Area Moment of Inertia for Goal Post
Area Moment of Inertia for Goal Post
(OP)
Hi All,
Here's the situation, I'm planning on installing a strain gauge on a goal post shaped steel beam (dimensions/drawing attached). We'll be applying load at the top of the goal post and reading strains down near the base (half way between the posts and the base actually) and we're trying to understand what kind of stress to strain relationship we can get out of it. Now I'm using Stress = MC/I for this (which may be wrong in itself for this shape of an object) because this is essentially a cantilever beam with a weight on the end..the problem is getting the "I", or second moment of inertia. I've tried using the parallel axis theorem and breaking the part up, but I'm not sure the correct way to do this so I've gotten a bunch of different answers and no way to know which is right.
The axis of rotation is through the center of the bottom of the goal post (where the strain gauge will be placed). A little bit of help to get me pointed in the right direction would be great.
Tyler
Here's the situation, I'm planning on installing a strain gauge on a goal post shaped steel beam (dimensions/drawing attached). We'll be applying load at the top of the goal post and reading strains down near the base (half way between the posts and the base actually) and we're trying to understand what kind of stress to strain relationship we can get out of it. Now I'm using Stress = MC/I for this (which may be wrong in itself for this shape of an object) because this is essentially a cantilever beam with a weight on the end..the problem is getting the "I", or second moment of inertia. I've tried using the parallel axis theorem and breaking the part up, but I'm not sure the correct way to do this so I've gotten a bunch of different answers and no way to know which is right.
The axis of rotation is through the center of the bottom of the goal post (where the strain gauge will be placed). A little bit of help to get me pointed in the right direction would be great.
Tyler






RE: Area Moment of Inertia for Goal Post
I don't see why Mc/I would be wrong. How I see it is you need to calculate I for each individual sections. For a rectangular plate with b=width d=depth I=bd^3/12. Section modulus (S) = bd^2/6 with stress breaking down to M/S. Be sure to keep your units straight.
RE: Area Moment of Inertia for Goal Post
BA
RE: Area Moment of Inertia for Goal Post
Would I have to use the parallel axis theorem to do this, since parts of the beam are further out from the axis of rotation? I'm more confused about how you're supposed to look at the shape when doing this equation...look at it as view 2, or view 3 (I've attached a modified drawing labeling the views).
Thanks again for the help.
Tyler
RE: Area Moment of Inertia for Goal Post
RE: Area Moment of Inertia for Goal Post
and applied acting down the leg, yes?
so the vertical leg is in compression, and the horizontal leg is in bending, yes?
the section looks to be 0.5" deep (=d) and 0.375" wide (=b) ... I = bd^3/12, c = d/2 ... yes?
Quando Omni Flunkus Moritati
RE: Area Moment of Inertia for Goal Post
I = bd^3/12
where d is the depth in the direction of the applied force and b is the total width of the rectangle(s).
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: Area Moment of Inertia for Goal Post
RE: Area Moment of Inertia for Goal Post
and as others are saying, the load is the load applied to a leg (so if you have a bar in the groove, loaded horizontally (as shown) then 1/2 the load will be applied to each leg ... yes?
Quando Omni Flunkus Moritati
RE: Area Moment of Inertia for Goal Post
Yes the load will be applied in the grooves of the goal post bending it back, so all the stress should be at the bottom (where we're placing the strain gauge). From your answers looks like I've got a couple options:
If it's just the rectangular section the gauge is on that you have to worry about you get I = 1/12 b d^3 = (1/12)*(0.5)*(.375^3) = .00219 in^4
If it's the whole width of the goal posts then I = (1/12)*(4)*(.375^3) = .0176 in^4
If it's each rectangle added together it's another number...
If I have to do the parallel axis theorem it's a whole other answer (a much larger one too).
This is where I'm getting confused...
Tyler
RE: Area Moment of Inertia for Goal Post
2) why, oh why, would you think the air between the two legs is reacting the load ?
3) see 1) (i think the load is evenly reacted by both legs, but what do i know)
4) this isn't a parallel axis problem, 'cause the neutral axis of the section is aligned with the bending axis. it's not like there's an area away from the axis ... it's simply bd^3/12
5) yes you are ... very (confused)
Quando Omni Flunkus Moritati
RE: Area Moment of Inertia for Goal Post
RE: Area Moment of Inertia for Goal Post
If the horizontal force on each upright is F, then the total horizontal force is 2F and the moment at the top of the pile cap is 2F*h where h is the height of the force above the cap. At the top of pile cap, the only section resisting moment is a rectangular section of width 0.5 and depth 0.375, so the stress is 2F*h/S where S is the section modulus, bd2/6.
Parallel axis theorem has nothing to do with the problem. What is this, a homework assignment?
BA
RE: Area Moment of Inertia for Goal Post
f=M/S where S = 1/6 x 0.5 x (0.372^2)
and
M = (distance from center of load application point to strain measurement centerline) x load
If the section is tube instead of a solid bar, the S value will change by subtracting out the hollow part using the hollow dimensions in the same formula.
RE: Area Moment of Inertia for Goal Post
you're matching with s/gauge ouput ... make sure you put the gauge on the right face !
Quando Omni Flunkus Moritati
RE: Area Moment of Inertia for Goal Post