hw2.sol.Fall06

hw2.sol.Fall06 - ME 452 Machine Design II Solution to...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
ME 452: Machine Design II Solution to Homework Set 2. Problem 1. (Problem 6-1, page 340) The minimum ultimate tensile strength can be found using Equation (2-17), page 37, as follows: 0.495 0.495 490 242.6 kpsi ut B S H = = × = (Note that even though the book indicates this correlation is only valid for H B between 200 and 450, it still gives a reasonable answer. See, for example, page 46 and 47 in Norton, where the correlation is .5*H B , giving S ut = 245 kpsi plus or minus 14.7 kpsi, with no limitations on the values of H B .) Because 200 kpsi ut S > , then from Equation (6-8), page 274 we get: 100 kpsi e S = From Table (6-2), page 280, for ground surface finish we have: 1.34 a = and 0.085 b = − Therefore from Equation (6-19), page 279, the surface condition modification factor is given by: 0.085 1.34 (242.6) 0.840 b a ut k aS = = × = For the given shaft diameter of d = 0.25 in, the size factor is given by Equation (6-20), page 280 as: 0.107 0.107 0.879 0.879 (0.25) 1.02 b k d = = × = For rotating bending case, the loading factor from Equation (6-26), page 282 is given by: 1 c k = Assuming that the operating temperature is 70 F ° the temperature factor from Table (6-4), page 283, is given by:
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern