https://file.aiquickdraw.com/mathbot/file/172695621617966eecj6g.jpg

The question revolves around the concept of deformation in continuum mechanics, specifically addressing the behavior of tensors under the deformation map \( \mathbf{x} = \phi(\mathbf{z}) = \mathbf{Q} \mathbf{z} \), where \( \mathbf{Q} \) is a constant orthogonal rotation tensor. Let’s examine each statement:

1. **Statement 1**: "Since \( \mathbf{Q} \) is not symmetric, in general, thus \( \mathbf{C} \) and \( \mathbf{E} \) will also not be symmetric."
   - **Analysis**: The tensor \( \mathbf{Q} \) represents a rigid body rotation, which is generally orthogonal but not necessarily symmetric. However, the right Cauchy-Green tensor \( \mathbf{C} = \mathbf{F}^\top \mathbf{F} \) and the Green-Lagrange strain tensor \( \mathbf{E} \) are derived from \( \mathbf{F} \), the deformation gradient tensor. Both \( \mathbf{C} \) and \( \mathbf{E} \) are symmetric tensors by definition. Therefore, this statement is **FALSE**.

2. **Statement 2**: "The Green-Lagrange strain tensor \( \mathbf{E} \) is the zero tensor \( \mathbf{0} \) everywhere."
   - **Analysis**: The Green-Lagrange strain tensor \( \mathbf{E} \) measures the strain due to deformation, and for a rigid body rotation (\( \mathbf{Q} \)), there is no actual strain (since distances between points do not change). Hence, \( \mathbf{E} \) would indeed be the zero tensor. This statement is **TRUE**.

3. **Statement 3**: "The Right Cauchy-Green deformation \( \mathbf{C} \) is the identity \( \mathbf{I} \) everywhere."
   - **Analysis**: The Right Cauchy-Green tensor \( \mathbf{C} = \mathbf{F}^\top \mathbf{F} \) will be the identity tensor if the deformation is purely rigid body rotation. Since \( \mathbf{Q} \) represents an orthogonal transformation (rotation), \( \mathbf{C} = \mathbf{I} \). This statement is **TRUE**.

4. **Statement 4**: "The deformation gradient tensor field \( \mathbf{F} \) is equal to \( \mathbf{Q} \)."
   - **Analysis**: The deformation gradient tensor \( \mathbf{F} \) describes the local deformation at a point, and for the given deformation map \( \mathbf{x} = \mathbf{Qz} \), \( \mathbf{F} = \mathbf{Q} \). This statement is **TRUE**.

### Conclusion:
The **FALSE** statement is the first one: "Since \( \mathbf{Q} \) is not symmetric, in general, thus \( \mathbf{C} \) and \( \mathbf{E} \) will also not be symmetric." \( \mathbf{C} \) and \( \mathbf{E} \) are symmetric, regardless of \( \mathbf{Q} \) being non-symmetric.

---

Do you have any questions or would you like more details on any of the concepts?

Here are 5 related questions:
1. What is the physical interpretation of the Green-Lagrange strain tensor?
2. Why is the right Cauchy-Green deformation tensor symmetric?
3. How does an orthogonal tensor represent rigid body rotation in mechanics?
4. What is the relationship between the deformation gradient tensor and strain?
5. Why does rigid body motion result in zero strain?

**Tip**: The Green-Lagrange strain tensor is useful in measuring finite strains, especially in non-linear deformation scenarios. Understanding it can be key for studying non-linear elasticity!

For a body that undergoes the deformation map: x = φ(z) = Qz, where Q is a constant orthogonal rotation tensor, which of the following is FALSE?

Analyze the deformation map and tensor symmetry in rigid body rotation using concepts like the Green-Lagrange strain tensor and Cauchy-Green deformation. Suitable for advanced continuum mechanics studies.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation