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7/28/2019 Plasticity - 2013
http://slidepdf.com/reader/full/plasticity-2013 1/25
PLASTICITY
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Dislocations and Materials Classes
• Covalent Ceramics (Si, diamond): Motion hard.-directional (angular) bonding
• Ionic Ceramics (NaCl): Motion hard.
-need to avoid ++ and - -neighbors.
+ + + + + + +
+ + + +
- - - - - - -
- - -
• Metals: Disl. motion easier. -non-directional bonding-close-packed directions
for slip. electron cloud ion cores
+ +
+ +
+ + + + + + + + + + + + +
+ + + + + + +
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Dislocation MotionDislocations and plastic deformation• Cubic & hexagonal metals - plastic
deformation by plastic shear or slip whereone plane of atoms slides over adjacentplane by defect motion (dislocations).
• If dislocations don't move,deformation doesn't occur!
Adapted from Fig. 7.1,
Callister 7e.
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Dislocation Motion• Dislocation moves along slip plane in slip
direction perpendicular to dislocation line• Slip direction same direction as Burgers
vector
Edge dislocation
Screw dislocation
Adapted from Fig. 7.2,Callister 7e.
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If a material is subjected to a load of sufficient magnitude it shows permanent(irrecoverable) deformation.It is result of the permanent displacement of atoms and molecules from their originalposition.
If the deformation is continuously increasingthen the phenemenon is called FLOW.
Slip plane
P
x’
x
P P n
P s
Slipdirection
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Slip System: Slip plane anddirection• Slip plane - plane allowing easiest
slippage- Highest planar densities
• Slip direction - direction of movement - Highest linear densities
Deformation Mechanisms
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FCC
FCC Slip occurs on {111} planes (close-packed) in<110> directions (close-packed)
=> total of 12 slip systems in FCC
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(111) (111) (111) (111)
(111)
Parallel
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(111), [101] or [101]--- 1
(111), [110] or [110]--- 2
(111), [011] or [011]--- 3
1
2
3
So, for (111) plane:
Therefore, for an FCC structure:{111} - <110> , there are 12 slip systems.
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To sum up:
Slip phenemenon is used to explain the plasticbehaviour of materials.Slip occurs along certain crystal planes anddirections.Slip planes & slip directions make slip systems.For an FCC structure {111} - <110> , there are 12slip systems.
For an BCC structure {110} - <111> → 12 slipsystems.For an HCP structure → 3 slip systems.
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The stress that initiates slip is known as thecritical resolved shear stress.
Ø: Angle between thenormal to the slip planeand the applied stressdirections.
λ: Angle between the slipand stress directions.
A s
A
F
τ R = σ . cos λ . cos Ø
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Stress and Dislocation Motion• Crystals slip due to a resolved shear stress, t R .
• Applied tension can produce such a stress.
slip plane
normal, ns
Resolved shear stress: t R
= F s / A s
AS
tR
t R
F S
Relation betweens and t R
t R = F S / AS
F cos l A /cos f
l F
F S
f nS
AS
A
Applied tensilestress: = F / A s
F
A
F
flst coscosR
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Fs= F cos λ (shear force along the slip direction)
τR =
σcos
λcos Ø
A s = A / cos Ø (shearing area)
cosλ
. cos ØF
A A / cos ØF cos λ Fs
A s τ
R = = =
σ
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Single Crystal Slip
Adapted from Fig. 7.8, Callister 7e.
Adapted from Fig.7.9, Callister 7e.
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Ex: Deformation of single crystal
So the applied stress of 65 MPa will not
cause the crystal to yield.
MPa65
coscos
s
f l s t
l =35 ° f =60 °
MPa30 MPa62.26
)41.0(MPa)65(
)60)(cos35cos(MPa)65(
crsst t
t
t crss = 30 MPa
a) Will the single crystal yield?
b) If not, what stress is needed?
s = 65 MPa
Adapted fromFig. 7.7,Callister 7e.
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Ex: Deformation of single crystal
MPa25.7341.0
MPa30coscos
crss
f l t
s y
What stress is necessary (i.e., what is the yieldstress, s y )? (!We will learn about yield stress later on!)
)41.0(coscosMPa30crss y y s f l s t
MPa25.73 ys s
So for deformation to occur the applied stress mustbe greater than or equal to the yield stress
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Ex: Consider a single crystal of BCC ironoriented such that a tensile stress is appliedalong [010] direction.
a) Compute the resolved shear stress along a(110) plane and in a [111] direction when atensile stress of 52 MPa is applied.
b) If slip occurs for the above plane and
direction, and the critical resolved shearstress is 30 MPa, calculate the magnitude of the applied tensile stress necessary toinitiate yielding.
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a) Ø=45 ° angle b/w normal[110] & stress [010]
σ= 52MPa(110)
[111]z
y x
σ= 52
cos Ø =1*0 + 1*1 + 0*0
(1 2+1 2+0 2) (0 2+1 2+0 2)= 1
√2 Ø = 45 °
λ: angle b/w slip direction [111] & stress direction [010]
cos λ = -1*0 + 1*1 + 1*0(1 2+1 2+1 2) (1)
= 1√3 λ = 54.7 °
τ
R = 52 * cos 45 * cos 54.7 = 21.3 MPa
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b) σy = τ CR
cos Ø cos λ = 30
cos 45 cos 54.7
σy = 73.4 MPa
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• Stronger - grain boundariespin deformations
• Slip planes & directions (l , f ) change from onecrystal to another.
• t R will vary from onecrystal to another.
• The crystal with the largest t R yields first.
• Other (less favorably oriented) crystalsyield later.
Adapted from Fig.7.10, Callister 7e. (Fig. 7.10 iscourtesy of C.
Brady, NationalBureau of Standards [now theNational Institute of Standards andTechnology,Gaithersburg, MD].)
Slip Motion in Polycrystalss
300 mm
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• Condition for dislocation motion: CRSStt R
• Crystal orientation can make it easy or hard to move dislocation
10 -4 GPa to 10 -2 GPa
typically
flst coscosR
Critical Resolved Shear Stress
t maximum at l = f = 45º
t R = 0
l =90°
s
t R = s /2
l =45° f =45°
s
t R = 0
f =90°
s
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DISLOCATIONS&
σ - ε CURVES
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I IIIII
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Elastic Region: σy is the stress required to startplastic deformation. It is called the yield stress.From σy & on → plastic deformation starts.
Stage I: dσ /d ε ≈ 0 (work hardening rate) is
low because only the primary slip systems areactive. The slip planes are parallel to eachother and only these parallel planes will slipand they do not intersect themselves, i.e.dislocations are moving along parallel planes.
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Stage II: Other dislocations will start to movealong intersecting planes (more than one slip
system becomes active). Therefore they formbarriers to one another’s motion. It becomesharder to further deform the material. This
stage is known as Work Hardening Stage .
Stage III: The geometry of the planes have
so changed that the planes will acceleratelyslip and failure will occur.