 Home Engineering  # Octahedral stress

An octahedral plane inclines equally against three mutually orthogonal axes. It is possible to form eight octahedral planes and an octahedron as a result, as shown in Fig. 2.19. FIGURE 2.19 Octahedral plane and octahedron along three mutually orthogonal axes.

Consider an octahedral plane ABC below. As an octahedral plane, the lengths OA, OB and OC are equal, as shown in Fig. 2.20.

Applying the relationship in Eq. (2.12) into Eq. (2.13) yields the following Simplification of the equation to express r in terms of the orthogonal axes gives us the following expression: x, у and z represent OC, OB and OA, respectively, and OA = OB = OC, therefore, the above equation can be written as: Simplifying the equation above results in the following: By applying the definition of r as per Eq. (2.49) and x = OA into Eq. (2.12), we can get: FIGURE 2.20 Stresses acting on an octahedral plane. Similarly, we can write the following equations for other direction cosine components: Stresses acting on an octahedron around point О will have the same intensity of normal and shear stresses at that point. These stresses are known as octahedral normal stress, aoc, and octahedral shear stress, zoc,, as indicated in Fig. 2.21.

By applying (2.39), P„ will be interchangeable with goc, if the plane is an octahedral plane, and this leads to the following: The following is obtained by applying relationships in Eqs. (2.50) and (2.51) into equation:  FIGURE 2.21 Stresses acting on an octahedron.

Simplifying the equation above leads to the following equation: By substituting Eq. (2.35) into the equation above, we can obtain Applying the relationships in Eqs. (2.50) and (2.51) into (2.40), and defining Pr as Gr.oci- which denotes resultant stress of octahedral normal and shear stresses, yields the follows: The following equation is obtained through simplification of the equation above: Using the Pythagoras theorem, we can write the following expression By substituting Eqs. (2.52) and (2.53) into equation above produces the following: Expanding the equation above yields the following equation Simplify the equation above and we get: Factorisation of the equation above produces the following Taking the square root on both sides of the equation leads to the following

# Plane stress

Consider the thickness, dz, of a body (say, along the z-axis) is much smaller than its length, dy and width, dx (along x and у axes). When force is exerted in a direction normal to the z-axis, stresses are developed only on the xy-plane. By inspecting the sectional planes normal to the z-axis, it can be observed that the thickness of body is too small for the stress to vary along the z-direction. Since thickness does not play any role in stress distribution within the body, this scenario can be simplified into plane stress scenario, as demonstrated in Fig. 2.22. FIGURE 2.22 Plane stress.

Under the plane stress scenario, the component of stresses as per Eq. (2.3) can be simplified as follows: Related topics