Multiaxial Deformations and Stress Analyses

Abstract

When a structure is subjected to uniaxial tension, the transverse dimensions decrease (the structure undergoes lateral contractions) while simultaneously elongating in the direction of the applied load. This was illustrated in the previous chapter through the phenomenon called necking. For stresses within the proportionality limit, the results of uniaxial tension and compression experiments suggest that the ratio of deformations occurring in the axial and lateral directions is constant. For a given material, this constant is called the Poisson’s ratio and is commonly denoted by the symbol ν (nu):

$$ \nu =-\frac{\mathrm{Lateral}\ \mathrm{strain}}{\mathrm{Axial}\ \mathrm{strain}}. $$

Consider the rectangular bar with dimensions a, b, and c shown in Fig. 14.1. To be able to differentiate strains involved in different directions, a rectangular coordinate system is adopted. The bar is subjected to tensile forces of magnitude F _x in the x direction that induces a tensile stress σ _x (Fig. 14.2). Assuming that this stress is uniformly distributed over the cross-sectional area (A = ab) of the bar, its magnitude can be determined using:

$$ {\sigma}_x=\frac{F_x}{A}. $$

Under the effect of F _x , the bar elongates in the x direction, and contracts in the y and z directions. If the elastic modulus E of the bar material is known and the deformations involved are within the proportionality limit, then the stress and strain in the x direction are related through the Hooke’s law:

$$ {\epsilon}_x=\frac{\sigma_x}{E}. $$

Equation (14.2) yields the unit deformation of the bar in the direction of the applied forces. Strains in the lateral directions can now be determined by utilizing the definition of the Poisson’s ratio. If ϵ _y and ϵ _z are unit contractions in the y and z directions due to the uniaxial loading in the x direction, then:

$$ \nu =-\frac{\epsilon_y}{\epsilon_x}=-\frac{\epsilon_z}{\epsilon_x}. $$

In other words, if the Poisson’s ratio of the bar material is known, then the strains in the lateral directions can be determined:

$$ {\epsilon}_y={\epsilon}_z=-\nu {\epsilon}_x=-\nu \frac{\sigma_x}{E}. $$

The minus signs in Eqs. (14.3) and (14.4) indicate a decrease in the lateral dimensions when there is an increase in the axial dimension. Strains ϵ _y and ϵ _z are negative when ϵ _x is positive, which is the case for tensile loading. These equations are also valid for compressive loading in the x direction for which σ _x and ϵ _x are negative, and ϵ _y and ϵ _z are positive.