Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations waross on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Circular Tank - Shear Transfer at Wall/Slab Interface

Status
Not open for further replies.

chavez1297

Civil/Environmental
Oct 5, 2009
4
I know that this has been explained in previous posts, but I still have some confusion. In designing a circular RC tank, using the PCA document, you get the shear/foot at the base of the wall from Tbl. A-12. You then transfer this shear to the slab via dowels. You then must account for this shear in the slab by providing ring reinforcement in the slab in the vicinity of the wall.

My question concerns the determination of the ring reinforcement that is needed. One post that I read indicated that you take the shear/foot * pi * radius to get the value for the amount of shear that needs to be resisted by the ring reinforcement. Could someone explain the development of that equation.

Thanks
 
Replies continue below

Recommended for you

It is the shear per foot by the radius in feet. it's basically the same thing as the hoop tension in your cylinder.

Michael.
Timing has a lot to do with the outcome of a rain dance.
 
Should pi be omitted in the equation in my original post?
 
You need to sum up the total hoop stress over one half the tank circumference. Since you have a force per foot, the total force over this length is force/ft * pi * radius. This is never explained in the PCA reference.
 
The one half of the circumference had me confused. I guess if you would sum the forces over the entire circumference, then they would cancel out? Is that the correct thinking?

 
One half pulls against the other half. If you were to fracture a tank (or pipe) under pressure, one half would go one way and the other the opposite way.
 
Thanks for the explanation. I appreciate it.

 
Pi should not be in the equation. The shear force is acting radially, outward, from the center. Pick a centerline and add up the components of the forces perpendicular to that centerline. The total is the diameter multiplied by the unit force divided by the two sides where the centerline crosses the wall.

Michael.
Timing has a lot to do with the outcome of a rain dance.
 
Can you scan and post T A-12 & associate graphic illustration? I kind of get lost in this base shear (for circular tank) business. Thanks.
 
From:
Hoop stress.

PressureVesselCylindricalC.gif


Michael.
Timing has a lot to do with the outcome of a rain dance.
 
Paddingtongreen is absolutely correct. Pi does not come into the picture. Ring tension is gamma*z*r where gamma is the density of liquid, z is the depth and r is the radius of tank.

That assumes no radial reaction at the bottom slab. If the walls are prevented from straining elastically by ties to the base slab or rubber pads at the base of the walls, then the ring tension in the vicinity of the base slab is reduced by an amount which depends on the stiffness of the slab and wall or the properties of the rubber pads. It becomes a problem of strain compatibility.



BA
 
I failed to see shear stress in the graph above, it is tensile stress to me. I am still puzzling over what is the unit base shear about.
 
Let me guess the "shear" in PCA's term:

For an unit element (cubical) subjected to surface pressure, there are out of plane shears on all 4 sides to resist the force resulted from the pressure.

If this is the case, I think it is wrong. Without concentrate force, a pressure vessel will experience tensile/compressive stress but shear. Or, I may have missed the fundamental.

 
Consider the cylinder, the pressure inside causes hoop stress and the resulting strain causes the cylinder to expand radially outward. The foundation doesn't want to expand with it, thus initiating a shear force between the bottom of the cylinder and the foundation. All of those dowels want to move outward so you put rings of reinforcing around to hold them in.

When you add up the components of those outward forces, in the same direction, the result looks just like the hoop stress diagram I posted, above.

cntw1953, this is the shear between the cylinder wall and the foundation.

Michael.
Timing has a lot to do with the outcome of a rain dance.
 
So, it is a strain compatibility phenomenon - the shear caused by the relative movement in between wall and the slab exists in two directions (circumferential and radial). If this is the case, how could pi not get into the mix?
 
Other than compatibility as talked above, all I can see is vertical shear at the wall-slab interface (same as beam-column interface) caused by weight of the water and the slab. Again, pi has something to do with this.
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor