Yousef ZAA
Structural
Hi All,
I am designing a field topped composite double tee and I have some doubts on how all engineering programs calculates the horizontal shear forces and stresses: For a composite member all commercial programs checks vertical shear for an expected shear crack through the thinnest part of the section, often the web(s) - bw, per ACI 318-19:22.5.6. It also checks for horizontal shear cracking across the full width of the composite interface, bv, per ACI 318-19:16.4.4.
In the field topped double tees design, when the program -that conforms with ACI!- checks the horizontal shear capacity in topped DTs, the program calculates the maximum horizontal shear force corresponding to the total double tee flange width (bv) and total effective depth (d) and compares this force with the vertical shear force being equal to the horizontal shear force, and I have doubts on this. The simplified elastic procedure has the vertical shear stress equal to the horizontal shear stress. Shouldn’t it be correct -especially with DTs- to have (bv) as the total webs width? Not the total flange width? I have a case attached, where I the vertical shear = 58kips, width of DTs = 144”, effective depth of DTs is around 24.8”, The program is doing the following: Vuh=Vu= 58kips, PhiVnh (Horizontal Shear Capacity) = 0.75*80psi*144”(width of flange)*24.8”(depth of DT)/1000 = 214 kips, compares it with Vuh of 58 kips. And the design is safe! Wouldn’t it be correct if we have instead, the following Vuh =Vu = 58kips, Max. horizontal shear stress = 58/ (13.7*24.8) = 170 psi, more than 80 psi! Not Safe. I have used the total web width 13.7” corresponding with the analogy of having the vertical shear stress equal the horizontal shear stress. In the case of beams with topping T-beams for instance, I am fine with what ACI is doing, but I am not sure about Double tees! My main concern in Double tees that usually it does not come with reinforcement crossing the contact surface, in case of beams usually projecting U-Bars are provided and Lv can be taken as half the span because of ductility. We don’t have that in DTs!!
I am designing a field topped composite double tee and I have some doubts on how all engineering programs calculates the horizontal shear forces and stresses: For a composite member all commercial programs checks vertical shear for an expected shear crack through the thinnest part of the section, often the web(s) - bw, per ACI 318-19:22.5.6. It also checks for horizontal shear cracking across the full width of the composite interface, bv, per ACI 318-19:16.4.4.
In the field topped double tees design, when the program -that conforms with ACI!- checks the horizontal shear capacity in topped DTs, the program calculates the maximum horizontal shear force corresponding to the total double tee flange width (bv) and total effective depth (d) and compares this force with the vertical shear force being equal to the horizontal shear force, and I have doubts on this. The simplified elastic procedure has the vertical shear stress equal to the horizontal shear stress. Shouldn’t it be correct -especially with DTs- to have (bv) as the total webs width? Not the total flange width? I have a case attached, where I the vertical shear = 58kips, width of DTs = 144”, effective depth of DTs is around 24.8”, The program is doing the following: Vuh=Vu= 58kips, PhiVnh (Horizontal Shear Capacity) = 0.75*80psi*144”(width of flange)*24.8”(depth of DT)/1000 = 214 kips, compares it with Vuh of 58 kips. And the design is safe! Wouldn’t it be correct if we have instead, the following Vuh =Vu = 58kips, Max. horizontal shear stress = 58/ (13.7*24.8) = 170 psi, more than 80 psi! Not Safe. I have used the total web width 13.7” corresponding with the analogy of having the vertical shear stress equal the horizontal shear stress. In the case of beams with topping T-beams for instance, I am fine with what ACI is doing, but I am not sure about Double tees! My main concern in Double tees that usually it does not come with reinforcement crossing the contact surface, in case of beams usually projecting U-Bars are provided and Lv can be taken as half the span because of ductility. We don’t have that in DTs!!
