5

Brittle Materials Modified I Mohr

( )

BAuc

B

A

BBA

uc

B

utuc

Autuc

A

BBA

BAut

A

nS

nSSSSS

nS

σσσ

σσ

σσσσ

σσ

σσ

σσσ

≥≥−=

>≥≥=−−

≤≥≥

≥≥=

0

1and01

1and0

0

5-11 Selection of Failure Criteria

Problem 5-15 & 16 ( )

( )( )

( )ksi175.9

2375.0375.0)4(190

ksi52.274375.0

375.0)6(190

4

4

===

===

πτ

πσ

JTcIMc

xz

x

967.0076.3332

32076.3354.16175.976.13

76.13252.27

22

==

=>=

=+=

==

n

SDR

C

yA

D

A σx

τxz

( ) ( )

007.178.3132

78.31175.9352.273 222/122

==

=+=+=

n

xyx τσσ

Extra Part of Problem 5-15

( ) ksi94.4547.11328.41:Misesvon

ksi23.4747.1164.2022;64.20228.41:Tresca

ksi47.11175.925.1175.9ksi28.4152.275.152.27

133.0/;33.1/

22

22

'

=+=

=+×=×===

=×==

=×==

==

req

reqy

txz

tsx

yS

RSC

KK

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

2max

rrryAQ

=

== ππ

( )ksi21.1630.1609.04/)5.0()5.0(1600

5.075

42

=+−=

+−=

+−=

ππ

σIcM

AR

x

c

yy

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

Mx=200(8) =1600 lb-in

A

B

Ty=200(3) =600 lb-in

A

( )( ) ( ) ksi84.2

2/5.0)5.0(600

)1(4/5.03/)5.0(2125

44

3

max

−=−

+=

+=

ππ

τJcM

ItQR yx

xy

B ( )ksi78.7

4/)5.0()5.0(775

5.075

42

=

+−=

+−=

ππ

σIcM

AR z

c

yy

( )( ) ( )ksi4.3

2/5.0)5.0(600

)1(4/5.03/)5.0(2200

44

3

max

=

+=

+=

ππ

τJcM

ItQR yz

yz

A

B

y

x

z

3’

8’

Fy=75lb Fx=125lb

Fz=200lb

d = 1”

π34r

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

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

=−⋅=

=+=

c

R

VV

32o

25.84o

( ) ( )ksi195.18095.01.18

5.075

45.05.082.1777

24

−=−−=

−⋅

−=−−=ππ

σAR

IcM yR

( )( )ksi095.3055.304.0

2600

24329.24

44

3

=+=

+⋅

=+=cc

ccc

JTc

IbVQ

ππτ

ksi34.214.52.922 22 =+== RD

Problem 5-5

1’ 1.1’

0.05’

D / d =1.1; r / d = 0.05σ 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 = 33.662

=16.85; Syreq = 2×R = 2× 16.852 +3.822 = 34.555ksi

vonMises : Sy

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

( )⎩⎨⎧

+=

=

⎟⎠

⎞⎜⎝

⎛ +=

⎟⎠

⎞⎜⎝

⎛ −=

yxz

Ixy

Iy

Ix

rKr

Kr

K

σσνσ

θθθπ

τ

θθθπ

σ

θθθπ

σ

023cos

2cos2

sin2

23sin

2sin1

2cos

2

23sin

2sin1

2cos

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

KKn =Factor of Safety

8

Problem 5-43

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−=

2

2

22

2

2

2

22

2

1

1

rr

rrpr

rr

rrpr

o

io

iir

o

io

iit

σ

σ