25_ch 05 Mechanical Design budynas_SM_ch05

# 25_ch 05 Mechanical Design budynas_SM_ch05 - δ d = 50 01...

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Chapter 5 139 (b) For a solid inner tube, p = 30(10 6 )(0 . 0005) 1 (1 . 5 2 1 2 )(1 2 ) 2(1 2 )(1 . 5 2 ) = 4167 psi Ans. ( σ t ) i = − p = − 4167 psi, ( σ r ) i = − 4167 psi σ i = [( 4167) 2 ( 4167)( 4167) + ( 4167) 2 ] 1 / 2 = 4167 psi Ans. ( σ t ) o = 4167 1 . 5 2 + 1 2 1 . 5 2 1 2 = 10 830 psi, ( σ r ) o = − 4167 psi σ o = [10 830 2 10 830( 4167) + ( 4167) 2 ] 1 / 2 = 13 410 psi Ans. 5-35 Using Eq. (3-57) with diametral values, p = 207(10 3 )(0 . 02) (50 3 ) (75 2 50 2 )(50 2 25 2 ) 2(75 2 25 2 ) = 19 . 41 MPa Ans. Eq. (3-58) ( σ t ) i = − 19 . 41 50 2 + 25 2 50 2 25 2 = − 32 . 35 MPa ( σ r ) i = − 19 . 41 MPa Eq. (5-13) σ i = [( 32 . 35) 2 ( 32 . 35)( 19 . 41) + ( 19 . 41) 2 ] 1 / 2 = 28 . 20 MPa Ans. Eq. (3-59) ( σ t ) o = 19 . 41 75 2 + 50 2 75 2 50 2 = 50 . 47 MPa, ( σ r ) o = − 19 . 41 MPa σ o = [50 . 47 2 50 . 47( 19 . 41) + ( 19 . 41) 2 ] 1 / 2 = 62 . 48 MPa Ans. 5-36 Max. shrink-fit conditions: Diametral interference
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Unformatted text preview: δ d = 50 . 01 − 49 . 97 = . 04 mm . Equa-tion (3-57) using diametral values: p = 207(10 3 )0 . 04 50 3 ± (75 2 − 50 2 )(50 2 − 25 2 ) 2(75 2 − 25 2 ) ² = 38 . 81 MPa Ans. Eq. (3-58): ( σ t ) i = − 38 . 81 ³ 50 2 + 25 2 50 2 − 25 2 ´ = − 64 . 68 MPa ( σ r ) i = − 38 . 81 MPa Eq. (5-13): σ ± i = µ ( − 64 . 68) 2 − ( − 64 . 68)( − 38 . 81) + ( − 38 . 81) 2 ¶ 1 / 2 = 56 . 39 MPa Ans....
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