Vc=0 and probable moments & shear strengths 2

[howarts]

From Mcgregor's "Reinforced Concrete Mechanics and Design 6E"

If you are familiar with the concept, jump to the bottom questions:

(briefing quoting Mcgregor)
"When the frame is displaced laterally through the inelastic deformations required to
develop the ductility of the structure, the reinforcement at the ends of the beam will yield
unless the moment strength is several times the moment due to seismic loads. The yielding
of the reinforcement sets an upper limit on the moments that can be developed at the ends
of the beam. The design shear forces, Ve are based on the shears due to factored dead and live loads (Fig. 19-19c) plus the shears due to hinging at the two ends of the beam for the
frame swaying to the right or to the left, as shown in Fig. 19-19a. Mpr is the probable
moment strength of the members, based on the dimensions and reinforcement at the joint
and assuming a tensile strength of 1.25 fy and alpha=1.0. For a rectangular beam without
axial loads, ACI Code Section 21.5.4.1 requires that beams be designed for the sum of
Vsway = (Mpr1 + Mpr2) / ln
Vg = wu ln / 2
Ve = Vg + Vsway"

It is stated that Vc shall be taken equal to zero if (a) the shear Vsway due to plastic hinging at the two ends of the beam exceeds half or more of the maximum shear Vu, within the span

Let say your Vsway = 88 kips (yeah Vsway is really huge)
Your Vg (1.2 LL + 1.6 DL) or shear from factored gravity load = 87kips
When Vsway is equal or more than Vg then you need to set Vc= 0
But here's the dilemma. If you add 2 kips to the Vg, then Vsway is less than Vg, and Vc=0 is no longer required.
There is a big difference in the beam to accomodate Vc=0. This means you need to design stirrups that will take on the entire Ve without any contribution from the Vc shear capacity of concrete. This would cause increase in section sizes as well as bigger diameter stirrups (can cause some congestion too). And if Vc is not zero, it simply means you can use the concrete shear capacity (the logic of having Vc=0 is "The damage to the hinging area due to repeated load reversals
greatly reduces the ability of the cross section to resist shear, requiring more transverse
reinforcements"

So in the event you encounter situation where your Vsway = 88 kips and Vg (1.2 LL + 1.6 DL) = 87kips.. you can simply add 2 kips to make Vsway less than Vg and Vc bcome no longer zero and make available the concrete shear capacity (saving a lot of member size increase or bigger diameter stirrups).

What would you do if you encounter the scenerio above. Add 2kips and make Vc not zero or make Vc=0 (but more costly and even attracting more moments from increased member sizes)?

 

[Deker]

Not sure what you mean by add 2 kips...the dead load is the dead load and should represent a reasonable estimate of the permanent loads imposed on the structure. You should be comparing the seismic shear to the gravity shear that is present under the seismic load combinations, not 1.2D+1.6L. For the purpose of this check, you should be comparing the seismic shear to (0.9-Sds)D. If the seismic shear isn't more than that value, you likely have sufficient safety factor built into your capacity by designing for 1.2D+1.6L.

 

[howarts]

Not sure what you mean by add 2 kips...the dead load is the dead load and should represent a reasonable estimate of the permanent loads imposed on the structure. You should be comparing the seismic shear to the gravity shear that is present under the seismic load combinations, not 1.2D+1.6L. For the purpose of this check, you should be comparing the seismic shear to (0.9-Sds)D. If the seismic shear isn't more than that value, you likely have sufficient safety factor built into your capacity by designing for 1.2D+1.6L.

I meant adding more dead load to make seismic shear Vpr less than Vg so hence fulfilling Vc not equal to 0.

 

[Deker]

I do understand what you mean, but I think you are missing the point. If the seismic shear is roughly equal to (0.9-0.2Sds)D, then the seismic load combination is not likely to control your shear design (would still need to satisfy detailing requirements). I suggest you create a sample problem for yourself and do the shear design using Vs only for the seismic load combinations and Vc+Vs for the gravity load combinations. Use reasonable beam proportions and gravity loads that you are likely to encounter in practice. Then compare the shear designs for the case where the seismic shear is just over half the total shear under the seismic load combination. Which load combination controls, and by how much?

 

[howarts]

But you should use 1.2 D + 1.6 D if its larger than the seismic load combination. That's what the book says. I'll give you actual example from the book example itself (only quoting the relevant parts). First your formula of 0.9(1-Sds)D comes from

E = Eh + Ev
E = Qe + 0.2 Sds D (ACI Eq. 19-4)

where
Qe = the horizontal load effect caused by seismic forces.
Ev = the vertical component of seismic forces, taken equal to 0.2 Sds D
Sds = the design spectral acceleration at a short period, such as 0.2 sec,
D = the dead load.

ACI Seismic load combination:
U = 0.9D + 1.0E = 0.9D + 1.0E (ACI Eq. 9-5)
Substituting Eq. (19-4) into the combination with E load (gives

U = 0.9D + 1.0 (Qe) + 0.2Sds D

Where Qe= horizontal component of E and 0.2Sds D is the vertical component.

This is the equation you gave. But this equation is only used when the dead load stabilizes a structure subjected to overturning loads or stress reversals. Therefore your (0.9-0.2Sds)D is not used in normal building. Or even if you use it, the book says the largest factored loads are used (in the case it's 1.2 D + 1.6 L).

Now let's go directly to the book example where he used your formula and others:

"The beam shown in Fig. 19-20a and b is a typical floor beam in the special moment resisting frame of an office building. The beam supports a uniform unfactored
dead load of 4.0 kips/ft and a uniform unfactored live load of 2 kips/ft. The normal weight concrete and reinforcement strengths are 4000 psi and 60,000 psi. Design the reinforcement.

Assume Sds = 0.3

Load combination 1
U = 0.9D - 0.2 Sds D (load combination you gave which book also shows)= 0.2 x 4.0 - 0.2 * 0.3 * 4.0 = 3.36 kips/ft

Load combination 2 (ACI Eq. 9-5)
U = 1.2D + 1.0 (Qe) + 1.0 (0.2 Sds D) + 0.5 L
U = 1.2 x 4.0 + 1.0 (0.2 x 0.3 x 4.0) + 0.5 x 2.0 = 6.04 kips/ft

Load combination 3 (ACI Eq. 9-2)
U = 1.2 D + 1.6 L = 1.2 x 4.0 + 1.6 * 2.0 = 8.0 kips/ft

The book uses 8.0 kips/ft because 1.2D + 1.6 L is the largest factored load compared even to seismic load (at least the vertical component).

Reactions due to gravity loads: wu ln /2 = 8.0 kips/ft x 22/2 = 88 kips upward at each end (see illustrations where all the numbers were shown)

The book computes for Mpr (see illustration) and reactions Vpr due to MPR at each end.

Reactions due to Mpr at each end, frame swaying to right:
(Mpr1 + Mpr2)/ln = (233 + 531)/ 22 = 34.7 kips down at left end

Total reaction at left end = 88 kips + (and -) 34.7 kips down
= 53.3 kips up at left end
Total reacdtion at right end = 123 kips up at right end.

You can see in the book example that the seismic shear is added to the factored load of 1.2 D + 1.6 L (which is 88 kips)

So your stirrups have to be designed for both the seismic shear and 1.2 D + 1.6 D (see illustrations)

In the book example. They use Vc (not zero) since Vpr is 34.7 kips is less than half of total shear of 123 kips.

Get it so far? Do you have any questions?

My original inquiry is. Lets say the book example uses lighter dead load and live load. Then the Vpr can be more than the load combination of 1.2 D + 1.6 L.. but supposed they are so close or Vpr is a bit more than the factored gravity load (again this controls since it's larger value than the seismic load), then you add a little bit more dead weight to make it more than Vpr, and your Vc is no longer zero. So you can control Vc = 0 or not by adding dead load or removing it (like putting a big concrete block on the floor). Does it make sense?

 

[Deker]

My McGregor text is at the office so I can't check, but if he compares Vpr to 1.2D+1.6L I believe that is a mistake. Take a look at the load combination shown in ACI 318-11 figure R21.5.4. The way you should think about the load combinations is that you design the entire building once for each combination. There is no load combination that combines seismic with 1.2D+1.6L, so it doesn't make sense to compare them for this check. What happens in your example above if you use the right load combinations?

To answer your question, yes I suppose you could add dead load in order to use Vc in your design for the seismic load combinations. But this is not common practice. It may also end up being counterproductive since it will create additional demand in your design for 1.2D+1.6L, which is kind of the point I'm trying to make.

 

[TehMightyEngineer]

I agree with Deker and have a copy of McGregor's text in front of me. I agree that he does use 1.2D+1.6L when calculating M[sub]pr[/sub], but I agree with Deker that this doesn't match my understanding of the intent of ACI 318-11 section 21.5.4.1.

In the definitions of ACI 318, V[sub]e[/sub] is specifically defined as "design shear force for load combinations including earthquake effects, lb, see 21.5.4.1 and 21.6.5.1, Chapter 21". Thus, the design shear force for the sway case given in Chapter 21, and shown in Fig. R21.5.4, is related to the shear associated with the probable moment strength ± the max. shear associated with the gravity load from the seismic load cases. This is given as M[sub]pr1[/sub] + M[sub]pr2[/sub] / ln ± w[sub]u[/sub]l[sub]n[/sub]/2 as shown in the figure.

This is not to say that the shear reinforcement might not be controlled by 1.2D+1.6L, but this is not related to seismic and would be designed as a normal beam (considering Vc).

Regarding your question, in reference to 21.5.4.2(a); yes I believe you are technically correct that if you force your gravity (either dead or live) shear load to be higher then you technically can include your concrete shear strength. However, read the commentary to 21.5.4.2 in ACI 318. In brief, for special moment frames the code wanted a strength reduction due to the high cyclic loads required to be resisted. So, what they're saying is if your beam's shear design is controlled by the cyclic seismic load (V[sub]sway[/sub] > V[sub]g[/sub]) then you need to have more shear strength. Increasing your V[sub]g[/sub] is roughly accomplishing the same thing but a 2 kip increase does not seem equivalent to the shear strength increase associated with setting V[sub]c[/sub] = 0. The commentary continues to describe how V[sub]c[/sub] is actually important and shear strength at the hinge regions is obtained partially by the confined concrete core. Thus, adding 2 kips gravity shear may meet the letter of the law but doesn't meet the intent in my opinion. Plus, you get all the issues Decker pointed out with adding excess dead or live load.

Professional Engineer (ME, NH, MA) Structural Engineer (IL)
American Concrete Industries

http://www.americanconcrete.com

 

[KootK]

I would say that the answer to your question is yes if you actually intend to increase the gravity induced shear on your beam during a seismic event by increasing the mass of your structure in the real world. This seems unlikely, however, as there are cost implications associated with adding mass frivolously.

I would say that the answer to your question is no if your proposal is simply to bump up your gravity load values numerically without truly adding gravity load to the system in the real world. If Ve < Vg under the shared load case considered then the beam shears do not reverse under seismic action and the condition is more favourable. Thus Vc can be relied upon. If you're just fudging Vg, then the shears can still be expected to reverse and Vc should be discounted.

This is something that could definitely be spelled out more clearly in the code.

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.

 

[KootK]

Regarding practical design, I'd be fine keeping Vc if Ve < 1.05 Vg. Two kips of reversible shear is nothing and structural engineering is rough science, particularly with respect to seismic.

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.

 

[howarts]

Kootk. What is your Vg? Is it 1.2 D + 1.6 L as Mcgregor used in his computations? Mcgregor logic is that during seismic activity, the plastic hinging itself can produce moments in addition to existing moments.. so you have to add the two especially in getting the shears. So the shears from Vpr is just related to seismic. You need to get the shear from gravity loads and what more complete load combination to get than 1.2 D + 1.6 L. Hence technically (and logically) you need to add the shear from Vpr and shear from 1.2 D + 1.6 L. The effect is like having (1.2 D + 1.6 L) x 2 that must be handled by the beam using very thick stirrups and if your set Vc=0. You will even have so much stirrups that you even need to increaes beam size and depth just to accomodoate Vc=0, shears from Vprs added to 1.2 D + 1.6 L.
In a colleague work. He added VPR shear to the shear from 1.2 D + 1.6 L and set Vc=0 (without regards if Ve is less or more than Vg). The result is very heavy shear requirement that can't accomodated in the beam so they ended up wrapping the beam with FRP carbon fiber.
For some guys who dont agree 1.2 D + 1.6 L should be used in addition to Vpr, what load combination do you use then and approved by ACI.?
And is still using 1.2 D + 1.6 L adding to Vpr wrong? But how can Mcgregor be wrong?

 

[KootK]

I agree with the interpretations of Decker and TME: Vg should be that from the seismic load case being considered. That said, MacGreggor is nothing short of a minor deity in my neck of the woods. I'll try to check out his stuff tonight. Granted, MacGregor's book is now in the editorial hands of others.

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.

 

[KootK]

I just finished reviewing MacGregor -- couldn't resist. As far as I can tell, MagGregor offers no rational explanation for his choice of load case for Vg in his example. Given:

1) the absence of such a rationale expressed by MacGregor;
2) the fact that I personally find the choice illogical and;
3) the corroborating opinions so far expressed by my esteemed ET colleagues...

I feel confident in saying that MacGregor's book likely contains an error in this regard. There are always a few.

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.

 

[howarts]

I just finished reviewing MacGregor -- couldn't resist. As far as I can tell, MagGregor offers no rational explanation for his choice of load case for Vg in his example. Given:

1) the absence of such a rationale expressed by MacGregor;
2) the fact that I personally find the choice illogical and;
3) the corroborating opinions so far expressed by my esteemed ET colleagues...

I feel confident in saying that MacGregor's book likely contains an error in this regard. There are always a few.

Do you find seismic load case always have smaller values than 1.2 D + 1.6 L? Maybe Mcgregor use the latter for conservative estimate.

Also let's be clear on something. Say you hold a rebar and you pull it with your bare hands (say you were Superman) until it yields, the strength to yield it by mere pulling is called the probable moment strength right? It is independent from the dead or live load.

This means for shear capacity. You need to use the most conservative gravity load of 1.2 D + 1.6 L and add it to the probable shear Vpr from the independent probable moment strength.

Maybe many of you think that probable moment strenght is just the normal load with additional 1.25 overstrength and without 0.85 factor? No, it's not like that! Ponder on it. And let me know exactly what you think (so I know what you think). Thank you.

 

[howarts]

The formula for Mpr is

Mpr = 1.25 fy As (d - a/2)
where a = 1.25 fy As/0.85 f'c b

and fy = 60,000 psi, f'c =4000 psi

Now the one dollar question is. In real life situation. Are the moments from this formula taken from the existing moments of the dead and live load that is pushed to 1.25 overstrength (hence yielding).. or are they separate independent moments created from the yielding process itself (independent of the loads)??

If it's the latter. Mcgregor is right. if it's the former. Then he uses so higher shear capacity. Thousands or millions of buildings are at stake here. Because most doesn't have Mcgregor standard of huge shear capacity.

 

[Deker]

The reinforcing in the beam is determined by making sure the design strength (including phi factor) is greater than the factored loads for all load combinations required by the building code. Once you know the reinforcing in the beam, the seismic demand is determined using a capacity design procedure assuming that the beam has yielded at each end. This seismic demand is now independent of the code loads, and if you were to arbitrarily increase the reinforcing in the beam, you would correspondingly increase the seismic demand even though the code loads did not change. The seismic loads need to be combined with the gravity loads to get the maximum load effect, but the load factors on the gravity loads are based on the load combinations that include seismic. You wouldn't combine seismic loads with the full factored gravity loads because the two are statistically unlikely to occur at the same time. It's the same reason we don't design for the maximum considered earthquake and a hurricane simultaneously...the chances of it occurring are highly improbable and to do so would be overly conservative. McGregor is great for concrete design, but for seismic design using capacity design procedures I suggest you read Paulay and Priestley, Moehle, and Englekirk. There is also a NEHRP Techinical Brief on concrete moment frame design that I suggest you read.

 

[TehMightyEngineer]

I have the 4th edition of MacGregor and the example appears only slightly changed from what howarts quoted; so I believe this particular example hasn't been touched in a while by MacGregor. Take that information for what you want.

Also let's be clear on something. Say you hold a rebar and you pull it with your bare hands (say you were Superman) until it yields, the strength to yield it by mere pulling is called the probable moment strength right? It is independent from the dead or live load.

Correct by my understanding. M[sub]pr[/sub] is similar to seismic design per AISC for steel; it's the probable moment strength used to establish an upward bound shear force associated with that probable moment. Essentially it's the highest moment (and associated shear force) you can expect to obtain from swaying of the frame due to the inelastic seismic demand as beyond that moment the frame rebar goes inelastic and the deformation associated with this will give you a limit to the moment and shear.

Are the moments from this formula taken from the existing moments of the dead and live load that is pushed to 1.25 overstrength (hence yielding).. or are they separate independent moments created from the yielding process itself (independent of the loads)??

The M[sub]pr[/sub] moments and the associated shears to develop those moments are independent of any loads.

Given that this is an upward bound limit for the sway moment; the added gravity load shear is conservative (as you mentioned) if you use the 1.2D + 1.6L. Perhaps this is the only reason MacGregor used it; conservative design. However, ACI 318 is plenty explicit in my mind that this is not required and I would agree. It seems excessive to design for the structure to see both a live load 60% above it's design live load while simultaneously undergoing a code-level seismic event. That odds of that seem so remote as to not be practical.

Professional Engineer (ME, NH, MA) Structural Engineer (IL)
American Concrete Industries

http://www.americanconcrete.com

 

[howarts]

The reinforcing in the beam is determined by making sure the design strength (including phi factor) is greater than the factored loads for all load combinations required by the building code. Once you know the reinforcing in the beam, the seismic demand is determined using a capacity design procedure assuming that the beam has yielded at each end. This seismic demand is now independent of the code loads, and if you were to arbitrarily increase the reinforcing in the beam, you would correspondingly increase the seismic demand even though the code loads did not change. The seismic loads need to be combined with the gravity loads to get the maximum load effect, but the load factors on the gravity loads are based on the load combinations that include seismic. You wouldn't combine seismic loads with the full factored gravity loads because the two are statistically unlikely to occur at the same time. It's the same reason we don't design for the maximum considered earthquake and a hurricane simultaneously...the chances of it occurring are highly improbable and to do so would be overly conservative. McGregor is great for concrete design, but for seismic design using capacity design procedures I suggest you read Paulay and Priestley, Moehle, and Englekirk. There is also a NEHRP Techinical Brief on concrete moment frame design that I suggest you read.

Do you know why the load factors on the gravity loads that is based on the load combination that include seismic have lower value than full factored gravity loads (1.2 D + 1.6 L)? Does it mean no seismic movement can exceed the strength from 1.2 D + 1.6 L? Then why use other seismic load combination when one can just use 1.2 D + 1.6 L?

 

[Deker]

The load factors on gravity loads are smaller in the seismic load combinations because they represent a more realistic estimate of the gravity loads that might be present during an earthquake. You could combine seismic loads with 1.2D+1.6L, but it is not required by code and as TME and I pointed out above, it would be overly conservative.

 

[howarts]

The above is from the book Reinforced Concrete (A Fundamental Approach ) by Edward G. Nawy.
In the above what load combination is wu= 1.2 D + 1.6 L + 1.4E? Where is it in ACI 318-11 or the latest? There seems to be codes that make use of it. Is it seismic design that is based on elastic strength or capacity design?

Anyway. Deker uses Vpr + (0.9-0.2Sds)D.. the ACI seems to use VPR + wu of 1.2D + 1L + 0.2S (timesl/2)
What is the standard usage that others really use?
Guys. What do you use in your code for seismic load combination with probable moment strength?

 

[Deker]

In practice I use (1.2+0.2Sds)D + f1L + Vpr, and I use Vc=0 due to high seismic loads. I was only trying to point out that you need to check all of the seismic load combinations when checking that provision. What is the governing building code where you practice?