Modeling bolt holes as stress concentrators in I beam bending

[foundry]

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 =\sum(I + Ad²)
I=\frac{1}{12}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 = \frac{Md}{I}

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 = \frac{P}{A}

now modeling the hole as a stress concentrator

σ_average = \frac{P}{(w-2r)t}

w = width of flange
r = radius of hole
t = thickness of flange

calculate \frac{r}{w} in order to find a value for K (tabled value)

σ_max = K σ_average

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.

 

[foundry]

I take it with the proper bolt preloading, these holes aren't an issue? Is that a correct assumption?

 

[Mech_Engineer]

Their effect on the beam will depend largely on the geometry and their location on the flange. It could be as long as the holes are far from an edge that you can just consider them to be located in a plate in tension, in which case there are well-documented stress concentration factors to calculate their stress rise. It will also depend on your loading conditions, but I wouldn't assume proper preload on a bolt in the hole will solve your problem.

Otherwise your best bet is FEA, but it brings its own set of limitations and challeges. Interestingly, many stress concentration factors (e.g. Peterson's Stress Concentration Factors) are calculated using parametric FEA models and curve fits to their outputs.