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Additional resources for Continuum and solid mechanics : concepts and applications
4): where A is a rotation matrix with components aij. 4 Transformation of the stress tensor Expanding the matrix operation, and simplifying some terms by taking advantage of the symmetry of the stress tensor, gives The Mohr circle for stress is a graphical representation of this transformation of stresses. Normal and shear stresses The magnitude of the normal stress component σn of any stress vector T(n) acting on an arbitrary plane with normal vector n at a given point, in terms of the components σij of the stress tensor σ, is the dot product of the stress vector and the normal vector: The magnitude of the shear stress component τn, acting in the plane spanned by the two vectors T(n) and n, can then be found using the Pythagorean theorem: where Equilibrium equations and symmetry of the stress tensor Figure 4.
The 11-component of stress on that interface is the sum of all pairwise forces between atoms on the two sides.... Stress modeling (Cauchy) In general, stress is not uniformly distributed over the cross-section of a material body, and consequently the stress at a point in a given region is different from the average stress over the entire area. 1). According to Cauchy, the stress at any point in an object, assumed to behave as a continuum, is completely defined by the nine components of a second-order tensor of type (0,2) known as the Cauchy stress tensor, : The Cauchy stress tensor obeys the tensor transformation law under a change in the system of coordinates.
Not imaginary due to the symmetry of the stress tensor. The three roots , , and are the eigenvalues or principal stresses, and they are the roots of the Cayley–Hamilton theorem. The principal stresses are unique for a given stress tensor. Therefore, from the characteristic equation it is seen that the coefficients , and , called the first, second, and third stress invariants, respectively, have always the same value regardless of the orientation of the coordinate system chosen. For each eigenvalue, there is a non-trivial solution for in the equation .