Hyperplasticity Basic continuum mechanics 1

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Equipped with the language of tensors, we are now in a position to revise some basic continuummechanics. I hope that much of this will be familiar to you, but it is worth going over a few topicsin the terminology I shall use.

Hyperplasticity Basic continuum mechanics 2

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I shall emphasise the four basic sets of equations of any mechanical problem

• The equilibrium equations that link forces and stresses

• The compatibility equations that link displacements and strains

• The boundary conditions that define the problem that we have in hand

• and the final set of equations, the constitutive model for the material, or materials, thatmake up our problem.

The remainder of this course will almost exclusively be dealing with the constitutive model, butit is important to know about the other equations too.

It is important too to realise that the first three sets of equations are not really subject todebate – they may be written in different ways, but they are treated as if they werefundamentally true. The last set of equations though has to be rooted in empirical observation– we want the constitutive model to represent some real material. A constitutive model is not“true” or “false” it simply represents real material behaviour with a satisfactory orunsatisfactory level of accuracy.

Hyperplasticity Basic continuum mechanics 3

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We start with displacements.

We consider some material point that was initially (at t = 0) at some position X_i and is currently,at time t, at position x_i. The displacement from its initial position is u_i = x_i – X_i

A very important quantity is the deformation gradient F_ij – the differential of the currentposition with respect to the initial position. Dx_i/dX_j

The displacement gradient is simply F_ij – delta_ij, where delta is the Kronecker delta

Small displacements will involve an appropriate norm of the displacement much less than unity,so F_ij is approximately equal to delta_ij. If we follow through the differentials, we can see thatin this case we do not need to distinguish between differentials with respect to X_i and withrespect to x_i, in other words we do not need to worry about the difference between our initialand current coordinate systems.

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We now turn to strains. There are a variety of strain definitions that are appropriate in differentcircumstances. They have to satisfy the condition that they are zero if the material does notdeform.

The two commonest large-strain definitions are the so called Green-Lagrange tensor, which isused when the initial coordinate system is used – this is called a Lagrangian approach

… and the Euler-Almansi tensor when the current coordinate system is used, which is called anEulerian approach

If displacements are small we can use the small strain tensor, as we do not need to distinguishbetween differentiation within the different coordinate systems. In this case the second orderterms disappear.

By definition the norm of the small strain tensor must always be small.

We can also write the small strain tensor in a particularly succinct way in tensor notation.

Note that all of the strain tensors are symmetric.

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It is worth thinking a little about the difference between small strain and small displacement.

The most general cases involve arbitrarily large displacements and strains.

Within that there is a class of problems for which the strains are small, but the displacementsmay be large. For instance if we bend elastically a thin metal bar by a large amount, at no pointin the bar are the strains large, yet the overall displacement certainly is. In these cases we needto use the appropriate large strain definition, but the norm of the strain will be small.

And then within the small strain problems is a further category in which the displacements aresmall as well as the strains. In these cases we can use the small strain definition.

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This table illustrates the same problem a different way.

We can have finite displacement and finite strain,

… or we can have finite displacement and small strain,

… or small displacement and small strain.

The one thing we cannot have is large strain but small displacement.

In order to keep things simple, I am going to talk almost entirely in terms of small displacement,small strain – the easiest case. In this case we can use the small strain tensor, and we do notneed to distinguish between the initial and current coordinate system.

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We now turn to equilibrium, and will illustrate this with just the 2-D case.

Consider a small element of the material. On one side of the element will be normal and shearstresses. We use double subscripts – the first subscript indicates the direction of the forcecomponent, and the second the normal to the face on which the stress component acts.

On the opposite sides of the element we take account of the fact that the stresses may bevarying, so for instance in the x direction on the vertical faces sigma_xx has changed to sigma_xxplus the rate of change of sigma_xx with respect to x, multiplied by the x-dimension of theelement.

If we sum the relevant forces we can write the horizontal and vertical equilibrium equations,including a gravity term in the vertical one.

The first order terms in the equations all cancel, and then there is a common factor of delta_x xdelta_y which we can remove, and we end up with the simple version of the equations at thetop right.

In three dimensions we can write these in tensorial form using the summation convention, andreduce this even more using the comma notation.

Incidentally, it is easy to show that the moment equilibrium equation just leads to tau_xy =tau_yx, so the stress tensor is symmetric.

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What about compatibility?

For some reason this sometimes causes a little confusion, but compatibility can be expressed in avery easy way. The six independent strain components (bearing in mind that the strain tensor issymmetric) can be derived from just three displacements. So the strains must in effect be linkedby three compatibility requirements, just as the stresses are linked by three equilibriumequations.

By far the easiest way to express these compatibility requirements is simply to write down thestrain definitions in terms of the displacements.

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Now we turn to boundary conditions.

There are two main types of boundary conditions. The first possibility is that we simply specifythe displacements at the boundary.

The alternative is that we define the tractions at the boundary. These are often, but incorrectly,referred to as stress boundary conditions, but we cannot in fact specify the stress on theboundary. Consider the two dimensional case. We can specify the normal and shear componentsof force per unit area on the boundary, and these are referred to as the tractions. They areclosely related to some of the stress components, but we cannot specify anything about thenormal stress in the direction parallel to the boundary.

The relationship between the internal stresses and the tractions at the boundary are given bythis simple equation which involves the unit normal to the surface. You should satisfy yourselvesthat this is correct.

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It is important to realise that stress and strain are not just chosen arbitrarily.

The stress and strain rate must form a work conjugate pair.

That is to say that the stress times the strain rate gives the rate of work input to the element ofmaterial per unit volume.

You can prove this by recognising that the rate of work input to a volume will be the integralover the boundary of the tractions multiplied by the corresponding velocities.

We can substitute the expression for the stresses instead of the tractions, and then use Green’stheorem to convert the integral over the surface to an integral over the volume. We expand thedifferential, and then recognise that the first term is zero because of the equilibrium equation.The second term involves the stress times the velocity gradient, but because the stress issymmetric we can just pick out the symmetric part of the displacement gradient, which is thestrain rate.

Hyperplasticity Basic continuum mechanics 11

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In solving any problem it is useful to check on the count of variables and equations just to makesure nothing has been missed out.

First of all in the continuum:We have 9 stresses, nine strains and three displacementsWe have already identified three equilibrium equations, and nine strain definitions whicheffectively capture the compatibility between displacement and strainWhat we are missing is nine equations – and these are the constitutive relations thatrelate the nine stresses to the nine strains. The rest of the course will be about thesemissing equations

At the boundary there are:Three tractions and three displacements – six variables in totalThere are three equilibrium equations that relate the tractions to stress components… and three other equations. On a traction controlled boundary they will be the threestresses, and on a displacement controlled boundary the three displacements. Of courseit is also possible to have more complex mixed boundary conditions, but in any casethere are three specified quantities.

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So in summary:

I have introduces the stress and strain definitions, emphasised that stress and strain must bework-conjugate. To keep things simple I am just going to use small strains.

In the background we must remember the equilibrium equations, the compatibility equationsand the boundary conditions…

… and then the vital set of equations that describe the material we are talking about. That iswhen things get interesting, and that is what this course is about.