AASHTO Nongravity cantilevered wall design

[coloradobridges]

AASHTO Figure 5.6.2A (in the 17th ed. standard spec) shows a simplified pressure distribution for wall elements embedded in rock. The formula for the passive resistance of the rock is given as:
Pp = s*D /(1+tan B)/(D+2^.5*b)

If s is shear strength of rock in psf, shouldn't the (D+2^.5*b) term be in the numerator to make the units work (i.e. give Pp in lbs or kips)?

Thanks in advance.

 

[PEinc]

You are looking at the 17th Edition AASHTO Figure 5.6.2A.b. The 16th Edition does not have the (D+2^.5*b) term in the Pp formula. I haven't yet figured out why AASHTO made the change.

I agree that it seems to be more appropriate in the numerator. The Pp formula is the total passive resistance for the soldier beam so the term in question should be related to the width of the available toe resistance. If so, it should be a numerator and the width should not be a function of D, embedment length. Seems to me the term would be more correct if it were a numerator and equaled (b+2^.5*b) = 2.414*b. Comment?

Many people blindly use these AASHTO earth pressure formulas but few understand their derivation. I don't have any trouble deriving the formulas except (1) the formula you questioned and (2) the Pa2 formula in Figure 5.6.2A.a. In the Pa2 formula, I get Pa2 = 0.5*Ka2*b*D*(etc.). For some reason, AASHTO has multiplied by 3 the active pressure behind the soldier beam toe similar to how the passive pressure is applied to 3b. Active pressure behind the soldier beam toe should be applied only to a beam width of b, not 3b.

 

[coloradobridges]

Thanks PEinc.

(b+2^.5*b) in the numerator makes sense to me or just 3b like the 'embedment in soil' case. AASHTO has issued errata for the 17th ed., but this figure was not changed.

As for Pa2, I generally ignore this force for temporary construction (I'm primarily using these formulas for excavation shoring systems). Comment? But why not an effective width of 3b on both the active and passive sides?

I'll take a look at the 16th ed. as well as the LRFD.

 

[PEinc]

For the temporary soldier beam walls, AASHTO will let you ignore the active pressure behind the soldier beam below subgrade. Refer to Note 1 on AASHTO 17th Edition, page 131 or Note 1 on page 106 of the 16th Edition.

As far as 3b, it is common to use 3 times the width on the passive side. This helps account for additional resistance provided by side shear of the passive wedge and distribution of the embedment force over an area wider than the soldier beam.

On the back or active side of the soldier beam, below subgrade, you can't have 3b wide of active pressure pushing on a beam that is only b wide. However, that is what AASHTO's Pa2 formula indicates in Figure 5.6.2A.a. In fact, if the soil is capable of arching, there should be no active pressure on the toe for the temporary or short term condition. I discussed this with FHWA's Jerry DiMaggio who I believe agreed with me but all he wanted to tell me was that I should be using the LRFD method rather than ASD. I'm not sure what LRFD has to do with the soldier beam width. The same width should be used with either design method.

 

[cap4000]

I am not familiar with the formula. What does the tan B represent? Based on your formula as shown, it does end up in the numerator as it gets flipped up into it and the units do check out. Good Luck.

 

[coloradobridges]

Cap,
Tan B accounts for sloping ground. For level ground it is zero.

The formula as shown in AASHTO has the following units:
pressure * length / length = pressure
Pp is a force not a pressure.

I checked the 16th ed, and as PEinc noted the formula is different and gives a force result. But I don't understand how the old formula was derived. It seems to take too much influence from the depth D and not enough from width b.

Thanks for your insight PEinc and Cap.

 

[cap4000]

I wonder how AASHTO looks at H-Piles vs.Walls in Rock?? In pure clay they use 2C (cohesion) as the Passive uniformly distrubuted load on an embedded pole. What I have done in the past is use the rock cohesion value in a similiar situation. For rock, its much to straight forward and can't be exactly right. Hopefully, further rock research will simplify these tricky conditions. Good Luck.

 

[coloradobridges]

The figure we are discussing is for cantilever walls with discrete vertical elements, ie. a solider pile wall. I believe many engineers use this pressure distribution for H piles in rock.

It is supposed to be straightforward and not exactly right (but conservative). In fact the figure is not in equilibrium as it neglects the passive pressure behind the wall (below the point of rotation).

I often see C (cohesion) given in soils reports and have assumed that is roughly equivalent to Sm, the shear strength of rock mass. Any comments on that assumption?

 

[cap4000]

ASCE Army Corps of Engineers Design Guide # 16 has typical rock cohesion values on the order of 2ksf to 10ksf pending the RQD of the Rock. The Shear Strength of Rock is much higher and not even close. Done from a Direct Shear Test and plotting several points of failure I connect the dots and at the y-intercept with no axial load comes the cohesion value and that's what I have used. Conservative yes, as typically my rock sockets are 3 to 4 feet deep for walls 10 to 14 feet high.

 

[PEinc]

cap4000,

Coloradobridges' formula in his original post above is not excatly written as it shows in AASHTO's Figure 5.6.2A.b. Coloradobridges showed a division sign between (1+ tanB) and (D+2^.5*b). AASHTO does not. Therefore, if AASHTO is correct, the term stays in the denominator where it was typed. I still feel that my original post, paragraph 2, is probably what AASHTO means but did not write. Unfortunately, we may never know for sure. Meanwhile, hundreds of consulting engineers and DOT engineers are using the formula blindly.

Also, I have yet to see a DOT highway project where any significant rock testing was performed. The testing information you would like to plot is just not available or done for most projects. Sometimes, we may be given "the" unconfined shear strength of the rock but never a series of plottable points.

With respect to the diagram not "being in equilibrium" consider the following. The diagram is solved for moment or overturning equilibrium. That is how the initial, minimum embedment depth is determined for the cantilevered sheeting wall. However, some engineers feel compelled to also sum the forces in the x direction. When this is done, the available passive resistance will always be greater than the driving pressures (active, surcharge, water, etc.). Then, because the available passive resistance is greater, engineers try to increase the embedment depth to try to add more active in an effort to get equilibrium in the x direction. But consider this - the passive resistance can not push back the wall or else the passive side would become the driving force and the active side would then have to provide the passive resistance. This just can't happen. The passive resistance is just that - a resistance that hopefully builds up to the level needed for moment equilibrium. It will not become large enough to push back a sheeting wall which has more dirt behind than in front.

In the overturning calculation, the moment arm for the passive resistance is always going to be shorter than the moment arm for the driving forces. At moment equilibrium, therefore the passive resistance has to be greater than the driving forces. Therefore, there will be a fairly good safety factor against sliding toward the excavation.

It is my humble opinion that engineers who increase the pile embedment to achieve x direction force equilibrium for cantilevered sheeting walls are playing number games and not seeing the bigger picture.

 

[coloradobridges]

Cap,
I agree with your thesis on equilibrium. I only use the pressure distributions to sum moments at the base and solve for D. There is actually an additional force at the base which does not contribute to the summation of moments. This force achieves equilibrium and does not require any additional D.

As for my statement of the AASHTO formula, I wrote it as I would in a calculator or Excel, ie. 4/2/2 = 1 (not 4).

I hope any engineers using the AASHTO formula blindly will follow the old rule "check your units" and realize that something's wrong.

Thanks again for your input.

 

[cap4000]

PEinc

You got me thinking again. 1st: Lets say TanB=0 for the sloping ground. Use b=1 for 12 inch HPile Flange. AASHTO Formula reduces to : (S)(D)/D+1.41. Use D=to 5ft, Pp=5(S)/6.41 or Pp=0.78S.

Your Formula: S(5)/2.414 or Pp=2.07S?? Seems very high?

With the RQD value known, I am sticking safely to only the "cohesion" value which assumes no rock overburrden or heavy mass near or onto the rock socket. Perhaps if it was deep into the rock mass I would go to Unconfined Shear Strength. Is your 3b arching effect actually valid for Rock?? As usual nice job .

 

[PEinc]

As I said, the 17th Edition formula makes no sense to me unless the questioned term is in the numerator and its D is actually a b. Then, using D = 5' and b = 1', Pp would = s*5*(b+(1.41*b)) = s*5*(1+(1.41*1)) = 12.05s. This Pp would be slightly less than if we had used 3b as we do for soil. I'm not saying that's what AASHTO is trying to say but it makes more sense than the formula that is written in the 17th Edition.

I have never used the formula from the 17th Edition. I have used the formula from the 16th Edition many times - sometimes, in the past, I even used Pp = s*D*(3b) but not lately because many engineers don't know anything about sheeting but they can read the AASHTO specs. Besides, with any reasonable s, you won't need to use 2b or 3b.

I've never had a problem with a cantilevered wall. The important things, in my opinion, are to not use too high of a shear strength and to consider how fractured the rock is. It is ridiculous to have a 10' high cantilever with a 1' toe in rock. The numbers may say it's OK but what if the rock is very fractured? How well will very fractured rock support the wall? Sometimes, for questionable rock, I'll calculate Pp as if the rock were a very good clay soil per Figure 5.6.2C and use 3b. Rarely will I use less than a 4 or 5 foot toe embedment in fractured rock for a cantilevered soldier beam. I will use less if the rock is solid but this is rarely the case around here.

 

[cap4000]

PEinc

I think AASHTO has got it right. They are trying to reduce the "s" value in the 17th edition. Your right about the fracturing of the rock. There is a paper out by To,Ernst and Einstein August 2003 published by ASCE that tries to explain the jointed rock "BLOCK WEDGE THEORY" by using 3D Matrices. It's complicated kinematics but seems to get to the point. Also the concrete say 2ft in diameter, that surrounds the HP may NOT be as strong as the diabase rock found here in NJ, thus I use a deeper toe to compensate for that.

 

[PEinc]

cap4000'

I don't think AASHTO has it right. Pp is the total passive resistance in lbs or kips. If AASHTO is correct, the numerator has a D (feet) and the denominator has a [D + 2^.5*b] (feet). These feet would cancel out and Pp would equal psf or psi which is not a total passive resistance. This would reduce the s a little but you would still have to multiply by D and by some width to get Pp into lbs or kips.

Even if the concrete around the toe is not as strong as the rock, it is still extremely strong - strong enough to give you a reasonable toe length. Even lean mix concrete or flowable fill would have 50 psi (7.2 ksf) to about 500 psi (72.0 ksf) compressive strength. That may not be hard rock but is extremely stronger than a very hard clay.

I still think AASHTO may be wrong. It would be nice if we knew who changed the formula and why.

 

[coloradobridges]

I sent an email to AASHTO. I'll let you know if I get a response.

 

[PEinc]

Please do that. Thank you.

 

[PEinc]

Further Developments (i.e. Confusion)

Ref. Temp. Soldier Beam Wall With Toe In Rock

AASHTO 16th Edition says Pp = (s*D)/(1+tanB')

AASHTO 17th Edition says Pp = (s*D)/[(1+tanB')*(D+2^.5*b)]

AASHTO LRFD 3rd Edition says Pp = [s*D*(D+2^0.5*b)]/(1-tanB') (unfactored)

3 books, 3 different formulas - no explanation

(1+tanB') vs. (1-tanB') ?????????
numerator vs. denominator ?????????
(D+2^.5*b) vs. no (D+2^.5*B) ?????????
incompatible Pp units for 17th and 3rd Editions ?????????

 

[cap4000]

Just came across a Caltrans pdf Bridge Design File 2003 with Pp=s(D+b*sqrt2)/(1-tanB'). Yet just another formula, however, the Pp represents a uniform load passive pressure(K/ft) on the embedded pile in the rock.Thankfully, at least that now makes some sense versus a total Passive Force. Unreal confusion still exists.WOW.

 

[coloradobridges]

I did advise AASHTO of the error and followed up for a couple months. They said they passed the info on to their technical committee. Hopefully, the issue will be addressed in the next errata.