Code Clarification CHBDC Shear

[Quade999]

Hello Everyone,

I'm looking for clarification on the clause below.

From what I take is that this clause only applies if my ultimate shear exceeds my concrete shear resistance. The only thing that threw me off is that the clause provides a maximum value for Vc. If you calculate the shear resistance using sectional design it gives a larger resistance than the one shown here. Am I correct in assuming that the equation for Vc is just an upper bound for when using this clause only to plug into the equation to find the required stirrup area? i.e. if my ultimate shear is 300kN, and from sectional analysis I calculate Vc = 320kN, but the equation present here gives Vc = 280kN, this equation doesn't apply since I am comparing the 320kN resistance to the 300kN ultimate shear, and not the 280kN resistance to the 300kN ultimate.

I believe this is the case, but just clarifying. From the commentary below it seems that Vc in this clause is just an upper bound for the stirrup equation.

Thanks

 

[Agent666]

I might be interpreting it slightly different to you, I read it as that equation for Vc is also used for the first comparison with Vu as well. Agree its an upper bound for determining if you need stirrups.

I'm not familiar with the code, but that's just my interpretation of the wording.

 

[Quade999]

From the commentary it seems that the equation provided for Vc provides a lower number in an effort increase required area of stirrups (as the higher your value of Vc, the less stirrup area you need), and the reason for this is because the section is assumed to be cracked. This is because the commentary specifies that this is the maximum value of Vc that can be used to reduce the stirrup area, and not the actual maximum of Vc for resistance of the section as a whole.

 

[Agent666]

Yeah, I think in re-reading it multiple times you are correct (explained below just to make sure I'm on the same page).

Determine if you need stirrups based on 0.25phi_cbd(f'c)^0.5 Vc limit, if your Vc>=Vu then you don't need stirrups and Vc is Vc = 0.25phi_cbd(f'c)^0.5.

If you need stirrups based on the above inequality, Vc is subsequently limited to 0.166phi_cbd(f'c)^0.5. Section shear capacity is this upper limit Vc + Vs determined form stirrups. I don't believe based on the commentary that you add Vc determined from 0.25 equation to the steel reinforcement shear contribution (presumably Vs in your code).