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Chapter 6
6-1 MSS: DE: σ1 − σ3 = S y /n n= Sy σ ⇒ n= Sy σ1 − σ3
1/2

2 2 σ = σ A − σ AσB + σB

1/2

2 2 2 = σx − σx σ y + σ y + 3τx y

(a) MSS:

σ1 = 12, σ2 = 6, σ3 = 0 kpsi 50 n= = 4.17 Ans. 12 σ = (122 − 6(12) + 62 ) 1/2 = 10.39 kpsi, 12 2
2

DE:

n=

50 = 4.81 10.39

Ans.

12 ± (b) σ A , σ B = 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. n= 50 = 2.73 18.33 Ans.

σ = (122 + 3(−82 )) 1/2 = 18.33 kpsi, −6 − 10 ± 2 −6 + 10 2
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.
A

σ = [(−6) 2 − (−6)(−10) + (−10) 2 + 3(−5) 2 ]1/2 = 12.29 kpsi 50 n= = 4.07 Ans. 12.29 12 − 4 2
2

12 + 4 ± (d) σ A , σ B = 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 50 = 4.66 10.72 Ans.

σ = [122 − 12(4) + 42 + 3(12 )]1/2 = 10.72 kpsi n= Ans.

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Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design

6-2 S y = 50 kpsi MSS: DE: (a) MSS: DE: σ1 − σ3 = S y /n
2 2 σ A − σ AσB + σB


1/2

n=

Sy σ1 − σ3 ⇒
2 2 n = Sy / σ A −σ A σ B + σ B 1/2

= S y /n

σ1 = 12 kpsi, σ3 = 0, n = n= [122

50 = 4.17 Ans. 12 − 0 Ans.

50 = 4.17 − (12)(12) + 122 ]1/2 50 = 4.17 12

(b) MSS: DE: (c) MSS: DE: (d) MSS: DE:

σ1 = 12 kpsi, σ3 = 0, n = n= [122

Ans. Ans.

50 = 4.81 − (12)(6) + 62 ]1/2

σ1 = 12 kpsi, σ3 = −12 kpsi, n = n= [122

50 = 2.08 Ans. 12 − (−12) Ans.

50 = 2.41 − (12)(−12) + (−12) 2 ]1/3

σ1 = 0,σ3 = −12 kpsi, n = n= [(−6) 2

50 = 4.17 Ans. −(−12)

50 = 4.81 − (−6)(−12) + (−12) 2 ]1/2

6-3 S y = 390 MPa MSS: DE: (a) MSS: DE: σ1 − σ3 = S y /n
2 2 σ A − σ AσB + σB


1/2

n=

Sy σ1 − σ3 ⇒
2 2 n = Sy / σ A − σ A σ B + σ B 1/2

= S y /n

σ1 = 180 MPa, σ3 = 0, n = n=

390 = 2.17 Ans. 180 Ans.

390 = 2.50 [1802 − 180(100) + 1002 ]1/2 180 2
2

180 ± (b) σ A , σ B = 2MSS: DE: n= n=

+ 1002 = 224.5, −44.5 MPa = σ1 , σ3

390 = 1.45 Ans. 224.5 − (−44.5) [1802 390 = 1.56 + 3(1002 )]1/2 Ans.

Chapter 6

151

160 ± (c) σ A , σ B = − 2 MSS: DE: n=

160 − 2

2

+ 1002 = 48.06, −208.06 MPa = σ1 , σ3

390 = 1.52 Ans. 48.06 − (−208.06) 390 n= = 1.65 Ans. 2 + 3(1002 )]1/2 [−160

(d) σ A , σ B = 150, −150 MPa = σ1 , σ3 380 = 1.27 Ans. n= MSS: 150 − (−150)390 n= = 1.50 Ans. DE: [3(150) 2 ]1/2 6-4 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 = 2.40 Ans. n= 91.65 n= 220 = 2.20 Ans. 100 220 = 2.31 Ans. 95.39 220 = 1.22 Ans. 100 − (−80)

(b) σ1 = 100, σ2 = 10, σ3 = 0 MPa MSS: DET: n=

σ = [1002 − 100(10) + 102 ]1/2 = 95.39 MPa n=

(c) σ1 = 100, σ2 = 0, σ3 = −80MPa MSS: DE: n=

σ = [1002 − 100(−80) + (−80) 2 ]1/2 = 156.2 MPa 220 = 1.41 Ans. n= 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

152

Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design

6-5 (a) MSS: DE: (b) MSS: DE: n= n= n= n= 2.23 OB= = 2.1 OA 1.08 OC 2.56 = = 2.4 OA 1.08 OE 1.65 = = 1.5 OD 1.10 OF 1.8 = = 1.6 OD 1.1
B

(a) C B

A Scale 1" 200 MPa O D E F (b)

A

J G K L (d) H I (c)

(c) MSS: DE: (d) MSS: DE:

n= n= n= n=

OH 1.68 = = 1.6 OG 1.05 OI 1.85 = = 1.8 OG 1.05 OK 1.38 = = 1.3 OJ 1.05 OL 1.62 = = 1.5 OJ 1.05

Chapter 6

153

6-6 S y = 220 MPa (a) MSS: DE: (b) MSS: DE: n= OB 2.82 = = 2.2 OA 1.3OC 3.1 n= = = 2.4 OA 1.3 OE 2.2 n= = = 2.2 OD 1 n= OF 2.33 = = 2.3 OD 1
B

(a) C B

A

1"

100 MPa D O

E F

(b)
A

G J H I (c)

K

L (d)

(c) MSS: DE: (d) MSS: DE:

n= n= n=

OH 1.55 = = 1.2 OG 1.3 OI 1.8 = = 1.4 OG 1.3

OK 2.82 = = 2.2 OJ 1.3 OL 3.1 n= = = 2.4 OJ 1.3

154

Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering...
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