Shearing Strain

The next step in coming up with a strain tensor is to look at the deformation caused by a shear force.  We start with a 2-dimensional description.  Imagine a square or rectangular region within a material body, with a coordinate system such that the x-axis is horizontal, and the y-axis is vertical.  Now add stress vectors acting on the rectangle, parallel to the top and sides.  These are the shear forces.  It helps to imagine this region as embedded in a beam, with one end attached to a wall, and the other end with a load pulling it down.  (We'll see later that tensile, compressive and shear forces are all operating within the beam).

The above drawing shows the rectangle before it deforms under stress.

Since we are looking for shearing strain, we consider the deformation in the x direction at two points along the y axis. Again making use of the Taylor series expansion, the x components of the displacements at the two points (x,y) and (x,y+b) are related by

The displacement is described by a vector u(x,y), but the x and y components of u are scalars.  These components are functions of two variables, x and y, but we can expand about one while holding the other fixed.  The prime refers to the partial derivative with respect to y, and for sufficiently small values of b, we have

This gives the variation in the y direction of the x-component of the displacement.  Now think of the z direction as coming out of the plane of the paper (or computer screen).  Then the top and bottom horizontal edges of the rectangle in the drawing are lines in planes which are parallel to the x-z plane.  So these displacements in the x direction take place in two nearby x-z planes.  The normal vector for the x-z plane is in the y direction, i.e. n = (0,1,0), so the displacement is perpendicular to the normal vector of the surface in which it occurs.  That's what makes it a shear force, instead of normal a force.

Now consider the vertical edges of the rectangle.  These lines are embedded in two nearby y-z planes, separated by a short distance a in the x direction.  We consider the difference in the y component of the displacement vector in these two planes, for a fixed value of y.  Proceeding in the same manner as previously, we find

We are now in a position to postulate a strain tensor, generalizing to three dimensions.  We are looking  a linear relationship between stress and strain, so since the stress tensor is symmetric, we want the strain tensor to be also.  It should contain both normal and shear strains.  The result is the following definition:

One reason for the 1/2 is so the diagonal components of the strain (xx, yy, and zz) will be given by their respective derivatives of the displacement.  Another is, so the trace (sum of diagonal terms) of the strain tensor is given by the sum of the normal strains, i.e.

                                                       Trace u = du/dx + dv/dy + dw/dz,

where u, v, and z are the displacements in the x,y and z directions of the point at (x,y,z).  (These are of course, partial derivatives, but I don't have a way of using Greek letters without going off into another program).

Using the results obtained from the Taylor series expansion in an earlier blog, we can show that this sum of normal strains is equal to the fractional change in a sufficiently small volume of the material, because the difference in the displacement of two nearby points, say at x and x+a, is simply the change in length of the line segment joining them.  Then a small volume with sides of lengths lx, ly and lz  in the x, y and z directions before deformation would have a new volume, with sides lx+(du/dx)lx, ly+(dv/dy)ly, and lz+(dw/dz)lz, .  The before and after volumes are the products of the  lengths of their three sides:

                                          New volume = lx ly lz (1+du/dx)(1+dv/dy)(1+dw/dz)

                                                             = Old volume (1+du/dx + dv/dy)(1+dw/dz)

                                                             = Old volume (1+du/dx + dv/dy + dw/dz)

where we have neglected the second-order terms.  From this we see that the sum of the normal strains is equal to the fractional change:  (New volume - Old volume)/Old volume.

We chose a coordinate system where all the strains were normal.  But the trace of a tensor, being a scalar, is invariant to an  arbitrary orthogonal rotation of the coordinates, so the result holds generally.  (Just think of a cube, which undergoes pure compression or expansion:  you can always choose a coordinate system with axes parallel to the three sides of the cube).

The other question to be answered is, why are the shear (off-diagonal) terms added together the way they are in the strain tensor?  It does give us a symmetric tensor, which is good since the stress tensor is symmetric, and we want the the two tensors to be linearly related.  But there are a couple of other reasons.  The first is that the pairs of shear terms in the stress tensor are additive. 

 Referring to the diagram of the rectangle at the beginning of this blog, the stress vectors at the two nearby vertical lines (representing y-z planes) are both driving the rectangle in a clockwise direction.  This is quite different from the normal stress vectors, which are equal and opposite, thereby cancelling each other out.  

Similarly for the stress vectors on the two horizontal x-z planes, which are also working together, to drive the rectangle counter-clockwise.  Of course the four vectors must sum to zero in equilibrium, but the ones on opposite sides need not be equal.

I'll save the second reason for the next blog.  It will use the drawing shown at the beginning of this one.