No, ##P_2=\sigma_{12}n_1## is the second component of a vector. That vector is the result of multiplying your original normal vector by a matrix, so can be in an arbitrary direction. Its second component is the component in the direction of your second basis vector.Seyn said:
Do you mean the point is,Ibix said:
Thank you for your kind explanation!Ibix said:
Thanks for the reply. Do you mean the term ##n_i##s are the component of arbitrary surface vector##\vec{n}##, and the author chose an unit vector in the 1 direction as the surface vector for the first example which is ##(1,0,0)##? As a result, because ##n_1=1##, ##n_2=0##, ##n_3=0##, the surface force vector is ##P_j=\sigma_{ij} n_i=(1 \cdot \sigma_{11}+0 \cdot \sigma_{21}+0 \cdot \sigma_{31},1 \cdot \sigma_{12}+0 \cdot \sigma_{22}+0 \cdot \sigma_{23},1 \cdot \sigma_{13}+0 \cdot \sigma_{23}+0 \cdot \sigma_{33})=(\sigma_{11},\sigma_{12},\sigma_{13})##??Chestermiller said:
yesSeyn said:
These are amazing. Thank you very much!Chestermiller said:
A stress tensor is a mathematical representation that describes the internal forces acting within a material. It quantifies how these forces are distributed over an area and is represented as a second-order tensor, which can encapsulate normal and shear stresses acting on different planes within the material.
Pressure is a specific case of the stress tensor where the stress is uniform in all directions. In a fluid at rest, the stress tensor is isotropic, meaning that the normal stress components are equal, and the shear stress components are zero. This uniform stress is often referred to as hydrostatic pressure.
A stress tensor in three dimensions has nine components: three normal stresses (σ_xx, σ_yy, σ_zz) acting on the respective axes and six shear stresses (σ_xy, σ_xz, σ_yx, σ_yz, σ_zx, σ_zy) acting on the planes defined by those axes. The normal stresses correspond to forces acting perpendicular to the surfaces, while the shear stresses correspond to forces acting parallel to the surfaces.
Under deformation, the components of the stress tensor can change depending on the material's properties and the type of deformation (e.g., tensile, compressive, or shear). The relationship between stress and strain is typically described by constitutive models, such as Hooke's law for elastic materials, which relates stress to strain through material-specific parameters like Young's modulus and Poisson's ratio.
Yes, the stress tensor can be used to predict failure in materials by applying failure criteria, such as the von Mises or Tresca criteria. These criteria relate the stress state represented by the stress tensor to the material's yield strength, allowing engineers to assess whether a material will yield or fracture under applied loads.