Masonry beam shear area 2

[ENGR_2321]

Hi everyone. I was wondering if anyone knows what is considered the "net shear area" in a masonry beam. I went through the current masonry code and searched online and it seems to be equal to An which is the cross sectional area. In other words Anv=An. however when im calculating the max allowable shear force (ASD), I get a big number in kips and I feel like maybe it's because these areas are actually different. I tried playing with different possibilities for Anv but I still won't get the right answer. any comment would be very helpful for me if anyone has a clue thank you. below is problem statement (from ncees practice workbook , answer is 12 kips)and then I put down pictures of my calculations

 

[WARose]

Just like RC design, I've taken it as the depth to the centroid of the tensile reinforcement. In this case it would be: 20" x 7.625"= 152.5 in[sup]2[/sup].

 

[dik]

and, I usually use 7" width to accommodate the tooled joint... not necessary, but, I do that.

Dik

 

[ENGR_2321]

I changed the area to 152.5 and it only decreased my answer to 216 kips it's still significantly bigger than the answer 12 kips though

 

[WARose]

By the way, I think the allowable you are using for Fv is way too high. For unreinforced flexural members, you'd be limited to Fv= (f[sub]m[/sub]')[sup]1/2[/sup]. Which cannot exceed 50 psi.

For shear reinforced flexural members it would be: Fv= 3*(f[sub]m[/sub]')[sup]1/2[/sup] <150 psi

 

[DETstru]

For beams where M/(V*d) >= 1.0, the allowable shear stress is Fv <= 2*sqrt(f'm)

So for your beam, the allowable shear is
V = Fv*b*d = 2*sqrt(f'm)*b*d = 2*sqrt(1500)*7.625*20 = 11.8 kips

 

[WARose]

For beams where M/(V*d) >= 1.0, the allowable shear stress is Fv <= 2*sqrt(f'm)

I think that is for shear walls. For flexural members see my post above.

I got about 18k for shear stress......and about 12k based on stirrup spacing. (Which would control.) See Sections 2.3.5.2.3 & 2.3.5.3.

 

[DETstru]

WARose, it's for flexural members.
See the commentary in 2.3.6.1.2 in 530-11.

 

[WARose]

WARose, it's for flexural members.
See the commentary in 2.3.6.1.2 in 530-11.

Interesting. But the problem statement says 530-05.....and I don't see that in the code commentary there.

 

[ENGR_2321]

I thought we were allowed to use M/Vd = 1.0 to simplify calculations (per the commentary). I'm using the most current code (2011) but I think I am looking at the same equations you have. The problem only has the beam dimensions and gives the fy and fm. We cant really know the value of M/Vd therefore right? We can only use 2(fm)^0.5 where M/Vd> 1.0 . Thats whats confusing too. The problem figure has shear reinforcement and thats why I used the general formula Fv=Fvm +Fvs (allowable) and then equated that answer to the fv= V/ Anv (calculated) to solve for V which is the force in kips (which is what they are asking for). I thought we were supposed to take into consideration both the strength provided by the masonry and the shear reinforcement

 

[WARose]

I thought we were allowed to use M/Vd = 1.0 to simplify calculations (per the commentary). I'm using the most current code (2011) but I think I am looking at the same equations you have.

The problem statement says 530-05. And I don't see it in the commentary there.

The problem figure has shear reinforcement and thats why I used the general formula Fv=Fvm +Fvs (allowable) and then equated that answer to the fv= V/ Anv (calculated) to solve for V which is the force in kips (which is what they are asking for). I thought we were supposed to take into consideration both the strength provided by the masonry and the shear reinforcement

No. At least as far as ACI 530-05 is concerned. (I am not familiar enough with 530-11 to say.) But with masonry, the stirrups are carrying all or nothing.

 

[msquared48]

Amrhein page 361 shows about 40 psi shear or just under 8K.

Mike McCann, PE, SE (WA)

 

[DETstru]

WARrose, sorry, I don't have a copy of '05 (I also failed to see the code reference, hopefully they don't rescind my license for this...)

Everything I'm saying is based on 530-11:
The requirements of 2.3.6.1.2 (a) and (b) put an effective cap on the shear capacity. Since you don't know M, you must take the conservative approach. It's also common to take M/Vd = 1.0 (as noted in the code)
Since your Fvs + Fvm is greater than the capped Fv (capped at 2*sqrt(f'm)), you cannot exceed that limit.

Fvs is based on 2.3.6.1.5 and is 43 psi
Fvm is based on 2.3.6.1.3 and is 44 psi
Fvs + Fvm = 87 psi

The cap of Fv <= 2*sqrt(f'm) = 77 psi .... so you must use that.

(77psi)*(20")*(7.625") = 11.8 kips

 

[WARose]

WARrose, sorry, I don't have a copy of '05 (I also failed to see the code reference, hopefully they don't rescind my license for this...)

I won't tell if you won't.

I point it out in large part not just to get the "right" answer....but to hammer the point home to the OP (if he is studying for/taking a NCEES exam): he'd better be using the code called out in the problem statement.

Everything I'm saying is based on 530-11:
The requirements of 2.3.6.1.2 (a) and (b) put an effective cap on the shear capacity. Since you don't know M, you must take the conservative approach. It's also common to take M/Vd = 1.0 (as noted in the code)
Since your Fvs + Fvm is greater than the capped Fv (capped at 2*sqrt(f'm)), you cannot exceed that limit.

Fvs is based on 2.3.6.1.5 and is 43 psi
Fvm is based on 2.3.6.1.3 and is 44 psi
Fvs + Fvm = 87 psi

The cap of Fv <= 2*sqrt(f'm) = 77 psi .... so you must use that.

(77psi)*(20")*(7.625") = 11.8 kips

Good info. I got exactly 12 kips based on stirrup spacing. (I.e. V=(Av*Fs*d)/s. That is rearranging eq. 2-26 in 530-05. )

 

[ENGR_2321]

Oh...I think I see what the problem is. the 2011 code says on the commentary that prior to 2011 edition, the shear resistance provided by the masonry was not added to the shear resistance provided by the shear reinforcement rebar. But now they do. so I think this question might be outdated

 

[DETstru]

WARose, you're definitely right! Can't get the right answer with the wrong code!!

I'm sure the SE exam is no longer based on '05. Must've been based on '11 when I wrote it.

 

[ENGR_2321]

in case anyone else sees this post, here is the revised solution to this problem so no one gets confused. solution here is modified to apply to the 2011 code , we still get the same answer 12 kips. ( tip for rookies like me: DON'T square root ksi !! it messes up your values tremendously. no wonder I was getting giant answers lol ). again, thanks everyone for your help !

 

[dik]

M^2...
Amrhein was the author of my first masonry design book... nearly 30 or 40 years back; at the time, it was likely the best masonry book published... nice to see the name again.

Dik

 

[ENGR_2321]

I've heard good things of that book. It's one of the books I plan to buy soon, need to get me a couple of good masonry design books

 

[wannabeSE]

With square roots, use f'm with psi units not ksi

sqrt(1500) = 38.8 psi.
When you use ksi sqrt(1.5) = 1.22 ksi which is not the correct answer