos(beta) of 23.11 is now combined with Conduto's Kp, you get Bowles' modified one. cos(beta only comes into play once.

 

[CWEngineer]

THANKS. That step in calculating Pp really makes sense. Here is a little bit of more info on what I am trying to figuere out.

Coduto:
Kp [Eq. 23.13] = 3.654
Pp/b [Eq. 23.11] = 27.204 (normal force between soil and wall)

Bowles:
Kp [Eq. 11-8a] = 3.598
Pp [Eq. 11-9] = 27.200 (Force at a wall angle Beta)
Pph [Bowles, page 603] = 26.78

1. Using Coduto's formulas to find Pp I get 27.204. And using Bowles formulas to get Pp I get 27.200. Both numbers agree.
2. However, Coduto is calling his Pp/b the normal force between soil and wall). His Figuere 23.9 shows Pp/b basically horizontal.
3. From his Figures Bowels is calling his Pp Force at wall angle of beta, then he goes ahead and calculates the horizontal component by multiplying Pp by cos beta.

I guess that is my greatest confusion that Coduto and Bowles Pp numbers agree based on the formulas they are using (Coduto using cos beta in his Pp formulas and Bowles using cos beta in his Kp formula). BUT Coduto is calling his Pp normal force between soil and wall and Bowles is calling his Pp Force at a wall angle of beta.

Would really appreciate if you can help me out in clarifying this. I have attached Coduto's figure 23.9.

 

[BigH]

First off - get the Kp values correctly. Kp = 3.654. End of discussion - Bowles says it too - it is just that he then modifies it by combining the cos(beta) term with Kp, let's call it Kp# which you call 3.598. Conduto does show Pp/b (why he puts in the damn b is beyond me as no one else does - it is understood that this is per unit length of wall). Anyway, Pp horizontal = gamma*H^2cos(beta)*Kp/2 for Conduto. Pp horizontal for Bowles = gamma*H^2cos(beta)*Kp/2, but Bowles has the modified Kp# so Pp Bowles = gamma*H^2*Kp#/2. Need to put Bowles' orange back to Conduto's apple. The fact that Bowles and Conduta give you the same answer proves this out (forget the 0.004)That's it for me!

 

[CWEngineer]

Okayyyy, last comments. Really appreciate one more feedback on this one.

1. In my example for Bowles Force to agree with Coduto's Pp Horizontal Forcce, cos(beta) is only used ONCE, in the Ka term.

2. But in Bowles Example on page 604, to find the horizontal force component, he is using the cos(beta) TWICE, once in the Ka term and then in the Pa,h term.

 

[BigH]

In the example, he is actually talking of Coulomb and the reason cos(beta) is used twice has to do with taking into account wall friction (Rankine assumes no wall friction). Here, Bowles has taken 10deg as both angle of slope AND wall friction angle.

 

[CWEngineer]

THANKS. BigH. I really appeciate your time in helping me clarifying this. You were VERY helpful.

 

[Panars]

gman1
To thank BigH, you should give him a star. That's how things work in this forum.

 

[CWEngineer]

Thanks for the information, didn't know about the star thingy. I just gave BigH a Star. THANKS.