Stress and Strain in Solids - WordPress.com · 2017. 9. 7. · Vincent Casey Solid State Physics:...

Stress and Strain in Solids

A solid is stressed by applying external forces such that both the net force andthe net torque are zero, i.e. static equilibrium.

Figure 1: (a) Zero net force and (b) zero net torque, applied to a cubic volume elementof face area A. The first suffix indicates the direction in which the stress (force) actsand the second suffix indicates the direction of the normal to the face upon which thestress acts.

Zero Net Force achieved by exerting forces equal in magnitude, but in oppositedirections, perpendicularly to opposite faces, tending tocompress/elongate the sample.

Zero Net Torque achieved by applying equal and opposite forces in a parallelfashion to opposite faces, tending to shear the sample

Vincent Casey Solid State Physics: Crystal Mechanics September 7, 2017 1 / 34

The Stress Tensor I

A force Fx applied in the x-direction to a plane with area Ax whose normal isalso in the x-direction produces a stress component:

σxx =FxAx

Likewise, the same force applied to a plane whose normal is in the y -directionproduces the stress component:

σxy =FxAy

Since stresses on opposite faces of the cube are equal and opposite we onlyneed to consider three faces. Thus, stress in general is described by asecond-rank tensor σ with nine components σij(i, j = x ,y ,z), convenientlywritten as a 3×3 array:


The Stress Tensor II

Figure 2: Representation of the ’three face stresses’.

σ =

σxx σxy σxzσyx σyy σyzσzx σzy σzz


Special cases I

However, only six of these stress components are independent because of theconstraint of zero torque, which implies that:

σxy = σyx ;σyz = σzy ;σzx = σxz

Figure 3: Uniform hydrostatic compressive stress and pure shear deformation (atconstant volume).


Special cases II

Special cases of stress have most of the six independent stress componentsequal to zero. For instance for uniaxial stress in the x-direction, the onlynon-zero component is σxx = σ.For hydrostatic stress, the non-zero components are the diagonalcomponents

σxx = σyy = σzz =−P

For pure shear, in which the shape but not the volume of the solid is changed,is represented by say

σxy = σyx = σ

being the only non-zero components.


Diagonalization I

A particular orientation of the spatial coordinates can always be found suchthat the general stress matrix has diagonal components only, i.e.

σ′=

σx 0 00 σy 00 0 σz

the six independent variables defining the stress system now being the threeprincipal stresses σi(i = x ,y ,z) acting along the principle axes, and the threevariables needed to determine the orientation of the principle axes with respectto the original coordinate system.


Matrix Transformation I

A general stress matrix, transformed to the principal axis system can always bewritten as

σx 0 00 σy 00 0 σz

= σ0 0 00 σ0 0

0 0 σ0

+ (σx −σ0) 0 00 (σy −σ0) 0

0 0 (σz−σ0)

where the new stress component is given by

σ0 = (σx + σy + σz)/3

The first term of the transformed stress matrix represents a purely hydrostaticterm, with

σ0 =−P = (σx + σy + σz)/3

which causes a change in volume (dilatoric), but not of shape, of an elasticallyisotropic solid.


Matrix Transformation II

The second term of the transformed stress matrix represents a pure shear,causing a change of shape (deviatoric), but not of volume, of a solid, since thesum of the diagonal components equals zero, i.e. ∑x ,y ,z σii = 0


Non uniform deformation: Strains

The application of an external force to a solid causes a deformation becausedifferent points in the material are displaced by different amounts. Consider apoint P initially at r and a neighbouring point Q initially at r + u, displaced underthe action of stress to P

′at (r + ∆r) and Q

′at (r + ∆r + u + ∆u) respectively.

Figure 4: 2D representation of infinitesimal homogeneous elastic strain.


Strain Components I

For small relative displacements (|∆u|

Strain Components II

being independent because of the requirement that the off-diagonalcomponents obey the condition eij = eji in order to exclude rigid rotations.Special cases of strain include uniaxial strain in the x-direction with only onenon-zero strain component being exx = e; uniform dilation or compression,resulting from hydrostatic stress, with eij = e (i = x ,y ,z); and pure shear witheij = eji = e (i = x , j = y) or other orthogonal pair.


General Strain

As with a general stress, a general strain can be separated into a dilation-straincomponent (volume change) and a pure strain (deviatoric-strain) component(shape change). For a general strain component

eij =e03

δij + (eij −e03

δij)

where the Kronecker delta symbol has the properties:

δij = 1, i = jδij = 0, i 6= j


Notation simplification

Since there are only six independent stress and strain components, aconvenient short-hand notation is to relabel the component indices as follows:

xx → 1yy → 2zz→ 3yz→ 4zx → 5xy → 6

Hooke’s law states that the stress and strain are directly proportional to eachother. Thus, in terms of elastic-stiffness coefficients, Cij , a stress coefficientcan be written as a function of a strain as

σi =6

∑j=1

Cijej


Stress-strain coefficients

The relationship between stress and strain coefficients can also be written inmatrix form as

σ1σ2σ3σ4σ5σ6

=

C11 C12 C13 C14 C15 C16C21 C22 C23 C24 C25 C26C31 C32 C33 C34 C35 C36C41 C42 C43 C44 C45 C46C51 C52 C53 C54 C55 C56C61 C62 C63 C64 C65 C66

e1e2e3e4e5e6


Compliance - a useful alternative! I

Alternatively, a strain coefficient can be expressed in terms of theelastic-compliance coefficients, Sij , as a function of the stress:

ei =6

∑j=1

Sijσj

where both elastic compliance and stiffness are quantities describing amaterial as an elastic continuum.


Compliance - a useful alternative! II

Figure 5: Stress Strain Matrix


Physical Meaning

Physical meaning of eij

e11 =∂ux∂x

e22 =∂uy∂y

e33 =∂uz∂z

Simple tensile strains along the x ,y ,z axes are positive for tension andnegative for compression. Deformation in the xy plane of a rectangular thin filmcorresponds to:

e12 6= 0;e21 6= 0

e11 = e22 = 0

e3i = ei3 = 0; i = x ,y ,zVincent Casey Solid State Physics: Crystal Mechanics September 7, 2017 17 / 34

Pure Shear

Figure 6: Pure Shear

With angles greatly exaggerated, this corresponds to:

e12 ≈∂ux∂y

e21 ≈∂uy∂x

which equals the angles respectively.Vincent Casey Solid State Physics: Crystal Mechanics September 7, 2017 18 / 34

Pure Rotation

Possible combinations of e12 and e21:

Pure Shear e12 = e21Pure Rotation e12 =−e21Simple Shear e12 6= 0;e21 = 0

Figure 7: A pure rotation.


Figure 8: Simple Shear

Figure 9: Comparing pure strain and rotation.


Generalization I

e =

0 γ/2 0γ/2 0 00 0 0

+ 0 γ/2 0−γ/2 0 0

0 0 0

eij = εij + ωij

εij = 1/2(eij + eji)

ωij = 1/2(eij −eji)

eij = eji - symmetric with respect to i and j : measures shape change!ωij =−ωji - antisymmetric with respect to i and j : measures rotation.ωij = 0(i = j)Therefore ωij has three independent elements implying that three independentrotations are possible, one about each axis.


Simplification of the Matrix I

Of the 36 elastic stiffness (or compliance) coefficients in the most general caseof triclinic crystals, 21 are independent and non zero as a result of the generalcondition

Cij = Cji

The presence of higher symmetry reduces the number of coefficients further.For cubic crystals, just three components are independent, C11,C12 and C44with

C11 = C22 = C33,C12 = C13 = C23,C44 = C55 = C66,

and all other coefficients being zero.


Matrix: Simple Cubic Crystal

C11 C12 C12 0 0 0C12 C11 C12 0 0 0C12 C12 C11 0 0 00 0 0 C44 0 00 0 0 0 C44 00 0 0 0 0 C44


Stiffness and Compliance Coefficients

These elastic stiffness constants are related to the corresponding elasticcompliance coefficients via the relationships

S11 =C11 + C12

(C11−C12)(C11 + 2C12)> 0,

S12 =−C12

(C11−C12)(C11 + 2C12)≤ 0,

S44 =1

C44as obtained from inversion of the stiffness matrix.


Isotropic Solids

In the case of elastically isotropic solids (amorphous solids), the number ofindependent elastic coefficients decreases to two since, in addition to theequalities applicable to cubic materials, the coefficients are further linearlyrelated to each other by the equation

C11 = C12 + 2C44

Lamé Constants: For such isotropic materials, the two independent elasticstiffness coefficients are conventionally referred to as the Lamé constants:

λ≡ C12and

µ = C44with

C11 ≡ (λ + 2µ)Vincent Casey Solid State Physics: Crystal Mechanics September 7, 2017 25 / 34

Uniaxial Stress: Isotropic Media I

Uniaxial Stress: For uniaxial stress,Young’s modulus is defined as

E =σxxexx

and so

E =1

S11=

(C11−C12)(C11 + 2C12)C11 + C12

= µ(3λ + 2µ)

(λ + µ)for the case of isotropic media. Under such stress loading, the sample alsodeforms in directions perpendicular to the stress direction. This behaviour isquantified by Poisson’s ratio, defined as

ν =|transverse strain|

normal strain=|eyy |exx

with

ν =−S12S11

=C12

(C11 + C12)=

λ2(λ + µ)


Uniaxial Stress: Isotropic Media II

where 0≤ ν≤ 0.5.


Hydrostatic Stress

Hydrostatic Stress: In the case of hydrostatic stress (pressure), the bulkmodulus B (equal to the inverse of the compressibility, κ = 1B relates thepressure to the dilation (fractional change in volume) e0,

B =Pe0

with

B =1

3(S11 + 2S12)=

(C11 + 2C12)3

=(3λ + 2µ)

3

The bulk modulus is connected to microscopic quantities relating to theinteratomic potential.


Pure Shear

Pure Shear: Finally, for the case of pure shear, the shear (or rigidity) modulusis defined as:

G =shear stressshear strain

=σxyexy

giving for isotropic solids

G =1

2(S11−S12)=

(C11−C12)2

≡ C44 = µ


Some Interrelationships

Some Interrelationships: The various elastic moduli are inter-related; forexample

G =E

2(1 + νand

B =E

3(1−2ν)


Deformation Modes of Cubic Crystals 1

(a) Dilation by hydrostatic stress.Hydrostatic: Under a hydrostatic pressure P:

σ1 = σ2 = σ3 =−P

σ4 = σ5 = σ6 =−0

and the dilation4= e1 + e2 + e3 is

4= e1 + e2 + e3without rotation. Using these relationships we can show that the bulk modulusB is given by:

B =−p4

=13

(C11 + 2C12)


Deformation Modes of Cubic Crystals 2

(b) Shear on a cube face parallel to a cube axis.

Figure 10: Two shear modes for a cubic crystal.

For example on the (010) plane in the [001] direction:

σ4 = C44e4C44 is a shear modulus, which by convention we name µ0.


Deformation Modes of Cubic Crystals 2 I

(c) Shear at 45◦ to a cube axisFor example, on the (110) plane in the [11̄0] direction. In this case the shearmodulus is:

µ1 =12

(C11−C12)

This allows us define an anisotropy factor,

µ0µ1

=2C44

(C11−C12)which of course will be one for an isotropic material.In terms of the Lamé constants:

λ = C12;µ = C44;λ + 2µ = C11

which explains why cubic crystals are not in general isotropic.


Deformation Modes of Cubic Crystals 2 II

W Al Fe Cu Naµ0µ1

1.0 1.2 2.4 3.2 7.5Structure bcc fcc bcc fcc bcc


Stress and Strain TensorsPhysical MeaningSimplification of the 'Matrix'Deformation Modes of Cubic Crystals

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