Recent content by JGrant

NVN: Wouldn't the P force be located in the center of the distributed load not at the end? Therefore, instead of using 7.1 mm, it would be 3.55mm? Thank you for helping me with this, Sorry if I ask to many questions.

NVN: Thank you for helping me on this. I am feeling like I have a much better grasp on it, but I am not sure how you calculated w. (i.e. w=213.7N/mm, in post 7). Once I know that I feel I can do these calculations on my own on any future shafts. Thank you

NVN: The shaft diameter at the larger section is 9.5275 mm. Do you calculate the force to bend the shaft the same as it is static even tho it is rotating? I thought I had to calculate a torsional force in addition to a load.

Let me know if this works.

NVN: I apologize for not knowing this, but I feel I'm dumber for pretending I know and not asking, then to shallow my pride and ask. Is there suppose to be a ratio used between the FSu and the FSy? Looking at the numbers you have listed Id say its 1.375. And where is that ratio derived from?

Hello, It's been a while since I did this in college and I cannot remember a few steps. I am trying to calculate the maximum force my designed shaft can handle before bending/shearing. My shaft material is C1018 steel rod, with a yield strength of 53.7 KSI and an ultimate strength of 63.8 KSI...

NVN: Upon doing some research and the calculations myself. The material I am using is brittle like a cast iron therefore I should be using the ultimate strength, correct? Due to the fact it will snap before bend. If this is the case, assuming the same numbers except the FSy=2 and for Yield...

Thank you NVN, This will help me out a lot. Thank you. I converted the Newtons to pounds to get 397.9 lbs and that sounds like a good number we can work with. We will be using a safety factor of 2 tho. The ultimate strength of the material is 82 KSI(566MPa). Does that change anything?

Sorry for the confusion, I thought I was making it simpler to understand. The image shows where the cross section is at and the slit is from the undercut which only goes into the shaft so far. The undercut is from a hole drilled out from the opposite side of the shaft. The part is actually...

NVN, The first pic is the direction of the force. The 2nd pic is the cross section view of the shaft. The undercut is not like in your pic. Sorry for the misunderstanding. Yes the force is located at distance 3.175 mm (.125 in), but the diameter is 8mm (.315in) prior to the undercut...

I know that if I pull straight out on this shaft that the shear force would be 3370.2 LBS [60,000PSI (ultimate str) X .05617 in sq (area)]. But I am trying to calculate the max force applied on the top of the shaft. Would that be: 3370.2LBS X .125 inches = 421.275 in-lbs? but this is in torque...

Hello, If I have a shaft that has a section area of .0562 in. sq. and a material yield strength of 60KSI. And I will have the force applied at the top of the shaft pushing down. The shaft will be in a cantilever configuration where the force will be applied .125" from the base of the shaft...