Shear ties compared to spiral reinforcement
Shear ties compared to spiral reinforcement
(OP)
Quick question: in a broad sense, wouldn't it take a larger bar size for spiral reinforcement on a 6" pitch to equal the shear strength of a column with shear ties spaced at 6"?






RE: Shear ties compared to spiral reinforcement
RE: Shear ties compared to spiral reinforcement
For a spiral at the same unit spacing the bars are equivalent to pi/2 bars worth of strength per unit spacing.
pi/4 is the 'efficiency' of a single bar in a circular spiral and you have two complete bars per spiral turn.
You generally need to reduce the spacing for the same overall size (say 600mm square vs 600mm diameter) but it depends on the member size obviously. Circular stirrups simply aren't as effective. But depends on the effective depth.
Just review your code equations and you will see this.
RE: Shear ties compared to spiral reinforcement
I don't see your argument. My reply was intended to be generic, but I would be interested to know what code equations you are referring to.
In ACI318-08, for example, the area of Vs for spirals is defined as 2 x bar (11.4.7.3).
RE: Shear ties compared to spiral reinforcement
RE: Shear ties compared to spiral reinforcement
check page 336 of the above link.
In New Zealand we have this formulation in our code, our code is based on ACI318 so I made the incorrect assumption that it was in there as well. But I see on checking ACI that its behind the times a bit in this respect.
Tomfh & hokie66, the point is that for a typical 45 degree shear crack/plane, not all bars in a circular member intersect the crack at 90 degrees (think of the bar crossing the 45 degree shear plane anywhere but the center of the member), this leads to a reduction in the 'efficiency' of the spiral when compared with typical ties in a rectangular member. In a rectangular member the bars cross the shear plane at 90 degrees, in a circular member this angle varies depending on where you take your slice through the section. Most codes including ACI have provisions for ties which are placed at some other angle than being vertical in a beam, they are not as effective if you tilt a stirrup over, its similar in concept to this and is the "correct" formulation.
I don't know what else to say if I haven't turned you into a believer!
RE: Shear ties compared to spiral reinforcement
RE: Shear ties compared to spiral reinforcement
RE: Shear ties compared to spiral reinforcement
@Hokie: before cracking, stirrups are most effective when oriented perpendicular to the shear cracks that they cross. However, once a crack has formed, the optimal orientation for a stirrup is vertical, or perpendicular to the longitudinal axis of the remember, correct? My point being that I believe that the pitch of the spirals is always a detriment to their ULS carrying capacity and never an asset.
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.
RE: Shear ties compared to spiral reinforcement
RE: Shear ties compared to spiral reinforcement
RE: Shear ties compared to spiral reinforcement
So it's not a matter of taking the full strength of each bar, some are simply less efficient.
RE: Shear ties compared to spiral reinforcement
ACI would seem to agree with you based on their inclined stirrup capacity equation. @!@@&?@!!#$%!@#%!!!. Just when I was gearing up for a nice, relaxing weekend. Now, for the bazillionth time, I'm back to not understanding concrete shear.
Can you explain to me why inclined stirrups would be more efficient in a post-cracked state where the principle stresses have already dissipated due to cracking? See my sketch "A" below. Conceived of in typical modified truss fashion, I would certainly expect vertical ties to be the most efficient. Help!
Sketch "B" below illustrates my understanding of the phenomenon that Agent666, has been describing.
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.
RE: Shear ties compared to spiral reinforcement
RE: Shear ties compared to spiral reinforcement
Placing the stirrups at an inclined angle simply makes the concrete strut intersect more stirrups. This means 'Av' in your equation increases (or 's' decreases, whichever way you want to look at it). Assuming a concrete strut at 45 degrees, placing stirrups at 45 degrees perpendicular to the strut would be the most efficient and increase your 'Vs' by sqrt(2). It is easier to visualize if you do the opposite and consider inclining the stirrups the other way to be parallel to the strut; this would obviously reduce your strut efficiency to 0.
RE: Shear ties compared to spiral reinforcement
RE: Shear ties compared to spiral reinforcement
At last, I get what Agent666 has been saying, but don't know how they came up with pi/2 bars (that might be more apparent if I had pages 335-337 of the reference). Regardless of the stirrup orientation, you can't have all the bars cross a given crack at the ideal position, which would be at the section centroid where the shear stress is greatest.
RE: Shear ties compared to spiral reinforcement
RE: Shear ties compared to spiral reinforcement
Reinforced concrete design will always involve many assumptions and approximations, but then so does steel design. It depends on which material you do most. Modern steel design mystifies me.
RE: Shear ties compared to spiral reinforcement
Yes, this is exactly it! Thanks for setting me straight.
@Hokie:
Using nonplussed's guidance, I've attempted my own derivation of the ACI inclined stirrups equation (below and attached). No doubt I could have just found it in a textbook if I had the right one. Anyhow, here are the conclusions that I've drawn from the derivation:
1) The sin term may be thought of as representing the stirrups that would have been available even if there were no incline. The capacity of those stirrups is multiplied by the sine of the incline angle to reduce their effectiveness.
2) The cos term may be thought of as representing the extra stirrups that cross the shear crack as a result of the incline. The capacity of these stirrups, too, are multiplied by the sine of the incline angle to reduce their effectiveness. This multiplication cancels out in the mathematics of determining the extra tie area available however.
3) The ACI equation for inclined stirrups shows an increase in capacity relative to perpendicular stirrups only because more stirrups become involved, not because individual stirrups are more effective when oriented more perpendicular to the anticipated angle of the shear crack.
4) The derivation of the ACI inclined stirrup equation suggests that the optimal orientation for stirrups is perpendicular to the longitudinal axis of the member. All other orientations, including perpendicular to the shear crack, are inferior and and that inefficiency is quantified by multiplying capacity by the sine of the angle that the stirrups make with the direction of the applied shear.
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.