SSG 516 Mechanics of Continua 2

8/2/2019 SSG 516 Mechanics of Continua 2

1/27

SSG 516 Mechanics of ContinuaLecture Two

Kinematics: Study of Displacements & Motionthe various possible types of motion in themselves, leaving out

the causes to which the initiation of motion may be ascribed constitutes the science of Kinematics.ET Whittaker

Helpful Reading: Bower pp 13-27, Fakinlede 146-177


2/27

Kinematics is an organized geometrical description of

displacement and motion. The emphasis here is the fact that displacement and

motion are independent of the material constitution.While we may use the terminology of solid mechanics

the concepts and the ensuing relationships are

independent of the particular materials involved.

We shall also define the concepts of strain and strainrates as deriving from displacements and motion.

Kinematics


3/27

Our cube, for simplicity will have the dimension of unity andwith one edge at the origin.

Let , represent the initial Cartesian coordinates in anundeformed body. We consider the transformation,

= 1 0

1

00

00

,01

01

,10

1

tan,

11

1

1 + t a n

And for small values of, we can easily see that,

00

00

,01

01

,10

1

,11

1

1 +

Simple Shear of a Cube


4/27


5/27

Similar to the above, the transformation,

=1 0 1

Transforms the edges as,

00

00

,01

1

,10

10

,11

1 +

1

The square element in the above diagram has therefore

undergone a shearing in the horizontal axes as shown.

Horizontal Shear


6/27

Mathematica Code


7/27

We have seen how a simple tensor can be used totransform a simple shape into a new shape. We are now togeneralise this concept using the figure (Bower) below:

General Deformation


8/27

As a result of the transformation, such that,(, ) = + (, )

Using the linear gradient operator we defined last term,we can take the gradient of the above equation andobtain,

grad (, ) = grad + , (, ) = + grad , [In component form,

=

+

= +

]

Recall that by definition, + , , = grad + , + ( ) So that as we can write that =

Or in full in full component form, =

Transformation equations


9/27

The equation, = transforms an element in the

original configuration to . Note that while we havesuppressed the dependency of the deformation gradient on the location and time, it is general a function of both. Itis therefore more accurate to write, = (, )

This tensor contains all the necessary information aboutthe deformation. Once the deformation gradient isspecified for all locations and time, we can always find thenew configuration based on an initial state and a giventime .

The Deformation Gradient


10/27

The mapping = is necessarily one to onebecause of the physical implications. It necessarily hasan inverse. Therefore, given a new configuration, wecan obtain the old configuration that resulted in it,as =

Furthermore, for two successive deformationgradients, and ,

= , and = we can easily see that,

= = or that =

Inverse & Multiple Transformations


11/27

Transformation of a line element


12/27

Consider the simple parallelepiped shown

here. The base area,A = sin is or = in indexnotation, this cross product is =

the perpendicular height is cos so that the

volume is = A cos =

= =

Volume as a triple product


13/27

Jacobian


14/27

Recall that in component form, = +

so that

the Jacobian determinant of the deformation,

= det = det( +

)

This is a measure of the change in volume as a resultof the deformation. If the three sides of a

parallelepiped deform as, = , = and = , then

= and =

The Jacobian of F


15/27

Clearly,

= = = det =

So that,

= [To understand the above expression, note that in two-D,

det F = and that = det F. It is easy to work

the 2-D version out manually so understand the 3-D]

The Jacobian


16/27

Furthermore, it is easily shown that mass density is

related as

=

This follows from the above derivation. Show this to

be true.

Mass density


17/27

= =1

3!

Differentiating wrt , we obtain,

=

1

3!

+

+

=1

3! + +

=1

3!

+ +

=1

2!

The cofactor of

Lagrangian Strain Tensor


18/27

The tensor Tplays and important role in

continuum mechanics. It is called the right Cauchy-Green Tensor.

The concept of strain is used to evaluate how much agiven displacement differs locally from a rigid bodydisplacement. Consequently, a strain function is apoint function that vanishes in a state consisting ofonly rigid body displacements.

The Right Cauchy-Green Tensor


19/27

We will introduce the two most common strain

functions and demonstrate what they mean in onedimension: Lagrangian and Eulerian Strains.

First the Lagrangian Strain: =

()

Notice that when there is no deformation, we can assumethat the deformation gradient is the identity tensor(Why?). In this case, T = .Clearly = , the zerotensor.

Strain Functions


20/27

First, show that

= 12 = 12

+

+

For example let = 1, = 1:1

2 =

1

2 + +

= 12

+

+

+

+

=1

22

+

+

+


21/27

=

where is the deformation gradient tensor and is its

inverse. If we take the coordinates , = 1, , 3 as the

material coordinates and , = 1, , 3 as spatial

coordinates, then simple chain rule implies,

=

=

Eulerian Strain


22/27

We can therefore write that,

=

If the unit vector along the deformed element is ,

then the length of the element is given by: =

Similarly, in the undeformed system, we have:

= with as the unit vector aligned with thematerial element . In this case, we can write,

= =

=

Eulerian Strain


23/27

The strain,

=

2=

1

2

=

where is the Eulerian strain tensor,

( ). In

component form, following the same arguments it may

similarly be proved that in one dimensional strain,

=

2 =

Eulerian Strain


24/27

Again, beginning with unit vectors ,, = and

= . Recall that with the deformation gradienttransformation,

= = squaring, we have that,

= =

from which we can see that

=

()

=

=

=

Strain of a line element


25/27


26/27

()

2=

+ ( )

2

2( )

2=

( )

Which is the elementary definition of strain. We can seefrom here that this definition is only approximatelycorrect when strain is small and will not be a valid strain

function if the difference between and isappreciable. When that happens, we have a situationcalledgeometric nonlinearity in the sense that a linearfunction no longer defines the strain-displacementrelation

Infinitesimal Strain


27/27

Another way to arrive at this same conclusion is to note

that for small strains, the second-order terms in theLagrangian function

=

=

+

+

Is small so that,

[

+

]

Both the Lagrangian as well as the Eulerian strain becomeindistinguishable from small strain.

Infinitesimal Strain

SSG 516 Mechanics of Continua 2

Documents

Transcript of SSG 516 Mechanics of Continua 2