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I really should have studied more when I was in school. A LOT more.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!
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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?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!
You must be registered for see images attach
Facts. I've often said that adding letters into my numbers ruined math for me....When the letters out number the numbers in a math equation, I'm lost.
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!
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.