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Nanoindentation, spherical tip, yield strength estimation
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ON THE MEASUREMENT OF YIELD STRENGTH BY SPHERICAL INDENTATION
E. G. Herbert 1, W. C. Oliver 1, and G. M. Pharr 2
1 University of Tennessee, Dept. of Materials Science and Engineering; & MTS Nano Instruments Innovation Center
2 University of Tennessee, Dept. of Materials Science and Engineering; & Oak Ridge National Laboratory, Metals and Ceramics Division
CONTEMPORARY INVESTIGATIONS
• 1995 Field and Swain
1996 Y d Bl h d• 1996 Yu and Blanchard
• 1998 Taljet, Zacharia, and Kosel
• 2001 Schwarzer et al.How well do they work?
• Test material: Al 6061‐T6• 2002 Durst, Goken, and Pharr
• 2003 Ma, Ong, Lu, and He
• 2004 Cao and Lu
• Test material: Al 6061 T6
• Diamond sphere, measured radius of 385 nm
( ll h i ll• 2004 Cao and Lu
• 2004 Mulford, Asaro, and Sebring
• 2004 Kogut and Komvopoulos
(smallest sphere commercially available)
• 2004 Lee and Lee
• 2004 Kwon et al.
UNIAXIAL TENSILE MEASUREMENTS
400
l
300
350
MPa
)
Al 6061‐T6
200
250
tres
s (M
E = 72.59 GPa ± 2.54%(confirmed ultrasonically)
273 MP ± 0 70%
100
150
True
St = 273 MPa ± 0.70%yσ
Literature Values:E = 69 GPa
0
50
0 0 02 0 04 0 06 0 08 0 1 0 12
T E 69 GPa= 275 MPayσ
0 0.02 0.04 0.06 0.08 0.1 0.12True Strain (-)
UNIAXIAL TENSILE MEASUREMENTS
400
l
300
350
MPa
)
Al 6061‐T6
200
250
tres
s (M
nkεσ =Power law fit:
100
150
True
St 093.00.432 εσ =
0
50
0 0 02 0 04 0 06 0 08 0 1 0 12
T
0 0.02 0.04 0.06 0.08 0.1 0.12True Strain (-)
CONFIRMATION OF MEASURED TIP RADIUS
3 280
0 20 40 60 80 100M
Contact Depth (nm)
2 4
2.8
3.2
60
70
Modulus from unload,(mN
)M
odulus
1.6
2
2.4
40
50R = 385 nm
Sam
ple
s of ElaFused silica
0.8
1.2
6
20
30
oad
on S
asticity (
0
0.4
0
10
0 20 40 60 80 100 120 140 160
Experimental DataHertz Theory, R = 385 nm
Lo(G
Pa)
0 20 40 60 80 100 120 140 160Displacement Into Surface (nm)
INDENTATION DATA
0.5N
)
0 3
0.4
ple
(mN
Controlled loading,
Al 6061‐T6
0 2
0.3
n Sa
mp g,
P/P = 0.05 s‐1.
0.1
0.2
oad
On
00 20 40 60 80 100
Lo
0 20 40 60 80 100Displacement Into Surface (nm)
EFFECTS OF SURFACE ROUGHNESS
Mechanically polished Al 6061‐T6Al 6061 T6
EFFECTS OF SURFACE ROUGHNESS
Mechanically polished Al 6061‐T6Al 6061 T6
Prevents us from accuratelydetermining small stressesgand strains.
Ma et al., J. Appl. Phys. 94, 1 (2003)
X
m hhPP ⎟⎟⎠
⎞⎜⎜⎝
⎛= ⎟⎟
⎠
⎞⎜⎜⎝
⎛Φ=
Rh
EEn
EERP miyH
Hm ,,,2
σ⎟⎟⎠
⎞⎜⎜⎝
⎛=
Rh
EEn
EX miyH
H ,,,σ
ψmh ⎠⎝ ⎠⎝ REEER ⎠⎝
Ma et al., J. Appl. Phys. 94, 1 (2003)
1.00 < X < 1.36
0 04
hm = 0.05R = 19.25 nm
hm = 0.025R = 9.63 nm
hm = 0.01R = 3.85 nm
0.001 < Pm / (ER2) < 0.01
Pm / (ER2):0 03
0.035
0.04
mN
) P = 0.03666(h/hm)1.8417
P = 0.0097747(h/h )1.7786
0.01R→ 0.0002110.025R→ 0.0009080.05R→ 0.003407
0 02
0.025
0.03
ampl
e (m (
m)
P = 2.2654E-3(h/hm)1.2192
0 01
0.015
0.02
d O
n Sa
This isn’t working!Presumably due to
0
0.005
0.01
Loa Presumably due to
roughness and contaminants on thesurface.0
0 0.2 0.4 0.6 0.8 1h/h
m (-)
Cao and Lu, Acta Materialia 52, (2004)n
fy
yE
⎟⎟⎠
⎞⎜⎜⎝
⎛+= εσ
σσ 1 ⎥⎦
⎤⎢⎣
⎡+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛= 43
22
31
2 lnlnln CECECEChPr
r
r
r
r
rgrg σσσ
σ
2
106.1435.000939.0 ⎟⎟⎞
⎜⎜⎛
−+=Rh
Rh gg
rε
n
⎟⎠
⎜⎝ RRr
n
ry
yrE
⎥⎥⎦
⎤
⎢⎢⎣
⎡+= 1,1, 1 εσ
σσ
n
ry
yrE
⎥⎥⎦
⎤
⎢⎢⎣
⎡+= 2,2, 1 εσ
σσy ⎦⎣
Cao and Lu, Acta Materialia 52, (2004)
⎥⎦
⎤⎢⎣
⎡+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛= 43
22
31
2 lnlnln CECECEChPr
r
r
r
r
rgrg σσσ
σn
fy
yE
⎟⎟⎠
⎞⎜⎜⎝
⎛+= εσ
σσ 1
hg, 1 = ~ 0.01R = 4.2 nm
P 1 = 2 18 µN
2
106.1435.000939.0 ⎟⎟⎞
⎜⎜⎛
−+=Rh
Rh gg
rεPg, 1 2.18 µN
hg, 2 = ~ 0.06R = 23.3 nm
P 52 98 N n
⎟⎠
⎜⎝ RRr
Pg, 2 = 52.98 µN
= 4.24E+7 Pa1,rσSolution does notconverge for 0 ≤ n ≤ 1
n
ry
yrE
⎥⎥⎦
⎤
⎢⎢⎣
⎡+= 1,1, 1 εσ
σσ
= 2.98E+8 Pa
= 0.0141
2,rσ
1,rε
0 ≤ n ≤ 1Presumably due toroughness and contaminants on the
n
ry
yrE
⎥⎥⎦
⎤
⎢⎢⎣
⎡+= 2,2, 1 εσ
σσ 0.0141
= 0.0316
,
2,rεcontaminants on thesurface.
y ⎦⎣
Yu & Blanchard, J. Mater. Res. 11, 9 (1996)
⎟⎠⎞
⎜⎝⎛ −=
RaC Ra 4921.0845.2, λ( )
⎪
⎪⎨
⎧≤≤
=arb
ar-p
rpm for 1
23
2
2
⎠⎝ R( )⎪⎩
⎨≤≤ brC yRa 0 for , σ
⎧
Yu & Blanchard, J. Mater. Res. 11, 9 (1996)
⎪⎪⎪⎪⎧
≤−
−<
−Ra
RaE
RaE y
y
14921.0845.2
)1(2
0 for )1(3
4 2
2
συπσυπ
⎪⎪⎪
⎪⎪⎪
⎨
∞<−
<⎟⎟⎟⎟⎞
⎜⎜⎜⎜⎛
⎟⎠⎞
⎜⎝⎛ −
⎟⎠⎞
⎜⎝⎛ −
=
RaE
Ra
a
RH
y
y
)1(2
1for 4921.0845.2
-14921.0845.2 2
2
2
συπσ
⎪⎪⎪⎪
⎩
−⎟⎟⎟
⎠⎜⎜⎜
⎝⎟⎟⎠
⎞⎜⎜⎝
⎛
−
⎟⎠
⎜⎝
Ra
RaER
y
4921.0845.2)1(
232
2 συπ
( ) ( ) ( )( )Ea
RaRaR
H
y
yyy
yy
2222242
2222222
021.520299.0
13459.011730.094.18 σ
σσυσυ
συσυ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−+−
+−−−−−
=REa
H 22=
Apply P, measure h and S, assume E to get A, @ h = 20 nm (full contact), H = 1186.12 MPa
= 424 MPa, relative error = 55.3%yσ
Field and Swain, J. Mater. Res. 10, 1 (1995)2
P ca'20hrh
rte hhh −=2e
rbh
hh +=
2
⎟⎠⎞
⎜⎝⎛
+−
=1412
252
nnc
PdP 3
32
⎟⎟⎠
⎞⎜⎜⎝
⎛=
s
t
PP
r
2)'( 8.2 caP
r πσ =
Rca
r 2.0=ε1−−
=r
hrhh ts
r
22' bb hRha −=ee h
PdhdP
23
=
Pt and ht
P and hPs and hs
Field and Swain (1993)
Field and Swain, J. Mater. Res. 10, 1 (1995)
0.50 20 40 60 80 100Al 6061-T6
0.4Field & SwainP/P
(mN
) .
0.3
Sam
ple
peak load
0.1
0.2
oad
on S
0
0.1
0 20 40 60 80 100
Lo 50% unloaded
0 20 40 60 80 100Displacement Into Surface (nm)
Field and Swain, J. Mater. Res. 10, 1 (1995)
• slope = Meyer’s index, m+ 2-0 6
Al 6061‐T6
•m + 2 = nwhere• n = 0.5491
il d 0 0931
-0.8
0.6
Y = ‐6.7719 + 2.5491X R = 0.99937nkεσ =
• n, tensile data = 0.093
•-1.2
-1
og (P
)
h = 100 nm ⎟⎟⎠
⎞⎜⎜⎝
⎛+−
=1412
252
nnc
• c = 0.893 → sink‐in-1.6
-1.4Lo ⎠⎝ +142 n
• The stiffness equation:
-2
-1.8
1 9 2 2 1 2 2 2 3 2 4 ASEr 2
π=1.9 2 2.1 2.2 2.3 2.4
Log (a')A2
Field and Swain, J. Mater. Res. 10, 1 (1995)⎞⎛2
P ca'20hrh
rte hhh −=2e
rbh
hh +=
2
⎟⎠⎞
⎜⎝⎛
+−
=1412
252
nnc
PdP 3
32
⎟⎟⎠
⎞⎜⎜⎝
⎛=
s
t
PP
r ASEr 2
π=
2)'(caP
r πσ =
Rca
r 2.0=ε1−−
=r
hrhh ts
r
22' bb hRha −=ee h
PdhdPS
23
==
Pt and ht
P and hPs and hs
FIELD AND SWAIN ‐ TABOR: TENSILE VS. IIT
8000 0.03 0.06 0.09 0.12 0.15 0.18
0.2 a(Es=72.59GPa)/R (-)
600
700
800
Pa) 135.02.651 εσ
εσ
=
= nkAl 6061‐T6
400
500
600
ess
(MP 093.00.432 εσ =
h = 100 nm
200
300
400
rue
Stre
0
100
200 Tensile dataF and S, (P
m via A(E
s=72.59GPa))/2.8
Tr
00 0.03 0.06 0.09 0.12 0.15 0.18
True Strain (-)
FIELD AND SWAIN ‐ TABOR: TENSILE VS. IIT
8000 0.03 0.06 0.09 0.12 0.15 0.18
0.2 a(Es=72.59GPa)/R (-)
600
700
800
Pa)
Al 6061‐T6135.02.651 εσ
εσ
=
= nk
400
500
600
ess
(MP 093.00.432 εσ =
200
300
400
rue
Stre
h = 100 nm
0
100
200 Tensile dataF and S, (P
m via A(E
s=72.59GPa))/2.8
Tr
00 0.03 0.06 0.09 0.12 0.15 0.18
True Strain (-)
P/P ‐ TABOR: TENSILE VS. IIT.
8000 0.03 0.06 0.09 0.12 0.15 0.18
0.2 a(Es=72.59GPa)/R (-)
600
700
800
Pa)
h = 17 nm εσ = nk
h = 100 nm
400
500
600
ess
(MP h 17 nm
093.0
189.0
04320.742εσ
εσ
εσ
=
=
= k
200
300
400
Tensile datarue
Stre 0.432 εσ =
0
100
200 Tensile dataP/P, (P
m via A(E
s=72.59GPa))/2.8
Tr .
00 0.03 0.06 0.09 0.12 0.15 0.18
True Strain (-)
PREVIOUS OBSERVATIONS: ISE
Indenters:Spherical indentation of Iridium
Swadener et al., J. Mech. Phys. Solids 50 (2002)
• sapphire spheres,R = 69,122,318 mm
• diamond “sphere”,3.00 0.01 0.02 0.03 0.04
Effective strain, 0.2a/R
R = 14 mm• hardened steel ball,R = 1600 mm2.0
2.5
(GP
a)
14 μm
122
69 μm
Features:• H and 3s similarfor large spheres1.0
1.5
ardn
ess
( 122 μm1600 μm
318μm g p• H increases as R decreases
• increase in H with0 0
0.5
Ha
3σ ρG =1
bR
a/R parallels workhardening
0.00 0.05 0.1 0.15 0.2
a/R
P/P ‐ TABOR: TENSILE VS. IIT.
6000 0.03 0.06 0.09 0.12 0.15 0.18
0.2 a(Es=72.59GPa)/R (-)
εσ = nk
500
600
Pa)
h = 100 nm
093.0
189.0
0.4322.561εσ
εσ
=
=h = 17 nm
300
400
ess
(MP 03 εσ
RE in k = 30%
RE in n = 103%
200
300
Tensile dataP/P, (P
m via A(E
s=72.59GPa))/3.7ru
e St
re
.⎟⎟⎠
⎞⎜⎜⎝
⎛=
n
yEkσ
σ
0
100Tr
Al 6061‐T6 MPa 8.180=∴
⎟⎠
⎜⎝
y
y
σ
σ
RE i 34%σ00 0.03 0.06 0.09 0.12 0.15 0.18
True Strain (-)RE in = ‐34%yσ
SUMMARY AND CONCLUSIONS
1. Assumption of a perfect sphere.
b d h d ld b ll l d2. FEA based methods could not be experimentally implemented: proscribed depths < roughness and/or contaminants. Techniques are not ideally suited to investigating volumes of material that require small spheres.
3. Yu’s theoretical pressure distribution overestimated σy by 55%.
4. Field & Swain’s procedure overestimated the plastic flow curve by ~40%.
5. Work‐hardening from the mechanical polishing cannot fully account for these discrepancies.
6. One possible explanation: Indentation Size Effect.