Solucionario Diseño Shigley Cap 5
Páginas: 30 (7499 palabras)
Publicado: 19 de mayo de 2012
budynas_SM_ch05.qxd
11/29/2006
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FIRST PAGES
Page 115
Chapter 5
5-1
MSS:
σ1 − σ3 = S y / n
DE:
n=
⇒
Sy
σ
2
2
σ = σ A − σ AσB + σB
(a) MSS:
DE:
Sy
σ1 − σ3
n=
1/2
2
2
2
= σx − σx σ y + σ y + 3τx y
1/2
σ1 = 12, σ2 = 6, σ3 = 0 kpsi
50
n=
= 4.17 Ans.
12
σ = (122 − 6(12) + 62 ) 1/2 = 10.39 kpsi,
12
±
(b) σ A , σ B =
212
2
n=
50
= 4.81
10.39
Ans.
2
+ ( −8) 2 = 16, − 4 kpsi
σ1 = 16, σ2 = 0, σ3 = −4 kpsi
MSS:
DE:
(c) σ A , σ B =
n=
50
= 2.5
16 − ( −4)
Ans.
σ = (122 + 3( −82 )) 1/2 = 18.33 kpsi,
−6 − 10
±
2
−6 + 10
2
n=
50
= 2.73
18.33
Ans.
2
+ ( −5) 2 = −2.615, −13.385 kpsi
σ1 = 0, σ2 = −2.615, σ3 = −13.385 kpsi
MSS:
DE:
n=
50
= 3.74
0 − (−13.385)
B
Ans.
σ = [( −6) 2 − ( −6)( −10) + ( −10) 2 + 3( −5) 2 ]1/2
= 12.29 kpsi
50
= 4.07 Ans.
n=
12.29
12 + 4
±
(d) σ A , σ B =
2
12 − 4
2
2
+ 12 = 12.123, 3.877 kpsi
σ1 = 12.123, σ2 = 3.877, σ3 = 0 kpsi
MSS:
DE:
n=
50
= 4.12
12.123 − 0
Ans.
σ = [122 − 12(4) + 42 + 3(12 )]1/2 = 10.72 kpsi
n=
50
= 4.66
10.72
Ans.
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Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
5-2 S y = 50 kpsi
MSS:
σ1 − σ3 = S y / n
⇒
2
2
σ A − σ AσB + σB
DE:
n=
DE:
= Sy / n
[122
n=
[122
⇒
2
2
n = Sy / σ A − σ A σ B + σ B
n=
DE:
[122
Ans .
50
= 4.81
− (12)(6) + 62 ]1/2Ans.
n=
50
= 2.08 Ans.
12 − ( −12)
50
= 2.41
− (12)( −12) + ( −12) 2 ]1/3
σ1 = 0, σ3 = −12 kpsi, n =
DE:
Ans.
50
= 4.17
12
σ1 = 12 kpsi, σ3 = −12 kpsi, n =
(c) MSS:
1/2
50
= 4.17 Ans.
12 − 0
50
= 4.17
− (12)(12) + 122 ]1/2
σ1 = 12 kpsi, σ3 = 0, n =
DE:
(d) MSS:
Sy
σ1 − σ3
σ1 = 12 kpsi, σ3 = 0, n =
(a) MSS:
(b) MSS:
1/2
n=[( −6) 2
Ans.
50
= 4.17 Ans.
−( −12)
50
= 4.81
− ( −6)( −12) + ( −12) 2 ]1/2
5-3 S y = 390 MPa
MSS:
σ1 − σ3 = S y / n
2
2
σ A − σ AσB + σB
DE:
(a) MSS:
DE:
⇒
1/2
n=
Sy
σ1 − σ3
= Sy / n
σ1 = 180 MPa, σ3 = 0, n =
n=
n=
DE:
n=
2
2
n = Sy / σ A − σ A σ B + σ B
390
= 2.17 Ans.
180
390
= 2.50
[1802 − 180(100) + 1002 ]1/2
180
±
(b)σ A , σ B =
2
MSS:
⇒
180
2
2
+ 1002 = 224.5, −44.5 MPa = σ1 , σ3
390
= 1.45 Ans.
224.5 − ( −44.5)
[1802
Ans.
390
= 1.56
+ 3(1002 )]1/2
Ans.
1/2
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Page 117
Chapter 5
160
±
(c) σ A , σ B = −
2
160
−
2
2
+ 1002 = 48.06, −208.06 MPa = σ1 , σ3
390
= 1.52 Ans.
48.06 − ( −208.06)390
= 1.65 Ans.
n=
2 + 3(1002 )]1/2
[−160
n=
MSS:
DE:
(d) σ A , σ B = 150, −150 MPa = σ1 , σ3
390
= 1.30 Ans.
n=
MSS:
150 − ( −150)
390
= 1.50 Ans.
n=
DE:
[3(150) 2 ]1/2
5-4 S y = 220 MPa
(a) σ1 = 100, σ2 = 80, σ3 = 0 MPa
MSS:
DET:
220
= 2.20 Ans.
100 − 0
σ = [1002 − 100(80) + 802 ]1/2 = 91.65 MPa
220
n=
= 2.40 Ans.
91.65
n=
(b) σ1 = 100, σ2 = 10, σ3 = 0 MPaMSS:
DET:
n=
220
= 2.20 Ans.
100
σ = [1002 − 100(10) + 102 ]1/2 = 95.39 MPa
n=
220
= 2.31 Ans.
95.39
(c) σ1 = 100, σ2 = 0, σ3 = −80 MPa
MSS:
DE:
n=
220
= 1.22 Ans.
100 − ( −80)
σ = [1002 − 100( −80) + ( −80) 2 ]1/2 = 156.2 MPa
220
n=
= 1.41 Ans.
156.2
(d) σ1 = 0, σ2 = −80, σ3 = −100 MPa
220
n=
= 2.20 Ans.
MSS:
0 − ( −100)
σ = [( −80) 2 − ( −80)(−100) + ( −100) 2 ] = 91.65 MPa
DE:
n=
220
= 2.40 Ans.
91.65
117
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Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
5-5
(a) MSS:
DE:
(b) MSS:
DE:
n=
2.23
OB
=
= 2.1
OA
1.08
n=
2.56
OC
=
= 2.4
OA
1.08
n=
OE
1.65
=
= 1.5
OD
1.10...
11/29/2006
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FIRST PAGES
Page 115
Chapter 5
5-1
MSS:
σ1 − σ3 = S y / n
DE:
n=
⇒
Sy
σ
2
2
σ = σ A − σ AσB + σB
(a) MSS:
DE:
Sy
σ1 − σ3
n=
1/2
2
2
2
= σx − σx σ y + σ y + 3τx y
1/2
σ1 = 12, σ2 = 6, σ3 = 0 kpsi
50
n=
= 4.17 Ans.
12
σ = (122 − 6(12) + 62 ) 1/2 = 10.39 kpsi,
12
±
(b) σ A , σ B =
212
2
n=
50
= 4.81
10.39
Ans.
2
+ ( −8) 2 = 16, − 4 kpsi
σ1 = 16, σ2 = 0, σ3 = −4 kpsi
MSS:
DE:
(c) σ A , σ B =
n=
50
= 2.5
16 − ( −4)
Ans.
σ = (122 + 3( −82 )) 1/2 = 18.33 kpsi,
−6 − 10
±
2
−6 + 10
2
n=
50
= 2.73
18.33
Ans.
2
+ ( −5) 2 = −2.615, −13.385 kpsi
σ1 = 0, σ2 = −2.615, σ3 = −13.385 kpsi
MSS:
DE:
n=
50
= 3.74
0 − (−13.385)
B
Ans.
σ = [( −6) 2 − ( −6)( −10) + ( −10) 2 + 3( −5) 2 ]1/2
= 12.29 kpsi
50
= 4.07 Ans.
n=
12.29
12 + 4
±
(d) σ A , σ B =
2
12 − 4
2
2
+ 12 = 12.123, 3.877 kpsi
σ1 = 12.123, σ2 = 3.877, σ3 = 0 kpsi
MSS:
DE:
n=
50
= 4.12
12.123 − 0
Ans.
σ = [122 − 12(4) + 42 + 3(12 )]1/2 = 10.72 kpsi
n=
50
= 4.66
10.72
Ans.
A budynas_SM_ch05.qxd
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FIRST PAGES
Page 116
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
5-2 S y = 50 kpsi
MSS:
σ1 − σ3 = S y / n
⇒
2
2
σ A − σ AσB + σB
DE:
n=
DE:
= Sy / n
[122
n=
[122
⇒
2
2
n = Sy / σ A − σ A σ B + σ B
n=
DE:
[122
Ans .
50
= 4.81
− (12)(6) + 62 ]1/2Ans.
n=
50
= 2.08 Ans.
12 − ( −12)
50
= 2.41
− (12)( −12) + ( −12) 2 ]1/3
σ1 = 0, σ3 = −12 kpsi, n =
DE:
Ans.
50
= 4.17
12
σ1 = 12 kpsi, σ3 = −12 kpsi, n =
(c) MSS:
1/2
50
= 4.17 Ans.
12 − 0
50
= 4.17
− (12)(12) + 122 ]1/2
σ1 = 12 kpsi, σ3 = 0, n =
DE:
(d) MSS:
Sy
σ1 − σ3
σ1 = 12 kpsi, σ3 = 0, n =
(a) MSS:
(b) MSS:
1/2
n=[( −6) 2
Ans.
50
= 4.17 Ans.
−( −12)
50
= 4.81
− ( −6)( −12) + ( −12) 2 ]1/2
5-3 S y = 390 MPa
MSS:
σ1 − σ3 = S y / n
2
2
σ A − σ AσB + σB
DE:
(a) MSS:
DE:
⇒
1/2
n=
Sy
σ1 − σ3
= Sy / n
σ1 = 180 MPa, σ3 = 0, n =
n=
n=
DE:
n=
2
2
n = Sy / σ A − σ A σ B + σ B
390
= 2.17 Ans.
180
390
= 2.50
[1802 − 180(100) + 1002 ]1/2
180
±
(b)σ A , σ B =
2
MSS:
⇒
180
2
2
+ 1002 = 224.5, −44.5 MPa = σ1 , σ3
390
= 1.45 Ans.
224.5 − ( −44.5)
[1802
Ans.
390
= 1.56
+ 3(1002 )]1/2
Ans.
1/2
budynas_SM_ch05.qxd
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FIRST PAGES
Page 117
Chapter 5
160
±
(c) σ A , σ B = −
2
160
−
2
2
+ 1002 = 48.06, −208.06 MPa = σ1 , σ3
390
= 1.52 Ans.
48.06 − ( −208.06)390
= 1.65 Ans.
n=
2 + 3(1002 )]1/2
[−160
n=
MSS:
DE:
(d) σ A , σ B = 150, −150 MPa = σ1 , σ3
390
= 1.30 Ans.
n=
MSS:
150 − ( −150)
390
= 1.50 Ans.
n=
DE:
[3(150) 2 ]1/2
5-4 S y = 220 MPa
(a) σ1 = 100, σ2 = 80, σ3 = 0 MPa
MSS:
DET:
220
= 2.20 Ans.
100 − 0
σ = [1002 − 100(80) + 802 ]1/2 = 91.65 MPa
220
n=
= 2.40 Ans.
91.65
n=
(b) σ1 = 100, σ2 = 10, σ3 = 0 MPaMSS:
DET:
n=
220
= 2.20 Ans.
100
σ = [1002 − 100(10) + 102 ]1/2 = 95.39 MPa
n=
220
= 2.31 Ans.
95.39
(c) σ1 = 100, σ2 = 0, σ3 = −80 MPa
MSS:
DE:
n=
220
= 1.22 Ans.
100 − ( −80)
σ = [1002 − 100( −80) + ( −80) 2 ]1/2 = 156.2 MPa
220
n=
= 1.41 Ans.
156.2
(d) σ1 = 0, σ2 = −80, σ3 = −100 MPa
220
n=
= 2.20 Ans.
MSS:
0 − ( −100)
σ = [( −80) 2 − ( −80)(−100) + ( −100) 2 ] = 91.65 MPa
DE:
n=
220
= 2.40 Ans.
91.65
117
budynas_SM_ch05.qxd
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FIRST PAGES
Page 118
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
5-5
(a) MSS:
DE:
(b) MSS:
DE:
n=
2.23
OB
=
= 2.1
OA
1.08
n=
2.56
OC
=
= 2.4
OA
1.08
n=
OE
1.65
=
= 1.5
OD
1.10...
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