towing with my lowered crew cab

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drewcrew

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So if you consider steel section .125" wide x 8" high and bending about the neutral axis and you need to reduce the depth to 4":

Case 1: The moment of inertia is 5.33 in^4. Area is 1.0 in^2

Case 2. Reducing the depth to half: .125" x 4" high. The moment of inertia is 0.67 in^4. About 1/8 of the one is Case 1.

Case 3. Reducing the depth half, but adding twice the thickness: .25" x 4" high. The same area Case 1 (1.0 in^2.) The moment of inertia is only 1.33 in^4.

To maintain the same moment of inertia of 5.33 in^4 in Case 1, the section needs to be 4" high by 2" wide. A solid 2x4 of steel! LOL

I forgot what this thread was about! :banana-mario:

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I really should have studied more when I was in school. A LOT more.
 

618 Syndicate

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So if you consider steel section .125" wide x 8" high and bending about the neutral axis and you need to reduce the depth to 4":

Case 1: The moment of inertia is 5.33 in^4. Area is 1.0 in^2

Case 2. Reducing the depth to half: .125" x 4" high. The moment of inertia is 0.67 in^4. About 1/8 of the one is Case 1.

Case 3. Reducing the depth half, but adding twice the thickness: .25" x 4" high. The same area Case 1 (1.0 in^2.) The moment of inertia is only 1.33 in^4.

To maintain the same moment of inertia of 5.33 in^4 in Case 1, the section needs to be 4" high by 2" wide. A solid 2x4 of steel! LOL

I forgot what this thread was about! :banana-mario:

You must be registered for see images attach
As I understand it those calculations are to return the point of failure of a given spot in the frame to what it was before right?
But we aren't talking about what it's gonna take to get the notched part TO fail, we're talking about what it's going to take to make sure it doesn't. I don't need to understand the math (and I absolutely do not understand it...) to know what we're looking for.
 

Erik the Awful

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I wish that my 9th grade math class hadn't been such a $#!+ show, and being frustrated and not taking a math class in 10th grade was a mistake. I like to think that if I'd understood it better, I'd have been an engineer.
 
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So if you consider steel section .125" wide x 8" high and bending about the neutral axis and you need to reduce the depth to 4":

Case 1: The moment of inertia is 5.33 in^4. Area is 1.0 in^2

Case 2. Reducing the depth to half: .125" x 4" high. The moment of inertia is 0.67 in^4. About 1/8 of the one is Case 1.

Case 3. Reducing the depth half, but adding twice the thickness: .25" x 4" high. The same area Case 1 (1.0 in^2.) The moment of inertia is only 1.33 in^4.

To maintain the same moment of inertia of 5.33 in^4 in Case 1, the section needs to be 4" high by 2" wide. A solid 2x4 of steel! LOL

I forgot what this thread was about! :banana-mario:

You must be registered for see images attach


I'll admit, I'm a mechanical engineer and very seldom do these calculations anymore. I totally agree with the strength calcs above. I was just picturing in my head the loads on the beam and the stresses. Then opened my mouth without thinking the whole thing through.
Ultimately, my point was supposed to be that the load from a bumper pull trailer is not centered on the c-notch and therefore it shouldn't affect towing weight. I guess I will have to find the time to draw the force diagrams and do the calcs to back up that thought.
 

stutaeng

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I'll admit, I'm a mechanical engineer and very seldom do these calculations anymore. I totally agree with the strength calcs above. I was just picturing in my head the loads on the beam and the stresses. Then opened my mouth without thinking the whole thing through.
Ultimately, my point was supposed to be that the load from a bumper pull trailer is not centered on the c-notch and therefore it shouldn't affect towing weight. I guess I will have to find the time to draw the force diagrams and do the calcs to back up that thought.

I was also think about that myself, but couldn't really visualize it in my mind.... Over the weekend I was thinking about it over morning coffee and came up with some concepts and sketches:
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Assuming the frame is a beam, on the case where the load is directly over the axle (gooseneck, 5th wheel), the load is transferred directly to the axle via the leaf springs. The only bending of the frame from the load is between those two leaf spring support.

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On the case where the load (assuming the same load) is from a bumper pull hitch, the frame actually cantilevers back to the location near the rear bumper itch ball is. So the same load actually causes a magnification reaction at the rear axle. And, since the stresses of the frame are related to the axle reaction, a bumper pull load would be slightly worse over a gooseneck in terms of stresses in question. This is general statement.


In this example, if P = 1000 lbs, P2 is greater than 1000 lbs!
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I never really wondered about the mechanics as to why you can almost always tow higher with a gooseneck vs pumper hitch, but this may be one of the factors? I'm sure there are other factors to consider like stability, sway, hitch rating, etc.

Again, using basic statics principles here. It's a lot more complicated I can imagine with things like dynamics, fatigue, ductility and things like that.
 
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