Problem In Frame

[haro0n]

Assalam u alaikum rana sir as u said that design manually so I made a simple frame structure and did it manually then I model it in etabs and compare the results and I was happy that most of them matched but there is some problem that for shorter beams my results are different as this is a single span so I did not solve it using ACI moments coefficients. First I solved it using direct formuals and also did it using Moment Distribution mtd but results are changed I want to know how these values came for shorter beam? and how software calculated this. as mine are different. And what I did that I first solve the slab then transfer its dead load and live load separately on the supporting long and short beams by tributory area then using ACI coefficient I solved the longer beams and results got matched but did not find way to solve the shorter beams so solved it using direct formulas nd other methods but results were change. Then from beams I transferred the loading and moments to colums and sloved the col also. E.g for end columns where longer and shorter beams is meetng I took the greater moment which was negative moment of longer beam and used it. For axial load I transferred the loading and multiplied it with half span of the beam in conection with that column.

image and file is attached.

Same for elevation B. its value is also change. The attached image is the values for Moment 3-3. If I use wl^2/8 for this shorter beam then Mmax is 12.62 kip ft. my calculated factored load per unit length is 1.01 k/ft for this shorter beam whhich is Shown above so If I select Moment 2-2 values then for this end column it shows the moment to be 12.24 close to my value. But here it shows totally change values. Similarly for center short beam where tributory area is doulbe then this. Solving it by wl^2/8 gives close value for moment to be transferd on the center column. As the negative moments from longer beams will be canceled out.

thanks in advance

edb file + imagel.rar

[WR1]

fella, come on skype this evening....ok? waseemrana87 is my id

[WR1]

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

[WR1]

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!

[haro0n]

Sir I did that and there are some problems

First u said 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)
   

how can I do this?? I did this like. Select all >assign>frame/line>end length offsets> there are two options automatic from conectivity and the second one is define lengths?? Previously I use this option at end before analyzing and select the first opiton that is automatic from conectivity is this right? And now for this problem how can I make the end offsets =0?? I did by 2^nd option placed 0 there.

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)

sir I am still confused with this shell, memberane and plate, where to use slab as a shell , where memberane and where plate?? And if we define slab as membrane then why we donot do meshing??

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...

I did this in another way like this, 10x20=200 so ¼ of this 200ft^2 will be for shorter beams that is 50ft^2 so 25 for one shorter beam and 25 for other so our live load is 125 so for shorter end beam 125 x 25/10=312.5 lb/ft udl for shorter end beam. For longer beam xied it with 75ft^2.. 625 lb/ft will be triangular load so we have to convert it into udl then use it. The way I am doing is right?? If found it very easy so doing by this.

you said results matched with moment distribution method...on ELEVATION 1 but not on ELEVATION A

no I solved mdm only for elevation A and for elevation 1 I use ACI moment coefficents. Can I use the moment coffients for this shorter beam?? As here I have only 1 span and in moment coeffients there is no such condition.

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...

solving for mdm for beams? In this case the loading will only be on the center beam so result will be same. As in this condition..

one more thing that previously u said that apply modifiers at end now I do by this way but after analyzing if I want to check those modifiers then I cannot see them as if I apply them while defining setions then after anayzing there are still shown there?

[WR1]

what about skype? your basic ideas need to be corrected ......u will be ready in 2 hours for skype?

[Waqar Saleem]

@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.

[haro0n]

sory rana sir i saw your msg so late can u give me some time for tomorrow?? as now i am jst leaving home to get to hostel from monday our final pprs are starting.. so if u can arrange some time for tomorrow that will be good.

[haro0n]

@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.

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.

[Waqar Saleem]

what loadings are you applying on the slab?what is the live load and what is the dead included all finshes and insulations or if you have taken only dead,plz let me know.and how is it possible that with same loading on the slab a beam with span of 10' and other with 20' carrying same load.even if one is carrying triangular portion and other is taking trapezoidal.you yourself calculated it above

"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...

I did this in another way like this, 10x20=200 so ¼ of this 200ft^2 will be for shorter beams that is 50ft^2 so 25 for one shorter beam and 25 for other so our live load is 125 so for shorter end beam 125 x 25/10=312.5 lb/ft udl for shorter end beam. For longer beam xied it with 75ft^2.. 625 lb/ft will be triangular load so we have to convert it into udl then use it. The way I am doing is right?? If found it very easy so doing by this."

[WR1]

its clear from pic...

etabs will do it this way even if the slab is 2000x10...

for strictly one way loading...you can click "ONE-WAY" option in ETABS

if you slab is supported by 2 sides only as in case of precase panels....you have to click "ONE-WAY" option otherwise your load transfer by 2 way method will be incorrect...

but if you have solid slab supported on all four sides and even it is one-way (theoretically e.g. 20x10) it will be considered 2 way by ETABS, its more rational...

but what i would do is to let ETABS do it, its way and I will increase the reinforcement by some margin in the longer beams....

for triangular loads (short side) : mulitply the intensity (w in kip/ft or kN/m) by 1.33, it will convert this loading to simple uniform loading over whole beam length...then use your conventional formulas to calculate moments and shear

for trapezoidal loads use factor of 1.30.

@waqar

yes moment coefficients are not 1/24 and 1/12 but i said these are coefficients for fully fixed end one beam span....similary 1/11 or 1/9 etc coefficients in ACI 8.3 approximate method are also approximate....by engineering judegement and your experience you can solve quickly some beam by these methods...some times the factor is not even 12 rather may be 20....see the limitations in ACI 8.3 for the method...

[Waqar Saleem]

thanx Rana it is clear how loads are distributed in two way slab at 45 degrees as you have shown the fig.why do we multiply the loads with factors 1.33 and 1.30 what is the reason?any reference to these factors.as you said we multiply these factors for converting the loads to simple uniform loading on beams from slab,even then the laod on the shorter side will be different from the load on longer side.also why the factor for shorter span is 1.33 and greater than the factor for longer side.in actual sense shorter span beam takes lesser load.

[haro0n]

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.

@waqar bro the fig(a) is for beams having more than 2 spans so i cannot use it for shorter beam as it doesnt fulfil this criteria. for 2 span i-e longer beam i used the 2nd fig. nd option d is also for more than 2 spans.

[haro0n]

@ rana bro what about my problems :( when u will be free ?? as from tomorrow our final pprs are also starting so what will be best time to to this as i am veryyyyyyyyyyyyyyyyyyyyyyyyyyy serious in learning this all

[Waqar Saleem]

@haroon

there is no option for single span beam fixed at ends so i mentioned earliar that check the moment diagram for that and BMD for one beam would be same as the one for fig d.so we can use the coefficents of that option.

[WR1]

yar masla kia ha....ku preshan ho.....its simple...haroon...for one beam resting on 2 columns...make those columns straight..(like 3 spans of a beam) and use moment coefficient method....) it will solve your problem....beam col connection is between a fixed / pin option....

For Skype sorry I got busy in writing a book....pm me in this forum whenever u get free....ok?

@waqar

factors of 1.33 and 1.3 are used for ease to convert triangular and trapz loads to equivalent udl to calculate moments...

Actually you can calculate moments without converting them to equivalent udl (this is just for ease)...

For example look at picture 1 showing your triangular loads

It states that "EQUIVALENT TOTAL LOAD = 4/3 of W = 1.33W" in kips

but other formulas are given to calculate moments...you will see

M = WL/6 where W is total load (W=0.5xLXW in kip/ft) so M = WL²/12 for simple beam having triangular load..ok

now this factor of 4/3 or 1.33 is calculated so that both formulas yield equal moments....but as i have said...you can ignore this short cut and solve by your usual triangular loaded beam formulas...

anyways...for your reference you can calculate exact moments for triangular loaded beam by formulas attached in above picture....for trapz loaded beam see the pic below...

i hope it helps!

[Waqar Saleem]

@RANA thanx sir but this is not the triangular load on the beam in the sense you described here.load on the triangular area of slab of slab is distributed over the beam but it is not the triangular load in exact sense.this is the tributry area for the beam which is triangular or trapezoidal.

Kindly consider the chapter 14 of Nilson 13e yieldline analysis for slabs fig 14.1 and fig 14.2

[WR1]

waqar i did not understand your question....yes it is the tributary area of slab which in turn makes loading on beam also triangular or trapezoidal...if the area is triangular (and the load on slab is uniform) then the loading on beam will also will be triangular...sure...and your are mentioning figures 14.1 and fig 14.2 of which edition? 13,14,10???

Please upload those pictures here....

[Waqar Saleem]

sorry sir that fig was misunderstood by me check this attahed fig. it will clear what i need to say

[WR1]

yeah i checked so whats wrong in that? this is trapezoidal loading from trapezoidal distributary area from slab...solve it by the last figure i attached for you in last post...(but that fig was for one span beam) you can find similar formulas for continuous beams online...or in Reynolds book....in continuous beams chapter....or simply convert the loading to equivalent udl by multiplying it by 1.30 and use moment coefficient method of ACI 8.3....is there any other confusion? tell me

[Waqar Saleem]

sir my point of view is that triangular/trapezoidal loading comes from the slab but on beam we take it as uniform loading,the picture a shered above is not continouos beam it is the two sides of one span beams taking loading from shorter and longer side of slab.hope you got my point

[WR1]

no we dont take it as uniform load...we take actual load...its triangular or trapezoidal or whatever shape...we take actual load...how can we take it uniform? I am sorry if i did not get your point...

but yeah we take this loading (in whatever shape) as distributed over the entire length of beam...(again this is not called uniformly distributed...this called distributed over the entire length)

[Waqar Saleem]

i am afraid sir i disagree at the point then why we do not take the load of slab on beam as cantilever in more exact form that slab is projecting on the beam,then we should take triangular cantilever portion of slab on the beam on shrter side and trapezoidal cantilever portion on the other longer side.it makes more sense as slab is cantilever from the beam not on the top side of the beam.

don't mind sir i am argueing to clarify myself.

[WR1]

i did not understand any thing

[WR1]

for your reference i have uploaded my concrete DDM assignment...please look at it..it might help