Ch2 Mechanical Properties of Matter 2013.pps

7/24/2019 Ch2 Mechanical Properties of Matter 2013.pps

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Engineering Physics Course: Lecture 3

Engineering PhysicsEngineering PhysicsMechanical Properties of MatterMechanical Properties of Matter

Ch-2 Mechanical Properties ofCh-2 Mechanical Properties ofMatterMatter

Dr. Mohamed Farhat Othman

Math. & Engineering Phys. DepartmentMath. & Engineering Phys. Department

Mansoura UniversityMansoura University

Faculty of EngineeringFaculty of Engineering


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Chapter OutlineChapter Outline

Introduction to the MechanicalProperties of Matter.

Stress and StrainElasticity and Plasticity

Brittle and Ductile Materials

Poissons atio

Safety !actors


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Questions to Think AboutQuestions to Think About

Stress and strainStress and strain: "hat are they and #hy arethey used$

Elastic behaviorElastic behavior: "hen loads are s%all& ho#%uch defor%ation occurs$ "hat %aterialsdefor% least$

Plastic behaviorPlastic behavior: 't #hat point do dislocationscause per%anent defor%ation$ "hat %aterials

are %ost resistant to per%anent defor%ation$

Brittle and ductilityBrittle and ductility: "hat are they and ho#do #e %easure the%$


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StressStress

Stress is a %easure of the internal resistance ina %aterial to an e(ternally applied load. !ordirect co%pressi)e or tensile loading the

stress is defined as:

oAareasectionalCross

Fload=stress


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tensile stress compressive stress

1- Normal Stress1- Normal Stress


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oA

F)(stressNormal =

1- Normal Stress1- Normal Stress

The normal stress

F The force in ( N )

AoThe original cross-sectional area (m2)

Units of N/m2=Pa

l/in2=Psi


7/45Engineering Physics Course: Lecture 3

The Cross Sectional AreaThe Cross Sectional Area

Cu*e

'+Lo,

FF

a

b

ectangular

'+a-*

FCylinder:

'+r,

'+! (d/2)2= ! d2/"



2- Shear Stress2- Shear Stress

=F/ AoThe force is parallel to

the #pper and lo$erfaces (N)%

Ao& The original cross-

sectional area (m2)

Units of N/m2=Pa

l/in2=Psi



1- Normal Strain1- Normal Strain

oo

o

'

'

'

'')(strainNormal

=

=

'o is the original length

$itho#t load

' is the instantaneo#slength%

Strain is

dimensionless



2- Shear Strain2- Shear Strain

h



2- Shear Strain2- Shear Strain

h

tan)(strain*hear ==

h

h

x



F

bonds

stretch

return toinitial

1. Initial 2. Small load 3. Unload

Elastic means reversible.

Elastic DeformationElastic Deformation



/. Initial ,. Load 3. 0nload

Plastic means permanent.

Plastic Deformation Metals!Plastic Deformation Metals!

planesstill

sheared

F

elastic + plastic

bondsstretch& planes

shear

plastic



"inear Elastic Properties"inear Elastic Properties

Moulus of Elasticity! E"

#$oung%s moulus&

Hooke's a!"

# $

F

Fsimpletensiontest

PsiorPa+

=

Load

0nload



#oo$%s "aoo$%s "a&The proportionalit, of stress and strain (#nder

certain conditions) is called Hooke's law

PsiorPa*trainNormal*tressNormal().od#l#so#ng

==

PsiorPa*train*hear*tress*hear(*).od#l#s*hear

==



E'ample 2-1 pa(e 2)!E'ample 2-1 pa(e 2)!' structure steel rod has a radius + 1.2 %% and

a length Lo+ / c%. ' force of 4., - /5//6stretches it a(ially and 7oung %odulus

7 +,.5 - /5//

68%,

.a9 "hat is the stress in the rod$

*9 "hat is the elongation of the rod under theload$

c9 "hat is the strain$



SolutionSolution

2

20

11

2m/N12%2

)m13%4(

N12%5

6

F

A

F=

=

==

-''

'''

oo

o ===

( ) ( )

( )

mm4%

m/N1%2

m1%m/N12%2

-

''

211

2

o =

=

=

0

0

o

11%1m1%

m14%

'

'

=

=

=



E'ample 2-2 pa(e 2*!E'ample 2-2 pa(e 2*!' steel rod ,.5 % long has a crosssectional area of 5.3 c%,. ;he

rod is no# hung *y one end fro% a support structure& and a 225



SolutionSolution

( ) ( )( )

Pa1%1m10%

s/m%47g33

A

F 2"

2

=

==

"

1

1%4Pa12

Pa1%1

-

=

=

=

( ) ( ) m11m%21%4'''

'

""

o

o

===

=



E'ample 2-+ pa(e 2,!E'ample 2-+ pa(e 2,!' piece of copper of cross section area >-/53 %,and

height of /.5 % is su*?ected to shearing force of

>55556. Calculate @a9 shear stress

@*9 shear strain @c9 linear displace%ent (.

;he shear %odulus is >.,5-/5/568%,



Pa1%1m1%"

N"

A

F 820 =

==

"

1

8

10%2Pa12%"Pa1%1

** =

====>=

h/tan ==

m10%2m%110%2h "" ===

The shear stress:

The shear strain:

The linear displacement x:



Poisson%s atioPoisson%s atio

E = ' S # () &

9

,

9

=

=

(

A

y!

!

L

D

'

'9

=

:

:

=

:

:,

=

9,9

9

,

9

aial

transverse

=

=

=



E'ample2-.E'ample2-.' tensile stress is to applied along the long a(is

of cylindrical *rass rod that has a dia%eter of

/5 %%. Deter%ine the %agnitude of the load

reuired to produce a ,.2-/5:3 %% change in

dia%eter if the defor%ation is elastic

@7+/5./-/5>Mpa & + 5.329


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solutionsolution

oo'

'

d

d =

"

0

o

1;1"%803%

)mm1/mm13%2('

'

=

=

d !"mm# d $%&!"()mm # F *

(x/

+

x (

+


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solutionsolution

'

'AAF

A

F


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E'ample 2-/E'ample 2-/' solid cylinder of dia%eter d carries an a(ial load

!. Sho# that its change in dia%eter is >!87d

#here 7 is the 7oungs %odulus of elasticity.


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SolutionSolution

,

aial

transverse

=

=

,= -

,=The c,linder is #nder compression

therefore is negative


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SolutionSolution

,

aial

transverse

=

=

A

F

- o

=

=

( )

-dF"d

"/d-

F

d

d

2,

=

=

=

( )

"/d

F

2

=

,=


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Stress-Strain CurveStress-Strain Curve,eckin- .oint

ield .oint

.roportional 0imit

1lastic

2eha3ior

.lastic

2eha3ior

Fract4re

point


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0ield Stren(th0ield Stren(th yy

;he yiel strength

*y& can *e deter%inedusing the 5.55, strain

offset%ethod#+here

there is noticeableplastic eformation&.

"%""$

5


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ensile Stren(th S!ensile Stren(th S!

Tensile Strength

#TS&" or ,ltimate

stress is the

%a(i%u% stresscorresponding to

the %a(i%u% point

on the stressstraincur)e.


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!ro% the tensile stressstrain *eha)ior for the*rass speci%en sho# in the !ig. deter%inethe follo#ing:

a9 ;he %odulus of elasticity.

*9 ;he yield strength at a strain offset of 5.55,.

c9 ;he %a(i%u% load that can *e sustained *y acylindrical speci%en ha)ing an original

dia%eter of /,.%%.d9 ;he change in length of a speci%en originally

,2> long #hich is su*?ected to a tensile stressof 3>2 MPa.

E'ample 2-3E'ample 2-3


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-.--(/

E'ample 2-3E'ample 2-3a9 ;he %odulus of elasticity.

E + slope+

psi10%1"

1"%

psi2

5=

=


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'0- MPa

*9;he yield

strength at a

strain offset of

5.55,.

y+34555 psi

or ,25 Mpa


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E'ample 2-3E'ample 2-3;he %a(i%u% load that can *e sustained *y a cylindricalspeci%en ha)ing an original dia%eter of /,.%%.

/0- MPa 10 --- psi;s+ 42555 psi or

>25 Mpa

N188%3

ATFA

FT

"

osma

o

ma

s

==

=


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2/0 MPa

-.-1

Lo+,2> %%

L+$ for

+3>2 Mpa

5%

.Pa0"3stressAt

=

=

5%''o

==

mm2%13'5%'o==


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Brittle and ductile MaterialsBrittle and ductile Materials

' %aterial that suffers

)ery little plastic

defor%ation is *rittle.

Ductilityis a %easureof the plastic defor%ation

that has *een sustainedat fracture:


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$n%ineerin% tensile strain

$n%ineerin%

tensilestress

smaller $(brittle i) $*,

lar%er $(ductile i)$-,

Ductility%ay *e e(pressed *y(3 percent elongation@ plastic strain at fracture9 o )

o)

1 xl

llEL

o

of

=

Ductility 4E"Ductility 4E"


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1areasectionalcross>nitial

areasectionalcrossFinal-areasectionalcross>nitial

=1A

A-A+'=

o

of =

$( .ercent 6ed4ction in Area:The red#ction in cross-sectional area ofa tensile specimen at fract#re=

o )

.o.)


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4actor of Safety4actor of Safety

!.S. is the safety factor

Fy is the yielding stress&

F#is the #or


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"hen a ru**er strip #ith cross section of 3%%-/.2 %% is suspended"hen a ru**er strip #ith cross section of 3%%-/.2 %% is suspended)ertically and )arious %asses are attached to it& a student o*tains the)ertically and )arious %asses are attached to it& a student o*tains thefollo#ing data for length )ersus load:follo#ing data for length )ersus load:

5oa gm - (-- '-- 2-- /-- 0--

5ength cm 0.- 0.1 1.' 1.6 7.8 (-.-

Lo+ 2c% & '+ 3-/.2+>.2-/54%,

i o

oo

F ' ' 'mg

A 'A '

= = = =

'oad 7g %1 %2 %0 %" %3

'ength m %3 %35 %52 %54 %8 %1

*tress . pa %2188 %"03 %5300 %811 1%

*train %12 %2" %0 %35 1


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.pa1"%112%

.pa2188%+ =

=

=


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If the load applied is >25 g%find the length of the strip.

5

o

mg (%"3)(4%)%4.pa

A "%3;1

From graph& at %4.pa

%88

%88 '/ '

' 0%38 cm

' 3 cm 0%38 cm = %38 cm

= = =

=

=

= =

=

= +


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Thanks for your attention

Documents

Ch2 Mechanical Properties of Matter 2013.pps