Elasticity/Equilibrium example 1: Difference between revisions

Latest revision as of 02:38, 5 October 2021

Given:

Euler's second law for the conservation of angular momentum

(1) \int_{\partial B} e_{i j k} x_{j} n_{l} σ_{l k} d S + \int_{B} ρ e_{i j k} x_{j} b_{k} d V = \frac{d}{d t} (\int_{B} ρ e_{i j k} x_{j} v_{k} d V)

The divergence theorem

(2) \int_{\partial B} n_{i} σ_{i j} d S = \int_{B} \frac{\partial σ_{i j}}{\partial x_{i}} d V

The equilibrium equation (Cauchy's first law)

(3) \frac{\partial σ_{i j}}{\partial x_{i}} + ρ b_{j} = \frac{d}{d t} (ρ v_{j})

Show:

(4) σ_{i j} = σ_{j i}

Let us first look at the first term of equation~(1) and apply the divergence theorem (2). Thus,

\begin{matrix} \int_{\partial B} e_{i j k} x_{j} n_{l} σ_{l k} d S & = \int_{\partial B} n_{l} (e_{i j k} x_{j} σ_{l k}) d S \\ = \int_{B} \frac{\partial (e_{i j k} x_{j} σ_{l k})}{\partial x_{l}} d V \\ = \int_{B} (e_{i j k} \frac{\partial x_{j}}{\partial x_{l}} σ_{l k} + e_{i j k} x_{j} \frac{\partial σ_{l k}}{\partial x_{l}}) d V \\ = \int_{B} (e_{i j k} δ_{j l} σ_{l k} + e_{i j k} x_{j} \frac{\partial σ_{l k}}{\partial x_{l}}) d V \\ = \int_{B} (e_{i l k} σ_{l k} + e_{i j k} x_{j} \frac{\partial σ_{l k}}{\partial x_{l}}) d V \end{matrix}

Plugging this back into equation~(1) gives

\int_{B} (e_{i l k} σ_{l k} + e_{i j k} x_{j} \frac{\partial σ_{l k}}{\partial x_{l}}) d V + \int_{B} ρ e_{i j k} x_{j} b_{k} d V = \frac{d}{d t} (\int_{B} ρ e_{i j k} x_{j} v_{k} d V)

Therefore, bringing all terms to the left hand side,

(5) \int_{B} [e_{i l k} σ_{l k} + e_{i j k} x_{j} (\frac{\partial σ_{l k}}{\partial x_{l}} + ρ b_{k} - \frac{d}{d t} (ρ v_{k}))] d V = 0

Using the equilibrium equations~(3), equation~(5) reduces to

(6) \int_{B} e_{i l k} σ_{l k} d V = 0

Since this holds for any $B$ , we have

(7) e_{i l k} σ_{l k} = 0

If you work this expression out, you will see that $σ_{i j} = σ_{j i}$ . Hence, the stress tensor is symmetric.