Guidance with regard to the detailing of beam in seismic zone

[Osama Anwar]

Assalaam o Alliakum Everyone,

Can Anybody guide me regarding the values of L1 and L2 in the image attached below? It is "Fig. 5—Typical seismic-resistant details: flexural members." of ACI-315-99 Detailing Manual. And also are the values same for edge beams and continuous beams?

[Simple Structures]

The attached may help Anchorage ACI 318 calc procedure.pdf

[Osama Anwar]

1 hour ago, Simple Structures said:

The attached may help Anchorage ACI 318 calc procedure.pdf

Thank you so much for the reply. 

I was curious if we will apply this method when we are taking steel areas from ETABs? Does ETABs give values according to Ld or simply after L/3 for continuous part or L/4 for discontinuous part? I knew the procedure to drive Ld but this thing is annoying me. If we calculate the Ld values manually they will be shorter than L/3 (most probably) and there would be some part (L/3-Ld) which may have greater moment than the flexural strength. So won't it be conservative to use L/3.

Can you elaborate for which lengths (Ld or L/3 and L/4) ETABs give the 3 top and 3 bottom longitudinal steel area values? 

 

[Simple Structures]

Analyse on ETABS (and then detail drawing to L/4)- the extra bit helps during an earthquake dynamic loading - L/4 should be fine.

I am not a ETABS user myself .... but am learning that its Pakistan's most favourite analysis programme....wow!

As an engineer you should be able to calculate the MINIMUM anchorage (from previous information i sent you) and then always provide a bit more. My rule of thumb sketched below.

 

 

 

[Osama Anwar]

As I said 

2 hours ago, Osama Anwar said:

I knew the procedure to drive Ld

I know how to calculate Ld for tension and compression and Ldh etc. Thing that confused me was that we do not have the Bending moment envelop. But I know Max +ive and -ive values from the software. And I have checked Longitudinal area of steel against them it is equal to manual calculations.

Suppose if my steel is as follows (which it is for one beam)

My B.M. diagram would be SOMEWHAT similar to this not necessarily.

Seismic zone is 3 so I am getting high negative moments at ends.

So my Reinforcement would be somewhat like this

Software must divide the beam in three pieces like this to get the value of highest moment in each section

ETABs do divide the section in three parts because sometime it would give highest value say at station 7.5". In short what I want to say is that the reinforcement values are not the values of moment at left end, middle and right end of beam but rather maximum moment values in the three respective sections. 

Suppose if steel area is 1.71 in left section then the highest moment (for which As came out to be 1.71) in that section may be lying at the start, end or in between of that section. What I am interested in is that what is the length of the section used by ETABs? In what proportion ETABs divide sections? Are all section L/3? Or first and last section is L/4 and middle is (L-2*L/4).

I calculated Ld and min length values as advised by you and got the following results

Where Ln is clear length. 

If I am not mistaken, by 0.15*L+h you meant Ld>=0.15*L+h. Right?

In that case, top length would be 3.8375' and bottom would be 2.05. Right?

What I am is saying that what if ETABs is dividing sections in L/3 (or Ln/3) or even L/4 (or Ln/4) and Max moment is lying at the very end of section 1 at L/3 (or Ln/3) then would not it be under design if we take value as calculated through Ld>=0.15*L+h? Should not we be using Max(L/3,Ld,0.15*L+h)? 

 

 

[Badar (BAZ)]

ETABS divides the region of member to report forces and perform design based on the option shown in attached screen shot.

You can divide the member into different regions either by specifying minimum number of stations(sections) or maximum spacing of stations(sections).

The default setting is that it will check the beam at every section located at 500 mm distance along the clear length of the member (max station spacing).

See the attached screen shot of the ETABS's manual (CFD ACI 318-14) as well.

[Simple Structures]

1. Apologies - I can only think in metric units - so ASSUMING all the above numbers to be metric units , for argument sake, 

Then, If L/4 = 2.75, then L = 11 m

Therefore, 0.15L = 1.65m + depth of beam ..... which you say then totals 3.835m : This would mean your beam is 2.185m deep (L/5) which cannot be right! Get an Engineer in your office to check you numbers!

With regards to rebar from fixed ends: Left or right , one would take 100% of the reinforcement for distance GREATER THAN 0.15L+h, then take 60% to distance 0.3L+4, and so on. You could decide to take 100% to 0.3L+h length. It depends how many bars you have in top. These are detailing rules are guidance and you as the engineer need to make safe choices based on that.

For example, If you have 2 bars on top as 100% reinforcement, at say 0.15L+h you can splice them with smaller bars (60%), than after 0.3L+h, you can splice them again with small bars to take 40% etc. With so many splices, this is not practical, increases tonnage of rebar ... so based on ones engineering judgement, one would take the two bars to say 0.3L+ha and splice there with two small bars and just have two bars in top..... remember you have to do this from both ends, so it may be better and practical to take the 100% bars all the way across the top of the beam.

If you had 5 bars on top, you can go continue with say 3 bars after the point beyond 0.15L+h (stop two there) and then you are continuing on with 60% etc.

With seismic loading you will get M reversal at bottom (-ve M) so you may need to follow same rules for bottom bars.

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2. Secondly, you must appreciate anchorage of top (& bottom) bars into the column needs to be sufficient to avoid joint failure. Calculate the anchorage length as shown "blue" on sketch below.... what the computer printout illustrates is inadequate! Your joint will fail under seismic loading .... or use long U-bars at ends in the  beam. You must provide sufficient anchorage of beam bars into the column, both top and bottom bars.... Calculate anchorage length from centre-line of the column !

Master "reinforcement curtailment into the beam &, anchorage into the column or wall".

 

 

3. The BM envelope will come from gravity loads + seismic loads. Most computer programme give the BM diagrams envelopes for various load combinations.  Like I say I am not a ETABs user, leave to that to others to advise.

Below is an example - based on Eurocode EC2, the code i use - for beam pinned at LH side and continuous on the right... showing typical reinforcement curtailment based on BM. This is a general illustration only, and shows how BM Envelopes and Curtailment of reinforcement works.

When I trained as a structural engineer I had to follow a training sequence and learn three things: 1. Work as reinforcement detailer/draughtsman for two years, then work out and draw bending moment diagrams by hand for another few years for beams and frames, and then design by computer! 

 

 

 

 

 

[Osama Anwar]

8 hours ago, Badar (BAZ) said:

ETABS divides the region of member to report forces and perform design based on the option shown in attached screen shot.

You can divide the member into different regions either by specifying minimum number of stations(sections) or maximum spacing of stations(sections).

The default setting is that it will check the beam at every section located at 500 mm distance along the clear length of the member (max station spacing).

See the attached screen shot of the ETABS's manual (CFD ACI 318-14) as well.

Thank you so much for the reply.

8 hours ago, Badar (BAZ) said:

The default setting is that it will check the beam at every section located at 500 mm distance along the clear length of the member

ETABs neither considers full length (c/c) nor clear length while dividing the member into sections but average of both full length (c/c) and clear length like this

My default is at Maximum Station Spacing and number of stations are 24. This means that ETABs divide each member in 24 parts and report values for each load combination. I have checked it.

When we design our model ETABs give out 3 values for top and bottom reinforcement. I am interested to know that if members are being divided into 24 parts/sections/stations then each of the 3 values is being reported from highest of the 8 parts/sections/stations? i.e. left value is the highest values reported in the 8 parts/sections/stations on the left side and middle value is the highest values reported in the 8 parts/sections/stations located in the middle part of the member and right value is the highest values reported in the 8 parts/sections/stations located on the right side of the member. Right?

[Osama Anwar]

On 7/10/2020 at 1:21 PM, Simple Structures said:

1. Apologies - I can only think in metric units - so ASSUMING all the above numbers to be metric units , for argument sake, 

Then, If L/4 = 2.75, then L = 11 m

Therefore, 0.15L = 1.65m + depth of beam ..... which you say then totals 3.835m : This would mean your beam is 2.185m deep (L/5) which cannot be right! Get an Engineer in your office to check you numbers!

Thanks for the reply

I deal with imperial units L/4 is not equal to 2.75. Ln/4 is equal to 2.75. And Ln=11' not L=11'. My beam depth is 24" and L is 12.25"

so 0.15*L+h = 0.15*12.25 + 24/12 = 3.875'

On 7/10/2020 at 1:21 PM, Simple Structures said:

 

As I said my beam depth is 24" and I am using #6 (0.75") bar. Hook length greater than 12db (12*0.75=8) does not contribute in holding concrete. I see no reason to increase my hook length unnecessarily.

Edited July 11, 2020 by Osama Anwar