Uniform Surcharge Retaining Wall Load 1

[Steve73MGB]

Hello,

I am studying for the April SE. I have a question on the calculation of the horizontal pressure due to a uniformly distributed load.

Using AASHTO section 3.11.6.2 and EM 1110-2-2502 return very different values, please see attachment.

If X1 is 4 ft, X2 is 14 ft and z is 4 ft with a p (or q) = 640 psf line load and the wall is non-yielding.

The AASHTO alpha = 0.785 rad, delta = 1.29 rad, angles are measured from the wall face to the line and the equitation returns a value of 900 plf

Using the EM 1110, alpha = 1.037, beta = 0.505 and the equation returns a value of 301 plf

Why is there a factor of 3 difference in the result?

Thank you for any help in answering this questions.

Cheers,

Steve

 

[RFreund]

I feel it is kinda tacky to reference my own blog, but I discus the elastic methods here:

http://howtoengineer.com/surcharge-analysis-elastic-methods-strip-load/

I will look at your example.


EIT

http://www.HowToEngineer.com

 

[RFreund]

A couple of possible mistakes:
Your delta should = beta as they are the same angle.
Also note that alpha in EM 1110 = (alpha from aashto + 0.5*beta). It is not halfway between the X2 and X1.


EIT

http://www.HowToEngineer.com

 

[RFreund]

One more note -

Your units are actually psf at that point. Then sun/integrate all your sections to find the force, plf, along the back of the wall.

EIT

http://www.HowToEngineer.com

 

[Steve73MGB]

Hi RFreund,

AASHTO Definition: α = angle between foundation wall and a line connecting the point on the wall under consideration and a point on the bottom corner of the footing nearest to the wall (rad.)

The solution manual for the problem defines it as: The angle α is the angle between the wall and a line connecting the point on the wall under consideration and the near side of the surcharge load.

For the AASHTO calculation α is not taken to the center of β.

RFreund, you are correct, I misread the answer and it is 0.90 kpsf

Given the AASHTO angle definition,
α = tan^-1(4/4) = 0.785 rad
δ = tan^-1(14/4) = 1.29 rad
ΔPH = 2(640 psf)/π (1.29 - sin(1.29)cos(0.785 + 2(1.29)) = 922.9 psf

Using the AASHTO angles to determine the EM1110 angles
β = δ - α = 1.29 - 0.785 = 0.507 rad
α = δ - β½ = 1.29 - ½(0.507) = 1.036 rad
ΔPH = 2(640 psf)/π (0.507 - sin(0.507)cos(2(1.036)) = 301.6 psf

The AASHTO answer is correct as given by the “16 Hour SE Practice Exam”

Any thoughts on what I am not seeing? ;-)

Cheers,

Steve

 

[Steve73MGB]

PS

I forgot to include the AASHTO definition of δ:

δ = angle between foundation wall and a line connecting the point on the wall under consideration and a point on the bottom corner of the footing furthest from the wall (rad.) (C3.9.5) (3.11.5.3) (3.11.6.2)

Cheers,

Steve

 

[dhoward26]

Haven't looked to in depth, but the first thing that caught my attention is your values are different by a factor of 3. Seams somewhat odd.

 

[RFreund]

Steve - I would need to see the AASHTO eqn in the code. I think there may be an error.
If you use delta as it is defined as the angle between the end and start of the surcharge (as it is in EM) than you get the same result as EM which I believe to be correct.
However what may be the case is that if this AASHTO equation is to be used for a basement wall condition, say pinned top and bottom, then you would most likely have an 'at-rest' soil pressure condition. Thus the high surcharge load (greater than 1!). However that is not typically what rigid refers to. Normally rigid means a cantilever concrete wall or something similiar, then you may have semi-rigid say soldier pile and lagging then you may have flexible- SRW type wall.

EIT

http://www.HowToEngineer.com

 

[Steve73MGB]

Hi RFreund,

The AASHTO equation and diagram are in the attachment, not the complete section. All references I have use the EM equation with the exception of AASHTO. I will contact someone at AASHTO and hopefully receive an answer to how they arrived at their equation. I do not believe soil acts differently when a bridge component is involved ;-)

Cheers,

Steve

 

[pegarrett]

Steve73MGB,

As RFreund stated above, but did not pursue, delta from the AASHTO equation and beta from the EM equation are equal to one another.

An error is being made during the calculation of the value of delta. The correct value, given your inputs and AASHTO section 3.11.6.2, is:

[TAN-1(14/4)-TAN-1(4/4)] = 0.5071 rad

With the other angles being:

alpha(AASHTO): [TAN-1(4/4)] = 0.7854 rad

alpha(EM): [alpha(AASHTO)+0.5*delta] = 1.0390 rad

beta: [delta] = 0.5071 rad

Given these inputs the AASHTO and EM equations both yield a result of 302.71 psf

-Philip

 

[Steve73MGB]

Hi Philip,

AASHTO defines delta as the angle between foundation wall and a line connecting the point on the wall under consideration and a point on the bottom corner of the footing furthest from the wall.

per AASHTO
delta = tan^-1(14/4) = 1.29 rad, measured from the wall to the far line
0.5*delta = 0.645 rad which does not equal beta(EM) = 0.507 rad.

I think the AASHTO diagram is misleading, but they do define the angles in Section 3.3 Notation. AASHTO delta does not equal EM beta per the definition and diagrams.


Cheers,

Steve

 

[Steve73MGB]

I have attached the problem, AASHTO and EM solutions. I think the problem is caused by the AASHTO angle definitions.

Cheers,

Steve