ACI 350 Flexure 2

[Rabbit12]

I've read a handful of past threads on this, but wanted some assurance I'm doing this correctly.


To determine the design moment we take: S_d*M_strength

This would then be compared to M=A_s*F_y*(d-a/2)


I'm checking a design that calculates S_d=f_y/f_s (assume phi and gamma are 1.0) and then take S_d*M_service. The design then compares that amplified moment to M=A_s*F_y*(d-a/2).

I seem to be able to get a little more out of my section if I actually calculate S_d using phi=.9 and gamma equal to my load factor.

 

[lemmiwinksownz]

I'm a bit confused. What is S_d? There is a value, S, know as the environmental factor. Your moment demand is always multiplied by S=1.3. Shear is also multiplied by S=1.3 if additional shear capacity from steel, Vs, is required.

You need to determine the controlling load case, U, and multiply it by S. Which version of ACI 350 are you using? I don't recall seeing S_d.

 

[Rabbit12]

This is ACI 350-06. To my knowledge this is the newest version of this code.

S_d is indeed the environmental durability factor. It is not always 1.3. The equation (9-8) is (phi*f_y)/(gamma*f_s).

 

[lemmiwinksownz]

How interesting. I was not aware of this change from '01 to '06. I wonder if my firm has continued with S=1.3 based off the commentary - see C.9.2.9.1.

Do you only have one load factor (i.e. soil/fluid load on a wall)?

 

[strucbells]

Yes, ACI 350-06 and Sd is the latest code approach in the US. Rabbit12 did you read the commentary next to that equation?

ACI 350-06 Commentary R9.2.6 seems to suggest that the design approach you are asking about listed above may be valid depending on what your "M" value represents

A couple of questions:

1. Assume this is for a tension-controlled section?
2. You are holding Sd >= 1.0 for your checks, right?
3. Are you including phi for LRFD checks on your capacity side? You didn't explicity list if your "M" value included any resistance factors.

 

[r13]

Something not right here. Assume M_service is non-factored moment,

M_strength = LF*M_service
M_u = As*fy(d-a/2)
M_u = S_d*M_strength = S_d*LF*M_service = (fy/fs)*M_service ??? The last expression is valid in this form," = S_d*(fy/fs)*M_service", in which fy/fs = LF, and fs is steel stress calculated using ASD.

For last expression to be valid, fy/fs must equal to S_d*LF, however, you told us that in the calculation, S_d = fy/fs, where is the LF then? And what is fs, how was it determined? Quite confusing here.

 

[strucbells]

In ACI 350-06 fs is a permissible tensile stress value that varies from 17 ksi - 24 ksi depending on whether there is one-way or two-way action and depending on the severity of the environmental exposure.

I'm confused by some of the nomenclature being used here: Rabbit12's vague "M" and retired13, I'd swap your Mu for phi*Mn, and your Mstrength for Mu. Where is your phi?

Usually we just divide phi*Mn by Sd and check against Mu. Use your typical phi and gamma values per the code to calc Sd. That may be a simpler approach

 

[Rabbit12]

structbells,

I have read the commentary. It's pretty confusing to be honest.

To answer your questions:

1) Yes, let's assume tension controlled.
2) Yes, S_d must be greater than or equal to 1.0 per equation 9-8.
3) Which M are you referring too? The M=A_s*f_y*(d-a/2)?

I really think this is just a general question more so than explicitly about "my" design. It seems to me if you want to use service level moments you need to take M_service*S_d and compare that to M=As*fs*(d-a/2) if you want to use f_y instead of f_s you'd need to add the load factor to the S_d*M_service calc.

 

[r13]

I dropped strength reduction factor for simplification. The only thing I wanted to point out is that there is a missing term in the whole mix - the equations do not equate.

 

[r13]

From earlier version of ACI350,

9.2.8 — Required strength U shall be multiplied by the following environmental durability factors (S) in portions of an environmental engineering concrete structure where durability, liquid-tightness, or similar serviceability are considerations.

So, ØM_n = ØAs*fy*(d-a/2) >= S_d*M_strength= S_d*LF*M_service

S_d = 1.65 for axial tension, 1.3 for flexural, and 1,3 for excessive shear carried by shear reinforcement.

 

[strucbells]

retired13, think I've got it. I'll use your LF designation instead of ACI 350's "gamma" for clarity:

ØMn = ØAs*fy*(d-a/2) >= Sd*Mstrength= Sd*LF*Mservice

If we substitute in Sd = Ø*fy/(LF*fs) into your expression on the right, we get:

ØAs*fy*(d-a/2)>= Ø*fy/(LF*fs)*LF*Mservice

Let the LFs and Øs drop out and you get:

As*fy*(d-a/2)>= fy/fs * Mservice

Does that clarify things Rabbit12 and explain the commentary section now?

 

[r13]

Clever! Well explained. Thanks.

 

[Rabbit12]

strucbells,

Thanks for that explanation. When I first looked at the calculation we were comparing a service load to a capacity calculated with Fy. That seemed like we were comparing apples to oranges, especially since we actually calculated Fs.