week-4_properties and behaviour

8/19/2019 Week-4_Properties and Behaviour

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En. Muhammad Hanaf BinAsril Rajo Mantari

Hanaf,JKM,PPD,2015

DAM 21102ENGINEERING MATERIAL

ELE!TI"N!HA#TER $ %MATERIAL ENGINEERING

&THEIR !HARA!TERITI!%

Mat'rial B'ha(iours


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ME!HANI!AL #R"#ERTIE ANDBEHA)I"*R

Hanaf,JKM,PPD,2015

1. INTR"D*!TI"N

2. !"N!E#T "+ TRE AND TRAIN$. ELATI! DE+"RMATI"N

,. #LATI! DE+"RMATI"N

-. T#E "+ TET/. A+ET +A!T"R


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Hanaf,JKM,PPD,2015

INTR"D*!TI"N Many material, when in service, are subjected toforces or loads.

It is necessary to know the characteristics of the

material from which it is made such that any resulting deformation will not beexcessive and fracture will not occur.

Mechanical behavior –the relationship between

its response or deformation to an applied load orforce.

Examples : strength, hardness, ductility &stiffness


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Hanaf,JKM,PPD,2015

Mechanical properties of materials areascertained –it need to perform laboratoryexperiments.

Standardized testing techniques –to ensure theconsistency in the tests conducted and in theinterpretation of results.

Examples : ASTM(American Standards forTesting and Materials), ISO


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Hanaf,JKM,PPD,2015

If the load is ;

• static or changes relatively slowly with time

•

applied uniformly over a cross section orsurface of a member

the mechanical behavior may beascertained by a simple stress-strain test

(conducted at room temperature)• Three principle ways of applied load ;

tensile, compression and shear(torsion)

!"N!E#T "+ TRE AND

TRAIN


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Hanaf,JKM,PPD,2015

Schematic

illustration of how

different types of load

produce an elongationand positive linear strain.

(a) tensile load.

(b) compression load.

(c) Shear strain.

(d) torsion deformation.


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Hanaf,JKM,PPD,2015

ELATI! DE+"RMATI"N

Degree of structure deforms or strains – depends on the magnitude of animposed stress.

ε σ E =

Stress

(N/m2= Pa)

Young’s Modulus

(modulus of elasticity)

(Pa)

Strain

• Elastic deformation – deformation in which stress andstrain are proportional.

when the applied load is released, the piecereturns to its original shape.


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Hanaf,JKM,PPD,2015

• Metals deform when subjected to uniaxial tensile force

– ELASTIC DEFORMATION• Regain original dimensions when force removed• Atoms return to original positions

Consider acylindrical rodsubjected to uniaxialforce

Engineering stress,σ:

σ = F/Ao Engineering strain,ε:

ε = l – lo = ∆l

lo l

lo

Ao

lo

___

Δl

___

A

F


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Hanaf,JKM,PPD,2015

• Example

A 1.25 cm diameter bar is subjected to a load of 2500kg. calculate the engineering stress on the bar inmegapascals (MPa). Acceleration of gravity is 9.81 ms-2.

• Solution

F = ma = (2500 kg)(9.81 ms-2) = 24,500 N

σ = F/Ao

= F/(π/4)(d2)

= 24,500 N / (π/4)(0.0125 m)2

= 200MPa


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Hanaf,JKM,PPD,2015

• Example

Pure aluminium 0.5 cm width, 0.040 cmthick and 8 cm length which hasgagemarkings 2 cm apart in the middle of thesample is strained so that the gagemarkings are 2.65 cm apart. Calculateengineering strain elongation which thesample undergoes

• Solution :

ε = l – lo

lo

= ∆l = (2.65 cm – 2.00 cm) lo 2.00 cm

= 0.65 cm = 0.325

2.00 cm

% elongation = 0.325 x 100% = 32.5 %

8 c m

2 c m

2 . 6

5

c m


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Hanaf,JKM,PPD,2015

• Most metallic materials, elastic deformation persists onlyto strains of about 0.005.

• PLASTIC DEFORMATION ;

Cannot fully recoverto original dimensions Atoms permanently displaced and take up new

positions.- involved the breaking of bonds with original

atom neighbors and then reforming bonds with

new neighbors, upon removal of the stress they

do not return to their original positions.

#LATI! DE+"RMATI"N


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Hanaf,JKM,PPD,2015

T#E

"+TET


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Hanaf,JKM,PPD,2015

T'nsil' t'st• Purpose – to evaluatestrength of materials

• Sample is pulled tofailure in a relativelyshort time at aconstant rate

• Plot of engineeringstress vs. engineeringstrain can be

constructed


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Hanaf,JKM,PPD,2015

• From a stress-straindiagram we canobtain

1) Modulus ofelasticity, E (whereE = σ/ε)

2) Yield strength at0.2%* offset, σ y

3) Ultimate tensile

strength, σUTS

4) Percent elongationat fracture

5) Percent reduction

Stress-strain diagram


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Hanaf,JKM,PPD,2015

• Typical stress-straine!a"i#r $#r ametal s!#%in&elastic ' plastic(e$#rmati#ns, t!eyiel( stren&t! σy as

(etermine( )sin&t!e 0.002 strain#*set met!#(.

• +tress-straine!a"i#r #$ s#mesteels(em#nstratin& t!eyiel( p#intp!en#men#n.


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Hanaf,JKM,PPD,2015

i&)re . / Typical en&ineerin& stress-strain e!a"i#r t# $ract)re, p#int . TS is t!etensile stren&t!. T!e circ)lar insets represent t!e &e#metry #$ t!e (e$#rme( specimenat "ari#)s p#ints al#n& t!e c)r"e.


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##m-Temperat)re lastic an( +!ear M#()li, an( P#iss#ns ati# $#r3ari#)s Metal 4ll#ys

E l

l E

o

∆== ε σ

where

∆l = elongation

Lo = original lengthHanaf,JKM,PPD,2015


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•

True stress : σt = F/Ai

• True strain εt = l∫ ilo (dl/l)

= ln (li/lo )

= ln (Ao/Ai )

(assume the volume isconstant)

lo- original gage length

li – instantaneous extendedgage length

Ao –original area

Ai – instantaneous minimumcross-sectional area ofsample True stress and Engineering stress

Hanaf,JKM,PPD,2015


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• Example :

Load applied to specimen, F = 170Kg

original specimen length,lo = 80mm

length of specimen under 170Kg load,li = 100mm

original specimen diameterd0 =10mm

Diameter of specimen under 170Kg load,di = 5mm

• Solution : Ao = (π/4)do2 = (π/4)(0.01)2 = 7.85x10-5 m2

Ai = (π/4)di2= (π/4)(0.005)2 =1.96 x10-5 m2

Eng. Stress,σE = F/ Ao = (170x9.81)/7.85x10-5= 212.446 M N/m-2

Eng. Strain,εE = ∆l/lo= 100-80/80= 0.25

True Stress,σ T = F/ Ai=(170x9.81)/1.96 x10-5 =850.867 M N/m-2

True Strain,ε T = ln(li/lo) = ln( 100/80)=0.22Hanaf,JKM,PPD,2015


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Hardness –the resistance of a metal to permanent(plastic) deformation.

Four common hardness test :

1.Brinell

2.Vickers

3.Knoop

4.Rockwell

Non-destructive test

Hardn'ss T'st


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ardness Testing Techni!ues


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• To ascertain thefracture characteristics ofmaterials.

• Fracture – separation of a solid under stress intotwo or more parts.

• Two types of impact test : IZOD or CHARPY.

Imat

T'st

FRACTURE

DUCTILE- Slow crack propagation

BRITTLE- Rapid crack propagation

DUCTILE + BRITTLE


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" schematic drawing of an impact specimens an testing apparatus

(b) #harpy specimen

(c) $%od specimen

(a) $mpact machine


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*ect #$ temperat)re #n t!e ener&y as#re( )p#n impact y(i*erent types #$ materials.


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Fatigue – failures which occur under dynamic orfluctuating stresses (repeated or cyclic stress).

Examples : in connecting rods, shafts and gears. Occurring very suddenly and without warning. Usually starts/originates at a point of stressconcentration such as a sharp corner or notch.

Schematic diagram of fatigue machine

+atiu' T'st


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+ +tress "s. n)mer #$ cycles7 c)r"es $#r $ati&)e$ail)re


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• Creep – deformation under static loador stress at elevated temperature.

• Examples : in high-pressure steamlines, turbine rotors in jet engines etc.

• Time-independent and permanentdeformation of materials.

• For metals – only important when theservice temperature> 0.4Tm (Tm =absolute melting temperature).

!r'' T'st


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Typical creepc)r"e #$ strain"ers)s time atc#nstant stressan( c#nstantele"ate(

temperat)re.

8n9)ence #$stress σ an(temperat)re T#n creep

e!a"i#r.

= #reep rate


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A+ET +A!T"R&se to predicting material failure and designing safe product

'actor of safety

N

y

w

σ

σ =

*hich is

σw = safety stress + wor,ing stressσ y = material yield stress = safety factors

(common used number . and /)


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:ample/

4 tensile apparat)s is t# e c#nstr)cte(%!ic! m)st stan( %it!stan( a ma:im)m

l#a( 220;. T!e (esi&n calls $#r t%#cylin(rical s)pp#rt p#sts, eac! #$ %!ic! ist# s)pp#rt !al$ #$ t!e ma:im)m l#a(.)rt!erm#re, plain car#n 10


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+>?@T8>

Ai"en B1.5, t!)s, t!e %#r;in& stress σ%

Mpa

Mpa

N

y

w

67.2065.1

310==

=

σ

σ

w

o

F d A

σ

π =

=

2

2

d is the diameter of the rod and ' is the force applied

m

xd

F d

w

026.0

)1067.206(

1100002

2

6

=

=

=

π

πσ Therefore the diameter

of rod should be2.6 x10-2m


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......88+H ......

THan; y#)

Hanaf JKM PPD 2015

week-4_properties and behaviour

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