ACI 350-06 - Environmental Factor for Flexure

[Jacst3]

Has anyone used the "Sd" factor when calculating the moment in a concrete tank wall? It seems that the "Sd" factor for flexural steel as defined in section 9.2.6 and 10.6.4 (ACI 350-06) lead to very high factored moment values. I am calculatng factors of 2.25, 1.9 and 1.7 which are used in addition to and based on the load combination factors defined in section 9.2.1.

ACI 350-01 only required the additional "Sd" fator of 1.3 for flexureal steel. Has anyone had any experience with this. Thanks for your insight !

Jason
Pittsburgh

 

[JAE]

Jacst3,
Take a look at the attached mini-spreadsheet. This is an example of what we have used on a recent tank design.

Note that you use [φ] = 1.0 with the increased moment value from Sd.

The steps we would use are:

1. Determine the SERVICE moment in the member.
2. Determine the distance "a" which is the depth of the compression block under service loads.
3. Calculate the moment arm between the reinforcement and the center of the compression block (half-way across dimension a).
4. Calculate fs as the moment divided by the moment arm, divided by the area of reinforcement.
5. Calculate the maximum reinforcement stress, fs(max) from ACI 350 section 10.6.4. This is based on equations 10-4 or 10-5.
6. Verify that actual fs is less than fs(max).
7. Calculate the maximum permissible fs (fs(permissible)) from 10.6.4.
8. Calculate Sd using the fs(permissible)
9. Your design factored moment is your service moment times the Sd factor. No other load factors should be used.
10. Use your SdMs value with phi = 1.0.


You do get fairly large Sd values. But the key is to use more, smaller bars rather than fewer, larger bars to minimize the Sd.

 

[JAE]

Also - check out this thread where I struggled a bit with the ACI verbiage:

thread507-268671

 

[Jacst3]

JAE,

Thanks alot for your help. Are you sure that the Sd factor is not used with any other factors? Section 9.2.6 states that:

"Required strength U for other than compression-controlled sections ....shall be multiplied by the following environmental durability factor (Sd)...."

U are factored load combinations. This makes me think that Sd is an additional multiplier to the other factors affecting flexural steel.

Also - fs permissable, according to 10.6.4, does not need be less than 24 ksi (two-way action, normal env. exposure). So if fs actual is less than fs permissable, I will just use fs permissable in the Sd eqation.

Thanks,

Jason

 

[JAE]

Also - fs permissable, according to 10.6.4, does not need be less than 24 ksi (two-way action, normal env. exposure). So if fs actual is less than fs permissable, I will just use fs permissable in the Sd eqation.

Yes, this was the hang up I was getting at in the other thread - I was initially using fs actual for the Sd calculation and getting HUGE numbers. After re-reading the language logic in Chapter 9 and in conjunction with 10.6.4 I realized that I should be using the fs permissible.

We used the Sd only for the 1.4F load combination as this was the controlling case. There are other load combinations of course and all should technically be checked for the appropriate Sd factor.

 

[Jacst3]

JAE,

Thanks again - I think I'm following this now. But I'm still not sure of the following: Are you applying both the Sd factor and the load combination factor to the service moment ?

Take the U = 1.4(D + F) Load Combination for example:

Ms = Service Moment
Mu = Factored Moment

Mu = 1.4*(Sd)*Ms

Thanks,

Jason

 

[JAE]

No - just Sd x M(service)

 

[JAE]

Sorry - wrong answer - it is 1.4 x Sd x M(service)

 

[JAE]

This is because the code states that Sd shall be taken times U. And U = service loads x load combinations.

But [φ] = 1.0

 

[JMarch]

JAE,

Still a little confused...

I have read your other threads as well. Can you provide an example using ACI 350-06 LC 9-6?

say fs = 20ksi, fy = 60ksi, phi = 0.9 (flexural case)

What is the Sd factor for LC 9-6?

Thanks

-JMarch

 

[JMarch]

Or anybody.

LC 9-6

U = 0.9D + 1.2F + 1.6W + 1.6H

Do I need to determine Sd for each load case within the load combination?

For dead load:

Sd = (phi*60ksi) / (load factor*20ksi)

Sd = (0.9*60) / (0.9*20) = 3.0

Therefore

Sd*U = (0.9)(3.0)D + 1.2*Sd*F + etc...

Am I thinking of this correctly?

 

[ARLORD]

It is my understanding that:

U = the factored loads per Section 9.2.1
Gamma = factored loads(U)/unfactored loads
fs = fs,max per section 10.6.4.1, or the actual fs
phi = max of 0.9 for tension controlled

Sd = phi(fy)/gamma/fs

Sd varies with gamma because of different U values.
---------------------------------------------------
I think you can only use phi = 1, if you multiply the unfactored loads by fy/fs instead of the factors in the U equations, per Section R9.2.6(1)and(2).

U = fy/fs(unfactored loads)
Sd = fy/fs
phi = 1

Required strength > Sd(U)

 

[ARLORD]

Correction to the above phi = 1 case:

For tension controlled zones:
Sd x U = fy/fs x (unfactored loads)
phi = 1

 

[Dennis59]

Thanks for starting this post Jacst3.
I've been struggling with this also.

I ran some hypothetical cases, and achieved the following results. If anyone can affirm or contest these results, I'd be glad to hear it:

All my cases use a retained liquid of 63 pcf, 21' deep.
All are for a pure cantilever wall holding back liquid.
All are an attempt to determine Mu and the adequacy of steel reinforcing at bottom inside face of wall.
All have 2" clear to face of vertical steel.
All use Fy = 60ksi, F'c = 4.5 ksi
All use a liquid (F) load factor of 1.4
All use Phi = 0.9 for flexure
Calculation is solely for lateral pressure load - ignoring everything else.

By varying the wall thickness and reinforcing, and by going from "normal" to "severe" exposure, the Sd number moves around from 1.13 to 1.49, and Mu changes accordingly.
That is what I have discovered.
Unfactored M at bottom of wall in all cases is 97.2 kip-ft/ft. (.063*21^3/6)


26" wall thickness, "normal" exposure, Verts #8 @ 6"
RESULTS: fs = 34.14, Sd = 1.13 Mu = 153.8 kip-ft/ft and PhiMn = 159.7

24" wall thickness, "normal" exposure, Verts #9 @ 6.5"
RESULTS: fs = 32.21, Sd = 1.20 Mu = 163.0 kip-ft/ft and PhiMn = 168.1

24" wall thickness, "severe" exposure, Verts #9 @ 5.5"
RESULTS: fs = 28.81, Sd = 1.34 Mu = 182.2 kip-ft/ft and PhiMn = 196.5

24" wall thickness, "severe" exposure, Verts #10 @ 6.5"
RESULTS: fs = 25.89, Sd = 1.49 Mu = 202.8 kip-ft/ft and PhiMn = 209.2


(Again, these are all hypothetical, just for the sake of discussing the fs, Sd and Mu calculation)

Thanks for considering.

 

[Jacst3]

Dennis59,

I just took a quick look at your case of "Severe exposure" and 24" thk. walls. I get mostly the same results as you for fs, Sd and Mu with the following exceptions:

For Sd, I use (fs acutal) in ACI equation (9-8), which came to be fs = 28.07 ksi. I believe you used fs, max. for servere exposure. Using fs,actual gives you a slightly higer value; i.e. Sd = 1.37 and Mu = 187.01 K-ft/ft, is what I got.

It seemes to me that the Phi factor is not required to be used when the Sd factor is applied. See ACI Commentary R9.2.6, near the end of the paragraph. Therefore your PhiMu would equal 182.2 Kip-ft/ft or 187.01 K-ft/ft if you use fs,actual.

Also, did you use a cover of 2" or 3" ? Your results seem to be based on 3" of cover; i.e. d = 20.44" for a 24" thick wall and #9 vert. bars.

Also, re-check you fs actual (ACI section 10.6.4.6), I find for the severe exposure and #10@6.5" produce a stress greater than fs,max - which would mean 6.5" spc is no good. Check your req'd spacing in ACI section 10.7, this is also based on fs,actual.

 

[Dennis59]

Jacst3,
Thank you for your response. If you are still in the mood, I'd like to discuss further... (anyone else out there looking at this, I'd appreciate a reply from you also)

I posted four scenarios, and I don't know which one of these you are remarking about with your 2nd, 3rd and 4th paragraphs. Please clarify.
####

For the last case (24" wall with #10 @ 6.5", severe exposure) - here is how I came to my answer:
As = 1.27*12/6.5 = 2.34in²
D = 24-2-1.27/2 = 21.4"

For equation 9-8, ACI 9.2.6.1 says I need to go to eqtn 10-5 to find "permissible" fs. Using 10-5, fs_max = 260/beta/sqrt(S²+4(2+db/2)². For beta I used the approximation of 10.6.4.4, so beta = 1.2. My fs_max = 260/1.2/sqrt{6.5²+4(2+ 1.27/2)²} = 25.89 ksi. Limits on fs for severe case: minimum = 17 (10.6.4.2) and maximum = 36 (10.6.4), so I am between those min and max. I don't understand your reference to ACI 10.7.

Going back to Eqtn 9-8, Sd = phy fy / gamma / fs_max
I am using load case 9-1 so Liquid load factor = 1.4 (I stated this in my assumptions)
so Sd = 0.9 * 60 / 1.4 / 25.89 = 1.49

Mu = M_unfact*LF*Sd = 97.2 * 1.4 * 1.49 = 202.8 kip-ft/ft

Phi Mn>> a = AsFy/.85/F'c/B = 3.06 in.
Phi Mn = 0.9 * 2.34 * 60 * (21.4 - 3.06/2) / 12 = 209.2


Please point out where you think I am going wrong.

Anyone else, please also feel free to chime in and let me know if/where I am going wrong.

For what it's worth, ACI 350-01 would have given me Mu = 97.2 * 1.7 * 1.3 = 214.8, so for this case 350-06 is less conservative than 350-01.
For a typical ~Normal~ case, 350-06 is significantly less conservative than 350-01, according to what I've found. Has anyone else found this to be true?


Thanks for considering.

 

[Jacst3]

Dennis59,

I re-evaluated my numbers and they do match yours alomst exactly. I calculate an Sd of 1.49 and an Mu of 202.82 K-ft. Reasons for the discrepancy with my initial analysis: I was using an f'c of 5 ksi. My calculation of "d" was based on 3" of cover instead of 2" and I was calculating the # 10 rebar area incorrectly. Once I made these changes my results compared to yours.

As an additional check, I do think you should also check (fs, actual) as per 10.6.4.6:
[M(unfac)K-in / As(d - a/2)]

For case #4:
fs = (97.2*12) / 2.34(21.4 - 3.06/2) = 25.08 ksi which is just under fs,max = 25.89 ksi - so all is good. I think (fs,actual) should be used in the calculation of "Sd" in Eq. (9-8), this way you will get a higher (more conservative) factor.

For the rebar spacing check, I meant to say ACI Secton 10.6.5 (Not 10.7)