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kawshik9 General Sponsor
Joined: 28 Oct 2009 Posts: 16

Posted: Thu Jul 01, 2021 10:25 am Post subject: SHEAR FORCE IN TWO WAY SLAB 


Hello SEFI Members, I have a query regarding shear force calculation in two way slab design. Kindly clarify my doubt.
I used to check the slab for maximum shear force of w(lxd)/2 , lx=shorter span. d= effective depth. But, in one of the documents i have seen that shear force is calculated as
Maximum shear force intensity in either direction = w.Lx . [r/(2r+1)] where r =ly/lx (longer span to shorter span ratio)
I would like to know the basis for above calculations and reference (if any) for the above equation. Please find the design calculations in the attachement for clarity.
Thank you
Ramakanth
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vikram.jeet General Sponsor
Joined: 26 Jan 2003 Posts: 3470

Posted: Thu Jul 01, 2021 12:04 pm Post subject: 


Dear Er
Two way slab  shear check
The load distribution in two way slab is in triangular (short span ) and trapezoidal (long span).
Shear check in slab is to be made at location of maximum load intensity , and the formula w(Lx  d)/2 followed by you is ok. In any such two way slab max load intensity is at longer beams and has to be as shown by you, i.e 0.50 Lx
The formula wlx(R / 2R+1) seems equivalent udL on beam for BM .
For R=1 , equivalent udL = wLx/3 which is equivalent UDL for square panel .
For R =2 , equivalent udL works to 0.40 Lx from this formula but it works to 0.458 wLx from exact formula.
However value of 0.50Lx is in place


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vikram.jeet General Sponsor
Joined: 26 Jan 2003 Posts: 3470

Posted: Thu Jul 01, 2021 1:57 pm Post subject: 


TWO WAY SLABS  EQUIVALENT UDL FORMULAS
Just to further elaborate :
Equivalent udL on beam for BM
The equivalent udL is the UdL which generate same simply suppoerted BM at center as the triangular / trapezoid load on beam
If m=ly/Lx , Lx is short span of panel of size Lx ×. Ly
The equivalent udL for BM
= wLx/3 on short span beam
= (wLx/6 )*(3  m^2) on long span beam
Equivalent udL for shear force on beams at support
= wLx/4 on short span beam
= (wLx/4)*(2  m) on long span beam of panel
But for checking shear force in slab , the above formulas are not required . Only from trapezoid load distribution in two way slabs , one can easily say that max SF in slab is wLx/2 at points of maximum load intensity.s.


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es_jayakumar General Sponsor
Joined: 24 Nov 2011 Posts: 1406 Location: Cochin

Posted: Thu Jul 01, 2021 4:06 pm Post subject: 


I hope, the attached material will be useful in this context :
E S Jayakumar
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vikram.jeet General Sponsor
Joined: 26 Jan 2003 Posts: 3470

Posted: Fri Jul 02, 2021 1:55 am Post subject: 


m = Lx/Ly. , Lx short span
Correction in this post
Inconvenience regretted
vikram.jeet wrote:  TWO WAY SLABS  EQUIVALENT UDL FORMULAS
Just to further elaborate :
Equivalent udL on beam for BM
The equivalent udL is the UdL which generate same simply suppoerted BM at center as the triangular / trapezoid load on beam
If m=ly/Lx , Lx is short span of panel of size Lx ×. Ly
The equivalent udL for BM
= wLx/3 on short span beam
= (wLx/6 )*(3  m^2) on long span beam
Equivalent udL for shear force on beams at support
= wLx/4 on short span beam
= (wLx/4)*(2  m) on long span beam of panel
But for checking shear force in slab , the above formulas are not required . Only from trapezoid load distribution in two way slabs , one can easily say that max SF in slab is wLx/2 at points of maximum load intensity.s. 


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vikram.jeet General Sponsor
Joined: 26 Jan 2003 Posts: 3470

Posted: Sat Jul 03, 2021 9:13 am Post subject: 


The content enlightened by Er E S Jayakumar is very exact as different panel edge conditions are covered.
Reynold's Handbook also provides coefficients for equivalent UDL on supporting beams for various edge conditions , for BM and Shear force separately.
es_jayakumar wrote:  I hope, the attached material will be useful in this context :
E S Jayakumar 


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es_jayakumar General Sponsor
Joined: 24 Nov 2011 Posts: 1406 Location: Cochin

Posted: Sun Jul 04, 2021 7:10 am Post subject: 


I analysed the square slab in question using plate elements of STAAD, for 3 edges discontinuous and one edge fixed condition. The findings are as below :
1. Total support reaction on the continuous (fixed) edge= 652.3 kN
2. Total support reaction on the discontinuous edge opposite to the fixed edge = 384.9 kN
3. The support reaction on the other two lateral discontinuous edges parallel to each other = 299.3 kN each
4. The total load = 652.3 + 384.9 + 2x299.3 = 1636 kN
5. The total load as per design data in the Excel sheet = 52.6 kN/m^{2} x 5.575m x 5.575m = 1635 kN
6. The load shared in the continuousdiscontinuous direction = 6520.3+384.9 = 1037.2 kN
7. The load shared in the discontinuousdiscontinuous direction = 2x 299.3 = 599 kN
8. The proportion of load shared in the continuousdiscontinuous direction = 1037.2 / 1636 = 0.63
9. The same, as per the formula given in the above table [5h^{4}/(2+5h^{4}) ] = 0.71 (in this case, h= 1)
10. Maximum (average) Shear force per metre ( at the continuous edge) = 652.3 kN/5.575m = 117 kN ( an SF of 97.8kN is seen considered in the Excel sheet)
ES Jayakumar


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vikram.jeet General Sponsor
Joined: 26 Jan 2003 Posts: 3470

Posted: Tue Jul 06, 2021 5:36 am Post subject: 


Very nice illustration by Er ES Jayakumar.
Please keep it up for benefit of one and all.
es_jayakumar wrote:  I analysed the square slab in question using plate elements of STAAD, for 3 edges discontinuous and one edge fixed condition. The findings are as below :
1. Total support reaction on the continuous (fixed) edge= 652.3 kN
2. Total support reaction on the discontinuous edge opposite to the fixed edge = 384.9 kN
3. The support reaction on the other two lateral discontinuous edges parallel to each other = 299.3 kN each
4. The total load = 652.3 + 384.9 + 2x299.3 = 1636 kN
5. The total load as per design data in the Excel sheet = 52.6 kN/m^{2} x 5.575m x 5.575m = 1635 kN
6. The load shared in the continuousdiscontinuous direction = 6520.3+384.9 = 1037.2 kN
7. The load shared in the discontinuousdiscontinuous direction = 2x 299.3 = 599 kN
8. The proportion of load shared in the continuousdiscontinuous direction = 1037.2 / 1636 = 0.63
9. The same, as per the formula given in the above table [5h^{4}/(2+5h^{4}) ] = 0.71 (in this case, h= 1)
10. Maximum (average) Shear force per metre ( at the continuous edge) = 652.3 kN/5.575m = 117 kN ( an SF of 97.8kN is seen considered in the Excel sheet)
ES Jayakumar 


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kawshik9 General Sponsor
Joined: 28 Oct 2009 Posts: 16

Posted: Tue Jul 06, 2021 11:33 am Post subject: 


Thank you very much Er. Vikram.jeet and Er.ES Jayakumar for sparing time to clear my query and sharing the concept , different references and even illustrating the same with Plate model for calculating load distribution in two way slab. It helped a lot in my work.
I could find that slab is under designed for shear force, As the shear force calculated in the excel gives the Load share on Beam but not the shear force per unit strip.
Ramakanth


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es_jayakumar General Sponsor
Joined: 24 Nov 2011 Posts: 1406 Location: Cochin

Posted: Fri Jul 09, 2021 4:33 pm Post subject: 


You are welcome, Er.Ramakanth....
N.Subramanian sir, in his famous book "Design of Reinforced Concrete Structures" has given a table (TABLE 10.5, page375) that lists coefficients to work out shear forces in twoway restrained slabs. This is table is very handy. Using the relevant coefficient of this table, the maximum Shear Force in the present case (at the continuous edge) = 0.45 x 52.63 x 5.575 = 132 kN
E S Jayakumar


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