Principal of Minimum Complementary Energy

[TheFerruccio]

Homework Statement

I am reading through this link:

http://en.wikiversity.org/wiki/Introduction_to_Elasticity/Principle_of_minimum_complementary_energy

I need to show that a certain thing follows from a previous thing. I will elaborate in the next section.

Homework Equations

I am stuck, specifically, at this point:

"We can also show that

Therefore,

The Attempt at a Solution

I know that I can represent $\int \mathbf{t '}\cdot\mathbf{u}dA$ as $\int \mathbf{\sigma '\cdot n}\cdot\mathbf{u} dA$ and then use divergence theorem to arrive at $\int\mathbf{\nabla\cdot\sigma '}\cdot\mathbf{u}dV$

But, I do not see where to go from here to arrive at where I need to be, the last statement at the end of line (2). Does anyone have any suggestions?

I've been mulling over this for at least 4 hours and I have gotten nowhere with it. The method shown here is one of my many attempts to get to a solution.

 

[KataKoniK]

Hi there,

Based on the information you provided, it seems like you are trying to show that $\int \mathbf{\nabla\cdot\sigma '}\cdot\mathbf{u}dV$ is equal to $\int \mathbf{t '}\cdot\mathbf{u}dA$. One possible approach to this problem is to use the divergence theorem again, but this time in reverse. The divergence theorem states that for a vector field $\mathbf{F}$ and a region $V$ bounded by a surface $S$,

$$\int_V \nabla \cdot \mathbf{F} dV = \int_S \mathbf{F} \cdot \mathbf{n} dS$$

where $\mathbf{n}$ is the outward unit normal vector to the surface $S$.

In your case, you can set $\mathbf{F} = \mathbf{\sigma '}\cdot\mathbf{u}$ and $V$ to be the entire volume bounded by your surface of interest. Then, using the divergence theorem, you can rewrite the integral as:

$$\int_V \nabla \cdot (\mathbf{\sigma '}\cdot\mathbf{u}) dV = \int_S (\mathbf{\sigma '}\cdot\mathbf{u}) \cdot \mathbf{n} dS$$

Now, since $\mathbf{u}$ and $\mathbf{n}$ are both unit vectors, we can write $\mathbf{u} \cdot \mathbf{n} = cos(\theta)$, where $\theta$ is the angle between them. Also, note that $\mathbf{\sigma '}\cdot\mathbf{n} = \mathbf{t '}$, as you have already mentioned. Therefore, the integral becomes:

$$\int_V \nabla \cdot (\mathbf{\sigma '}\cdot\mathbf{u}) dV = \int_S \mathbf{t '}\cdot cos(\theta) dS$$

Now, recall that $dS = dA cos(\theta)$, where $dA$ is the differential area of the surface. Substituting this into our equation, we get:

$$\int_V \nabla \cdot (\mathbf{\sigma '}\cdot\