metal thickness for boxing frame

[76zimmer]

I used Kerts boxing kit, and its holding up great...my frame was in pretty good condition, and I'm also fully caged front to rear. I think it makes the vehicle so much more solid...the twisting and bending are not an issue...it makes the suspension do all of the work, which is what it is supposed to do.

 

[howdiy]

I have a stupid question. I'm always hearing about people boxing their frame. The profile is a rectangle which isn't the strongest shape. Why not make the profile a triangle, the strongest shape. Basically weld the plate from the upper inside corner of the frame to the lower outside edge. I'm sure there's a reason but my brain's not working right now.

You would be correct if you are talking pure torsion and keeping the cross sectional area the same. But when you make it a triangle you cut the area in half it doesn't work.

The torsion formula for a triangle is 4.81T/a^3 and for a rectangle it is 20T/a^3. That is for an equilateral triangle, a right angle would be less.

T being torque and a being area

When these are solved for max torque while keeping the cross sectional area the triangle can resits about 4 times more torque.

But when you change the area of the triangle to half of what the rectangle is [which you do be doing what you said] the rectangle can resits about twice as much torque as the triangle. Probably close to 3 or 4 times if I used the correct formula.

Hopefully that helps

 

[rampage]

Nice! I like things explained with formulas and #'s
Thanks for clearing that up.

 

[ChrisPerry]

But when you change the area of the triangle to half of what the rectangle is [which you do be doing what you said] the rectangle can resits about twice as much torque as the triangle. Probably close to 3 or 4 times if I used the correct formula.

I think the formula must be off. If you take a rectangle and put in a diagonal brace it becomes two triangles and according to your figures it would be weaker than the rectangle you started with.

 

[imiceman44]

I think the formula must be off. If you take a rectangle and put in a diagonal brace it becomes two triangles and according to your figures it would be weaker than the rectangle you started with.

To get 2 triangles you will need to put the diagonal AND box, if you don't box you end up with half the area.

 

[ChrisPerry]

To get 2 triangles you will need to put the diagonal AND box, if you don't box you end up with half the area.

I understand that you would have more strength with a rectangle frame than a triangle frame half the size, but the formula he posted says the larger rectangle was 3-4 times stronger than the triangle that is half the size. I was just curious how that works. Maybe bigger IS so much better?


Edit: I'm just trying to get some learning from those smarter than me...

 

[howdiy]

For the frame to become two triangles you would have had to add a diagonal and a regular box piece

Those formulas are for for solid members, and the area is the cross section area. Basically trying to optimize the amount of material used

The triangle that is half the size is half the size in area, same height and width.

The area being cubed has a big impact

I'm not that far along with my engineering courses yet, just trying to apply what I've learned so far lol

 

[owenst7]

I don't think us engineers are very good at the explaining lol. This may help anyone out that wants to understand the concept of the shape/dimension and resistance to torsion (bending and compression/tension is also a very similar mathematical and practical concept to torsion, and much simpler actually)

http://www.engineeringtoolbox.com/torsion-shafts-d_947.html
Shear Stress in the Shaft

When a shaft is subjected to a torque or twisting, a shearing stress is produced in the shaft. The shear stress varies from zero in the axis to a maximum at the outside surface of the shaft.
The shear stress in a solid circular shaft in a given position can be expressed as:

σ = T r / Ip (1)
where
σ = shear stress (MPa, psi)
T = twisting moment (Nmm, in lb)
r = distance from center to stressed surface in the given position (mm, in)
Ip = "polar moment of inertia" of cross section (mm4, in4)
​

The "polar moment of inertia" is a measure of an object's ability to resist torsion.






Ip is essentially the idea of the distance of mass from the center. In a rotating shaft, the stress from torque at the exact center is zero, and the maximum is at the surface of the outside diameter. You could somewhat think of this by visualizing a breaker bar on a pinion. In order to turn the pinion, the most effective place to apply force is very far from the pinion (a longer lever).


The same essentially happens with the atoms in a material (metal in our case). In our case of talking about a frame rail in torsion, the farther away the iron atoms are from the centerline of the frame rail, the less force they need to apply to that center line in order to produce the same resisting torque. There's also a large increase in surface area, which distributes force over a larger area (stress is force divided by area).

These are some flash demonstrations that I think are fantastic at helping visualize what we're talking about. If you click on Chap. 6 Torsion in the left column, there are several presentations demonstrating gears and torque tubes and stuff.
http://web.mst.edu/~mecmovie/