{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

23_ch 05 Mechanical Design budynas_SM_ch05

# 23_ch 05 Mechanical Design budynas_SM_ch05 - − 25 333 −...

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

Chapter 5 137 The screw equation is F i = T 0 . 2 d (2) From Eqs. (1) and (2) p i = F w r i = T 0 . 2 d w r i dF x = f p i r i d θ F x = 2 π o f p i w r i d θ = f T w 0 . 2 d w r i r i 2 π o d θ = 2 π f T 0 . 2 d Ans. 5-32 (a) From Prob. 5-31, T = 0 . 2 F i d F i = T 0 . 2 d = 190 0 . 2(0 . 25) = 3800 lbf Ans. (b) From Prob. 5-31, F = w r i p i p i = F w r i = F i w r i = 3800 0 . 5(0 . 5) = 15 200 psi Ans . (c) σ t = r 2 i p i r 2 o r 2 i 1 + r 2 o r r = r i = p i ( r 2 i + r 2 o ) r 2 o r 2 i = 15 200(0 . 5 2 + 1 2 ) 1 2 0 . 5 2 = 25 333 psi Ans . σ r = − p i = − 15 200 psi (d) τ max = σ 1 σ 3 2 = σ t σ r 2 = 25 333 ( 15 200) 2 = 20 267 psi Ans . σ = ( σ 2 A + σ 2 B σ A σ B ) 1 / 2 = [25 333 2 + ( 15 200) 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: − 25 333( − 15 200)] 1 / 2 = 35 466 psi Ans . (e) Maximum Shear hypothesis n = S sy τ max = . 5 S y τ max = . 5(63) 20 . 267 = 1 . 55 Ans. Distortion Energy theory n = S y σ ± = 63 35 466 = 1 . 78 Ans. dF x p i r i d ±...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern