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Everything posted by Rana

  1. What about if columns at base are fixed and lateral resisting system becomes a cantilever column system? First of all in flat slab system, connection is not hinged with shear walls or columns. Else it becomes a cantilever system for lateral loads. See above comments. Yes but would you design the system as cantilever columns? and make sure base is always fixed? Consider flat slab as wide shallow beam spanning between columns. Lateral load is resisted by axial and bending stiffness of beams and columns. Beams and columns will bend in double curvature developing the frame action. However in case of cantilever system (beams pinned), there will be no shear force and hence no moment in beams only the axial force and hence axial deformations. In columns there will be only bending deformations not axial. Several definitions; Aspect ratios Walls resist in-plane bending and shear, columns usually resist biaxial moments Rigid body rotation at base in case of walls poisson effects in shell elements as far modelling and fea is concerned Columns are usually compression elements, walls develop tension compression over large area.
  2. What do you mean by load will not be transferred to node in second option?
  3. Apply the moment manually or draw dummy cantilever line.
  4. Dear Umar, First-off, many thanks for the explanation and your interest in the topic. Highly appreciated. Perhaps, the RSA image added to the confusion than doing anything good, i guess. We are in the same boat as far as T within the range of maximum spectral acceleration is concerned, however, kernel of the discussion revolves around the "scaling of dynamic base shear with respect to ELF base shear". As such, analysis time period (used in RSA graph), even if lies within the maximum acceleration range might give dynamic base shear less than ELF V. For this, and as required by certain codes (American e.g.), for scaling up the dynamic base shear, we need to establish a "Time Period" for static procedures and calculate static base shear. So I wanna twist our discussion in another way. Let me ask, if T(dynamic) and T(static) are same, do we get same base shear from both methods? Static procedures V depend on R; a compound factor for ductility, damping, inelastic/elastic response etc. On the other hand, spectral acceleration from RSA is a function of Ca, Cv and damping but not specifically R. Dynamic base shear is the product of mass matrix, acceleration and participation factor of each mode then we add up all the modes by some type of combination, right. Now, even if time periods were same, and even if the structure behaved in fundamental mode (SDOF concept of RSA) with very rigid torsion response, base shears from both methods would still be little different due to partial active mass in dynamic analysis (maybe 90%?). Hence (and correct me), saying that T corresponding to Vmax of static procedures if used in RSA would result in same base shear would not be appropriate, i guess. Moreover, in my opinion, RSA analysis (although based on actual earthquake records), is a tool where you need to put up some boundaries; upper and lower. Whereas the empirical time period limitations make more sense as they were calculated based on actual buildings and earthquakes through statistical analysis. That might be the reason FEMA recommends using Ta.
  5. Yes, consider the axial loads and out-of-plane moment for analysis.
  6. Please use search feature of this forum before repeating the same question. Refer to
  7. Abid, the link I have shared, explains the load combinations to be used in ETABS file where static seismic, temperature, fluid and hydrostatic pressure is present. It also contains edb file that contains all the load combinations already.
  8. Some background can be found in chapter 3 "lateral load resisting systems" in the book "reinforced concrete design of tall buildings" by Tranath.
  9. Well, it depends on what softwares, the firm is using, in which you are working. ROBOT is still not used widely in middle east atleast. For REVIT id agree, spend some time on REVIT but as a structural designer, you dont need to concentrate on production of drawings in REVIT, just an overall view,
  10. ETABS Load Combo Creation Have a look at it...may be u find it useful...its for manually creating combos in excel and then importing back to etabs..as it allows the flexibility to create combinations in table format and less chances for errors as compared with that of ETABS unfriendly load combo interface...Report back any bugs or suggestions...thanks To download click on the link above with blue color or www.4shared.com/file/THoXlkzv/ETABS_Load_Combo.html
  11. SAFE can include these inverted beams as line elements or shell elements.
  12. Eq will likely not govern, as a rough value 1kpax1.6 is the wind load. You get mu=wl2/2 and vu=1.6kn. T=C on each side will be mu/arm between bolts. For each bolt tension will be T/2 and shear will be Vu/2 (only 2 bolts resist shear). Remember there will be tension shear interaction for bolt design plus use Appendix D of ACI. This is a simple example. For complex base plate patterns use RisaBase software for biaxial design or make a model with assumption that base plate is stiff (use very thick shell) and frame elements as bolts. Or use the formula to calculate tension compression on bolts from applied moment and P which depends on distance of bolt from centroid divided by group x2 or y2. See any foundation book.
  13. Ok, what about these; #1 2015 NEHRP Recommended Seismic Provisions: Design Examples FEMA P-1051/July 2016 https://www.fema.gov/media-library-data/1474320077368-125c7a1d1a3b864648554198526d671f/FEMA_P-1051.pdf "Calculation of a period based on an analytical model of the structure is encouraged, but limits are placed on the results of such calculations. These limits prevent the use of a very flexible model in order to obtain a large period and correspondingly low acceleration." Page: 2-17 and "A lower limit to the base shear determined from the modal analysis procedure is specified based on the static procedure, and the approximate periods specified in the static procedure." Page: 2-18 Also consider the example in section 13.2.6 and I have attached the sketch explaining that example for easy understanding here; (Ta was used instead of Tc). #2 See ASCE publication seismic design guidelines based on ASCE 7-10, Example 17 on page; 117 & 118; and
  14. Yes, both give an indication, but i was studying where both methods make a difference. Usually in books, for example in Elements of Earthquake Engineering and Structural Dynamics By André Filiatrault https://books.google.ae/books?id=3FRzzDLlQwYC&pg=PA457&lpg=PA457&dq=drift+vs+displacement+to+check+torsional+irregularity&source=bl&ots=oEgfxDf-eS&sig=agIoyGCOOwtUo301UpQnrpCTRNo&hl=en&sa=X&ved=0ahUKEwiMnbDA9afRAhXBJ8AKHdzrCXI4ChDoAQhDMAg#v=onepage&q=drift vs displacement to check torsional irregularity&f=false it is calculated based on drifts and not by displacements, so the rotation of lower floor is also important.
  15. In order to amplify accidental torsion, as per UBC-97 and ASCE 7-05, maximum and average displacements at corners of floor plates must be compared. However, in csi wikki, they use the word "DRIFT" not "DISPLACEMENT", so drifts should be compared instead of displacement. Drift: Difference of displacements at a specific point of upper floor and lower floor Displacement: Abs elastic or in-elastic displacement with reference to base I went through ASCE 7-10 but it was more confusing, as in the text they use word "DRIFT" but in the formulas they talk about "DISPLACEMENT". In FEMA examples, they are sometimes using "Inter-story drift" and sometimes "displacements". Anything I am missing here?
  16. Dear Umair, I understand what you said, but, that is only one of the condition I asked about (see below). Ok let me put this in following way; lets call T-computed simply as T, there are following 4 cases; a. T < Ts < Ta b. T > Ts but < Ta c. T > Ta but < CuTa {not my concern} d. T > CuTa {not my concern} Now lets focus on condition a & b; Let's say; Ts=0.3s and Ta=0.7s Condition a Assume T=0.1s which is less than both Ts and Ta In order to jump to V, we need to first set the time period. a1. If we use Ts, then obviously maximum base shear would govern. This is what you are referring to. a2. if we use Ta (which is higher than Ts), base shear would be less than Max. V so we get some saving. Now the question is what T you'd use? Ts or Ta? As per FEMA (moment frame example) use Ta. As per research papers I have mentioned for tunnel construction you should use actual T (or in other words Ts as Max V would govern, as you rightly pointed out). Condition b Assume T=0.5s which is more than Ts but less than Ta. Still, if we go with FEMA approach, we should use Ta which will give us lesser base shear. And use of actual T would give us higher base shear. So the question remains, what value of T to be used? 0.1 x (no of floors) is for certain moment frame buildings as per ASCE 7-10 seismic design examples book (example 17), that might not be correct for buildings with large no/area of shear walls. Anyway, the intention, as i see, is to use approximate method (Ta) if analysis T is less than Ta, which is in-line with what Baz said. However, after going through the above research papers, I am again inclined towards using T from analysis instead of Ta for such buildings. For moment frames, yes we could use Ta.
  17. I have another point, (Source: http://www.tdmd.org.tr/TR/Genel/pdf/TDMSK010.pdf) Lower bound and upper bound limits of regression analysis are given as in building codes. More exact solutions (such as in analytical models) give higher T than Ta due to absence of partition and infill walls etc. But my concern is for buildings having large shear walls areas such as tunnel construction, very very dynamically stiff (e.g. 0.1s). In such cases (T-computed < Ta) and as mentioned by following paper, we should not use Ta (as per discussion in posts above) and should use T-computed. Paper can be found here: http://erolkalkan.com/Pubs/12.pdf Also see this: http://www.sciencedirect.com/science/article/pii/S1687404815000565
  18. Wow. But there is no clear reference for this in ubc or asce. I got the sense and could save some structures from over-design but would be very difficult to convince authorities without a clear overt, explicit reference to building codes.
  19. No way Ilyas. That will reduce the total base shear. Wrong approach.
  20. Do you mean "There is no need to use the value of time period that is LESS than Ta" or "NOT LESS than Ta"? As you quoted FEMA 451, it says Ta is lower-bound limit (more conservative for base-shear as T-computed would be higher) CuTa is the upper-bound limit (can be ignored for drift calculations) But, There is no guidance on if T-computed is less than Ta, not even in ETABS manuals. All the documents highlight the upper limit but not the lower one. For example, consider a bearing wall building, (density of walls/floor is large), T-computed is ~ 0.10s, however Ta=0.5s; huge difference. ETABS only checks the upper limit (lets say 1.0s) and not the lower one and selects 0.10 automatically and calculates base shear for this T. However, if I am getting you right or as per; 2009 NEHRP Recommended Seismic Provisions; Chapter 4, page-41, it says; if T-computed is less than Ta, then use Ta. https://www.fema.gov/media-library-data/1393888487069-c071850c3050c34da135376baa478a1b/P-752_Unit4.pdf Please shed some light on this issue.
  21. Building Drift in ETABS waseemrana.com 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. ranawaseem.com
  22. Waqas, just reverse calculate; In books you will find for singly reinforced; a = Asfy/0.85f'cb and Phi Mn = 0.9Asfy(d-a/2) re-arrange, and you will find your answer.
  23. 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.