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well u have membrane slab and you are meshing it? remove the meshing and see the results you will find 625 lb/ft on your longer and shorter beams for live load... (Display>Show Loads>Live) remember it will distribute load by 45 degree method you despite that your slab is 10x20 (1way and your short beams will carry nothing) but if you do it by 45 degree your short beams will also have 625lb/ft load for live loading... when your loading is okay then your shear and moments will also be ok.... you said results matched with moment distribution method...on ELEVATION 1 but not on ELEVATION A i think it should also match ...you have single frame 1 beam and 2 columns...just make the columns straight and you will have 3 span continous beam...thats it..solve it by moment distribtuion method...i hope you will get the desired results....ask again if you are still in difficulty!

thnx waqar bro i will consult nilson as i get to my hostel then will reply back and rana sir is right that load on both shorter and longer beam will be 625lb/ft but the difference is that for shorter beam it will be triangular and for longer beam it will be trapiziodal so will solve separately for them.

@Rana sir moment coefficients are not 1/12 and 1/24 here . @haroon , moments in the continous beam need to be caculated by coefficient method and where you are taking them wl^2/8 that is incorrect.for beam use Nilson pg#396,for shorter beams fig(a) use column option as support i.e 1/16,1/14/,1/10and for 2span longer beams fig ( also the column option 1/16,1/14.in case of yours only one beam fixed at both ends use the fig d coefficients look at the BMD of the d option you will get my point. @rana sir when slab is 10x20 then it can be considered one way or two way ,etabs take it two way and distributes the loads in that way on both side beams by yield line theory method of distributing loads at 45 degrees.you said that the loads on both shorter and longer span both be same 625lbs so that would not be possible. @haroon PLATE beahaves like a beam it resist by bending only no horizontal stresses are distributed,they can bend in two direction i.e up and down and twist. MEMBRANE take only horizontal stresses and can not bend.this is known as membrane action also SHELL have both the charactersistics of membrane and plate. this is the simplest difference among the three.the idea is hard to aplly in practical.

well i just the image u attached....it should give you proper results as compared with moment distribution method.....you cannot get moment (+) by wl²/8...because its a continuous frame.....here idealize the beam as fixed end...and take the moment wl²/12(-) and wl²/24 (+) these factors 12 & 24 will vary depending upon the stiffness of the beam/column joint..if they are fully fixed then this formula will give u the required results... anyways, first of all make sure that.... 1) end offsets for all frames are = 0 (because you need moment at c/c of beam & col not at face or distance d as in design) 2) make sure no dynamic analysis is being done 3) make sure no earthquake forces are there 4) use same section to col and beam for equal stiffness, use square sections lets say 12 in x 12 in

look! first you need to study code for proper load combinations. This is very very critical. All the design results and serviceability criteria is based on it. You need to understand which load combination to use and when to use. Read chapter 9 of ACI and PCA Notes..Read its commentary..Read UBC...Try to know what is the difference between 1.3W of UBC and 1.6W of ACI. Get to know why the factor in E is 1.0 in strength load combinations. Know why omega factor is used. 2nd...Research what are the options in softwares which affect automatic load combination creation. For example in ETABS what is the impact of code you select for concrete or steel design and the seismic design category? Why there is SPECIAL SEISMIC LOADS option in ETABS under define menu? I would recommend you to print the page of ACI where load combinations are defined and also for UBC and keep it in front of you!..Take a paper and write all the appplicable load combinations based on your load cases. Then compare ACI vs UBC I would say you dont use automatic load combinations rather make your own combinations. Once you are done and confident about your understanding of load combinations then use the AUTOMATIC load combinations. You should know which combination is appropriate for Seismic drift limit or wind drift limit. What is the meaning of probability or exceedence of seismic or wind forces? If still this all stuff seem boring to you! do this Goto options in ETABS> define design code to be ACI 2008..Select appropriate seismic design category. If you are not familiar with SDC open chapter 1 of ACI and see what is the equivalent SDC of UBC in Table 1.1 of ACI Then goto define menu>Special seismic forces. If your SDC is C or less it will not have an impact. So you select it or not...doesnt matter then create automatic load combinations and verify them one by one..Do this atleast in your first few models. There is a trick to make service load combinations automatic too...if you are interested i will tell you later!