I study in Germany for mechanical engineering. I would like to thank you because you explain things clearly, and it really helps. Even though some of the formulas are different here i follow your whole lectures. Many thanks from Stuttgart.

@TheBomPE

5 жыл бұрын

I'm glad you find them helpful! Best of luck in your studies!

If you found this video useful, consider helping me upgrade the old tablet PC I use to create these videos! Thanks! www.gofundme.com/help-replace-my-2011-tablet-pc

Very underrated lecture videos! Highly helpful, thorough, and insightful. Thank you so much

@TheBomPE

3 жыл бұрын

Glad you liked it! Thanks for watching!

Thank you for your videos it is very helpful to learn. I have been riding my mountain bike for a few years and it uses a coil shock. I use your method and the same sheet to try to understand better what was happening for my bike: I have a coil spring of a free length of 151mm coupled with a 65mm stroke from the shock. It is labelled as 96N/mm, the wire diameter is around 9mm and the mean diameter is between 45 and 50mm. The coil is squarred and grounded and made of some kind of steel because it is coated sothat it doesnot rust. With the value I could find online, I decided to go for 2000Mpa for Ultimate tensile strength and G 81Gpa. It makes the allowable shear strength 900Mpa. When I apply Hooke's Law I get at 65mm of stroke 6240N Force (the bike linkage multiplies the forces applied on the rear tire by around 3 and I am around 90kg so it seems good to me) and also 1362Mpa shear stress which is ENORMOUS and should deform my coil at bottom out for sure. I took a look at mine and it seems 100% good. I suspect that when the diameter is growing, one cannot simply calculate the allowable shear stress with 0,45. Also my spring only retails at around 30$ after market that is to say it cannot be made of any superalloys with special characteristics. On top of that I try the same exercise with a titanium grade 5 coil (I have one as spare with the same K as the steel one (96N/mm)). It measured 140mm with a 10mm wire and I found online that the titanium grade5 ultimate tensile strength is around 950Mpa. Again I have a hard time comparing the shear strength for my case (around 1035Mpa at bottom out) and the max shear stress allowed before plastic deformation: 427,5Mpa I would have understand if I am not far of due to the security factors and the difference between 0,45 and 0,56 (Samonov) but I am doing something else wrong but i am too far off. Been scratching my head for 3or4 hours now but could not find the mistake. Thanks in advance for your help, Greetings from France, Arthur

Thank you very much for your valuable contribution Professor! your teaching has given me the common sense of being a mechanical engineer clearing the concepts. Could you just help me with the book that you have been referring to in the lectures? it will be greatly helpful. keep up the good work! thanks again.

@TheBomPE

4 жыл бұрын

I'm very glad my videos have been helpful! The text is called Shigley's Mechanical Engineering Design, 10th edition. Thanks for watching!

Is there a reason you are interested here in maximum shear and not say principle stresses? Is it because we are using ductile materials, so it's the main driving parameter of plastic deformation?

@TheBomPE

Жыл бұрын

Typical coil springs primarily carry shearing stresses, not really some complicated combined stress scenario. Therefore it makes sense to evaluate factors of safety based on shearing stresses that cause failure and "working" shearing stresses. Thanks for watching!

Can this spread sheet be used to design the spring buffers in the link below? www.thyssenkruppelevator.com/webapps/classroom-on-demand/LessonViewer.aspx?lesson=16049