Waqas Haider

Is it from some document? Kindly give reference. I would like to study more about it. thanks.

Dear waqas,
Calculation of Ie for live load only doesn't make sense as live load will always act on the section after the application of dead loads and will act simultaneously with dead loads.
Therefore , Ie must be calculated for possible combinations that could be 1.0D,1.0D+0.5L & 1.0D+1.0L as per ACI-435.
Detailed calculations of deflection for 1-way NP flexural members as per ACI 435 is shown in attachment.
 
 
 
 

There is no lower limit for T but there is an upper limit for Maximum V, so even if the time period calculated by ETABS is significantly lower than Ta, code provisions for maximum base shear would govern and the calculated value of T would be ignored.
Similarly, no vertical distribution of force would take place for a small time period hence your design would still be in reasonable FOS limits.

For the exact quote, Google SK Gosh Associates and start of this quote. The statement reiterates the basis of strength design earthquake design being followed. Read the codes that you are using. They all refer to strength level earthquake forces.
Thanks.

Really very nice gathering.
it was my first meeting with all the members but due to our discussions on this forum for the last couple of years, I didn't even feel once for the 3-4 hours meeting that we were meeting for the first time.
Thank you all....
 

Great..!!! I m only familiar with Engr Junaid because he is my class fellow, with Umar Makhzumi Bhai, and With Waqar Saleem Brother. Today I came to know what is full form of BAZ  And he has changed his looks also. By the way, I used to think Engr Uzair Sir quite younger but he is quite senior to us. Mostly people after getting senior does not use social media forums much.. especially for responding people.. Really nice to know the members. This forum is really great.!!

Modifiers are related to cracking and loss of stiffness before reaching ultimate limit state (ULS) which happens in concrete. You wouldn't provide stiffness modifiers for steel or glass as there is no cracking before and change in stiffness. Cracking of glass is failure.
Thanks.

Great..!!! I m only familiar with Engr Junaid because he is my class fellow, with Umar Makhzumi Bhai, and With Waqar Saleem Brother. Today I came to know what is full form of BAZ  And he has changed his looks also. By the way, I used to think Engr Uzair Sir quite younger but he is quite senior to us. Mostly people after getting senior does not use social media forums much.. especially for responding people.. Really nice to know the members. This forum is really great.!!

If you are having a geotechnical report with you, there must be clearly stated value of spring constant with the name Modulus of Subgrade Reaction. Incase soil investigation have not been done on site for any reasons, you can use a tentative value for k by following relation.
k = 3 x qa x 12 (in FPS in k/ft3)
k = 3 x qa x 40 (in SI in KN/m3)
where qa = allowable bearing capacity. and 3 is Factor of safety applied on soil while determining bearing capacity of soil.  For the reference of this formula, please read the attached document.
Correlation_BC_and_K.pdf

You must export combos related to all loads i.e. gravity and lateral loads because ultimately effects due to all the loads is going to be transferred to foundation. Designing foundation only for gravity loading is not correct approach. But remember, since mostly bearing capacity provided by Geotechnical consultant is safe bearing capacity instead of ultimate bearing capacity (safe bearing capacity means full factor of safety has been applied to bearing capacity as we do in ASD approach). Hence if we use this bearing capacity for design of foundations, loading combos of ASD or unfactored loading combos must be used. UBC and IBC provides these laod combos and footing must be designed for all this loading combos instead of autogenerated loading combos of ETABS which are factored. Either define these loading combos in etabs and export to safe or directly define then in safe.
 
Increase of bearing capacity upto 33% is not allowed in all the cases. It can only be used where in loading combos having two or more variable/Transient loads togather, we have not reduced the intensity of transient loads. If we have reduced these loads by 0.75 factor then bearing capacity can not be increased by 33%. Basically it is not the case with bearing capacity only. Any strength determined by using ASD approach (i.e. full factor of safety applied on strength of member, ) can be increased by this approach but only if in loading combos transient loads have not been reduced. UBC and IBC proposes two types of loading combos for ASD. in first type of combos you may find for example a combo D+L+W and in 2nd type you can find the same combo as D+0.75L+0.75W. if you use first type of combo you can increase by 33% and if u r using 2nd type of combo, note that the loads have been reduced by 1/1.33 = 0.75 times instead of increasing strength. so same factor of 1.33 has been applied on load side so in this case no increase can be made in strength.
For a detailed study refer to following document.
1-3rd stress increase AISC.pdf

Great..!!! I m only familiar with Engr Junaid because he is my class fellow, with Umar Makhzumi Bhai, and With Waqar Saleem Brother. Today I came to know what is full form of BAZ  And he has changed his looks also. By the way, I used to think Engr Uzair Sir quite younger but he is quite senior to us. Mostly people after getting senior does not use social media forums much.. especially for responding people.. Really nice to know the members. This forum is really great.!!

Many engineers are confused with stiffness modifiers stuff in ETABS including me. Let me add my cent here.
 
1. Any reasonable set of stiffness can be used as explained in ACI 318.
 
2. You have to determine what type of frame you are analyzing; braced or un-braced
 
3. Whats the analysis type? Strength, service?
 
4. Code allows use of 0.35,0.7 factors on inertia in chapter 10 of ACI 318 for slenderness effects.
 
5. Also according to code, you can use the above same model in step 4 for lateral deflections.
 
6. Now for lateral deflections, if the lateral load is service (like Wind load of ASCE 7-05 and previous) multiply above factors by  1.43 or 1.40 (see code). For strength lateral load like Earthquake, no need to multiply by this factor.
 
7. Some times you really dont need to apply modifiers at all. For example for strength design.
 
8. For a two-way frame with membrane slabs at top, you just need to apply 0.5 factor to beam elements. Code allows this which says for strength design you can either
 a. use the same modifiers as used for slenderness
 b. use 0.5 for beam stems only
 
 
In simple words, modifiers are factors to reduce inertia for cracked sections. But if you are doing strength design, why you need the cracked inertia.
 
And as far as serviceability is concerned, authors like Nilson even argue that a factor of 0.5 as we applied in step 8 can also be skipped because of the following reasons.
 
 a. For positive moment, beam is designed as T section but in ETABS we use rectangular section, Stiffness of T = approx 2 x stiffness of rect. So no need to apply 0.5 inertia factor to beams because T compensates for that. Just use the rectangular section in ETABS.
 
b. For negative moments, beam is designed as rectangular, and we also model the rectangular section in ETABS. Now the cracking in this section is offset by the continuation of bottom bars into support which have stiffening effect.
 
I hope that helps. Let me put it as a summary here;
 
 
1. Strength & service design with membranes (slab on rigid beams as compared to slabs).
 
Beams = 0.5
 
or beams = 0.35 & columns = 0.7 (or whatever ratios just keep it constant).
 
or no modifiers at all
 
walls = 0.7 for un-cracked
            0.35 for cracked
 
2. Strength & service design with shells (flat plates etc).
 
beams and walls same as above
slabs  = 0.25 for out-of-plane m factors.
 
just multiply above factors by 1.43 for service lateral deflections.

Great..!!! I m only familiar with Engr Junaid because he is my class fellow, with Umar Makhzumi Bhai, and With Waqar Saleem Brother. Today I came to know what is full form of BAZ  And he has changed his looks also. By the way, I used to think Engr Uzair Sir quite younger but he is quite senior to us. Mostly people after getting senior does not use social media forums much.. especially for responding people.. Really nice to know the members. This forum is really great.!!

You must export combos related to all loads i.e. gravity and lateral loads because ultimately effects due to all the loads is going to be transferred to foundation. Designing foundation only for gravity loading is not correct approach. But remember, since mostly bearing capacity provided by Geotechnical consultant is safe bearing capacity instead of ultimate bearing capacity (safe bearing capacity means full factor of safety has been applied to bearing capacity as we do in ASD approach). Hence if we use this bearing capacity for design of foundations, loading combos of ASD or unfactored loading combos must be used. UBC and IBC provides these laod combos and footing must be designed for all this loading combos instead of autogenerated loading combos of ETABS which are factored. Either define these loading combos in etabs and export to safe or directly define then in safe.
 
Increase of bearing capacity upto 33% is not allowed in all the cases. It can only be used where in loading combos having two or more variable/Transient loads togather, we have not reduced the intensity of transient loads. If we have reduced these loads by 0.75 factor then bearing capacity can not be increased by 33%. Basically it is not the case with bearing capacity only. Any strength determined by using ASD approach (i.e. full factor of safety applied on strength of member, ) can be increased by this approach but only if in loading combos transient loads have not been reduced. UBC and IBC proposes two types of loading combos for ASD. in first type of combos you may find for example a combo D+L+W and in 2nd type you can find the same combo as D+0.75L+0.75W. if you use first type of combo you can increase by 33% and if u r using 2nd type of combo, note that the loads have been reduced by 1/1.33 = 0.75 times instead of increasing strength. so same factor of 1.33 has been applied on load side so in this case no increase can be made in strength.
For a detailed study refer to following document.
1-3rd stress increase AISC.pdf

In regular non inverted beams, the negative steel of slab rests on the upper portion of beam which is in compression and hence fully available for end bearing load of slab.
But in inverted beams, this negative steel rests on the lower portion of beam which is in tension (in middle span) and hence will be having flexural cracks (though up to minor crack widths), I m amazed if this cracking affects the bearing strength in any way.

As you said, according to R7.12.1.2, first of all, reduce the modulus of rupture by half i.e. from 0.62√f'c to 0.31√f'c.
Second, increase the reinforcement as required for axial tension (T=phi As fy) in addition to what required for moment.

Waqas,
The statement is asking you not to rely on temp and shrinkage steel values provided for cases where analysis shows higher amount is required. You can carry out the temperature analysis and add to that the amount of steel required due to shrinkage. You can calculate extra steel due to shrinkage based on how much strain would be caused by shrinkage or any other method if you are aware of.
Thanks.
Thanks.

@Ahsan Kazmi
right click the column and go to flexure details There you can find the design combo Check if it includes lateral loads It will also tell you the design moments for this reinforcement

The beam with larger depth should be the main beam, and the beam smaller depth should be the secondary beam. By main beam, I mean that the beam has ability to attract forces from adjoining members because of its larger stiffness.
Keeping that thing in mind, secondary beams should have been transferring the reaction directly to main beams, and main beams should have been transferring the loads of secondary beams, which are carrying the load of metal deck and the live load, to concrete beams.
In the figure, they have placed larger and smaller size beams in the same directions. If smaller size are able to carry load, as is evident from the standing structure, why use larger beams. They have wasted material.

W salam,
In monolithic construction , T or rec beam is not a matter of choice of designer but its indicated by analysis that if depth of compression block lies below flange (flange = slab thk in beam-slab system) then the compressive force in concrete is balanced by a certain width of flange + total width of web, otherwise only web balance compressive force i.e example of rec section carrying slab load in monolithic construction.
Example of rec beam sections also exists in some specific cases of construction for eg RC beams supporting metal deck system,hollow core slab panels and precast construction system.
ETABS checks the eqn "a < or > ds" for maximum analyzed Mu-bott and design positive R/F accordingly.
 
 

Yes. How? Go through the following paragraphs.
In case of ASD, service loads are used, and both dead and live loads have same load factor of 1. On the other hand, considering the the basic load combination of 1.2 D + 1.6 L, dead load (DL) has a load factor of 1.2 and live load (LL), a load factor of 1.6. Now consider two simple examples.
For the first example, assume that a certain beam has to carry a DL of 30 kips, and a LL of 10 kips. For this beam, total design service load for ASD method will be DL+LL=30+10 = 40 kip. For the Strength Design, basic design load will be 1.2 DL+1.6LL = 1.2(30)+1.6(10) = 52 kips. Strength Design load in this case is  (52-40)/40*100 = 30% larger than the Design Load for ASD.
Now, for the second example, assume that there is another beam, similar to that in the first example, except that the DL & LL values are reverse of those in first example i.,e., now DL = 10 kips, & LL = 30 kips. Total Design Load for ASD method is now  DL+LL=10+30 = 40 kip (same as in first example). However, design load for Strength design changes and it is now 1.2 (10)+1.6(30) =60 kips, which is (60-40)/40*100 = 50 % larger.
Thus, it is very clear that for both the examples, Design load is the same when using ASD method, and resultantly the member size & reinforcement will also be the same. However, since Design load for Strength Design method is larger in second example (when LL is larger than DL), a comparatively larger member size or reinforcement will be required. Thus, ASD method is generally economical (as compared to Strength Design method) when the live load on a member is larger than the dead load.
Regards.

Some portion of slab is used by many engineers to take into the account the effect of line load on the slab. Many call it a beam, because the reinforcement arrangement resembles like a beam. But it cannot technically be called a beam; beam is a member that attracts force by virtue of its stiffness. Since, the thickness of so called beam is equal to that a slab, it cannot attract forces.
It is a simplified way of taking into the account the effect of line load on slab; by arranging reinforcement in this way, one assumes that the effect of line load will be limited within the width of so called beam. So the width of the beam will be decided by the design moment.
The beam will be only designed for the load of the supported wall, or the load of supported wall + the load transferred by the wall from upper levels. The so called beam cannot take any load from adjacent slab area, as it does not have the stiffness to attract any load. So tributary area thing is not applicable to that member.
If you understand what is written above, you should conclude that there is no need to check the deflection.
In many cases, this way of locally reinforcing the slab ( so called concealed beam) will not be able to support the load transferred from above levels, unless you increase the thickness of slab.

Assalam o alaikum Dear friends,
I have two questions. I am to detail SMRF for zone 3. I was reading ACI 318-11 chapter 21. Now i have some confusion.
1) 21.1.1 says, ........................................For which, design forces , related to earth quack forces, have been determined on the bases of ENERGY DISSIPATION IN NONLINEAR RANGE OF RESPONSE.
My question is What is meant by energy dissipation in non linear range of response??
What i have get through my research and understanding is,
When Earth quack comes, due to inertia of building, it produces lateral forces in building. Due to these forces, structure deflect. If members of structure have big amount of reinforcement and cross sections, the force (or in other words energy) exerted by earth quack to the building will not yield the members and structure and members will restore to initial positions. In this case, members will not yield, and will remain in ELASTIC RANGE. But this makes structure too costly. On the other hand, if members have ability to yield but not fail, this energy will be used to yield the members and hence much of this energy will be dissipated in non linear range of members response. And we design structures for this because this is does not make structures too costly. So we design structure in the 2nd way. But due to this energy dissipation , structure is no longer in use at service stage and needs some or much requirement. We compromise over serviceability stage upto some extent and does not compromise over limit stage.. Is my concept right???? Is this the mean of energy dissipation in non linear range of response??
2) Commentary of R 21.1.1 says,
The integrity of the structure in the inelastic range of response should be maintained because the design earth quack forces, defined in documents such as ASCE/SEI 7, the IBC, the UBC and NEHRP provisions are considered less than those corresponding to linear response at the anticipated earthquack intensity.
What does it mean??
Jazak Allah...

When ever we say Allowable bearing capacity, It can be either Gross allowable bearing capacity or net allowable bearing capacity. And it is the most important thing to decide when to add over burden, when to subtract and when not to. 
More over keep this equation before you 
for equilibrium
q(all.gross) > P/A +Y.Df
where 
q(all.gross) = gross bearing capacity
Y.Df = Backfill load
P/A = Applied pressure
 
q(all.net) = q(all.gross)-Y.df = (P/A +Y.Df) - Y.Df 
hence for equilibrium
q(all.net) > P/A
where 
q(all.net) = net allowable bearing capacity
Y.Df = Backfill load
P/A = Applied pressure
  So gross bearing capacity is capacity available to bear pressure of existing soil as well as applied loads and net bearing capacity is capacity available to bear just applied pressures because pressure caused by existing soil has already been deducted.  
Case when NOT to deduct over burden
 
Lets say in my geotechnical report i have been provided by NET bearing capacity of 1 tsf at a depth of 5 ft. Now it means at 5 ft depth, soil is capable to bear pressure of 1 tsf IN EXCESS of existing soil. Because it is NET allowable bearing pressure and pressure of existing soil has already been subtracted/balanced from bearing capacity. Now only extra available bearing capacity is reported as net. The soil can bear 1 tsf of EXTERNAL/APPLIED pressure which includes load of above structure as well as load of footing. Load of footing is also external and applied. In this case, you need not to deduct over burden because over burden is already deducted and you are provided with NET bearing capacity.
But If you have bearing capacity at 5 ft depth but in return you have plan to backfill upto 7 ft, then obviously extra 2 ft soil is applied pressure. which you need to either add in applied pressure or you should deduct it from the bearing capacity.
 
Case when to deduct over burden
But if my geotech report says me that at 5 ft depth, i have 1.25 tsf GROSS allowable bearing capacity. Now it means my soil is capable to bear 1.25 tsf pressure including pressure caused by existing soil. Because it is gross allowable bearing pressure and i have not subtracted pressure of existing soil. The soil can bear 1.25 tsf of total pressure which includes EXTERNAL/APPLIED pressure caused by load of above structure including load of footing, and pressure caused by existing soil . In this case, for the sake of sizing of footing, you need to to deduct weight of over burden soil so that so may find out net bearing capacity available to resist applied load of structure and footing. Note that, after deducting the pressure caused by existing soil i.e. Y.Df = 110*5 = 550 psf = 0.25 tsf, we again get the same 1tsf bearing capacity available for external/applied loads.
 
Case when to ADD over burden
Now Let say you are given with any of capacity. Let say Net allowable capacity of 1tsf at 5 ft depth. But you know that you are not going to backfill the foundation (in case of basement etc), then NOT ONLY this 1 tsf is available for applied pressure but also that 0.25tsf available which was going to be consumed by existing soil backfill. Now you can add this 0.25tsf to your net bearing capacity to get bearing capacity, available for external loads and you need to find out size of foundation according to this available bearing capacity.

Building Drift in ETABS
Drift is a very complex topic in structural engineering. It involves too many factors to arrive at a suitable decision. It involves engineering judgment, the phenomenon fresh engineers might not feel. In this article, I have tried to explain what is building drift, allowable limits, ways and means to check in ETABS models and to control the excessive drift. Please keep in mind, this article is not about the building drift as far as structural science is concerned, rather this topic of drift is related to ETABS software.
First of all you must be familiar with the term story drift. For convenience, I am quoting here the definitions from UBC-97 code:-
STORY DRIFT is the lateral displacement of one level relative
to the level above or below.
STORY DRIFT RATIO is the story drift divided by the story
height.
1) Maximum Limits
Now what for story drift limits? What is the maximum permissible value? Well it depends upon the type of drift. Is it seismic or wind?
For seismic, I will refer to UBC-97 code which in section 1630.10.2 talks about drift limits for earthquake.
 

 
Now in simple words, the maximum limit for seismic drift is:-
delta M shall not exceed 0.025 x story ht (if building seismic period is less than 0.7)
delta M shall not exceed 0.020 x story ht (if building seismic period is equal or greater than 0.7)
Important to note here is that it talks about SEISMIC drift so SEISMIC building period not the WIND period.
Now delta M = Max inelastic response displacement = 0.7R delta S
where R = from Table 16-N
delta S = displacement from static, elastic analysis
this value is read from ETABS.
you multiply this value by 0.7R to get delta M

This was all about seismic drift, but for wind drift code is mute. I will refer you to ASCE 2005 commentary CC.1.2

So we can understand that the limit for wind drift is "on the order of l/600 to l/400" for "common usage". This is common thing, however, in reality this figure can be up or down depending upon the ductility of cladding material and finishes. However for common usage value of l/400 is thought to be well satisfactory. Here l means story ht.
The concept of drift limits is same throughout all the governing codes, and the typical limits of story height by some number is same, but obviously you have to take care of the process of calculating the wind force or seismic forces. You should not calculate wind force from one code and apply limits of another code.
2) Load Combinations
Once the drift limit has been determined separately for seismic and wind forces, now is the need to check the actual drift vs the limit. Determination of actual drift depends on the load combination and the period of recurrence. If not properly calculated, this may dramatically increase or decrease the accepted drift values in model.
Seismic force E is always already factored so that's the reason its factor is always 1.0 in load combinations of ACI/ASCE code. The recurrence period for seismic force is 50 years. In seismic drift we do not convert it into service seismic force. Seismic drift is checked against the direct load case of EQx, EQy etc in ETABS.
For wind drift, we need to convert 50 year wind to service wind force. It has been recommended by ASCE commentary CC.1.2

To convert 50 year service wind force to 10 year service wind force it is multiplied by 0.7, as the equation says, and other gravity loads; D and 0.5L are also added.
So in a nutshell we create following load combinations in ETABS to check our drift:-
DRIFTWx1 = D+0.5L+0.7Wx
DRIFTWx2 = D+0.5L-0.7Wx
DRIFTWy1 = D+0.5L+0.7Wy
DRIFTWy2 = D+0.5L-0.7Wy
For seismic drift, as discussed earlier, we do not need any combination, drift will be checked just on EQx and EQy load cases only.
3) How to check in ETABS
Now we have obtained both the actual drift and the drift limit, but how can we do this in ETABS easily?
Well, after creating the drift combinations as discussed in step 2, we need to do as below:-
For seismic drift goto File>Print Tables>Summary Report

Select the file name

Scroll down to the end of the page, you will find out a section about drifts, similar to this one:-

It displays the max drift for each lateral load case for each story. As we want the drift for wind to be on drift load combinations and not on wind load cases, so we will not compare this wind drift without limits. In this table we are going to check just the drift values of our ETABS model for individual seismic load cases; EQx and EQy.
As you noticed, this table shows us values in fraction format. For example 1/105 that becomes 0.009523809524. This 1/105 value is story drift divided by story ht. It means delta S / story ht.
Now this value is delta S. First we need to convert it to delta M by multiplying it with 0.7R. Assume R here is 3.5 so
delta M = 0.7 x 3.5 x 1/105 = 7/300 = 0.023333 which is less than 0.025 so safe ( if T<0.7).
So instead of calculating every time by 0.7R we can check these limits in other way.
If our limit is 0.025 then the limit we get is 0.025/R/0.7. Assume R=3.5. Now the values in ETABS are inverse so our limit is 0.7x3.5/0.025 = 98.
In ETABS the drift is reported as 1/x where x is some number. Now as long as x (some number) is greater than 98 our limit of 0.025 x story ht is being satisfied. This way you can quickly check and compare seismic drifts.

Now for the wind drifts, goto Display>Show tables, select Point displacements>Story drifts and then select only drift combinations for results. Click on and then copy the table to EXCEL.




To save time you can right click on EXCEL taskbar and select maximum and minimum. Then just select the column H or I and see the maximum value that should be less than H/400 to H600 limit (0.0025 t0 0.00167). Again the values reported in ETABS are divided by story ht.
http://4.bp.blogspot.com/-9qv8XKHgL8Q/UALNKflmVsI/AAAAAAAAAEQ/AwKBYWt2iys/s320/image022-773193.jpg
4) Controlling Excessive Drift Values
sometimes you may face problem of excessively large values in drift tables in ETABS. Well we are not going to talk about different measures and modeling techniques to control the drift values. We are going to talk about large numbers in drift tables. Sometimes it happens that a point or node is free in the model or is connected to a NULL line or very flexible section. Drift tables for example the story drift table in wind captures the maximum displaced points. Obviously the displacement of several meters in tables is not what we are looking for. Drift values (relative) may be still okay for these points, but it requires you to check the displacement values too before checking directly the drift. Unlock the model and remove all free points, check for any discontinuity and modify your models to remove all the errors.