16_ch 06 Mechanical Design budynas_SM_ch06

16_ch 06 Mechanical Design budynas_SM_ch06 - 2 i = Mc i Aer...

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162 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Since the midrange stress is negative, n y = ± ± ± ± S y σ min ± ± ± ± = ± ± ± ± 54 10 . 67 ± ± ± ± = 5 . 06 S a = S e = 24 . 4 kpsi n f = S a σ a = 24 . 4 15 . 2 = 1 . 61 Thus the design is controlled by the threat of fatigue equally at the fillet and the hole; the minimum factor of safety is n f = 1 . 61 . Ans. 6-24 (a) Curved beam in pure bending where M =− T throughout. The maximum stress will occur at the inner fiber where r c = 20 mm, but will be com- pressive. The maximum tensile stress will occur at the outer fiber where r c = 60 mm. Why? Inner fiber where r c = 20 mm r n = h ln( r o / r i ) = 5 ln (22 . 5 / 17 . 5) = 19 . 8954 mm e = 20 19 . 8954 = 0 . 1046 mm c i = 19 . 8954 17 . 5 = 2 . 395 mm A = 25 mm
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Unformatted text preview: 2 i = Mc i Aer i = T (2 . 395)10 3 25(10 6 )0 . 1046(10 3 )17 . 5(10 3 ) (10 6 ) = 52 . 34 T (1) where T is in N . m, and i is in MPa. m = 1 2 ( 52 . 34 T ) = 26 . 17 T , a = 26 . 17 T For the endurance limit, S e = . 5(770) = 385 MPa k a = 4 . 51(770) . 265 = . 775 d e = . 808[5(5)] 1 / 2 = 4 . 04 mm k b = (4 . 04 / 7 . 62) . 107 = 1 . 07 S e = . 775(1 . 07)385 = 319 . 3 MPa For a compressive midrange component, a = S e / n f . Thus, 26 . 17 T = 319 . 3 / 3 T = 4 . 07 N m T T...
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