Area Moment of Inertia for Goal Post

[TylerJO]

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

 

[SteelPE]

What direction do you planning on applying the load?

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.

 

[BAretired]

The stress will be P/A ± My/I in the rectangular cross section above the base. If you are mounting the strain gauge on the surface, then stress is P/A ± M/S where S is the section modulus. For a rectangular section, I = bd³/12 and S = bd²/6. For these derivations, you can consult any strength of materials text.



BA

 

[TylerJO]

Thanks for the input. I know the equations, I'm just not sure which dimensions are b and d. The load will be applied at the top of the goal post (see view 1 in attachment).

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

 

[Ron]

Place the strain gauge as low as possible on the 0.375 x 0.5 member, on the 0.5 face. The strain gauge will read the extreme fiber stress at the surface which will correlate to the moment at that location. For that stress, "f", the equation f=M/S applies, with the "S" being the section modulus of the rectangular section.

 

[rb1957]

the load is going to be applied in the two semi-circular grooves, yes?
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

 

[IDS]

The second attachment shows that the load is applied in the horizontal direction, so it is a cantilever with five different cross sections along its length. Since each cross section is either one or two rectangles the second moment of area is simply given by:

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/

 

[SteelPE]

Is the load on one arm of the goal post or both?

 

[rb1957]

so it's bending about the other axis ... b = .5" d = .375",
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

 

[TylerJO]

All,

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

 

[rb1957]

1) is one leg supporting the load ? (is the load evenly divided between the two legs, or is more on one side ?)

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

 

[TylerJO]

Yes, it's evenly distributed. There will be a rod sitting across the goal posts that will be applying the load. Stress should be it's highest down near the base. Since I'm measuring stress at the base and you say it's not a parallel axis problem is it just bd^3/12 for the base section and that's it? Just I = (0.5)(.375^3)/12?

 

[BAretired]

You should have shown the load on View 3 as well as View 1. Then we wouldn't have to ask how the load was distributed.

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

 

[Ron]

Tyler...your analysis is straightforward. It is only the 0.5 x 0.375 section you are concerned with. For the sketch you provided and assuming your section is a solid bar not a tube, then your member stress will be:

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.

 

[rb1957]

"Just I = (0.5)(.375^3)/12?" ... if you're looking at one leg (carrying 1/2 the load)

you're matching with s/gauge ouput ... make sure you put the gauge on the right face !

Quando Omni Flunkus Moritati

 

[SLTA]

Out of curiosity, what will you be doing with this problem?