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### Transcript of Problem 5-15 & 16 Extra Part of Problem 5-15 pkwon/me471/Lect 5.2.pdf · PDF file 5!...

• 5

Brittle Materials

Modified I Mohr

( )

BA uc

B

A

B BA

uc

B

utuc

Autuc

A

B BA

BA ut

A

n S

nSSS SS

n S

σσσ

σ σ

σσ σσ

σ σ

σσ

σσσ

≥≥−=

>≥≥=− −

≤≥≥

≥≥=

0

1and01

1and0

0

5-11 Selection of Failure Criteria

Problem 5-15 & 16

( )

( ) ( )

( ) ksi175.9

2375.0 375.0)4(190

ksi52.27 4375.0

375.0)6(190

4

4

===

===

π τ

π σ

J Tc I Mc

xz

x

967.0 076.33 32

32076.33 54.16175.976.13

76.13 2 52.27

22

==

=>=

=+=

==

n

SD R

C

yA

D

A

σx

τxz

( ) ( )

007.1 78.31 32

78.31175.9352.273 222/122

==

=+=+=

n

xyx τσσ

Extra Part of Problem 5-15

( ) ksi94.4547.11328.41:Misesvon

ksi23.4747.1164.2022;64.20 2 28.41:Tresca

ksi47.11175.925.1175.9 ksi28.4152.275.152.27

133.0/;33.1/

22

22

'

=+=

=+×=×===

=×==

=×==

==

req

req y

txz

tsx

y S

RSC

K K

drdD

τ

σ

¾”

1”

0.1”

Determine the required yield strength

of a material after considering the stress

concentration factors.

• 6

21

Problem 5-5

3/2

)3/4)(2/( 3

2 max

r rryAQ

=

== ππ

( ) ksi21.1630.1609.0 4/)5.0( )5.0(1600

5.0 75

42

=+−=

+−=

+−=

ππ

σ I cM

A R

x

c

y y

Mz=125(8)-75(3) =775 lb-in

Mx=200(8) =1600 lb-in

A

B

Ty=200(3 ) =600 lb-in

A

( ) ( ) ( ) ksi84.22/5.0

)5.0(600 )1(4/5.0 3/)5.0(2125

44

3

max

−= −

+=

+=

ππ

τ J cM

It QR yx

xy

B ( ) ksi78.7

4/)5.0( )5.0(775

5.0 75

42

=

+−=

+−=

ππ

σ I cM

A R z c

y y

( ) ( ) ( ) ksi4.3

2/5.0 )5.0(600

)1(4/5.0 3/)5.0(2200

44

3

max

=

+=

+=

ππ

τ J cM

It QR yz

yz

A

B

y

x

z

3’

8’

Fy=75lb Fx=125lb

Fz=200lb

d = 1”

π3 4r

Problem 5-5

Mz=125(8)-75(3) =775 lb-in

Mx=200(8) =1600 lb-in

A

B

Ty=200(3) =600 lb-in

75lb

125lb

200lb

775 lb-in

1600 lb-in

125lb

200lb

inlb82.1777

1600775 22

−=

+=RM

( ) lb9.2484.2532sin85.235 lb85.235200125 22

=−⋅=

=+=

c

R

V V

32o

25.84o

( ) ( ) ksi195.18095.01.18

5.0 75

45.0 5.082.1777

24

−=−−=

− ⋅

−=−−= ππ

σ A R

I cM yR

( )( ) ksi095.3055.304.0

2 600

24 329.24

44

3

=+=

+ ⋅

=+= c c

cc c

J Tc

Ib VQ

ππ τ

ksi34.214.52.922 22 =+== RD

Problem 5-5

1’

1.1’

0.05’

D / d =1.1; r / d = 0.05 σ x ' = Kts27.52+Kt 0.095=1.85×18.1+1.87×0.095= 33.66ksi

τ xz = Kt9.175=1.25×3.055= 3.82ksi

Tresca :C = 33.66 2

=16.85; Sy req = 2×R = 2× 16.852 +3.822 = 34.555ksi

vonMises : S y

req = 33.662 +3 3.82( )2 = 34.3ksi

Determine the required yield strength of a material after considering the stress concentration factors.

Rectangular plate with hole subjected to axial load. (a) Plate with cross-sectional plane; (b) one-half of plate with stress distribution; (c) plate with elliptical hole subjected to axial load.

Rectangular Plate with Hole

5-12 Introduction of Fracture Mechanics

• 7

(a) Mode I, opening; (b) mode II, sliding; (c) Mode III, tearing.

Three Modes of Fracture

Infinite Plate with Crack

( )⎩ ⎨ ⎧

+ =

=

⎟ ⎠

⎞ ⎜ ⎝

⎛ +=

⎟ ⎠

⎞ ⎜ ⎝

⎛ −=

yx z

I xy

I y

I x

r K r

K r

K

σσν σ

θθθ π

τ

θθθ π

σ

θθθ π

σ

0 2 3cos

2 cos 2

sin 2

2 3sin

2 sin1

2 cos

2

2 3sin

2 sin1

2 cos

2

ToughnessFracture:ICI KaK ≥= πσ Crack Propagate when

Yield stress and fracture toughness data for selected engineering materials at room temperature.

Fracture Toughness

propagateCrackICI KaK ≥= πβσ

aKI πβσ=Other geometries

I

IC

K Kn =Factor of Safety

• 8

Problem 5-43

⎟⎟ ⎠

⎞ ⎜⎜ ⎝

⎛ −

− =

⎟⎟ ⎠

⎞ ⎜⎜ ⎝

⎛ +

− =

2

2

22

2

2

2

22

2

1

1

r r

rr pr

r r

rr pr

o

io

ii r

o

io

ii t

σ

σ