1.033/1.57 Q#2: Elasticity Bounds Conical Indentation Test

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1 1.033/1.57 Q#: Elasticity Bounds Conical Indentation Test November 14, 003 MIT 1.033/1.57 Fall 003 Instructor: Franz-Josef UM Instrumented nano-indentation is a new technique in materials science and engineering to determine material properties at very fine scales. A typical indentation test is composed of a loading and an unloading part. The loading part is used to extract strength properties, the unloading portion is used to extract the elasticity properties of the indented material. We have already studied the link between strength properties and hardness measurements for a flat indenter (Homework Set #) and for a conical indenter (Quiz #1). This exercise deals with the link between elastic properties and the indentation result upon unloading. An indentation test is a surface test, but its effect is felt in a bulk volume of characteristic size around the indentation cone. Within this bulk zone the material undergoes deformation as a consequence of the indentiation, while the material situated outside this zone will not feel the localized indentation (no deformation). The focus of this exercise is to estimate this characteristic size associated with the elastic unloading by means of the upper bound displacement approach. To this end, we consider a rigid conical indenter of half-apex angle α, which during the loading phase has penetrated into the material to an indentation depth h (see figure 1 TOP). At this stage, we consider an infinitesimal unloading s << h. The slope of the unloading, df /ds > 0 (see figure 1. BOTTOM), is related to the stiffness properties of the indented material by: df ds = AM (1) π where A = πr is the projected contact area at the surface z =0 (see figure), and M is the indentation modulus. In this exercise, we will first establish an upper bound for the relation between M and the elasticity constants of a homogeneous, linear elastic isotropic material, characterized by the amé constants λ, G (= const). Then, we will use this solution to evaluate the characteristic length scale. Throughout this exercise we will assume quasi-static conditions (inertia effects neglected), and we will neglect body forces. NOTE: PEASE HAND IN THE FIGURE BEOW WITH THE REQUESTED DRAWINGS TOGETHER WITH YOUR SOUTION. THANKS & GOOD UCK.

2 November 14, 003 page 1. Kinematically Admissible Displacement Field: For purpose of analysis, [ we consider two subdomains of finite size, noted respectively Ω 1 = πr and Ω = πr (1 + /R) 1 (see figure 1). Both domains are assumed to extend by a length in both the vertical and the horizontal direction. The displacement is assumed to be non-zero only within Ω 1 and Ω. The displacement in the rest of the halfspace, i.e. situated outside Ω=Ω 1 Ω,is assumed to be zero. In Ω 1 and Ω, we consider vertical displacement fields of the form: in Ω i : = f i (z, r) e z ; i = 1, () i (a) For both domains, Ω 1 and Ω, and their common interface, specify the conditions which functions f i (z, r) (i = 1, ) need to satisfy, so that is kinematically admissible. HINT: The indenter is a rigid cone, and the test is displacement driven. (b) We consider a linear form of the displacement in Ω 1 and Ω, that is: z h (1 r/r) 1 = f 1 (z, r) e z = s 1 e z (3) r R = f (z, r) e z = s (1 z/) 1 e z (4) Sketch this displacement field in Figure 1, and show that it is kinematically admissible in Ω =Ω 1 Ω. (c) Determine the non-zero strain components, in Ω 1 and Ω. HINT: Go back to the geometrical interpretation of the linear strain components. By means of very precise mechanics argument, show that the displacement field is NOT the linear elastic solution (ξ, σ) of the problem.. Potential Energy: We will employ the upper bound energy method, E pot (ξ) E pot ( ) [notation as in lecture notes]: (a) For the displacement driven indentation test, show that the potential energy of the solution of the problem reads E pot (ξ) = 1 df s. Display this result in the force ds displacement curve in Figure 1 (BOTTOM). (b) Using the displacement field i defined by (3) and (4), show that the upper bound of the potential energy E pot ( ) reads: [ ] E pot ( )=πr s 1 (5λ +1G)+(λ +4G)(/R)+6G (h/r) (5) 1 (Hint: If you run out of time, just continue working in what follows with the result (5). (c) With the results of a. and b. develop an upper bound M + for the indentation modulus: M M + (λ, G, α, R/) (6) (d) Finally, the exact solution for the indentation stiffness is M = E/ (1 ν ).From the previous results, determine an upper bound for the characteristic length scale /R of the indentation test, and determine /R for α = π/4 and ν = 0.

3 November 14, 003 page 3 Rigid Indenter F h α R s r ξ =0 Ω Ω Ω 1 ξ =0 ξ =0 z )RUFH>µ1 1@ max F oading Unloading df dh 000 ds 1 0 s <<h ,QGHQWDWRQ'HSWK>QP P@ Figure 1: TOP: Conical indentation test and domains considered in the exercise. BOTTOM: Indentation Test Results: oading and Unloading. In this exercise we are interested only in the linear elastic unloading part of the problem. 3

4 November 14, 003 page Kinematically Admissible Displacement Field A kinematically admissible displacement field is a displacement field which satisfies the displacment boundary conditions on Ω ξ d. For a rigid conical indenter, the displacement boundary is situated (a) along the indenter-material interface, along which: on Ω ξ d (z = h (1 r/r); r [0,R])) : ξ d = se z and (b) at the boundary of the considered bulk domain, where ξ d =0,inparticular: The given displacement field, satisfies these boundary condition, i.e. on Ω ξ d (z = h (1 r/r); r [0,R])) : ξ d =0 z h (1 r/r) 1 = s 1 e z = f 1 (z, r) e z r R = s (1 z/) 1 e z = f (z, r) e z 1 (z = h (1 r/r); r [0,R])) = se z 1 (z = h (1 r/r); r [0,R])) = 0 (z = ) = 0 (r = R + ) = 0 Furthermore, the displacement field is continuous at r = R: (r = R) ξ 1 (r = R) =0 That is, the chosen displacement field is kinematically admissible. For a displacement field of the form = f z (z, r) e z, the non-zero strains are: f z (z, r) ε zz = z 1 f z (z, r) 1 ε zr = = θ (e z, e r )[half distortion] r Thus the strains and the strain invariants read: In domain Ω 1 : [ ] s 1 z h h r sh ε zz = ; ε = s 1 + = zr r R R ( ( s s h ) s ) 1 h trε = ;tr(ε ε )= 1+ = 1+ R R 4

5 November 14, 003 page 5 In domain Ω : z r R s r R ε zz = s 1 1 = 1 z 1 z r R s z ε zr = s = 1 z [ s ( r R ) s r R ] 1 z trε = 1 ;tr(ε ε )= Note that ε zz is continuous over the surface r = R, while ε zr is not. The associated stress, σ =Gε is not continuous over r = R, as it would be required if zr zr was the solution of the elasticity problem (i.e. T (n = e r )+T(n = e r )=0 [[σ zr ]] = 0). This shows that is not the solution of the problem, but just an approximation, appropriate to be used in an upper bound solution procedure. 0. Potential Energy We first need to evaluate the potential energy of the solution, which is conveniently evaluated from Clapeyron s formula: 1 E com (σ) =E pot (ξ) = [Φ (σ) Φ(ξ)] In this displacement driven test, Φ(ξ) =0; and the only contribution to the solution potential energy results from the displacement boundary condition: 1 1 E com (σ) =E pot (ξ) = Φ (σ) = ξ d T (n) da = 1 Fs Ω ξ d Next, we will address the potential energy associated with : E pot (ξ) E pot ( )=W ( ) Φ( ) In the same way as for the solution, Φ( )=0in this displacement driven test so that the potential energy can be entirely evaluated from the internal free energy in Ω 1 and Ω : Thus, E pot ( )=W Ω1 ( 1)+W Ω ( ) In domain Ω 1 : [ ] λ 1 h W Ω1 ( 1 )= (trε ) + Gtr (ε ε s λ ) dω =Ω 1 + G 1+ Ω 1 R where Ω 1 is +h(1 r/r) Ω 1 =π dz h(1 r/r) rdr = πr 5

6 November 14, 003 page 6 In domain Ω : [ )] ( s λ r R ( r R ) 1 z W Ω ( ) = 1 + G dω Ω [ z= ( s r=r+ λ r R ) ] G z = π + G rdrdz z=0 r=r [ ] s λ G 1 1 = π + G (4R + )+ + R 1 3 [ ] s λ 1 R G 1 R = π 3 + G Hence, 1 h λ 1 G 1 E pot ( ) = πr s λ + G G R 6 R 3 R [ ] = πr s 1 10λ + λ (/R)+4G +6G (h/r) +4G (/R) 1 [ ] = πr s 1 λ (10 + (/R)) + G (h/r) +(/R) 1 This potential energy is greater or equal than the solution potential energy: E pot (ξ) E pot ( ) 1 1 df [ ] Fs = s πr s 1 λ (10 + (/R)) + G (h/r) +(/R) ds 1 The previous inequality allows us to determine an upper bound for the indentation stiffness, + [ ( ) ( )] df df 1 R R R = πr λ G 1 +3 (h/r) + ds ds 6 Remarks: 3. If we use the solution, df ds = AM =RM π we obtain the sought upper bound of the indentation stiffness: [ ( ) ( )] π R R R M M + = λ G 1 +3 (h/r) Using M = E/ (1 ν ), we obtain an upper bound for /R: π 14ν 6ν +1+3(h/R) (ν 3ν +1) R 1 π 3πν 4ν +5πν 6

7 November 14, 003 page 7 For ν = 0,we have: R π ( h R ) 1 π That is for h/r = 0...1, /R It is also useful to note that /R for ν = (which clearly shows some limitations of the solution). 5. It would now be appropriate to complement the upper bound approach with a lower bound solution. 7