Just a Random Sample Question in Machine Design (Major Subject for Mechanical Engineering Students)

When I was still studying, I am usually approached by my fellow Mechanical Engineering students in our school especially those in the lower years when they experience difficulty in a certain subject or certain problem wherein they have been trying to figure out how to solve that problem. And I do help them as much as I can and as much as possible. Well today, I received a message from one of the graduates for the school year 2017-2018, a good friend of mine, a fellow Steemian and at the same time a fellow contributor in Utopian, no other than Hive account@philiparniebinag. He started his self-review while the review classes in the review centers had not started during this time and also for his preparation in the upcoming Mechanical Engineers Licensure Examination. He sent a messaged to me in Facebook regarding on this problem in Machine Design to be specific in the aspect of shaftings and pins; and the question goes like this:

Determine the minimum mean diameter of a taper pin for use to fix a lever to a shaft, if it is to transmit a maximum torque of 700 in-lbs. The shaft diameter is 2 inches and the material's allowable stress is 15,000 psi. Use a factor of safety of 2.

Upon reading the question, I take a look at it and read it slowly in order for me to understand and think how this works and how can I solve this certain problem. I have to be honest, of the three major subjects of the Mechanical Engineering discipline namely the: Mathematics, Power and Industrial Plant Engineering (PIPE), and lastly, Machine Design; my weakness is on Machine Design primarily on the terms and some of the numerous empirical formulas it has; which would incur me a lot of me or more time in solving as compared to PIPE or Mathematics. My forte is the Power and Industrial Plant Engineering.

So I tried solving the problem.

• First I obtained the torsional shearing stress of the shaft which can be assumed and obtained by dividing the material’s allowable stress to its factor of safety which is 15,000 divided by 2; that is 7,500 psi.
• Second, I obtain the torque of the shaft from the torsional shearing stress formula S_s = (16 x T) / (pi x d³; wherein:
  o S_s is the torsional shearing stress,
  o T is the torque and d is the diameter.
  And I obtained a torque of 11,780.97245 in-lbs.
• Third, I obtained the force of the shaft by using the formula, T = Fr; wherein
  o T is the torque,
  o F is the force,
  o and r is the radius of the shaft which is equal to D/2.
  And I obtained a force of 11,780.97245 lbs.
• Then, I used the formula for the torque in obtaining the diameter of the pin and I obtained a value of 0.12 inches and the nearest size is 0.125in or 1/8 inch. And since the answer found in the answer key is 6.2mm, I converted my answer into millimeters and I was upset to know that my answer is 3.02mm.

Well, I did try other ways to solve that problem but unfortunately I always arrived on my first answer. Maybe I am wrong and missed the point of the question. If any of you got the solution and answer for this problem, let me know. It appears that two years without practice solving makes you forget how to solve or maybe forgot some formulas.

I guess that would be all for tonight.

Thank you for passin’ by and reading this post.

Much love and respect.

Ace | Hive account@josephace135