Allowable shear stress from Steel reinforcement in CMU 1

[shaneelliss]

I am reading through the ACI 530-11 masonry code and don't understand how the reinforcement shear stress is calculated. The formula is Fvs = 0.5(Av*Fs*d/(An*s)). I am looking at this specifically for a shear wall, not a beam, but I think the formula is the same for both.

Av is defined as the area of steel that is resisting the shear. If I have two #5 bars spaced at 48" horizontally, I believe this area is just the area of two #5, or 0.62 sq. in. and the spacing is dealt with later. Is that right?

Fs is just the allowable yield strength of the reinforcement. If I am using 60ksi bars, is the allowable something less than 60 because this is ASD? I am guessing it is, but I have never used ASD before except for wood, so I am not sure.

d is the depth to the steel, and on a shearwall I remember from school that this can be approximated with 0.8*length of wall. Is that correct?

An is the net cross sectional area of the member. This is the term that confuses me the most. All of the other terms make sense to me in a general sense, but I can't figure out why this term is in the formula. The formula above is almost the same as for steel shear strength in concrete except for this term. I don't get why the steel strength has anything to do with the area of the wall, and why I would be dviding the strength of the steel by such a large number is beyond me, so if someone can explain that I would appreciate it. For a shearwall that is say 120" long and 7.625" wide (8" block) and for simplicity assuming fully grouted, I think this term would be 120" x 7.625". Is that correct?

s is the spacing of the reinforcement, so if I have two #5 bars at 48" horizontally, this would be 48". Correct?

Any help is appreciated.

 

[IsaacStructural]

Fs is absolutely less than 60ksi. Check out section 2.3.3 of ACI 530-11.



M.S. Structural Engineering
Licensed Structural Engineer and Licensed Professional Engineer (Illinois)

 

[shaneelliss]

Thanks. Oh, and I figured out why the An term is in there. I just needed to look at the units and remembered this is giving me a psi, not a lb answer. So I think I have the rest right if I use the correct Fs, but can anyone just give me some validation?

 

[IsaacStructural]

I believe you are doing it correctly. I glanced over this section. You are computing the allowable strength contributed by the Reinforcing, which is Fvs. The total allowable strength of the wall section is Fv = Fvm + Fvs , so you still need to compute Fvm, which is in the code right above the equation you've been using. Finally, when you have this value you compare it withe required shear stress, fv, which is computed on the previous page.

using d=0.8*width seems appropriate and something I've seen in the past, though I don't know where/if that is specifically given in the ACI 530-11.

M.S. Structural Engineering
Licensed Structural Engineer and Licensed Professional Engineer (Illinois)

 

[shaneelliss]

Thanks.

 

[DST148]

Ref ACI 530-05. An is net cross-sectional area of masonry. (to account for hollow cores in masonry)
d is distance from extreme compression fiber to centroid of tension reinforcement. d does not necessarily have to be 0.80 x length of shear wall.
For design of concrete structural walls for shear, ACI 318-11 section 11.9.4 gives d = 0.80 * length of wall, which most designers use even for masonry shear walls.