Horizontal Loads on Foundation Wall 2

[DMWWEngr]

I'm looking at placing a piece of equipment on grade adjacent to a building. This equipment will be placed near (about 4-5' away) from the existing building.

What is the most accurate method for determining the horizontial load distribution on the foundation (basement) wall??

I have a foundations book but it is only calculating the resultant force and it is for retaining walls. My wall serves as a basement and then extend past this as a foundation wall. Hence, I would like to have an accuarte distribution of the forces along the wall.

Thanks in advance
---
Andrew

 

[ishvaaag]

Against a strong wall midterm the backfill attains pressure at rest. On the other hand, for cohesionless backfills it may be acceptable to think the friction angle at the interface is close to the inner friction angle. One way of seeing it for a pressure caused for an area load is to interpret that the stresses at such interface would be those there caused by the area load in an elastic halfspace, that is, by the integration of the effects of elements of area load in different points of interest in the wall through the Bpussinesq equation.

I have the case of the infinite strip load at a distance such way in Mathcad Collaboratory, which has only values varying along the depth in the interface.

Unfortunately I have not it for the finite rectangle in which also would vary along the length of the wall, but to make a sheet to such purpose wouldn't be as difficult starting from the formulation of the stresses for a point load or modifying other sheets.

By the way note that as interesting is then to determine the tangential loads in the wall, that help in stabilization and give the inclination of the resultant in a particular element.

Now a formulation in card OCE 5

Numerator= b x c x s x k

Divisor=
((a+b) x tan ((fi/4)+(pi/2)) - a x tan (fi)) x (a + c))

Unitary push= Numerator / Divisor

a shortest distance in plan of load to wall
b dimension of load perpendicular to wall
c dimension of load parallel to wall
s uniform load per unit surface
k active pressure coefficient
fi angle of inner friction
pi 3.1416

This unitary load affects to the region

in plan
spreading 1 along the wall to 2 along the normal to the wall
(i.e. spreads at 26.66 deg from closest vertices to wall)

in section
from the closest point in the load, draw a line at fi with the horizontal downwards and towards the wall.
The region above gets excluded from loading.

from the farther point to the wall, draw a line at
(fi/4)+(pi/2) with the horizontal downwards and towards the wall.
The region below gets excluded from loading.

 

[DMWWEngr]

Thanks for your help ishvaaag. My line of thought agrees with what you are saying!!

The book I have only gives the TOTAL reaction so your method should work great for me (load is sitting on sand fill). I have one question on your method......

The angle from the furthest point of the load, (fi/4)+(pi/2); I'm not getting a "good" angle for that. At this point I'm assuming fi=45 deg. When I try to calculate the angle above I'm getting up around 100 deg. That doesn't seem correct because it would never intersect the wall. What am I doing wrong.....fi angle too high??

To everyone:
Based on experience does anyone THINK a 40,000 lb peice of equipment (8'x12'), placed 5' away from a basement wall, will cause the wall to collapse?? The basement wall is 15" thick and doubly reinforced, As= #5's @ 12" and As'=#5's @ 12". Wall is 21' tall, laterally supported at grade and 10' down from grade. My calcs are saying that it will fail....and fail by a longshot!!

Thanks for any advice you can offer. You guys are my structural mentors )
---
Andrew

 

[ishvaaag]

You have detected well an errata

((pi/4)+(fi/2))

is what there must be in the formula AND for drawing the downwards and towards the wall line. It is the classical inclination for the active pressure wedge. Sorry.

I was thinking in making a Mcad sheet for this since I only copied it, have not checked even the dimensionality.

The K the method uses in the reference is that for the active push, not the at rest.

 

[ishvaaag]

But note the drawings show walls in cantilever free atop. Do may be using Ko for at rest pressure is not unreasonable, since a wall for a cellar is likely to develop such if the wall stays rigidly restrained atop.

 

[DMWWEngr]

Me again...

What about the FIRST angle in the divisor. Which form should that be in, ((pi/4)+(fi/2)) OR ((fi/4)+(pi/2))??

It makes sence that the angle should be the one for the active pressure wedge, ((pi/4)+(fi/2)), but I'm getting a negative divisor using that one. What does a negative number indicate?? Or is there another error??

Any help would be appreciated. Thanks,
---
Andrew

 

[ishvaaag]

DMWWEngr, it is ((pi/4)+(fi/2)) but I will check tomorrow by doing an easy Mathcad sheet. Post an e-mail address and will send you either it or a rendering of it to .doc format. At least you will see for sure the equation well portraited, if there's nothing mathematically wrong in it, what I will detect. No idea on what will happen with the sign. Maybe I will add the sketch if all goes well.

 

[DMWWEngr]

ishvaaag,

You are the best!! E-mail is lundy@dmww.com

Thanks for all your help. The building's wall will appreciate it....but not near as much as I will!! )
---
Andrew