Shigleys mechanical engineering design 11th edition solutions pdf chapter 5

y = 350 MPa. MSS: σ 1 − σ 3 = S y /n ⇒ () 1 3 y S n σ σ = − DE: () () 1/2 1/2 2 2 2 2 2 3 A A B B x x y y xy σ σ σ σ σ σ σ σ σ τ ′ = − + = − + + y S n σ = ′ (a) MSS: σ 1 = 100 MPa,σ 2 = 100 MPa,σ 3 = 0 350 3.5. 100 0 n Ans = = − DE: 2 2 1/2 350 (100 100(100) 100) 100 MPa, 3.5. 100 n Ans σ ′ = − + = = = (b) MSS: σ 1 = 100 MPa,σ 2 = 50 MPa,σ 3 = 0 350 3.5. 100 0 n Ans = = − DE: 2 2 1/2

y = 350 MPa. MSS: σ 1 − σ 3 = S y /n ⇒ () 1 3 y S n σ σ = − DE: () () 1/2 1/2 2 2 2 2 2 3 A A B B x x y y xy σ σ σ σ σ σ σ σ σ τ ′ = − + = − + + y S n σ = ′ (a) MSS: σ 1 = 100 MPa,σ 2 = 100 MPa,σ 3 = 0 350 3.5. 100 0 n Ans = = − DE: 2 2 1/2 350 (100 100(100) 100) 100 MPa, 3.5. 100 n Ans σ ′ = − + = = = (b) MSS: σ 1 = 100 MPa,σ 2 = 50 MPa,σ 3 = 0 350 3.5. 100 0 n Ans = = − DE: 2 2 1/2

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Shigley’s MED, 10

th

edition Chapter 5 Solutions, Page 1/52

Chapter 5

5-1S

y

= 350 MPa.

MSS:

σ

1

−

σ

3

= S

y

/n ⇒

()

y

S

n

=−

DE:

222 22

3

AABBxxyyxy

σσσσσσσσστ

′=−+=−++

n

=

(a)

MSS:

σ

1

= 100 MPa,

σ

2

= 100 MPa,

σ

3

= 0

350

==

−

DE:

221/2350

(100100(100)100)100 MPa, 3.5 .

σ

′=−+ ===

(b)

MSS:

σ

1

= 100 MPa,

σ

2

= 50 MPa,

σ

3

= 0

350

==

−

DE:

221/2

350

(100100(50)50)86.6 MPa, 4.04 .

σ

′=−+===

(c)

2

2

100100

22

AB

σσ



=±+−=−





123

σσσ

== =−

MSS:

350

140(40)

==

−−

DE:

()

1/2

22

350

σ



′=+−===



(d)

2

2

50 755075

22

AB

σσ

−−−+



=±+−=−−





123

σσσ

==−= −

MSS: 350

0(114.0)

==

−−

DE: 2221/2

[(50)(50)(75)(75) 3(50)]109.0MPa

σ

′=−−−−+−+−=

350

==

(e)

()

2

2

10020100 20

22

AB

σσ

+−



=±+−=



