{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Solution_tutorial_7_2011_bearing& fracture

Solution_tutorial_7_2011_bearing& fracture - We have...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Solution: tutorial 7, plain bearings and fracture 1 . p = N/(DL) hence L = N/(Dp) = 0.2MN/(0.25m x 4MPa) = 0.2 m = 200 mm 2. Power dissipated = frictional torque x angular velocity = f N R ϖ = 0.015 x 0.2E6 N x 0.125 m x (150/60 x 2 π ) rad/s = 5890.48 Nm/s = 5.89 kW 3. (a) Oil viscosity at 70 0 C is μ = 17.9 cp = 0.0179 Ns/m 2 (b) Frictional torque T f = f N R The Petrov equation gives f = π(μϖ /p)(R/c) ϖ = 800/60 x 2 π = 83.776 rad/s p = N/(DL) , so f = 2 π ( μϖ L/N)R 2 /c and T f = 2 πμϖ LR 3 /c or T f = 2 π x 0.0179 x 83.776 x 0.2 x 0.09 3 /0.00007 = 19.625 Nm (c) Energy dissipation = T f ϖ =19.625 x 83.776 = 1644.1 W 4. IC o With a large rectangular sheet we have no geometry effect, hence Y=1
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: We have =95 ksi and a=0.8 in. K 1.8 (1.8)(95 ) 0.800 . 152.947 . For the sheet with the 1.9 in. crack, 152 1.8 o o IC a ksi in ksi in K a σ = = = = = 3 o .947 87.17798 1.8 1.9/ 2 p= (2 ) 87.17798(10 )(12)(0.5) 523067.9 ksi wt lb = = = 5. s 75 44.82107 2.0 2.0 0.70 ( )( ) 44.82107(6)(0.06) 16135.59 lb 16135.19 With F 2.5, 4610.167 lb 3.5 IC o o app K ksi a P W t P = = = = = = = = = 6. o 3 o 85 2.0 2.0 1.75 224.8889 . 224.8889 69.40221 2.0 2.0 2.625 P= (W)(t)=69.40221(10 )(50)(0.05)=173505.5 lb IC IC o critical IC IC K K ksi c K ksi in K ksi c = = = = ⇒ = = = =...
View Full 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