WEBVTT
Kind: captions
Language: en
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The previous lecture we started talking about
nonlinear finite element model development.
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We started by discussing sources of non-linearity
and we saw that the main source of non-linearity
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could be due to strain displacement relations
being nonlinear. That is called geometric
00:00:45.500 --> 00:00:50.990
non-linearity and the stress-strain relations
could be nonlinear that is material non-linearity.
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There could be non-linearity associated with
boundary conditions or energy dissipation
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mechanisms.
So various schemes of classification are possible.
00:01:01.580 --> 00:01:07.250
All these nonlinearities can coexist in a
problem. So in this case what is shown here
00:01:07.250 --> 00:01:13.970
is small displacements and small strains but
there is material non-linearity here, whereas,
00:01:13.970 --> 00:01:20.079
in this case there is material could be linear
or non-linear but there are large rotations
00:01:20.079 --> 00:01:25.600
but small strains.
Here material again could be linear or non-linear.
00:01:25.600 --> 00:01:30.950
There is large rotation and large strains
and this is a schematic of a situation where
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there could be non-linearity associated with
boundary conditions. So this spring will come
00:01:36.090 --> 00:01:40.859
into action only when this displacement here
exceeds this threshold in which case the stiffness
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of the system increases. Here the loading
and unloading path will be tracing each other
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whereas in material non-linearity the unloading
path will be different from the loading path.
00:01:56.149 --> 00:02:02.049
After reviewing few details about qualitative
feature of nonlinear system response and how
00:02:02.049 --> 00:02:07.229
it differs from response of linear systems
we started talking about elements of continuum
00:02:07.229 --> 00:02:13.200
mechanics and I briefly started talking about
kinematics that is study of motion and deformation
00:02:13.200 --> 00:02:20.959
without concerning the causes of motion and
deformation. So we have a configuration at
00:02:20.959 --> 00:02:26.560
time equal to 0. It is here and this could
be the reference configuration and this is
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a Cartesian coordinate system in which the
position of particles in the material points
00:02:31.920 --> 00:02:39.810
in this body are described. So P, the position
vector of P is the X and after deformation
00:02:39.810 --> 00:02:45.810
this point P with position vector X gets mapped
to point P with position vector lower X through
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this transformation. And P-P is the displacement
vector defined as U of X minus X. So in Lagrangian
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descriptions we treat the coordinates of capital
P that is capital X1, X2, X3 as the independent
00:03:04.100 --> 00:03:10.611
variables whereas in Eulerian system we treat
the current position, the position of particle
00:03:10.611 --> 00:03:15.120
P in the current configuration that is lower
case x1, x2, x3 as the independent variables.
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That is Eulerian system. So in solid mechanics
problem often Lagrangian coordinate system
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is used.
Now let us consider a line segment PQ in the
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body be at in the reference configuration
and due to this deformation, this point PQ
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moves to this position pq as shown here and
the initial length is d capital X, this is
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d lowercase x. So various position vectors
of PQ in the T equal to 0 and at some time
00:03:48.420 --> 00:03:56.610
T are shown here.
Now the position vector in the current configuration
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is a function of the position in the original
configuration and I can write X1, X2, X3 in
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the long hand in this form and from this I
deduce dX1 is dX1 by dX1 into dX1 etcetera
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that is shown here. So this set of equations
where I get dX1,dX2 dX3 which are components
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of position vector in the deformed configuration
which are related to the components of position
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vector in undeformed configuration through
this matrix and this matrix is known as deformation
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gradient tensor.
So we get this equation in matrix notation
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it is dx is equal to F dX in in the indicial
notation dXi is Fidxj or in tensile notation
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is f.dX. The determinant of this matrix F
is known as Jacobian.
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By inverting this relation I can also here
the components of line segments in the deformed
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configuration are related to component of
position vector in the original configuration.
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This can be inverted and I can get this relation
and dX is f inverse dx. We have displacement
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vector as U is X minus x and from which I
get to use du =dx-dX and du I can write therefore
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as for dX I will write FdX so this is (F-I)dx
and this matrix -I I call it as capital G
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and this is a displacement gradient with respect
to reference configuration.
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Alternatively I can express dX in terms of
dx through this relation and I get this is
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I minus F inverse into dX and this matrix
is known as J naught. So these are displacement
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gradient with respect to current configuration.
So this G and J naught matrices are shown
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here. This is with respect to displacements
whereas this is with respect to the position
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coordinates of the position vector.
So G is this. J naught is this but for small
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deformation G and J naught related through
this relation. So small deformation this relations
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apply.
We can consider a few examples. Suppose I
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have the displacement field given by lambda
e1X1, lambda e2X2, lambda e3X3 so that x1,
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x2, x3 are respectively lambda X1, lambda
X2 and lambda X3. The displacement e1, e2,
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e3 will be lambda-X1, X1 and lambda- X2, lambda-1
X3. From this I get the deformation this matrix
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F as this and this type of deformation is
called pure dilatation.
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Now another example I will consider x as 1+alpha
e1X1+e2X2+e3X3 so that X1 is 1+alphaX1 from
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which I get e1 is alphaX1 and X2 and X3 are
such that e2 and e3 are zero from which I
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get F to be given by this.
So this deformation is called pure extension.
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Now I can consider X to be A into X plus C
from this it follows u is (A - I)X plus C
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and F is simply A. when the matrix F is independent
of X, we say that the deformation is homogeneous.
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A slightly more involved another example here
x1 is X1 plus gamma X2, x2 is lambda X2, x3
00:07:42.490 --> 00:07:49.340
is lambda X3 so that the F matrix will be
given by this and this deformation is called
00:07:49.340 --> 00:08:02.440
pure shear.
Now here I have a displacement field where
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there are non-linear terms X1 into X2 and
so on and so forth. So u1 will be gamma 1
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X1X2 u2 is X2 is gamma 2 X1X2 and X3 is gamma
3 X2X3 from which I get F to be this which
00:08:14.960 --> 00:08:22.980
is a function of now X1X2X3 and such deformation
are called non-homogeneous. So there are simple
00:08:22.980 --> 00:08:29.389
illustrations.
Now this Phi of X comma T which maps the position
00:08:29.389 --> 00:08:40.979
vector X to lowercase x it has to satisfy
certain conditions. This function is continuously
00:08:40.979 --> 00:08:47.500
differentiable and this is one-to-one and
that means F matrix can be inverted so dX
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is FdX and J is the Jacobian determinant of
F. We impose the condition J to be greater
00:08:54.000 --> 00:09:00.480
than 0. You can see that the determinant can
be expressed in terms of displacements as
00:09:00.480 --> 00:09:08.430
shown here if all displacements are 0 then
I get J equal to one. So for F to be invertible
00:09:08.430 --> 00:09:12.940
J must not be equal to zero because it is
determinant of F, therefore F inverse to exist.
00:09:12.940 --> 00:09:17.810
Determinant of F must not be zero.
So upon deformation J cannot become negative
00:09:17.810 --> 00:09:22.610
without crossing J equal to zero. So hence
we impose the condition that J must be greater
00:09:22.610 --> 00:09:26.420
than or equal to zero and these are such motions
are admissible motions.
00:09:26.420 --> 00:09:33.350
Now I can - an example of a motion which is
not not admissible is shown here you can verify
00:09:33.350 --> 00:09:38.640
that this is not admissible.
I mentioned in the previous lecture that rotations
00:09:38.640 --> 00:09:46.570
play a crucial role in analysis of nonlinear
systems. So let us consider the displacement
00:09:46.570 --> 00:09:50.920
field where there is a rigid body, translation,
and a rotation.
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This R matrix is a rotation matrix. So this
relation is expressed in matrix form here,
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indicial form here and a tensorial form as
shown here. This R of T is a rotation matrix,
00:10:02.990 --> 00:10:08.680
XT of T is a rigid body translation. Now from
this equation I find dX it will be RdX plus
00:10:08.680 --> 00:10:16.070
dxt of t dxt of t is 0 because rigid body
translation implies no change in length. Therefore
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this is RdX. Now if you find the length dxt
square of the length dxtdx and if for dxt
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if I write this equation I get dX transpose
R transpose RdX. Since R transpose R is a
00:10:32.880 --> 00:10:39.650
identity matrix because R is a rotation matrix
I get this. So this length remains unchanged
00:10:39.650 --> 00:10:46.790
therefore I mean I have to verify that R transpose
R is I. Therefore the argument is the length
00:10:46.790 --> 00:10:52.880
in a rigid body rotation, and translation
length does not change. Therefore dxt transpose
00:10:52.880 --> 00:11:02.280
dX must be equal to dxtdx therefore this is
true for any dX. Consequently I get the relation
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that R transpose R is I or in other words
R transpose R inverse that is R is orthogonal.
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Now how does pictorially it looks like? Suppose
you'd consider a triangular domain upon this
00:11:14.000 --> 00:11:22.740
transformation RX+xT this R, effect of R is
to rotate this as shown here and xT the effect
00:11:22.740 --> 00:11:31.120
of xT is to translate. So this is how the
element looks upon undergoing this deformation.
00:11:31.120 --> 00:11:37.511
Now if we consider two coordinate systems
xj and xj prime such that a position vector
00:11:37.511 --> 00:11:45.630
in xj system is xjej and position vector in
xj prime coordinate system xjej prime where
00:11:45.630 --> 00:11:54.120
ei unit vectors in xj system and ej prime
are the unit vectors in xj prime system. Now
00:11:54.120 --> 00:12:02.510
repeated indices imply summation here. Now
clearly this dot product of eiej is Delta
00:12:02.510 --> 00:12:15.250
ej and since this is equal I get this relation.
I will dot product with ei and I get this
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and from this denote ej prime dot ei Rji and
this gives me the relationship between xi
00:12:24.860 --> 00:12:32.300
and xj prime and similarly I consider this
equation and again do another dot product.
00:12:32.300 --> 00:12:41.070
This time with the ei prime I get this relation.
So from this analysis we get that X is R transpose
00:12:41.070 --> 00:12:49.779
x Prime and x prime is Rx. So this is how
a vector undergoes transformation due to coordinate
00:12:49.779 --> 00:12:55.620
transformation. By using similar arguments
we can show that a tensor like stress undergoes
00:12:55.620 --> 00:13:05.959
transformation following this role.
We introduce another quantity known as angular
00:13:05.959 --> 00:13:12.290
velocity. So we consider again the rigid body
translation and rotation as shown here and
00:13:12.290 --> 00:13:18.830
I differentiate this with respect to time
I get X dot is R.X plus X.T Twitter. Now for
00:13:18.830 --> 00:13:29.970
X I will write using this relation, it is
X- xT of t into R inverse of that and R inverse
00:13:29.970 --> 00:13:36.670
is RT and I get this relation. So from this
I get this equation. I rewrite in this form
00:13:36.670 --> 00:13:42.300
by denoting R. R transpose by capital Omega
and this quantity is known as angular velocity
00:13:42.300 --> 00:13:48.440
tensor.
Now we can show that this angular velocity
00:13:48.440 --> 00:13:54.870
tensor is a skew symmetric matrix to show
that we consider D by DT of RR transpose since
00:13:54.870 --> 00:14:00.371
RR transpose is identity matrix this must
be equal to 0 so that means r dot r transpose
00:14:00.371 --> 00:14:05.720
plus RR dot transpose must be 0 from which
i get the relation Omega must be equal to
00:14:05.720 --> 00:14:14.800
minus Omega transpose therefore Omega is skew
symmetric and this will have this form.
00:14:14.800 --> 00:14:23.330
A simple example I consider x equal to 2X
plus 3Yt and y is 2Xt plus 3Y. So the question
00:14:23.330 --> 00:14:27.310
is examine the admissibility of the motion
sketch the configuration of the element at
00:14:27.310 --> 00:14:33.220
T equal to 0.5 second and determine the displacement
velocity and acceleration fields. So the domain
00:14:33.220 --> 00:14:41.490
is OAB is described here. This length is 1.
this length is also 1.
00:14:41.490 --> 00:14:47.990
So deformation gradient you can easily evaluate
it to be this and the Jacobian will be 6 minus
00:14:47.990 --> 00:14:53.650
6t square and this has to for motion to be
admissible this has to be greater than 0 therefore
00:14:53.650 --> 00:15:00.860
this motion is admissible only for t less
than 1 at T equal to 1/2 I will put for T
00:15:00.860 --> 00:15:09.370
1/2 I get this displacement field and I can
map the points, the three points here O, A,
00:15:09.370 --> 00:15:15.670
B, and I will plot them in the transform coordinate
and find out the displacement field and velocity
00:15:15.670 --> 00:15:21.540
field and acceleration field.
And this triangle is the O prime, B prime
00:15:21.540 --> 00:15:30.180
is the - this is how the triangle looks upon
deformation at T equal to half. Okay.
00:15:30.180 --> 00:15:36.010
Now there are other few things that I am stating
without proof. Some of this we have done but
00:15:36.010 --> 00:15:43.600
other couple of things I am not doing. Upon
deformation dX becomes FdX this we have shown.
00:15:43.600 --> 00:15:48.790
Similarly area element and volume element
undergo transformation as shown here da is
00:15:48.790 --> 00:15:56.110
JF transpose inverse to da and volume is dv
is JdV. Fij we already discussed duX by duXJ
00:15:56.110 --> 00:16:03.810
and inverse of this is this and J is the Jacobian
--so is the determinant of F called the Jacobian.
00:16:03.810 --> 00:16:09.150
So these two results I am stating without
proof. You can with some effort you will be
00:16:09.150 --> 00:16:13.320
able to show that.
There is an important concept known as polar
00:16:13.320 --> 00:16:21.960
decomposition theorem and that leads to notion
of stretch and rotation tensors. The main
00:16:21.960 --> 00:16:26.820
idea is this the motion of a line segment
can be expressed as a pure deformation followed
00:16:26.820 --> 00:16:32.320
by a rigid body rotation or a rigid body rotation
followed by a pure deformation. That is F
00:16:32.320 --> 00:16:38.610
can be written as R into U or V into R where
R is a rotation matrix and U and V are symmetric
00:16:38.610 --> 00:16:45.279
positive definite matrices. So for pure rigid
body rotations F is R and consequently U will
00:16:45.279 --> 00:16:54.420
be I and V will be I. Now for other situations
we can work out how you can be determined
00:16:54.420 --> 00:16:59.600
is as follows. We consider F equal to RU and
from this I get F transpose F is F transpose
00:16:59.600 --> 00:17:06.089
RU and for F transpose I write U transpose
R transpose and this becomes U transpose U
00:17:06.089 --> 00:17:12.809
which is U square. So from this I get U to
be square root of F transpose F. Similarly
00:17:12.809 --> 00:17:20.039
if I consider the second relation F equal
to VR I get post-multiplying by F transpose
00:17:20.039 --> 00:17:28.710
I get FF transpose is VRF F transpose and
again by making the substitutions we show
00:17:28.710 --> 00:17:39.870
that we square root of FF transpose.
So how to find R from a given F? U is first
00:17:39.870 --> 00:17:45.400
we find U and V by square root of F transpose
F and FF transpose square root of a of transpose
00:17:45.400 --> 00:17:53.880
F. F is RU therefore R is F U inverse or alternatively
F is VR, R is V inverse F. Now this quantity
00:17:53.880 --> 00:18:00.600
F transpose F is known as denoted by CR and
it is known as right Cauchy green stress tensor
00:18:00.600 --> 00:18:05.940
and CL which is FF transpose is left Cauchy
green stress tensor. Clearly they are symmetric
00:18:05.940 --> 00:18:10.460
and positive-definite. So that can be verified
by inspection here.
00:18:10.460 --> 00:18:17.720
Now how do you find a square root of a symmetric
positive definite matrix? Quickly we can recall.
00:18:17.720 --> 00:18:24.230
Let A be a n by n symmetric positive definite
matrix. Let us consider the eigenvalue problem
00:18:24.230 --> 00:18:29.550
a phi equal to lambda phi. Let Phi be the
matrix of eigenvectors such that it is Phi
00:18:29.550 --> 00:18:36.140
transpose Phi is capital lambda and Phi transpose
Phi is I. So this lambda is a diagonal matrix
00:18:36.140 --> 00:18:42.190
of eigenvalues obtained by solving this problem.
And we know that these lambdas are non-positive
00:18:42.190 --> 00:18:49.351
and Phi is real valued. So from this I can
write A as Phi lambda phi transpose. Now let
00:18:49.351 --> 00:18:55.370
B be square root of A. That's what I wish
to find out. So and I consider B equal to
00:18:55.370 --> 00:19:02.940
Phi D Phi transpose. I do not know what is
D. So from this I get B square is Phi D transpose
00:19:02.940 --> 00:19:08.950
Phi D Phi transpose into Phi D Phi transpose
but Phi transpose Phi is I that is how we
00:19:08.950 --> 00:19:13.510
have normalized. This will therefore will
be Phi D square Phi transpose. This is equal
00:19:13.510 --> 00:19:19.850
to A because B square is A and consequently
by comparing these two relations I get lambda
00:19:19.850 --> 00:19:26.360
to be D square it will be equal to lambda
capital lambda and therefore D square root
00:19:26.360 --> 00:19:32.260
lambda. So B is therefore obtained as Phi
square root lambda phi transpose.
00:19:32.260 --> 00:19:39.280
So this gives interpretation of what is the
square root of A matrix.
00:19:39.280 --> 00:19:48.170
Now we will now talk about measures of strain.
Now there are two requirements that we need
00:19:48.170 --> 00:19:54.670
to satisfy. First is the strains must vanish
when body undergoes rigid body motion. Secondly
00:19:54.670 --> 00:19:59.720
the strains coincide with the infinitesimal
strains in the limit of strains becoming small.
00:19:59.720 --> 00:20:05.500
So any measure of strain that we develop should
conform to this and we can ask - begin by
00:20:05.500 --> 00:20:10.040
asking the question why do we need new measures
of strain, why not be content with linear
00:20:10.040 --> 00:20:14.790
measures of strain. That is infinitesimal
strain. Now the problem with infinitesimal
00:20:14.790 --> 00:20:19.760
strains is that linear measure of strain does
not lead to zero strains for structures undergoing
00:20:19.760 --> 00:20:25.210
rigid body rotations. How do we see that?
We can consider a small example. You consider
00:20:25.210 --> 00:20:29.800
a two-dimensional example of an element which
is rotated by angle theta. It is a rigid body
00:20:29.800 --> 00:20:38.600
rotation. So the deformation is given by X
is equal to x1x2 is cos theta minus sine theta
00:20:38.600 --> 00:20:45.000
sine theta cos theta x1x2. So for this deformation
for no any value of theta we expect that there
00:20:45.000 --> 00:20:50.880
won't be any - the strains would be zero.
Now we can find the displacement field u1
00:20:50.880 --> 00:20:59.020
is X - and this and u2 is this and from which
I compute the strain infinitesimal strain
00:20:59.020 --> 00:21:05.610
component du1 by du x1 is cos theta minus
one, epsilon YY is du2 by du2X which is cos
00:21:05.610 --> 00:21:12.510
theta minus 1 and shear strain is 0. Now you
look at epsilon XX and epsilon YY they are
00:21:12.510 --> 00:21:19.280
not 0 but of course as theta goes to 0 this
goes to 0 that is okay but for large theta
00:21:19.280 --> 00:21:25.720
epsilon XX and epsilon YY do not vanish. This
is why we need to redefine strain when you
00:21:25.720 --> 00:21:32.190
are considering large problems with large
displacements. Now to examine this in a slightly
00:21:32.190 --> 00:21:37.340
greater detail we can consider cos theta minus
1 this can be shown to be equal to minus theta
00:21:37.340 --> 00:21:41.970
square by 2 plus order of theta to the power
of 4 and approximately we can take it as theta
00:21:41.970 --> 00:21:47.419
square by 2. So based on this we will be able
to judge when to abandon linear strain measures.
00:21:47.419 --> 00:21:53.240
So when theta is large we need to abandon.
So just to quickly see that suppose the strain
00:21:53.240 --> 00:21:58.929
being measured is about point naught 1 and
acceptable accuracy is about 1% of that. So
00:21:58.929 --> 00:22:03.410
this is accuracy. Now the term that we are
ignoring in linear strain approximation is
00:22:03.410 --> 00:22:08.380
of the order theta square by 2. We assume
theta to be small so that theta square by
00:22:08.380 --> 00:22:13.950
2 can be ignored. So for approximation to
be acceptable theta has to be less than this
00:22:13.950 --> 00:22:19.179
which is about point naught one radians. So
if you want if you are dealing with strains
00:22:19.179 --> 00:22:23.450
of about point naught one and you want to
characterize with 1% accuracy the rotation
00:22:23.450 --> 00:22:28.390
should not cross this. Similarly if the strain
being measured is about ten to the power of
00:22:28.390 --> 00:22:33.000
minus four and acceptable accuracy is ten
to the power of minus 6 that is again 1% percent
00:22:33.000 --> 00:22:38.190
error. The linear measure of strain is acceptable
if theta is less than point naught naught
00:22:38.190 --> 00:22:46.950
radian. Now obviously if theta exceeds this
the major infinitesimal strain measures are
00:22:46.950 --> 00:22:48.623
not acceptable.
Now it is important to note that if the structure
00:22:48.623 --> 00:22:55.309
is on the verge of losing stability small
strains can cause large rotations. So we cannot
00:22:55.309 --> 00:23:00.470
use linear measures of strain in buckling
analysis that is why we, if you recall, we
00:23:00.470 --> 00:23:04.929
use nonlinear strain displacement relations
when we did buckling analysis.
00:23:04.929 --> 00:23:11.090
Now equipped with this we can introduce the
first strain measure that is Green-Lagrange
00:23:11.090 --> 00:23:15.630
strain measure. we have seen this earlier
but in a slightly different notation. dX is
00:23:15.630 --> 00:23:21.580
F into dX and therefore length of an element
dX square is dX transpose dX. This is in the
00:23:21.580 --> 00:23:26.190
deformed configuration and in the original
configuration it is this. So the change in
00:23:26.190 --> 00:23:31.380
square of the length say ds square minus dS
square and this I can write in this form now.
00:23:31.380 --> 00:23:37.990
For dX if I use the relation FdX, I can rewrite
this as DX transpose F transpose FdX minus
00:23:37.990 --> 00:23:44.080
dX transpose dX. So I will write this quantity
in this form dX transpose F transpose F minus
00:23:44.080 --> 00:23:53.160
I into dX. This quantity in the parentheses
I call it as 2 into tensor E defined as half
00:23:53.160 --> 00:24:00.720
of F transpose F minus I. This quantity is
known as Green-Lagrange strain measure.
00:24:00.720 --> 00:24:08.500
Now we can relate this to displacement, gradients
of the displacement. So we have U is x minus
00:24:08.500 --> 00:24:14.930
X that is du by duX is dux by duX minus I
and we have G is equal to F minus I and for
00:24:14.930 --> 00:24:20.720
F if I write now I plus G, I will be able
to get that and upon slight simplification
00:24:20.720 --> 00:24:29.760
I get E as this. Now one of the required that
we stipulated is under rigid body motions
00:24:29.760 --> 00:24:35.020
the strain measure should go to zero. So we
can verify whether that is true here. So I
00:24:35.020 --> 00:24:40.640
again consider rigid body motions as R into
X plus X T of T . So this is translation.
00:24:40.640 --> 00:24:46.520
This is rotation. R is a rotation matrix.
So F is R of T in this case and if I substitute
00:24:46.520 --> 00:24:54.679
that into this I get F transpose is R transpose.
Here it is R transpose R minus I and R transpose
00:24:54.679 --> 00:25:00.950
R is I, therefore, this is 0. So unlike the
infinitesimal strain measures this is 0 for
00:25:00.950 --> 00:25:05.940
any rotation any rigid body translation and
rotation.
00:25:05.940 --> 00:25:12.120
Now how about the other requirement that when
strains are small we should recover the infinitesimal
00:25:12.120 --> 00:25:18.240
strain components. So to be able to do that
we expand this and write in terms of all the
00:25:18.240 --> 00:25:24.610
terms we write in longhand and quantities
that are shown in the red are the nonlinear
00:25:24.610 --> 00:25:30.280
terms and the quantities in black are the
infinitesimal strengths. So for small strains
00:25:30.280 --> 00:25:36.669
you can clearly see that quadratic terms can
be ignored. So we recover back the - all the
00:25:36.669 --> 00:25:41.960
terms in the red vanish and we recover the
infinitesimal strain components. So this is
00:25:41.960 --> 00:25:46.010
- so therefore this definition is acceptable
by the two [Indiscernible] [0:25:46] that
00:25:46.010 --> 00:25:52.190
we stipulated.
Now we can also show that the magnification
00:25:52.190 --> 00:26:11.240
of a line segment that is in the - if there
is a line segment PQ with direction cosines
00:26:11.240 --> 00:26:21.520
N alpha upon deformation if I define a quantity
known as magnification factor as ds by dS
00:26:21.520 --> 00:26:26.539
whole square minus 1 that you can see that
this is nothing but ds minus dS whole square
00:26:26.539 --> 00:26:31.159
divided by dS square and this is defined as
a magnification factor of a line element.
00:26:31.159 --> 00:26:40.429
We can show that in terms of the Green-Lagrange
strain tensor this magnification factor is
00:26:40.429 --> 00:26:46.370
given by this. So clearly here if the line
segment is such that it aligns along with
00:26:46.370 --> 00:26:55.549
X1 axis this is a repeated index imply summation.
So the line segment that is lying along X
00:26:55.549 --> 00:27:02.679
axis is magnified by the quantity E11 and
a line segment which is aligned with X2 axis
00:27:02.679 --> 00:27:10.110
is magnified by E22 and the line segment along
this is magnified by E33. Now similarly if
00:27:10.110 --> 00:27:16.060
you take two line segments which bear an angle
theta before deformation and deform to this
00:27:16.060 --> 00:27:21.120
configuration, we can show that a measure
of shearing strain this again we have discussed
00:27:21.120 --> 00:27:27.590
in the previous one of the previous lectures,
is given by this and the strain E appears
00:27:27.590 --> 00:27:37.430
here. So here again if theta is PI by 2 and
a line segment is aligned with X-axis and
00:27:37.430 --> 00:27:45.230
Y-axis here there are 2 direction cosines
N alpha is direction cosines of PA and M alpha
00:27:45.230 --> 00:27:52.529
direction cosines of PB. So if PA aligns along
one of these axis and PB aligns follow along
00:27:52.529 --> 00:28:01.590
one of this axis then for example epsilon
e12 will be the shearing strain as per this
00:28:01.590 --> 00:28:08.540
definition between these two line segments.
So the Green-Lagrange strain has this interpretation.
00:28:08.540 --> 00:28:19.490
There's another strain measure known as Almansi-Hamel
or Eulerian strain. Here instead of eliminating
00:28:19.490 --> 00:28:32.400
capital dX we eliminate - here in this case
if you see here we obtained the difference
00:28:32.400 --> 00:28:35.929
in square of the length in terms of dX in
the original configuration. A similar equation
00:28:35.929 --> 00:28:42.650
can be derived by using dX in the current
configuration. So that definition takes us
00:28:42.650 --> 00:28:50.650
to the Almansi-Hamel Eulerian strain and this
is defined with notation small e and here
00:28:50.650 --> 00:28:58.220
ds square minus dS square is written in terms
of lowercase dX and we get in this form and
00:28:58.220 --> 00:29:08.320
this quantity I minus F inverse transpose
F inverse is defined as e. This is the Almansi-Hamel
00:29:08.320 --> 00:29:13.130
strain measure.
Here again if you take a rigid body rotation
00:29:13.130 --> 00:29:21.000
we can show that we can first derive the strain
components in terms of displacement and we
00:29:21.000 --> 00:29:26.419
get in terms of J naught matrix is expressed
in terms of J naught matrix as shown here.
00:29:26.419 --> 00:29:33.100
This can be verified.
If you consider now rigid body motions X as
00:29:33.100 --> 00:29:39.020
RX plus XT of T again we can show that E becomes
0 and by expanding the terms we can again
00:29:39.020 --> 00:29:44.210
show that it can be verified that for small
strain the strain measures agree, measures
00:29:44.210 --> 00:29:51.300
agree with results from infinitesimal strains.
Two definitions of strain measure.
00:29:51.300 --> 00:29:56.559
We also talk about what is known as the rate
of deformation. We call capital L as velocity
00:29:56.559 --> 00:30:05.700
gradient where Lij is defined as duvi by duxj.
So that means dvi is Lij dxj. This L matrix
00:30:05.700 --> 00:30:11.900
furthermore we write it as sum of a symmetric
matrix and an anti-symmetric matrix and this
00:30:11.900 --> 00:30:19.670
symmetric component of that is known as rate
of deformation tensor and W is given by this.
00:30:19.670 --> 00:30:25.429
Now if you consider the rate of change of
the line segment ds square rate of change
00:30:25.429 --> 00:30:31.660
of square of the length of infinitesimal line
element if you consider this you can begin
00:30:31.660 --> 00:30:36.250
by noting that ds squared dX1 square plus
dX2 squared plus dX3 square and by writing
00:30:36.250 --> 00:30:43.940
this in this form we will be able to see that
it is 2dxt dx transpose dx by dt.
00:30:43.940 --> 00:30:53.940
And we can rearrange these terms and use this
identity and we can actually show that the
00:30:53.940 --> 00:31:05.130
F matrix and d matrix are related through
this. So this F dot matrix. So this is some
00:31:05.130 --> 00:31:10.360
discussion on rate of deformation.
The relationship between D and derivative
00:31:10.360 --> 00:31:14.721
of the Green-Lagrange tensor can also be derived.
I have indicated the steps here and we can
00:31:14.721 --> 00:31:21.809
show that E dot is F transpose DF. So I leave
it as an exercise for you to verify this.
00:31:21.809 --> 00:31:28.940
Now how about measures of stress? We have
talked about measures of strains. While defining
00:31:28.940 --> 00:31:36.799
stress there are two alternative perspectives.
In the first perspective we think of an internal
00:31:36.799 --> 00:31:42.659
force and an area over which this force acts.
In Cauchy stress tensor with which we are
00:31:42.659 --> 00:31:48.100
all familiar, the force is a deform -- force
is reckoned with respect to deform configuration
00:31:48.100 --> 00:31:58.610
and area is also reckoned with respect to
deform configuration. Now this Cauchy stress
00:31:58.610 --> 00:32:02.330
tensor is difficult to use because beforehand
we will not know the properties of the deformed
00:32:02.330 --> 00:32:07.620
configuration. So that is what makes us to
think of alternative measures of stress. So
00:32:07.620 --> 00:32:12.330
in first The first Piola-Kirchhoff stress
tensor the force is measured with respect
00:32:12.330 --> 00:32:16.470
to a deformed configuration but the area is
transformed back to the undeformed configuration.
00:32:16.470 --> 00:32:23.600
So it is force internal force reckoned with
respect to deformed configuration and expressed
00:32:23.600 --> 00:32:31.080
with respect to area, the distress is expressed
with respect to area in the undeformed configuration.
00:32:31.080 --> 00:32:35.430
In the second Piola-Kirchhoff stress tensor
the force is also transformed back to the
00:32:35.430 --> 00:32:39.720
undeformed configuration. Area is also transformed
back to the undeformed configuration. So this
00:32:39.720 --> 00:32:47.820
is one way of looking at stress but the other
alternative is to look at stress and strain
00:32:47.820 --> 00:32:53.890
measures as conjugate pairs which combine
together to produce an expression for internal
00:32:53.890 --> 00:33:02.880
work done. So in terms of virtual work concept
we have seen that a strain energy stored in
00:33:02.880 --> 00:33:09.260
a body is expressed as product of integral
of a product of stress and strain. So we can
00:33:09.260 --> 00:33:14.630
think of stress as something that is a conjugate
of a strain measure so that along with the
00:33:14.630 --> 00:33:18.610
associated with strain measure it leads to
a proper definition of internal work done
00:33:18.610 --> 00:33:25.090
due to deformation.
Now let's quickly recall the Cauchy stress.
00:33:25.090 --> 00:33:32.390
Definition of Cauchy stress. So we have an
object in the initial configuration body B
00:33:32.390 --> 00:33:38.270
I call it as B naught and this is a coordinate
system and this is acted upon by body forces
00:33:38.270 --> 00:33:41.220
that is forces which are proportional to the
volume and surface tractions which are forces
00:33:41.220 --> 00:33:47.600
which are proportional to the area, and upon
application of these forces the body deforms
00:33:47.600 --> 00:33:51.890
and the process of deformation is opposed
by an internal set of forces set up in the
00:33:51.890 --> 00:33:59.899
body. And that internal set of forces is what
creates stress in the body. To characterize
00:33:59.899 --> 00:34:05.390
that what we do is we consider the body in
the current configuration that is after the
00:34:05.390 --> 00:34:10.730
application of these body forces and surface
tractions the body has deformed and internal
00:34:10.730 --> 00:34:16.760
force system has been developed.
So what I do is I consider an imaginary region
00:34:16.760 --> 00:34:26.690
C and I cut this out from this configuration.
So this picture represents the current configuration
00:34:26.690 --> 00:34:33.889
with inner part of C removed and this configuration,
this figure represents the current configuration
00:34:33.889 --> 00:34:43.159
with outer part removed. So what I do is at
a point P I consider an area element say Delta
00:34:43.159 --> 00:34:54.069
a and that element has a in this figure it
has an outer unit normal n and the internal
00:34:54.069 --> 00:35:01.150
force acting on this Delta a produces a vector
and that is Delta f. It need not coincide
00:35:01.150 --> 00:35:09.589
with the N nor it should be need to be parallel
to the surface area. This force system when
00:35:09.589 --> 00:35:15.900
imagined for this part there is a hole here
and n is a unit outward normal and Delta f
00:35:15.900 --> 00:35:26.049
is the force acting on the elementary area
Delta a at P. Now the fact that such internal
00:35:26.049 --> 00:35:33.720
force system exists is the Cauchy-Euler hypothesis.
So what it says is material occupying the
00:35:33.720 --> 00:35:38.779
interior of C exerts a force field on the
material exterior to C. Similarly material
00:35:38.779 --> 00:35:43.959
exterior to C exerts a force field on material
interior to C. these two force fields are
00:35:43.959 --> 00:35:48.490
equal and opposite. The interaction is free
of any moment. There are no couples. Okay.
00:35:48.490 --> 00:35:55.269
It's only, the Delta f is only a force. There
is no moment there. Okay. This is an assumption
00:35:55.269 --> 00:36:02.019
that we make and under these conditions we
define stress at P I call it as t tilde n
00:36:02.019 --> 00:36:07.599
there is N is a unit outward normal tilde
is a denotes that is a vector. This is limit
00:36:07.599 --> 00:36:13.829
of Delta a going to zero Delta f by Delta
a. Delta a is defined in the current configuration
00:36:13.829 --> 00:36:20.599
and t tilde n is a vector. Normal stresses
components of t tilde n along n and shear
00:36:20.599 --> 00:36:27.420
stress components of t tilde n perpendicular
to n. t tilde n clearly depends on n. That
00:36:27.420 --> 00:36:32.680
means passing through this point P I can select
so many area segments that means this the
00:36:32.680 --> 00:36:38.000
way I have cut this is not the only way. I
can cut it in many ways. So the direction
00:36:38.000 --> 00:36:42.739
of unit outward normal can vary. So passing
through point P I can draw an infinity of
00:36:42.739 --> 00:36:50.819
planes with unit outward normal n and we need
to if you want to define state of stress at
00:36:50.819 --> 00:36:56.089
point P I need should be able to specify what
is t tilde n for any choice of the orientation
00:36:56.089 --> 00:37:02.700
of n. So complete specification of state of
stresses at P requires all these t tilde n
00:37:02.700 --> 00:37:09.700
to be specified for any choice of n.
Now stress analysis is determination of stress
00:37:09.700 --> 00:37:17.250
analysis consists of to determine state of
stress at all points in B. So this looks like
00:37:17.250 --> 00:37:21.781
a tall order at any one point I need some
infinity of vectors and there are infinite
00:37:21.781 --> 00:37:29.269
points in B. So how do we proceed?
So here what we do is we select a cardinal
00:37:29.269 --> 00:37:34.640
coordinate system and erect three planes which
are mutually perpendicular passing through
00:37:34.640 --> 00:37:41.539
P and define the stress vector on these three
planes and knowing that we will be able to
00:37:41.539 --> 00:37:48.630
specify stress on any plane that is inclined
to this Cardinal plane. So according to Cauchy
00:37:48.630 --> 00:37:56.799
stress formula this sigma is the stress tensor.
This is t tilde n is the stress vector with
00:37:56.799 --> 00:37:59.930
the unit outward normal n and this is given
by this.
00:37:59.930 --> 00:38:08.759
So this sigma is a second order tensor and
this is symmetric because there are no interacting
00:38:08.759 --> 00:38:13.769
moments and if you change coordinate system
sigma prime is given by C sigma C transpose
00:38:13.769 --> 00:38:18.770
where C is the transformation matrix and this
I am quickly recalling. I expect that you
00:38:18.770 --> 00:38:23.140
have - this is not the first time you are
hearing about all this. So this leads to the
00:38:23.140 --> 00:38:28.800
concept of principal stresses and principal
axis then stress invariants and we will be
00:38:28.800 --> 00:38:34.700
able to find out maximum normal and shear
stresses and the planes for which they act
00:38:34.700 --> 00:38:40.109
and when writing stress in finite element
formulations as you have seen stress can be
00:38:40.109 --> 00:38:45.440
written either as a three by three matrix
which is symmetric or as a column vector by
00:38:45.440 --> 00:38:51.019
using what is known as white convention. So
we select elements in this order. These diagonals
00:38:51.019 --> 00:38:58.329
this and this and I have sigma 1-1, 2-2, 3-3
then 2-3, 1-3 and 1-2. So this is how we arrange
00:38:58.329 --> 00:39:03.569
the column vector.
Now Cauchy stress is the most natural measure
00:39:03.569 --> 00:39:08.430
of stress because it finds how - it considers
the body in the deformed configuration when
00:39:08.430 --> 00:39:13.380
the internal force system has been set up
and it describes the state of force per unit
00:39:13.380 --> 00:39:20.009
area in some sense. That area is also recorded
with respect to deformed configuration. But
00:39:20.009 --> 00:39:23.540
that itself leads to a certain difficulties
that is it is defined with respect to deform
00:39:23.540 --> 00:39:29.849
geometry which would not be no during the
solution process. So in Lagrangian description
00:39:29.849 --> 00:39:35.420
equations are written with respect to the
known reference configuration. So this - there
00:39:35.420 --> 00:39:42.730
is a contradiction between these two and consequently
treats the requirement that we need alternative
00:39:42.730 --> 00:39:48.690
definitions of stress measures.
So this leads to a couple of definitions for
00:39:48.690 --> 00:39:56.769
stress. The first is known as first Piola-Kirchoff
stress. So here what we do is this is - the
00:39:56.769 --> 00:40:04.450
description is quite similar to what I talked
about. Now this stress is defined what we
00:40:04.450 --> 00:40:12.079
do is this is the unit outward normal here
and this is the force vector and I consider
00:40:12.079 --> 00:40:19.920
this area Delta a I map it back to in the
-- what it would be in the undeformed configuration.
00:40:19.920 --> 00:40:25.400
So I will consider this Delta f and this area
in the undeformed configuration and set up
00:40:25.400 --> 00:40:32.040
a definition for stress. How do I do that?
So we have seen that the rule for transformation
00:40:32.040 --> 00:40:40.829
of areas is dA JF inverse transpose dA and
we have df is t tilde n da t tilde n itself
00:40:40.829 --> 00:40:47.220
is sigma n. These results are known. Now what
I do is we introduce capital T tilde n I call
00:40:47.220 --> 00:40:54.831
it a stress vector acting on element dA. That
is this in this. Okay. It is introduced in
00:40:54.831 --> 00:41:03.960
such a way that it produces the force df that
force is this df. So I have df as T tilde
00:41:03.960 --> 00:41:12.729
n da and this must be equal to capital T tilde
n dA. So this and consistent with this definition
00:41:12.729 --> 00:41:18.549
T tilde n is sigma n I introduce another matrix
P such that T tilde n is P into N where capital
00:41:18.549 --> 00:41:23.869
N is the vector of unit outward normal. So
this quantity capital P is known as first
00:41:23.869 --> 00:41:32.529
Piola-Kirchoff stress tensor. This is a current
force per unit undeformed area. So we need
00:41:32.529 --> 00:41:36.550
not know the deformed configuration to work
with the first Piola-Kirchoff stress. There
00:41:36.550 --> 00:41:45.489
is a problem here and you can relate the the
first Piola-Kirchoff stress to the Cauchy
00:41:45.489 --> 00:41:54.130
stress through this relation using the relation
between you know da and capital dA as shown
00:41:54.130 --> 00:42:00.849
here. If you observe this matrix carefully
we see that P is not symmetric and it has
00:42:00.849 --> 00:42:06.359
9 independent components and when working
with constitutive laws with symmetric strain
00:42:06.359 --> 00:42:13.240
matrices this becomes inconvenient. So this
is not going to be convenient for our modeling
00:42:13.240 --> 00:42:17.119
purposes.
So that leads us to introduction of an another
00:42:17.119 --> 00:42:22.959
stress measure known as second Piola-Kirchoff
stress. Now here what we do is we introduce
00:42:22.959 --> 00:42:29.309
a pseudo force vector fashioned after the
relation dX is F inverse dx. I define DP cap
00:42:29.309 --> 00:42:39.269
as F inverse df. See I have here this df and
I define with respect to the undeformed configuration
00:42:39.269 --> 00:42:46.769
another force vector, see a line segment which
is again a vector gets transformed through
00:42:46.769 --> 00:42:52.259
this relation. So using - this is a vector
and force is also a vector using the same
00:42:52.259 --> 00:43:00.940
transformation I define a force vector DP
cap as F inverse df. So this DP cap is F inverse
00:43:00.940 --> 00:43:12.109
df and using our relations or definition of
df I can write this as F inverse for df I
00:43:12.109 --> 00:43:19.789
will write t tilde n da and again for t tilde
n if I write sigma nda I can rearrange the
00:43:19.789 --> 00:43:26.269
terms and I get a quantity know SndA where
S is given by jf F inverse sigma F inverse
00:43:26.269 --> 00:43:31.859
transpose. This quantity is known as second
Piola-Kirchoff stress tensor. And as you can
00:43:31.859 --> 00:43:37.079
see S will be symmetric here. Sigma is symmetric
and there is a F inverse and F inverse transpose
00:43:37.079 --> 00:43:45.089
coming here. So if you find S transpose of
S it will be same as S. So this is the second
00:43:45.089 --> 00:43:51.160
Piola-Kirchoff stress. So the relationship
between second Piola-Kirchoff stress Cauchy
00:43:51.160 --> 00:43:58.729
stress and the first Piola-Kirchoff stress
is through these three relations. Okay.
00:43:58.729 --> 00:44:10.900
Now to proceed further we need to set up the
physical laws which are expressed as what
00:44:10.900 --> 00:44:14.930
are known as balanced laws that is principle
of conservation of mass, principle of conservation
00:44:14.930 --> 00:44:20.359
of linear momentum, principle of conservation
of angular momentum, and principle of conservation
00:44:20.359 --> 00:44:25.880
of energy. So these form the backbone of our
mathematical formulation of problems of continuum
00:44:25.880 --> 00:44:30.829
mechanics and these basically relate the field
variables like displacement, velocity, acceleration,
00:44:30.829 --> 00:44:37.309
stresses and strains, and to the body geometry
applied surface, tractions and body forces,
00:44:37.309 --> 00:44:45.039
boundary conditions, etc. and they lead to
the governing equations to be solved.
00:44:45.039 --> 00:44:53.670
Now we need to elaborate on that but before
that we can make some observations. When introducing
00:44:53.670 --> 00:45:01.400
the notion of stress I talked about conjugate
pairs of stress and strain. We can show that
00:45:01.400 --> 00:45:06.140
the second Piola-Kirchoff stress tensor and
the Green-Lagrange strain tensor form conjugate
00:45:06.140 --> 00:45:11.290
pairs so that we can compute the strain energy
stored due to deformation using this relation.
00:45:11.290 --> 00:45:17.089
This is S, double dot, E dot dv, where v is
the volume. So W is the internal work done
00:45:17.089 --> 00:45:23.690
per unit time per unit volume in the reference
configuration. So we start with this expression
00:45:23.690 --> 00:45:29.839
such as this and use either principle of virtual
work or variational approaches and we will
00:45:29.839 --> 00:45:35.109
be able to express S and E in terms of the
displacement fields and those displacement
00:45:35.109 --> 00:45:44.989
fields will be interpolated within an element
and we derive the governing structural matrices
00:45:44.989 --> 00:45:51.900
and vectors. So that we need to do.
To begin our discussion what we will do is
00:45:51.900 --> 00:45:57.309
we will focus on linear relationship between
conjugate stress and strain measures. So for
00:45:57.309 --> 00:46:03.690
example we will take that the PK-2 that is
Piola-Kirchoff second stress tensor and the
00:46:03.690 --> 00:46:08.280
Green-Lagrange strain tensor are linearly
related. It's a like Hooke's law between stress
00:46:08.280 --> 00:46:14.479
and strain but the stress is now not the Cauchy
stress and strain is not the Green-Lagrange,
00:46:14.479 --> 00:46:21.049
this Green-Lagrange but stress is second Piola-Kirchoff.
So this is what we will start doing and if
00:46:21.049 --> 00:46:26.170
there is material non-linearity of course
this will be more involved.
00:46:26.170 --> 00:46:35.140
Now in our development of finite element formulations
there are three alternative kinematic descriptions
00:46:35.140 --> 00:46:40.400
that are possible. So they are known as total
Lagrangian approach, updated Lagrangian approach,
00:46:40.400 --> 00:46:44.709
and co-rotational formulation.
So I will just explain what these are then
00:46:44.709 --> 00:46:49.719
we will consider the more details in the following
lectures. In the total Lagrangian approach
00:46:49.719 --> 00:46:54.839
the base and reference configurations coincide
and it is taken to remain fixed and the current
00:46:54.839 --> 00:47:01.029
configuration is this. So we describe the
reference configuration to be the undeformed
00:47:01.029 --> 00:47:04.450
or the initial configuration. This is total
Lagrangian approach.
00:47:04.450 --> 00:47:11.059
In updated Lagrangian approach the reference
configuration is updated at each increment
00:47:11.059 --> 00:47:21.190
of loading while solving the equilibrium equations.
So this is a current configuration and this
00:47:21.190 --> 00:47:25.119
is the reference configuration. This is the
undeformed configuration. So this reference
00:47:25.119 --> 00:47:30.859
state gets updated at every time as the load
is incremented.
00:47:30.859 --> 00:47:37.299
In co-rotational formulation we start with
the base configuration and it is used as a
00:47:37.299 --> 00:47:43.539
reference to measure rotations and co-rotated
configuration is used as a reference to measure
00:47:43.539 --> 00:47:50.700
current stress and strength. So it will - this
this - all these figures are very exaggerated.
00:47:50.700 --> 00:47:55.289
Here the co-rotated configuration undergoes
rigid motion. For example, the CG of these
00:47:55.289 --> 00:48:01.849
two configurations will coincide. So it is
a rigid body motion and then from this rotated
00:48:01.849 --> 00:48:11.749
configuration we characterize a current configuration.
So what we will do is in the following lectures
00:48:11.749 --> 00:48:20.589
we will elaborate on this and try to develop
finite element formulations with whatever
00:48:20.589 --> 00:48:26.930
time that is left probably we will deal with
total Lagrangian formulation for simple line
00:48:26.930 --> 00:48:31.640
elements like bars and beams and I will also
outline how to proceed for two-dimensional
00:48:31.640 --> 00:48:35.849
and other problems. So that we will take up
in the following lectures. So at this stage
00:48:35.849 --> 00:48:37.789
we will close this lecture