Intermediate Moment Frame- Design Requirements as per ACI 318 1

[SBRC]

Greetings,
Dear Engrs,

I am designing a building in Seismic Design Category "C", lateral force resisting system is Intermediate moment resistant frames.As per section 21.3.3 of ACI 318, which states that shear strength shall be taken as smaller of the two methods.

I have following questions:

1. Shall I add gravity load case 1.2D+L and take shear from this combo only or I can use 1.2D+1.6L or any other gravity load combo which is governing in shear.

2. After calculating shear from step 1, then I will calculate shear from factored end moments at both sides of beam due to combo which has maximum earthquake forces(E).

3. In step 3, I ll add both of them, and compare it with another combo which contains double earthquake as specified in 21.3.3.b,and consider smaller of these values.

4.What is the difference in nominal moment arising at ends due to sway and shall we will take plastic moments at both ends and how to take plastic moment capacity in ETABS?

5.[/b] AS per my CSI knowledge, ETABS already do all these checks once I have assigned frame design preference to Category "C". AM i correct in understanding ETABS CFD in Seismic design category C as shown below, Otherwise it will be a never ending process to check each and every beam and column.

Please clarify the above points

Thank you

 

[HTURKAK]

As per you info, SDC =C ,Frame Type = IMF (Intermediate Moment Frame) ACI 318 Requirements Sec. 21.3.3. The Performance Objectives is to avoid shear failures in beams and columns and plastic hinge development.

I just want to clarify the concept of ACI 21.3.3 . I copied and pasted the same picture below ,

Ø Vn shall not be less than the smaller of (a) and (b)

Requirement (a) ; Ø Vn ≥ ( ((Mnl + Mnr )/ ln ) +,- wu*ln/2 ) where wu= 1.2D + 1.0L + 0.2S , Mnl and Mnr are nominal flexural strength. Mnr and Mnl are calculated with fs= 1.25 fy and Ø = 1.0

OR,

Requirement (b) ; Design load combination with 2*E U=1.2D + 1.0L + 0.2S + 2.0E
Calculate Vul and Vur for this combination, Ø Vn ≥ Vu

I think the commentary R 21.3 is clear....pls read again..

 

[r13]

Other than load combination U = 0.9D + 1.6W + 1.6H, the load combination in the example produces the least gravity load shear forces at the beam end, which meets the code intent - "ØVn...shall not be less than the smaller of (a) and (b):", and in (a) "....and the shear calculated for factored gravity load". The code is clear in its intention to assure a least shear capacity for members subject to seismic loadings, however, I think this clause is poorly stated, and one might take the wrong impression, that if condition (b) yields a larger shear than (a), the factored nominal shear capacity only needs to satisfy the shear force derived from (a), which is the "smaller" of the two. Am I missing something here? Or, is the provision really necessary? I appreciate any comments.

 

[HTURKAK]

[quote="ØVn...shall not be less than the smaller of (a) and (b):", and in (a) "....and the shear calculated for factored gravity load". The code is clear in its intention to assure a least shear capacity for members subject to seismic loadings, however, I think this clause is poorly stated, and one might take the wrong impression, that if condition (b) yields a larger shear than (a), the factored nominal shear capacity only needs to satisfy the shear force derived from (a), which is the "smaller" of the two. Am I missing something here? Or, is the provision really necessary? I appreciate any comments.][/quote]

The requirement (a) in general more stringent than (b). For SMF, this is the only requirement and SMF is more stringent than IMF.
The requirement (a) is calculated with the formula Vu= ( ((Mnl + Mnr )/ ln ) +,- wu*ln/2 ) where, Mnl and Mnr are nominal flexural strength and calculated with fs= 1.25 fy and Ø = 1.0

Apparently the requirement (b) is contributed for IMF to relax the requirement for beam shear strength.

if condition (b) yields a larger shear than (a), the factored nominal shear capacity only needs to satisfy the shear force derived from (a), which is the "smaller" of the two. That is true !...

 

[r13]

Thanks.

1) If (b) is smaller than (a), then the code says ØVn > (b), in this case, who will be insane enough to design shear capacity per (b) but (a)?
2) If (b) is larger than (a), as you stated correctly, (b) fits the requirement as the "smaller".

Here is my confusion - is there any case that (a) will not be the default minimum ØVn threshold? To me, this provision seems unnecessary, but cause confusion.

 

[cristiano.ronaldo]

ACI 318-2008 is outdated. Please get new one (2019).

 

[HTURKAK]

IMO , the code is clear, if ( a ) < ( b ), the threshold is ( a )....otherwise the threshold will be ( b ).
There will be cases where (b) < (a) depending on the longitudinal reinf. provided, the span length and magnitude of earthquake forces ,E.

I have API 318-14 The wording somehow different ;

18.4.2.3 ϕVn shall be at least the lesser of (a) and (b):[/b]

(a) The sum of the shear associated with development of
nominal moment strengths of the beam at each restrained
end of the clear span due to reverse curvature bending and
the shear calculated for factored gravity loads

(b) The maximum shear obtained from design load
combinations that include E, with E taken as twice that
prescribed by the general building code]

 

[r13]

Stay on 318-08, the code is clear on selection (a) or (b) to be the minimum (greater than the smaller of the two), but I suspect the hidden intention is setting the maximum as graphically shown below.

318-14 has only changed condition (a) from the least gravity load combination to any gravity load combination plus the shear from lateral load effect. The hidden message stays - ØVn < Shear from any load combination + 2E (the ceiling of shear capacity for any members involving seismic effect).