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a121s General Sponsor
Joined: 13 Jun 2011 Posts: 48

Posted: Wed Jun 19, 2013 12:48 pm Post subject: shear strength of plain concrete 


Hello Sefians,
I would like to know how to calculate the permissible shear strength of concrete. I tried by IS 456 cl 31.6.3 but its takes into account the percentage of steel into the formula, the same goes with the formula given in cl 39 of SP24. But I want to calculate the shear strength of plain concrete. How to go ahead. Is it correct to put beta = 1 in formula given in SP24 and get the result which works out to be 0.184sqrt(fck) or go as per IS 456 as o.16sqrt(fck). Kindly guide me in this regard. _________________ SHISHIR S. KADAM
M.Tech(Structures),A.M.I.E 

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Dr. N. Subramanian General Sponsor
Joined: 21 Feb 2008 Posts: 5341 Location: Gaithersburg, MD, U.S.A.

Posted: Thu Jun 20, 2013 6:37 pm Post subject: Re: shear strength of plain concrete 


Dear Er. Kadam,
ACI code formula does not contain reinforcement percentage (of course there is another formula in ACI) and it is sq.root (f_{c})/6, if you take f_{c }=0.8f_{ck}, you will get it as 0.15 sq.root(f_{ck}). Thus for M25 concrete it will be 0.75 N/mm^{2}.
Since the term p_{t} appears in the denominator of Beta, you can not find out the value of Tauc as per IS 456 with zero reinforcement. As you have said, if we assume Beta =1, we will get a result of 0.184 sq.root(f_{ck}), which is slightly higher than ACI result. But compare it with the values given in Table 19 of IS 456 for p_{t} <0.15 (minimum reinforcement) it is 0.291 MPa, that is 2.58 times less than the strength specified in ACI! Only when pt is equal to 1.55% we get a value of 0.75 N/mm^{2} ! Hence you can not use the formula in IS 456 to get the tauc value of plain concrete.
In fact, even a minimum longitudinal steel can increase the shear strength considerably due to the dowel action. Shear strength is a complicated area in RC design. Hundreds of papers have been published in America alone on shear strength of concrete.
Note that the Americans were using beams without shear reinforcement till the failure of Wilkins Air Force Depot in Shelby , Ohio, on Aug 17, 1955. A second warehouse roof collapse took place at Robins Air Force Base near Macon , Georgia , early on the morning of September 5, 1956. In both cases, the design, materials, and workmanship were up to the codes and standards of the day. However, the failures had still occurred. But both were subjected to wide temperature variations.
The temperature stresses, unrelieved by the locked expansion joints, combined with shrinkage and shear effects caused high tensile stress. These two cases illustrate the importance of providing at least some minimum amount of reinforcement, called temperature steel, to resist tension forces due to thermal, shrinkage, and other effects. The ACI code was subsequently changed to include minimum shear reinforcement!
Best wishes,
NS
References:
 Feld, J., and Carper, K. (1997). Construction Failure. 2nd Ed., John Wiley & Sons, New York, N. Y.
 McKaig, T. (1962). Building Failures: Case Studies in Construction and Design. McGrawHill, New York, N. Y
a121s wrote:  Hello Sefians,
I would like to know how to calculate the permissible shear strength of concrete. I tried by IS 456 cl 31.6.3 but its takes into account the percentage of steel into the formula, the same goes with the formula given in cl 39 of SP24. But I want to calculate the shear strength of plain concrete. How to go ahead. Is it correct to put beta = 1 in formula given in SP24 and get the result which works out to be 0.184sqrt(fck) or go as per IS 456 as o.16sqrt(fck). Kindly guide me in this regard. 


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a121s General Sponsor
Joined: 13 Jun 2011 Posts: 48

Posted: Fri Jun 21, 2013 4:01 am Post subject: 


Dear Dr N. Subramanian
Thank you a lot for clearifying my doubts, I will go through the reference and ACI code. Sir, I would like to ask you one more thing, In shotcrete (SFRS) the shear strength is higher than plain concrete, so is their any formula to calculate the permissible shear strength of shotcrete. In IS 15026:2202 (Tunnelling methods in rock masses guidlines), they have given the shear strength of SFRS as 8 t0 10 MPa and asked to adopt 5.5 MPa ( They have not mentioned whether it is ultimate strength) So in my calculation I have used the shear strength as follows,
example:
1. Shotcrete shear strength as suggested = 10 MPa
2. Partial Safety Factor = 1.5
3 Factor of safety for loads = 1.5
therefore, permissible shear strength = 10/(1.5* 1.5) = 4.44 MPa
I would like to know whether I am going in correct direction ? and how I can I find the shear strength of Shotcrete in Working Stress method?, Is the same above applicable for WSM?
With Regards,
SHISHIR S. K. 

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sknsrinivasan ...
Joined: 15 Apr 2013 Posts: 281

Posted: Fri Jun 21, 2013 4:49 am Post subject: 


Dear Er Kadam
permissible shear strength of concrete in Flat slab ks*tc IS 456 cla 36.6.3.1 Page no 59
where ks=(0.5+beta C) ,beta C is the ratio of short side to long side of column capital
tc= 0.25sqrt(fck) in Limitstate method
0.16sqrt(fck) in working stress method
The above Formula Applicable only permissble shear strss in Flat slab
For Beam or column ( Limit state method) Given Below
tc=(.85*sqrt(0.8*fck)*(sqrt(1+5)*beta1))/6*beta
where beta=0.8(fck)/6.89pt This also available in SP16 cl 4.1 page175 All the Formula also available in IS code Plz check it onceagain
Thanks
N. Srnivasan 

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sknsrinivasan ...
Joined: 15 Apr 2013 Posts: 281

Posted: Fri Jun 21, 2013 6:28 am Post subject: 


cont. above Mail To find the permissible shear stress of plain concrete can we use the above mentioned formula in flat slab shear stress I request Dr NS sir plz clarify
As per cal 31.6.3.1 The permissible shear stress ksxtc If your cocrete size is 1mx1m beta C= 1/1=1, ks= 0.5+1=1.5 should not grater than 1 if the value ks grater than 1 consider 1Here I consider 1, so permissible shear stress is 1x0.25xsqrt(fck) awaiting for your Reply
Thanks
N.Srinivasan 

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Dr. N. Subramanian General Sponsor
Joined: 21 Feb 2008 Posts: 5341 Location: Gaithersburg, MD, U.S.A.

Posted: Fri Jun 21, 2013 3:32 pm Post subject: 


Dear Er Shishir,
Shotcrete typically has a density and compressive strength similar to normal and highstrength concrete, but hardened properties truly are operatordependent.
According to the American Shotcrete Association (ASA), in residential applications, shotcrete mixtures can in fact exceed the compressive strength of most mixtures used for placed walls. The compaction that occurs during application of shotcrete helps achieve improved compressive strength and durability, and low watercementitious material ratios of shotcrete mixtures produce other benefits, including reduced shrinkage and lower permeability.
On the safe side, you can take the shear strength as that of normal concrete, according to the strength. You may visit American Shotcrete Association archived article pages to know more: http://www.shotcrete.org/pages/archivesearch/ArchiveSearch.asp
You can also read the current and old issues of Shotcreat magazine.
best wishes,
NS
a121s wrote:  Dear Dr N. Subramanian
Thank you a lot for clearifying my doubts, I will go through the reference and ACI code. Sir, I would like to ask you one more thing, In shotcrete (SFRS) the shear strength is higher than plain concrete, so is their any formula to calculate the permissible shear strength of shotcrete. In IS 15026:2202 (Tunnelling methods in rock masses guidlines), they have given the shear strength of SFRS as 8 t0 10 MPa and asked to adopt 5.5 MPa ( They have not mentioned whether it is ultimate strength) So in my calculation I have used the shear strength as follows,
example:
1. Shotcrete shear strength as suggested = 10 MPa
2. Partial Safety Factor = 1.5
3 Factor of safety for loads = 1.5
therefore, permissible shear strength = 10/(1.5* 1.5) = 4.44 MPa
I would like to know whether I am going in correct direction ? and how I can I find the shear strength of Shotcrete in Working Stress method?, Is the same above applicable for WSM?
With Regards,
SHISHIR S. K. 


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Dr. N. Subramanian General Sponsor
Joined: 21 Feb 2008 Posts: 5341 Location: Gaithersburg, MD, U.S.A.

Posted: Fri Jun 21, 2013 3:38 pm Post subject: 


Dear Er Srinivasan,
What you are quoting is the punching shear strength!
Best wishes,
NS
sknsrinivasan wrote:  cont. above Mail To find the permissible shear stress of plain concrete can we use the above mentioned formula in flat slab shear stress I request Dr NS sir plz clarify
As per cal 31.6.3.1 The permissible shear stress ksxtc If your cocrete size is 1mx1m beta C= 1/1=1, ks= 0.5+1=1.5 should not grater than 1 if the value ks grater than 1 consider 1Here I consider 1, so permissible shear stress is 1x0.25xsqrt(fck) awaiting for your Reply
Thanks
N.Srinivasan 


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sknsrinivasan ...
Joined: 15 Apr 2013 Posts: 281

Posted: Sat Jun 22, 2013 3:50 am Post subject: 


Dear Dr NS
Thank you very much for your clarification
Thanks
N.Srinivasan 

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a121s General Sponsor
Joined: 13 Jun 2011 Posts: 48

Posted: Mon Jun 24, 2013 4:26 am Post subject: 


Dear Dr N.S.
I clearly agree with your Dr. N.S. statement that, the formula stated as 0.16sqrtfck and other formulas as said above are used to find the punching shear in flat slab and in shear strength in beams, and as we provide the shear reinforcemnt in a perticular direction the formulas can be feasible, But in case of shotcrete the reinforcement is random thus the above formulas cannot be used. For shotcrete we need the pull out test results because shotcrete not only fails in shear but also in adhesion. I will go through the link provided and continue the discussions with you sir. Its a great pleasure to receive comments from a renowned person like Dr. N.S., there is lot to learn from you sir.
With Regards
SHISHIR S. K. 

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