Question

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)

Short answer

The surface force at (1, 1, 1) is \( \left[ \frac{2}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}} \right] \).

Step-by-step solution

Calculate Surface Normal Vector

To determine the surface normal vector, we start by finding the gradient of the surface equation \( x^2 + y^2 + 2z^2 = 4 \). The gradient, \( abla F \), at any point \((x,y,z)\) is given by the vector \([2x, 2y, 4z]\). Evaluating this at the point \((1, 1, 1)\), we have:\[abla F = [2 \cdot 1, 2 \cdot 1, 4 \cdot 1] = [2, 2, 4]\] This vector \([2, 2, 4]\) is the gradient, and consequently, the normal vector to the surface at the point \((1, 1, 1)\).

Calculate Surface Normal Unit Vector

The surface force requires a unit normal vector, which we can obtain by normalizing the normal vector found in Step 1. To normalize a vector \([a, b, c]\), divide each component by the vector's magnitude. The magnitude of the vector \([2, 2, 4]\) is:\[ \|abla F\| = \sqrt{2^2 + 2^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}\]So, the unit normal vector \(\mathbf{n}\) is:\[ \mathbf{n} = \left[ \frac{2}{2\sqrt{6}}, \frac{2}{2\sqrt{6}}, \frac{4}{2\sqrt{6}} \right] = \left[ \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}} \right]\]

Evaluate Stress Tensor at Point (1, 1, 1)

Substitute \(x = 1\), \(y = 1\), and \(z = 1\) into the given stress tensor equation \(\mathbf{\Sigma}\):\[\mathbf{\Sigma} = \begin{bmatrix} xz & z^2 & 0 \ z^2 & 0 & -y \ 0 & -y & 0 \end{bmatrix} = \begin{bmatrix} 1 \times 1 & 1^2 & 0 \ 1^2 & 0 & -1 \ 0 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 0 \ 1 & 0 & -1 \ 0 & -1 & 0 \end{bmatrix}\]Thus, the stress tensor at \((1, 1, 1)\) is:\[ \mathbf{\Sigma} = \begin{bmatrix} 1 & 1 & 0 \ 1 & 0 & -1 \ 0 & -1 & 0 \end{bmatrix}\]

Calculate Surface Force Using Stress Tensor

The surface force \( \mathbf{f} \) can be found by multiplying the stress tensor \( \mathbf{\Sigma} \) by the unit normal vector \( \mathbf{n} \). That is, \( \mathbf{f} = \mathbf{\Sigma} \cdot \mathbf{n} \).Given \( \mathbf{n} = \left[ \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}} \right] \) and \( \mathbf{\Sigma} \) is:\[\begin{bmatrix} 1 & 1 & 0 \ 1 & 0 & -1 \ 0 & -1 & 0 \end{bmatrix}\cdot \begin{bmatrix} \frac{1}{\sqrt{6}} \ \frac{1}{\sqrt{6}} \ \frac{2}{\sqrt{6}} \end{bmatrix}= \begin{bmatrix} 1 \times \frac{1}{\sqrt{6}} + 1 \times \frac{1}{\sqrt{6}} + 0 \times \frac{2}{\sqrt{6}} \ 1 \times \frac{1}{\sqrt{6}} + 0 \times \frac{1}{\sqrt{6}} + (-1) \times \frac{2}{\sqrt{6}} \ 0 \times \frac{1}{\sqrt{6}} + (-1) \times \frac{1}{\sqrt{6}} + 0 \times \frac{2}{\sqrt{6}} \end{bmatrix}= \begin{bmatrix} \frac{2}{\sqrt{6}} \ \frac{1}{\sqrt{6}} - \frac{2}{\sqrt{6}} \ -\frac{1}{\sqrt{6}} \end{bmatrix}= \begin{bmatrix} \frac{2}{\sqrt{6}} \ -\frac{1}{\sqrt{6}} \ -\frac{1}{\sqrt{6}} \end{bmatrix}\]Thus, the surface force \( \mathbf{f} \) is \( \left[ \frac{2}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}} \right] \).

Summarize the Result

The surface force at point \((1, 1, 1)\) on the given surface is:\[ \mathbf{f} = \left[ \frac{2}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}} \right] \]

Key concepts

Surface Force

In the realm of physics and engineering, understanding the concept of surface force is pivotal when evaluating how different materials and structures interact with forces applied to them. A surface force is typically defined as a force that is spread over a surface area, influencing the points on that surface. It examines how forces, whether from contact with another object or tension within a structure, apply stress across a defined area.
The surface force will usually depend on two main factors:

• The stress tensor, which gives the distribution of internal forces within a body.
• The normal vector, which indicates the orientation of the surface at a given point.

The surface force can then be calculated by taking the dot product of the stress tensor and the unit normal vector at a specific point on the surface.
This results in a force vector that tells us not only the magnitude of the force but also its direction in relation to the surface. For example, in the exercise, surface force calculations reveal the stress and tension experienced by the surface at a point, guiding engineers and scientists in understanding material behavior under stress conditions.

Normal Vector

A normal vector is a vector that is perpendicular to a given surface at a specific point. It is crucial in physics and engineering contexts because it helps to define how surfaces interact with external forces and provides insight into the surface's orientation.
The process of finding a normal vector involves calculating the gradient of the surface's mathematical equation. The gradient is a vector operation that points in the direction of the steepest ascent and its magnitude gives the rate of increase of the function's value.

• In the original exercise, the surface equation, given by \(x^2 + y^2 + 2z^2 = 4\), needs a gradient calculation to find the normal vector.
• The gradient vector \([2x, 2y, 4z]\) evaluated at the point \((1,1,1)\) yields the normal vector \([2, 2, 4]\).

However, for simplicity and to make calculations easier in further steps, the normal vector is often normalized to a unit vector, providing both direction and a vector of magnitude one. This unit vector is important in calculations involving force interactions on the surface, such as in stress tensor evaluations.

Stress Tensor Evaluation

A pivotal concept in understanding material behavior is the stress tensor evaluation. This concept helps us analyze how forces are distributed internally in a material, especially when subjected to external loads.
In mathematics and physics, a stress tensor is a 3x3 matrix that describes the stress applied to a point within a medium across three dimensions. Each element of the matrix represents a component of force experienced by an area element in the material.
The process involves these steps:

• Identify the components of the stress tensor for the specific point of interest. In the given exercise, the stress tensor \(\Sigma\) represents these components at point \((1,1,1)\).
• To evaluate how these stresses affect the surface, the tensor is multiplied by the unit normal vector derived from the surface equation. This multiplication gives the actual surface force, which can be interpreted to understand the material's response at that point.

Stress tensors thus allow engineers and scientists to design structures and materials that can withstand specific stress levels by providing detailed descriptions of stress distributions. This information is crucial in fields such as structural engineering, materials science, and geophysics, where understanding internal stress is essential for safety and efficiency.