Dr. N. Subramanian General Sponsor
Joined: 21 Feb 2008 Posts: 5236 Location: Gaithersburg, MD, U.S.A.

Posted: Fri May 25, 2018 4:10 am Post subject: Re: Maximum allowable shear stress 


Dear Er. Shah
It is because shear failure is in fact a diagonal tension failure. You may be aware that concrete is weak in tension and hence the value is small. As per ACI 318, this value is 0.75×0.15×sq.root(fck)(after converting cylinder strength to cube strength). This value is unchanged for several years and us based on numerous expts. There is another equation to take into account the percentage reinforcement.
As I have mentioned in several previous postings, the Indian code IS 456 equation(it is given in the explanatory handbook) and Table is based on the expts. conducted by Prof. Vijaya Rangan in the 1960s. Now that the cement and strength of concrete being used is different from that used in 1960s, we need to use a different equation. Prof. Collins and his colleagues in Canada have done extensive tests on shear and developed the modified compression field theory, and based on that simplified approach is used in the Canadian code, which may be adopted in the Indian code. This also incorporated the size effect.
More discussion on shear strength may be found in my RC Design book.
Best wishes
NS
shah wrote:  In different code shear stress is limited to certain value despite how much shear reinforcement is provided. This is in general for avoiding over reinforced design. In IRS CBC it is given as 0.75 sqrt (fck). In IS 456 it is given directly in tabular (Table 20 )form. (slightly conservative than 075 sqrt (fck))
Now my question is that why that much low capacity is considered though compression capacity of concrete is equal to fck? How does assumed compression strut will crush in that much low stress.?
Thank you
Parshva Shah 

