WR1

yes Mu is +2.61 and -5.1 kip-ft, but you have to make sure + sign represents inner or outer face of wall. It will depend upon the direction of local 3-3 axis. Furthermore, you dont need to design (Mu) for peak/max values shown in status bar of ETABS/SAP/SAFE. This should be enough for longhorns but to be a veteran you need to look at other things too. Read a good book about FEM. It will really be an eye opening and advantageous for you. Read how to take design values from contours, what are stress concentrations, what Guass points and how meshing affects stress concentrations. In short for a simple problem, yes, use M22 moments for vertical steel design and M11 for horizontal steel of a wall (if default orientation of local axis is used)

Interesting discussion. I am in to have your more detailed views on this. So Umar you mean we dont need to apply EQ on 2 basements because of stiff diaphragm and walls? Asad, yes 0.7 on all is perfect.

No need for expansion joint in footings...But dont consider it final answer..Discuss with others too... Regarding 5.3 inelastic displacement at roof..yes you need 7.5in gap but first ask why 5.3 inch? That means 132.5mm. Improve your framing. Is your building like 120m high? How many stories? And how much is W and base shear in terms of W?

lol...not even started...but...that does not mean it wont!

Dear Engr. Waqas, I dont think 2" expansion joint will be sufficient for a building in seismic zone 3. But other things do matter as well for example; how many stories? etc In ASCE/UBC/Australian codes, seismic separation should be SRSS of inelastic displacements. that is delta = [ (inelastic detla of building 1)² + (inleastic delta of building 2)² ] ^ 0.5 Remember that these delta values should be at the same level. For example if a building is 9 story high and adjacent building 3 story high. You have to calculate expansion joint at level 3 of both buildings. Ofcourse you need to check expansion joints at all other stories too (Level 2, Level 1 and basements) if displacement on those levels is more than level 3 (due to high seismic weights or anything). Please also remember that usually in softwares you consider X or -X directions for seismic analysis (and Y / -Y) take the max from X and -X and multiply it with Cd/I (for ASCE) or 0.7R (for UBC) to calculate seismic movement of that building. Do this for the second building and then SRSS will give you final value. HTH

Yes

Ofcourse, see ACI chapter 9 for calculating minimum thickness of two way slabs and one way slabs. It is Ln/36 etc. Where Ln is the span. Working load is the load, a structure can carry safely without failing. Service load corresponds to the serviceability limit state. The load at which a structure or a building or an equipment continues to function properly. For example for your home, ultimate load could be very high which your house could sustain, but some percentage of this load might cause one of the doors in your home to stop working. So the serviceability is disturbed. Load less than that was the serviceability limit state of that door. I did not get it. There are many things to relate Thickness of slab must be greater than s/2 where s is the main reinforcement spacing in two way slabs or s/3 in one way slabs you can also relate it to the beam stiffness Section modulus of beams and columns? Section modulus if a shape property. Check your question again

If beams fails in shear and torsion combined, you should increase the concrete section. Engr Uzair, ties will not help because the combined stresses are more than 8xf'c^0.5 (imperial units). You need to increase concrete section size. Will update later. This happen sometimes. Dependence of foundation on bearing capacity of soil? I did not get it. But in SAFE there is no bearing capacity. You have to put modulus of sub grade reaction (you can relate bearing capacity with k as k=120 times the bearing capacity in SI units). See Bowels book or use k from geotechincal report. Two way shear or punching shear can be checked in SAFE from the menu: Design>Punching check One way shear can be checked by showing contours of V13 and V23. V = SRSS of both these components. And phi Vc = 2xf'c^0.5x0.75 in Imperial units or 0.17xf'c^0.5x0.75 in SI units Mass is used in dynamic and seismic analysis. You should study about excitation of mass and lumped mass from any book on dynamics. Weight is the normal weight of materials acting as loads on the structure. For example dead, live etc. W=mg as Engr Uzair said. Excessive gravity load buckles the building or column sideways by delta. This delta creates additional moments M=P (gravity) x delta. Thats why this moment is called P-delta moments. If you have not designed for this additional moment, your structural gonna collapse. Ofcourse this happens in very slender columns or building or very heavily loaded columns. Weight or load is involved in calculations. In dynamic analysis (usually done for irregular buildings) you try to capture higher modes. This is done by lumping mass at story levels. There are SDOF (single degree of freedom) and MDOF systems. Read from any book on dynamic analysis. Mass is involved here. Strength of concrete depends on proper curing and mix ratio of ingredients. Punching ratio=1 means, Shear stress is exactly equal to the shear capacity of concrete section. Shear capacity of concrete for punching is calculated from 3 formulas given in code. The least value is used. Open any book on concrete and see punching shear topic. In ACI-318; There is no storey drift limitations. However, you can find these limits in ASCE/IBC. Its the same thing. Cracking limit in ACI can be found in chapter 9 (i think) or chapter 10. For water tight structures you need to refer to ACI 350. For detailed cracking limits you need to refer to ACI 224-06. Beaware that ACI 224-06 and 224-01 or 02 and ACI-95 differ in methods

What is the course name for which you are taking the exam...btw?

Links updated

Sure, why not! Welcome to the forum

page 84, of reinforced concrete design of tall buildings bungale s. taranath

Well I used LUSAS too. But dont worry, the formula you have written above is sum of shear and flexural deformation. I just wrote it separately. No worries. Well I am not sure why the deflection in your case matched well with top/bottom supports. But just make your report and write your results. At the end the instructor wants to see how well you can compare FEA results.

What FEA software are you using? How are you checking FEA results manually? Calculate flexural plus shear deflection manually and compare with FEA. (Not just flexural) flexural defl = 5wl^4/384EI shear defl = 0.15wl²/GA Regarding supports; If you apply supports at top or bottom of beam, results will be same but with opposite sign. Tensile, compressive and shear stresses will be reversed. Middle supports will produce linear distribution of flexural longitudinal stresses at the center of the beam. You could use this to compare FEA results with hand calculations, and you will not find much difference. Then use this as benchmark to compare the results of other supports.

haha it reminds me of my old assignment during masters. Are you studying?

Check deflected shape, there should be step numbers under W. These step numbers are different directions of wind under W load case.

There is no need for any reference for that statement. Its simple. Flanges will resist moment and and when there are no bolts around flanges, there will be no couple (tension, compression) to resist the moment. Only axial load will be resisted by bolts near the web.

Yeah! its actually not the envelope. But in ETABS 2013 EQ and WIND has just one load case. That single load case includes every thing. For example EQ in ETABS 2013 contains all 6 directions and W in ETABS 2013 contains all 12 loadings. This is much rational approach they adopted in 2013 ETABS. So you dont have to make now hundreds of load combinations.

You mean there are 5 load combinations governing at the same time? Please check it again. Anyway for your loads, 49x49" column will work at higer reinforcement (7.9%). After that every inch increase in column cross section will decrease the reinforcement by considerable amount. 56x56" column will give quite reasonable amount of reinforcement around 3.5%. But the column size even 48x48" is very unreasonable. For a 33 storey, a square bay of about 9m (30' approx) with ultimate loading of 20kPa (420 psf approx) will give this much axial load reaction at ground floor column. So do you have 30feet spans? Sure you need improved framing for 33 storey height building. You could use composite columns for example?

Well, I have not checked your model. But what is the governing combination for bottom columns? It should be gravity. If not then please update us. If it is gravity loading then what is the axial load and Mx and My for any of the 48x48" columns failing? What is the f'c? and fy?

Well, yeah I verified that if shell element has more than 1 support it again takes about 95% of total shear.

Thanks, yeah thats a good point about reduced beam length in case of shell element or wall. No I no. of columns is not more than these shell elements. They are almost equal. Ayway I am using frame resisting system but the question now is between shell or frame. For more than 2 supports to capture wall action, let me verify this in a an example. And for a simple cantilever shell vs frame, the result is same. I have done it previously many times, the only difference is that a shell is more stiffer than a frame of the same size because of possions ratio. As for shells flexural rigidity is EI/v instead of EI in case of frames.

Thanks Imran Zafar, I will study this in detail and will update if I find something new. Sir Umar, unfortunately what we discussed (that shell and frame should give same base shear in those elements) is not matching with my models. I checked models again with Poisson's ratio = 0 so shell = frame stiffness. But still I am surprised to see that those few elements when modelled as lines carry only 329 kN shear and when modelled as shell carry 922 kN shear. What could be the reason? Line elements have one pin support at base. Shell elements have 3 pin supports at base. Now if I change shell pin supports from 3 to 1 (keeping the middle pin support) the shear on shells is almost the same as that of frames. It is now 342 kN (very near to 329 kN of frame elements) So the base fixity is doing something here? What you say?

Well I dont believe in aspect ratios thing. ACI also clearly mentions in this clause commentary that it is the behaviour that determines if its a column or a wall not just merely the aspect ratios. Anyway one good thing to check is the moment diagram.

Umar, there is a difference between stiffness of shell and a line element due to Poisson ratio although negligible. I am getting higher forces because the line element was pinned and when modelled as shell...it is also pinned but now has 3 pin supports instead of 1 in line...so these 3 pin supports making the shell more stiff at base. PS: This explanation is not mine..so I am not satisfied with this too.. because as I know the lateral force in rigid diaphragm should depend on stiffness of vertical elements and stiffness of line and shell dont differ much so the the base shear in these columns should not jump from 70% to 95%. Agreed. Btw does the above explanation about base fixity has any effect? I dont think so but just asking?