Hooke's Law requires only one material property to describe shear, i.e. one constant, which is G, called the modulus of rigidity, or modulus of elasticity in shear:
One example of this would be the strain due to the force in the x direction on the y plane, that is
In the previous blog we asked why the force on one plane would cause displacement in two planes - in this case the force on the x plane causes displacement in the x and y planes. In an earlier blog (the one about shearing strain and the change in angle), we used a 2-dimensional drawing which I've modified as shown below:
Hooke's Law (in the first equation above) says that the total change in angle for these two planes is determined by the x-y component of the stress tensor, i.e. the force applied to the upper and lower horizontal y-z planes (whose normal vectors are in the + y and - y directions).
But we know from symmetry that the x-y and y-x components of the stress and strain tensors are equal. For the strain tensor, this is an identity, but for the stress, symmetry is a consequence of conservation of angular momentum. So in Hooke's Law, it doesn't matter which stress component you use - the strain shown by the total angle in the drawing may be thought of as caused by either the x-y (horizontal) or y-x (vertical) stresses.
If you imagine the action of these forces on the block of material, it makes sense that Hooke's law would only apply to homogeneous, isotropic materials - those in which the material properties are constant throughout.
In contrast, suppose the upper layer of the block was much "stiffer" than the rest of the body. The response of the vertical sides to the stresses applied to them would be quite different than the horizontal ones. The two tensors would still be symmetric, but the simple linear relationship shown at the beginning of this blog wouldn't apply. For one thing, you would need at least two material constants.
In the next blog, we'll look at the normal components of Hooke's law.
No comments:
Post a Comment