User:EGM6321.f10.team6.yoshioka

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= Homework #1 = [Due Wed, 15 Sep 10]

Problem 1 - Additional
Developing the Reynolds’ Transport Theorem (taken from Nonlinear Solid Mechanics - A Continuum Approach for Engineering, Gerhard A. Holzapfel) is an application of the first derivative of $$\Phi (\text{x},t)$$, and it is presents as follows:

Suppose we have a spatial scalar field $$\Phi (\text{x},t)$$ describing some physical quantity (for example, mass, internal energy, entropy, heat, or entropy sources) of a particle in space per unit volume at time t.

$$I(t)=\int\limits_{\Omega }{\Phi (\text{x},t)\text{d}\upsilon }$$ (9)

Assume $$\Phi (\text{x},t)$$ to be smooth, so that it is continuously differentiable. Hence, the present status of a continuum body in some three-dimensional region Ω with volume $$\upsilon $$ at given time t may be characterized by the scalar-valued function

The aim is now to compute the material time derivative of the volume integral $$I(t)$$ Since the region of integration Ω depends on time $$t$$, integration and time differentiation do not commute. Therefore, as a first step $$I(t)$$ must be transformed to the reference configuration. By changing variables using the motion $$\text{x}=\chi (\text{X},t)$$ and the relation $$\text{d}\upsilon \text{= }J(\text{X},t)\text{d}V$$ we find the time rate of change of $$I(t)$$ to be

$$\dot{I}(t)=\frac{\text{D}}{\text{Dt}}\int\limits_{\Omega }{\Phi (\text{x},t)\text{d}v=}\frac{\text{D}}{\text{Dt}}\int\limits_{\Phi (\chi (\text{X},t),t)J(\text{X},t)\text{d}V}$$ (10)

Since the region of integration is now time-independent, integration and differentiation commute. Hence, as a second step, from (10)we obtain, using the product rule of differentiation,

$$\frac{\text{D}}{\text{Dt}}\int\limits_{\Omega }{\Phi (\text{x},t)\text{d}\upsilon =}\int\limits_{\left[ \dot{\Phi }(\chi (\text{X},t),t)J(\text{X},t)+\Phi (\chi (\text{X},t),t)\dot{J}(\text{X},t) \right]}\text{d}V$$ (11)

where $${\dot{\Phi }}$$ denotes the material time derivative of the spatial scalar field $$\Phi $$. In a last step we undo the change of variables and convert the volume integral back to the current configuration. By means of the material time derivative of the scalar J, $$\text{d}\upsilon \text{= }J(\text{X},t)\text{d}V$$ and motion $$\text{x}=\chi (\text{X},t)$$, we find finally that

$$\begin{align} & \frac{\text{D}}{\text{Dt}}\int\limits_{\Omega }{\Phi (\text{x},t)\text{d}\upsilon =}\int\limits_{\left[ \dot{\Phi }(\chi (\text{X},t),t)+\Phi (\chi (\text{X},t),t)\frac{\dot{J}(\text{X},t)}{J(\text{X},t)} \right]J(\text{X},t)}\text{d}V \\ & =\int\limits_{\Omega }{\left[ \dot{\Phi }(\text{x},t)+\Phi (\text{x},t)\frac{\dot{J}(\text{X},t)}{J(\text{X},t)} \right]\text{d}\upsilon } \\ & =\int\limits_{\Omega }{\left[ \dot{\Phi }(\text{x},t)+\Phi (\text{x},t)\text{div}\nu (\text{x},t) \right]\text{d}\upsilon } \\ \end{align}$$ (12)

In the following the arguments of the tensor quantities are dropped in order to simplify the notation. However, in cases where additional information is needed, they will be employed. Hence, relation (12) reads as

$$\frac{\text{D}}{\text{Dt}}\int\limits_{\Omega }{\Phi \text{d}\upsilon =}\int\limits_{\Omega }{\left( \dot{\Phi }+\Phi \text{div}\nu \right)}\text{d}\upsilon $$ (13) Where we have assumed smoothness of the spatial velocity field v.

Other forms of the time rate of change of the integral (9) result from (13) by means of the material time derivative of the spatial scalar field $$\Phi $$ can be developed, i.e. we can write the first derivative of the homework equation as

$$\dot{\Phi }(\text{x},t)={{\left( \frac{\partial \Phi (\text{x},t)}{\partial t} \right)}_{\text{x}}}+{{\left( \frac{\partial \Phi (\text{x},t)}{\partial \text{x}} \right)}_{t}}\cdot {{\left( \frac{\partial \chi (\text{X},t)}{\partial t} \right)}_{\text{X}={{\chi }^{-1}}(\text{x},t)}}$$ (14)

$$v(\text{x},t)=\frac{\partial \chi (\text{X},t)}{\partial t}$$ (15)

$$\text{grad}\Phi (\text{x},t)=\frac{\partial \Phi (\text{x},t)}{\partial \text{x}}$$ (16)

$$\text{div(}\Phi \text{u) = }\Phi \text{divu + u}\cdot \text{grad}\Phi $$ (17)

We can write the integral as

$$\dot{\Phi }(\text{x},t)=\frac{\text{D}\Phi (\text{x},t)}{\text{Dt}}=\frac{\partial \Phi (\text{x},t)}{\partial t}+\text{grad}\Phi (\text{x},t)\cdot \nu (\text{x},t)$$

$$=\int\limits_{\Omega }{\left( \text{div}\left( \Phi \nu \right)+\frac{\partial \Phi }{\partial t} \right)}\text{d}\upsilon $$ (18)

And finally, using the divergence theorem

$$\int\limits_{S}{\Phi \text{u}\cdot \text{nd}s=\int\limits_{\upsilon }{\text{div(}\Phi \text{u)d}}}\upsilon $$ (19)

The first term on the right-hand side of eq.(19) characterizes the rate of transport (or the outward normal flux) of $$\Phi \nu $$ across the surface $$\partial \Omega $$ out of region $$\Omega $$, which is assumed to be fixed. This contribution arises from the moving region. The second term denotes the local time rate of change of the spatial scalar field $$\Phi $$ within region $$\Omega $$. In (19) n denotes the outward unit normal field acting along $$\partial \Omega $$. Relation (18) is referred to as Reynolds’ transport theorem.