JAE - Question about an old durability post 1

[tngolfer]

JAE, I have a question about a post you made a couple of years ago. It is now closed so I can't put this question there. Can you help me understand your reasoning in Step 9 below on multiplying Sd x Ms and not fy/fs x Ms while using Phi=1? This has me hung up as I try to understand ACI 350 and how it bounces back and forth between service levels and factored levels.

Thanks.

JAE (Structural) 4 Jun 10 9:14
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]

tngolfer - You asked:

help me understand your reasoning in Step 9 below on multiplying Sd x Ms and not fy/fs x Ms while using Phi=1?

Sd should not be multiplied by Ms but rather by Mu.

The Sd factor is intended to be applied to U which is a factored effect.
U has load factors of varying degrees such as 1.2D + 1.6H, etc.
The gamma factor ([γ]) is essentially the effective or weighted average of the load factors.

Sd is defined as [φ]fs / [γ]fy

So Sd x U for moments is Sd x Mu = [φ]fs / [γ]fy x Mu

So since Mu = [γ]Ms we get:

Sd x Mu = [φ]fy / [γ]fs x Mu

Per the code we use [φ] = 1.0.

So all this reduces to Sd x Mu = fy/fs x Ms


This means that the Sd factor removes the load factors and phi factor and instead of those uses Sd to factor up the service moment Ms to a higher level than the ACI 318 amounts.

 

[tngolfer]

So Step 9 should be Sd x gamma x Ms?

Am I doing two checks for each section?
1. Service level check against fs
2. Ultimate level check using fy

Thanks for your help.

 

[JAE]

No - Step 9 is Sd x Mu

which is the same as

[φ]fy / [γ]fs x Mu

which is the same as

fy/fs x Ms with [φ] = 1.0

 

[JAE]

...and one design check using Sd x Ms as your moment.

 

[tngolfer]

Sorry for beating a dead horse here. but I want to clarify this for others that are reading this sometime in the future. I think we are on the same page but you have me confused. Maybe it was a typo. Two posts up you say:

Sd x Mu (which I agree with. )

The post above says:

Sd x Ms. (I think this should be fy/fs x Ms)

This is the main part of my confusion. If Sd x Mu = fy/fs x Ms then am I correct in saying that I can basically neglect the term Sd by determing my steel rebar size and spacing by meeting the fs requirements of Ch. 10 and multiplying fy/fs by my service moment, Ms, to determine my ultimate moment, Sd x Mu?

(You deserve more than the 1 star I can give your post for getting me through this. Again, I appreciate the help.)

 

[JAE]

Sorry - two posts up should have been: ...and one design check using Sd x Mu as your moment.

 

[tngolfer]

If I'm checking my design capacity ratio (SdMu / Phi Mn), won't it always be 1.0 between fs,min and fs,max?

Sd*Mu = (fy/fs)Ms
= (fy/fs)As*fs*(d-a/2), fs in numerator and denominator cancel eachother out leaving
= As*fy*(d-a/2)

Phi*Mn = Phi*As*fy*(d-a/2) phi=1.0 therefore
= As*fy*(d-a/2)

As*fy*(d-a/2) = 1.0
As*fy*(d-a/2)

 

[JAE]

The fs of fy/fs is not the actual fs from your service moment but rather the "permissible" fs from 10.6.4. (see step 8 from my list above that you quoted)

Attached is an example calculation spreadsheet that we developed. Most of the variables are indicated in it - perhaps you can check through the calcs by hand.

From above with some emphasized portions:

1. Determine the FACTORED AND THE SERVICE moments 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 the actual fs as the service 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 and is called the PERMISSIBLE STRESS fs(max)
6. Verify that actual fs is less than the permissible stress fs(max) .
7. Calculate the maximum permissible fs (fs(permissible)) from 10.6.4. (this item is redundant as step 5 already had it)
8. Calculate Sd using the Permissible fs(max), your phi factor, your [γ] load factor, and fy
9. Your design factored moment is your factored moment times the Sd factor. No other load factors should be used. (I previously had this as service moment and was incorrect - Sd is applied to U and U is which is a factored effect such as Mu)
10. Use your SdMu value with phi = 1.0 and design as you would any other concrete member. Sd varies from a low of 1.0 to perhaps upwards of 3.0 should you use large bars with large spacings.

Sorry for the confusion as the list you referred to was incorrect. Thanks for keeping at this as I'm glad I could correct it.