2.3. Introducing Graphical Parameters

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 as (12)

In this part, we are identifying key components from equations (10) and (11) that describe stress (σ) and shear (τ) in a material. Both of these parameters are influenced by cosine and sine functions related to the angle α at which we are examining the planes. The introduction of the scalar R simplifies our calculations by allowing us to convert the trigonometric expressions into a more manageable form, depicting the relationship graphically.

Think of R as the length of a string that you attach to a point (representing an angle) on a circle. By knowing R, you can determine the height (or vertical distance) and width (horizontal distance) to that point from the center of the circle, which helps visualize how stress and shear behave on different planes.

We can now think of a right angled triangle with hypotenuse R and the two perpendicular arms as σ and τ as shown in Figure 3. Let us denote the angle between the hypotenuse and the base as 2φ.

Here, we transition to a geometric representation of the stress and shear components. By imagining a right triangle where the sides represent σ and τ, we can better visualize the relationships between these parameters. The angle 2φ is essential as it relates how rotations in our system affect these stress components.

Imagine you’re at a playground where a swing can pivot. The swing represents R, the angle at which it swings up relates to 2φ, and the height of the swing (vertical arm) represents σ while the distance it swings horizontally (the horizontal arm) represents τ. Understanding the geometry of the swing helps predict how high or far it will go based on the angle it’s at.

Using equation (13) in equations (10) and (11), we get (14) Upon further using following trigonometric identities: cos(a−b)=cos(a)cos(b)+sin(a)sin(b), (15) sin(a−b)=sin(a)cos(b)−cos(a)sin(b), (16) we get σ + σ xx yy σ = + Rcos(2φ− 2α) 2 τ = Rsin(2φ− 2α)

This segment dives deeper into the mathematical derivation involving trigonometric identities. By applying equations (14), we can express stress and shear in terms of R and angles. This transformation captures the geometric essence of stress behavior and allows for a clearer understanding of their interdependence on the rotation of the plane.

Consider cooking pancakes. The batter spreads out, forming a circular shape (like our triangle). Depending on how you tilt or turn the pan (rotating around an angle), the thickness and spread of the batter will change, similar to how stress (σ) and shear (τ) vary with angles in our equations.

These are the formulae obtained for getting σ and τ on a plane making an angle α with e1 axis. For a given stress matrix, we can find out R and φ using equations (12) and (13) respectively. Then, using equation (17), we can find σ and τ on the plane which is obtained by rotating e1 by angle α about e3 axis.

Finally, we summarize the results: For any given stress matrix, we can determine the scalar R and angle φ. This information allows us to apply the derived formulas (17) to find specific stress and shear values for any tilted plane. It's a systematic approach that serves as the foundation for analyzing complex stress states in materials.

Think of navigating a city using Google Maps. As you rotate or change the angle of your view (like rotating the axes), you adjust the center (R) and direction (φ) to find the best route (σ and τ) to your destination. This process illustrates how understanding the relationship between stresses helps find optimal solutions in engineering and physics.

--