17_ch 06 Mechanical Design budynas_SM_ch06

17_ch 06 Mechanical Design budynas_SM_ch06 -...

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Unformatted text preview: budynas_SM_ch06.qxd 11/29/2006 17:40 FIRST PAGES Page 163 163 Chapter 6 Outer fiber where rc = 60 mm rn = 5 = 59.96526 mm ln(62.5/57.5) e = 60 − 59.96526 = 0.03474 mm co = 62.5 − 59.96526 = 2.535 mm σo = − Mci −T (2.535)10−3 =− (10−6 ) = 46.7 T Aeri 25(10−6 )0.03474(10−3 )62.5(10−3 ) Comparing this with Eq. (1), we see that it is less in magnitude, but the midrange component is tension. 1 σa = σm = (46.7T ) = 23.35T 2 Using Eq. (6-46), for modified Goodman, we have 23.35T 23.35T 1 + = ⇒ T = 3.22 N · m Ans. 319.3 770 3 (b) Gerber, Eq. (6-47), at the outer fiber, 3(23.35T ) 3(23.35T ) + 319.3 770 reduces to T= 2 =1 T 2 + 26.51T − 120.83 = 0 1 −26.51 + 2 26.512 + 4(120.83) = 3.96 N · m Ans. (c) To guard against yield, use T of part (b) and the inner stress. 420 ny = = 2.03 Ans. 52.34(3.96) 6-25 From Prob. 6-24, Se = 319.3 MPa, S y = 420 MPa, and Sut = 770 MPa (a) Assuming the beam is straight, 6M 6T = 48(106 ) T σmax = 2 = 3 bh 5 [(10−3 ) 3 ] 24T 1 24T + = 319.3 770 3 Goodman: ⇒ T = 3.13 N · m Ans. 3(24) T 3(24) T + 319.3 770 (b) Gerber: 2 =1 T 2 + 25.79T − 114.37 = 1 T= 1 −25.79 + 2 25.792 + 4(114.37) = 3.86 N · m Ans. ...
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This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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