Integral railroad abutment and retaining wall
Integral railroad abutment and retaining wall
(OP)
I'm curious if someone could explain to me the railroad live load interaction of a railroad abutment sitting on drilled shaft piles that are part of a drilled shaft retaining wall whose tracks intersect the wall at a 90 degree angle. Rather than the tracks running parallel to this retaining wall, they are crossing over it so that the wall and the tracks are perpendicular to one another. I was wondering if someone could explain to me how to convert the railroad live loading (UPRR cooper E-80) into a surcharge to be distributed behind the wall if the abutment is sitting on top of a portion of this wall.





RE: Integral railroad abutment and retaining wall
On the line load, you can find that equation in FHWA publications online or in the Elastic Solutions book by Poulos & Davis, now also online. Keep in mid, if your line load has a finite width much larger than say 0.5 m, you should use a trip load. But this strip load will be of finite length, like a rectangular strip.
On the E80 strip load, you can use the modified Boussinesq method (I have attached an example) or you can take the distributed lateral stress and use a horizontal subgrade reaction method, like the Winkler approach. Since the static and other surcharges are using lateral pressures, it may be easier to use the strip load result of lateral stresses and superimpose all the lateral pressures. One decision you will need to make is, do you stop the E80 strip load only after 10 ft of embedment or do you carry it all the way down to bottom of the wall? You can decide that.
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RE: Integral railroad abutment and retaining wall
www.PeirceEngineering.com
RE: Integral railroad abutment and retaining wall
http://www.up.com/cs/groups/public/@uprr/@customer...
RE: Integral railroad abutment and retaining wall
In looking at the sheet you sent with the modified equation, I don't see how this is applicable. I see that it's the boussinesq equation but that would be for a track that is parallel to a wall face. I guess what I'm seeing if someone could explain would be how the load is transferred from the track to the approach slab sitting on the abutment which in turn is sitting on drilled shafts that are part of a retaining wall (like a secant pile wall but not quite).
PEinc,
I do have access to the AREMA 2014 manual and it gives two equations for the train surcharge hitting a wall parallel/perpendicular but is this still the case with an approach slab or does it get reduced? Is the abutment itself taking the majority/all of the surcharge? Wouldn't that mean that it all changes to vertical loads and bearing?
I guess my main question is at a railroad bridge abutment, how is the load from the train transferred to all of the components? (approach slab, abutment, drilled shafts)