91Ó°ÊÓ

The shear modulus is a key concept when understanding how materials deform under shear stress. It is a measure of a material's rigidity or stiffness. Think of it as the "stiffness constant" for shear deformation. The larger the shear modulus, the stiffer the material and the less it deforms when shear force is applied.
\[ G = \frac{F}{A} \cdot \frac{1}{\Delta x / L} \]
In this formula, \( G \) represents the shear modulus, \( F \) the force applied, \( A \) the area over which the force is applied, \( \Delta x \) the resulting deformation, and \( L \) the original length or thickness of the material. In our exercise, the cartilage disc has a shear modulus of \( 1.2 \times 10^7 \, \mathrm{N/m}^2 \), indicating that it is reasonably stiff—which helps maintain spinal stability under various loads.

Cartilage discs play a crucial role in our spine, acting as cushions between vertebrae. These discs handle various forces, including shearing forces. They are special because they allow slight movement between the bones, which is essential for flexibility and shock absorption.
These discs are like small, thick pancakes with a radius. In this exercise, the disc has a radius of \( 3.0 \times 10^{-2} \, \mathrm{m} \) and a thickness of \( 7.0 \times 10^{-3} \, \mathrm{m} \). These dimensions are important as they directly relate to the area and the shear deformation calculation when a force is applied. Understanding the physical properties of the cartilage disc is essential for calculating how much it deforms under stress.

To calculate how much the top surface of a disc moves under a shearing force, we use the displacement formula for shear. The equation is:
\[ \Delta x = \frac{F \cdot L}{A \cdot G} \]
Here, \( \Delta x \) stands for the displacement, and it’s what we’re solving for. Substituting the known values from the cartilage disc problem, such as force \( F = 11 \, \mathrm{N} \), thickness \( L = 7.0 \times 10^{-3} \, \mathrm{m} \), area \( A = 2.827 \times 10^{-3} \, \mathrm{m}^2 \), and shear modulus \( G = 1.2 \times 10^7 \, \mathrm{N/m}^2 \), we follow through the calculation steps to determine the displacement. This calculation helps in understanding how the structure mechanically behaves under force.

A shearing force is a type of force that causes shearing action—an effect where parts of a material slide past each other. In the context of our cartilage disc, this type of force is applied parallel to the surface of the disc, causing one side to move relative to the other.
The exercise gives us a shearing force of \( 11 \, \mathrm{N} \). This force's direction and magnitude affect how much deformation occurs. Different materials have different responses to shearing forces; thus, knowing the exact force and material properties allows us to predict and calculate the deformation accurately. Understanding shearing forces is essential because it explains how materials behave in real-world applications, especially in biological systems like the human spine.