Chapter 11: Problem 3
A displacement field is given by $$ \begin{aligned} u_{x} &=\alpha(x+2 y)+\beta x^{2} \\ u_{y} &=\alpha(y+2 z)+\beta y^{2} \\ u_{z} &=\alpha(z+2 x)+\beta z^{2} \end{aligned} $$ where \(\alpha\) and \(\beta\) are 'small'. (a) Calculate the divergence and curl. (b) Calculate Cauchy's strain tensor. (c) Calculate the stress tensor in a linear elastic medium with Lamé coefficients \(\lambda\) and \(\mu\) for the special case \(\beta=0\)
Short Answer
Step by step solution
Calculate the Divergence
Calculate the Curl
Calculate Cauchy's Strain Tensor
Calculate the Stress Tensor
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Divergence
For the displacement field given by the set of equations:
- \(u_{x} = \alpha(x+2 y)+\beta x^{2}\)
- \(u_{y} = \alpha(y+2 z)+\beta y^{2}\)
- \(u_{z} = \alpha(z+2 x)+\beta z^{2}\)
Curl
To find the curl of the displacement field, we use:\[abla \times \mathbf{u} = \left( \frac{\partial u_z}{\partial y} - \frac{\partial u_y}{\partial z}, \frac{\partial u_x}{\partial z} - \frac{\partial u_z}{\partial x}, \frac{\partial u_y}{\partial x} - \frac{\partial u_x}{\partial y} \right)\]Calculating each term for our displacement gives:
- \(\frac{\partial u_z}{\partial y} - \frac{\partial u_y}{\partial z} = 0\)
- \(\frac{\partial u_x}{\partial z} - \frac{\partial u_z}{\partial x} = 0\)
- \(\frac{\partial u_y}{\partial x} - \frac{\partial u_x}{\partial y} = 0\)
Cauchy's Strain Tensor
The strain tensor \(\varepsilon\) is formulated as:\[\varepsilon_{ij} = \frac{1}{2}\left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right)\]For the given displacement field, this tensor reveals:
- Normal strains: \(\varepsilon_{xx} = \alpha + 2\beta x\), \(\varepsilon_{yy} = \alpha + 2\beta y\), \(\varepsilon_{zz} = \alpha + 2\beta z\)
- Shear strains: \(\varepsilon_{xy} = \varepsilon_{yx} = \varepsilon_{yz} = \varepsilon_{zy} = \varepsilon_{zx} = \varepsilon_{xz} = \alpha\)
Stress Tensor
For a linear elastic medium, the stress tensor \(\sigma\) interlinks directly with the strain tensor \(\varepsilon\) through the expressions:\[\sigma_{ij} = \lambda \delta_{ij} \varepsilon_{kk} + 2\mu \varepsilon_{ij}\]where \(\lambda\) and \(\mu\) are Lamé's coefficients. Assuming \(\beta = 0\), the strain reduces to constant terms where:
- \(\varepsilon_{xx} = \varepsilon_{yy} = \varepsilon_{zz} = \alpha\)
- \(\varepsilon_{xy} = \varepsilon_{yz} = \varepsilon_{zx} = \alpha\)
- Normal stresses: \(\sigma_{xx} = \sigma_{yy} = \sigma_{zz} = (3\lambda + 2\mu)\alpha\)
- Shear stresses: \(\sigma_{xy} = \sigma_{yz} = \sigma_{zx} = 2\mu\alpha\)
Linear Elasticity
Key assumptions in linear elasticity include:
- Material linearly relates stress and strain, meaning that stress is directly proportional to strain.
- Deformations are small so that linear approximations remain valid.
- It applies in the elastic region of the material, where no permanent deformation occurs.