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AXIAL DEFORMATION OF COLUMN IN TALL STRUCTURES
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bijay sarkar
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PostPosted: Sun Dec 02, 2012 4:56 am    Post subject: Reply with quote

Dear Sir,

If you have an example of axial deformation calculation based on Prof. Samra's Method + ACI , can you please upload the same ??

with regards,

bijay sarkar
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P.K.Mallick
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PostPosted: Sun Dec 02, 2012 8:54 am    Post subject: Reply with quote

NUMERICAL EXAMPLE BASED ON METHOD DEVELOVED BY PROF.SAMRA.

Assumed an inside column 50 stories below the roof.
Floor to floor height is 3.5mm.
The size of column is 750*1500mm.
The column is reinforced with 4% of reinforcement.
n=Es/Ec  is taken as 8.
Es=2*100000Mpa.
The column is subjected to load of 16, 8000 Newton per floor.     
     Every fifth floor the reinforcement and size of column changes.  The details of column sizes and reinforcement are as follows-


                                                  SIZE               REINFORCEMENT
Ground to 5th floor               750x1500              4%
5TH to 10th floor                   650x1400             3.5%
10th to 15th floor                 550x1300              3%
15th to 20th floor               450x1200               3%
20th to 25th floor               450x1100               3%
25th to 30th floor               450x1000               3%
30th to 35th floor               450x9000               3%
35th to 40th floor               450x800               3%
40th t o 45th floor               450x700               2.5%
45th to 50th floor               450x600               2%

a)     Planned construction two floors per week
b)     Concrete is moist cured.
c)     Relative humidity 40%.
d)     Slump of concrete 34%.
e)     Fine contents 34%.
f)     Air contents 5%
g)     Cement content 356kg/m3
h)     Age at loading 28 days.
SOLUTION     
STEP-1: Calculation of creep Coefficient.
     Since planned construction is quite fast i.e., two floors per week, the effect of incremental loading can be neglected.
Ø∞ (t,to) = 2.35 K1 K2 K3 K4 K5 K6
     K1 = 1.25 t0 -0.118
                         
           = 1.25(28-0.118 )= 0.843
     K2 = 1.27 – 0.006(40) = 1.03
     v/s = (750x1500)/(2(750+1500)) = 250
     k3 =2/3 [1+1/13 e-0.0212(v/s) ]
         = 2/3 [1+1.13 e -0212(250) ]
             = 0.67
     K4 = 0.82+0.00264 (100)]
                   = 1.084
         K5 = 0.88+0.0024(34)
         = 0.9616
     K6 = 0.46+0.09(5) = 0.91<1
     Hence k6 = 1
Ø∞(t,to) = 2.35 (0.843) (1.03) (0.67) (1.084) (0.9616) (1)
     = 1.425= Øt
Step-2: Calculation of Ultimate Shrinkage
     Ultimate shrinkage, Sh∞
               = 780X10 –6 K1 K2 K3 K4 K5 K6 K7
          K1 = 1 (assumed 7 days curing)
          K2= 1.4-(0.01)(40)= 1
          K3 = 1.2 e -0.00743(v/s)
                  = 1.2 e-0.00743(v/s) =0.3678
          K4 = 0.89+0.00264(100) =1.154
          K5 = 0.30+0.014(34)=0.776
          K6 = 0.75+0.00061 (356) =0.967
          K7 = 0.95+0.008(5) = 0.99
Sh∞ = 780x10 -6(1) (1) (0.3678) (1.154) (0.776) (0.967) (0.99) = 2.459x10 -4 = єsh  

APPLICATION OF PROF SAMRA’S EQUATION.
As = 4/100X750X1500 = 45000mm2
Steel ratio with respect to gross section=0.04= Sr
n=8
fci = P/ (Ac+nAs) = P/ (Ag + (n-1)As  )  
= 50X16, 8000/1440000 = 5.8333N/mm2
fct  = fci(((1-Sr)+n(1-0.2 Øt)Sr)/((1-Sr)+n(1+0.8 Øt)Sr))
By substituting
fci =5.8333N/mm2
n=8
Øt=1.425
Sr=0.4
We get
fct =4.216 N/mm2
fst=nfci((0.2)Øt+ ((fct/ fci)(1+0.8 Øt)))
Substituting all the values ,we get
fst=85.4778 N/mm2
Hence deformation due to creep= fst/ES =85.4778/(2*100000)=4.27 *(1/ 10000) mm
Stress in steel due to shrinkage
= єsh/ ( (1+ Øt)/Ec + (Ac/AsEs))
=27.69 N/mm2
Strain due to shrinkage=27.69/(2*100000)=1.3845 *(1/10000) mm.
Hence deformation for 3.5m high column=
(1.3845+4.27)(1/10000) (3500) =1.979 mm
Hence height deformation of column 50 storied below the roof = 1.979 mm.
Going by same procedure axial deformation of each storied column i.e. column 50 storied below the roof to 1st storied below the roof can be calculated. Summation of deformation of each segments of column is equal to total axial deformation of the column.

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Last edited by P.K.Mallick on Sun Dec 02, 2012 12:20 pm; edited 2 times in total
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baldevprajapati
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PostPosted: Sun Dec 02, 2012 10:31 am    Post subject: Reply with quote

hallo sir r  u know the  design of tall building by Etabs ?
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thirumalaichettiar
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PostPosted: Sun Dec 02, 2012 10:57 am    Post subject: Reply with quote

Er.P.K.Mallick:

I appreciate your hard work.

Er.Baldevprajapati.


Do you post this beleiving that it relates the prexent topic AXIAL DEFORMATION OF COLUMN IN TALL STRUCTURES? or does it relate to TALL BUILDINGS?

If it is different please repost it under General discussion and delete it.

T.RangaRajan.
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baldevprajapati
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PostPosted: Sun Dec 02, 2012 12:18 pm    Post subject: Reply with quote

sir thirumalaichettiar
    i have ETAbs file my work .  i give u to show the axial deform column in the result after analysis, so give me  you email id so i send the file
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P.K.Mallick
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PostPosted: Sun Dec 02, 2012 12:25 pm    Post subject: Reply with quote

baldevprajapati wrote:
sir thirumalaichettiar
    i have ETAbs file my work .  i give u to show the axial deform column in the result after analysis, so give me  you email id so i send the file


I think ETABS calculate only elastic shortening not

inelastic shortening. Will you please confirm?

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thirumalaichettiar
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PostPosted: Sun Dec 02, 2012 1:12 pm    Post subject: Reply with quote

baldevprajapati wrote:
sir thirumalaichettiar
    i have ETAbs file my work .  i give u to show the axial deform column in the result after analysis, so give me  you email id so i send the file


Er.Baldevprajapati,

My e mail id is
thirumalaichettiar@yahoo.com

T.RangaRajan
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bijay sarkar
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PostPosted: Sun Dec 02, 2012 1:52 pm    Post subject: Reply with quote

Thank you Mr. Mallick for the example. Now it appears fine.


with regards,

bijay sarkar
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PostPosted: Sun Dec 02, 2012 4:35 pm    Post subject: Reply with quote

COMPUTER PROGRAM
From the numerical example based on Prof.Samra's method,it is evident that the procedure involves lot of arithmetic computation and highly repetitive as for each column segment the procedure has to be repeated.
Back in 1997, I had developed computer  program to calculate inelastic axial shortening based on Prof.Samra's method. The computer program was developed using Q basic language( please do not laugh because that is the only programing language I knew and I know,may be forgotten now)
The basic outlines of the program are as follows:
a) The program calculates the deformation of column due to creep and shrinkage for any number of storied building.
b)The program is in interactive mode and asks information to user, one by one such as no of stories,story height,relative humidity etc. Hence it can be operated by any person who understands basic engineering terminology.
c) The program calculates creep coefficient and ultimate shrinkage by methods outlined in ACI 209R.Then the program uses the calculated creep coefficient and ultimate shrinkage in Prof.Samra's equations. The deformation for each segment of column is calculated and then cumulative inelastic axial deformation is calculated.
d)The program assumes that floor to floor height of column and load coming per floor remain constant and the size and reinforcement of column change every fifth floor.This has been done to keep the inputs to a minimum but the program can be easily modified to suit a particular problem.
e) The validity of the program has been checked by solving various problems and comparing computer output with manual calculation.
---------------------------------------
CONCLUDED.
-------------------------------------------------------------

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bijay sarkar
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PostPosted: Sun Dec 02, 2012 5:21 pm    Post subject: Reply with quote

Dear Mr Mallick,

Two questions only as follows :

1.  In case of K4 calculation, what is 100 used, when slump is 34% ?

2.  Can you please confirm the result of K3 in case of shrinkage as 0.3678?  The equation is 1.2e(-0.00743*v/s) ?

with regards,

bijay sarkar
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