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ME111 Lecture 9 1

1

Final Exam

ME111 Stress, Strain and Strength

Monday, December 6, 1999

Three hours

Open Notes and Textbook (Only)

A machined shaft is subjected to a fully reversed torque of 8.5 in-klb.

Determine the diameter of the shaft needed to provide a factor of safety

of 2.5 if a life of 105 cycles is required. The material properties are:

)(100,9.0

,67,220

correctedfullykpsiSSS

kpsiSkpsiS

eutm

yut

==

==

Problem 1 (20 points)

SOLUTION

2

( ) ( ) ( )

indd

Sd

S

Now

S

abSbSa

S

S

S

Sb

aNS

SCompute

SNNow

d

dd

d

I

Tc

ff

f

utm

e

ut

e

m

b

f

f

ff

aa

a

14.1;493.157.125

5.298.74

57.125;98.74

;5.2

,

57.125000,1004.392

04.3925933.239.0log3loglog

09889.09.0

log3

1log

3

1

:

5.2

98.7433

129.43

32/

2/5.8

3

3

09889.0

101010

1010

32

34

==

=

===

==

====

=

=

=

=

==

===

=

==

ss

s

tts

pt

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ME111 Lecture 9 2

3

In a machine component subject to a fluctuating multiaxial stress state,

the von Mises effective alternating and mean stress are:

The relevant material properties are:

(a) Determine whether or not the element is safe for infinite life fatigue

(hint: compute the appropriate factor of safety assuming a fixedratio of stresses),

(b) If it is not safe for infinite life, estimate the number of cycles to

failure.

kpsikpsima

40,24 == ssss

)(40,9.0

,67,80

correctedfullykpsiSSS

kpsiSkpsiS

eutm

yut

==

==

Problem 2 (20 points)

05.14024

67::

.909.0

:

=+

=+

=

=+

=

ma

yy

emuta

utef

SNyieldingbyfailurenoNote

fatiguelifeinfiniteforsafenotSS

SSN

fatiguelifeinfiniteFor

ss

ss

Part (a)

4

( ) ( ) ( )

cyclesNNGiving

a

bSbSa

SS

SSb

aNSNow

utm

e

ut

e

m

bf

341,1176.1290.48:

6.129

1126.239.0log3loglog

085.09.0log31log

31

085.0

101010

1010

==

=

===

=

=

=

=

0.481

:

=+

== ffmuta

utf

f SSS

SSN

HCFinfailureFor

ss

Part (b)

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ME111 Lecture 9 3

5

Problem 3 (20 points)

The figure below shows a schematic drawing of an automobile jack

which is subject to a load of W = 10 kN. The opposite-handed threads

on the two ends of the screw are cut to allow the link angle to vary

from 15o to 70o. The links are rectangular in cross-section with

dimensions 25 x 7 mm. They are made from steel with E = 206.8 GPaand S

y= 290 MPa. If the load is to be supported over the full range

of the jack, what is the factor of safety against buckling of the links?

Assume that the axial load in the links is central (i.e. there is no

eccentricity) and that the links can be modeled as pinned-pinned

about both axes.

mm220

007015: toanglelink

kNW 10=

mmthickness 7:

mm25

6

( )

52.1318.19

382.29

382.292

1

6.1182

87.10802.2

220

)(02.212

7

12

12/

318.1915sin2

15sin2

:

2

2/

3

00

===

=

=

==

===

====

=

==

link

crb

crry

ycr

yyr

r

linklink

P

PN

kNPSS

ES

db

P

columnJohnson

S

ES

LS

axisweakusingd

bd

bd

kNW

PWP

staticsfromlinkinloadaxialCompute

p

p

r

r

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ME111 Lecture 9 4

7

A thin-walled cylindrical pressure vessel has an inner radius of r = 10 in.

and a wall thickness of t = 0.8 in. The pressure vessel supports an

internal pressure of p = 5 kpsi and an axial compressive load of

N = 9000 klb.

Consider the following:

(a) The pressure vessel is made of a ductile material with a yield

strength of Sy= 250 kpsi. Determine the factor of safety against

yielding under the applied load using:

(i) The von Mises (i.e. distortion energy) criterion,

(ii) The maximum shear stress criterion.

(b) The material is brittle with an ultimate tensile strength ofS

ut= 180 kpsi. What is the minimum ultimate compressive

stress Suc

that is required of the material to provide a factor of

safety of 1.5 against brittle failure? Use the modified Mohr

theory and assume plane stress conditions.

int 8.0=

inr 10=

kpsip 5=

klbF 000,9=

Problem 4 (20 points)

8

23.14.203

250),,max(

:

39.145.179

250:

45.179

9.1402

0.5

5.62

13133221

13322123

22

21

3

2

1

==

=

=

==

=

=++=

==

==

==

ssssssss

s

ssssssssss

s

s

s

yyy

y

y

SSNTresca

SNMisesvon

A

P

t

pr

p

t

pr

( )

( )kpsiS

S

Sor

SS

SSN

andquadrantfourthSince

uc

uc

uc

utuc

ucutb

5.2455.19.1405.621805.62

180

5.1

,

311

13

==

+==

>

sss

ss

Part (a)

Part (b)

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ME111 Lecture 9 5

9

The beam shown has depth of50mm and a thickness of 20 mm. A

crack through the thickness of the beam is detected in the bottom face

as indicated. The crack is measured to be 10 mm long and it needs

to be determined if the beam can safely support the two point loads

P = 30 kN. The beam is fabricated from AISI 4340 steel with a yield

strength Sy = 800 MPa and a fracture toughness Kc = 185 MPa(m)1/2.

(a) Using LEFM, determine the value of the load P that will cause

the crack to propagate.

(b) Repeat (a) taking plasticity at the crack tip into account.

(c) Can the beam support the design load of P = 30 kN?

Problem 5 (20 points)

kNP 30=kNP 30=

mm10

mm20

mm50

mm200mm200

10

( ) kNPP

Now

P

I

McaKK

d

a

C

5.431000

10

12/5020

252000.1185

12/5020

2/50200;185

0.12.0

3

3

=

=

=====

=

p

spbs

b

Part (a)

Part (b)

( ) kNPP

Now

aKK

d

a

mmaaKK

effC

eff

meters

effC

y

C

S

K

6.151000

27

12/5020

252007.1185

185

7.154.0

0.271000

3

21

=

=

===

=

=+==

p

pbs

b

p

Part (c)

No, fails when plasticity at the crack tip is considered.