03-BDeformasi Plastis Dan Penguatan

1

1

Deformasi (plastis)

dan

Mekanisme Penguatan

ISSUES TO ADDRESS...ISSUES TO ADDRESS...ISSUES TO ADDRESS...ISSUES TO ADDRESS...

• What are the forces on and between dislocations?

• Dislokasi menimbulkan deformasi

melalui rusaknya ikatan secara

bertahap

• Jika dislokasi tidak merambat, maka

deformasi platis tidak terjadi

3

Deformasi plastis dapat terjadi melalui

4

2

5

Dislokasi & Klasifikasi Bahan

• Keramik kovalen(Si, diamond): Motion hard.

- very hard, Tm>3550C

-directional bonding

• Keramik ionik (NaCl):Motion hard.

-need to avoid ++ and - -

neighbors.

- non-directional bonding

+ + + +

+++

+ + + +

- - -

----

- - -

• Logam: Disl. motion easier.-non-directional bonding

-close-packed directions

for slip. electron cloud ion cores

+

+

+

+

+++++++

+ + + + + +

+++++++

Jenis dislokasi : edge and screw

PD corresponds to the motion of large numbers of dislocations6

Yang Mana lebih Mudah

Faktor yg berpengaruh :

1. Densitas tumpukan

2. Jarak antar bidang

Perambatan Dislokasi

Dislokasi & deformasi plastis

• Logam kubik & heksagonal – deformasi plastis oleh geseran

plastis or slip dimana satu bidang atom bergeser ke bidang

disekitarnya dengan perambatan cacat (dislokasi).

7

• Jika dislokasi tidak bergerak, deformasi tidak terjadi

Mekanisme deformation plastis :1. slipping 2. twinning

• Dislokasi bergerak sepanjang bidang slip dalam arah slip

tegak lurus terhadap garis dislokasi ( slip system)

• Arah slip sebagai Burgers vector

8

Dislokasi sisi

Dislokasi ulir

Slip adalah suatu proses dimana deformasi plastis dihasilkan oleh

perambatan dislokasi

Perambatan Dislokasi

3

9

Slip: adalah suatu proses dimana deformasi plastis dihasilkan oleh perambatan

dislokasi

The slip system tergantung pada struktur kristal dari logam dan is such that the

atomic distortion that accompanies the motion of a dislocation is a minimum

dislocation density: (The number of dislocations) dinyatakan sbg panjang

dislokasi total per unit volum, atau , jumlah dislokasi yang memotong satuan

luas dari suatu penampang.

Satuan dari kepadatan dislokasi (density dislokasi) adalah millimeters of

dislocation per cubic millimeter or just per square millimeter. Dislocation

densities as low as 10 3 mm -2

Asemua logam mengandung dislokasi yang terbentuk selama : 1.solidifikasi ,

2. deformasi plastis, dan sbg konsekuensi dari 3. tegangan thermal dari proses

pendinginan cepat.

Mekanisme Slipping

Sistem Slip

– Bidang Slip – bidang dimana garis dislokasi berjalan

• Bidang memungkinkan slip paling mudah

• Kerapatan bidang paling tinggi

– Arah Slip – arah dari pergeseran - Highest linear densities

– Slip pada FCC Slip terjadi pada bidang {111} (close-packed) di arah

<110> (close-packed)

=> total of 12 slip systems in FCC

– Pada BCC & HCP slip systems yang lain terjadi.

10

11

Bidang slip biasanya di bidang tumpuk yang paling

padat/rapat.Slip terjadi di bidang tumpukan yang padat

karena dibutuhkan tegangan geser yang lebih rendah untuk

pergeseran atom-atom dibandingkan dengan bidang

tumpukan yang kurang padat.

Slip di arah tumpukan yang padat juga lebih mungkin terjadi

karena energi yang dibutuhkan untuk menggeser atom-atom

dari satu posisi ke posisi (jika atom2 bedampingan).

Bidang dan arah Slip pada Bidang dan arah Slip pada Bidang dan arah Slip pada Bidang dan arah Slip pada BCCBCCBCCBCC

{110} bidang pd arah

Slip systems: 6 x 2 =12

< 1 11>

{211 bidang pada arah


< 1 11>

{321} bidang pada arah


< 1 11>

Sistem slip pada BCC

αFe, K

αFe, Mo,

W, Na

αFe, Mo,

W, β brass

4

Bidang dan arah Slip pada Bidang dan arah Slip pada Bidang dan arah Slip pada Bidang dan arah Slip pada HCP HCP HCP HCP

{0001} planes in the direction of

Slip systems: 1 x 3 = 3

<112 0 >

planes in the direction of


Slip sistem dpt tergantung pada c/a dan relative

orientation dari beban thd bidang slip

{101 0} <112 0 >

planes in the direction of


{101 1} <112 0 >

c/a ≥ 1.6333 (ideal)

c/a ≤ 1.6333 (ideal)

hcp Zinc

single crystal

Adapted from Fig.

7.9, Callister 6e.

Adapted from Fig.

7.8, Callister 6e.

Cd, Zn, Mg, Ti, Be …

Ti

Mg, Ti

14

Apakah dimungkinkan secara fisi, slip di suatu kristal fcc terjadi

slip di arah [110] di bidang (a) (111) (b) (111)?

16

5

Single Crystal Slip

17

Tegangan dan Perambatan Dislokasi

φλσ=τ coscosR

18

• Slip pada kristal terjadi disebabkan oleh resolved shear stress, τR.

• Beban menimbulkan tegangan tertentu.

slip plane

normal, ns

Resolved shear stress: τR =Fs/As

AS

τR

τR

FS

Relation between σ and τR

τR=FS/AS

Fcos λ A/cos φ

λF

FS

φnS

AS

A

Applied tensile stress: = F/Aσ

FA

F

Single crystals: critical resolved shear stress

Shear force:

Area:

λcosF

θ

θ

cos/

/cos

01

10

AA

AA

=⇒

=

Resolved shear stress:

m*coscoscos/A

cosF00

0RSS σ=θλσ=

θ

λ=τ

20

Critical Resolved Shear Stress• Syarat terjadi geseran dislokasi: ( ) CRSSmax ττ >R

• Orientasi kristal dpt membuatnya lebih mudah atau sulit untuk menggeser dislokasi.

10-4 GPa to 10-2 GPa

typically

φλσ=τ coscosR

τ maximum at λ = φ = 45º

τR = 0

λ=90°σ

τR = σ/2λ =45°φ =45°σ

τR = 0

φ=90°σ

Critical resolved shear stress: adalah tegangan geser

minimum yang dibutuhkan untuk menginisiasi slip dan merpk

satu sifat dr material yg menentukan kapan yielding terjadi

6

21

crssy

y

crssy

ycrss

then

whenoccursyieldingroduceint

tonecessarstressminimumthe

stressyielding

τσ

λφ

σ

φλ

τσ

φλστ

2

45

)cos(cos

)cos(cos

max

max

=

==

−

=

=

For 0=crssτ

The crystal ordinarily fractures rather than deforming

plastically

Contoh : Deformasi pada kristal tunggal

Shg, tegangan 6500 psi belum menyebabkan kristal tunggal tersebut yield.

22

τ = σ cos λ cos φ

σ = 6500 psi

λ=35°φ=60°

τ = (6500 psi) (cos35o)(cos60o)

= (6500 psi) (0.41)

τ = 2662 psi < τcrss = 3000 psi

τcrss = 3000 psi

a) Apakah kristal tunggal akan yield? b) Jika tidak, pada tegangan berapa yield?

σ = 6500 psi

23

psi 732541.0

psi 3000

coscoscrss ==

φλ

τ=σ∴ y

Berapa tegangan yg dibutuhkan,(berapateg yield , σy)?

)41.0(cos cos psi 3000crss yy σ=φλσ==τ

psi 7325=σ≥σ y

Shg, agar deformasi platis terjadi, maka diperlukan teg yg lebih besar atau sama dengan teg. yield

Contoh : Deformasi pada kristal tunggal

24

7

25 26

Slip Motion in Polycrystals• Stronger metals- batas butir

menghambat deformasi

• Bidang dan arah slip berubah

(λ, φ) berubah dari satu kristal ke kristal yang lain.

• τR bervariasi dari satu kristal ke kristal yg lain.

• Kristal dengan τR paling besar akan luluh pertama.

• Kristal lain (kristal dg orientasi yang lain) akan luluh kemudian.

σ

300 µm

27

Anisotropy in σy

• Anisotropis dapat dihasilkan dengan pengerolan pada logam

polikristalin

- before rolling

235 µm

- isotropic

since grains are

approx. spherical& randomly

oriented.

- after rolling

- anisotropic

since rolling affects grain

orientation and shape.

rolling direction

Isotropis. Mempunyai sifat –sifat dengan arga yang identik pada semua arah

kristalography.

28

Anisotropi pada Deformasi

side view

1. Cylinder of

Tantalum

machined

from a

rolled plate:

rolli

ng

dire

ctio

n

2. Fire cylinder

at a target.

• The noncircular end view shows

anisotropic deformation of rolled material.

endview

3. Deformed

cylinder

plate

thickness

direction

8

Twinning mechanism

• A part of the atomic lattice is deformed so that is forms

a mirror image of the un-deformed lattice next to it.

• Twinning plane: is the plane between the un-deformed

and deformed parts of the metal lattice

29 30

slipping twinning

1. the atoms in one side of the slip plane

all move equal distances

1. the atoms move distances proportional

to their distance from the twinning

plane

2. Slip Leaves a series of

steps (lines)

2. Twinning leaves small but well defined

regions of the crystal deformed

3. Most for FCC and BCC structure,

they have more slip systems

3. Is most important for HCP structure ,

because its small number of slip

system

4. normally slip results in relatively large

deformations

4. only small deformations result for

twinning

Comparison between slip and twinning

mechanisms

Mekanisme penguatan pada Logam

• Kekerasan dan kekuatan mrpk kemampuan logam utk

berdeformasi plastis tergantung thd kemampuan dari

dislokasi tsb merambat.

• Menghambat perambatan dislokasi menghasilkan

material lebih keras dan kuat.

• Dibutuhkan gaya mekanis yang lebih besar utk

menginisiasi deformasi plastis lebih lanjut.

• Keduktilan dikorbankan ketika logam diperkuat.

• Mekanisme penguatan utk logam phase tunggal : (a)

penurunan ukuran butir, (b) pemaduan larut-padat (c)

pengerasan regang / pengerjaan dingin, (d) pengerasan

dg presipitasi/pengendapan.31

4 Strategies for Strengthening:

1: Reduce Grain Size

21 /yoyield dk −+σ=σ

32

• Batas butir merupakan penghambat thd slip

1. Mengubah arah

2. Diskontinyuitas bidang slip

• Barrier "strength"

increases withIncreasing angle of

mis-orientation.• Smaller grain size:

more barriers to slip.

• Hall-Petch Equation:

d = average grain diameter

σy - yield strength

σ0, ky are constant for

particular material

9

33

Material dapat berupa material berbutir halus (berbutir kecil-

kecil) dan material berbutir kasar. Material berbutir halus lebih

keras dan kuat dibanding material berbutir kasar.

Ukuran butir dapat ditentukan oleh :

1. Kecepatan pembekuan dari phase cair, dan

2. Juga oleh deformasi plastis yg diikuti dg perlakuan panas yg

sesuai.(tgt dr waktu dan suhu). Semakin besar suhu dan waktu

maka ukuran butir menjadi semakin besar/kasar.

34


2: Solid Solutions (alloying)• Atom ketidakmurnian mendistorsi lattice & men-

generate tegangan.

• Tegangan menghasilkan penghambat terjadinya perambatan dislokasi.

• Smaller substitutional

impurity

Impurity generates local stress at A

and B that opposes dislocation motion to the right.

A

B

• Larger substitutional

impurity

Impurity generates local stress at C

and D that opposes dislocation motion to the right.

C

D

Konsentrasi tegangan pada dislokasi

35

Strengthening by Alloying

• small impurities tend to concentrate at dislocations

• reduce mobility of dislocation ∴ increase strength

36

10

Strengthening by alloying

• large impurities concentrate at dislocations on low density

side

37

Ex: Solid Solution

Strengthening in Copper

38

• Tensile strength & yield strength increase with wt% Ni.

• Empirical relation:

• Alloying increases σy and TS.

21 /y C~σ

Ten

sile

str

en

gth

(M

Pa

)

wt.% Ni, (Concentration C)

200

300

400

0 10 20 30 40 50 Yie

ld s

tre

ng

th (

MP

a)

wt.%Ni, (Concentration C)

60

120

180

0 10 20 30 40 50


3: Cold Work (%CW)

39

• Deformasi pada temperatur kamar.

• Biasanya prosesnya akan merubah luas permukaan penampang

-Forging

Ao Ad

force

die

blank

force-Drawing

tensile force

Ao

Addie

die

-Extrusion

ram billet

container

containerforce

die holder

die

Ao

Adextrusion

100 x %o

do

A

AACW

−=

-Rolling

roll

Ao

Adroll

Strain hardening/pengerasan regang : suatu logam duktil menjadi lebih keras

dan kuat karena dideformasi plastis(effect dari pengerjaan dingin tsb dapat

dihilangkan dengan perlakuan panas anil.

40

Dislocations During Cold Work

• Ti alloy after cold working:

• Dislocations entanglewith one another

during cold work.• Dislocation motion

becomes more difficult.

0.9 µm

11

Result of Cold Work

Dislocation density =

– Carefully grown single crystal

� ca. 103 mm-2

– Deforming sample increases density

� 109-1010 mm-2

– Heat treatment reduces density

� 105-106 mm-2

41

• Yield stress increasesas ρd increases:

total dislocation length

unit volume

large hardening

small hardening

σ

ε

σy0

σy1

Effects of Stress at Dislocations

42

Impact of Cold Work

43

• Yield strength (σy) increases.• Tensile strength (TS) increases.• Ductility (%EL or %AR) decreases.

As cold work is increased

44

Cold Work Analysis

• What is the tensile strength &

ductility after cold working?

%6.35100 x %2

22

=π

π−π=

o

do

r

rrCW

% Cold Work

100

300

500

700

Cu

200 40 60

yield strength (MPa)

σy = 300MPa

300MPa

% Cold Work

tensile strength (MPa)

200

Cu

0

400

600

800

20 40 60

ductility (%EL)

% Cold Work

20

40

60

20 40 6000

Cu

Do =15.2mm

Cold Work

Dd =12.2mm

Copper

340MPa

TS = 340MPa

7%

%EL = 7%

12

45

σ-ε Behavior vs. Temperature• Results for

polycrystalline iron:

• σy and TS decrease with increasing test temperature.

• %EL increases with increasing test temperature.

• Why? Vacancies

help dislocations

move past obstacles.

2. vacancies replace atoms on the disl. half plane

3. disl. glides past obstacle

-200°C

-100°C

25°C

800

600

400

200

0

Strain

Str

ess (

MP

a)

0 0.1 0.2 0.3 0.4 0.5

1. disl. trapped by obstacle

obstacle

46


4: Precipitation Strengthening

• Hard precipitates are difficult to shear.Ex: Ceramics in metals (SiC in Iron or Aluminum).

• Result:S

~y

1 σ

Large shear stress needed

to move dislocation toward

precipitate and shear it.

Dislocation “advances” but precipitates act as “pinning” sites with

S.spacing

Side View

precipitate

Top View

Slipped part of slip plane

Unslipped part of slip plane

Sspacing

47

Application:

Precipitation Strengthening

• Internal wing structure on Boeing 767

• Aluminum is strengthened with precipitates formedby alloying.

1.5µm48

Note:

Effect penguatan dengan pengurangan

ukuran butir dan pengerasan regang

dapat dieliminasi atau diperkecil

melalui perlakuan panas pada suhu

tertentu. Sebaliknya, penguatan

dengan larut-padat tidak dipengaruhi

oleh perlakuan panas.

13

Recovery, Recrystallization, and

Grain Growth

• These properties and structures may revert

back to the precold-worked states

by appropriate heat treatment (sometimes

termed an annealing treatment). Such

restoration results from two different

processes that occur at elevated temperatures:

recovery and recrystallization, which may be

followed by grain growth.

49 50

Effect of Heating After %CW

• 1 hour treatment at Tanneal...decreases TS and increases %EL.

• Effects of cold work are reversed!

• 3 Annealingstages to

discuss...

tensile

str

eng

th (

MP

a)

ductilit

y (%

EL)tensile strength

ductility

600

300

400

500

60

50

40

30

20

annealing temperature (ºC)200100 300 400 500 600 700

51

Recovery

Annihilation reduces dislocation density.

• Scenario 1Results from diffusion

• Scenario 2

4. opposite dislocations meet and annihilate

Dislocations annihilate and form a perfect atomic plane.

extra half-plane of atoms

extra half-plane of atoms

atoms diffuse to regions of tension

2. grey atoms leave by vacancy diffusion allowing disl. to “climb”

τR

1. dislocation blocked; can’t move to the right

Obstacle dislocation

3. “Climbed” disl. can now move on new slip plane

52

Recrystallization• New grains are formed that:

-- have a small dislocation density

-- are small

-- consume cold-worked grains.

33% cold

workedbrass

New crystals

nucleate after3 sec. at 580°C.

0.6 mm 0.6 mm

14

53

Further Recrystallization• All cold-worked grains are consumed.

After 4

seconds

After 8

seconds

0.6 mm0.6 mm

54

TR

º

º

TR = recrystallization temperature

Recrystallization Temperature, TR

TR = recrystallization temperature = point of

highest rate of property change

1. Tm => TR ≈ 0.3-0.6 Tm (K)

2. Due to diffusion � annealing time� TR = f(t) shorter

annealing time => higher TR

3. Higher %CW => lower TR – strain hardening

4. Pure metals lower TR due to dislocation movements

• Easier to move in pure metals => lower TR

55 56

grain growth: After recrystallization is complete, the strain-free grains will

continue to grow if the metal specimen is left at the elevated temperature this

phenomenon is called grain growth.

•Grain growth does not need to be preceded by recovery and recrystallization; it may

occur in all polycrystalline materials, metals and ceramics alike

• As grains increase in size, the total boundary area decreases, yielding an attendant

reduction in the total energy; this is the driving force for grain growth.

• Grain growth occurs by the migration of grain boundaries. Obviously, not all

grains can enlarge, but large ones grow at the expense of small ones that shrink.

• The directions of boundary movement and atomic motion are opposite to each

other

growth via atomic diffusion

Grain Growth

15

57

Grain Growth• At longer times, larger grains consume smaller ones. • Why? Grain boundary area (and therefore energy)

is reduced.

After 8 s,

580ºC

After 15 min,

580ºC

0.6 mm 0.6 mm

• Empirical Relation:

Ktdd no

n =−elapsed time

coefficient dependenton material and T.

grain diam.at time t.

exponent typ. ~ 2

Coldwork CalculationsA cylindrical rod of brass originally 0.40 in (10.2 mm) in

diameter is to be cold worked by drawing. The circular

cross section will be maintained during deformation. A

cold-worked tensile strength in excess of 55,000 psi (380

MPa) and a ductility of at least 15 %EL are desired.

Further more, the final diameter must be 0.30 in (7.6 mm).

Explain how this may be accomplished.

58

Coldwork Calculations Solution

If we directly draw to the final diameter what

happens?

59

%843100 x 400

3001100 x

4

41

100 1100 x %

2

2

2

..

.

D

D

xA

A

A

AACW

o

f

o

f

o

fo

=

−=

π

π−=

−=

−=

Do = 0.40 in

Brass

Cold Work

Df = 0.30 in

Coldwork Calc Solution: Cont.

• For %CW = 43.8%

60

Adapted from Fig. 7.19, Callister 7e.

540420

– σy = 420 MPa

– TS = 540 MPa > 380 MPa

6

– %EL = 6 < 15

• This doesn’t satisfy criteria…… what can we do?

16

Coldwork Calc Solution: Cont.

61

380

12

15

27

For %EL < 15

For TS > 380 MPa > 12 %CW

< 27 %CW

∴ our working range is limited to %CW = 12-27

Coldwork Calc Soln: Recrystallization

Cold draw-anneal-cold draw again

• For objective we need a cold work of %CW ≅ 12-27

– We’ll use %CW = 20

• Diameter after first cold draw (before 2nd cold draw)?

– must be calculated as follows:

62

100

%1 100 1%

2

02

2

2

2

02

2

2 CW

D

Dx

D

DCW ff =−⇒

−=

50

02

2

100

%1

.

f CW

D

D

−= 50

202

100

%1

.

f

CW

DD

−

=⇒⇒⇒⇒

m 3350100

201300

50

021 ..DD

.

f =

−==Intermediate diameter =

Coldwork Calculations Solution

Summary:

1. Cold work D01= 0.40 in � Df1 = 0.335 m

2. Anneal above D02 = Df1

3. Cold work D02= 0.335 in � Df 2 =0.30 m

Therefore, meets all requirements

63

20100 3350

301%

2

2 =

−= x

.

.CW

24%

MPa 400

MPa 340

=

=

=σ

EL

TS

y

⇒⇒⇒⇒

%CW1 = 1−0.335

0.4

2

x 100 = 30

Fig 7.19

Rate of Recrystallization

• Hot work � above TR

• Cold work � below TR

• Smaller grains

– stronger at low temperature

– weaker at high temperature 64

t/R

T

BCt

kT

ERtR

1:note

log

logloglog 0

=

+=

−=−=

RT1

log t

start

finish

50%

17

65

Summary

• Dislocations are observed primarily in metals

and alloys.

• Strength is increased by making dislocation

motion difficult.

• Particular ways to increase strength are to:

--decrease grain size

--solid solution strengthening

--precipitate strengthening

--cold work

• Heating (annealing) can reduce dislocation density

and increase grain size. This decreases the strength.

problem

The lower yield point for an iron that has

an average grain diameter of 1x10-2 mm

is 230 MPa (33,000 psi). At a grain diameter of

6x10-3 mm, the yield point increases to 275

MPa (40,000 psi). At what grain diameter will

the lower yield point be 310 Mpa (45,000 psi)?

66

67

We are asked to determine the grain diameter for an iron which will give a yield

strength of 310 MPa (45,000 psi). The best way to solve this problem is to first

establish two simultaneous expressions of Equation 7.7, solve for σ0 and ky, and

finally determine the value of d when σy = 310 MPa. The data pertaining to this

problem may be tabulated as follows:

problem

• The critical resolved shear stress for copper

is 0.48 MPa (70 psi). Determine the

maximum possible yield strength for a

single crystal of Cu pulled in tension.

• Answer:

In order to determine the maximum possible yield strength for a

single crystal of Cu pulled in tension, we simply employ Equation 7.5

as

68

18

problem

Two previously un-deformed cylindrical

specimens of an alloy are to be strain hardened by

reducing their cross-sectional areas (while

maintaining their circular cross sections). For one

specimen, the initial and deformed radii are 15 mm

and 12 mm, respectively. The second specimen,

with an initial radius of 11 mm, must have the same

deformed hardness as the first specimen; compute

the second specimen’s radius after deformation.

69 70

In order for these two cylindrical specimens to have the same

deformed hardness, they must be deformed to the same percent

cold work. For the first specimen

For the second specimen, the deformed radius is computed

using the above equation and solving for rd as

Documents

03-BDeformasi Plastis Dan Penguatan