28_ch 05 Mechanical Design budynas_SM_ch05

# 28_ch 05 Mechanical Design budynas_SM_ch05 - σ 1 and σ 2...

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142 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design The inner radius is in plane stress: σ x = 8 . 38 kpsi, σ y =− 22 . 59 kpsi, τ xy = 5 . 59 kpsi σ ± i = [8 . 38 2 (8 . 38)( 22 . 59) + ( 22 . 59) 2 + 3(5 . 59 2 )] 1 / 2 = 29 . 4 kpsi n i = S y σ ± i = 60 29 . 4 = 2 . 04 Ans. Outer radius experiences a radial stress, σ r σ ± o = 1 2 ± ( 19 . 60 + 2 . 997) 2 + ( 2 . 997 9 . 78) 2 + (9 . 78 + 19 . 60) 2 + 6(6 . 52) 2 ² 1 / 2 = 27 . 9 kpsi n o = 60 27 . 9 = 2 . 15 Ans. 5-40 σ p = 1 2 ³ 2 K I 2 π r cos θ 2 ´ ² µ ³ K I 2 π r sin θ 2 cos θ 2 sin 3 θ 2 ´ 2 + ³ K I 2 π r sin θ 2 cos θ 2 cos 3 θ 2 ´ 2 1 / 2 = K I 2 π r µ cos θ 2 ² ³ sin 2 θ 2 cos 2 θ 2 sin 2 3 θ 2 + sin 2 θ 2 cos 2 θ 2 cos 2 3 θ 2 ´ 1 / 2 = K I 2 π r ³ cos θ 2 ² cos θ 2 sin θ 2 ´ = K I 2 π r cos θ 2 ³ 1 ² sin θ 2 ´ Plane stress: The third principal stress is zero and σ 1 = K I 2 π r cos θ 2 ³ 1 + sin θ 2 ´ , σ 2 = K I 2 π r cos θ 2 ³ 1 sin θ 2 ´ , σ 3 = 0 Ans. Plane strain:
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Unformatted text preview: σ 1 and σ 2 equations still valid however, σ 3 = ν ( σ x + σ y ) = 2 ν K I √ 2 π r cos θ 2 Ans. 5-41 For θ = 0 and plane strain, the principal stress equations of Prob. 5-40 give σ 1 = σ 2 = K I √ 2 π r , σ 3 = 2 ν K I √ 2 π r = 2 νσ 1 (a) DE: 1 √ 2 [( σ 1 − σ 1 ) 2 + ( σ 1 − 2 νσ 1 ) 2 + (2 νσ 1 − σ 1 ) 2 ] 1 / 2 = S y σ 1 − 2 νσ 1 = S y For ν = 1 3 , · 1 − 2 ³ 1 3 ´¸ σ 1 = S y ⇒ σ 1 = 3 S y 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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