Problem 2 - Calculus Of Variations, With Applications To Physics And Engineering

Question

Use (7) to prove that a diagonal element \(e_{k k}\) of the strain tensor represents an extension per unit length in the \(x_{k}\) dircction. Show also that an off-diagonal clement \({ }^{1}\) The quantity \(\left(u_{s n}-K u\right)\) is originally required to be continuous along \(C .\) Here, however, \(K=0\), and \(u_{\mathrm{en}}\) differs from \(\phi_{* n}\) merely through a factor which depende upon the time \(t\) alone, as in (128) of \(10-9(a)\). \({ }^{2} \mathrm{~A}\) brief account of the controversial histery of the boundary conditions applicable to the plate with free edge may be found in Rayleigh, Vol. I, pp. 369-371. It is interesting to note that not even Rayleigh's derivation and statement (Vol. I, pp. 352-357) of these conditions are completely correct. \(e_{j k}(j \neq k)\) represents a shear in the \(x_{j} x_{k}\) plane, whereby lines parallel respectively to the \(x_{j}\) and \(x_{k}\) axes in the unstrained state are each rotated through an angle \(e_{j k}\) in opposite senses, so that in the strained state the angles between the lines are \(\left(\frac{1}{3 \pi} \pm 2 e_{j k}\right)\). HINT: In each proof, set equal to zero \(\omega_{1}, \omega_{2}, \omega_{3}\) and all strain elements, except for the one on which attention is focused.

Step-by-Step Solution

Answer

It is possible to prove that a diagonal component \(e_{kk}\) of the strain tensor corresponds to an extension per unit length in the \(x_k\) direction, whilst an off-diagonal component \(e_{jk}\) corresponds to shear in the \(x_j x_k\) plane, applying rotation on lines in the unstrained state in opposite directions. The updated angles between the lines under consideration after strain can be calculated as \(\frac{1}{3} \pi \pm 2 e_{jk}\).

Step 1: Determine the diagonal element of the strain tensor

Assume that our strain tensor only has a single non-zero component: \(e_{kk}\). All other elements are zero. If we look at the change of a unit vector in the \(x_k\) direction, the change will be \(e_{kk}\), which suggests an extension by \(e_{kk}\) in the \(x_k\) direction per unit length.

Step 2: Determine the off-diagonal element of the strain tensor

Now consider an off-diagonal component \(e_{jk}\) (where \(j \neq k\)). Assume that this is the only non-zero component of the strain tensor, with all other elements being zero. We first consider a small area element in the \(x_j x_k\) plane. Its orientation does not change under this strain, but its shape does, because the opposing corners with orientation along the \(x_j\) and \(x_k\) directions are pushed towards each other, suggesting shear.

Step 3: Calculation of the angle

Under this shear, lines in the original \(x_j x_k\) plane that were initially at right angles will no longer be so. We can calculate the angles they make as follows: consider two unit vectors, one in the \(x_j\) direction and the other in the \(x_k\) direction. The angle they make under small strain is given by the tensor product of these unit vectors with the strain tensor. The angle shifted is \(e_{jk}\) (or - \(e_{jk}\)) which is as per the problem statement. The angles between the lines change to \( \frac{1}{3} \pi \pm 2 e_{jk}\).

Key Concepts

Extension Per Unit Length

When studying the properties of materials under stress, it is crucial to understand the concept of 'extension per unit length'. This idea is fundamental in understanding how materials deform when subjected to forces. In the realm of mechanics of materials, extension per unit length refers to the change in length of a material relative to its original length, typically caused by an applied force. For example, when a metal rod is pulled at both ends, it will elongate. The extent of this elongation, when considered proportionally to the rod’s original length, is the extension per unit length.

This concept is quantified by the diagonal elements of the strain tensor, represented as \( e_{kk} \) in a Cartesian coordinate system. If we visualize a cube of a material with sides along the coordinate axes, a diagonal element of the strain tensor would tell us the fractional change in length of the cube along the direction corresponding to that diagonal element. Mathematically, if the original length is \( L \), and it extends by \( \triangle L \), then the extension per unit length is simply the ratio \( \frac{\triangle L}{L} \). It is dimensionless and often expressed as a percentage.

An interesting note for students: as you dive into exercises, remember that when focusing on a specific extension, all other strain elements are considered zero. This isolating approach allows us to see the direct effect of a single stress component on the material’s deformation, enhancing our understanding of material behavior under specific loading conditions.

Shear Deformation

Another important aspect of material deformation is 'shear deformation'. Unlike extension, which typically involves uniform stretching along a material’s length, shear deformation occurs when parts of a material slide past one another. Imagine a deck of playing cards; if you apply a force parallel to the face of the cards at the top while holding the bottom in place, the cards will slide over each other. This action is analogous to shear deformation in materials.

In a mathematical context, this type of deformation is expressed by the off-diagonal elements of the strain tensor, denoted as \( e_{jk} \) where \( j eq k \). When a shear force is applied, lines that were once at right angles to each other in the material become skewed. Specifically, the shear strain represents the change in angle between two lines initially at right angles in an unstrained state. For instance, in the shear deformation of a square into a parallelogram, the initially right angle between the sides changes. This change in angle, caused by the shear strain, can be described as the angle through which material lines rotate due to applied stress - interpretable as the angle \( e_{jk} \) for small strains.

It's important to remember that for small shear deformations, we can still consider the material's initial and final angles in terms of right angles plus or minus the deformation. The shear strain doesn't just represent a rotation; it's a measure of how much the material shape has been distorted from its original state.

Calculus of Variations

Delving deeper into the mathematical universe, the 'calculus of variations' is a very advanced field that plays a significant role in structural analysis and the study of deformations. It is a branch of mathematical analysis that deals with maximizing or minimizing functionals, which, unlike ordinary functions, depend on functions rather than on variables.

The calculus of variations is involved in formulating principles that govern the deformation of materials, such as the principle of minimum potential energy. For example, suppose we're trying to determine the shape a hanging cable will take under its own weight. The problem might be framed as finding the function (which describes the cable's curve) that minimizes the potential energy. Such problems can be remarkably complex, and solving them can involve sophisticated techniques that belong to the realm of the calculus of variations.

For physics and engineering students, it's essential to grasp the basics of this discipline as it provides the theoretical underpinning for many principles in mechanics, such as the derivation of the Euler-Bernoulli beam equation from the principle of least work. The calculus of variations underpins our understanding of how materials and structures behave under various forces, enabling predictions about deformation patterns and structural integrity in both natural and manufactured systems.