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Deflection of waffle slab
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akjhacpwd
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PostPosted: Thu Jan 13, 2011 8:21 am    Post subject: Deflection of waffle slab Reply with quote

Dear friends
I am designing a waffle slab 20 m by 20 m which will house library. Ribs are at 1 m spacing in both directions . I have taken ribs width as 250mm. The depth of rib including topping is 900 mm. The waffle slab is supported on peripherial beams which in turn are supported on columns of 400 mm by 900 mm . The columns are spaced at 5 m in one direction and at 4m in the other. The size of peripherial beam is 400 by 900. I am facing problem of deflection . Short term deflection is 62 mm and total defn incl creep is 112 mm. deflection has been calculated considering discontinuous ends by Rankin theory ( taking w/2 in each direction and using 5/384 wl^4/EI). Effective MIhas been calculated according to IS 456. Deflection by plate formula is also coming almost same.
Now permissible deflection is 80 mm (Span/250). If I increase the depth of ribs, headroom is getting reduced as storey ht is 3.6m. What to do? shall I provide camber? Varyani Saheb has recommended camber in his book.

Best Wishes,
A K JHA

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vikram.jeet
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PostPosted: Thu Jan 13, 2011 9:59 am    Post subject: Deflection of waffle slab Reply with quote

I think Camber can be provided in all long span structures.An Arch is nothing but a
heavily camber-ed beam .However additional effect of providing camber need to be
taken in the supporting beams and columns assuming hipped structure with small
rise(=camber) at centre especially for this 20mX20m waffle.

best regards

vikramjeet
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gpsarathyy
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PostPosted: Thu Jan 13, 2011 10:03 am    Post subject: Reply with quote

Dear Sefian,

Deflection is also a function of stress in steel, hence increase the Tension steel and compression steel and check as per the IS 456 code. The allowable deflection shall be modified with some factor based on the stress in steel.

Regards,
G.Parthasarathy.
Chennai
Email: gpsarathyy@gmail.com
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hemal
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PostPosted: Thu Jan 13, 2011 10:15 am    Post subject: Deflection of waffle slab Reply with quote

Dear Mr. JHA,

You can try Post-tensioning.

Regards



--- On Thu, 13/1/11, akjhacpwd <forum@sefindia.org> wrote:
Quote:

From: akjhacpwd <forum@sefindia.org>
Subject: [SEFI] Deflection of waffle slab
To: general@sefindia.org
Date: Thursday, 13 January, 2011, 2:30 AM

Dear friends
I am designing a waffle slab 20 m by 20 m which will house library. Ribs are at 1 m spacing in both directions . I have taken ribs width as 250mm. The depth of rib including topping is 900 mm. The waffle slab is supported on peripherial beams which in turn are supported on columns of 400 mm by 900 mm . The columns are spaced at 5 m in one direction and at 4m in the other. The size of peripherial beam is 400 by 900. I am facing problem of deflection . Short term deflection is 62 mm and total defn incl creep is 112 mm. deflection has been calculated considering discontinuous ends by Rankin theory ( taking w/2 in each direction and using 5/384 wl^4/EI). Effective MIhas been calculated according to IS 456. Deflection by plate formula is also coming almost same.
Now permissible deflection is 80 mm (Span/250). If I increase the depth of ribs, headroom is getting reduced as storey ht is 3.6m. What to do? shall I provide camber? Varyani Saheb has recommended camber in his book.

Best Wishes,
A K JHA
     



     



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B.V.Harsoda
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PostPosted: Thu Jan 13, 2011 11:30 am    Post subject: Reply with quote

Respected Sir,

My best Friend & Our Member Er. Lakhlani has been designed such structure as linked below:-

http://www.lakhlani.com/sabhagruh.php


for more details, You should contact him, his website is as below:-

http://www.lakhlani.com/index.php


Best Regards,

B. V. Harsoda
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anshugoel
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PostPosted: Thu Jan 13, 2011 4:28 pm    Post subject: Re: Deflection of waffle slab Reply with quote

akjhacpwd wrote:
Dear friends
I am designing a waffle slab 20 m by 20 m which will house library. Ribs are at 1 m spacing in both directions . I have taken ribs width as 250mm. The depth of rib including topping is 900 mm. The waffle slab is supported on peripherial beams which in turn are supported on columns of 400 mm by 900 mm . The columns are spaced at 5 m in one direction and at 4m in the other. The size of peripherial beam is 400 by 900. I am facing problem of deflection . Short term deflection is 62 mm and total defn incl creep is 112 mm. deflection has been calculated considering discontinuous ends by Rankin theory ( taking w/2 in each direction and using 5/384 wl^4/EI). Effective MIhas been calculated according to IS 456. Deflection by plate formula is also coming almost same.
Now permissible deflection is 80 mm (Span/250). If I increase the depth of ribs, headroom is getting reduced as storey ht is 3.6m. What to do? shall I provide camber? Varyani Saheb has recommended camber in his book.

Best Wishes,
A K JHA

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Why not increase the rib width and reduce the spacing - that won't hurt.
May be increase the slab depth by couple of inches (50 mm)
The depth of 900 mm for span of 20 metre seems to be less.
If that does not work, may be use higher strength concrete - to get the E up.
Pre-post tensioning will be little expensive, I think.

On another note - providing camber to take care of deflection - would it not introduce horizontal thrust at supporting columns ?
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rg.gupta
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PostPosted: Fri Jan 14, 2011 1:36 pm    Post subject: deflection in waffle slab Reply with quote

Dear A K Jha
Your problem has been understood and a solution is within the
preview of clause 23.2.1 of IS-456:2000 to control the deflection
by the increase in either tension steel or compression steel is explained
as under:
For the explanation of your problem,The following tentative datas
have been worked out and taken up assumingly. The design of waffle
slab is no other than  a slab except hollow portions surrounded by ribs.
As well as toppings.
DL+LL for slab =2.0 t/m2(unfactored)
FRCK=25 n/mm2
EFFECTIVE COVER=4.0 cms
SUPPORT CONDITION =9 (AS PER IS-456:2000)
D:\FORTRAN\BINB>slabrg
A PROGRAM FOR ONE WAY , TWO WAY & CANTILEVER SLABS
BY Er R.G.GUPTA             *BASED ON LIMIT STATE METHOD*
FOR FY=415.0 N/MM2 AND FCK IN N/MM2
LX =SHORT SPAN IN MTRS
LY =LONG SPAN IN MTRS
OD =OVERALL DEPTH IN CMS
W  =LOAD IN TONNES/METER2(UN-FACTORED)
N  =SUPPORT CONDITION AS PER IS-456
IF SUPPORT CONDITION IS CORRECT THEN C=1 OTHERWISE C=0
FOR ONE WAY SLABS
USE K =1,IF SLAB IS DISCONTINUIOUS OVER SUPPORTS, OTHERWISE
USE K =2, FOR SLAB IS CONTINUIOUS OVER SUPPORTS
USE K =3, FOR SLAB IS CANTILEVER
USE J = 0 IF DESIGN OF SLABS ARE TO BE CONTINUED OTHERWISE USE J= 1

FCK  =25.
LX   =20.0
LY   =20.0
OD   =90.
W    =2.
ECOVER=4.
SLAB IS TWO WAY SLAB
N    =9
SELECTED CONDITION IS FOUR EDGES DISCONTINUED
C    =1
SLAB IS SAFE IN DEFLECTION & AMF=  1.57 RMF=  1.16
DIA1 = 20.
DIA2 = 20.
.056  44.74 22.61  DIA= 20.  SPACING= 14.
.056  44.80 22.64  DIA= 20.  SPACING= 14.

AS PER ABOVE RESULTS
MAX. MOMENT IN WAFFLE SLAB=44.80 tm
D:\FORTRAN\BINB>beamrg
A PROGRAM FOR SINGLY & DOUBLY REINFORCED BEAMS
BY Er R.G.GUPTA         *BASED ON LIMIT STATE METHOD*
FOR FY IN N/MM2 AND FCK IN N/MM2
B  =WIDTH IN CMS
OD =TOTAL DEPTH IN CMS
D1 =EFFECTIVE COVER IN CMS
M  =MOMENT IN TONNES METERS
V  =SHEAR IN TONNES
DIA=DIA. OF V/STIRRUP IN MM
USE J1 = 0 IF DESIGN OF BEAMS TO BE DISCONTINUED FOR SHEAR
OTHERWISE USE J1= 1
USE J = 0 IF DESIGN OF BEAMS ARE TO BE CONTINUED OTHERWISE USE J= 1
FY =  415.
FCK =  25.
B  =  25.
OD =  90.
D1 =  4.
M  =  44.80
VALUES OF M IS UNFACTORED,SAY YES OTHERWISE NO
yes
k=  3.634  Pt=  1.211  Pc=   .265

           At= 26.032   Ac=  5.707
AS PER ABOVE RESULTS
TENSION STEEL          =26.03 cm2
COMPRESSION STEEL= 5.70 cm2
RELATIVE MODULAR RATIO =112/80=1.40(AS ACTUALLY REQUIRED)
D:\FORTRAN\BINB>defrg
         A PROGRAMME FOR FINDING MODIFICATION FACTOR FOR STEEL

IF N = 0, Modification for TENSION reinforcement
IF N = 1, Modification for COMPRESSION reinforcement
ALFA = Modification Factor
    D = Overall Depth in Cms.

    B = Width in Cms.
COVER = EFFECTIVE COVER

   AT = Area of Tension Steel
   AC = Area of Compression Steel
N  =0
AT =26.03
COVER=4.
B  =25.
D  =90.

  % STEEL AREA OF STEEL  SER.STRESS  MOD. FACTOR

       1.211      26.030     240.700        .956
       1.311      28.180     222.336       1.014
       1.411      30.330     206.575       1.069
       1.511      32.480     192.901       1.121
       1.611      34.630     180.925       1.172
       1.711      36.780     170.349       1.221
       1.811      38.930     160.941       1.268
       1.911      41.080     152.518       1.313
       2.011      43.230     144.932       1.356
       2.111      45.380     138.066       1.398
       2.211      47.530     131.820       1.439
       2.311      49.680     126.116       1.478
       2.411      51.830     120.884       1.515
       2.511      53.980     116.069       1.552
       2.611      56.130     111.623       1.587
       2.711      58.280     107.506       1.621
       2.811      60.430     103.681       1.654
       2.911      62.580     100.119       1.686
Stop - Program terminated.
                     OPTION 1
FROM ABOVE RESULTS
TENSION STEEL REQUIRED TO OVER COME THE DEFLECTION OF 112mm
MF=(1.3131+1.356)/(2X0.956)=1.40
AT=1.95X25X(90-4)/100 =41.92 cm2 AGAIST 26.03
D:\FORTRAN\BINB>defrg
         A PROGRAMME FOR FINDING MODIFICATION FACTOR FOR STEEL
IF N = 0, Modification for TENSION reinforcement

IF N = 1, Modification for COMPRESSION reinforcement
ALFA = Modification Factor
    D = Overall Depth in Cms.
    B = Width in Cms.
COVER = EFFECTIVE COVER
   AT = Area of Tension Steel
   AC = Area of Compression Steel
N  =1
AC=5.70
COVER=4.0
B  =25.
D  =90.

    %  STEEL   AREA OF STEEL   MOD. FACTOR
         .265        5.700        1.078
         .365        7.850        1.102
         .465       10.000        1.139
         .565       12.150        1.166
         .665       14.300        1.185
         .765       16.450        1.203
         .865       18.600        1.222
         .965       20.750        1.243
        1.065       22.900        1.264
        1.165       25.050        1.284
        1.265       27.200        1.303
        1.365       29.350        1.320
        1.465       31.500        1.335
        1.565       33.650        1.349
        1.665       35.800        1.361
        1.765       37.950        1.373
        1.865       40.100        1.386
        1.965       42.250        1.401
        2.065       44.400        1.419
        2.165       46.550        1.443
        2.265       48.700        1.470
        2.365       50.850        1.499
Stop - Program terminated.
OPTION 2
FROM ABOVE RESULTS
COMPRESSION  STEEL REQUIRED TO OVER COME THE DEFLECTION OF 112mm
MR=1.51/1.078=1.40
AC=2.37X25X(90-4)/100 =50.96cm2 AGAIST 5.70
OPTION 3
FROM ABOVE RESULTS
BUT IF WE PROVIDE 10 NOS 20 TOR AT TOP & BOTTOM
AREA OF STEEL AT TOP & BOTTOM=31.40cm2
MF1 DUE TO TENSION STEEL=(1.069+1.121)/(2X0.956)=1.145
MF2 DUE TO COMPRESSION STEEL=1.335/1.078=1.238
MF=MF1XMF2=1.145X1.238 =1.418
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knsheth123
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PostPosted: Sat Jan 15, 2011 3:34 pm    Post subject: Deflection of waffle slab Reply with quote

Provide Larger widths (400) for Ribs spanning between Columns & To carry out analysis as Space Frame.
Will result in reduce deflections.

Use of Higher Conc Grade, Some Addl., Imposed loads Loads applied after 3 to 6 month duration and provision of Camber 25mm.

All these put together should work to bring deflection under Control.

With Regards

K. N. Sheth

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PostPosted: Sun Jan 16, 2011 7:22 am    Post subject: Re: Deflection of waffle slab Reply with quote

akjhacpwd wrote:
Dear friends
I am designing a waffle slab 20 m by 20 m which will house library. Ribs are at 1 m spacing in both directions .
Now permissible deflection is 80 mm (Span/250). If I increase the depth of ribs, headroom is getting reduced as storey ht is 3.6m. What to do? shall I provide camber? Varyani Saheb has recommended camber in his book. Best Wishes, A K JHA Posted via Email

     Dear Mr. Jha,
     I stand to be corrected.The floor being designed has to carry loading of library or is the roof of library block?
     Eitherway. Is it necessary to provide orthogonal grid for the slab. Can it be diagonal? In case of diagonal grids maximum span of beam is 14.12M and not 20m,hence beam moments, shear and deflection will come down drastically. In that case instead of ribbed theory use grid theory and increase distance between ribs to 1.25m.
     Regards Umesh Rao
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PostPosted: Sun Jan 16, 2011 8:45 am    Post subject: Reply with quote

Respected Er. Rao Sir,

I am not clear about how diagonal beams would have a smaller span than perpendicular ones. The maximum span of beam would be 20*sqrt(2) = 28.3m.

Regards,

A S Oundhakar
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