2. Deriving formulas for normal and shear components on such planes

Let us draw a cuboid element at the point of interest with its face normals along the coordinate system. As discussed earlier, the third coordinate axis (e3) is along the third principal direction. We will call e1, e2, and e3 axes as x, y, and z axes, respectively. On the e1 plane, we have normal component of traction denoted as σxx and shear component of traction denoted as τxy pointing towards y axis. The third component (τzx) is absent as per the stress representation.

This chunk introduces the concept of a cuboidal element used to analyze stress states. We are looking at three axes of a coordinate system: x (e1), y (e2), and z (e3). In this specific setup, the z-axis is aligned with the principal stress direction, meaning it experiences no shear stress (τzx). The normal stress (σxx) acts directly on the e1 plane, while the shear stress (τxy) acts parallel to the e1 axis towards the y-axis. Understanding these components is fundamental as they set the stage for further analyzing stress on different planes.

Think of a cuboidal box filled with water sitting on a table. The box represents our cuboidal element. The weight of the water pushing downwards on the bottom of the box is analogous to the normal stress (σxx). If we were to push one side of the box towards the other, the force we apply parallel to that side represents shear stress (τxy). In our analysis, we want to understand how these forces affect the box in different orientations.

There is another simpler way to draw such a state of stress for whom there are no shear components in the third direction: we can just draw a square instead of a cube where the sides of the square denote the faces of the cuboid. The right edge of the square represents the (+e1) face. On the edges of the square, we have both the shear and the normal component of traction.

This chunk provides an alternative visualization for the cuboidal element. Instead of a cube, we consider a 2D square to simplify the analysis, especially when the third direction (e3) has no shear components. Each side of the square corresponds to a face of the cuboid. The right side represents the face where the normal stress σxx is acting, and we can depict shear stress τ as acting on the square’s edges.

Imagine a flat pizza. Each slice represents a side of our square. The toppings and sauce act like normal stress on the surface of the pizza. If you were to push down on the pizza, the force you apply would represent the shear stress. By simplifying our view to just the pizza (the square), we make calculations easier while still understanding how the stress is distributed.

Now, our goal is to calculate the normal and shear components of traction on planes whose normals are perpendicular to e3. Let n be the normal to this plane making an angle α with e1. The column representation of n will be specified. Using this, σ will be given by the derived formula and similarly for τ.

In this segment, we focus on deriving the formulas for shear and normal stresses when rotated planes make angles with the principal directions. We assume the normal vector n makes an angle α with the e1 axis and will need to represent its orientation geometrically. By applying trigonometric functions and manipulating equations, we find expressions for σ and τ, which are the normal and shear components acting on the plane.

Picture a street sign standing vertically. If a wind blows against it at an angle (just as n makes an angle with e1), the force acting on the sign can be decomposed into two parts: one straight pushing against the post (normal stress) and another causing it to sway or twist (shear stress). By knowing the angle of the wind, we can mathematically break down these components similar to how we analyze the stresses here.

If we look at equations (10) and (11), we notice that σ and τ both have cos(2α) and sin(2α) terms. Let us define a scalar R and relate it to a right-angled triangle formed with σ and τ.

This chunk introduces the concept of defining a scalar R based on the geometric relationship between normal and shear stress components visualized as the sides of a right-angled triangle. By identifying these relationships, we simplify the process of calculating stresses across various orientations. The trigonometric identities will help us connect these relationships to our overall formulas for stress.

Think of a triangle formed by a ramp leading up to a platform. The height of the platform could represent total stress while the distance along the base could model shear stress. By establishing this triangle using R as the hypotenuse, we can navigate through our stress calculations just like measuring the incline of a ramp to ensure we can safely bring items up to the platform.

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