- #1
foundry
- 2
- 0
curious as to the proper way to model this.
beam is symmetric so centroid is in the middle (used to calculate distance from centroid d in bending moment of inertia equation)
bending moment of inertia is
I =[itex]\sum[/itex](I + Ad2)
I=[itex]\frac{1}{12}[/itex]bh
b=base
h=height
d=distance of the area's center to total centroid
A=area of section
using max moment from a FBD the stress is calculated
σ = [itex]\frac{Md}{I}[/itex]
say a bolt hole is placed on the top flange, to me it makes sense to model it this way...
take just the top flange use stress calculated from above equation to find the compressive load P via this equation and the cross sectional area of the flange A
σ = [itex]\frac{P}{A}[/itex]
now modeling the hole as a stress concentrator
σaverage = [itex]\frac{P}{(w-2r)t}[/itex]
w = width of flange
r = radius of hole
t = thickness of flange
calculate [itex]\frac{r}{w}[/itex] in order to find a value for K (tabled value)
σmax = K σaverage
like I said this makes sense to be but when I run through a sample calculation a beam originally having a factor of safety of 2 ends up not being able to withstand the load just by introducing a bolt hole. thanks for your time.
beam is symmetric so centroid is in the middle (used to calculate distance from centroid d in bending moment of inertia equation)
bending moment of inertia is
I =[itex]\sum[/itex](I + Ad2)
I=[itex]\frac{1}{12}[/itex]bh
b=base
h=height
d=distance of the area's center to total centroid
A=area of section
using max moment from a FBD the stress is calculated
σ = [itex]\frac{Md}{I}[/itex]
say a bolt hole is placed on the top flange, to me it makes sense to model it this way...
take just the top flange use stress calculated from above equation to find the compressive load P via this equation and the cross sectional area of the flange A
σ = [itex]\frac{P}{A}[/itex]
now modeling the hole as a stress concentrator
σaverage = [itex]\frac{P}{(w-2r)t}[/itex]
w = width of flange
r = radius of hole
t = thickness of flange
calculate [itex]\frac{r}{w}[/itex] in order to find a value for K (tabled value)
σmax = K σaverage
like I said this makes sense to be but when I run through a sample calculation a beam originally having a factor of safety of 2 ends up not being able to withstand the load just by introducing a bolt hole. thanks for your time.