Solutions-HW-Chapter5

# Solutions-HW-Chapter5 - budynas_SM_ch05.qxd 18:16 Page 115...

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Chapter 5 5-1 MSS: σ 1 σ 3 = S y / n n = S y σ 1 σ 3 DE: n = S y σ 0 σ 0 = ( σ 2 A σ A σ B + σ 2 B ) 1 / 2 = ( σ 2 x σ x σ y + σ 2 y + 3 τ 2 xy ) 1 / 2 (a) MSS: σ 1 = 12, σ 2 = 6, σ 3 = 0 kpsi n = 50 12 = 4 . 17 Ans. DE: σ 0 = (12 2 6(12) + 6 2 ) 1 / 2 = 10 . 39 kpsi, n = 50 10 . 39 = 4 . 81 Ans. (b) σ A , σ B = 12 2 ± s µ 12 2 2 + ( 8) 2 = 16, 4 kpsi σ 1 = 16, σ 2 = 0, σ 3 =− 4 kpsi MSS: n = 50 16 ( 4) = 2 . 5 Ans. DE: σ 0 = 2 + 3( 8 2 )) 1 / 2 = 18 . 33 kpsi, n = 50 18 . 33 = 2 . 73 Ans. (c) σ A , σ B = 6 10 2 ± s µ 6 + 10 2 2 + ( 5) 2 2 . 615, 13 . 385 kpsi σ 1 = 0, σ 2 2 . 615, σ 3 13 . 385 kpsi MSS: n = 50 0 ( 13 . 385) = 3 . 74 Ans. DE: σ 0 = [( 6) 2 ( 6)( 10) + ( 10) 2 + 3( 5) 2 ] 1 / 2 = 12 . 29 kpsi n = 50 12 . 29 = 4 . 07 Ans. (d) σ A , σ B = 12 + 4 2 ± s µ 12 4 2 2 + 1 2 = 12 . 123, 3 . 877 kpsi σ 1 = 12 . 123, σ 2 = 3 . 877, σ 3 = 0 kpsi MSS: n = 50 12 . 123 0 = 4 . 12 Ans. DE: σ 0 = [12 2 12(4) + 4 2 + 3(1 2 )] 1 / 2 = 10 . 72 kpsi n = 50 10 . 72 = 4 . 66 Ans. s B s A budynas_SM_ch05.qxd 01/29/2007 18:16 Page 115

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116 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 5-2 S y = 50 kpsi MSS: σ 1 σ 3 = S y / n n = S y σ 1 σ 3 DE: ( σ 2 A σ A σ B + σ 2 B ) 1 / 2 = S y / n n = S y / ( σ 2 A σ A σ B + σ 2 B ) 1 / 2 (a) MSS: σ 1 = 12 kpsi, σ 3 = 0, n = 50 12 0 = 4 . 17 Ans. DE: n = 50 [12 2 (12)(12) + 12 2 ] 1 / 2 = 4 . 17 Ans. (b) MSS: σ 1 = 12 kpsi, σ 3 = 0, n = 50 12 = 4 . 17 Ans . DE: n = 50 [12 2 (12)(6) + 6 2 ] 1 / 2 = 4 . 81 Ans. (c) MSS: σ 1 = 12 kpsi, σ 3 =− 12 kpsi, n = 50 12 ( 12) = 2 . 08 Ans. DE: n = 50 [12 2 (12)( 12) + ( 12) 2 ] 1 / 3 = 2 . 41 Ans. (d) MSS: σ 1 = 0, σ 3 12 kpsi, n = 50 ( 12) = 4 . 17 Ans. DE: n = 50 [( 6) 2 ( 6)( 12) + ( 12) 2 ] 1 / 2 = 4 . 81 5-3 S y = 390 MPa MSS: σ 1 σ 3 = S y / n n = S y σ 1 σ 3 DE: ( σ 2 A σ A σ B + σ 2 B ) 1 / 2 = S y / n n = S y / ( σ 2 A σ A σ B + σ 2 B ) 1 / 2 (a) MSS: σ 1 = 180 MPa, σ 3 = 0, n = 390 180 = 2 . 17 Ans. DE: n = 390 [180 2 180(100) + 100 2 ] 1 / 2 = 2 . 50 Ans. (b) σ A , σ B = 180 2 ± s µ 180 2 2 + 100 2 = 224 . 5, 44 . 5MPa = σ 1 , σ 3 MSS: n = 390 224 . 5 ( 44 . 5) = 1 . 45 Ans. DE: n = 390 [180 2 + 3(100 2 )] 1 / 2 = 1 . 56 Ans. budynas_SM_ch05.qxd 01/29/2007 18:16 Page 116
Chapter 5 117 (c) σ A , σ B =− 160 2 ± s µ 160 2 2 + 100 2 = 48 . 06, 208 . 06 MPa = σ 1 , σ 3 MSS: n = 390 48 . 06 ( 208 . 06) = 1 . 52 Ans. DE: n = 390 [ 160 2 + 3(100 2 )] 1 / 2 = 1 . 65 Ans. (d) σ A , σ B = 150, 150 MPa = σ 1 , σ 3 MSS: n = 390 150 ( 150) = 1 . 30 Ans.

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Solutions-HW-Chapter5 - budynas_SM_ch05.qxd 18:16 Page 115...

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