Interpretation of AASHTO 5.8.2.9, definition of dv

[JHart_11]

The current AASHTO Bridge Design Spec. defines dv, effective shear depth, as the distance between the resultants of the tensile and compressive forces due to flexure. The commentary says this can be determined as dv=Mn(As*fy+Aps*Aps). For a non-prestressed rectangular section with no compression steel, this formula can be whittled down to, dv=d–a/2. In section 5.8.2.9, the definition of dv states, "it need not be less than 0.9de or 0.72h (in.)". My question is, if dv=d-a/2 is larger than 0.9de or 0.72h, do I use that value or the greater of 0.9de vs. 0.72h? I am thinking that the 0.9de or 0.72h statement is just a limiting value and I should use the dv=d-a/2 if larger, but am looking for confirmation from the experts . Thanks!

 

[Lomarandil]

That's correct, in the case where your physical distance is greater than 0.9d or 0.72h, you can use the true physical distance.

 

[JHart_11]

Thank you for the quick response.