www.www.buonovino.com

[vikram.jeet]

Er Ramu

Calculations for deflection Chek purposes:

As per calcs fm u , pt = 0.128 for LF =1.0

Refer to chart figure 3 , the multiplier is 2.0

The effect of continuity cannot be taken twice , the % reinforcement worked is based on continuity and there fore Gross value of L/d = 2*20= 40

Hence eff depth = 327/40 = 8.2cm
Overall depth = 8.2+ 2= 10.2 cm cf 11.5 ok - - - This is as per clause 22.2.1

Also as per clause 23.1 for slabs as already discussed , L/D = 32
Hence D , the o/all depth = 327/32 = 10.22 cm , just the same

Calculation for reinf provision :
Actual reinf to be provided based on LF= 1.5
Hence Mu/bd^2 = 0.426*1.5 = 0.639 pt = 0.187
Ast 0.187*9.5= 1.78 cm2/m

[vikram.jeet]

One thing must be borne in mind that while making deflection check using load factor = 1.0 , pl calculate pt and see modification factor and do not further calculate Ast , Since Ast to be provided has to be based on Pt using Load factor =1.5.

Just my view after seeing yr calcls

[vikram.jeet]

Er Ramu

Cantilever slab

Your calcs are OK and depth rightly revised . Cantilever slabs must be provided with liberal depth as reinf is at top and chances of reinf getting little down at site are very much there inspite of placing chairs precautions . Even some designers always take view of this and provide some additional reinf at Cantilever slabs , though many designers who are very rigid may not encourage this . But while providing we have to judge site situations and should not act as theoretical tigers.

[ramu_se]

Thank you sir for giving such a brief explanation.

Why do we have to use the load factor as 1 for % of reinforcement calculation to find out the modification factor?

Is there any specfic codal provison for this?

To calculate fs we need to know Ast required and Ast provided, so this Ast is based on load factor 1 or load factor 1.5?

Thanks & Regards,
Ramu.

[vikram.jeet]

Deflection is serviceability requirement and as per IS 456 this needs to be checked at working loads as evident from clause 35.4 on partial safety factors for Loads Table 12 ( relating clauses 18.3 and 35.4.1 of IS 456 1978).

The Load factor for serviceability limits is( 1.0 DL + 1.0 LL)

[spsvasan]

Dear Er.Ramu & Er.Vikramjeet

I am generally in agreement with Er.Vikramjeet. However, I differ in one aspect only

fs = 0.58 x fy x Ast(required) / Ast(provided)
fs = 0.58 x 415 x 122 / 250 = 117.46 N/mm2
As per Fig.4 in IS-456:2000
M.F = 2.0

在上面的计算,恕我直言,Ast(必需)是to be the limit state requirement only, ie 178 sqmm/m.

In the expression f_s= 0.58 f_y(As reqd / As provided), the factor 0.58 lowers the limit state stress to the service stage level

In this particular case, the modification factor still remains as 2.0

S.P.Srinivasan

[vikram.jeet]

[BVRAO]

Slab Design with deflection check