13_ch 06 Mechanical Design budynas_SM_ch06

13_ch 06 Mechanical Design budynas_SM_ch06 - ( σ i ) a =...

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Chapter 6 159 Curved beam: r n = h ln( r o / r i ) = 3 ln(6 / 3) = 4 . 3281 mm r c = 4 . 5mm , e = r c r n = 4 . 5 4 . 3281 = 0 . 1719 mm σ i =− Mc i Aer i F A =− (0 . 1015 F )(1 . 5 0 . 1719) 54(0 . 1719)(3)(10 3 ) F 54 =− 4 . 859 F MPa σ o = Mc o Aer o F A = (0 . 1015 F )(1 . 5 + 0 . 1719) 54(0 . 1719)(6)(10 3 ) F 54 = 3 . 028 F MPa ( σ i ) min =− 4 . 859(150 . 9) =− 733 . 2MPa ( σ i ) max =− 4 . 859(50 . 3) =− 244 . 4MPa ( σ o ) max = 3 . 028(150 . 9) = 456 . 9MPa ( σ o ) min = 3 . 028(50 . 3) = 152 . 3MPa Eq. (2-17) S ut = 3 . 41(490) = 1671 MPa Per the problem statement, estimate the yield as S y = 0 . 9 S ut = 0 . 9(1671) = 1504 MPa. Then from Eq. (6-8), S ± e = 700 MPa ; Eq. (6-19), k a = 1 . 58(1671) 0 . 085 = 0 . 841 ; Eq. (6-25) d e = 0 . 808[18(3)] 1 / 2 = 5 . 938 mm ; and Eq. (6-20), k b = (5 . 938 / 7 . 62) 0 . 107 = 1 . 027 . S e = 0 . 841(1 . 027)(700) = 605 MPa
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Unformatted text preview: ( σ i ) a = ± ± ± ± − 733 . 2 + 244 . 4 2 ± ± ± ± = 244 . 4 MPa ( σ i ) m = − 733 . 2 − 244 . 4 2 = − 488 . 8 MPa Load line: σ m = − 244 . 4 − σ a Langer (yield) line: σ m = σ a − 1504 = − 244 . 4 − σ a Intersection: σ a = 629 . 8 MPa, σ m = − 874 . 2 MPa (Note that σ a is more than 605 MPa) Yield: n y = 629 . 8 244 . 4 = 2 . 58 244.4 488.4 ± m ± a ± 1504 605 1504 MPa...
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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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