AASHTO Reinforcing for Crack Control 1

[MWPC]

In the 2002 AASHTO Standard Specification for Highway Bridges, Eqn. 16-27 the thrust Ns times a factor is added to the moment Ms in determining the stress in the reinforcing. This stress is used to check for crack control. By definition these terms are added when the trust is causing compression. It would seem that compression would help close the cracks and reduce the tensoin in the reinforcing. Can anyone shed any light on this?

 

[Silverroad75]

I am not a Structural Engineer, I ever been a engineer. I have however, been in the installation side; for deformed reinforcing steel (ASTM A-615/706) as 10 year vet from local 416 in So.CA. and the last ten years as a inspector: in Concrete (ACI), CALTRANS CERT, Reinforced Concrete, and Structural Masonry.

Putting a wooded stake to your knee and breaking it - the side closest to your knee is in compression, and the side furthest from your knee is in tension, Right! So, the concrete cracks; Do they have broken rocks inside or on top of cracks? Are the crack(s) clean of debris? Compression cracks show signs of a crush, tension cracks have the appearance of a tear.

A few Q.

How long has the concrete been in placed?

If, placed recently
How hot of a day was it?

How long was the pour?

Was delivery an issue? Reason, Cold joints! if delivery was unbearable than you will have countless cold joints. This is frequent in Los Angeles, in surrounding areas, traffic jams are concrete killers.

What was the daily avg. slump?

Nonetheless it is something to look at, temperature steel is the prevention of expansion and contraction. I can't see rebar pulling together concrete and making it brake. Unless, of coarse, there is post tension cables installed, but you not mention that.

After all I am not engineer, but I have been around concrete for a long time. I clearly see things different than you because I do not have the education you have - and just like I tell my kids "I've been your age you haven't been mine" experience accounts for something.

I hope I helped, though I feel you are looking for answers I can't provide.

"Work Safe, and never get got with your head down"

 

[henri2]

MWPC,

This does not answer your question but you might find the link below interesting. It delves into the background behind AASHTO crack control design methodology as it relates to bridge structures and suggests a more simplified approach.

CRACK CONTROL IN CONCRETE BRIDGE STRUCTURES

http://www.dot.state.fl.us/structur...ural Crack Control in Concrete Structures.pdf

As to whether compression closes cracks and reduces tension, would it not depend on how the compressive force is applied? Eccentricty came to mind....increasing compressive stresses on one face but increasing flexural streses on the opposite face. But nontheless I
catch your general drift.

 

[miecz]

It has to do with the calculation of "i". As the Compressive force, Ns, increases, so does "i". As "i" is in the denominator, the tensile stress goes down, as you would expect.

 

[MWPC]

All, thanks for the replies.

henri2, thanks for the link. I hope the simplified method is adopted. The 0.6fy is what I would normally check for. I also found the Ns factor used on circular concrete pipe but could not find any commentary.

jmiec, thanks for the insight. We will run through the numbers. It just seems counter-intuitive and the local DOT did not know the history of the factors. Any history on how this came about?

 

[miecz]

This is just basic working stress theory for beams subjected to axial force. It was well explained in a document I no longer have. I believe it was titled SP3 or SP15, but I don't even remember who published it. If you run through a few examples, the rational will start to click. The nice part about this working stress approach is that it gives a simple and direct way to analyze bending and axial force. Ultimate strength theory doesn't handle this common situation as easily.

 

[MWPC]

jimiec,

It shook out just like you said. I did not think the factor in the denominator would overcome the addition in the numerator. I guess since Ns is multiplied by d-h/2 and that term is so small, it has little increasing effect.

Thanks again.