In this post we are looking at the change in angle of one of the vertices after a deformation, for the rectangle discussed in the previous blog. (Click on the drawing to enlarge it). The original angle for the vertex at (x,y) is, of course, 90 degrees, or pi/2 radians. After deformation, all four vertices may be displaced, as shown below. As before, we use the vector u to represent the displacement.
The y-component of the displacement of the point (x,y) is in the positive y direction, as shown on the right side of the figure. Similarly the y-component of the point (x+a,y) is in the negative y direction. Thus the difference in displacement between the two points is
Now consider the angle of the deformed figure for the vertex which was initially at (x,y). It is larger than 90 degrees, due to the combined displacements of three of the vertices of the deformed figure: the ones at (x,y), (x+a,y), and (x,y+b).
The change in the original 90 degree angle due to the above expression is the one formed by the dashed horizontal line and the lower side of the deformed figure. The sine of that angle is given by
provided the displacements are small enough (see footnote below). As we've seen in earlier blogs, this expression is simply the derivative of the y-component of the with respect to x, i.e., one part of the x-y (or y-x) shear component of the strain tensor.
To get this result, we have to keep the signs straight, bearing in mind that the displacement at x+a is in the negative y-direction. Strictly speaking, the change in angle is also negative, but we are looking for magnitudes here, so the minus sign that would otherwise be in front of the expression above was dropped.
Finally, we note that for a small angle, the sine of the angle is nearly equal to the angle itself, measured in radians. I'll discuss this useful approximation in the next blog. But for now, we need to examine the other portion of the change in the angle at (x,y), i.e. the x-displacement of the points at (x,y) and (x,y+b).
These displacements are also in opposite directions, so we have to remember that the displacement of the upper vertex is in the negative x-direction. By the same process as used above, we find the sine of this angle to be
which is the other piece of the x-y component of the strain tensor. As before, this is also the sine of the angle being considered, and equal to the angle itself, provided the angle is small enough.
The total change in angle is the sum of these two terms, i.e., twice the x-y component of the strain tensor. This is the second reason for defining the strain tensor the way it is. (The factor 1/2 in the strain tensor will hopefully become clear later).
The above discussion is an elaboration of a similar one in the first chapter of Theory of Elasticity, by Timoshenko and Goodier.
After a small detour to talk about the approximation that the sine of a small angle is approximately equal to the magnitude of the angle, in radians, we'll tackle the question of relating stress and strain.
Footnote: The use of a in the denominator of the expression for the sine of the angle above is an approximation which can be justified as follows: The actual length for the long side of the angle is given in the denominator of the expression below, for the sine of the angle (which replaces the previous equation for the sine):
Using the definition of a derivative as before, the sine becomes
After cancelling a in the fraction, the denominator has the form 1/(1+x), which can be expanded in a Taylor series (or see Pierce and Foster, A Short Table of Integrals):
Applying this to the previous expression gives
where we have neglected higher order terms, as usual. Multiplying out, we see that the second term, a product of two strain terms, will be small compared to the first term, with the result that the sign of the angle is one part of the x-y strain tensor, as stated in the main part of this blog.
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