Projects

(a) Pantano, et al. discuss several previously reported estimates of CNT elastic properties. Explain how the cited works in this review inferred the elastic properties of these CNTs from resonance of the structure (how did they do it, and how does resonance relate quantitatively to E?).

(b) On p. 795, the authors state that the CNT can be described with three elastic constants, but they also state that the wall material is isotropic. Isotropic materials require only 2 independent elastic constants, so demonstrate in terms of material and crystal symmetry why three are employed.

(b) Pantano, et al. extensively model the elastic buckling instabilities of CNTs, and essentially find that computationally expensive MD and FEM simulations predict the same critical loads as Timoshenko’s continuum elasticity theory for shells. In Pantano’s model of cantilevered CNT bending, they completely neglect the possibility of plastic deformation at the fixed end of the CNT cantilever. However, if the stresses get high enough in the CNT “beam”, plastic deformation will result. State the yield criteria / surface that you think best captures the onset of plastic deformation in a CNT (single or multiwalled; your choice), graph it as a yield surface under biaxial loading.

(c) Determine the deflection required to induce plastic deformation under cantilevered point loading. Be sure to clearly state the assumed fixed-end boundary conditions, and cite primary sources for any required data.

(d) Poncharal, et al. report the resonance of CNTs and demonstrate a frequency- and diameter-dependent response. Although this is structural resonance, it is similar in origin to the linear viscoelastic responses used to describe polymers. From this and other available sources, justify whether the CNT dynamics are best captured by a Maxwell, Kelvin-Voigt, or SLS model. Express the resonance of the CNTs in terms of the elements of the phenomenological spring/dashpot model you have chosen, keeping in mind that the elastic properties of the CNTs are well-known.

Plasticity and fracture of microelectronic thin films/lines
Effects of multidimensional defects on III-V semiconductor mechanics
Defect nucleation in crystalline metals
Role of water in accelerated fracture of fiber optic glass
Carbon nanotube mechanics | Problem Set 2 | Problem Set 3 | Problem Set 5
Superelastic and superplastic alloys
Mechanical behavior of a virus
Effects of radiation on mechanical behavior of crystalline materials

(b) The papers you have chosen discuss the elastic properties of CNTs. Of course, the applications people are excited about are related to the high tensile strength of these organic fibers. We know they’ll easily buckle in tension, and it’s hard to think of apps that would benefit from that. Read the 2 papers cited by Pantano, et al. that discuss nanotube fracture to review this concept.

Then compute the number of carbon bonds that must be broken to achieve tensile fracture of a single CNT (not a multiwall). It [elastic energy stored at fracture] should be the number of carbon bonds transversing the fracture plane, multiplied by the carbon-carbon bond strength, but is it?

Final Presentation

“Carbon Nanotube Mechanics.” (PDF)

(a) Pantano, et al. [1] discuss several previously reported estimates of CNT elastic properties. Explain how the cited works in this review inferred the elastic properties of these CNTs from resonance of the structure (how did they do it, and how does resonance relate quantitatively to E?).

Several different groups have performed experiments to determine elastic properties using vibration and resonance of the CNT structure:

Treacy, et al. (1996) [2] used thermally induced vibration of cantilevered multi-walled nanotubes (MWNTs) in a TEM to determine a value of Young’s modulus. Their technique involved placing MWNTs on the edge of TEM grid. With one end of the MWNT fixed on the edge of the TEM grid, the other end of the MWNT was free to move. Focusing the TEM on the free end of the tube was not possible because the tube was vibrating, due to thermal energy, in-and-out of the TEM focal plane. By measuring the width of the blurred tip, relative to the constrained end, the amplitude of the vibration could be inferred. The temperature of the MWNT was varied and a linear relationship was observed between temperature and mean-squared amplitude of the vibration. The Young’s Modulus was determined from this information.

The vibration energy, W_n, is the thermal energy in the system equal to kT, which obeys the Boltzmann distribution, and u_n is the horizontal vibration amplitude of the tip.

The effective spring constant is given by

where Y is the Young’s Modulus, L is the tube length, a and b are the outer and inner diameters of the tube respectively, and the β term is related to the vibration mode of the MWNT, which is assumed to be analogous to that of a clamped cantilever. It is given by

where ρ is the density of the CNT wall material. For CNTs at room temperature a frequency of 100 MHz can be expected. Using the above relations the Young’s Modulus was inferred to be roughly 1.8 TPa on average.

Using the same technique as Treacy, Krishnan, et al. (1998) [3] determined the Young’s modulus to be between 0.9 and 1.7 TPa for a single-walled nanotube (SWNT).

Poncharal, et al. (1999) [4] used a slightly different technique to infer the Young’s modulus. This group mounted MWNTs on a specially designed TEM grid such that they could be biased. A grounded counter electrode was mounted within 20 um of the ends of the MWNTs. The application of a bias to the nanotubes caused a deflection towards to the grounded counterelectrode. By applying a time dependent oscillating voltage to the nanotubes, the resonance frequency could be found. The frequency was evident when large deflections were observed for relatively small voltages. Using essentially the same method and governing equation as in the previous two papers,

based on the assumption that nanotubes can be modeled as resonantly excited cantilever beams, allowed the calculation of Young’s modulus. Poncharal, et al. confirmed previous measurements of modulus to be roughly 1 TPa.

(b) On p. 795 [1], the authors state that the CNT can be described with three elastic constants, but they also state that the wall material is isotropic. Isotropic materials require only 2 independent elastic constants, so demonstrate in terms of material and crystal symmetry why three are employed.

Two elastic constants are required for materials that are isotropic in space. The reference to the isotropic elastic properties of a two dimensional hexagonal structure as being isotropic indicates isotropy only in its two in-plane dimensions. Consider a graphene sheet: it is isotropic with respect to in-plane stretching and can be described fully with a stretching stiffness C and a Poisson’s ratio v, the two constants we expect for an isotropic material.

The analogy between space isotropy and in-plane isotropy dissolves when we consider applying a load normal to the plane. In a spatially isotropic material, this can be viewed as an in-plane load on a different plane, identical to the first by spatial symmetry. In a two-dimensional sheet, however, an out-of-plane load causes a fundamentally different form of deformation. Consider the example of the graphene sheet: an in-plane load causes bond stretching and rotates bond angles in plane, increasing some and decreasing others. Now, consider a hexagonal junction in the form of a Y loaded out-of-plane: all bond angles are reduced. This unique deformation is governed by a separate elastic constant D, the flexural rigidity.

All three constants are required to describe a nanotube, a rolled graphene sheet. The in-plane or membrane stretching stiffness and the Poisson’s ratio govern the behavior of the nanotube loaded axially, though this stiffness differs from the stiffness of the equivalent, unrolled sheet due to the change in geometry. The flexural rigidity governs the deflection of the nanotube. Unlike in the case of a spatially isotropic materially, it is not possible to describe this behavior in only two constants.

References

[1] Pantano, Antonio, David M. Parks, and Mary C. Boyce. “Mechanics of Deformation of Single- and Multi-wall Carbon Nanotubes.” Journal of the Mechanics and Physics of Solids 52 (April 2004): 789-821.

[2] Treacy, M. M. J., T. W. Ebbesen, and J. M. Gibson. “Exceptionally High Young’s Modulus Observed for Individual Carbon Nanotubes.” Nature 381 (June 20, 1996): 678-680.

[3] Krishnan, A., E. Dujardin, T. W. Ebbesen, P. N. Yianilos, and M. M. Treacy. “Young’s Modulus of Single-walled Nanotubes.” Physical Review B 58 (1998): 14013-14019.

[4] Poncharal, P., Z. L. Wang, D. Ugarte, and W. A. de Heer. “Electrostatic Deflections and Electromechanical Resonances of Carbon Nanotubes.” Science 283 (March 1999): 1513-1516.

Plasticity and fracture of microelectronic thin films/lines Effects of multidimensional defects on III-V semiconductor mechanics Defect nucleation in crystalline metals Role of water in accelerated fracture of fiber optic glass Carbon nanotube mechanics | Problem Set 2 | Problem Set 3 | Problem Set 5 Superelastic and superplastic alloys Mechanical behavior of a virus
Effects of radiation on mechanical behavior of crystalline materials

(b) Pantano, et al. [1] extensively model the elastic buckling instabilities of CNTs, and essentially find that computationally expensive MD and FEM simulations predict the same critical loads as Timoshenko’s continuum elasticity theory for shells. In Pantano’s model of cantilevered CNT bending, they completely neglect the possibility of plastic deformation at the fixed end of the CNT cantilever. However, if the stresses get high enough in the CNT “beam”, plastic deformation will result. State the yield criteria / surface that you think best captures the onset of plastic deformation in a CNT (single or multiwalled; your choice), graph it as a yield surface under biaxial loading.

Modeling the CNT as a cantilevered beam the maximum stress at the support due to bending can be found:

where the moment, M = [Force applied to the cantilever tip in the y-direction] x [Length of the nanotube], and y is distance from the neutral axis. The area moment of inertia for a hollow tube is:

where the D variables are the outer and inner diameter respectively. The highest stress occurs on the top and bottom surfaces of the tube (orthogonal to the direction of loading) at the support. When the stress in these regions reaches some critical value, bonds will be broken and the material will be irreversibly distorted. We propose a Tresca yield criteria to model this behavior. We are working under the assumption that graphene will respond similarly in tension and compression. Carbon nanotubes will not plastically deform, so instead of normalizing with respect to a yield stress, we will normalize with respect to tensile strength as experimentally determined in the literature. A range of values have been observed, but we will use the results from Yu, et al. who found an average tensile strength of roughly 45 GPa. [2]

The bending/buckling behavior of SWNTs can be solved accurately by numerical methods; to work by hand, we’ll have to make a number of assumptions. We will take Pantano’s effective modulus/effective thickness pair of 4.84 TPa and 0.075 nm, respectively [1]. The bending stiffness is now calculable; as an example for this problem we’ll choose an SWNT of R = 1 nm. Since the buckling behavior on the compressive side of the beam is beyond our ability to calculate, let’s consider the tensile region. The tensile stress generally experienced by a beam is

with a maximum at:

We will get plastic deformation when this stress reaches the tensile yield stress. We can rearrange to see that yielding occurs when

This gives us information regarding the critical curvature to induce yielding.

Let us specify the cantilever loading boundary conditions for a beam with an end fixed at x = 0 as follows:

These are the standard fixed-end boundary conditions. We need to find the deflection D which induces the yield stress in the beam, by finding the curvature generated by a specific displacement (specified for a general nanotube of length L, as D/L) of the free end. Working from the general case,

Given our boundary condition on the first derivative of u with respect to x taken at x= 0, we can see that this constant B is zero. Integrating again,

Given our boundary condition on u at x = 0, we see that this constant is also zero. Setting the stress equal to the yield stress and taking the value of c at R_max to find the very earliest possible yield (minimum value of deflection), we get a minimum free-end deflection of

where L is the length of the nanotube between the free and fixed ends and R is its radius, with values of E and t (effective wall thickness) from Pantano, et al. [1] as mentioned above, and using a minimum tensile yield strength of 11 GPa from Yu, et al. [2] as referenced in Pantano, et al. [1]. This is not an ideal yield stress value, as Yu, et al. focused on multiwall nanotubes; Yu, Files, Arepelli and Ruoff report breaking strengths for SWNTs between 13 and 52 GPa, with a mean of 30 [3], and these values may be more appropriate.

(d) Poncharal, et al. [4] report the resonance of CNTs and demonstrate a frequency- and diameter-dependent response. Although this is structural resonance, it is similar in origin to the linear viscoelastic responses used to describe polymers. From this and other available sources, justify whether the CNT dynamics are best captured by a Maxwell, Kelvin-Voigt, or SLS model. Express the resonance of the CNTs in terms of the elements of the phenomenological spring/dashpot model you have chosen, keeping in mind that the elastic properties of the CNTs are well-known.

We propose using a Kelvin-Voigt model to express the resonance of a carbon nanotube. This choice is justified because several groups have been able to model cantilevered beams using Kelvin-Voigt [5], [6], finding good agreement with the Bernoulli-Euler beam analysis technique, which was used successfully by Poncharal, et al. [4] to study electromechanical resonance of CNTs.

In the case of Gurgoze, et al. [5] the following physical model was chosen:

Courtesy Elsevier, Inc., Science Direct. Used with permission.

Along with the corresponding phenomenological model:

Courtesy Elsevier, Inc., Science Direct. Used with permission.

The resulting eigenstate of the system according to the calculations of Gurgoze, et al. is given by:

Where d-bar is given by:

References

[1] Pantano, A., Parks, D. M., and Boyce, M. C. “Mechanics of Deformation of Single- and Multi-wall Carbon Nanotubes.” Journal of the Mechanics and Physics of Solids 52 (2004): 789-821.

[2] Yu, M. F., Lourie, O., Dyer, M. J., Moloni, K., Kelly, T. F., and Ruoff, R. S. “Strength and Breaking Mechanism of Multiwalled Carbon Nanotubes Under Tensile Load.” Science 287 (2000): 637-640.

[3] Yu, M. F., Files, B. S., Arepalli, S., and Ruoff, R. S. “Tensile Loading of Ropes of Single Wall Carbon Nanotubes and their Mechanical Properties.” Physical Review Letters 84 (2000): 5552-5555.

[4] Poncharal, Philippe, Z. L. Wang, Daniel Ugarte, and Walt A. de Heer. “Electrostatic Deflections and Electromechanical Resonances of Carbon Nanotubes.” Science 283 (March 5, 1999): 1513-1516.

[5] Gurgoze, M., Dogruoglu, A. N., and Zeren, S. “On the Eigencharacteristics of a Cantilevered Visco-elastic Beam Carrying a Tip Mass and its Representation by a Spring-damper-mass System.” Journal of Sound and Vibration 301 (2007): 420-426.

[6] Mahmoodi, S., Jalili, N., and Khadem, S. “An Experimental Investigation of Nonlinear Vibration and Frequency Response Analysis of Cantilever Viscoelastic Beams.” Journal of Sound and Vibration 311 (2008): 1409-1419.

Plasticity and fracture of microelectronic thin films/lines Effects of multidimensional defects on III-V semiconductor mechanics Defect nucleation in crystalline metals Role of water in accelerated fracture of fiber optic glass Carbon nanotube mechanics | Problem Set 2 | Problem Set 3 | Problem Set 5 Superelastic and superplastic alloys
Mechanical behavior of a virus
Effects of radiation on mechanical behavior of crystalline materials

(b) The papers you have chosen discuss the elastic properties of CNTs. Of course, the applications people are excited about are related to the high tensile strength of these organic fibers. We know they’ll easily buckle in compression, and it’s hard to think of apps that would benefit from that. Read the 2 papers cited by Pantano, et al. [1] that discuss nanotube fracture to review this concept.

Then compute the number of carbon bonds that must be broken to achieve tensile fracture of a single CNT (not a multiwall).

Let’s consider the case of a zigzag nanotube, as Belytschko, et al. [2] discuss. In this case, fracture occurs around the circumference, when a circumferential set of bonds aligned axially with the tube are broken. Consider a sheet of graphite, specifically a set of parallel bonds. The equilibrium bond spacing given by Zhang, et al. [3] is 0.1315 nm, and all bond angles are 120 degrees. Simple geometry is sufficient to determine the normal spacing between perpendicular bonds as [0.1315*√(3)] nm, or, generally, [bond length*√(3)]. The circumference of a nanotube of diameter d is simply π*d, and we can divide this by the bond spacing to approximate the number of bonds along (and perpendicular to) the circumference that must be broken.

Note that r is the equilibrium bond spacing.

It [elastic energy stored at fracture] should be the number of carbon bonds transversing the fracture plane, multiplied by the carbon-carbon bond strength, but is it?

First, consider the general elastic energy stored per unit volume, and the simply transformed total elastic energy stored. (We are implicitly assuming there is no plasticity aside from fracture.)

The modeling values of Belytschko, et al. will be used throughout. For stress purposes, the effective area over which the stress is applied is taken as the circumference of the nanotube multiplied by a value h, a wall thickness, which Belytschko, et al. take to be the interlayer separation of graphite h = 0.34 nm, and which is given as a standard value for this sort of calculation. Morse potential simulations [2] give a failure strain of 15.7%, or a failure stress of 93.5 GPa. For the other constants, E = 1.16 TPa, A₀ = circumference*h, L₀ = r = 0.1315 nm = equilibrium bond spacing. While the first bond fails before the fracture strain is reached, we will take the failure strain to be the strain at which the first bond breaks, and assume the difference is minimal since no plasticity is expected. Next, consider the number of atomic bonds over which this energy is distributed, given in Eq. 1 for a zigzag nanotube. Once a single bond is broken, fracture follows readily at little additional displacement [2]. The expression simplifies as follows; note that the result does not depend on the nanotube size.

Using the previously given values, the energy applied to each bond at fracture strain is 0.91 eV; the dissociation energy for a single bond is given as 5.62 eV [2]. Though this is within the expected order of magnitude, fracture is observed at a significantly lower energy investment than calculated, and Belytschko, et al. note that fracture has been observed experimentally at stresses significantly lower than even these simulations. It is worthy of mention that in the simulations of Belytschko, et al. one bond was weakened by 10%, guaranteeing the site of fracture nucleation, but this is not at all sufficient to explain this discrepancy. Using the Brenner rather than the Morse results in [2] gives a result of 2.89 eV per atom, which is still significantly off.

These results can be extended to armchair or chiral nanotubes by appropriate modification of the function giving the number of bonds broken; Zhang, et al. [3] show in Fig. 6 that fracture proceeds by the stretching and breaking of axial or semi-axial bonds, rotating the fracture surface about an axis normal to the tube such that it becomes angular/ellipsoidal.

References

[2] Belytschko, T., S. P. Xiao, G. C. Schatz, and R. S. Ruoff. “Atomistic Simulations of Nanotube Fracture.” Physical Review B 65 (2002): 235430.

[3] Zhang, P., Y. Huang, H. Gao, and K. C. Hwang. “Fracture Nucleation in Single-Wall Carbon Nanotubes Under Tension: A Continuum Analysis Incorporating Interatomic Potentials.” Transactions of the ASME 69 (2002): 454-458.

Group Members

Tim Rupert
Aparna Singh
Hyunjung Yi
ShinYoung Kang

References

Li, Ju. “The Mechanics and Physics of Defect Nucleation.” MRS Bulletin 32 (2007): 151-159.

Minor, Andrew M., et al. “A New View of the Onset of Plasticity During the Nanoindentation of Aluminum.” Nature Materials 5 (2006): 697-702.

Van Vliet, Krystyn J., et al. “Quantifying the Early Stage of Plasticity Through Nanoscale Experiments and Simulations.” Physical Review B 67 (2003): 104105.

Wiki Assignments

(a) What is the average Young’s elastic modulus of the Au nanopillar in Fig. 1A of Li? From comparison of these data with E of Au reported in the literature, do you infer that elastic properties of Au are size-dependent down to these sample diameters?

(b) Derive the expression used several times by Li relating the maximum shear stress generated within an indented material as a function of applied load, Young’s elastic modulus, and indenter radius. You may wish to consult K. Johnson’s Contact Mechanics.

(b) The papers you have selected study the onset of plasticity via the applied contact load of indentation. Explain and illustrate why the location of homogeneous dislocation nucleation under an indenter is expected to occur below the indented material free surface, rather than at the indenter/material interface.

(c) Derive the location of maximum shear stress (i.e., depth) for indentation with a Hertzian elastic sphere for the case of 2-D indentation (i.e., the sphere is really a cylinder). Johnson’s Contact Mechanics and the PRB (2003) and its cited papers are good references.

(d) Minor, et al. show interesting experiments that enable visualization of dislocations in a thin foil of Al during indentation, similar in goal to the MD simulations of Al discussed in your other two papers. Discuss how well the authors justified the assumption that the load-drops in the load-displacement data correspond to homogeneous dislocation nucleation, as opposed to heterogeneous dislocation nucleation from grain boundaries or motion of pre-existing dislocations. This analysis should include consideration of images, reported data, and expected values of required stresses, dislocation spacing, and sufficient “perfect crystal” size comparable to the elastic strain field of an indenter.

(b) The papers you have chosen focus on the initial stages of plastic deformation, or incipient plasticity, in ductile metals. Consider Fig. 9 in paper 3, illustrating a dislocation loop passing through a single crystal of Al under indentation pressure. Consider a bulk metal vs. a thin film of the same material on a rigid substrate, explain how you would expect the load required to keep plastically deforming the material t a depth of 1 micron would differ.

(c) Compute the fracture stress of this Al according to the Griffith criterion, assuming an initial crack size at the resolution of the TEM used in Minor’s indentation experiments. Compare this to the published fracture strength of single crystal Al, and discuss.

(d) Delamination of films can also be treated as fracture, but this time at an interface between two dissimilar materials. Using Griffith again, compute the tensile stress required to delaminate an Al film from a Si substrate, if there is an initial blister at the film/substrate interface of width 1 micron. Then state whether you’d expect the film to first fracture due to opening of initial cracks within the film (of initial size stated in c), or to delaminate from the substrate when loaded in tension normal to the film/substrate interface.

Final Presentation

“Defect Nucleation in Crystalline Metals.” (PDF)

Plasticity and fracture of microelectronic thin films/lines
Effects of multidimensional defects on III-V semiconductor mechanics
Defect nucleation in crystalline metals | Problem Set 2 | Problem Set 3 | Problem Set 5
Role of water in accelerated fracture of fiber optic glass
Carbon nanotube mechanics
Superelastic and superplastic alloys
Mechanical behavior of a virus
Effects of radiation on mechanical behavior of crystalline materials

(a) What is the average Young’s elastic modulus of the Au nanopillar in Fig. 1A of Li [1]? From comparison of these data with E of Au reported in the literature, do you infer that elastic properties of Au are size-dependent down to these sample diameters?

The average Young’s elastic modulus of the Au nanopillar is the slope of the unloading curve which is equal to 48.5 GPa. The E of Au reported in literature is around 43 GPa along the direction which is the same direction in which the compression was done on the Au nanopillar. We conclude that the elastic modulus is not dependent on specimen size as it is a property that is related to the internal energy change associated with bond stretching or compression and thus does not depend on the size of the sample.

(b) Derive the expression used several times by Li [1] relating the maximum shear stress generated within an indented material as a function of applied load, Young’s elastic modulus, and indenter radius. You may wish to consult K. Johnson’s Contact Mechanics [2].

We are interested in modeling the maximum shear stress a material experiences when being indented. This is done by considering the elastic contact of two smooth surfaces. Among other things, we need to know how contact area changes with increased applied load. Contact mechanics allows us to do this so that we can also predict the deformation (stress, strain) in the region of the material affected by the contact.

We begin by describing the geometry of two surfaces being brought into contact. This allows us to define our problem in terms of critical variables.

Consider 2 surfaces in contact described by Eq. 1 and Eq. 2 below. Here, we assume that we have two smooth surfaces with surfaces which can be described by an equation which must be continuous up to the second derivative.

Surface 1:

Eq. 1

Surface 2:

Eq. 2

Therefore, the separation is h = z₁ - z₂. However, this separation currently requires two different coordinate systems for its definition.

Switching to common set of axes: x and y

Eq. 3

Since these axes are arbitrary, we can pick them intelligently to make our description of the problem easier. Here, we choose our axes such that C = 0,

Eq. 4

where A and B are positive constant; R’ and R" are the principal relative radii of curvature.

We can get relations for A and B using geometrical considerations. This is done in Appendix 2 of [2], but omitted here for clarity.

Eq. 5

Eq. 6

where α is the angle between the axes of principal curvature of each surface (x₁ and x₂)

We will now consider two surfaces of general shape, centered at O, deformed with a normal load P applied (Figure 1).

Image removed due to copyright restrictions. Please see Fig. 4.2 in [2].

Figure 1 Schematic showing the geometry of the problem [2].

We now must utilize Hertzian contact theory to determine displacements, which then allow us to find strains and stresses. Hertzian theory makes some assumptions which must be declared at this juncture. It is assumed that each body can be regarded as an elastic half-space which is loaded over an elliptical area on its surface plane. This is done because well-developed solutions exist for solving boundary value problems for an elastic half-space. It is also assumed that the surfaces are continuous and non-conforming. The strains experienced by the half-space area small. These last two considerations are represented by the assumptions that the contact radius is much smaller than the radius of curvature of each body. Finally, it is assumed that the surfaces are frictionless.

During compression, distant points T₁ and T₂ move towards O by displacements δ₁ and δ₂.

Due to contact pressure, the surface of each body is displaced parallel to O in the z direction by u_z1 and u_z2 relative to T₁ and T₂.

Looking at points S₁ and S₂, which are coincident within the elastic surface:

Eq. 7

If we write δ = δ₁ + δ₂ and use Eq. 6, we get elastic displacements:

Eq. 8

At this point, we can consider the specific geometry of our problem of interest to simplify the equations which describe the evolutions of the contacting bodies.

For this situation, we are interested in the contact of two solids of revolution. A solid of revolution is a solid body which is obtained by revolving a plane figure about some axis.

(R₁’ = R₁" = R₁ ; R₂’ = R₂" = R₂)

Therefore: A = B = ½(1/R₁ + 1/R₂)

The circular symmetry of the problem means that the contact area will be circular (with a radius we will call a).

Eq. 9

Where (1/R) = (1/R₁ + 1/R₂) is the relative curvature.

A pressure distribution which gives displacements satisfying Eq. 9 is given in Johnson Section 3.4 [2]. This gives the stress resulting from an applied pressure over a circular area for an elastic half-space (Hertzian contact).

Eq. 10

This pressure distribution gives the displacement:

Eq. 11

Since the pressure acting on the second body is equal to the first, we can write:

Eq. 12

Substituting Eq. 11 into Eq. 9 for u_z1 and u_z2, we are left with:

Eq. 13

Which give us the radius of circle contact, a, and the mutual approach of distant points, δ:

Eq. 14

Eq. 15

The total load compressing the solids is:

Eq. 16

Substituting this into Eq. 14 and Eq. 15, we get:

Eq. 17

Eq. 18

We also get an equation for the maximum pressure, p_o, if we compare our new expressions for a and δ with the old ones.

Eq. 19

We now have expressions for contact size, compression, and maximum pressure.

For this pressure distribution, the stresses beneath the surface along the z-axis are (refer to Johnson Section 3.4 [2]):

Eq. 20

Eq. 21

These are the principal stresses. Therefore, our principal shear stress is:

Eq. 22

We now plug in the values for our principal stresses to get an expression for the maximum shear.

Eq. 23

This principal shear stress is maximized at a depth of z = 0.48 a (for ν = 0.3). This step was evaluated in Maple. Substitution gives us an expression for the maximum shear stress under an indenter.

Eq. 24

P is the indentation load. R is the radius of curvature of the indenter tip. E* is the effective elastic modulus.

This equation tells us that the maximum shear stress occurs below the contact surface and is equal to 0.31 times the maximum pressure. This can also be expressed in terms of variables which can be easily measured in an experimental situation. From this equation, we would expect critical plastic events to occur underneath the surface of an indented material.

References

[1] Li, Ju. “The Mechanics and Physics of Defect Nucleation.” MRS Bulletin 32 (February 2007): 151-159.

[2] Johnson, K. L. Contact Mechanics. Cambridge, UK: Cambridge University Press, 2008. ISBN: 9780521347969.

The homogeneous dislocation nucleation occurs at the maximal shear stress state position. In the nano-indentation experiment in our papers, the stress field generated by an indenter can be calculated and the maximum shear stress position is the place where the homogeneous dislocation nucleation takes place. In problem set 2, we calculated the stress beneath the indented surface along z-axis (depth) in 3-D case. Based on problem set 2, we know that the principal shear stress is

Eq. 1

where σ_z and σ_θ are principal stresses expressed as

Eq. 2

Eq. 3

Here p₀ is the maximum pressure,

Therefore

Eq. 4

I plotted the (eq. 4) by using Excel and Origin Pro 7.0 (Here I assumed v = 0.3 and set a = 1) and the resulting graph is presented below. Now we see that the maximum shear stress state is inside the indented materials not the interface. In 3-D case plot, the maximum depth is z/a = 0.48.

Therefore we can say that homogeneous dislocation nucleation under an indenter is expected to occur below the indented material free surface, rather than at the indenter/material interface. Also, we can learn that nano-indentation experiments probe the bulk property, not the surface property of the material.

Another illustration is possible. The nano-indentation experiments can be thought of such that the indenter tip works somewhat like a lens, projecting the applied force to focus the maxima shear stress to an internal point inside the sample, away from the surface. [1]

In 2-D case, the plotting equation can be different from 3-D and this calculation is done at problem (c). Also we found an experimental report about the 2-D case. In that case the maximum shear stress position is z/a = 0.78. [2]

Image removed due to copyright restrictions. Please see Fig. 1 in [2].

(c) Derive the location of maximum shear stress (i.e., depth) for indentation with a Hertzian elastic sphere for the case of 2-D indentation (i.e., the sphere is really a cylinder). Johnson’s Contact Mechanics [3] and the PRB (2003) [4] and its cited papers are good references.

Handwritten Solutions (PDF)

Supplemental Calculations

1. Dummy variable s in equation (b)

As can be seen in the below figure, when a pressure distribution is applied on a xy plane, at x=s point, x is shifted by amount of s, thus x is replaced by (x-s). Then the pressure applied at x=s point is written as p(s) instead of p(x).

In this 2-D indentation case with same radius R(= R₁ = R₂), a = b.

2. The distribution of pressure

When a function takes the form below,

the solution F(x) has this form.

where

Thus (equation (c)) has a solution p(x) as seen in equation (d).

3. Procedure from the step (h) to the final answer

When we take derivative of 1 with respect to z,

Thus,

When we calculate this through a calculator, we can get z = 0.78a, and there τ_max = 0.30 p₀

(d) Minor, et al. [5] show interesting experiments that enable visualization of dislocations in a thin foil of Al during indentation, similar in goal to the MD simulations of Al discussed in your other two papers [1, 4]. Discuss how well the authors justified the assumption that the load-drops in the load-displacement data correspond to homogeneous dislocation nucleation, as opposed to heterogeneous dislocation nucleation from grain boundaries or motion of pre-existing dislocations. This analysis should include consideration of images, reported data, and expected values of required stresses, dislocation spacing, and sufficient “perfect crystal” size comparable to the elastic strain field of an indenter.

Minor, et al. [5] claim that the load drops in their nanoindentation data correspond to the homogeneous nucleation of dislocations within the crystal. In determining this, it is important to consider the possibility that these load jumps could be caused by heterogeneous dislocation nucleation from grain boundaries or the movement of pre-existing dislocations.

First, we consider the possibility of dislocations nucleated from grain boundaries. From inspection of the TEM images in Fig. 2 (and the corresponding movie in the online Supplemental section), we can see dislocations originating in the interior of the grain, underneath the indenter. This is most easily seen in the later parts of the movie (closer to load drop 3), where an increased amount of dislocation activity can be seen in this region. The possibility of grain boundary nucleated dislocations can also be ruled out by considering the size of the grain used to model a perfect crystal with the elastic stress field under the indenter. Our previous analysis (work done in PSet 2b and PSet 3c) has shown that the shear stress drops significantly and becomes close to zero at distances of greater than three contact radii away from the contact point. By looking at Supplementary Image 2 [5], we can see that the contact radius is approximately 37.5 nm while the grain boundaries are hundreds of nanometers away from the contact point. Since these boundaries are all more than five times the contact radius away, the shear stress acting on the boundaries will be insignificant. Since a large stress must be concentrated on a grain boundary for a dislocation to be emitted, this possibility can be ruled out. This combination of visual evidence and consideration of the applied stress field show that grain boundary nucleation cannot explain the load drops in the nanoindentation data.

Next, we look at the possibility of the load drops corresponding to critical events involving the movement of preexisting dislocations. The images of the 1st load drop show that this cannot be the case here, where a dislocation-free crystal becomes filled with dislocation debris. Images of the 2nd load drop show a greatly increased dislocation density, pointing to the fact that dislocation nucleation is occurring. Finally, as mentioned earlier, a higher level of dislocation nucleation and movement can be observed under the indenter contact when compared to the rest of the crystal at the 3rd load drop. The possibility of existing dislocation movement can also be analytically considered by looking at the stress necessary to move dislocations through an existing dislocation structure. The existing dislocation density in Fig. 2e is estimated in the article as ρ ~ 10¹⁴ m^-2. Using the Taylor work hardening law (shown below), we can estimate the shear stress needed to cause slip in a crystal with a given dislocation density. This theory is based on the stress needed to pull two dislocations past each other. Here, the spacing between dislocations, l, is equivalent to the inverse square root of the dislocation density, ρ. By choosing α to be unity (the high end of the usual 0.05 to 1 range) and using material properties for aluminum (G = 25 GPa; b = 0.286 nm), we can calculate that a shear stress of 71.5 MPa would correspond this mechanism. Since the slip event occurs at a much higher stress than this, the motion of existing dislocations moving through an existing dislocation network cannot be the mechanism responsible for the load drop.

Eq. 1

The TEM images and a calculation of stress under the indenter points to the fact that homogeneous dislocation nucleation is occurring, corresponding to the load drops in the indentation data. The dislocation events can be observed to occur at a position underneath the contact surface in a region where elastic contact theory predicts the shear stress will be highest. The magnitude of shear stress in this region is of the same order (~2 GPa) as that predicted to be necessary to nucleate a dislocation in a dislocation-free aluminum crystal [6, 7].

Image removed due to copyright restrictions. Please see Supplementary Fig. 2 in [5]. (PDF)

References

[1] Li, Ju. “The Mechanics and Physics of Defect Nucleation.” MRS Bulletin 32 (2007): 151-159.

[2] Gouldstone, Andrew, Krystyn J. Van Vliet, and Subra Suresh. “Simulation of Defect Nucleation in a Crystal.” Nature 411 (June 2001): 656.

[3] Johnson, K. L. Contact Mechanics. Cambridge, UK: Cambridge University Press, 2008. ISBN: 9780521347969.

[4] Van Vliet, Krystyn J., et al. “Quantifying the Early Stage of Plasticity Through Nanoscale Experiments and Simulations.” Physical Review B 67 (2003): 104105.

[5] Minor, A. M., et al. “A New View of the Onset of Plasticity During the Nanoindentation of Aluminum.” Nature Materials 5 (2006): 697-702.

[6] Gouldstone, A., H. J. Koh, K. Y. Zeng, A. E.Giannakopoulos, and S. Suresh. “Discrete and Continuous Deformation During Nanoindentation of Thin Films.” Acta Materalia 48 (2000): 2277-2295.

[7] Kramer, D. E., K. B. Yoder, and W. W. Gerberich. “Surface Constrained Plasticity: Oxide Rupture and the Yield Point Process.” Philosophical Magazine A 81 (2001): 2033-2058.

(b) The papers you have chosen focus on the initial stages of plastic deformation, or incipient plasticity, in ductile metals. Consider Fig. 9 in paper 3 [1], illustrating a dislocation loop passing through a single crystal of Al under indentation pressure. Consider a bulk metal vs. a thin film of the same material on a rigid substrate, explain how you would expect the load required to keep plastically deforming the material t a depth of 1 micron would differ.

The load required to plastically deform the thin film of Aluminum and the bulk metal at a depth of 1 micron below the free surface would differ considerably. The stress required in the case of the aluminum film would be much greater as compared to the bulk metal. This is because plastic deformation is facilitated by dislocation movement and it would be obstructed more in the thin film than in the bulk metal. This is attributed to the fact that when a stress is applied the dislocation loop would tend to move but would not be able to move easily as it would be repelled by the hard substrate as it has a higher modulus than aluminum so a higher stress is needed to move the dislocation loop close to the substrate.

The energy of a dislocation goes as (Gb²)/2 so it would be more in the hard substrate and so also would be repelled by the hard substrate as the dislocation energy increases. We can also draw analogy to the fact that the dislocation would tend to be attracted to a free surface whose G = 0 because the dislocation energy becomes 0. Going on the same lines it would be repelled by a substrate whose G is greater than the G of the metal.

In the case of the bulk metal a dislocation finds itself in the vicinity of the bulk metal only at a depth of 1 micron so it would move with no additional resistance.

(c) Compute the fracture stress of this Al according to the Griffith criterion, assuming an initial crack size at the resolution of the TEM used in Minor’s indentation experiments [2]. Compare this to the published fracture strength of single crystal Al, and discuss.

Here we assume an initial crack the size of the resolution of the microscope used by Minor in this study [2]. From the NCEM website [3], we find the minimum point-to-point resolution of the JEOL 3010 to be 1.7 Å, which we set equal to a crack of width 2_a_. The surface energy of aluminum was found to be 0.057 eV _Å_² [4]. The Young’s Modulus for aluminum is 68 GPa [5]. We now have all the necessary parameters to calculate the required fracture stress for a crack of this size.

This is a very high fracture stress, being near the theoretical fracture stress given for aluminum in class (20 GPa). A molecular dynamics simulation of a single crystal of aluminum gives a yield stress of 2 GPa, corresponding to the initiation of a plastic event in the material [6]. No direct experimental data of fracture stress could be found, but it should be even lower than this due to the presence of preexisting defects such as cracks and pores. Prof. Van Vliet gave the fracture stress of aluminum to be 0.7 GPa in class.

The high fracture stress for a crack at the TEM resolution limit points tells us that this is not a very reasonable explanation for failure in this system. Either the theoretical shear stress will be reached and a dislocation will be initiated or preexisting macroscopic defects will propagate cracks within the material before a crack on the Angstrom level would propagate.

Griffith’s criterion:

Griffith’s criterion considers crack propagation in single material. To apply Griffith’s equation to thin film/substrate system, we need to consider both Al film and Si substrate.

Surface energy: In the thin film case, the change of surface energy (Γ) because of the crack is the sum of increased surface energy of Al (γ_Al) and Si (γ_Si) and the decrease of interface energy (γ_Al/Si) between thin film and substrate.

= 8 N/m (from [5])

This value corresponds to 2γ in the Griffith’s equation.
Young’s modulus: In a delamination problem, Young’s modulus (E) should also be modified by using effective Young’s modulus (E_eff) since the release of elastic energy comes from both the thin film and substrate.
The effective energy release rate is given in [7],

where D₁ is given as and is the plane strain modulus for the film and substrate.

= 80 GPa. (E_f is 70 GPa and ν_f is 0.35)
= 141 GPa. (E_s is 130 GPa and _ν__s is 0.28)
D₁ value for our Al/Si system is therefore -0.28.
Combining 1) and 2), we can get the tensile stress required to delaminate an Al film on Si substrate with 1 µm blister,
From problem (c), we know that the fracture stress for 0.17 nm crack is 21.5 GPa. Since the fracture stress for fracture is much higher than that for delamination, we would expect the film to delaminate first.

References

[1] Van Vliet, Krystyn J., et al. “Quantifying the Early Stage of Plasticity Through Nanoscale Experiments and Simulations.” Physical Review B 67 (2003): 104105.

[2] Minor, Andrew M., et al. “A New View of the Onset of Plasticity During the Nanoindentation of Aluminum.” Nature Materials 5 (2006): 697-702.

[3] National Center for Electron Microscopy.

[4] Needs, Richard J. “Calculations of the Surface Stress Tensor at Aluminum (111) and (110) Surfaces.” Physical Review Letters 58 (1987): 53-56.

[5] MatWeb.

[6] Yuan, Lin, et al. “Molecular Dynamics Simulation of Tensile Deformation of Nano-single Crystal Aluminum.” Journal of Materials Processing Technology 184 (2007): 1-5.

[7] Freund, L. B., and S. Suresh. “Delamination and Fracture.” Chapter 4 in Thin Film Materials. New York, NY: Cambridge University Press, 2004. ISBN: 9780521822817.

Group Members

Nan Yang
Sal Barriga
Sahil Sahni
Vivek Raghunathan

References

Hjort, Klas, Jan Soderkvist, and Jan-Ake Schweitz. “Gallium Arsenide as a Mechanical Material.” Journal of Micromechanics and Microengengineering 4 (1994): 1-13.

Navamathavana, R., D. Arivuolib, G. Attolinic, C. Pelosic, and Chi Kyu Choia. “Mechanical Properties of Some Binary, Ternary and Quaternary III-V Compound Semiconductor Alloys.” Physica B 392 (2007): 51-57.

Yonenaga, Ichiro, Koji Sumino, Gunzo Izawa, Hisao Watanabe, and Junji Matsui. “Mechanical Property and Dislocation Dynamics of GaAsP Alloy Semiconductor.” Journal of Materials Research 4 (March/April 1989): 361-365.

Wiki Assignments

(a) Draw the crystal structure of GaAs, and a separate graphic which is the closest-packed plane and direction on that plane. Discuss the major differences in this structure, as compared to Si.

(b) From the data provided in Hjort’s Table 1, determine the Poisson’s ratio of GaAs, as compared to Si and AlAs. Discuss comparison of these predictions with experimentally reported values of nu, from open literature and databases such as matweb.com.

(c) Hjort’s Fig. 2 seems to indicate a different number of independent elastic constants than your crystal structure in (a) would suggest; explain. (It’s a simple explanation.)

(d) Based on your analysis of GaAs elastic anisotropy, explain (graphically and/or text) which film orientation you would aim to achieve to create GaAs films or wafers that would be maximally resistant to bending.

(b) Why do Yonenaga, et al. claim that understanding of dislocation dynamics is relevant to GaAsP studies (basic science and applications)? Please do not restate their arguments, but rather justify them through a more detailed analysis than they could consider in a paper intro.

(c) Compare the observed GaP, GaAsP and GaAs critical shear stress (followed by a sharp drop in applied stress in Yonenaga’s Fig. 1) to the corresponding theoretical shear strength of these materials, and discuss possible reasons for the relative discrepancies between theoretical and experimentally observed values among these three compounds.

(d) Given the stated impurity concentrations for the compounds considered by Yonenaga, et al., determine the maximum (average) glide distance a dislocation would require to encounter a point defect.

(b) Eq. 2 of Navamathavan, et al. states an equivalence between fracture surface energy and K_IC, the plane strain fracture toughness of the semiconductor films inferred from indentation cracking. From the definitions of K_IC and Griffith’s fracture criterion, prove whether this equation is true for a brittle material.

(c) Although they report these data in Table 3, the authors do not appear to put the results in context by comparing the calculated K_IC and surface energy with values reported for these (or similar) materials via other experiments. Do this, and discuss whether you feel the characterization of these fracture properties and surface energies of the materials is valid.

(d) Why do you think the authors observed a film thickness dependence of K_IC and fracture surface energy (in Table 3), if these are supposed to be properties of the material?

Final Presentation

“Multidimensional Defects in III-V Semiconductors.” (PDF)

Plasticity and fracture of microelectronic thin films/lines
Effects of multidimensional defects on III-V semiconductor mechanics | Problem Set 2 | Problem Set 3 | Problem Set 5
Defect nucleation in crystalline metals
Role of water in accelerated fracture of fiber optic glass
Carbon nanotube mechanics
Superelastic and superplastic alloys
Mechanical behavior of a virus
Effects of radiation on mechanical behavior of crystalline materials

(a) Draw the crystal structure of GaAs, and a separate graphic which is the closest-packed plane and direction on that plane. Discuss the major differences in this structure, as compared to Si.

GaAs Zinc-Blende Structure [1]

Image removed due to copyright restrictions.

Please see: http://www-vrl.umich.edu/project2/miller/111plane.JPG

GaAs closest packed direction and plane [2]

Silicon Diamond Structure [1]

The GaAs structure is a zincblende structure whose basis consists of two different atoms, gallium and arsenic. Silicon has a similar structure except that it has a one atom basis, which is Silicon. This implies the GaAs structure does not have inversion symmetry whereas the silicon structure does. In addition, in GaAs, “the electron clouds [tend] to shift towards the Arsenide atoms, and results in a dipole moment along the [111] axis” [3]; the material exhibits piezoelectric characteristic similar to quartz. The zinc blende structure also offers alternating (111) Gallium and Arsenic planes resulting in very different properties depending on the surface-terminating plane. The asymmetry also results in such properties as hardness, anisotropy, and different dislocation velocities. [3]

(b) From the data provided in Hjort’s Table 1 [3], determine the Poisson’s ratio of GaAs, as compared to Si and AlAs. Discuss comparison of these predictions with experimentally reported values of ν, from open literature and databases such as matweb.com.

Image removed due to copyright restrictions. Please see Table 1 in [3].

The Poisson’s ratio (ν) calculated (by taking the ratio -S₁₂/S₁₁ , values of which are given in table 1) for GaAs, Si, and AlAs are respectively 0.3162, 0.2727, and 0.325. Thus we find that the Poisson’s ratio for AlAs > GaAs > Si. From literature, the Poisson’s ratio for bulk GaAs is 0.31 [4], for Si it ranges between 0.22 for bulk isotropic, and 0.27 for <111> crystalline Si [4], and 0.328 for AlAs [5]. The calculated values are in close agreement with the values got from literature.

(c) Hjort’s Fig. 2 [3] seems to indicate a different number of independent elastic constants than your crystal structure in (a) would suggest; explain. (It’s a simple explanation.)

Image removed due to copyright restrictions. Please see Fig. 2 in [3].

As expected, the matrix in Fig. 2 shows three independent elastic constants, C₁₁, C₁₂ and C₄₄ . However, this matrix has 9x9 elements because it also accounts for the interactions between the mechanical and electromagnetic fields. It displays the relationship between orientation, piezoelectricity, d, and dielectricity, ε and the corresponding responses: S^E as the mechanical response to a mechanical force, d^T as the mechanical response to an electric field, d as the electrical response to a mechanical force and finally ε^S as the electrical response to an electrical field.

Image removed due to copyright restrictions. Please see Fig. 3a in [3].

To resist bending, we want to choose the orientation with the highest Young’s elastic modulus. This is the <111>, as can be seen in Fig. 3a and in our calculations, attached. (PDF)

However, it is important to understand the exact orientation of this the (111) planes with respect to film-substrate interface, which would minimize bending. To do so let us first look at a bi-material strip as shown in the figure below. The two materials can be characterized by different elastic moduli, and thus will exhibit different strains under the same state of stress. This leads to the bending of the strip to incorporate this difference, as understood more clearly by the figure.

Effect of the introduction of stresses on a bi-material strip.

For a GaAs film on a substrate (of a different material, and thus different elastic modulus, usually much greater), the lattice-mismatch, or a change of environment temperature, etc., generates stresses oriented parallel to the plane of the interface, leading to bending of the thin film, as described above. Also, the elastic modulus of the substrates used, is generally greater than that of the thin film (for example, sapphire, a commonly adopted substrate, has a Young’s Modulus of 400 GPa [6]). Thus, in order to resist bending, it is desired to have the generated stresses experience maximum stiffness. Hence, the (111) planes (greatest modulus) should be oriented such that the normal to these planes are parallel to the generated stresses (in the plane of the interface).

References

[1] Wikimedia Commons.

[2] Miller Indices in VRML.

[3] Hjort, Klas, Jan Soderkvist, and Jan-Ake Schweitz. “Gallium Arsenide as a Mechanical Material.” Journal of Micromechanics and Microengineering 4 (1994): 1-13.

[4] MEMSnet.

[5] Leavitt, R. P., and F. J. Towner. “Determination of the Lattice Parameter and Poisson Ratio for AlAs via High-resolution x-ray-Diffraction Studies of Epitaxial Films.” Physical Review B 48 (1993): 9154-9157.

[6] Moody, N. R., et al. “Substrate Composition Effects on the Interfacial Fracture of Tantalum Nitride Films.” Journal of Materials Research 14 (June 1999): 2306-2313.

(b) Why do Yonenaga, et al. [1] claim that understanding of dislocation dynamics is relevant to GaAsP studies (basic science and applications)? Please do not restate their arguments, but rather justify them through a more detailed analysis than they could consider in a paper intro.

An understanding of dislocation dynamics is extremely important for understanding the GaAsP system. First of all, it is very important to understand the dynamics of dislocations because they contribute to basic science and help verify theory.

From an engineering perspective, dislocation dynamics also pervade many aspects of GaAsP device processing and operation. Dislocations can act as electronic recombination sites for electrons and holes as well as short circuit channels. Thus, they must be minimized in the active regions of a GaAsP device. Device operation often results in a temperature increase, which activates diffusion processes. Dislocations which were not active at “normal” room temperatures, may all of a sudden be activated and climb at higher temperatures, under driving forces such as the stress fields from neighboring dislocations, leading to device degradation, as stated in Yonenaga, et al. [1] Understanding dislocations is also important during processing, as device processing often subjects materials to high temperatures and therefore activates diffusion processes such as climb.

Dislocations are also important for the substrate. If a GaAsP single crystal could be used as a substrate, dislocations are important because any dislocation that terminates on the substrate frontside surface will extend into epitaxially grown device regions and affect device operation. As the dislocation density also affects material strength, it is important to understand dislocation dynamics in order to engineer strong substrates that are resistant to mechanical deformation, especially bow. If GaAsP single crystals are not used (which is more often the case due to cost considerations and the difficulty in fabricating single crystal GaAsP) and a graded buffer on GaAs or Si is employed for the substrate, dislocation dynamics are just as critical. Lattice mismatch induces dislocations which, as mentioned, will affect device performance if they extend into the active region of the device.

(c) Compare the observed GaP, GaAsP and GaAs critical shear stress (followed by a sharp drop in applied stress in Yonenaga’s Fig. 1 [1]) to the corresponding theoretical shear strength of these materials, and discuss possible reasons for the relative discrepancies between theoretical and experimentally observed values among these three compounds.

There is a large discrepancy between the theoretical and experimental critical shear stresses. The shear moduli of GaAs and GaP are 32.85 GPa and 39.2 GPa, respectively. The theoretical critical shear stresses, approximately G/2 (or G/10 depending on the model used) are therefore 16.425 GPa (or 3.285 GPa) and 18.6 GPa (or 3.92 GPa) respectively. For GaAsP, which is an alloy of GaAs and GaP, we could expect an intermediate theoretical critical shear stress. Yonenaga, however, reports much lower critical shear stress, on the order of MPa for all three materials. The upper yield stresses are approximately 17.5 MPa, 12.5 MPa and 4 MPa for GaP, GaAsP and GaAs whereas the lower yield stresses are 11 MPa, 11.5 MPa and 3 MPa for the three materials. The discrepancy between the theoretical and actual critical shear stress can be attributed to dislocation motion. The theoretical shear stress assumes a perfect crystal and during shear, bonds along one atomic plane break and the atoms must overcome a potential barrier in order to align with and form bonds with the next row of atoms. However, in actuality, shear is caused by dislocation motion, which requires much less energy. Thus, the actual critical shear stress is orders of magnitude lower than the theoretical.

(d) Given the stated impurity concentrations for the compounds considered by Yonenaga, et al. [1], determine the maximum (average) glide distance a dislocation would require to encounter a point defect.

Consider a volume, say V cm³, of the material.

Let the impurity concentration be x cm^-3 .

Thus, the number of impurity atoms in the referred volume = V * x

Assuming that the impurity atoms are homogeneously distributed in the crystal, studying the impurity distribution along a single dimension,

Considered Length = V^(1/3)

Number of impurity atoms = (V * x)^(1/3)

Average spacing between the impurity atoms along that length = (Considered Length) / (Number of impurity atoms) = x^(-1/3) = L cms

For GaAs, with a Si impurity concentration of 10¹⁶ cm^-3, L = 4.64 * 10^-6 cm

For GaP, with a S impurity concentration of 3.3*10¹⁷ cm^-3, L = 1.44 * 10^-6 cm

This is the maximum distance a dislocation needs to glide before encountering an impurity atom. Hence, the average distance a dislocation needs to glide before encountering an impurity atom = L/2.

It should be understood that we have assumed that in gliding through this distance, a dislocation does not encounter another dislocation first, that is, the average spacing between the dislocations is greater than L. However this can be easily seen from the fact that the density of the grown-in dislocations, in both GaAs and GaP, is about 10⁶ cm^-2 . From the definition of dislocation density, this is the number of dislocations per unit area. In other words, when you take a cross section of a plane perpendicular to dislocation length direction, then you see 10⁶ dots uniformly distributed in that plane. So average spacing between two neighboring dislocations would be 1/(10⁶ )^0.5 = (10)^-3 cm. Thus, the avg distance between to dislocations is orders of magnitude higher than the avg distance between 2 impurities. In all probability, a dislocation would encounter 2 impurities before it encounters another dislocation.

Reference

[1] Yonenaga, Ichiro, Koji Sumino, Gunzo Izawa, Hisao Watanabe, and Junji Matsui. “Mechanical Property and Dislocation Dynamics of GaAsP Alloy Semiconductor.” Journal of Materials Research 4 (March/April 1989): 361-365.

(b) Eq. 2 of Navamathavan, et al. [1] states an equivalence between fracture surface energy and K_IC, the plane strain fracture toughness of the semiconductor films inferred from indentation cracking. From the definitions of K_IC and Griffith’s fracture criterion, prove whether this equation is true for a brittle material.

The Griffith fracture criterion is for brittle materials and K_IC does not depend on whether the material deforms plastically or elastically. Thus, there is no reason why the Griffith fracture criterion and K_IC cannot both be used to describe a brittle material.

However, strictly speaking, fracture surface energy for nanoindentation tests and K_IC are not equivalent. K_IC typically refers to the plane strain fracture toughness in Mode I loading. However, the loading behavior in nanoindentation need not be Mode I. Nanoindentation comprises of a more complicated stress state that implies a mixed-Mode loading situation. The K_IC defined here, however, is inferred from the indentation test and therefore is the stress intensity factor for this particular type of mixed mode loading (calculated using equation (1) in the paper).

However, given the newly defined K_IC from the paper, and

K_IC = f*σ*√[π*a] (although for Mode I)

If we take f = 1,

K_IC = σ*√[π*a]

and the Griffith’s fracture criterion

σ = √[2E*(γ)/(π*a)]

we can solve for the fracture surface energy:

σ = √[2E*(γ)/(π*a)]

σ*√[π*a] = √[2E γ]

K_IC = √[2E γ]

γ_f = (K_IC² )/(2E).

Thus, we obtain Eq. (2), from Navamathavan, et al. under the given conditions, and with this definition of K_IC.

(c) Although they report these data in Table 3 [1], the authors do not appear to put the results in context by comparing the calculated K_IC and surface energy with values reported for these (or similar) materials via other experiments. Do this, and discuss whether you feel the characterization of these fracture properties and surface energies of the materials is valid.

The fracture surface energy and the K_IC values for InGaAs/InP epilayers reported in Table 3 range from 0.035-0.332 J/m² and 0.303-0.977 MPa(m)^1/2 respectively. The only reference that we got that gave a measure of Fracture surface energy is Shi, et al. [2] They report of a value of 0.525J/m² for the fracture surface energy of GaAs/GaAs (and using the equation derived above we get a K_IC of 0.3667 MPa m^1/2). We see that this value is of the same range as that reported in Table 3 [1]. However the difference in values is obviously due to different compounds dealt in both the papers. Also, as pointed out before, fracture surface energy for nanoindentation tests and K_IC are not equivalent. K_IC typically refers to the plane strain fracture toughness in Mode I loading nanoindentation comprises of a more complicated stress state that implies a mixed-Mode loading situation. However these approximations seem to be giving values that are of the same range as that reported in a different experiment.

(d) Why do you think the authors observed a film thickness dependence of K_IC and fracture surface energy (in Table 3), if these are supposed to be properties of the material?

K_IC is a material property but it only becomes constant regardless of sample geometry at very large dimensions, on the order of meters, in the plane strain state. Here, authors are measuring a thin film. That, coupled with the fact that nanoindentation comprises mixed loading Mode, results in some dependence of K_IC on thickness.

In addition, the films tested are grown epitaxially on InP. Thus, the films are stressed laterally and in thicker films, some of this residual stress is relaxed, as shown by [3]. Thus, some of the variation in K_IC with respect to film thickness may be due to stress relaxation in the thicker films.

Image removed due to copyright restrictions. Please see Fig. 1 in [3].

References

[1] Navamathavana, R., D. Arivuolib, G. Attolinic, C. Pelosic, and Chi Kyu Choia. “Mechanical Properties of Some Binary, Ternary and Quaternary III-V Compound Semiconductor Alloys.” Physica B 392 (2007): 51-57.

[2] Shi, Frank, et al. “Hybrid Integrated GaAs-GaAs and InP-GaAs.” Journal of Applied Physics 93 (2003): 5750-5756.

[3] Attolini, G., et al. “Raman Scattering Study of Residual Strain in GaAs/InP Heterostructures.” Journal of Applied Physics 75 (April 1994): 4156-4160.

Group Members

Hannah Brice
Michael Short
Jerod Ketcham

References

Matthews, J. R., and M. W. Finnis. “Irradiation Creep Models - An Overview.” Journal of Nuclear Materials 159 (1988): 257-285.

Ando, M., et al. “Creep Behavior of Reduced Activation Ferritic/Martensitic Steels Irradiated at 573 and 773 K at up to 5 dpa.” Journal of Nuclear Materials 367-370 (2007): 122-126.

Garner, F. A., and D. S. Gelles. “Irradiation Creep Mechanisms: An Experimental Perspective.” Journal of Nuclear Materials 159 (1988): 286-309.

Wiki Assignments

(a) Jitsukawa, et al. summarizes steel’s mechanical properties by starting with tensile yield and fracture strength. Elastic behavior is not mentioned. Explain why there is no extensive database on the variation in elastic properties of steel as a function of composition, processing, or radiation exposure.

(b) Given that you know linear elastic behavior (expected of alloys like steel) is terminated at the start of plastic/permanent deformation, regraph Jitsukawa’s Fig. 1 in terms of yield strain as a function of operating temperature. You could consider Fig. 8 as a representative source of elastic deformation data, or glean the other papers/literature.

(c) Based on Matthews & Finnis’ speculation about where vacancy clusters and damage cascades occur within the crystal structure (Fig. 1), graphically represent the location and orientation of such vacancy clusters in steel. Here, you can assume a simpler alloy than the one discussed in Jitsukawa, Fe containing a (thermodynamic) equilibrium concentration of carbon and a supersaturation of vacancies.

(b) Garner, et al. claim that vacancies such as those induced by radiation damage produce an isotropic strain field. Prove that this is so, and explain whether this strain contribution is expected to contribute to plastic deformation via dislocation glide.

(c) Garner, et al. then discuss that ion bombardment at metal surfaces can lead to plastic deformation, e.g., via blistering bubbles of subsurface trapped gases. Treat such a blister as a spherical pressure vessel surrounded by your choice of metals discussed in the other two papers (this will set the mechanical properties of the material surrounding the gas bubble. Assume a subsurface bubble size of 0.5 micrometer radius, and determine the gas pressure required to initiate plastic deformation and consequent roughness for a bubble centered 1 micrometer beneath the initially flat surface.

(d) Others have noted the superposition of lateral stress that contributes to plastic yielding at the surface. Explain the source and expected contribution of lateral stress for this metal.

Plasticity and fracture of microelectronic thin films/lines
Effects of multidimensional defects on III-V semiconductor mechanics
Defect nucleation in crystalline metals
Role of water in accelerated fracture of fiber optic glass
Carbon nanotube mechanics
Superelastic and superplastic alloys
Mechanical behavior of a virus
Effects of radiation on mechanical behavior of crystalline materials | Problem Set 2 | Problem Set 3 | Problem Set 5

(b) Aldo’s Fig. 3 shows relation between effective creep stress and creep strain. Which of the materials/conditions considered exhibit a linear viscoelastic type creep response under this temperature and radiation dosage?

(c) The authors give a creep model in Eq. 3. Restate this in terms of a steady state creep rate, and determine the creep mechanism in terms of those we have discussed in class. Which of the models we discussed best captures the trend observed by the authors? Assuming the appropriate energy barrier and diffusivity for your mechanism of choice, would you predict final creep strains comparable to the ones they measure at these stresses (where creep strain = strain rate*time)?

(d) The authors state the following: “The HFIR irradiated specimens may have a larger Burgers vector anisotropy because they were exposed at higher stress levels and lower irradiation temperatures than the FFTF irradiated specimens. Detailed investigation of these microstructures following irradiation will be performed in future studies.”

What do they mean by “Burgers vector anisotropy”? And did they yet report any follow up work that correlate the microstructure of these steels with the observed creep phenomena?

Final Presentation

“Effects of Radiation on Mechanical Behavior of Crystalline Materials.” (PDF)

Plasticity and fracture of microelectronic thin films/lines
Effects of multidimensional defects on III-V semiconductor mechanics
Defect nucleation in crystalline metals
Role of water in accelerated fracture of fiber optic glass
Carbon nanotube mechanics
Superelastic and superplastic alloys
Mechanical behavior of a virus
Effects of radiation on mechanical behavior of crystalline materials | Problem Set 2 | Problem Set 3 | Problem Set 5

(a) Jitsukawa, et al. [1] summarizes steel’s mechanical properties by starting with tensile yield and fracture strength. Elastic behavior is not mentioned. Explain why there is no extensive database on the variation in elastic properties of steel as a function of composition, processing, or radiation exposure.

There is no extensive database on the variation in elastic properties of steel as a function of composition, processing, or radiation exposure because these values vary very little inside different classifications of steels, as can be seen in the following graph:

Data compiled from [2].

The yellow bars represent the variation of the Young’s Modulus in each category, while the red and blue lines show the minima and maxima in each category. As one can see, there isn’t much variation across the entire gamut of steels, with the notable exception of tool steels, though these are rarely used for large structural components such as nuclear pressure vessels due to their low ductility.

The fact that elastic properties do not vary much is because properties such as the Young’s Modulus is primarily as a result of the interatomic potential, U(r) and equilibrium atomic separation, r₀ :

Equation for the Young’s Modulus from material covered in class.

These shall not vary much for different steels due to the fact that they are majority-iron based alloys with relatively small percentages of carbon and other alloying elements. Tool steels tend to have much larger carbon concentrations as well as significant amounts of Mo and W to obtain the desired hardness and hence have the greatest variation in Young’s Modulus.

On the contrary other properties such as the yield stress and ultimate tensile strength DO vary significantly and are due to differences in composition and processing. For example 308 Stainless Steel has a yield stress of 205 MPa when fully annealed, yet this is increased by a factor 8 to a yield stress of 1660 MPa when quenched. [2]

(b) Given that you know linear elastic behavior (expected of alloys like steel) is terminated at the start of plastic/permanent deformation, regraph Jitsukawa’s Fig. 1 [1] in terms of yield strain as a function of operating temperature. You could consider Fig. 8 as a representative source of elastic deformation data, or glean the other papers/literature.

Fig. 2 below is a plot of yield strain as a function of temperature:

Data to create this graph taken from Fig.’s 1 and 8 of S. Jitsukawa, et al. [1]. Radiation levels for the irradiated specimen were 0.32 dpa at 300C.

In order to explain why irradiated steel would have a higher yield strain than unirradiated steel (counter-intuitive, isn’t it?) one must look at the energy required for dislocations motion. In a non-irradiated crystal, one must only apply enough stress to overcome lattice friction, or at least lattice friction with a smaller number of vacancies, interstitials and voids. The entire lattice can elastically and uniformly deform only so much before dislocations get moving. In an irradiated material, there are more radiation-induced defects to pin dislocations, and therefore one must continue to apply stress until the dislocations can be unpinned. This results in more elastic deformation of the lattice required to get to the strain where dislocations will begin to move.

(c) Based on Matthews & Finnis’ [3] speculation about where vacancy clusters and damage cascades occur within the crystal structure (Fig. 1), graphically represent the location and orientation of such vacancy clusters in steel. Here, you can assume a simpler alloy than the one discussed in Jitsukawa [1], Fe containing a (thermodynamic) equilibrium concentration of carbon and a supersaturation of vacancies.

First of all, assume that the steel is BCC iron with a normal amount of carbon atoms on octahedral interstitial site at a low concentration - typically around 0.5% for a medium carbon steel. Normally it is very difficult to form vacancies on the Fe lattice, but radiation can knock just about any atom off its lattice site, so we assume a supersaturation of Fe vacancies. In addition, there would likely be a large number of vacancies from the steel having been quenched, since the equilibrium number of vacancies at a given temperature is given by (C_v = n*e^(-Q/RT), where Q is the activation energy and T is the temperature. Since Q does not change, and the concentration (C_v ) is frozen in place at a higher T, there are many more vacancies than usual.

We have learned from 3.21 that forming an interstitial is more difficult than forming a vacancy. However, once an interstitial is formed, it tends to diffuse much faster than a vacancy. Furthermore the high amount of radiation present in a reactor would eliminate that barrier, so a damage cascade from a fast neutron (for example) will include a number of vacancies and interstitials. The interstitials will tend to diffuse away, while the vacancies are more likely to stay put as well as become trapped by other vacancies, forming voids in regions of high vacancy concentration, as shown in Fig. 1 of Matthews & Finnis, et al. [3]

In the voids, there could be as many as two vacancies per unit cell, as a BCC unit cell contains two full atoms. In vacancy rich regions not encompassed by voids, there could be any number of vacancies per unit cell (below two), where the vacancy concentration drops with distance away from any vacancy-rich core.

References

[1] Jitsukawa, S., et al. “Recent Results of the Reduced Activation Ferritic/Martensitic Steel Development.” Journal of Nuclear Materials 329-333 (2004): 39-46.

[2] MatWeb.

[3] Matthews, J. R., and M. W. Finnis. “Irradiation Creep Models - An Overview.” Journal of Nuclear Materials 159 (1988): 257-285.

(b) Garner, et al. [1] claim that vacancies such as those induced by radiation damage produce an isotropic strain field. Prove that this is so, and explain whether this strain contribution is expected to contribute to plastic deformation via dislocation glide.

“If microscopy is employed to measure swelling, the only major strain detected is that due to void formation, a process known to be completely isotropic in its distribution of strain.” – Garner, et al.

Assuming a void is a spherical section of missing material, and assuming the void is not near any surface, there will be equal pressure from all sides from lattice atom repulsion in a cubic material. In the case of a single radiation-induced vacancy, it is likely that once the vacancy-interstitial pair is produced the interstitial easily diffuses away because the activation barrier for interstitial diffusion is much lower than that for vacancy diffusion. This leaves a vacancy in a (perfect) lattice. Each of its nearest neighbors is an equal distance away from the core of the vacancy, and therefore the lattice exerts equal pressure on the vacancy from all sides. This will strain the lattice isotropically in the direction of the vacancy.

As far as dislocation glide is concerned, one must look at the interactions of the stress fields produced by both the dislocation and the vacancy. A vacancy’s stress field is isotropically tensile, because the nearest neighbors (and next and so on) will relax into the vacant site a bit, stretching the lattice locally. A vacancy in the path of a dislocation may impede its glide motion, because a vacancy interacting with a dislocation can reduce the overall energy of the system by being absorbed by the half plane, producing one atom’s worth of dislocation climb. This correlates with experimental evidence of irradiation at low temperatures causing materials to become stiffer, for at a low temperature vacancies cannot easily diffuse from the sites in which they were created. This results in an increased fracture stress, implying that the motion of dislocations has been impeded.

(c) Garner, et al. [1] then discuss that ion bombardment at metal surfaces can lead to plastic deformation, e.g., via blistering bubbles of subsurface trapped gases. Treat such a blister as a spherical pressure vessel surrounded by your choice of metals discussed in the other two papers (this will set the mechanical properties of the material surrounding the gas bubble. Assume a subsurface bubble size of 0.5 micrometer radius, and determine the gas pressure required to initiate plastic deformation and consequent roughness for a bubble centered 1 micrometer beneath the initially flat surface.

Assumptions

Material around gas bubble can be modeled as a spherical pressure vessel. Outer radius of pressure vessel, r_o, taken to be the center of gas bubble beneath the material surface

Since r/t < 10, must use “thick wall pressure vessel theory

RVE: Radial and hoop stresses are the principal stresses

From von Mises yield criteria:

(d) Others have noted the superposition of lateral stress that contributes to plastic yielding at the surface. Explain the source and expected contribution of lateral stress for this metal.

The lateral stress would have the effect of squeezing the FCC and BCC grains in the steel. This lateral stress is due to thermal expansion, and is hydrostatic everywhere except near the surface, where it is essentially a two dimensional stress. The grains will therefore expand in the direction normal to the surface, and Fig. 1 in Garner, et al. shows that the FCC austenite grains are strained more than the BCC ferrite grains.

In addition, lateral stresses arise from voids created by ion bombardment. These are often due to gas pressure or to gas-free void nucleation, and are isotropically distributed throughout the plane parallel to the surface assuming constant normal irradiation. All these lateral stresses would have the effect of squeezing all the grains from four of its six sides, and these lateral stresses will cause the grains to expand into the free surface.

Mathematically, if the bulk of a crystal experiences a hydrostatic stress S from a combination of thermal expansion, void creation, gas bubble formation, and other processes, than at the surface σ₁ = σ₂ = S, and σ₃ = 0. Therefore the lateral strain in the direction normal to the surface would be twice the Poisson’s ratio times the hydrostatic stress in the crystal bulk.

Reference

[1] Garner, F. A., and D. S. Gelles. “Irradiation Creep Mechanisms: An Experimental Perspective.” Journal of Nuclear Materials 159 (1988): 286-309.

Plasticity and fracture of microelectronic thin films/lines
Effects of multidimensional defects on III-V semiconductor mechanics
Defect nucleation in crystalline metals
Role of water in accelerated fracture of fiber optic glass
Carbon nanotube mechanics
Superelastic and superplastic alloys
Mechanical behavior of a virus
Effects of radiation on mechanical behavior of crystalline materials | Problem Set 2 | Problem Set 3 | Problem Set 5

(b) Aldo’s Fig. 3 [1] shows relation between effective creep stress and creep strain. Which of the materials/conditions considered exhibit a linear viscoelastic type creep response under this temperature and radiation dosage?

Courtesy of Elsevier, Inc., Science Direct. Used with permission.

Without seeing the unloading data, we have no way of knowing whether these materials are elastic-plastic or linear-visco-elastic. However, assuming that deviation from linear elasticity in the above figure is indeed viscoelastic behaviour then it is clear from the plot that mainly the JLF-1 and JLF-1 MN(b) materials exhibit LVE behavior, as these are the only materials that deformed quickly under a certain stress and therefore would have any hysteresis under their stress/strain curves. The F82H IEA sample might also exhibit some LVE behavior to a small extent, but it’s difficult to tell.

(c) The authors [1] give a creep model in Eq. 3. Restate this in terms of a steady state creep rate, and determine the creep mechanism in terms of those we have discussed in class. Which of the models we discussed best captures the trend observed by the authors? Assuming the appropriate energy barrier and diffusivity for your mechanism of choice, would you predict final creep strains comparable to the ones they measure at these stresses (where creep strain = strain rate*time)?

Restating the authors’ creep model in Eq. 3 as a steady state creep rate requires one to take a time derivative of both sides, so de/dt = 1.15(√(2)/3)*(dD/dt)/D_o

Since the change in diameter would be due to a radial stress, it should be linearly proportional to that hoop stress. This leads to a strain rate linearly proportional to the imposed hoop stress, and therefore it is a diffusional creep mechanism. Finally, these steels are quenched & tempered Ferritic/Martensitic steels, so the grain size is likely to be very small, and Coble Creep is therefore the dominant creep mechanism.

(d) The authors [1] state the following: “The HFIR irradiated specimens may have a larger Burgers vector anisotropy because they were exposed at higher stress levels and lower irradiation temperatures than the FFTF irradiated specimens. Detailed investigation of these microstructures following irradiation will be performed in future studies.”

What do they mean by “Burgers vector anisotropy”? And did they yet report any follow up work that correlate the microstructure of these steels with the observed creep phenomena?

By “Burgers Vector Anisotropy” the authors mean a larger variance in observed Burgers vector. At a lower irradiation temperature, less of the created defects would be able to diffuse away because there is less thermal energy available to overcome the activation energy barrier. This results in more imperfections in the lattice structure and therefore more variations in the Burgers vector, assuming it’s defined in the usual way of traveling a fixed distance along a path in 4 directions about a dislocation. The more vacancies and oddball atoms about near the dislocation, the more likely this path will result in a distance other than the average Burgers vector.

As yet, we have found no further reports by these authors correlating the microstructure of these steels with the observed creep phenomena.

Reference

[1] Ando, M., et al. “Creep Behavior of Reduced Activation Ferritic/Martensitic Steels Irradiated at 573 and 773 K at up to 5 dpa.” Journal of Nuclear Materials 367-370 (2007): 122-126.

Plasticity and fracture of microelectronic thin films/lines
Effects of multidimensional defects on III-V semiconductor mechanics
Defect nucleation in crystalline metals
Role of water in accelerated fracture of fiber optic glass
Carbon nanotube mechanics
Superelastic and superplastic alloys
Mechanical behavior of a virus
Effects of radiation on mechanical behavior of crystalline materials | Problem Set 2 | Problem Set 3 | Problem Set 5

Group Members

Agustin Mohedas
Kevin Huang
Heechul Park
Jing Shan

References

Zandi, R., and D. Reguera. “Mechanical Properties of Viral Capsids.” Physical Review E 72 (2005): 021917.

Ivanovska, I. L., P. J. Pablo, B. Ibarra, G. Sgalari, F. C. MacKintosh, J. L. Carrascosa, C. F. Schmidt, and G. J. L. Wuite. “Bacteriophage Capsids: Tough Nanoshells with Complex Elastic Properties.” PNAS 101 (2004): 7600-7605.

Buenemann, M., and P. Lenz. “Mechanical Limits of Viral Capsids.” PNAS 104 (2007): 9925-9930.

Wiki Assignments

(b) Zandi, et al. note that virus capsid behavior can be considered as a competition between bending and strain energies at the continuum level. State the magnitude of these two mechanical energy components on a flat plate of thickness equal to the viral capsid wall, of width and length equal to the CCMV capsid facet, and under a state of hydrostatic stress (pressure) plus a superposed normal point load at the center of the capsid facet.

(c) The authors rely on a thermodynamic definition of stresses and a Lennard-Jones inter-capsomer potential to predict the dependence of stress on capsid size. State their definition of the LJ potential in terms of the symbols we used in class, and explain why the radial and tangential stresses they compute from this potential (under the defined stress state) monotonically decrease with increasing capsid radius.

(d) Briefly summarize the implications of this trend in terms of virus behavior.

(b) Considering Ivanovska, et al.’s Fig. 1E rendering of a bacteriophage prohead “shell”, compute the stress required to initiate plastic deformation if the prohead were considered to be a short cylindrical shell under uniaxial compression.

(c) These authors state that imaging of the proheads in contact mode atomic force microscopy (AFM) allows for the study of prohead deformation under “uniaxial pressure.” This is incorrect at several levels. Develop a brief, justified objection to this claim, considering the design of the experiment in detail.

(d) These authors also claim that the mechanical testing of such proheads enables prediction of the elastic properties of the bacteriophage, by recourse to nonlinear elastic bucking of shells. Based on the data they present in this paper and known continuum mechanical analysis of shell elasticity, is this claim justified? Timoshenko is an excellent resource on mechanical analysis of shells.

Plasticity and fracture of microelectronic thin films/lines
Effects of multidimensional defects on III-V semiconductor mechanics
Defect nucleation in crystalline metals
Role of water in accelerated fracture of fiber optic glass
Carbon nanotube mechanics
Superelastic and superplastic alloys
Mechanical behavior of a virus | Problem Set 2 | Problem Set 3 | Problem Set 5
Effects of radiation on mechanical behavior of crystalline materials

(b) One of the cool things about a virus is that after the prohead shoots in the genetic material into the host cell (say, a bacterium), the genetic material is replicated until there is so much pressure that the host cell fractures. Assuming this is an E. coli bacterium with an initial crack size equal to the width of the prohead, find the Griffith fracture strength of the E. coli. You’ll have to assume some properties to do this, so cite and justify these.

(c) Ivanovska, et al. says the virus capsids are tough, but we said in class that this is the strain energy density up to the point of fracture. Assuming the fracture strength of a virus is 3 times the computed yield strength from PS3, and that the virus deforms in a linear elastic-linear strain hardening plastic mode up to the point of fracture, compute the toughness U_T . Discussion on p. 7604 of their paper is also helpful.

Final Presentation

“Mechanical Behavior of a Virus.” (PDF)

Plasticity and fracture of microelectronic thin films/lines
Effects of multidimensional defects on III-V semiconductor mechanics
Defect nucleation in crystalline metals
Role of water in accelerated fracture of fiber optic glass
Carbon nanotube mechanics
Superelastic and superplastic alloys
Mechanical behavior of a virus | Problem Set 2 | Problem Set 3 | Problem Set 5
Effects of radiation on mechanical behavior of crystalline materials

(b) Zandi, et al. [1] note that virus capsid behavior can be considered as a competition between bending and strain energies at the continuum level. State the magnitude of these two mechanical energy components on a flat plate of thickness equal to the viral capsid wall, of width and length equal to the CCMV capsid facet, and under a state of hydrostatic stress (pressure) plus a superposed normal point load at the center of the capsid facet.

The strain and bending energy are directly proportional to the elastic modulus and the bending rigidity respectively.
Thus by calculating the moduli we can determine the relative contributions to stress from strain and bending energies.

Assumptions: (from cited paper)

- Filled CCMV capsid with internal pressure of approximately 1 atm

- Capsid modeled as a thin shelled sphere with an applied point load stress

- CCMV: radius ~ 12 nm

- Föppl-von Kármán number (γ) describes the elastic properties of thin shelled spheres. γ ≡ R²(K_e/K_b), where R is the equatorial radius of the virus, K_e is the elastic modulus, and K_b is the bending rigidity. (It has been shown that γ ~ 300 for CCMV)

- Internal pressure of a filled CCMV is approximated by the following: π = 0.01 * K_e/R ~ 1 atm

Calculations:

- K_e = (1 atm*12 nm) / (0.01) = .122 N/m²

- K_b = (R²*K_e) / γ = 5.84E-20 Nm²

- K_e/K_b proportional to γ by γ ≡ R²(K_e/K_b)

- γ = 300, thus strain energy contributes 300 times more to the stress than the bending energy

As shown in Fig. 1 of [2], the elastic response of the capsid to the deformation strongly depends on the ratio of bending and elastic energy characterized by the FvK number.

(c) The authors [1] rely on a thermodynamic definition of stresses and a Lennard-Jones inter-capsomer potential to predict the dependence of stress on capsid size. State their definition of the LJ potential in terms of the symbols we used in class, and explain why the radial and tangential stresses they compute from this potential (under the defined stress state) monotonically decrease with increasing capsid radius.

ε₀ = 4*A_LJ

σ_ij = B₁

(2)^(1/6)*σ_ij = B₂

The convention used in the paper [1] is for compressive stress to be positive and tensile stress to be negative. Thus, as the capsid radius, R, increases (and likewise, the center-to-center separation between two capsomers in the capsid shell), the capsomers experience less and less compressive stress until the virus reaches an optimal radius, past which the capsomers begin to experience tensile stress. The stress monotonically decreases with increasing R. As you decrease the center-to-center capsomer separation (and virus radius) below the equilibrium value, a compressive stress is required to keep them from going back to their equilibrium. Likewise, if you increase the center-to-center capsomer separation beyond the equilibrium value, a tensile stress is required to hold them there. This is supported by the LJ potential, which describes that a repulsive force dominates at small separations (below some equilibrium separation) and an attractive force dominates beyond the equilibrium separation. Therefore at center-to-center capsomer distances below the equilibrium value, the aforementioned compressive stress is required to balance this repulsive force between capsomers while at separations beyond the equilibrium value, an attractive force is required to balance the tensile stress. Because the repulsive and attractive contributions to the LJ potential have different dependencies on R, the energy minimum is asymmetric. At low R below the equilibrium value, the repulsive potential rises faster than the attractive potential does at high R. Thus, the compressive stress at low center-to-center capsomer separations decreases with R faster than the tensile stress does at separations beyond the equilibrium value. This has implications on virus mechanical behavior as described below. (Please see attached graphs).

Images removed due to copyright restrictions. Please see [3] and Fig. 5 and 6 in [2].

(d) Briefly summarize the implications of this trend in terms of virus behavior.

This trend indicates that because there exists an equilibrium separation between capsomer centers (as shown by the minimum in the LJ potential), there is an optimal value for virus capsid radius for a given number of capsomers. Additionally, the asymmetry in the LJ potential indicates that the tensile stress required to stretch a virus capsid by a given ΔR is much less than the compressive stress required to compress the capsid by the same ΔR. As a result, viruses more easily fail by bursting or rupture due to swelling than they do by compression.

References

[1] Zandi, R., and D. Reguera. “Mechanical Properties of Viral Capsids.” Physical Review 72 (2005): 021917.

[2] Buenemann, M., and P. Lenz. “Mechanical Limits of Viral Capsids.” PNAS 104 (2007): 9925-9930.

[3] Wikimedia Commons.

(b) Considering Ivanovska, et al.’s Fig. 1E [1] rendering of a bacteriophage prohead “shell”, compute the stress required to initiate plastic deformation if the prohead were considered to be a short cylindrical shell under uniaxial compression.

Ivanovska, et al. [1] states on p. 7604 of the manuscript that proheads responded linearly to indentation forces up to approximately 2.8 nN. Define this as F_yield. From this, we can calculate the stress required to initiate plastic deformation, σ_y, based on the cross-sectional area A. In other words, we’re given F_yield = 2.8 nN; and σ_y = F_yield/A.

Finding cross-sectional area A: The manuscript states that the cantilever used was OMCL-RC800PSA from Olympus. Based on the manufacturer specifications [2], the probe is a sharpened pyramidal tip, but we’ll approximate it as a cone here with a tip radius r = 15 nm, tip angle θ = 45°, tip height h = 2.9 μm (Fig. 1). Because the prohead and cantilever are of comparable sizes, we cannot model the prohead surface area being acted upon as a flat plane. Thus we cannot simply take the tip area to be our cross sectional area. Our actual cross-sectional area is the projected contact area (shaded region in Fig. 1). Therefore, using cone geometry, we can find A via A = π(r₁)², where r₁ is as indicated in Fig. 3. To calculate this, we need to know the strain at yield point, which is provided by the manuscript (pg. 7604): strain_yield = maximum linear displacement = 12 nm. Using trigonometry, we’re able to find the following value: r₁ = 19.97 nm. Therefore, A = 1252.2 nm².

Fig. 1. Cantilever Probe Tip. Dimensions based on manufacturer specifications of OMCL-RC800PSA from Olympus. Tip is approximated as a cone with tip radius r, tip height h and tip angle theta. Shaded region indicates cross sectional area used for yield stress calculation.

Finding σ_y: σ_y = 2.8 nN / (1252.2 nm²) = 2.2 MPa

(c) These authors [1] state that imaging of the proheads in contact mode atomic force microscopy (AFM) allows for the study of prohead deformation under “uniaxial pressure.” This is incorrect at several levels. Develop a brief, justified objection to this claim, considering the design of the experiment in detail.

The authors used scanning force microscopy (SFM) to image the proheads under different maximal loading forces and assumed this to be “uniaxial pressure”. However, given the imaging modality rasters an AFM tip, with a contact region of approximately 5 nm, across the virus prohead, which itself has an inherent curvature, the applied force is clearly not uniaxial, see figure below. Along the curvature of the virus prohead the applied force is resolved into various components. Also, the size of AFM tips (< 20 nm) are on the order of the size of a prohead and thus cannot be considered as a point load when forces are applied to the prohead.

(d) These authors also claim that the mechanical testing of such proheads enables prediction of the elastic properties of the bacteriophage, by recourse to nonlinear elastic bucking of shells. Based on the data they present in this paper and known continuum mechanical analysis of shell elasticity, is this claim justified? Timoshenko [5] is an excellent resource on mechanical analysis of shells.

The authors do nanoindentaion of the bacteriophage φ 29 shells and then model the shells using a continuum approach, assuming it is homogeneous and a thin shell. However, as we know the shell’s microstructure is inhomogeneous and they even admitted there is a bimodal distribution of elastic moduli. Also, the authors considered the shells as spherical thin shells, assuming h/R = 0.1. In order to be regraded as flat plates (thin shells), the shells may have the thickness (h) which is less than one hundredth of the least radius (R) of curvature [3]. This means that the nonlinearity of the shells can be reduced to the solvable linear shells [4]. If the ratio of the thickness to the radius is comparable to the one tenth like this case, the shells may be considered as the moderately thick shells. In thick shells, the interaction due to bending term and stretching term cannot be evaded. Therefore, their claim would not be justified.

References

[1] Ivanovska, I. L., P. J. Pablo, B. Ibarra, G. Sgalari, F. C. MacKintosh, J. L. Carrascosa, C. F. Schmidt, and G. J. L. Wuite. “Bacteriophage Capsids: Tough Nanoshells with Complex Elastic Properties.” PNAS 101 (2004): 7600-7605.

[2] Micro Cantilever.

[3] Cox, H. L. The Buckling of Cylindrical Plates and Shells. New York, NY: Pergamon, 1963, pp. 71-72.

[4] Gould, Phillip L. Analysis of Shells and Plates. Upper Saddle River, NJ: Prentice Hall, 1998, pp. 461-463. ISBN: 9780133749502.

[5] Timoshenko, S., and S. Woinowsky-Krieger. Theory of Plates and Shells . New York, NY: McGraw-Hill, 1959.

Plasticity and fracture of microelectronic thin films/lines
Effects of multidimensional defects on III-V semiconductor mechanics
Defect nucleation in crystalline metals
Roleof water in accelerated fracture of fiber optic glass
Carbon nanotube mechanics
Superelastic and superplastic alloys
Mechanical behavior of a virus | Problem Set 2 | Problem Set 3 | Problem Set 5
Effects of radiation on mechanical behavior of crystalline materials

Assumptions (from literature):

crack size, a = 70 nm [1]
Elastic modulus, E = 0.4 GPa [2, 3]
Surface energy, γ = 52 mN/m [4, 5]

Computation:

σ = (2Eγ/πa)^(1/2) = 13.8 MPa

(c) Ivanovska, et al. [1] says the virus capsids are tough, but we said in class that this is the strain energy density up to the point of fracture. Assuming the fracture strength of a virus is 3 times the computed yield strength from PS3, and that the virus deforms in a linear elastic-linear strain hardening plastic mode up to the point of fracture, compute the toughness U_T. Discussion on p. 7604 of their paper is also helpful.

From PS3 we calculated the yield stress to be 2.2 MPA and thus we can assume the fracture stress to be 3 times this value or 6.6 MPA.

From the following figures, taken from Ivanovska, et al., we see that the prohead dimension along the axis of stress/strain is 38 nm (Fig. 1). From Fig. 3, we see that the total strain distance the AFM tip moved from contact to fracture to be 62 nm - 29 nm = 33 nm. Thus the strain experienced by the prohead at fracture (rupture) is 33 nm/38 nm = .85.

Applying these values and the assumption of linear elastic and linear strain hardening we get a stress/strain curve as shown below.

Courtesy of National Academy of Sciences, USA. Used with permission.
Source: Ivanovska, I. L., P. J. Pablo, B. Ibarra, G. Sgalari, F. C. MacKintosh, J. L. Carrascosa, C. F. Schmidt, and G. J. L. Wuite. “Bacteriophage capsids: Tough nanoshells with complex elastic properties.” PNAS 101 (2004): 7600-7605. Copyright 2004 National Academy of Sciences, USA.

The toughness can be computed from the figure below by finding the area under the stress/strain curve.

1/2*.29*2.2 MPa + (.85 - .29)2.2 MPa + 1/2(.85 - .29)(6.6 MPa - 2.2 MPa) = 2.783 MPa

References

[2] Ramos, Daniel, et al. “Measurement of the Mass and Rigidity of Adsorbates on a Microcantilever Sensor.” Sensors 7 (2007): 1834-1845.

[3] Yao, X., et al. “Thickness and Elasticity of Gram-Negative Murein Sacculi Measured by Atomic Force Microscopy.” Journal of Bacteriology 181 (November 1999): 6865-6875.

[4] Lopez-Montero, Ivan, et al. “High Fluidity and Soft Elasticity of the Inner Membrane of Escherischia coli Revealed by the Surface Rheology of Model Langmuir Monolayers.” Langmuir 24 (2008): 4065-4076.

[5] van Oss, C. J., A. Docoslis, and R. F. Giese. “Role of the Water-air Interface in Determining the Surface Tension of Aqueous Solutions of Sugars, Polysaccharides, Proteins, and Surfactants.” Contact Angle, Wettability, and Adhesion 2 (2002): 3-12.

Plasticity and fracture of microelectronic thin films/lines
Effects of multidimensional defects on III-V semiconductor mechanics
Defect nucleation in crystalline metals
Role of water in accelerated fracture of fiber optic glass
Carbon nanotube mechanics
Superelastic and superplastic alloys
Mechanical behavior of a virus | Problem Set 2 | Problem Set 3 | Problem Set 5
Effects of radiation on mechanical behavior of crystalline materials

Group Members

Xing Sheng
Liang-Yi Chang
Hang Yu
Lin Jia

References

Celler, G. K., and Sorin Cristoloveanu. “Frontiers of Silicon-on-Insulator.” Journal of Applied Physics 93 (May 1, 2003): 4955-4978.

Miller, David C., Brad L. Boyce, Michael T. Dugger, Thomas E. Buchheit, and Ken Gall. “Characteristics of a Commercially Available Silicon-on-Insulator MEMS Materials.” Sensors and Actuators A 138 (2007): 130-144.

Ghasemi-Nejhad, Mehrdad N., Chiling Pan, and Hongwei Feng. “Intrinsic Strain Modeling and Residual Stress Analysis for Thin-Film Processing of Layered Structures.” Journal of Electronic Packaging 125 (March 2003): 4-17.

Wiki Assignments

(a) Miller, et al. noted the use of wafer curvature to infer stresses within thin films such as SOI. Derive the stress-curvature relationship given in Eq. 1, also known as the Stoney formula in the thin-film limit.

(b) Graphically represent the structure of Si in SOI, at the unit cell and micrometer-scale level, and separately indicate the closest packed plane and direction in that plane for this structure.

(c) During wafer bonding, blisters can occur at the wafer interfaces (Celler, Fig. 9). State the full tensorial stress state at the location of such a blister, and quantify the strains you would expect to correspond to this state in the adjacent Si.

(b) Nejhad, et al. present an analysis of retained internal stresses, and an associated elastic analysis in Eq. (8). Justify the form of Eq. (8) in terms of crystal structure of the two film materials of interest, and the origin of the strain terms.

(c) After Eq. (10), the authors state that there are so many contributors of intrinsic strain that it is difficult to consider them quantitatively. Rank-order the sources they list in terms of relative amount of stress contribution for this system, and justify your answer.

(d) Determine the % of residual stress in terms of the tensile and compressive failure strength of these two film materials. In other words, how closely does the residual stress in films or lines approach the failure stress of the bulk forms of these same materials?

Plasticity and fracture of microelectronic thin films/lines | Problem Set 2 | Problem Set 3 | Problem Set 5
Effects of multidimensional defects on III-V semiconductor mechanics
Defect nucleation in crystalline metals
Role of water in accelerated fracture of fiber optic glass
Carbon nanotube mechanics
Superelastic and superplastic alloys
Mechanical behavior of a virus
Effects of radiation on mechanical behavior of crystalline materials

(b) Silicon on insulator delamination can be modeled by fracture between two dissimilar interfaces: Si and SiO₂, for example. Assume that the interface blister you considered earlier is now an initial crack at the interface, and compute the fracture stress required to delaminate the film under Mode I loading. You can cite any required properties to implement a simple fracture theory such as Griffith’s.

(c) In section 4 of the Frontiers paper, the authors discuss Bruel’s innovation as leading to the realization that fracture could be induced by shearing. What kind of mode of loading is this, and will the stress required to initiate fracture for a given material be lower in this mode than in Mode I?

Final Presentation

“Plasticity and fracture of microelectronic thin films [SOI - Silicon on Insulator.]” (PDF)

- Plasticity and fracture of microelectronic thin films/lines | Problem Set 2 | Problem Set 3 | Problem Set 5
- Effects of multidimensional defects on III-V semiconductor mechanics
- Defect nucleation in crystalline metals
- Role of water in accelerated fracture of fiber optic glass
- Carbon nanotube mechanics
- Superelastic and superplastic alloys
- Mechanical behavior of a virus
- Effects of radiation on mechanical behavior of crystalline materials

(a) Miller, et al. [1] noted the use of wafer curvature to infer stresses within thin films such as SOI. Derive the stress-curvature relationship given in Eq. 1, also known as the Stoney formula in the thin-film limit.

See the definitions of the symbols in Fig. 1, where l is the cantilever length. L is the arc length after bending, α is its corresponding center angle. θ is the angle between the original length and the bended length. R is the radius of the arch L. From the geometry

Relations we know:

θ + β = π/2

2 β + α = π

thus α = 2θ

On the other hand, in the right-angle triangle l L δ

l² + δ² = L²

In the limitation that δ<<L, θ = δ/L

The length of the arch L corresponds to an center angle of α by

α = L/R

thus κ = 1/R = α/L = 2θ/L = 2δ/L² = 2δ/(l² + δ²)

(b) Graphically represent the structure of Si in SOI, at the unit cell and micrometer-scale level, and separately indicate the closest packed plane and direction in that plane for this structure.

At the unit cell scale level, Si has diamond structure. The unit cell is FCC, with Si atoms filling half the tetrahedron interstitial sites. The closest packed plane is {111}, and the closest packed direction in this plane is plane is <110>.

At micrometer-scale level, silicon thin film is single crystalline, with (100) oriented normal to the surface. SiO₂ is amorphous. The substrate can be single silicon wafer, sapphire, etc.

(c) During wafer bonding, blisters can occur at the wafer interfaces (Celler, Fig. 9 [2]). State the full tensorial stress state at the location of such a blister, and quantify the strains you would expect to correspond to this state in the adjacent Si.

Image removed due to copyright restrictions. Please see Fig. 9 in [2].

1. We use a sphere to simulate the blister as in Fig. 1.

2. The stress within the film equals σ.

3. In the Smart Cut process, the thickness of the blister is relatively small compared to the radius of curvature of the blister. Therefore, we can use thin-wall approximation to approach this problem.

Stress:

From Fig. 2,

P = pressure in the blister

R = radius of curvature of the blister

t = film thickness

4. Radius of curvature of the blister:

From Fig. 3,

a = radius of the circular base of the blister

h = height at the center of the blister

5. Strain:

Before forming a blister, the profile of the film is along X_LX_R in Fig. 3. However, after forming a blister, the film profile is along X_LX_HX_R. Therefore, the length of film along x or y direction is stretched from |X_LX_R| to |X_LX_HX_R|. As a result, the strain along x or y direction is

6. Stress state:

7. Strain state:

ν = Poisson’s ratio of the film

E = elastic modulus of the film

For instance, a = 1000 nm; h = 40 nm; ν = 0.16 [3]

References

[1] Miller, David C., Brad L. Boyce, Michael T. Dugger, Thomas E. Buchheit, and Ken Gall. “Characteristics of a Commercially Available Silicon-on-Insulator MEMS Materials.” Sensors and Actuators A 138 (2007): 130-144.

[2] Celler, G. K., and Sorin Cristoloveanu. “Frontiers of Silicon-on-Insulator.” Journal of Applied Physics 93 (May 1, 2003): 4955-4978.

[3] Piatkowska, A., G. Gawlik, and J. Jagielski. “AFM Studies of Hydrogen Implanted Silicon.” Applied Surface Science 141 (March 1999): 333-338.

(b) Nejhad, et al. [1] present an analysis of retained internal stresses, and an associated elastic analysis in Eq. (8). Justify the form of Eq. (8) in terms of crystal structure of the two film materials of interest, and the origin of the strain terms.

Image removed due to copyright restrictions. Please see Fig. 7 in [1].

Let us consider the two-film structure model above; we can refer to ZnO as material 1, and Si₃N₄ as material 2.

And as shown by the paper, we have the table below:

Table removed due to copyright restrictions. Please see Table 1 in [1].

We can separate the compliance matrix to two parts: the first one is related to the 1, 2, 3 dimension, and the other is related to the 4, 5, 6 dimension below:

For Si₃N₄ as material 2,

For ZnO as material 1,

For the two-layer composite, we can predict that the matrix has the below form,

Now we need to determine P, Q, J, L, K and M from the matrices of materials 1 and 2. Let us determine them one by one.

We set the volume fraction of material 1 as a and the volume fraction of material 2 as b.The equation is shown below,

= extrinsic strain = thermal strain in this work

= intrinsic strain =

dopants
“atomic peening” which causes compressive stress
microvoids which cause tensile stress
gas entrapment which causes compressive stress
shrinkage during process which causes tensile stress
grain boundaries

(c) After Eq. (10) [1], the authors state that there are so many contributors of intrinsic strain that it is difficult to consider them quantitatively. Rank-order the sources they list in terms of relative amount of stress contribution for this system, and justify your answer.

The contributors of intrinsic strain mentioned in the literature are:

dopants
“atomic peening” which causes compressive stress
microvoids which cause tensile stress
gas entrapment which causes compressive stress
shrinkage during process which causes tensile stress
grain boundaries

For this system, the Si₃N₄ and ZnO films are deposited through PECVD.

Among the 6 contributors above;

No dopants are introduced in the process of PECVD method, so this effect is very small.
The “atomic peening” mechanism refers to the reflected neutral atoms bombarding the growing film at low sputtering pressures. The paper does not use sputtering method, so this effect is also small.
Microvoids are caused by poor quality of film during process, which depends on the material, the environment, and the growth rate during PECVD.
Since plasma (Ar, for example) is used in PECVD, it is very possible that gas will be entrapped in the film. Therefore this contribution is relatively significant.
Shrinkage often happens in ceramics during sintering. There is no sintering in the process, so this contributor is small.
For Si₃N₄, it is an amorphous film, so there are no grain boundaries. For ZnO, this effect is possible.

Integrating and rank-ordering all the effects, I get the conclusion below:

4 > 3 > 6 > 5 > 1 > 2

Use the material properties give in Table I,

Given that residual strain is 10^-12

Substituting S matrix and residual strain into the residual stress tensor equation given above, we have residual stress,

Provided yield stress of silicon nitride and zinc oxide is 900 and 412 MPa respectively.

Therefore, the residual stress is very small compared to the yield stress of silicon nitride or zinc oxide.

Reference

[1] Ghasemi-Nejhad, Mehrdad N., Chiling Pan, and Hongwei Feng. “Intrinsic Strain Modeling and Residual Stress Analysis for Thin-Film Processing of Layered Structures.” Journal of Electronic Packaging 125 (March 2003): 4-17.

Image removed due to copyright restrictions. Please see Fig. 9 in [5], with the dimensions a (width of the crack) and t (length of the crack) marked.

(c) In section 4 of the Frontiers paper [5], the authors discuss Bruel’s innovation as leading to the realization that fracture could be induced by shearing. What kind of mode of loading is this, and will the stress required to initiate fracture for a given material be lower in this mode than in Mode I?

The crack propagating in the direction normal to the shear stress, so the load is Mode III.

The stress required to initiate fracture for a given material will be higher in this mode than in Mode I. Mode I fracture is preceded only by linear elastic deformation. Mode III fracture is preceded by macroscopic yielding. The G_c values (fracture toughness) for Mode III fracture are typically higher than for Mode I and Mode II fracture as a result of increased crack tip blunting (which raises ρ) and greater values of ε.

References

[1] Tong, Q.-Y., Q. Gan, G. Hudson, G. Fountain, and P. Enquist. “Low-Temperature Hydrophobic Silicon Wafer Bonding.” Applied Physics Letters 83 (8 December 2003): 4767-4769.

[2] Asay, David B., and Seong H. Kim. “Effects of Adsorbed Water Layer Structure on Adhesion Force of Silicon Oxide Nanoasperity Contact in Humid Ambient.” Journal of Chemical Physics 124 (2006): 174712.

[3] Tu, Yuhai, and J. Tersoff. “Structure of the Silicon-Oxide Interface.” Thin Solid Films 400 (2001): 95-100.

[4] MatWeb.

[5] Celler, G. K., and Sorin Cristoloveanu. “Frontiers of Silicon-on-Insulator.” Journal of Applied Physics 93 (May 1, 2003): 4955-4978.

[6] Henley, F. J., and N. W. Cheung. “Method for Controlled Cleaving Process.” U.S. Patent No. 6,033,974.

[7] Courtney. Mechanical Behavior of Materials. 2nd ed. Long Grove, IL: Waveland Press Inc, 2005. ISBN: 9781577664253.

Group Members

Matthew Smith
Sophie Poizeau
Michiel Vanhoutte
Noémie Chocat

Image removed due to copyright restrictions.
Please see: https://web.archive.org/web/20121213052722/http://www.rtcfiber.com/layout/multiflex3/images/fiber-optics.jpg

References

Kurkjian, C. R., J. T. Krause, and M. J. Matthewson. “Strength and Fatigue of Silica Optical Fibers.” Journal of Lightwave Technology 7 (1989): 1360-1370.

The above review article is a thorough review of the mechanics of silica optical fibers, but it was published in 1989. Here is a more recent but less focused review, in case 1989 is not sufficiently recent:

Suhir, E. “Fiber Optic: Structural Mechanics and Nanotechnology Based New Generation of Fiber Coatings.” Proceedings of the SPIE 6126 (2006): 612606.

Mauro do Nascimento, E., and C. M. Lepienski. “Mechanical Properties of Optical Glass Fibers Damaged by Nanoindentation and Water Ageing.” Journal of Non-Crystalline Solids 352 (2006): 3556-3560.

Michalske, T. A., and B. C. Bunker. “Slow Fracture Model Based on Strained Silicate Structures.” Journal of Applied Physics 56 (1984): 2686-2693.

Wiki Assignments

a) What are the (short-range ordered) structures and elastic moduli of E-glass (aluminum-barium borosilicate glass) and silica glass in the bulk form?

b) Would you expect the stiffness of these two types of glass to change as the physical dimensions were reduced to the diameters typical of current (2008) fiber optic applications? How about for nanoscale fibers? Why or why not? See also Silva, et al.

c) State the Cij matrix you feel best captures the elastic constants of these glasses, and justify your answer in terms of the material structure.

d) Kurkjian, et al. noted that Sakaguchi had intentionally added dust particles to increase the failure stress of such glass fibers. What would the impact on isotropic elastic properties be for the fibers Sakaguchi created, for the sizes and volume fractions of particles he used?

(b) Explain the physical mechanisms and mechanical consequences of aging in such fiber optic silica.

(c) Explain how do Nascimento and Lepienski calculated the fracture stress for glasses subjected to Berkovich indentation in Table 1. What was the point of using the indenter in this experiment? If they were instead to indent the fiber sufficiently to attain the fracture stress by loading the fiber surface with a Berkovich diamond indenter, state the corresponding indentation loads and depths for the pristine and maximally aged fibers.

(d) Graphically represent a cross-sectional view of this proposed fracture stress experiment on the fibers, considering the relative lengthscales of the fiber, the depth of indentation corresponding to fracture noted in your calculations for (c), and the finite radius of diamond Berkovich indenter probes.

[2] Composition for Ceramic Substrate and Ceramic Circuit Component.

(b) Michalske predicted that water molecules assisted bond breaking in silica long before it was observed directly in atomistic simulations. In his 1984 paper, Fig. 6 is a little hard to understand at first. Make this the inset in a larger figure that shows a macroscopic crack, the direction of loading to open that crack in Mode I, and the direction of propagation of that crack.

(c) Explain chemically assisted fracture so that anyone in 3.22 would understand it, using Michalske’s Fig. 7 type graphics to illustrate the interaction of water with silica at the molecular level.

(d) Michalske’s argument necessarily implies that the amount of chemical in the environment affects the speed of crack growth. Assume an initial, through-thickness crack of width 150 nm at the surface of a sheet of silica that is 1 cm thick. The crack is under Mode I loading. From Michalske’s Fig. 2, how long would it take for that crack to grow another 150 nm under an applied stress of 100 MPa in an environment of 100% water?

(e) At a partial pressure of 50%, how many water molecules would be sitting on the initial crack faces?

Final Presentation

“Water-ageing of Silica Optical Fibers.” (PDF)

Plasticity and fracture of microelectronic thin films/lines
Effects of multidimensional defects on III-V semiconductor mechanics
Defect nucleation in crystalline metals
Role of water in accelerated fracture of fiber optic glass | Problem Set 2 | Problem Set 3 | Problem Set 5
Carbon nanotube mechanics
Superelastic and superplastic alloys
Mechanical behavior of a virus
Effects of radiation on mechanical behavior of crystalline materials

a) What are the (short-range ordered) structures and elastic moduli of E-glass (aluminum-barium borosilicate glass) and silica glass in the bulk form?

Silica (SiO₂) glass is amorphous. This means that there is no long-range order, but there is a short-range order. This local ordering consists of a tetrahedral arrangement of O and Si atoms (see Fig. [1]).

E glass still consists for >50% of SiO₂. SiO₂ and B₂O₃ are the predominant network formers. BaO acts as a network modifier and Al₂O₃ is a glass network intermediate, which together with BaO has the capacity to form a network [2]. Just as for silica, there is no long-range order in E glass, which makes it an amorphous material. The specific short-range order in E glass will depend on its constituent atoms.

As for the elastic moduli, for SiO₂ we find E = 72 GPa [3] and for E glass we find E = 80 GPa [4].

Image removed due to copyright restrictions. Please see Fig. 3 in [3].

According to Silva, et al. [3], the stiffness of silica does not depend on the physical dimensions for diameters above 280 nm. The core of single-mode optical fibers used for near-infrared applications such as telecommunications is typically >5 μm. This is safely above the 280 nm limit. As for nanoscale fibers, Dikin, et al. [5] report a substantial but unexplained decrease of the Young’s elastic modulus E below 100 nm. Around diameter of the order of 10 nm, Silva, et al. [3] report that the Young’s modulus increases again. This elastic stiffening can be understood on the basis of surface-constrained elasticity. Tensile stresses generated at the material surface contract the wire and increase the stiffness of the core material within the wire, due to nonlinear elasticity effects. Overall, there is an increase of 30% in stiffness as the diameter decreases from 100 nm to 10 nm due to the increased surface-area-to-volume ratio of the network solid.

c) State the C_ij matrix you feel best captures the elastic constants of these glasses, and justify your answer in terms of the material structure.

Amorphous materials, having no long range structure, are inherently isotropic. The shape of an isotropic the C_ij matrix is found on page 141 of “Physical Properties of Materials,” J. F. Nye [6].

d) Kurkjian, et al. [7] noted that Sakaguchi [8] had intentionally added dust particles to increase the failure stress of such glass fibers. What would the impact on isotropic elastic properties be for the fibers Sakaguchi created, for the sizes and volume fractions of particles he used?

According to Chapter 6 in Courtney [9], the elastic modulus E_c of a two-phase (glass matrix + dust particles) composite lies between the values of two extreme cases: a laminar stack of two materials alternating each other, with the force applied in the direction of the stacking on the one hand, and the same composite but with the force applied orthogonal to the stacking direction on the other hand.

In the first case, each layer in the stack experiences an equal stress, and the elastic modulus E_c is calculated to be (Courtney 6.8):

with index g for the glass matrix and p for the dust particles. The last step is a Taylor expansion for the case that V_p << 1.

In the second case, each layer in the stack experiences an equal strain, and the elastic modulus is found to be (Courtney 6.12):

When Sakaguchi introduces dust particles with E_p > E_g, such as alumina (E[Al₂O₃]= 300 Gpa) or carbon (E[SiC]= 420 GPa), both of the cases predict an increase of the Young’s modulus of the glass fibers. The real value of E_c, which lies in between the two extrema, will thus also increase upon introduction of dust particles. This real value is sometimes described by an empirical relationship (Courtney 6.13):

where K_c < 1 is empirically determined.

To find a numerical value for the elastic modulus E_c, we first need to estimate the volume fraction V_p. In the most simple approximation, we can imagine the dust particles of diameter d to make up a mono-atomic shell of thickness d around the fiber of radius r. This corresponds in fact to the equal strain case described above. The volume fraction V_p is then given by the ratio of the cross-sectional area of the shell to the area of the fiber:

where the last approximation is valid for d << r. For the diameter of the silica fiber, we can use the value of 2*r = 125 μm mentioned in the article by do Nascimento [10].

For the alumina dust particles of d = 0.03 μm, 0.3 μm and 1 μm, the approximation d << r is valid and we find:

d = 0.03 μm → V_p = 0.00096 → E_c = 72.22 GPa = 1.003*E_g
d = 0.3 μm → V_p = 0.0096 → E_c = 74.2 GPa = 1.030*E_g
d = 1 μm → V_p = 0.032 → E_c = 79.4 GPa = 1.102*E_g

For the carbon dust particles of d = 20 μm (we don’t use the approximation d << r), we find: V_P = 0.74 ≥ E_c = 330.4 GPa = 4.6*E_g. Note the increase by a factor ~5 in this simplified model!

We can elaborate on the validity of our assumption that the dust particles form a shell of thickness d around the fiber. In reality, the dust particles “surface density” may be not high enough to consider the dust particles as a dense shell around the fiber. A surface density factor in the expression for V_p can account for this. Also, we cannot use the equal strain expression (K_c = 1) anymore, but we have to work with a value for K_c < 1.

References

[1] Wikimedia Commons.

[3] Silva, Emilio C. C. M., et al. “Size Effects on the Stiffness of Silica Nanowires.” Small 2 (2006): 239-243.

[4] The A to Z of Materials.

[5] Dikin, D. A., et al. “Resonance Vibration of Amorphous SiO₂ Nanowires Driven by Mechanical or Electrical Field Excitation.” Journal of Applied Physics 93 (2003): 226-230.

[6] Nye, J. F. Physical Properties of Crystals. New York, NY: Oxford University Press, 1985.

[7] Kurkjian, C. R., J. T. Krause, and M. J. Matthewson. “Strength and Fatigue of Silica Optical Fibers.” Journal of Lightwave Technology 7 (1989): 1360-1370.

[8] Sakaguchi, Shigeki, and Yoshinori Hibino. “Fatigue in Low-strength Silica Optical Fibers.” Journal of Materials Science 19 (October 1984): 3416-3420.

[9] Courtney. Mechanical Behavior of Materials. 2nd ed. Long Grove, IL: Waveland Press Inc., 2005. ISBN: 9781577664253.

[10] Mauro do Nascimento, Eduardo, and Carlos Mauricio Lepienski. “Mechanical Properties of Optical Glass Fibers Damaged by Nanoindentation and Water Ageing.” Journal of Non-Crystalline Solids 352 (September 15, 2006): 3556-3560.

Plasticity and fracture of microelectronic thin films/lines
Effects of multidimensional defects on III-V semiconductor mechanics
Defect nucleation in crystalline metals
Role of water in accelerated fracture of fiber optic glass | Problem Set 2 | Problem Set 3 | Problem Set 5
Carbon nanotube mechanics
Superelastic and superplastic alloys
Mechanical behavior of a virus
Effects of radiation on mechanical behavior of crystalline materials

(b) Explain the physical mechanisms and mechanical consequences of aging in such fiber optic silica.

Over time, the mechanical properties of optical fibers are degraded via chemical attacks from the environment. The most prevalent example of this is the aging effect of water on silica fibers. H₂O molecules can react with Si-O-Si bonds on the surface of the glass and at the crack tip, thus increasing the crack size with time and creating new surface flaws.

Image removed due to copyright restrictions. Please see Fig. 1 in [1].

This is considered a type of zero-load aging (do Nascimento, et al. [2]), because it can occur even in the absence of an applied stress. The addition of new surface flaws and the gradual increase in crack length with time both strongly influence the bulk mechanical properties. The general equation for crack propagation and failure length is [2]:

Following that the fracture toughness is a material constant, as the crack length increases with time the crack propagation stress will decrease and the mechanical strength of the fiber will follow. For example, do Nascimento and Lepienski show a decrease in fiber resistance from 4.1 GPa to 2.9 GPa over four weeks at 85°C and 85% humidity.

(c) Explain how do Nascimento and Lepienski [2] calculated the fracture stress for glasses subjected to Berkovich indentation in Table 1. What was the point of using the indenter in this experiment? If they were instead to indent the fiber sufficiently to attain the fracture stress by loading the fiber surface with a Berkovich diamond indenter, state the corresponding indentation loads and depths for the pristine and maximally aged fibers.

The fracture stresses in Table 1 [2] were obtained using a tensile test. The indented fibers were subjected to a tensile load of 100N, with displacement rates of 0.5, 5, 50 and 500 mm/min until failure. The failure probability was then plotted versus the tensile stress. The fracture stress in Table 1 is then taken to be equal to the tensile stress corresponding to a 50% failure probability. For example, for 10 mN Berkovich indented fibers, the fracture stress is found to be 0.35 GPa:

Courtesy of Elsevier, Inc., Science Direct. Used with permission.

The point of using an indenter in this experiment is to be able to make cracks of known and controlled depth. Using a 10 mN load with a Berkovich indenter, do Nascimento and Lipienski obtained an indentation depth of 150 nm, as shown in this load vs. displacement curve:

Courtesy of Elsevier, Inc., Science Direct. Used with permission.

These 10 mN Berkovich indentations were then used as a reference to calculate the indentation depths for the other fibers and estimate the crack size for the water aged fibers. Using the equation K/ψ = σ√c, with a fracture load σ = 0.35 GPa and an indentation depth c = 150 nm (values for the reference indentations), the authors of the paper find the ratio K/ψ to be equal to 0.13 MPa√m. Since this ratio is constant for a given material, they are then able to estimate the typical flaw size in other fibers from the fracture stress.

Let’s now suppose that they performed another experiment, where they indented the fiber sufficiently to attain the fracture stress by loading the fiber surface with a Berkovich diamond indenter. To calculate the applied stress of a Berkovich indenter, we need to know the applied force and the applied area. In nano-indentation the depth of penetration is measured as the load is applied. Knowledge of the geometry of the indenter allows for the calculation of the contact area as a function of penetration depth. This calculation is done below, using the standard Berkovich indenter geometry:

Image removed due to copyright restrictions. Please see slide 28 in Kim, Do Kyung. “Nanoindentation: Lecture 1 Basic Principles.” (PPT)

From the relation σ = F/A and the above expression of A as a function of h, we can deduce the following equation for the indentation depth as a function of the fracture stress and the indentation load:

Assuming an indentation load of 10 mN, we can estimate that the indentation depth for pristine fibers is 315 nm (with σ = 4.1 GPa) and that for 4 weeks aged fibers it is 381 nm (with σ = 2.8 GPa).

Image made using Photoshop. Finite tip radius as calculated by [3].

References

[1] Michalske, T. A., and B. C. Bunker. “Slow Fracture Model Based on Silicate Fracture Models.” Journal of Applied Physics 56 (November 15, 1984): 2686-2693.

[2] do Nascimento, E. Mauro, and C. M. Lepienski. “Mechanical Properties of Optical Glass Fibers Damaged by Nanoindentation and Water Ageing.” Journal of Non-Crystalline Solids 352 (2006): 3556-3560.

[3] Yu, Ning, Andreas A. Polycarpou, and Thomas F. Conry. “Tip-radius Effect in Finite Element Modeling of Sub-50 nm Shallow Nanoindentation.” Thin Solid Films 450 (March 1, 2004): 295-303.

Plasticity and fracture of microelectronic thin films/lines
Effects of multidimensional defects on III-V semiconductor mechanics
Defect nucleation in crystalline metals
Role of water in accelerated fracture of fiber optic glass | Problem Set 2 | Problem Set 3 | Problem Set 5
Carbon nanotube mechanics
Superelastic and superplastic alloys
Mechanical behavior of a virus
Effects of radiation on mechanical behavior of crystalline materials

(b) Michalske [1] predicted that water molecules assisted bond breaking in silica long before it was observed directly in atomistic simulations. In his 1984 paper, Fig. 6 is a little hard to understand at first. Make this the inset in a larger figure that shows a macroscopic crack, the direction of loading to open that crack in Mode I, and the direction of propagation of that crack.

Below is a larger figure that shows the macroscopic crack. The figure 6 is the one from Michalske’s 1984 paper. It is an atomic view of the crack front separating severed from intact bonds.

The direction of loading during mode 1 is normal to the crack faces.

Portion of image removed due to copyright restrictions. Please see Fig. 6 in [1].

(c) Explain chemically assisted fracture so that anyone in 3.22 would understand it, using Michalske’s Fig. 7 type graphics [1] to illustrate the interaction of water with silica at the molecular level.

Image removed due to copyright restrictions. Please see Fig. 7 in [1].

The diagram above shows an example of a chemically assisted fracture mechanism; in this case, how an active molecule (such as H₂O or NH₃) can assist the breaking of strained Si-O-Si bonds, ie. the kink on fig. 6 in question b. The steps are as follows:

The initial silica surface, in the absence of applied strain.
Under applied tension, silica initially responds by changing its conformation but retaining tetrahedral bond angles around the Si molecules.
As the tension increases the Si-O-Si bonds elongate to form an angle of 180° around the bridging oxygens. This step is almost costless (4 kcal/mol) and has little effect on the bond reactivity.
Finally, certain silicon atoms are forced to give up their tetrahedral molecular orbital configuration. In this strained state, the bridging Si-O-Si bond has a strongly increased reactivity.
The Si-O-Si bond reacts with H₂O and is cleaved into 2-OH terminal groups.

(d) Michalske’s argument [1] necessarily implies that the amount of chemical in the environment affects the speed of crack growth. Assume an initial, through-thickness crack of width 150 nm at the surface of a sheet of silica that is 1 cm thick. The crack is under Mode I loading. From Michalske’s Fig. 2, how long would it take for that crack to grow another 150 nm under an applied stress of 100 MPa in an environment of 100% water?

Image removed due to copyright restrictions. Please see Fig. 2 in [1].

In class we saw that the stress intensity K_I in Mode I loading is calculated by

where c is the crack size and f is a factor close to unity. Using the crack size and stress values given, we get a value K_I = 0.069 MPa m^1/2. This value is out of the range of the data plots on the figure above, but by extrapolation we can estimate the crack propagation velocity to be 5 nm/s. It would thus take 30 s for the crack to grow another 150 nm.

(e) At a partial pressure of 50%, how many water molecules would be sitting on the initial crack faces?

Michalske [1] states that Si-O bonds are incapable of adsorbing environmental chemicals until the strain exceeds a certain critical level. We can therefore assume that at the initial stage, molecules will sit on the crack faces without any preference. At a partial pressure of 50%, half of the molecules sitting on the crack faces will be water molecules. The molar volume of an ideal gas at room temperature is V_m = 24.8 L/mol. Therefore, the number of molecules per unit area is: (N_a /V_m)^2/3 = 8.39*1016 /m² where N_a is the Avogadro number. On a crack surface of width 150 nm and thickness 1 cm, the number of water molecules is: 6.29*10⁷ molecules.

Reference

[1] Michalske, T. A., and B. C. Bunker. “Slow Fracture Model Based on Silicate Fracture Models.” Journal of Applied Physics 56 (November 15, 1984): 2686-2693.

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Group Members

Danielle Issa
Anonymous
Sunyoung Kim

References

van Humbeeck, J. “Shape Memory Alloys: A Material and a Technology.” Advanced Engineering Materials 3 (2001): 837-850.

Taillard, K., et al. “Phase Transformation Yield Surface of Anisotropic Shape Memory Alloys.” Materials Science and Engineering A 438-440 (2006): 436-440.

Calloch, S., et al. “Relation Between the Martensite Volume Fraction and the Equivalent Transformation Strain in Shape Memory Alloys.” Materials Science and Engineering A 438-440 (2006): 441-444.

Wiki Assignments

(a) Explain the physical basis of superelasticity.

(b) You are asked to quantify the Young’s elastic modulus of a shape memory alloy such as Ni-Ti. Based on your understanding from the van Humbeeck review, how would you respond?

(d) What in terms of material structure and continuum level mechanical properties determines the maximum mechanical energy possible to obtain from a superelastic alloy in an actuator application?

(b) Taillard, et al. summarize Bouvet, et al.’s attempt at a yield surface for superelastic alloys; this is a modification of the von Mises equivalent stress and criterion that we have discussed. Restate this criterion and its components in terms of the symbols used in class.

(c) Taillard, et al.’s Fig. 1 and 2 demonstrate the yield surface in biaxial tension/compression and torsion for xCu-yAl-cBe, respectively. They claim Fig. 3 shows a difference between this response and that of NiTi. Graph the biaxially stressed yield surface for NiTi.

(d) Contrast the Bouvet yield surface with the other non-asymmetric yield surfaces considered in PS3. Is this equivalent to one of them or different, and how? What is the atomistic origin of the asymmetry justifying the Bouvet yield criteria?