Stress Transformation and Principal Stresses in Three Dimensions
Transformation of stress and strain in three dimensions and the principal axes and stresses and strains will be presented here without proof. The full development may be found in books on the theory of elasticity.
We show two sets of axes – one rotated with respect to the other – in Figure The components of the vector F are shown with respect to both the axes xyz and xyz.
To simplify the notation we define a set of direction cosines in the following table.
where, for example, l1 =cos xx, where xx signifies the angle between the x axis and the x axis. Similarly, m3 = cos yz, where yz signifies the angle between the y axis and the z axis. We define a transformation matrix
You can verify that the components of the force F are given by the matrix equation
Thus given the force components with respect to a particular orientation of the axes we can find the components with respect to any rotated axes. For the transformation of stress we must recall the definition of stress must specify both the direction of the force and the orientation of the surface to which it is referred. It can be shown that the transformation of stress is given by
where [σ] is written in the form
It can also be shown that there are three orthogonal planes on each of which the stress at a point consists of a normal stress and no shearing stress. These are called the principal stresses and the rotated coordinate axes that are normal to each of these planes are called principal axes.
Let us look at the stresses at a point on a small tetrahedron as shown in Figure The face of the tetrahedon is oriented so that the normal stress σp is the only stress acting on that face, thus it is a principal stress. It follows from Equation that
where [Tp] is the appropriate set of direction cosines for the pincipal axes and [σp] are the principal stresses.
The orientation of the principal axes for the normal stress in Figure is unknown. Let us identify them as
It can be shown that
This can be rewritten as a classical eigenvalue problem
One possible solutions is
however, equations of this form have solutions other than zero for discrete values of s determined by
When expanded it yields the following cubic equation
The three roots of Equation are the three principal stresses. The direction of the principal stresses are found by substituting one of the principle stresses into Equation
Since this is a set of homogenous equations we can only solve for two of the direction cosines in terms of the third. The third one is found from the property of direction cosines that
The software products, such as Mathematica, Maple, Matlab, and Mathcad, among others have eigenvalue solvers for exactly this form of the equations.