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At any given point \(P\) of a continuous medium, the surface forces are given by the stress tensor, which is a real symmetric matrix \(\mathbf{\Sigma}\). It is a well-known theorem of linear algebra (see the appendix) that any such matrix can be brought into diagonal form by a suitable rotation of the Cartesian coordinate axes. Use this to prove that at any point \(P\) there are three orthogonal directions (the principal stress axes at \(P\) ) with the property that the surface force on any surface normal to one of these directions is exastly normal to the surface.

Short Answer

Expert verified
Three orthogonal principal stress axes exist at point \( P \).

Step by step solution

01

Understanding the Stress Tensor

The stress tensor \( \mathbf{\Sigma} \) is a symmetric matrix. This means that for a matrix of size \( n \times n \), all elements across the main diagonal are symmetric: \( \mathbf{\Sigma}_{ij} = \mathbf{\Sigma}_{ji} \). In our context, this is a \( 3 \times 3 \) matrix since we are in three-dimensional space.
02

Principal Axes Theorem Application

According to the principal axes theorem, any real symmetric matrix can be diagonalized by an orthogonal transformation, i.e., \( \mathbf{\Lambda} = \mathbf{Q}^T \mathbf{\Sigma} \mathbf{Q} \), where \( \mathbf{Q} \) is an orthogonal matrix and \( \mathbf{\Lambda} \) is a diagonal matrix. The columns of \( \mathbf{Q} \) are the eigenvectors of \( \mathbf{\Sigma} \).
03

Finding Eigenvectors and Eigenvalues

The diagonal of \( \mathbf{\Lambda} \) contains the eigenvalues of \( \mathbf{\Sigma} \), and the corresponding eigenvectors given by \( \mathbf{Q} \) represent the principal directions. These directions are orthogonal and correspond to principal stress axes where the stress vector's direction coincides with the direction of its direct surface normal.
04

Relating Eigenvectors to Principal Stress Directions

For any surface at point \( P \) with a normal direction parallel to one of the eigenvectors, the stress tensor simplifies due to its diagonal form. As such, the stress vector on this surface becomes parallel to the surface normal, ensuring no shear components act on it.
05

Conclusion

The existence of these three orthogonal directions (eigenvectors of \( \mathbf{\Sigma} \)) as principal stress axes means that at point \( P \), the force on any surface normal to these directions is indeed normal to the surface due to the symmetry and transformation properties of the stress tensor.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Principal Axes Theorem
The principal axes theorem is an essential principle in linear algebra, particularly for analyzing symmetric matrices such as the stress tensor. This theorem states that a real symmetric matrix can be converted into a diagonal matrix through an orthogonal transformation. In simpler terms, it implies that we can "rotate" our coordinate system to align with certain special directions, known as the principal axes. These directions align with the eigenvectors of the matrix. When applying this theorem to the stress tensor, it allows for the identification of the principal stress axes at a given point. These axes are critical because they align with the directions where shear stress is zero, meaning the stress vector acts only in the normal direction. This property is indispensable for understanding the mechanical behavior of materials under stress.
Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors are fundamental concepts in the study of linear transformations. For a given matrix, an eigenvector is a non-zero vector that changes by only a scalar factor when that matrix is applied to it. The corresponding scalar is known as the eigenvalue. For the stress tensor, which is a symmetric matrix, the task is to find its eigenvectors and eigenvalues.
  • Eigenvectors: Provide directions known as principal stress directions.
  • Eigenvalues: Represent the magnitude of the principal stresses.
The eigenvalues appear on the diagonal after diagonalization, and together with their eigenvectors, they give a complete spectral decomposition of the matrix. These eigenvectors indicate directions in which the stress does not deviate, while eigenvalues indicate how much stress is acting along these eigenvectors. This means the surface force on any plane normal to these directions has no shear component but remains purely normal to the surface.
Symmetric Matrix
A symmetric matrix is a matrix that is equal to its transpose. This characteristic means that the values are mirrored along the main diagonal. For the stress tensor, being a symmetric matrix is key.
  • Properties of a Symmetric Matrix:
  • All eigenvalues are real numbers.
  • Eigenvectors corresponding to different eigenvalues are orthogonal.
These are crucial because they lead us directly to the principal stress axes' definition, where the properties ensure that the eigenvectors form an orthogonal basis. In physical terms, it means the directions of principal stresses are at right angles to each other. This orthogonality guarantees the separation of normal and shear components of the stress, making it possible to align our analysis seamlessly with the material's mechanical behaviors at any given point.
Diagonalization
Diagonalization is a process by which a matrix is expressed in its diagonal form. For a symmetric matrix like the stress tensor, this involves rotating the coordinate axes so that the stress tensor behaves simply along these new directions. The process can be broken down into these steps:
  • Determine eigenvalues and eigenvectors.
  • Form an orthogonal matrix from eigenvectors.
  • Apply the transformation to achieve diagonal form.
The result of diagonalization is a matrix where all non-diagonal elements are zero, and diagonal elements are the eigenvalues. This means the stress vector on any plane normal to an eigenvector is parallel to the vector, confirming no shear stress is present. This transformation simplifies analysis, making the behavior of stress straightforward to manage, particularly useful in stress analysis for material science and mechanical engineering.

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Most popular questions from this chapter

At a certain point \(P\) (which you can choose to be your origin) in a continuous solid, the strain tensor is \(\mathbf{E}\). Assume for simplicity that whatever displacements have occurred left \(P\) fixed and the neighborhood of \(P\) unrotated. (a) Show that the \(x\) axis near \(P\) is stretched by a factor of \(\left(1+\epsilon_{11}\right)\) (b) Hence show that any small volume around \(P\) has changed by \(d V / V=\operatorname{tr}\) E. This shows that any two strains that have the same trace dilate volumes by the same amount. In the decomposition \(\mathbf{E}=e \mathbf{1}+\mathbf{E}^{\prime}\) \((16.88),\) the spherical part \(e 1\) changes volumes by the same amount as \(\mathbf{E}\) itself, while the deviatoric part \(\mathbf{E}^{\prime}\) doesn't change volumes at all.

Let \(\mathbf{n}_{1}\) and \(\mathbf{n}_{2}\) be any two unit vectors and \(P\) a point in a continuous medium. \(\mathbf{F}\left(\mathbf{n}_{1} d A\right)\) is the surface force on a small area \(d A\) at \(P\) with unit outward normal \(\mathbf{n}_{1}\), so \(\mathbf{n}_{2} \cdot \mathbf{F}\left(\mathbf{n}_{1} d A\right)\) is the component of that force in the direction of \(\mathbf{n}_{2}\). Prove Cauchy's reciprocal theorem that \(\mathbf{n}_{2} \cdot \mathbf{F}\left(\mathbf{n}_{1} d A\right)=\mathbf{n}_{1} \cdot \mathbf{F}\left(\mathbf{n}_{2} d A\right)\)

[Computer] A taut string of length \(L=1\) is released from rest at \(t=0\), with initial position \(u(x, 0)=\left\\{\begin{array}{ll}2 x & {\left[0 \leq x \leq \frac{1}{2}\right]} \\ 2(1-x) & {\left[\frac{1}{2} \leq x \leq 1\right]}\end{array}\right.\) Take the wave speed on the string to be \(c=1\). (a) Sketch this initial shape and find the coefficients \(B_{n}\) in its Fourier sine series ( 16.31 ). (b) Make plots of the sum of the first several terms for several closely spaced times between \(t=0\) and \(\tau,\) the period. Animate your plots and describe the motion.

It is found that the stress tensor at any point \((x, y, z)\) in a certain continuous medium has the form (with an unspecified, convenient choice of units) \(\mathbf{\Sigma}=\left[\begin{array}{ccc}x z & z^{2} & 0 \\ z^{2} & 0 & -y \\ 0 & -y & 0\end{array}\right]\) Find the surface force on a small area \(d A\) of the surface \(x^{2}+y^{2}+2 z^{2}=4\) at the point (1,1,1)

An important tool in the development of the strain tensor was the decomposition (16.79) of a matrix \(\mathbf{M}\) into its antisymmetric and symmetric parts. Prove that this decomposition is unique. [Hint: Show that if \(\mathbf{M}=\mathbf{M}_{\mathrm{A}}+\mathbf{M}_{\mathrm{S}}\) where \(\mathbf{M}_{\mathrm{A}}\) and \(\mathbf{M}_{\mathrm{S}}\) are respectively antisymmetric and symmetric, then \(\left.\mathbf{M}_{\mathrm{A}}=\frac{1}{2}(\mathbf{M}-\tilde{\mathbf{M}}) \text { and } \mathbf{M}_{\mathrm{S}}=\frac{1}{2}(\mathbf{M}+\tilde{\mathbf{M}}) .\right]\)

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