Force to bend a real solid round aluminum bar fixed at one end

[blackbear3]

I know this is a fairly straight forward calculation, but I keep messing something up in my units or formula. My question is a real world one.
I have a 10ft aluminum alloy 7075-T651 solid rod 1.625in in diameter. It is fixed at one end and free at the other. The yield strength test was 75.6KSI. I have elongation and UTS numbers as well. Total mass of the rod is 25.125lb Sorry for not keeping significant digits even. ASTM spec is B211-12E1
If my understanding is correct, Stress S=My/I and in this case Inertia I=(ML^2)/3.
I think I keep getting Inertia incorrect as I come up with a very high Force when I solve M=Fd, where d=9.5ft, solve for F. OR I'm off base entirely and in the wrong chapter of mechanics/physics. Should I be looking at bending moment instead? I am seeking to understand what force F at 9.5ft it would take to begin the permanent bending of the rod, beyond spring back. I am hoping to keep the force F to somewhere between 150-175<=lb in the elastic region. May have to change the application requirement/load. Can you please help point me in correct direction? Much thanks!

 

[Erich]

Your equation for Inertia is incorrect. It should be I = π r² × d/2
Moment is Force * distance from force to wall.

 

[blackbear3]

Ah, that makes sense. So, running with that I get something like 1375lbf. Seems high but it is second strongest aluminum alloy vs 7078, which was not available. And the diameter is large at 1.625". Hmmmm, must ponder my math. Thank you for your help. Seems far more reasonable than the previous number I was getting which defaulted to putting the Inertia value way to high. Any additional advice would be much appreciated. My imposed force will be approximately 150lbf. at the tip of rod over the span of 9.50ft. Many thanks!

 

[blackbear3]

To the Moderator Erich,
Not to be picky, but wouldn't the above inertia equation give me units that don't add up. In other words I get inches^3. Shouldn't it be Inches^4, thus meaning the equation would be more like I=πd^4/64 or rewritten as I=πr^4/4? If I use this it changes the force value down significantly to the 260lbf level. This is still livable as it's the theoretical point at which yielding goes plastic. I'm still safe at 150lbf. Just wanting to double check the inertia equation one more time for a solid circular cross section and sanity. As professors used to say, "if the numbers and/or units don't add up or make sense, question it." Again many thanks for your help.

 

[Erich]

Yes unit analysis is a key tool in verifying that things are making sense.
I'm not sure what happened there. I NEVER use bold and fancy fonts so that equation is something that the website converted.
Your equations are correct.