Material characterization by indentation

Erland NordinProject Work in Contact Mechanics

2013-02-08

1 Introduction

Contact between a sphere and a plane is a classical and relatively simple prob-lem due to the solutions by Hertz, published 1882. There are four simplifyingassumptions: [1]• The surfaces must be continuous and non-conforming.• The strains must be small.• Each solid could be considered as elastic half-spaces.• The surfaces should be frictionless.The first three assumptions all means that the contact radius should be smallin comparison with the other dimensions of the problem, particularly the radiusof the sphere. Considering the problem of a rigid sphere indenting an elastichalf-space the contact radius can be calculated with equation 1.

a =(

3PR4E∗

) 13

(1)

Here P is the force pressing the sphere into the elastic plane, R the radius ofthe sphere and E∗ is the effective elastic modulus (se equ. 4) of the half-space.Equation 2 calculates the indentation depth δ for a rigid ball against an elastichalf-space. For two elastic bodies δ will be the approach of distant points oneach body.

δ =a2

R=(

9P 2

16RE∗2

) 13

(2)

The maximum contact pressure is denoted p0 (equ. 3) and is related to themean contact pressure by pm = 2

3p0 = Pπa2 .

p0 =3P

2πa2=(

6PE∗2

π3R2

) 13

(3)

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If the indenting ball is not rigid but elastic the effective elastic modulus iscalculated from the two bodies elastic properties using equation 4. For the casewhen the ball is rigid E1 or E2 is infinite and the corresponding term cancel.

1E∗

=1− ν2

1

E1+

1− ν22

E2(4)

If two spheres are pressed together an equivalent radius is calculated with equa-tion 5. The case of a rigid sphere against an elastic half-space considered aboveis a special case where the radius of the half-plane is infinite and the equivalentradius is thus the same as the radius of the rigid sphere.

1R

=1R1

+1R2

(5)

The surface displacements and stresses due to the Hertzian pressure canbe calculated when the maximum pressure p0 and contact radius a has beencalculated using the previous equations.The displacements in radial direction iscalculated by equation 6.

ur(r) = − (1− 2ν)(1 + ν)3E

a2

rp0

[1− (1− r2

a2)

32

]r ≤ a (6)

= − (1− 2ν)(1 + ν)3E

p0a2

rr > a

Here E and ν are the elastic properties for the body of interest. The radius ris the distance from the center of the contact to the point of interest along thesurface. The normal displacements is calculated with equation 7.

uz(r) =1− ν2

E

πp0

4a(2a2 − r2) r ≤ a (7)

=1− ν2

E

p0

2a

[(2a2 − r2) arcsin(

a

r) + r2

a

r

(1− a2

r2

) 12]

r > a

The stresses on the surface is first divided into points lying inside the contactradius:

σrp0

=1− 2ν

3

(a2

r2

)[1− (1− r2

a2)

32

]−(

1− r2

a2

) 12

(8a)

σθp0

= −1− 2ν3

(a2

r2

)[1− (1− r2

a2)

32

]− 2ν

(1− r2

a2

) 12

(8b)

σzp0

= −(

1− r2

a2

) 12

(8c)

The last equation, 8c, is of course the Hertzian pressure but with a minus sign.Outside the contact zone, but still restricted to the surface, the stresses are:

σrp0

= − σθp0

= (1− 2ν)a2

3r2(9)

σz = 0

2

Inside the material, beneath the surface, the stress equations are restricted tobe along the centerline beneath the contact. Letting the z -axis start at thecenter of the contact at the surface and be directed into the body of interrestthe stresses are:

σrp0

=σθp0

= −(1 + ν)[1− z

aarctan(

a

z)]

+12

(1 +

z2

a2

)−1

(10a)

σzp0

= −(

1 +z2

a2

)−1

(10b)

The σr, σθ and σz stresses along the z -axis are principal stresses. The principalshear stress τ1 can therefore be calculated as

τ1 =12|σz − σθ| (11)

Similarly the octahedral shear stress is calculated with equation 12 and

τoct =13[(σr − σθ)2 + (σθ − σz)2 + (σr − σz)2

] 12 (12)

= {σr = σθ} =13[(σθ − σz)2 + (σr − σz)2

] 12

the von Mises equivalent stress is then according to equation 13.

σe =1√2

[(σr − σθ)2 + (σθ − σz)2 + (σr − σz)2

] 12 (13)

= {σr = σθ} =1√2

[(σθ − σz)2 + (σr − σz)2

] 12

Figure 1 shows how the surface stresses look if the stresses are normalized withthe maximum pressure p0 and the radial distance is normalized with the contactradius a. The only tensile stress is the radial stress sr outside the contact. Figure2 shows the corresponding stresses along the centerline beneath the contact.Assuming Poisson’s ratio ν = 0.3 the maximum shear stress will be τ1,max =0.31p0, the maximum octrahedral shear stress is τoct = 0.29p0 and maximumvon Mises is σe = 0.62p0. All occurs at a depth of z = 0.48a.Therefore initialplasticity will start at a depth of about half the contact radius under the surface.

Figure 3 shows a comparison of the stresses and displacements between aload that is just below start of plasticity (elastic region) with a load after plas-ticity had started (corresponding to a max equivalent plastic strain of 0.5%).Figure 4 shows the amount of equivalent plastic strain corresponding to thechosen plastic region. The plastic deformation has not reached the surface andis still constrained by elastic material. Although the relatively large plasticdeformation beneath the surface the difference in surface displacements for acompletely elastic deformation at the same load is very small. Figure 5 showsthe force-indentation depth response from an finite element model and analyti-cally. There is practically no difference in measured force between the elastic and

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Figure 1: Normalized stresses on surface. Radial stress (sr), circumferentialstress (sth) and surface normal stress (sz).

Figure 2: Normalized stresses along centerline beneath centrum of contact. Ra-dial stress (sr), circumferential stress (sth), stress along centeraxis beneath con-tact (sz), maximum shear stress (tau1), octahedral shear stress (octShear) andvon Mises equivalent stress.

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plastic solution below 3–4 times the indentation depth that cause first yielding.Figure 6 shows the mean pressure normalized with yield stress plotted versus theJohnson parameter. The Johnson parameter is indentation size normalized withball radius times the ratio of elastic modulus to the yield stress. It can also herebe seen that there is very little difference a good bit above the yield point beforean elastic solution diverges from the plastic results. It is therefore very difficultto detect the yield point by practical indentation experiments. The initial sug-gestion by Hertz that the hardness should be defined as the contact pressure atinitial yield is not used today for that reason. Instead the Vickers pyramid forexample gives the hardness at approximately 8% strain and Brinell indentationsare always presented with the corresponding ball material/diameter and load.

2 Material parameters for aluminum

In this section an aluminum piece of unknown material properties is investi-gated. The goal is to determine elastic modulus and the stress-strain curveusing indentation techniques.

2.1 Vickers indentation

The Vickers indenter is a diamond pyramid which creates a self-similar stressfield. This means that the stress field looks the same regardless of the scaleof the indentation. This applies as long as the indent size is significant largerthan the grain size of the material so that discrete grains are averaged out [2].Figure 7 shows an example of a Vickers indentation on the aluminum used. Theforce used to press the diamond pyramid down into the material was 500 gf,i.e. the same force a 500 g weight would press down with due to gravity. Thetwo diagonal measurements are averaged and used to calculate the hardnessnumber with equation 14. P is the force in kgf, L is the mean of two diagonalmeasurements on the indent in mm and θ is the angle between the diamond tipsfaces (θ = 136◦). In order to check if there were any size dependences for thisaluminum several measurements at different forces were made. Figure 8 showshow different Vickers indents compare to each other when different forces areused. The measured diagonals of the indents and the corresponding hardnessis presented in Table 1. There were only natural scatter in the results and asa final Vickers hardness number for this aluminum an average was used, whichgives 189 HV.

HV =2P sin( θ2 )

L2(14)

2.2 Brinell indentation

Using the Vickers indenter can give an easily measured hardness but due tothe self-similar feature the result correspond to only one point in a stress-strain

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(a) Surface stress elastic region (b) Surface stress plastic region

(c) Bulk stress elastic region (d) Bulk stress plastic region

(e) Surface displacements elastic region (f) Surface displacements plastic region

Figure 3: Comparison of stresses and surface displacements in the elastic region(left column) and plastic region (right colummn). The plastic region had anmaximum equivalent plastic strain of 0.5%. Dashed lines are analytical resultsand point results are from a finite element model.

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Figure 4: Finite element results for equivalent plastic strain for plastic region.

Table 1: Vickers indentation with varying force. All measurements except 2 kgfwas measured with a 40x objective.

Force L1 [um] L2 [um] HV25 gf 15.72 15.13 19550 gf 22.29 22.49 185100 gf 31.84 30.85 189200 gf 44.58 43.99 189300 gf 54.53 53.54 191500 gf 70.46 69.46 1891 kgf 98.32 97.72 1932 kgf (10x objective) 143.12 142.33 182

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Figure 5: Force versus indentation depth for a rigid ball against a steel plate

Figure 6: Hardness versus Johnson parameter.

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Figure 7: Vickers indentation with 0.5 kgf force.

Figure 8: Vickers indentation with different forces. Left 2 kgf, upper middle 50gf, lower middle 25 gf and to the right 500 gf.

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curve, namely at approximately 8% strain. The Brinell indentation method usea sphere as the indent body instead of a sharp pyramid. The indent for a spherewill not be self-similar and while that might be a draw-back it can also be usedto measure the stress-strain curve at different equivalent strains. The Brinellmeasurements was made on a Wolpert machine that had an indent sphere witha diameter of 2.5 mm. Table 2 shows the Brinell measurements made alongwith the calculated Brinell hardness number, calculated by equation 15.

HB =P

πD2

(D −

√D2 − d2

) (15)

Here P is the force on the ball in kgf, D is the ball diameter in mm and d is theremaining indent diameter in mm.

Table 2: Brinell indentation with varying force.Force [kgf] d [mm] HB150 0.94 208100 0.78 20460 0.59 20940 0.53 17930 0.48 164

2.3 Approximating stress-strain curve

The stress-strain curve can be approximated from hardness values by convertinghardness to a corresponding flow stress. For a Vickers indentation the calculatedflow stress will always correspond to a strain of 8% because of the self-similarstress-field. For a Brinell (sphere) indentation a representative strain is approxi-mated which depend on the relative size of the indent. Using a power law modelfor the stress-strain relationship, (equation 16)

σ(εpl) = κεmpl (16)

Tabor [3] found that the relation,

H = C1κ(C2

a

D

)m(17)

is valid when the indentation is in the fully plastic region. Tabor determinedthe constants to:

C1 = 2.8 (18)C2 = 0.4 (19)

The exponent m and κ are material parameters. H is the hardness definedas the indentation force divided by the projected area. Therefore this is not

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the same hardness as reported from Vickers och Brinell indentations which useactual contact area. For the Brinell indentation the correct hardness value touse is:

HB =Pg

πa2(20)

and for Vicker indentation the conversion is:

HV =HV g

sin( θ2 )=

2PgL2

(21)

The conversion from kgf used in the Brinell and Vickers hardnesss to forceis performed by multiplying with a standardized gravitational constant (g =9.80665). Comparing equations 16 and 17 it is recognized that the representativestrain for a sphere indentation is approximated as:

εR = 0.4a

D(22)

and that the flow stress is approximated as:

σ =H

2.8(23)

For a Vickers indentation the representative strain is always 8% and the flowstress is also approximated with equation 23. Figure 9 shows the Brinell andVickers indents converted to corresponding strains and stresses. Equation 16 isfitted to the Brinell measurements and shown as the fitted curve. The values ofthe material parameters for this case is:

κ = 1939 MPam = 0.35

Using equation 16 outside the interval for the measured points, strains 0.04 to0.08 in this case, should of cause be made with caution.

2.4 Measuring elastic modulus

To estimate the elastic modulus a ball with 12.5 mm diameter was pressedinto the aluminum block while the force and displacement was recorded. Theball was loaded up to 1900 N, held for stable for 8 s and then unloaded at aquasi-static rate. When a ball is pressed beyond the elastic limit the pressuredistribution changes from the hertzian to a more flat distribution. The maxload is therefore chosen to give a clear and large indent. The indent mark inthis case is shown in Figure 10. The indent radius is 0.67 mm and thereforeabout 10% of the ball radius. Since the pressure distribution at max load isapproximately flat and the unloading is elastic a relation between the load anddisplacement can be calculated by the flat punch equation 24.

δ = πp0a1− ν2

E=

P

2aE∗(24)

11

Figure 9: Stress-strain points calculated from the Brinell and Vickers indenta-tions.

Solving for the effective elastic modulus, changing to an incremental version andinserting a constant that will be determined by FEM-simulations we get

E∗ =1c

1√Aproj

∆P∆δ

(25)

where Aproj = πa2 is the projected indent area. The slope of the curve atunloading is changing so ∆P/∆δ is evaluated at a chosen position, close to thestart of unloading. To determine the constant an fem-simulation of the processis made. As indenter a sphere with the same diameter and elastic properties,E = 206 GPa and ν = 0.3, as in the experiment is chosen. The target is chosento have E = 70GPa and ν = 0.3 and an ideal-plastic material with yield stress714 MPa (compare with Figure 9). No friction is included in the analysis. Figure11 shows the simulated force versus indentation depth. The thick line part isthe region used to evaluate the slope. In the simulation the contact diameter is1.267 mm at 1900 N force which is quite similar to the measured indent diameterof 1.34 mm. Using equation 25, with the known elastic modulus, the constantis determined to c = 1.18.

Figure 12 shows the measured force versus displacement. In the measure-ment the machine compliance has been accounted for so the displacement shouldcorrespond to the indentation depth. The thick red part of the line is the partwhere the slope is evaluated. To avoid some dynamic effects and letting the

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ball settle before unloading a wait time of 8 s is used before unloading starts.The force versus time procedure is shown in Figure 13. The evaluated elasticmodulus of the aluminum, assuming Poisson’s ratio to be ν = 0.3, is E = 84GPa. This is a reasonable value compared to the elastic modulus of aluminumof 70 GPa that is usually assumed.

Figure 10: Indent caused by a 12.5 mm diameter ball pressed into the aluminumwith 1900 N. The indent diameter is 1.34 mm.

3 Conclusions

In the first part of this report it was shown that the measurable properties(forces, displacements) in en indent test is very little influenced by the startif yielding. This is because the yielding starts beneath the surface and is con-strained by elastic material around it. The second part deals with how an un-known material can be characterized by indentation methods. The stress-curvecan be approximated with Brinell and Vickers indentation techniques and bothgave similar result. At 8% representative strain the flow stress was 714 MPawhich is quite large for aluminum. The elastic modulus could be reasonablewell measured with a ball indentation if force and displacement is continuouslymeasured at unloading. The resulting 84 GPa is a little bit higher than what isusually used for aluminum but still a good value for a relatively simple experi-ment.

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Figure 11: Force versus displacement for simulated ball indentation.

Figure 12: Force versus displacement for ball indenting aluminum plate.

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Figure 13: Force versus time for the ball indentation process.

References

[1] Johnson , K.L., Contact mechanics, Cambridge University Press, 1985

[2] Elmustafa, A.A., Eastman, J.A., Rittner, M.N., Weertman, J.R., Stone,D.S., Indentation size effect: Large grained aluminum versus nanocrystallinealuminum-zirconium alloys, Scripta Materialia 43 (2000), 951-955

[3] Tabor, D., The Hardness of Metals, Oxford University Press, 1951

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