aligorsi
Structural
- Apr 28, 2016
- 8
Hello Everyone,
I was checking the limits of drilling holes in flanges of beams without affecting the flexural capacity of the section and I found 2 pieces of contradictory information.
1. In section F 13.1 (Strength Reductions for Members with Holes in the Tension Flange), the code says that, When FuAfn ≥ YtFyAfg, the limit state of tensile rupture does not apply.
2. In section J 4.1 (Strength of Elements in Tension), for tensile yielding of connecting elements is Rn = FyAg to be divided by Ω = 1.67 (ASD), and For tensile rupture of connecting elements Rn = FuAe to be divided by Ω = 2 (ASD). So when you equate the two values of Rn, you get FyAg/1.67 = FuAe/2
The only difference between the two is that section F13.1 is equating the functions of tensile yield and tensile rupture without considering the safety factor while J 4.1 is. However, once you calculate the %age of the tension flange which can be drilled without any reduction in capacity you can drastically different values from the two approaches. E.g for Grade S275 steel (Fy = 275 MPa, Fu = 410 MPa, Yt = 1) you get 32.9% (F 13.1) vs 19.7% (J 4.1)
Assumptions:
This exercise was done because sometimes, we need to drill holes in the beams on site to support non structural components e.g. the foot of a electrical panel bolted to top flange of beam or fire fighting pipes connected to the bottom flange with bolts. Therefore, I have considered the value of U for calculation of effective net area (Ae) as 1 which makes it the same as An.
I was checking the limits of drilling holes in flanges of beams without affecting the flexural capacity of the section and I found 2 pieces of contradictory information.
1. In section F 13.1 (Strength Reductions for Members with Holes in the Tension Flange), the code says that, When FuAfn ≥ YtFyAfg, the limit state of tensile rupture does not apply.
2. In section J 4.1 (Strength of Elements in Tension), for tensile yielding of connecting elements is Rn = FyAg to be divided by Ω = 1.67 (ASD), and For tensile rupture of connecting elements Rn = FuAe to be divided by Ω = 2 (ASD). So when you equate the two values of Rn, you get FyAg/1.67 = FuAe/2
The only difference between the two is that section F13.1 is equating the functions of tensile yield and tensile rupture without considering the safety factor while J 4.1 is. However, once you calculate the %age of the tension flange which can be drilled without any reduction in capacity you can drastically different values from the two approaches. E.g for Grade S275 steel (Fy = 275 MPa, Fu = 410 MPa, Yt = 1) you get 32.9% (F 13.1) vs 19.7% (J 4.1)
Assumptions:
This exercise was done because sometimes, we need to drill holes in the beams on site to support non structural components e.g. the foot of a electrical panel bolted to top flange of beam or fire fighting pipes connected to the bottom flange with bolts. Therefore, I have considered the value of U for calculation of effective net area (Ae) as 1 which makes it the same as An.
