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User:Egm6936.s09/Elastic stability

=Principles of Mechanics=

Principle of virtual work
In Lurie 2005, p.58, it is written that "The elementary work done by all external and internal forces due to virtual displacement of the continuum particles from their equilibirum position is equal to zero." The elementary work of the external forces, in Lurie 2005, p.130, is expressed in the form:

 Remember to give a lot of references with page numbers immediately as you write so that you don't have to go back to add references, since you would forget where the page numbers were after a while. So as soon as you write about a new concept, or a new equation, immediately give the references. See the article Gradient of vector: Two tensor conventions for examples of how references with page numbers. Egm6321.f11 14:40, 17 June 2011 (UTC)

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$$ \displaystyle \begin{align} \delta A_e & = \int\limits_{S_C} {\mathbf t \cdot \delta \mathbf udS} + \int\limits_{V_C} {\rho \mathbf b \cdot \delta \mathbf udV} = \int\limits_{S_C} {\boldsymbol \sigma  \cdot \mathbf n \cdot \delta \mathbf udS}  + \int\limits_{V_C} {\rho \mathbf b \cdot \delta \mathbf udV}\\ & = \int\limits_{V_C} {\boldsymbol \nabla \cdot \left( \boldsymbol \sigma  \cdot \delta \mathbf u \right)dV}  +  \int\limits_{V_C} {\rho \mathbf b \cdot \delta \mathbf udV}  = \int\limits_{V_C} {\left( \boldsymbol \nabla  \cdot \boldsymbol \sigma  \right) \cdot \delta \mathbf udV}  + \int\limits_{V_C} {tr\left( \boldsymbol \sigma  \cdot \boldsymbol \nabla \delta \mathbf u \right)dV}  + \int\limits_{V_C} {\rho \mathbf b \cdot \delta \mathbf udV}  \\ & = \int\limits_{V_C} {\underbrace {\left( \boldsymbol \nabla \cdot \boldsymbol \sigma \mathbf + \rho \mathbf b \right)}_{\mathbf 0} \cdot \delta \mathbf udV}  + \int\limits_{V_C} {tr\left( \boldsymbol \sigma  \cdot \boldsymbol \nabla \delta \mathbf u \right)dV}  = \int\limits_{V_C} {tr\left( \boldsymbol \sigma  \cdot \boldsymbol \nabla \delta \mathbf u \right)dV} \\ \end{align} $$  (1)
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 You can define directly here the virtual internal work $$\delta A_i := \int\limits_{V_C} {tr\left( {\boldsymbol \sigma \cdot \boldsymbol \nabla \delta \mathbf u} \right)dV} \equiv \delta U$$, instead of defining $$\delta A_i := - \int\limits_{V_C} {tr\left( {\boldsymbol \sigma \cdot \boldsymbol \nabla \delta \mathbf u} \right)dV}$$ in Eq.(8) and Eq.(9). This definition is consistent with the definition of $$\delta A_i$$ in a comment box in section on The Lagrange's variational equation below. Eml5526.s11 14:48, 10 June 2011 (UTC)

Remember that the gradient operator acts in the spatial frame, so

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$$   \displaystyle \nabla \delta \mathbf u = \frac{\partial \delta \mathbf u}{\partial \mathbf x} \ne \delta \frac{\partial \mathbf u}{\partial \mathbf x}$$
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$$   \displaystyle \nabla _{\mathbf X}\delta \mathbf u = \frac{\partial \delta \mathbf u}{\partial \mathbf x} = \delta \frac{\partial \mathbf u}{\partial \mathbf x} $$
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 Refer to Gradient of vector: Two tensor conventions, and mention which convention used. Eml5526.s11 14:13, 3 June 2011 (UTC)

Decompose

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$$   \displaystyle \boldsymbol \nabla \delta \mathbf u = \frac{1}2\left[ \boldsymbol \nabla \delta \mathbf u + \left( \boldsymbol \nabla \delta \mathbf u \right)^T \right] + \frac{1}2\left[ \boldsymbol \nabla \delta \mathbf u - \left( \boldsymbol \nabla \delta \mathbf u \right)^T \right] = \delta \mathbf D + \delta \mathbf W $$ (2) then
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$$   \displaystyle \delta A_e = \int\limits_{V_C} {tr\left( \boldsymbol \sigma \cdot \delta \mathbf D \right)dV}  + \int\limits_{V_C} {\underbrace {tr\left( \boldsymbol \sigma  \cdot \delta \mathbf W \right)}_0dV}  = \int\limits_{V_C} {tr\left( \boldsymbol \sigma  \cdot \delta \mathbf D \right)dV} $$  (3) It is obvious that in p.45
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$$ \displaystyle \delta \mathbf F = \delta \frac{\partial \mathbf x}{\partial \mathbf x} = \delta \frac{\partial \left( {\mathbf u + \mathbf X} \right)}{\partial \mathbf x} = \delta \left( \frac{\partial \mathbf u}{\partial \mathbf x} + \mathbf I \right) = \delta \frac{\partial \mathbf u}{\partial \mathbf x} = \frac{\partial \delta \mathbf u}{\partial \mathbf x} = \frac{\partial \delta \mathbf u}{\partial \mathbf x} \cdot \frac{\partial \mathbf x}{\partial \mathbf x} = \boldsymbol \nabla \delta \mathbf u \cdot \mathbf F $$
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$$   \displaystyle \begin{array}{l} {\left( \delta \mathbf F \right)^T} = \delta \mathbf F^T = \mathbf F^T \cdot \left( \boldsymbol \nabla \delta \mathbf u \right)^T \end{array}$$ (4) So
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$$   \displaystyle \begin{array}{l} \delta \mathbf C = \delta \left( \mathbf F^T \cdot \mathbf F \right) = \delta \mathbf F^T \cdot \mathbf F + \mathbf F^T \cdot \delta \mathbf F = \mathbf F^T \cdot \left( \boldsymbol \nabla \delta \mathbf u \right)^T \cdot \mathbf F + \mathbf F^T \cdot \boldsymbol \nabla \delta \mathbf u \cdot \mathbf F \\ = \mathbf F^T \cdot \left[ \left( \boldsymbol \nabla \delta \mathbf u \right)^T + \boldsymbol \nabla \delta \mathbf u \right] \cdot \mathbf F = 2\mathbf F^T \cdot \delta \mathbf D \cdot \mathbf F \\ \end{array}$$ (5) but
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$$   \displaystyle \mathbf C = \mathbf I + 2\mathbf E \to \delta \mathbf C = 2\delta \mathbf E = 2\mathbf F^T \cdot \delta \mathbf D \cdot \mathbf F \leftrightarrow \delta \mathbf E = \mathbf F^T \cdot \delta \mathbf D \cdot \mathbf F \leftrightarrow \delta \mathbf D = \mathbf F^{-T} \cdot \delta \mathbf E \cdot \mathbf F^{-1} $$  (6) So the elementary work of the external forces (work done by surface forces and body forces) is:
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$$   \displaystyle \begin{array}{l} \delta A_e = \int\limits_{V_C} {tr\left( \boldsymbol \sigma \cdot \delta \mathbf D \right)dV}  = \int\limits_{V_C} {tr\left( \boldsymbol \sigma  \cdot \mathbf F^{-T} \cdot \delta \mathbf E \cdot \mathbf F^{-1} \right)dV}  \\ = \int\limits_{V_0} {tr\left( \mathbf F^{-1} \cdot \boldsymbol \sigma \cdot \mathbf F^{-T} \cdot \delta \mathbf E \right)\left( JdV_0 \right)}  = \int\limits_{V_0} {tr\left( {\underbrace {J\mathbf F^{-1} \cdot \boldsymbol \sigma  \cdot \mathbf F^{-T}}_\boldsymbol{\tilde \sigma} \cdot \delta \mathbf E} \right)dV_0}  = \int\limits_{V_0} {tr\left( \boldsymbol{\tilde \sigma} \cdot \delta \mathbf E \right)dV_0}  \\ \end{array}$$ (7) From the virtual principle work, the elementary work of the internal forces (work done by elastic forces) is:
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$$   \displaystyle \delta A_i = - \delta A_e =  - \int\limits_{V_0} {tr\left( \boldsymbol{\tilde \sigma} \cdot \delta \mathbf E \right)dV_0} $$  (8) Now assume that internal elastic forces have a potential, $$\displaystyle \delta A_i =  - \delta U$$ with
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$$   \displaystyle \delta A_i = - \delta A_e =  - \int\limits_{V_0} {tr\left( \boldsymbol{\tilde \sigma} \cdot \delta \mathbf E \right)dV_0} $$  (9)
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where $$\displaystyle u $$ is internal energy per unit mass (stored elastic energy)

$$   \displaystyle \to \delta U = \delta \int\limits_{V_C} {\rho udV} = \delta \int\limits_{V_0} {\rho _0udV_0 = } \int\limits_{V_0} {\delta \left( \rho _0 u \right)dV_0} $$  (10) From this we refer:
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$$   \displaystyle \delta U = \int\limits_{V_0} {\delta \left( \rho _0 u \right)dV_0} = \int\limits_{V_0} {\delta WdV_0}  = \int\limits_{V_0} {tr\left( \boldsymbol{\tilde \sigma} \cdot \delta \mathbf E \right)dV_0}  \leftrightarrow \delta W = tr\left( \boldsymbol{\tilde \sigma} \cdot \delta \mathbf E \right) $$  (11) where  $$\displaystyle W = \rho _0u $$ is the energy per unit volume. From the definition:
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$$   \displaystyle \delta W = tr\left( \frac\partial W\partial \mathbf E \cdot \delta \mathbf E \right) = tr\left( \boldsymbol{\tilde \sigma} \cdot \delta \mathbf E \right) \to \boldsymbol{\tilde \sigma} = \frac{\partial W}{\partial \mathbf E} $$ (12) If we now consider infinitesimal displacement gradients $$\displaystyle W = \rho _0u$$ then
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$$   \displaystyle \boldsymbol{\tilde \sigma} = \frac{\partial W}{\partial \mathbf E} = \left. \frac{\partial W}{\partial \mathbf E} \right|_{\mathbf E = \mathbf 0} + \left. \frac{\partial ^2 W}{\partial \mathbf E^2}\right|_{\mathbf E = \mathbf 0}\mathbf E + \mathbf O\left( \left| \mathbf E \right|^2 \right) $$  (13) Assume the unstrained body possesses no residual stress and define fourth-rank elastic modulus tensor
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$$   \displaystyle \mathbf C = \left. \frac{\partial ^2 W}{\partial \mathbf E^2} \right|_{\mathbf E = \mathbf 0} \to \boldsymbol{\tilde \sigma} = \boldsymbol \sigma = \frac{\partial W}{\partial \mathbf E} = \mathbf{C:E}$$ (14) Having established the fobold of above linear relations between stresses and strains, we can integrate to get:
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$$   \displaystyle W = \frac{1}2\mathbf{E:C:E} = \frac{1}2 tr\left( \boldsymbol \sigma \cdot \mathbf E \right) = \frac{\lambda }2tr{\left( \mathbf E \right)^2} + \mu tr\left( \mathbf E^2 \right) = \frac{1}{4\mu}\left[ tr\left(  \right) - \frac{\nu }{1+\nu}tr{\left( \boldsymbol \sigma  \right)^2} \right]$$ (15)
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The Lagrange’s variational equation
Consider a certain state of equilibrium or of small elastic vibrations of a body, which is charaterized by the stresses $$\displaystyle \boldsymbol \sigma $$ and the displacements $$\displaystyle u $$. From
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$$   \displaystyle \boldsymbol \nabla \cdot \boldsymbol \sigma + \rho \mathbf b = \rho \frac{\partial ^2 \mathbf u}{\partial t^2} \to \int\limits_{V_C} {\left( \boldsymbol \nabla  \cdot \boldsymbol \sigma  + \rho \mathbf b - \rho \frac{\partial ^2 \mathbf u}{\partial t^2} \right) \cdot \mathbf u^' dV}  = 0 $$  (16)
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 Actually, the most general form of the Cauchy equation of motion that is also valid for fluids is written as follows: $$\boldsymbol \nabla \cdot \boldsymbol \sigma \mathbf + \rho \mathbf b = \rho \frac$$, where $$\mathbf v$$ is the velocity field, and the acceleration is the material time derivative of the velocity, written as $$\frac$$. See Eq.(38) in Kolmogorov scales.

Eml5526.s11 14:21, 3 June 2011 (UTC)

where $$\displaystyle \mathbf u^'$$ are arbitrary functions which have continuous partial derivatives of the first and second order with respect to spatial variables and time. Using divergence theorem to convert:
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$$   \displaystyle \begin{array}{l} \int\limits_{V_C} {\left( \boldsymbol \nabla \cdot \boldsymbol \sigma  \right) \cdot \mathbf u^'dV}  = \int\limits_{V_C} \left[ \boldsymbol \nabla  \cdot \left( \boldsymbol \sigma  \cdot \mathbf u^' \right) - \boldsymbol \sigma  \cdot  \cdot \boldsymbol \nabla \mathbf v \right]dV = \int\limits_{S_C} {\mathbf n \cdot \boldsymbol \sigma  \cdot \mathbf u^'dS}  - \int\limits_{V_C} {\boldsymbol \sigma  \cdot  \cdot \boldsymbol \nabla \mathbf vdV}  \\ = \int\limits_{S_C} {\mathbf t \cdot \mathbf u^'dS - \int\limits_{V_C} \boldsymbol \sigma \cdot  \cdot \boldsymbol \nabla \mathbf u^'dV }  = \int\limits_{S_C} {\mathbf t \cdot \mathbf u^'dS - \int\limits_{V_C} tr\left( {\boldsymbol \sigma  \cdot \mathbf E^'} \right)dV }  \\ \end{array}$$ (17)
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 Define the "horizontal" double contraction "$$\cdot\cdot$$", and give reference, e.g., Malvern 1969. Eml5526.s11 14:21, 3 June 2011 (UTC)

So,
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$$   \displaystyle \int\limits_{S_C} {\mathbf t \cdot \mathbf u^{'} dS} + \int\limits_{V_C} {\rho \left( \mathbf b - \frac{\partial ^2 \mathbf u}{\partial t^2} \right) \cdot \mathbf u^{'} dV} = \int\limits_{V_C} {tr\left( \boldsymbol \sigma \cdot \mathbf E^' \right)dV} $$ (18)
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 The above equation is essentially $$\delta A_e - \delta A_m = \delta A_i$$, where the external virtual work is $$\delta A_e := \int\limits_{S_C} {\mathbf t \cdot \mathbf u^'dS + \int\limits_{V_C} {\rho \mathbf{b} \cdot \mathbf u^'dV}}$$, the inertial virtual work $$\delta A_m := \int\limits_{V_C} {\rho \frac{\partial ^2 \mathbf u}{\partial t^2}} \cdot \mathbf u^'dV$$, and the internal virtual work $$\delta A_i := \int\limits_{V_C} {tr\left( {\boldsymbol \sigma \cdot \mathbf E^'} \right)dV}$$. Usually, in FEM, the above "weak form" is written as follows $$\delta A_m + \delta A_i = \delta A_e$$, which leads to the discrete weak form $$c^T \left( M \ddot d + K d \right)= c^T F, \ \forall c$$ and then to semi-discrete equation $$M \ddot d + K d = F$$.

Eml5526.s11 14:21, 3 June 2011 (UTC)

where
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$$   \displaystyle \mathbf E^' = \frac{1}2\left[ \boldsymbol \nabla \mathbf u^' + \left( \boldsymbol \nabla \mathbf u^' \right)^T \right]$$ (19) Now let the displacements $$\displaystyle \mathbf u^'$$ be the actual displacements $$\displaystyle {\mathbf u}$$ and assume that the state of stress corresponds to the equilibrium of the body, then the equality reduces:
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$$   \displaystyle \int\limits_{S_C} {\mathbf t \cdot \mathbf udS + \int\limits_{V_C} \rho \mathbf b \cdot \mathbf udV} = \int\limits_{V_C} {tr\left( \boldsymbol \sigma  \cdot \mathbf E \right)dV}  = 2\int\limits_{V_C} {WdV} $$ (20)
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 The inertial force disappeared here; there is no need to remove it. Mention that you are now dealing with static equilibrium, not dynamic equilibrium, so you ignore the inertial term. Also, $${\rm tr} ( \boldsymbol \sigma \cdot \boldsymbol e ) = \boldsymbol \sigma : \boldsymbol  e$$. Eml5526.s11 14:21, 3 June 2011 (UTC)

Now let the displacements $$\displaystyle \mathbf u^'$$ be the virtual displacements $$\displaystyle \delta \mathbf u$$ and assume that the state of stress corresponds to the equilibrium of the body, then the equality reduces to principle of virtual work:
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$$   \displaystyle \int\limits_{S_C} {\mathbf t \cdot \delta \mathbf udS} + \int\limits_{V_C} {\rho \mathbf b \cdot \delta \mathbf udV}  = \int\limits_{V_C} {tr\left( \boldsymbol \sigma  \cdot \delta \mathbf E \right)dV}$$ (21) where
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$$   \displaystyle \delta \mathbf E = \frac{1}2\left[ \boldsymbol \nabla \delta \mathbf u + \left( \boldsymbol \nabla \delta \mathbf u \right)^T \right]$$ (22) Remember that in deriving above formular, we fixed quantities $$\displaystyle \mathbf t,\boldsymbol \sigma $$ and $$\displaystyle \rho \frac{\partial ^2 \mathbf u}{\partial t^2}$$ while varying the displacements $$\displaystyle {\mathbf u}$$ and corresponding strains. So,
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$$   \displaystyle \begin{array}{l} \int\limits_{S_C} {\mathbf t \cdot \delta \mathbf udS} + \int\limits_{V_C} {\rho \mathbf b \cdot \delta \mathbf udV}  = \delta \left( \int\limits_{S_C} {\mathbf t \cdot \mathbf udS}  + \int\limits_{V_C} {\rho \mathbf b \cdot \mathbf udV}  \right) \\ = \int\limits_{V_C} {tr\left( \boldsymbol \sigma \cdot \delta \mathbf E \right)dV}  = \int\limits_{V_C} {\delta WdV}  = \delta \int\limits_{V_C} {WdV}  \\ \end{array}$$ (23) or
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$$   \displaystyle \delta \left( \int\limits_{S_C} {\mathbf t \cdot \mathbf udS} + \int\limits_{V_C} {\rho \mathbf b \cdot \mathbf udV}  - W \right) = \delta A =  - \delta \Pi  = 0$$ (24) where $$\displaystyle A$$ is the total work done by external forces (surface and body forces) and internal forces (elastic forces) and $$\displaystyle \Pi$$is the total potential of the system.
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$$   \displaystyle \int\limits_{S_C} {\mathbf t \cdot \mathbf udS} + \int\limits_{V_C} {\rho \mathbf b \cdot \mathbf udV}  - W = A =  - \Pi$$ (25) This equality means that the potential energy has an extremum value.
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The first law of thermodynamics
First calculate the power input:
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$$ \displaystyle P_{input} = \int\limits_{S_C} {\mathbf t \cdot \mathbf vdS} + \int\limits_{V_C} {\rho \mathbf b \cdot \mathbf vdV}  = \int\limits_{V_C} {\underbrace {\left( \boldsymbol \nabla  \cdot \boldsymbol \sigma \mathbf + \rho \mathbf b \right)}_{\rho \frac{D\mathbf v}{Dt}} \cdot \mathbf vdV}  + \int\limits_{V_C} {tr\left( \boldsymbol \sigma  \cdot \boldsymbol \nabla \mathbf v \right)dV} $$
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$$ \displaystyle = \int\limits_{V_C} {\rho \frac{D\mathbf v}{Dt} \cdot \mathbf vdV} + \int\limits_{V_C} {tr\left( \boldsymbol \sigma  \cdot \mathbf D \right)dV}  = \int\limits_{V_C} {\frac{1}2\rho \frac{D}{Dt}\left( \mathbf v \cdot \mathbf v \right)dV}  + \int\limits_{V_C} {tr\left( \boldsymbol \sigma  \cdot \mathbf D \right)dV} $$
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$$   \displaystyle \begin{array}{l} = \frac{D}{Dt}\int\limits_{V_m\left( t \right)} {\frac{1}2\rho \mathbf v \cdot \mathbf vdV} + \int\limits_{V_C} {tr\left( \boldsymbol \sigma  \cdot \mathbf D \right)dV}  = \frac{DK}{Dt} + \int\limits_{V_C} {tr\left( \boldsymbol \sigma  \cdot \mathbf D \right)dV} \end{array}$$ (26) Here $$\displaystyle K$$is the kinetic energy of the system and$$\displaystyle \mathbf{D}$$ is the rate of deformation tensor: :{| style="width:100%" border="0" $$   \displaystyle \mathbf D = \frac{1}2\left[ \left( \boldsymbol \nabla \mathbf v \right)^T + \boldsymbol \nabla \mathbf v \right]$$ (27) Second, calculate the heat input:
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$$   \displaystyle Q_{input} = - \int\limits_{S_C} {\mathbf q \cdot \mathbf ndS}  + \int\limits_{V_C} {\rho q_s dV}$$ (28) where $$\displaystyle \mathbf q$$ is the heat flux vector per unit area and $$\displaystyle {q}_{s}$$ is the internal heat supply per unit mass. The first law of thermodynamics said that when a closed system is carried through a cycle and returned to its initial state:
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$$   \displaystyle \int\limits_{cycle} {\left[ P_{input} + Q_{input} \right]dt} = \int\limits_{cycle} {dE}  = 0 \to dE = \left[ P_{input} + Q_{input} \right]dt \leftrightarrow \frac{dE}{Dt} = \left[ P_{input} + Q_{input} \right]$$ (29) Where $$\displaystyle E$$ is the total enery of the system. Generally, decompose the total energy into kinetic energy plus internal energy, so
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$$   \displaystyle E = \int\limits_{V_C} {\left( \frac{1}2\rho \mathbf v \cdot \mathbf v + \rho u \right)dV} = K + \int\limits_{V_C} {\rho udV}$$ (30) with $$\displaystyle u$$ is internal energy per unit mass (stored elastic energy) We are now in a position to write the first law of thermodynamics:
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$$   \displaystyle \begin{array}{l} \frac{dE}{Dt} = \frac{D}{Dt}\int\limits_{V_m\left( t \right)} {\left( \frac{1}2\rho \mathbf v \cdot \mathbf v + \rho u \right)dV} = \frac{DK}{Dt} + \int\limits_{V_C} {\rho \frac{Du}{Dt}dV} \end{array}$$
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$$   \displaystyle \begin{array}{l} = P_{input} + Q_{input} = \frac{DK}{Dt} + \int\limits_{V_C} {tr\left( \boldsymbol \sigma \cdot \mathbf D \right)dV}  - \int\limits_{S_C} {\mathbf q \cdot \mathbf ndS}  + \int\limits_{V_C} {\rho q_sdV} \end{array}$$ (31)
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$$\displaystyle \to \int\limits_{V_C} {\rho \frac{Du}{Dt}dV} = \int\limits_{V_C} {tr\left( \boldsymbol \sigma  \cdot \mathbf D \right)dV}  - \int\limits_{S_C} {\mathbf q \cdot \mathbf ndS}  + \int\limits_{V_C} {\rho q_sdV} $$
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$$   \displaystyle \begin{array}{l} \leftrightarrow \rho \frac{Du}{Dt} = tr\left( \boldsymbol \sigma \cdot \mathbf D \right) - \boldsymbol \nabla  \cdot \mathbf q + \rho q_s \end{array}$$ (32)
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The second law of thermodynamics
For reversible processes,
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$$   \displaystyle \int\limits_{cycle} {\frac{Q_{input}}{\theta }dt} =  0 \to dS = \frac{Q_{input}}{\theta }dt \leftrightarrow \frac{dS}{Dt} = \frac{Q_{input}}{\theta }$$ (33) Where $$\displaystyle S$$ is the entropy of the system and $$\displaystyle \theta$$ is the absolute temperature. Expressing entropy of the system through its specific entropy $$\displaystyle \eta $$ (entropy per unit mass)
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$$   \displaystyle S = \int\limits_{V_C} {\rho \eta dV} \to \frac{D}{Dt}\int\limits_{V_m\left( t \right)} {\rho \eta dV}  =  - \int\limits_{S_C} {\frac{\theta }dS}  + \int\limits_{V_C} {\frac\rho q_s{\theta }dV}$$ (34) for reversible processes In irreversible processes, there is also internal entropy production from dissipative processes like internal friction, hence for general processes:
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$$   \displaystyle \frac{D}{Dt}\int\limits_{V_m\left( t \right)} {\rho \eta dV} \ge  - \int\limits_{S_C} {\frac{\mathbf q \cdot \mathbf n}{\theta }dS}  + \int\limits_{V_C} {\frac{\rho q_s}{\theta }dV}  \leftrightarrow \rho \frac{D\eta}{Dt} \ge \frac{\rho q_s}{\theta } - \boldsymbol \nabla  \cdot \left( \frac{\mathbf q}{\theta} \right)$$ (35)
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Thermoelasticity postulates
The second law of thermodynamics is
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$$   \displaystyle \rho \frac{D\eta }{Dt} \ge \frac{\rho q_s}{\theta } - \boldsymbol \nabla \cdot \left( \frac{\mathbf q}{\theta } \right) = \frac{\rho q_s}{\theta } - \frac{\boldsymbol \nabla  \cdot \mathbf q}{\theta } + \frac{\mathbf q \cdot \boldsymbol \nabla \theta}{\theta ^2}$$ (36) The last tebold in the above equation corresponds to the part of entropy production rate due to irreverible heat conduction in the presence of a temperature gradient (irreversible since heat only spontaneously flows from higher to lower temperature or $$\displaystyle \mathbf q \cdot \boldsymbol \nabla \theta  \le 0$$) Now assume that purely elastic behavior is fully reversible, then the inequality becomes an equality and$$\displaystyle \mathbf q \cdot \boldsymbol \nabla \theta = 0$$ :
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$$   \displaystyle \rho \frac{D\eta}{Dt} = \frac{\rho q_s}{\theta } - \frac{\boldsymbol \nabla \cdot \mathbf q}{\theta } \to \rho \frac{Du}{Dt} = tr\left( \boldsymbol \sigma  \cdot \mathbf D \right) - \boldsymbol \nabla  \cdot \mathbf q + \rho q_s = tr\left( \boldsymbol \sigma  \cdot \mathbf D \right) + \rho \theta \frac{D\eta }{Dt}$$ (37) Hence, internal energy is a function of entropy and deformations
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$$   \displaystyle u = u\left( \mathbf E,\eta \right)$$ (38) Consider two cases: i) When deformation is isentropic, $$\displaystyle \frac{D\eta }{Dt} = 0$$, which is equivalent to adiabatic (no heat flow, valid in rapid deformations)
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$$   \displaystyle \rho \frac{Du}{Dt} = tr\left( \boldsymbol \sigma \cdot \mathbf D \right) $$  (39) but
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$$   \displaystyle \mathbf D =\mathbf F^{-T} \cdot \frac{D\mathbf E}{Dt} \cdot \mathbf F^{-1}$$ (40)
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$$   \displaystyle \to \rho \frac{Du}{Dt} = tr\left( \boldsymbol \sigma \cdot \mathbf F^{-T} \cdot \frac{D\mathbf E}{Dt} \cdot \mathbf F^{-1} \right) = tr\left( \mathbf F^{-1} \cdot \boldsymbol \sigma  \cdot \mathbf F^{-T} \cdot \frac{D\mathbf E}{Dt} \right) = tr\left( \frac{1}{J}\boldsymbol{\tilde \sigma} \cdot \frac{D\mathbf E}{Dt} \right) = \frac{1}{J}tr\left( \boldsymbol{\tilde \sigma} \cdot \frac{D\mathbf E}{Dt} \right)$$ (41)
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$$   \displaystyle \leftrightarrow J\rho \frac{Du}{Dt} = \rho _0\frac{Du}{Dt} = tr\left( \boldsymbol{\tilde \sigma} \cdot \frac{D\mathbf E}{Dt} \right) \leftrightarrow d\left( \rho _0 u \right) = dW = tr\left( \boldsymbol{\tilde \sigma} \cdot d\mathbf E \right) $$  (42) Because
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$$   \displaystyle \eta = const \to u = u\left( \mathbf E,\eta \right) = u\left( \mathbf E \right)$$ (43)
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$$   \displaystyle \to W = W\left( \mathbf E \right) \to dW = tr\left( \frac{\partial W}{\partial \mathbf E} \cdot d\mathbf E \right) = tr\left( \boldsymbol{\tilde \sigma} \cdot d\mathbf E \right) \to \boldsymbol{\tilde \sigma} = \frac{\partial W}{\partial \mathbf E}$$ (44) ii) When deformation is isotheboldal, then $$ \displaystyle \frac{Dt} = 0 $$ (no temperature change, slow deformations)
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For reversible processes we still have:
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$$   \displaystyle \leftrightarrow \rho _0\frac{D\left(u - \theta \eta \right)}{Dt} = tr\left( \boldsymbol{\tilde \sigma} \cdot \frac{D\mathbf E}{Dt} \right) \leftrightarrow d\left[ \rho _0\left( u - \theta \eta \right) \right] = tr\left( \boldsymbol{\tilde \sigma} \cdot d\mathbf E \right)$$ (45) Define the Helmholz free energy per unit mass:
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$$   \displaystyle \psi \left( \mathbf E,\theta \right) = u - \theta \eta \to d\left( \rho _0\psi \right) = dW = tr\left( \boldsymbol{\tilde \sigma} \cdot d\mathbf E \right)$$ (46) Similarly, due to
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$$   \displaystyle \theta = const \to \psi  = \psi \left( \mathbf E,\theta \right) = \psi \left( \mathbf E \right)$$ (47) and we end up with the same result
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$$   \displaystyle \boldsymbol{\tilde \sigma} = \frac{\partial W}{\partial \mathbf E}$$ (48) Thus in the cases of very rapid or very slow deformations, the elastic constitutive equations take the fobold …., with the strain energy density function $$\displaystyle W$$ being a function depending on $$\displaystyle \mathbf{E}$$only (for other types of deformation, $$\displaystyle W$$ would also have to depend on $$\displaystyle \eta$$ or $$\displaystyle \theta$$.
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 Refer to Pearson's book (give detailed ref and page number) who said that for processes with intermediate deformation rates, i.e., between very slow and very fast, one can still use (48) as a good approximation.

Eml5526.s11 16:26, 3 June 2011 (UTC)

=General theory of elastic stability= Consider an arbitrary elastic body which is free from stress (state I.) By application of load or of heat, the body changes its shape and position (state II.) The material particlee has moved from the point, initially $$\displaystyle \mathbf X$$ to the point $$\displaystyle \mathbf X$$. The displacement vector is:
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$$   \displaystyle \mathbf v = \mathbf X - \mathbf X$$ (49) The Lagrangian strain tensor is:
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$$   \displaystyle \mathbf E = \frac{1}2\left[ \nabla \mathbf v + \left( \nabla \mathbf v \right)^T + \nabla \mathbf v \cdot \left( \nabla \mathbf v \right)^T \right]$$ (50)
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Let $$\displaystyle U$$ be the internal energy per unit mass and assume constant entropy. Then, $$\displaystyle U = U(\mathbf{E})$$ and

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$$ \tilde {\boldsymbol\sigma} = \frac{\partial (\rho_0 U)}{\mathbf E} = \rho _0 \frac{\partial U}{\partial \mathbf E} \to \boldsymbol \sigma = \frac{1}{J}\mathbf F \cdot \tilde {\boldsymbol \sigma} \cdot \mathbf F^T = \frac{\rho _0 \mathbf F}{J} \cdot \frac{\partial U}{\partial \mathbf E} \cdot \mathbf F^T = \rho \mathbf F \cdot \frac{\partial U}{\partial \mathbf E} \cdot \mathbf F^T $$     (51)
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 Remember to define $${\tilde {\boldsymbol \sigma} }$$ as the second Piola-Kirchhoff stress, which is usually denoted as $${\boldsymbol S}$$.

Eml5526.s11 15:55, 3 June 2011 (UTC)

It is now required to analyze the stability of the body in its defobolded state II. The stability condition is “for each infinitesimal displacement which is compatible with the boundary conditions, the work done by the surface and body forces does not exceed that absorbed as an increase in internal energy.” The body forces per unit mass assumed to be constant. Consider two cases:

The surface foces are dead loads
The surface fores vary neither in magnitude nor in direction during trial displacement. It means that the total traction vector is the same irrespective of the direction of the area. The work done by the body and surface forces in a trial displacement $$\displaystyle \mathbf u$$ from state II is:

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$$ \displaystyle A_e = \int\limits_{S_{II}} {\mathbf t \cdot \mathbf udS} + \int\limits_{V_{II}} {\rho \mathbf b \cdot \mathbf udV}  = \int\limits_{V_{II}} {tr\left( \mathbf{\boldsymbol \sigma} \cdot \nabla \mathbf u \right)dV}  = \int\limits_{V_{II}} {tr\left( \rho \mathbf F \cdot \frac{\partial U}{\partial \mathbf E} \cdot \mathbf F^T \cdot \nabla \mathbf u \right)dV} $$
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$$\displaystyle = \int\limits_{V_{II}} {\rho tr\left( \frac{\partial U}{\partial \mathbf E} \cdot \mathbf F^T \cdot \nabla \mathbf u \cdot \mathbf F \right)dV} = \int\limits_{V_{II}} {\rho tr\left( \frac{\partial U}{\partial \mathbf E} \cdot \mathbf F^T \cdot \frac{\partial \mathbf u}{\partial \mathbf x} \cdot \frac{\partial \mathbf x}{\partial \mathbf x} \right)dV} $$
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$$   \displaystyle = \int\limits_{V_{II}} {\rho tr\left( \frac{\partial U}{\partial \mathbf E} \cdot \mathbf F^T \cdot \frac{\partial \mathbf u}{\partial \mathbf x} \right)dV} $$  (52) The increase in internal energy is:
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$$   \displaystyle \Omega = \int\limits_{V_{II}} {\rho \left( U^' - U \right)dV} $$  (53)
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 To be consistent with the notation used above, $$U$$ should be written in lowercase $$u$$ (which is energy per unit volume per unit mass, i.e., the specific energy). In standard books like Malvern 1969, the notation "$$e$$" is used. Eml5526.s11 15:38, 10 June 2011 (UTC)

The stability condition is:
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$$   \displaystyle A_e - \Omega = \int\limits_{V_{II}} {\rho \left[ tr\left( \frac{\partial U}{\partial \mathbf E} \cdot \mathbf F^T \cdot \frac{\partial \mathbf u}{\partial \mathbf x} \right) - \left( U^' - U \right) \right]dV}  \le 0$$ (54)
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 That is, stability is maintained if the external work does NOT surpass (less than or equal to) the increase in internal work. In other words, instability occurs when the external work is greater than the increase in internal work.

Eml5526.s11 15:48, 3 June 2011 (UTC)

Here the internal energy is:
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$$   \displaystyle U^' = U\left( \mathbf E^' \right),U = U\left( \mathbf E \right) $$ (55) where
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$$   \displaystyle \begin{align} \mathbf E^' & = \frac{1}2\left[ \frac{\partial }{\partial \mathbf x}\left(\mathbf v + \mathbf u \right) + \left( \frac{\partial }{\partial \mathbf x} \left( \mathbf v + \mathbf u \right) \right)^T + \left( \frac{\partial }{\partial \mathbf x}\left( \mathbf v + \mathbf u \right) \right)^T \cdot \frac{\partial }{\partial \mathbf x}\left( \mathbf v + \mathbf u \right) \right] \\   & = \mathbf E + \frac{1}2\left[ \frac{\partial \mathbf u}{\partial \mathbf x} + \left( \frac{\partial \mathbf u}{\partial \mathbf x} \right)^T + \left( \frac{\partial \mathbf u}{\partial \mathbf x} \right)^T \cdot \frac{\partial \mathbf u}{\partial \mathbf x} + \left( \frac{\partial \mathbf v}{\partial \mathbf x} \right)^T \cdot \frac{\partial \mathbf u}{\partial \mathbf x} + \left( \frac{\partial \mathbf u}{\partial \mathbf x} \right)^T \cdot \frac{\partial \mathbf v}{\partial \mathbf x} \right] \\	& = \mathbf E + \frac{1}2\left[ {\left( {\underbrace {\mathbf I + {{\left( {\frac{\partial \mathbf x}} \right)}^T}}_{\mathbf F^T}} \right) \cdot \frac{\partial \mathbf u}{\partial \mathbf x} + \left( \frac{\partial \mathbf u}{\partial \mathbf x} \right)^T \cdot \underbrace {\left( {\mathbf I + \frac{\partial \mathbf x}} \right)}_\mathbf F + \left( \frac{\partial \mathbf u}{\partial \mathbf x} \right)^T \cdot \frac{\partial \mathbf u}{\partial \mathbf x}} \right] \\	& = \mathbf E + \frac{1}2\left[ \mathbf F^T \cdot \frac{\partial \mathbf u}{\partial \mathbf x} + \left( \frac{\partial \mathbf u}{\partial \mathbf x} \right)^T \cdot \mathbf F + \left( \frac{\partial \mathbf u}{\partial \mathbf x} \right)^T \cdot \frac{\partial \mathbf u}{\partial \mathbf x} \right] = \mathbf E + sym\left( \mathbf F^T \cdot \frac{\partial \mathbf u}{\partial \mathbf x} \right) + \frac{1}2\left( \frac{\partial \mathbf u}{\partial \mathbf x} \right)^T \cdot \frac{\partial \mathbf u}{\partial \mathbf x}	\\ \end{align} $$   (56)
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Using Taylor’s expansion of internal energy around state II to get:
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$$ \displaystyle \begin{align} U^' - U & = U\left( \mathbf E^' \right) - U\left( \mathbf E \right) = \left. \frac{\partial U}{\partial \mathbf E} \right|_{II} \left( \mathbf E^' - \mathbf E \right) + \frac{1}2\left. \frac{\partial ^2 U}{\partial \mathbf E^2} \right|_{II}\left( \mathbf E^' - \mathbf E \right)^2 + \cdots  \\ & = tr\left( \left. \frac{\partial U}{\partial \mathbf E} \right|_{II} \cdot \left( \mathbf E^' - \mathbf E \right) \right) + \frac{1}2\left. \frac{\partial ^2 U}{\partial \mathbf E^2} \right|_{II}\left( \mathbf E^' - \mathbf E \right)^2 + \cdots \\ \end{align} $$  (57)
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$$ \displaystyle \begin{align} & = tr\left( \left. \frac{\partial U}{\partial \mathbf E} \right|_{II} \cdot sym\left( \mathbf F^T \cdot \frac{\partial \mathbf u}{\partial \mathbf x} \right) \right) + \frac{1}2tr\left( \left. \frac{\partial U}{\partial \mathbf E} \right|_{II} \cdot \left( \frac{\partial \mathbf u}{\partial \mathbf x} \right)^T \cdot \frac{\partial \mathbf u}{\partial \mathbf x} \right) + \frac{1}2\left. \frac{\partial ^2 U}{\partial \mathbf E^2} \right|_{II}\left( \mathbf E^' - \mathbf E \right)^2 + \cdots \\ & = tr\left( {\left. \frac{\partial U}{\partial \mathbf E} \right|_{II} \cdot \left( \mathbf F^T \cdot \frac{\partial \mathbf u}{\partial \mathbf x} \right)} \right) + \frac{1}2tr\left( {\left. \frac{\partial U}{\partial \mathbf E} \right|_{II} \cdot \left( \frac{\partial \mathbf u}{\partial \mathbf x} \right)^T \cdot \frac{\partial \mathbf u}{\partial \mathbf x}} \right) + \frac{1}2\left. \frac{\partial ^2 U}{\partial \mathbf E^2} \right|_{II}\left( \mathbf E^' - \mathbf E \right)^2 + \cdots \\ \end{align} $$  (58)
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The stability condition reduces:
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$$   \displaystyle \int\limits_{V_{II}} {\rho \left[ tr\left( \left. \frac{\partial U}{\partial \mathbf E} \right|_{II} \cdot \left( \frac{\partial \mathbf u}{\partial \mathbf x} \right)^T \cdot \frac{\partial \mathbf u}{\partial \mathbf x} \right) + \left. \frac{\partial ^2 U}{\partial \mathbf E^2} \right|_{II} \left( \mathbf E^' - \mathbf E \right)^2 + \cdots  \right]dV \ge } 0$$ (59) If we write
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$$   \displaystyle \frac{\partial \mathbf u}{\partial \mathbf x} = \frac{\partial \mathbf u}{\partial \mathbf x} \cdot \frac{\partial \mathbf x}{\partial \mathbf x} = \frac{\partial \mathbf u}{\partial \mathbf x} \cdot \mathbf F $$ (60)
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and convert
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$$\displaystyle \rho tr\left( \left. \frac{\partial U}{\partial \mathbf E} \right|_{II} \cdot \left( \frac{\partial \mathbf u}{\partial \mathbf x} \right)^T \cdot \frac{\partial \mathbf u}{\partial \mathbf x} \right) = \rho tr\left( \left. \frac{\partial \mathbf u}{\partial \mathbf x} \cdot \frac{\partial U}{\partial \mathbf E} \right|_{II} \cdot \left( \frac{\partial \mathbf u}{\partial \mathbf x} \right)^T \right) $$
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$$   \displaystyle = \rho tr\left( \left. \frac{\partial \mathbf u}{\partial \mathbf x} \cdot \mathbf F \cdot \frac{\partial U}{\partial \mathbf E} \right|_{II} \cdot \mathbf F^T \cdot \left( \frac{\partial \mathbf u}{\partial \mathbf x} \right)^T \right) = tr\left( \frac{\partial \mathbf u}{\partial \mathbf x} \cdot \mathbf{\boldsymbol \sigma } \cdot \left( \frac{\partial \mathbf u}{\partial \mathbf x} \right)^T \right)$$ (61) then alternative fobold of stability condition is:
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 The intermediate step is: $$ A_e - \Omega = - \frac{1}2 \int\limits_{V_{II}} {\rho \left[  tr\left( \left. \frac{\partial U}{\partial \mathbf E} \right|_{II} \cdot \left( \frac{\partial \mathbf u}{\partial \mathbf x} \right)^T \cdot \frac{\partial \mathbf u}{\partial \mathbf x} \right) + \left. \frac{\partial ^2 U}{\partial \mathbf E^2} \right|_{II}\left( \mathbf E^' - \mathbf E \right)^2 + \cdots  \right] dV } \le 0 $$

Eml5526.s11 16:12, 3 June 2011 (UTC)

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$$   \displaystyle \int\limits_{V_{II}} {\left[ {tr\left( \frac{\partial \mathbf u}{\partial \mathbf x} \cdot \mathbf{\boldsymbol \sigma} \cdot \left( \frac{\partial \mathbf u}{\partial \mathbf x} \right)^T \right) + \left. {\rho \frac{\partial ^2 U}{\partial \mathbf E^2}} \right|_{II}\left( \mathbf E^' - \mathbf E \right)^2 + \cdots } \right]dV \ge } 0 $$  (62) Calculating the second term explicitly:
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$$   \displaystyle \frac{\partial ^2 U}{\partial \mathbf E^2}\left( \mathbf E^' - \mathbf E \right)^2 = \left( \mathbf E^' - \mathbf E \right)\mathbf{:}\frac{\partial ^2 U}{\partial \mathbf E^2}\mathbf{:}\left( \mathbf E^' - \mathbf E \right) $$
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$$   \displaystyle \approx sym\left( \mathbf F^T \cdot \frac{\partial \mathbf u}{\partial \mathbf x} \right)\mathbf{:}\frac{\partial ^2 U}{\partial \mathbf E^2}\mathbf{:}sym\left( \mathbf F^T \cdot \frac{\partial \mathbf u}{\partial \mathbf x} \right) = \left( \mathbf F^T \cdot \frac{\partial \mathbf u}{\partial \mathbf x} \right)\mathbf{:}\frac{\partial ^2 U}{\partial \mathbf E^2}\mathbf{:}\left( \mathbf F^T \cdot \frac{\partial \mathbf u}{\partial \mathbf x} \right) $$  (63) And the final expression for stability condition is:
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$$   \displaystyle \int\limits_{V_{II}} {\left[ tr\left( \frac{\partial \mathbf u}{\partial \mathbf x} \cdot \mathbf{\boldsymbol \sigma } \cdot \left( \frac{\partial \mathbf u}{\partial \mathbf x} \right)^T \right) + \rho \left( \mathbf F^T \cdot \frac{\partial \mathbf u}{\partial \mathbf x} \right)\mathbf{:}\frac{\partial ^2 U}{\partial \mathbf E^2}\mathbf{:}\left( \mathbf F^T \cdot \frac{\partial \mathbf u}{\partial \mathbf x} \right) \right]dV \ge } 0 $$  (64) In indicial notation, it is
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$$   \displaystyle \int\limits_{V_{II}} {\left[ \sigma _{ij}\frac{\partial u_k}{\partial x_i}\frac{\partial u_k}{\partial x_j} + \rho \frac{\partial ^2 U}{\partial E_{ij} \partial E_{pq}}\frac{\partial x_k}{\partial x_i}\frac{\partial u_k}{\partial x_j}\frac{\partial x_l}{\partial X_p} \frac{\partial u_l}{\partial X_q} \right]dV \ge } 0 $$  (65) Here all quantities are calculated for state II. Using Taylor’s expansion again to write the second tebold:
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$$\displaystyle \frac{\partial ^2 U}{\partial \mathbf E^2} = \left. \frac{\partial ^2 U}{\partial \mathbf E^2} \right|_I + \left. \frac{\partial ^3U}{\partial \mathbf E^3} \right|_I\left( \mathbf E - \mathbf E_I \right) + \cdots  \approx \left. \frac{\partial ^2 U}{\partial \mathbf E^2} \right|_I $$
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$$   \displaystyle \rho \frac{\partial ^2 U}{\partial \mathbf E^2} \approx \rho _0 \left. \frac{\partial ^2 U}{\partial \mathbf E^2} \right|_I = \left. \frac{\partial ^2 \left( \rho _0 u \right)}{\partial \mathbf E^2} \right|_I = \mathbf C_I$$ (66) Moreover, assuming the defoboldation between states I and II are small (the displacements $$\displaystyle \mathbf u $$ are not necessarily small), so there are no differences between positions $$\displaystyle \mathbf x $$ and $$\displaystyle \mathbf X $$:
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$$   \displaystyle \mathbf F = \frac{\partial \mathbf x}{\partial \mathbf x} \approx \mathbf I,\frac{\partial \mathbf u}{\partial \mathbf x} \approx \frac{\partial \mathbf u}{\partial \mathbf x},\mathbf C_I \approx \mathbf C_{II} \equiv \mathbf C $$ (67) Thus, stability condition becomes:
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$$\displaystyle \int\limits_{V_{II}} {\left[ tr\left( \frac{\partial \mathbf u}{\partial \mathbf x} \cdot \mathbf{\boldsymbol \sigma } \cdot \left( \frac{\partial \mathbf u}{\partial \mathbf x} \right)^T \right) + \frac{\partial \mathbf u}{\partial \mathbf x}\mathbf{:c:}\frac{\partial \mathbf u}{\partial \mathbf x} \right]dV \ge } 0 $$
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$$\displaystyle \leftrightarrow \int\limits_{V_{II}} {\left[ tr\left( \frac{\partial \mathbf u}{\partial \mathbf x} \cdot \mathbf{\boldsymbol \sigma } \cdot \left( \frac{\partial \mathbf u}{\partial \mathbf x} \right)^T \right) + sym\left( \frac{\partial \mathbf u}{\partial \mathbf x} \right)\mathbf{:c:}sym\left( \frac{\partial \mathbf u}{\partial \mathbf x} \right) \right]dV \ge } 0 $$
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$$   \displaystyle \leftrightarrow \int\limits_{V_{II}} {\left[ tr\left( \frac{\partial \mathbf u}{\partial \mathbf x} \cdot \mathbf{\boldsymbol \sigma } \cdot \left( \frac{\partial \mathbf u}{\partial \mathbf x} \right)^T \right) + \mathbf{e:c:e} \right]dV \ge } 0$$ (68) where $$\displaystyle \mathbf e$$ is defined (in state II):
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$$   \displaystyle \mathbf E = \frac{1}2\left[ \nabla \mathbf u + \left( \nabla \mathbf u \right)^T \right] = \frac{1}2\left[ \frac{\partial \mathbf u}{\partial \mathbf x} + \left( \frac{\partial \mathbf u}{\partial \mathbf x} \right)^T \right] $$  (69) or in the indical notation,
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$$   \displaystyle \int\limits_{V_{II}} {\left[ \sigma _{ij}\frac{\partial u_k}{\partial x_i}\frac{\partial u_k}{\partial x_j} + c_{mnpq}e_{mn}e_{pq} \right]dV \ge } 0$$ (70) For isotropic media,
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$$   \displaystyle c_{mnpq} = \lambda \delta _{mn} \delta _{pq} + \mu \left( \delta _{mp}\delta _{nq} + \delta _{mq}\delta _{np} \right) $$  (71) $$\displaystyle \lambda$$ and $$\displaystyle \mu$$ are Lame’s constants. This relation reduces the stability criterion to:
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$$   \displaystyle \int\limits_{V_{II}} {\left[ \sigma _{ij}\frac{\partial u_k}{\partial x_i}\frac{\partial u_k}{\partial x_j} + \lambda e_{mm}e_{nn} + 2\mu e_{mn}e_{mn} \right]dV \ge } 0 $$   (72)
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 Can you deduce from (72) that $$\displaystyle \lambda + \mu > 0$$ ? Since this condition is used later on to show that the Poisson's ratio of a stable composite material can be negative. Egm6321.f11 15:12, 24 June 2011 (UTC)

The surface foces are pressure loading
We assume that the system is still conservative so the total work done by pressure forces is independent of the path. Hence, let the trial displacements $$\displaystyle \mathbf u$$ grow at a constant rate, i.e at the time $$\displaystyle t$$, displacements are $$\displaystyle \mathbf u t$$. The work done in the interval from $$\displaystyle t = 0$$ to $$\displaystyle t =1$$ is:
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$$   \displaystyle A_e^1 = \int\limits_0^1 Dt \int\limits_{S_{III}} \left( - P\mathbf n_t \right) \cdot \frac{d}{Dt}\left( \mathbf ut \right)ds_t =  - P\int\limits_0^1 Dt \int\limits_{S_{III}} \mathbf u \cdot \mathbf n_tds_t$$ (73)
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 You need to define the surface $$S_{III}$$, which is the perturbed configuration from the surface $$S_{II}$$ at equilibrium. Eml5526.s11 14:26, 10 June 2011 (UTC)

 There are 3 configurations: Reference configuration $$\mathcal B$$, the current equilibrium configuration $$\mathcal B_0$$, the perturbed configuration $$\mathcal B_t$$. The modern notation for different quantities, e.g., stresses, strains, etc. is as follows. In the reference configuration $$\mathcal B$$, uppercase letters are used, e.g., $$\boldsymbol P$$ (first Piola-Kirchhoff stress tensor), $$\boldsymbol E$$ (Green-Lagrange strain tensor), $$\boldsymbol N$$ (normal to boundary surface $$\partial \mathcal B$$), etc. In the current equilibrium configuration $$\mathcal B_0$$, lowercase letters are used, e.g., $$\boldsymbol \sigma_0$$ (Cauchy stress tensor), $$\boldsymbol \epsilon_0$$ (small strain tensor), $$\boldsymbol n_0$$ (normal to boundary surface $$\partial \mathcal B_0$$), etc. In the perturbed configuration $$\mathcal B_t$$, lowercase letters are used, e.g., $$\boldsymbol \sigma_t$$ (Cauchy stress tensor), $$\boldsymbol \epsilon_t$$ (small strain tensor), $$\boldsymbol n_t$$ (normal to boundary surface $$\partial \mathcal B_0$$), etc. Eml5526.s11 14:26, 10 June 2011 (UTC)

where $$\displaystyle \mathbf {n}_{t}$$ is the normal vector of the area element  $$\displaystyle ds_{t}$$ in the spatial configuration and is related to the normal vector   $$\displaystyle n$$ of the area element  $$\displaystyle ds$$ in the reference configuration as follows:
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$$   \displaystyle \mathbf n_t ds_t = \det \left( \mathbf F_t \right)\mathbf F_t^{-T} \cdot \mathbf nds $$  (74) where  $$\displaystyle {\mathbf {F}_{t}}$$ is the deformation gradient
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$$   \displaystyle \mathbf F_t = \frac{\partial \mathbf X_t}{\partial \mathbf x} = \frac{\partial \left( \mathbf X + \mathbf ut \right)}{\partial \mathbf x} = \mathbf I + t\frac{\partial \mathbf u}{\partial \mathbf x} = \mathbf I + t\mathbf u\overleftarrow \nabla$$ (75) and
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$$   \displaystyle \mathbf u\overleftarrow \nabla  = \frac{\partial \mathbf u}{\partial \mathbf x}$$ (76) From this, we can determine:
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$$   \displaystyle \det \left( \mathbf F_t \right) = 1 + tI_1 + t^2 I_2 + t^3 I_3 $$  (77) where
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$$   \displaystyle I_1 = tr\left( \mathbf u\overleftarrow \nabla \right),I_2 = \frac{1}2\left[ tr^2\left( \mathbf u\overleftarrow \nabla \right) - tr\left( \mathbf u\overleftarrow \nabla   \cdot \mathbf u\overleftarrow \nabla \right) \right], I_3 = \det \left( \mathbf u\overleftarrow \nabla  \right)$$ (78) We assume infinitesimal deformation,
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$$   \displaystyle \left| \mathbf u\overleftarrow \nabla \right| <  < 1$$ (79) so
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$$   \displaystyle \mathbf F_t^{-1} = \left( \mathbf I + t\mathbf u\overleftarrow \nabla \right)^{-1} \approx \mathbf I - t\mathbf u\overleftarrow \nabla  + t^2 \left( \mathbf u\overleftarrow \nabla  \right)^2 - t^3 \left( \mathbf u\overleftarrow \nabla  \right)^3 +  \cdots$$ (80) or
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$$   \displaystyle \mathbf F_t^{-T} = \left( \mathbf I + t\mathbf u\overleftarrow \nabla \right)^{-T} \approx \mathbf I - t\nabla \mathbf u + t^2\left( \nabla \mathbf u \right)^2 - t^3\left( \nabla \mathbf u \right)^3 + \cdots $$  (81) here
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$$   \displaystyle \nabla \mathbf u = \left( \frac{\partial \mathbf u}{\partial \mathbf x} \right)^T$$ (82) The integrand becomes:
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$$   \displaystyle \mathbf u \cdot \mathbf n_tds_t = \det \left( \mathbf F_t \right)\mathbf u \cdot \mathbf F_t^{-T} \cdot \mathbf nds \approx \left( 1 + tI_1 + t^2I_2 + t^3I_3 \right)\mathbf u \cdot \left( \mathbf I - t\nabla \mathbf u + t^2\left( \nabla \mathbf u \right)^2 +  \cdots \right) \cdot \mathbf nds$$
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$$   \displaystyle = \mathbf u \cdot \begin{Bmatrix} \mathbf I + t\left( I_1\mathbf I - \nabla \mathbf u \right) + t^2\left[ \left( \nabla \mathbf u \right)^2 - I_1\nabla \mathbf u + I_2\mathbf I \right] + \cdots \end{Bmatrix} \cdot \mathbf nds$$ (83)
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And the work done by pressure forces is: :{| style="width:100%" border="0" $$   \displaystyle \begin{align} A_e^1 & =  - P\int\limits_0^1 Dt \int\limits_{S_{II}} {\det \left( \mathbf F_t \right)\mathbf u \cdot \mathbf F_t^{-T} \cdot \mathbf nds}  =  - P\int\limits_{S_{II}} {\mathbf u \cdot \int\limits_0^1 \begin{Bmatrix} \mathbf I + t\left( I_1\mathbf I - \nabla \mathbf u \right) + t^2\left[ \left( \nabla \mathbf u \right)^2 - I_1\nabla \mathbf u + I_2\mathbf I \right] + \cdots \end{Bmatrix} dt \cdot \mathbf nds} \\ & = -P\int\limits_{S_{II}} {\mathbf u \cdot \begin{Bmatrix} \mathbf I + \frac{1}2\left( I_1\mathbf I - \nabla \mathbf u \right) + \frac{1}{3}\left[ \left( \nabla \mathbf u \right)^2 - I_1\nabla \mathbf u + I_2\mathbf I \right] + \cdots \end{Bmatrix} \cdot \mathbf nds} \approx  - P\int\limits_{S_{II}} {\mathbf u \cdot \left[ \mathbf I + \frac{1}2\left( I_1\mathbf I - \nabla \mathbf u \right) \right] \cdot \mathbf nds}  \\ \end{align} $$  (84)
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Note that, Pearson got the above expression exactly as follows,, p. 141-142, using indical notation, , p. 16:
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$$   \displaystyle \left( \mathbf n_i \right)_t \left( ds \right)_t = \frac{1}2\epsilon _{jki}\epsilon _{mnp}\frac{\partial \left( x_j + tu_j \right)}{\partial {x_m}}\frac{\partial \left(x_k + tu_k \right)}{\partial x_n}n_p ds = \frac{1}2\epsilon _{jki}\epsilon _{mnp}\frac{\partial \left( x_j + tu_j \right)}{\partial x_m}\frac{\partial \left( x_k + tu_k \right)}{\partial x_n}n_pds$$
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$$   \displaystyle = \frac{1}2\epsilon _{jki}\epsilon _{mnp} \left( \delta _{jm} + t\frac{\partial u_j}{\partial x_m} \right)\left( \delta _{kn} + t\frac{\partial u_k}{\partial x_n} \right)n_p ds = u_i\left( \delta _{ip} + t\frac{\partial u_k}{\partial x_k}\delta _{ip} - t\frac{\partial u_p}{\partial x_i} + \frac{1}2t^2\epsilon_{jki}\epsilon _{mnp}\frac{\partial u_j}{\partial x_m}\frac{\partial u_k}{\partial x_n} \right)n_pds $$
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$$   \displaystyle = \left( u_i n_i + t u_i n_i \frac{\partial u_k}{\partial x_k} - tu_i\frac{\partial u_p}{\partial x_i}n_p + \frac{1}2t^2\epsilon _{jki}\epsilon _{mnp}u_i\frac{\partial u_j}{\partial x_m}\frac{\partial u_k}{\partial x_n}n_p \right)ds$$ (85) Plugging these formular into expression (73) to obtain:
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$$   \displaystyle A_e^1 = - P\int\limits_0^1 Dt \int\limits_{S_{III}} u_i \left( \mathbf n_i \right) _t \left( ds \right)_t =  - P\int\limits_{S_{II}} ds \int\limits_0^1  \left( u_i n_i + tu_i n_i\frac{\partial u_k}{\partial x_k} - tu_i\frac{\partial u_p}{\partial x_i}n_p + \frac{1}2t^2\epsilon _{jki}\epsilon _{mnp}u_i\frac{\partial u_j}{\partial x_m}\frac{\partial u_k}{\partial x_n}n_p \right) Dt$$
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$$   \displaystyle = - P\int\limits_{S_{II}} {\left( u_i n_i + \frac{1}2u_i n_i\frac{\partial u_k}{\partial x_k} - \frac{1}2u_i\frac{\partial u_p}{\partial x_i}n_p + \frac{1}{6}\epsilon _{jki}\epsilon _{mnp}u_i\frac{\partial u_j}{\partial x_m}\frac{\partial u_k}{\partial x_n}n_p \right)ds}  $$ (86)
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Convert back to direct notation:
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$$   \displaystyle A_e^1 = - P\int\limits_{S_{II}} {\left\{ \mathbf u \cdot \mathbf n + \frac{1}2\left[ \left( \nabla  \cdot \mathbf u \right)\mathbf u \cdot \mathbf n - \mathbf u \cdot \left( \nabla \mathbf u \right) \cdot \mathbf n \right] + \frac{1}{6}\left(\mathbf u \times \mathbf u\overleftarrow \nabla  \mathbf{:u}\overleftarrow \nabla   \times \mathbf n \right) \right\} ds}$$ (87)
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 You may want to refer to Pearson's notation as the "component form" to distinguish from the "tensor form" in Eq.(87) (which you called the "direct" notation) Eml5526.s11 15:19, 10 June 2011 (UTC)

 Eq.(87) is to be used in the stability condition expressed in Eq.(54). It is not clear yet how .... You want to explain this point clearly. Eml5526.s11 15:19, 10 June 2011 (UTC)

=Elastic stability of composite materials including an inclusion and a coating= Consider the stability of an elastic body intially free of stresses. The formular (72) reduces:

 Give references. Egm6321.f11 14:46, 17 June 2011 (UTC)

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$$   \displaystyle \int\limits_V {\left[ 2\mu \boldsymbol\epsilon :\boldsymbol\epsilon + \lambda \left( tr\boldsymbol\epsilon \right)^2 \right]dV>}0 $$  (88) Integrating the whole domain of the body will be calculated through the inclusion $$\displaystyle V_{1}$$ and the coating $$\displaystyle V_2$$ :
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$$   \displaystyle \int\limits_{V_1} {\left[ 2\mu _1\boldsymbol\epsilon :\boldsymbol\epsilon + \lambda _1\left( tr\boldsymbol\epsilon \right)^2 \right]dV}  + \int\limits_{V_2} {\left[ 2\mu _2\boldsymbol\epsilon :\boldsymbol\epsilon + \lambda _2\left( tr\boldsymbol\epsilon  \right)^2\right]dV >} 0 $$  (89) Introducing an arbitrary parameter $$\displaystyle \beta$$,  $$\displaystyle 0 \le \beta  \le 1$$ to vary the inclusion moduli which will turn out to be the full range for composite stability from positive definite to strongly elliptic.
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$$   \displaystyle \mu _1 = \beta \mu _1 + \left( 1 - \beta \right)\mu _1 $$  (90)
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 In the composite literature, the above equation is called the "rule of mixture". In mathematics, the above equation is called the "convex combination" of $$\displaystyle \mu_1$$ and $$\displaystyle \mu_2$$. Why not doing also a convex combination of the lambda's, i.e., $$\displaystyle \lambda_1$$ and $$\displaystyle \lambda_2$$. OK, the above is not the rule of mixture of $$\displaystyle \mu_1$$ and $$\displaystyle \mu_2$$, but only rewritting $$\displaystyle \mu_1$$ in terms of the parameter $$\displaystyle \beta$$. Why ? Egm6321.f11 14:51, 17 June 2011 (UTC)

Assume the coating is thin, so the strains are constanst through it and the second volume integration becomes the product of the surface one and the thickness $$\displaystyle t$$ of the coating. The stability condition is:
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$$   \displaystyle \int\limits_{V_1} {\left[ 2\mu_1\left( 1 - \beta \right)\boldsymbol\epsilon :\boldsymbol\epsilon + 2\mu_1\beta \boldsymbol\epsilon :\boldsymbol\epsilon + \lambda _1\left( tr\boldsymbol\epsilon \right)^2 \right]dV}  + \int\limits_{S_2} {t\left[ 2\mu_2\boldsymbol\epsilon :\boldsymbol\epsilon + \lambda_2\left( tr\boldsymbol\epsilon \right)^2 \right]dS > } 0 $$  (91) Decompose the strain tensor into the deviatoric tensor $$\displaystyle \boldsymbol\epsilon^'$$ and use the Kelvin’s identity as follows
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$$   \displaystyle \boldsymbol\epsilon = \boldsymbol\epsilon^' + \frac{1}{k}tr\left( \boldsymbol\epsilon \right)\mathbf I $$ (92)
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$$   \displaystyle \boldsymbol\epsilon :\boldsymbol\epsilon = \boldsymbol\epsilon ^' \mathbf{:}\boldsymbol\epsilon ^' + \frac{1}{k}\left( tr\boldsymbol\epsilon \right)^2 $$  (93)
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$$   \displaystyle \boldsymbol\epsilon :\boldsymbol\epsilon = \boldsymbol\omega :\boldsymbol\omega + \left( tr\boldsymbol\epsilon \right)^2 + \boldsymbol\nabla \cdot \left[ \left( \mathbf u\boldsymbol\nabla \right) \cdot \mathbf u - \mathbf u\left( \boldsymbol\nabla  \cdot \mathbf u \right) \right] $$  (94) Introducing the first identity into the first term, the second into the second, all in the first integrand, to get (after resorting)
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$$ \begin{matrix} \displaystyle \int\limits_{V_1} {\left\{ 2\mu_1\left[ \left( 1 - \beta \right)\boldsymbol\epsilon ^'\mathbf{:}\boldsymbol\epsilon^'+ \beta \boldsymbol\omega :\boldsymbol\omega \right] + \left[ \lambda _1 + \frac{2\mu_1}{k}\left( 1 + \left(k - 1 \right)\beta \right) \right]\left( tr\boldsymbol\epsilon \right)^2 \right\} dV} \\    \displaystyle \, + \, 2\mu_1\beta \int\limits_{V_1} {\boldsymbol\nabla \cdot \left[ \left( \mathbf u\boldsymbol\nabla \right) \cdot \mathbf u - \mathbf u\left( \boldsymbol\nabla  \cdot \mathbf u \right)\right]dV}  + \int\limits_{S_2} {t\left[ 2\mu_2\boldsymbol\epsilon :\boldsymbol\epsilon + \lambda_2\left( tr\boldsymbol\epsilon \right)^2\right]dS >} 0 \end{matrix} $$   (95)
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Using the divergence theorem and the following identity
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$$   \displaystyle \mathbf n \cdot \left[\left( \mathbf u\boldsymbol\nabla \right) \cdot \mathbf u - \mathbf u\left(\boldsymbol\nabla \cdot \mathbf u \right) \right] = tr\left[ \mathbf u \times \left( \mathbf u\boldsymbol\nabla \right) \times \mathbf n \right] $$  (96)
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to convert the second volume integration into the surface one and remember that the inclusion and the coating share the same surface $$\displaystyle S$$ to get the final expression for stability condition:
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$$ \begin{matrix} \displaystyle \int\limits_{V_1} {\left\{ 2\mu_1\left[ \left( 1 - \beta \right)\boldsymbol\epsilon ^'\mathbf{:}\boldsymbol\epsilon^'+ \beta \boldsymbol\omega :\boldsymbol\omega \right] + \left[ \lambda _1 + \frac{2\mu_1}{k}\left( 1 + \left(k - 1 \right)\beta \right) \right]\left( tr\boldsymbol\epsilon \right)^2 \right\} dV} \\    \displaystyle \, + \, \int\limits_{S} {\left\{t\left[ 2\mu_2\boldsymbol\epsilon :\boldsymbol\epsilon + \lambda_2\left( tr\boldsymbol\epsilon \right)^2\right] + 2\mu_1\beta tr\left[ \mathbf u \times \left( \mathbf u\boldsymbol \nabla \right) \times \mathbf n \right] \right\} dS} > 0 \end{matrix} $$   (97)
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Now we are in a position to analyze two cases.

Plane strain coated cylinder problem
In this case $$ \displaystyle k = 2 $$ and express the surface integral in terms of polar coordinate $$ \displaystyle r, \theta $$ and $$ \displaystyle dS = a d\theta $$.

Remember in polar coordinate we have the followings:

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$$ \displaystyle \boldsymbol\epsilon = \epsilon _{rr}\mathbf E_r\mathbf E_r + \epsilon _{r\theta} \mathbf E_r\mathbf E_\theta + \epsilon _{r\theta }\mathbf E_\theta\mathbf E_r + \epsilon _{\theta\theta}\mathbf E_\theta \mathbf E_\theta $$ (98)
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$$ \displaystyle \overleftarrow \nabla  = \frac{\partial}{\partial r}\mathbf E_r + \frac{1}{r}\frac{\partial}{\partial \theta}\mathbf E_\theta $$ (99)
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$$ \displaystyle \mathbf u = u_r\mathbf E_r + u_\theta \mathbf E_\theta $$ (100)
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$$ \displaystyle \mathbf u\overleftarrow \nabla  = \frac{\partial u_r}{\partial r}\mathbf E_r\mathbf E_r + \left( \frac{1}{r}\frac{\partial u_r}{\partial \theta} - \frac{u_\theta }{r} \right)\mathbf E_r\mathbf E_\theta + \frac{\partial u_\theta}{\partial r}\mathbf E_\theta\mathbf E_r + \left( \frac{1}{r}\frac{\partial u_\theta}{\partial \theta } + \frac{u_r}{r}\right)\mathbf E_\theta\mathbf E_\theta $$ (101)
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so

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$$ \displaystyle \boldsymbol \epsilon : \boldsymbol \epsilon = \epsilon _{rr}^2 + \epsilon _{\theta \theta }^2 + 2\epsilon _{r\theta }^2 $$ (102)
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$$ \displaystyle tr\left( \boldsymbol \epsilon \right) = \epsilon _{rr} + \epsilon _{\theta \theta} $$ (103)
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$$ \displaystyle \mathbf u\times \mathbf u\overleftarrow {\boldsymbol\nabla} \times \mathbf n = \left( u_r \mathbf E_r + u_\theta \mathbf E_\theta \right)\times \left[ \frac{\partial u_r}{\partial r}\mathbf E_r\mathbf E_r+\frac{1}{r}\left( \frac{\partial u_r}{\partial \theta} - u_\theta \right)\mathbf E_r\mathbf E_\theta + \frac{\partial u_\theta}{\partial r}\mathbf E_\theta \mathbf E_r + \frac{1}{r} \left( \frac{\partial u_\theta}{\partial \theta} + u_r \right)\mathbf E_\theta \mathbf E_\theta \right]\times \mathbf E_r $$
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$$ \displaystyle = - \frac{1}{r}\left( u_r^2 + u_\theta ^2 + u_r\frac{\partial u_\theta}{\partial \theta } - u_\theta \frac{\partial u_r}{\partial \theta } \right)\mathbf E_z\mathbf E_z $$ (104)
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$$ \displaystyle tr\left( \mathbf u \times \mathbf u\overleftarrow \nabla  \times \mathbf n \right) =  - \frac{1}{r}\left( u_r^2 + u_\theta ^2 + u_r\frac{\partial u_\theta}{\partial \theta } - u_\theta \frac{\partial u_r}{\partial \theta } \right) $$ (105)
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Plugging those expressions above into the stability condition to get:

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$$ \begin{matrix} \displaystyle \int\limits_{A_1} {\left\{ 2\mu_1\left[ \left( 1 - \beta \right)\boldsymbol \epsilon ^'\mathbf{:}\boldsymbol \epsilon ^' + \beta \boldsymbol{\omega :\omega } \right] + \left[ \lambda _1 + \left( {1 + \beta } \right)\mu_1 \right] \left( tr\boldsymbol \epsilon \right)^2 \right\}dA}   \\ \displaystyle + \int\limits_0^{2\pi } {\left\{ ta\left[ 2\mu_2\left( \epsilon _{rr}^2 + \epsilon _{\theta \theta }^2 + 2\epsilon _{r\theta}^2 \right) + \lambda_2 \left( \epsilon _{rr} + \epsilon _{\theta \theta } \right)^2 \right] - 2\beta \mu_1\left[ \left( u_r \right)^2 + \left( u_\theta \right)^2 + u_r u_{\theta ,\theta} - u_\theta u_{r,\theta } \right] \right\} d\theta} > 0 \end{matrix} $$ (106)
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Employ a Fourier series representation for $$ \displaystyle u_{\theta} $$

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$$ \displaystyle u_\theta = \sum_{n = -\propto}^\propto c_n e^{in\theta} $$ (107)
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 Below is a reformatting of the above equation. Note that the symbol $$\displaystyle \propto$$ means "proportional to", whereas the symbol $$\displaystyle \infty$$ is infinity. Egm6321.f11 15:02, 17 June 2011 (UTC)

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$$  \displaystyle u_\theta = \sum_{n = -\infty}^\infty c_n e^{i n \theta } $$     (108)
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where $$ \displaystyle c_{-n}= {\bar c}_{n} $$ (overbar is the complex conjugate) and no rigid body motion requires $$ \displaystyle c_{0} = c_{1} =0 $$. These and orthogonality of Fourier series show on the boundary $$ \displaystyle r = a $$ that

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$$ \displaystyle \int\limits_0^{2\pi } {\left( u_\theta \right)^2 d\theta = 4\pi \sum\nolimits_{n = 2}^ \propto  c_n \bar {c_n} \int\limits_0^{2\pi } \left( u_\theta ,\theta \right)^2 d\theta  = 4\pi \sum\nolimits_{n = 2}^ \propto  n^2 c_n \bar {c_n}} $$ (109)
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Hence,

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$$ \displaystyle 4\int\limits_0^{2\pi } {\left( u_\theta \right)^2 d\theta \le \int\limits_0^{2\pi} \left( u_{\theta ,\theta} \right)^2 d\theta } $$ (110)
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Using the integration by part as below

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$$ \displaystyle \int\limits_0^{2\pi} {u_\theta u_{r,\theta}d\theta = \left. u_\theta u_r \right|_0^{2\pi} - \int\limits_0^{2\pi} u_r u_{\theta ,\theta } d\theta} = - \int\limits_0^{2\pi} {u_r u_{\theta ,\theta} d\theta } $$ (111)
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to write the integration of last bracket

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$$ \displaystyle \int\limits_0^{2\pi} {\left[ \left( u_r \right)^2 + \left( u_\theta \right)^2 + u_r u_{\theta ,\theta} - u_\theta u_{r,\theta } \right] d\theta}  \le \int\limits_0^{2\pi} {\left[ \left( u_r \right)^2 + \frac{1}{4}\left( u_{\theta ,\theta } \right)^2 + 2u_ru_{\theta ,\theta} \right]d\theta}  = \int\limits_0^{2\pi} {\left[ \left( u_r + u_{\theta ,\theta} \right)^2 - \frac{3}{4}\left( u_{\theta ,\theta} \right)^2 \right]d\theta} $$ (112)
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 *

 * }

Note that

 * {|style="width:100%" border="0"

$$ \displaystyle \epsilon _{\theta \theta} = \frac{u_r + u_{\theta ,\theta}}{r} $$ (113)
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 *

 * }

one can write the stability condition as

 * {|style="width:100%" border="0"

$$ \begin{matrix} \displaystyle \int\limits_{A_1} {\left\{ 2\mu_1\left[ \left( 1 - \beta \right)\boldsymbol \epsilon ^'\mathbf{:}\boldsymbol \epsilon ^' + \beta \boldsymbol{\omega :\omega } \right] + \left[ \lambda _1 + \left( {1 + \beta } \right)\mu_1 \right] \left( tr\boldsymbol \epsilon \right)^2 \right\}dA}   \\ \displaystyle + \int\limits_0^{2\pi } {\left\{ ta\left[ 2\mu_2\left( \epsilon _{rr}^2 + \epsilon _{\theta \theta }^2 + 2\epsilon _{r\theta}^2 \right) + \lambda_2 \left( \epsilon _{rr} + \epsilon _{\theta \theta } \right)^2 \right] + 2\beta \mu_1\left[ \frac{3}{4}\left( u_{\theta ,\theta} \right)^2 - a^2\epsilon _{\theta \theta}^2 \right] \right\} d\theta } > 0 \end{matrix} $$ (114)
 * style="width:95% |
 * style="width:95% |
 *

 * }

From this, Drugan in concluded that:

 * {|style="width:100%" border="0"

$$ \displaystyle \mu_1 > 0 ,\quad \lambda_1 > -(1+\beta) \mu_1, \quad \mu_2> \beta \frac{a}{t}\mu_1 > 0 , \quad \lambda_2 > 0 $$ (115)
 * style="width:95% |
 * style="width:95% |
 *

 * }

 For each inequality, justify where it came from. For example, the requirement that $$\displaystyle \mu_{1} > 0$$ is to force the first integrand in the first term in (115), i.e., $$\displaystyle 2\mu_1\left[ \left( 1 - \beta \right) \boldsymbol \epsilon ^' \mathbf{:}\boldsymbol \epsilon ^'  + \beta \boldsymbol{\omega :\omega } \right] $$ to be positive. Egm6321.f11 15:25, 24 June 2011 (UTC) Weaker sufficient restrictions for stability are obtained by Drugan, as follows. Ignoring the terms $$ \displaystyle \epsilon_{r\theta}^2, \  \left( u_{\theta, \theta} \right) ^2 $$ the integrand in the second integration of (117) becomes:

 You need to justify why you want to ingnore the terms $$ \displaystyle \epsilon_{r\theta}^2, \ \left( u_{\theta, \theta} \right) ^2 $$; the reason is not clear. The reason is because the terms $$\displaystyle 4 ta \mu_2 \epsilon_{r \theta}^2$$ and $$\displaystyle \frac{3}2 \beta \mu_1 \mu_{\theta, \theta}^2$$ in (115) are already positive; so ignoring these two terms and enforcing the remaining terms to be positive is a stronger requirement. So for now, such stronger requirement led to the inequality (stability condition) $$\displaystyle \lambda_{1} > - (1+\beta) \mu_1$$, which means that $$\displaystyle \lambda_1$$ can be negative. But to conclude that $$\displaystyle \nu_1$$ (Poisson's ratio for material 1) can be negative, you need to look at the relationship between $$\displaystyle \lambda_1$$ and $$\displaystyle \nu_1$$. Now, we have $$\displaystyle \nu_1 = \frac{\lambda_1}{2 (\lambda_1 + \mu_1)}$$ (see Lame' parameters (constants) (wikipedia); verify !), then we need to have the condition $$\displaystyle (\lambda_1 + \mu_1) > 0$$, which is the condition for positive definiteness of the elastic moduli tensor $$\displaystyle \boldsymbol C$$; verify ! Can you deduce from the general stability condition (72) for elastic materials that $$\displaystyle (\lambda_1 + \mu_1) > 0$$ ? Egm6321.f11 15:25, 24 June 2011 (UTC)

 For $$\displaystyle \nu_1 <0 $$ (negative Poisson ratio), you need $$\displaystyle \lambda_1 < 0$$ and $$\displaystyle (\lambda_1 + \mu_1) > 0 $$. How do you ensure this condition ? Egm6321.f11 14:16, 10 October 2011 (UTC)

 This paper by Drugan 2007 is not about negative Poisson's ratio material, but about constructing a composite material made up of a core material and a coating material. A negative bulk-modulus material is used in the core, while a positive bulk-modulus material is used in the coating to make stable composite materials (with positive effective bulk modulus). The next question would be how to increase the effective bulk modulus as high as possible.

For a (single) homogeneous material: Recall that the relationship between the bulk modulus $$\displaystyle K$$ and the Young's modulus $$\displaystyle E$$ and the Poisson's ratio $$\displaystyle \nu$$ is $$\displaystyle K = \frac{E}{3 (1 - 2 \nu)}$$. Thus when $$\displaystyle \nu \to 0.5$$ then $$\displaystyle K \to \infty$$.

The relationship between the bulk modulus $$\displaystyle K$$ and the Lame' constant $$\displaystyle \lambda$$ and the Poisson's ratio $$\displaystyle \nu$$ is $$\displaystyle K = \frac{\lambda (1 + \nu)}{3 \nu}$$. Thus when $$\displaystyle \nu \to -1$$ then $$\displaystyle K \to 0 $$.

Hence, for homogeneous materials with free boundary (i.e., without support), the bulk modulus $$\displaystyle K$$ and the shear modulus $$\displaystyle \mu$$ must be positive (i.e., $$\displaystyle K \ge 0$$ and $$\displaystyle \mu \ge 0$$ ). The conditions lead to positive definite elastic tensor $$\displaystyle \mathbf C$$. So stable homogeneous materials can have negative Poisson's ratio since $$\displaystyle \nu \in [-1, 0.5]$$.

Note that homogeneous materials with boundary support (e.g., set the displacements to zero at the boundary) can also be stable with negative bulk modulus (because of the boundary support). The elastic tensor $$\displaystyle \mathbf C$$ is strongly elliptic.

Reference: F. de Veubeke et al. 1979, A course in elasticity, p.??? .

Egm6321.f11 14:49, 17 October 2011 (UTC)

 * {|style="width:100%" border="0"

$$ \displaystyle F = ta[2\mu_2(\epsilon_{rr}^2 + \epsilon_{\theta \theta}^2) + \lambda_2(\epsilon_{rr} + \epsilon_{\theta\theta})^2]- 2a^2\beta\mu_{1}\epsilon_{\theta\theta}^2 $$ (116)
 * style="width:95% |
 * style="width:95% |
 *

 * }

We require that this expression is positive or

 * {|style="width:100%" border="0"

$$ \displaystyle G = \frac{F}{ta} = 2\mu_2(\epsilon_{rr}^2+\epsilon_{\theta \theta}^2) + \lambda_2(\epsilon_{rr} + \epsilon_{\theta\theta})^2 - 2\alpha\epsilon_{\theta\theta}^2 > 0 $$ (117)
 * style="width:95% |
 * style="width:95% |
 *

 * }

where $$ \displaystyle \alpha = \beta\frac{a}{t}\mu_{1} $$

(118) is equivalent to

 * {|style="width:100%" border="0"

$$ \displaystyle (\lambda_2+2\mu_2)\epsilon_{rr}^2 + 2\lambda_2\epsilon_{rr}\epsilon_{\theta\theta} + (\lambda_2+2\mu_2-2\alpha)\epsilon_{\theta\theta}^2 > 0 $$ (118)
 * style="width:95% |
 * style="width:95% |
 *

 * }

or

 * {|style="width:100%" border="0"

$$ \displaystyle \begin{matrix}\lambda_2+2\mu_2>0 \\ \lambda_2^2\epsilon_{\theta\theta}^2 - (\lambda_2+2\mu_2)(\lambda_2+2\mu_2-2\alpha)\epsilon_{\theta\theta}^2 < 0 \end{matrix} $$ (119)
 * style="width:95% |
 * style="width:95% |
 *

 * }

Solving this system of two inequalities to get the final condition:

 * {|style="width:100%" border="0"

$$ \displaystyle \mu_2>\frac{\alpha}{2}, \lambda_2>-\mu_2\frac{\mu_2-\alpha}{\mu_2-\frac{\alpha}{2}} = -\mu_2\frac{1-\beta\frac{a}{t}\frac{\mu_1}{\mu_2}}{1-\beta\frac{a}{t}\frac{\mu_1}{\mu_2}} $$ (120)
 * style="width:95% |
 * style="width:95% |
 *

 * }

 This condition (122) is "better" than the condition in (117) since it provides a larger stability domain (and hence more choices of material properties) than that from (117). Note that if $$\displaystyle \mu_1 > \alpha$$, and since $$\displaystyle \mu_1 > 0$$, then $$\displaystyle - \mu_1 \frac{\mu_1 - \alpha}{\mu_1 - \frac{\alpha}2} < 0$$ and thus $$\displaystyle \lambda_1$$ can be negative. Egm6321.f11 15:25, 24 June 2011 (UTC)

= References =

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:Egm6936.s09/Elastic stability


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User talk:Skol fir12

Hi. Stop deleting the template that I added to Skol fir. Leave it to an admin to decide if it should be deleted or not. -- Tegel 23:29, 31 May 2011 (UTC)

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The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

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User talk:Homeeducation

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Thanks for your help!

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