# Solucionario Diseño Shigley Cap 5

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budynas_SM_ch05.qxd

11/29/2006

15:00

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 =
2

12
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 =
…ver más…

(5-30a)

n=

BCM: Eq. (5-31a )

Sut
30
=
= 1.5 Ans. σx 20

n=

30
= 1.5 Ans.
20
30 n= = 1.5 Ans.
20

MM: Eq. (5-32a )
(b) σx = 12 kpsi, τx y = −8 kpsi σ A, σB =

12
2

12
±
2

2

+ ( −8) 2 = 16, −4 kpsi

30
= 1.88 Ans.
16
16 ( −4)
1
=

⇒ n 30
100
30
= 1.88 Ans. n= 16 n= MNS: Eq. (5-30a)
BCM: Eq. (5-31b)
MM: Eq. (5-32a)

n = 1.74 Ans.

(c) σx = −6 kpsi, σ y = −10 kpsi, τx y = −5 kpsi σ A, σB =

−6