Shear in a Basement Wall

[DaveAtkins]

Assume Vu at the bottom of a basement wall exceeds phi*Vn. I contend that the only way to make the wall adequate is to make the wall thicker. But I have seen more than one engineer try to justify a wall that is too thin by checking shear friction in the wall, using the available vertical steel.

I think you must prove shear works based on the thickness of the wall--shear friction is a separate concept.

Thoughts?

DaveAtkins

 

[KootK]

Yeah, I've seen it accompanied by both plan and elevation details.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[MacGruber22]

I tend to avoid shear reinforcement in walls. Thicker wall is less work and less chance the contractor will muff it up.

"It is imperative Cunth doesn't get his hands on those codes."

 

[hokie66]

Count me in the thicker wall camp. The only retaining wall cases I have encountered where shear controlled were deep basement walls, and these are normally some type of deep foundation system walls such as slurry diaphragm walls.

 

[RFreund]

OK, I think I have one logical argument that suffices my wondering. If you have a section "x" inches away from the support that fails in shear i.e. Vc<Vu then that length of wall would be theoretically cracking at all sections between x and the support thus the shear friction reinforcement turns into some dowel bending type thing. Although I suppose you could still argue that the shear friction bar keeps these cracked pieces in compression as so as it starts to deform. I mean the shear friction reinforcement is crossing the same crack as the stirrups just in the orthogonal direction and not parallel to the direction of force. hmmm... I'm just trying to to prove to myself that mechanics of it don't work... not sure this argument is sufficient...

EIT

http://www.HowToEngineer.com

 

[KootK]

Oh, there's definitely a post Vc path for the shear in the form of dowel action. It's just a lousy path. Given normal placement of wall vertical reinforcing, there would be a tendency to spall behind the bars on both faces.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[manstrom]

That's some heavy water.

Of course, geotech engineers round pi up to 5.

When I am working on a problem, I never think about beauty but when I have finished, if the solution is not beautiful, I know it is wrong.

-R. Buckminster Fuller

 

[dcarr82775]

95 pcf isn't unreasonable for a poor soil. I typically end up with loads in the 80-85pcf range for submerged conditions. You need to design for more than just the unit weight of water.

 

[Wallache]

And if it were wet concrete, you would be designing for 150 pcf. Water can get heavier if you add heavy stuff to it.

 

[hokie66]

Yes, the pressure in submerged conditions has to include the water plus the buoyant soil pressure. The buoyant soil pressure is based on the density of the soil minus the density of the water. I think it is poor practice for a geotechnical engineer to just specify a single number like 95 pcf. While the water pressure always varies linearly with depth, the soil pressure is often trapezoidal rather than triangular.

 

[MacGruber22]

Yes, the pressure in submerged conditions has to include the water plus the buoyant soil pressure. The buoyant soil pressure is based on the density of the soil minus the density of the water.

That is sad we have to remind people of such a rudimentary concept.

"It is imperative Cunth doesn't get his hands on those codes."

 

[MacGruber22]

Feel free to correct any of this:

I think it is important to realize that the concrete shear we design stirrups for is not a purely horizontal shear - the stirrups are really resisting a principal stress. The direct shear (being resisted by shear-friction in this discussion) is just that, and does not have a principal tensile stress. Pure beam shear is complementary, but it is clear that it is only part of the total stress. If you could rely solely on the complementary nature of shear, than your direct shear demand *would* equal your horizontal shear demand. But, it doesn't because of principal stresses.

As I have recently learned from KootK and the ACI-318 commentary, you can utilize the main tension bars for shear-friction without adding more area, as long as the area provided satisfies both separately.

"It is imperative Cunth doesn't get his hands on those codes."