Kinematics of deformation (Continuum mechanics)

[Jinjolee]

Homework Statement

Calculating the deformation gradient, velocity field and acceleration field of a cavity motion

Relevant Equations

F=grad(x)
J=det(F)
v=dx/dt
A=dv/dt

Question is extracted from "Ellad B Tadmor, Ronald E Miller, Ryan S Elliott - Continuum mechanics and thermodynamics From fundamental concepts to governing equations".
I just got stuck at part (a). I think if part(a) is solved, I may be able to do the other parts.

 

[Chestermiller]

What have you done so far?

 

[Jinjolee]

What have you done so far?

I tried to do grad(x) = $$grad(\frac {f(R,t)}{R} X)$$
and by product rules:
=$$\frac{f(R,t)}{R} ∇X + X ⋅ ∇(\frac {f(R,t)}{R})$$
If I am correct, ∇X should be I, so
=$$\frac{f(R,t)}{R} I + X ⋅ ∇(\frac {f(R,t)}{R})$$
This is by far what I have done.

 

[Chestermiller]

Let's try this a little differently. Suppose, in the initial configuration, you have two material points at R and R+dR connected by a differential position vector $$\mathbf{dX}=dR\mathbf{i_r}$$ In terms of f(R,t), what is the differential position vector between these same two material points at time t?

 

[Jinjolee]

Let's try this a little differently. Suppose, in the initial configuration, you have two material points at R and R+dR connected by a differential position vector $$\mathbf{dX}=dR\mathbf{i_r}$$ In terms of f(R,t), what is the differential position vector between these same two material points at time t?

I this it should apply $$dx=x(X+dX)-x(X)$$
But I am not sure about the answer. Is it$$\frac{f(R+dR{\mathbf{i_r}},t)}{R+dR\mathbf{i_r}}(R\mathbf{i_r}+dR\mathbf{i_r})-\frac{f(R,t)}{R}R\mathbf{i_r}$$?

 

[Chestermiller]

I this it should apply $$dx=x(X+dX)-x(X)$$
But I am not sure about the answer. Is it$$\frac{f(R+dR{\mathbf{i_r}},t)}{R+dR\mathbf{i_r}}(R\mathbf{i_r}+dR\mathbf{i_r})-\frac{f(R,t)}{R}R\mathbf{i_r}$$?

No. You may be overthinking it. The quantity $\frac{\mathbf{X}}{R}$ is the unit vector in the radial spherical coordinate direction $\mathbf{i_r}$ (which is constant). So the differential position vector between the same two material points at time t is $$\mathbf{dx}=\mathbf{i_r}df=\frac{\partial f}{\partial R}(\mathbf{i_r}dR)$$What does that tell you about the r-r component of the deformation gradient tensor $\mathbf{F}$?
Next, let's consider a differential position vector between two material points that are initially oriented in the $\theta$ polar coordinate direction: $$\mathbf{dX}=(Rd{\theta})\mathbf{i_{\theta}}$$In terms of f(R,t), what is the differential position vector between this same pair of material points at time t?