zzzz -- fem coursematerials mtech cad cam 1st sem1

7/27/2019 Zzzz -- Fem Coursematerials Mtech Cad Cam 1st Sem1

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Finite Element Method1.

Introduction2. Computational Mechanics Modeling with Finite elementsWhat is Finite Element Method ?

To a mathematician it is something different.To engineers it is a versatile method of computing stresses strains, displacements or

pressures , velocities , temperatures etc. for structural and fluid dynamical problems.

Obviously it solves some equations .

But what are those equations and how are they derived?

This is a very powerful method of solving transformed but equivalent integral forms of

PDEs which are at the base of analysis of engineering systems , processes and their design.But this is not a direct solution method of partial differential equations which governs the

physical problem at hand ( the direct solutions of the PDEs are called the strong solutions.

Finite element method is an integral formulation of the original PDE as we will see verysoon. Whereas the partial differential equations are established at every point of the domain

where the problem is defined, in the finite element method we take those partial differential

equations multiply it with suitable functions then integrate in the whole domain. Then we usesome integral relations such as the Gauss theorems and transform it to a suitable integral

form. We have multiplied the partial differential equations by suitable arbitrary functions.

So this will give us a system of equations which do not depend on these arbitrary functions.

The equations thus formed are solved. This is in essence the finite element method. In the

process of integral transformation we use integration by part and the Gauss theorems etc..The great Hilbert commented that for many physical problems only these integral

formulations make sense. In structural mechanics we are lucky that this integral formulation

is equivalent to the energy formulation and the method is similar to principle of minimizationof total energy of the system. In the principle of virtual energy we get the system of

equations from the total virtual internal deformation energy and the virtual work done by the

external forces in the virtual displacement fields. This equivalent integral form of solution isalso called a weak solution. We will call it , finite element method solution . As it can

be derived from the principle of virtual energy or minimum of energy, it is also equivalent to

the weak solution formulation as we will show very soon. The differentials and their orders

are not the same in the strong solution method and the weak solution method. For second

order coupled partial differential equations the second order partial differentials are involvedin the strong solution methodology whereas as it will be shown soon the weak solution

involves the evaluation of the integrals of the terms containing product of the first derivativesonly.

From mathematical point of view its huge difference.

Some examples of finite element meshes, computations and industrial application of the

method to solve complex engineering problems:


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2.1 The Basic ingredients of Computational Mechanics Modeling

PreliminariesAs we have already mentioned the fundamentals of modern design processes uses knowledge

from various branches of applied mathematics and physics. As a result the subject is highly

applied mathematics and physics oriented. As we are going to deal with the behavior of moreand more sophisticated materials ( Composite materials, Structures having dimensions in the

nano range etc. ) naturally the tools are becoming very much sophisticated and advanced. As a

result of this and also to be able to follow the recent advanced books, research works andmanuals of advanced codes like LS- DYNA, MARC etc. , we have to get our students and

readers conversant with the modern ways of writing scientific texts using tensorial (indicial)

notations. We remind our readers that in the last 40 years lot of results of research works have

trickled down in the mechanical engineering design processes particularly through thecommercial codes like LSDYNA, MARC, ANSYS, NASTRAN and others which have changed

the roles of design engineers in industries completely. To cope with that the readers must also

swim in that stream. In principle we will be using some very elementary part of it. But we canassure you that a deeper knowledge of tensor and differential calculus will only help you

understanding the fundamental principles of mechanics including computational fluid mechanics

( CFM/ CFD) and computational structural mechanics (CSM).


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Indicial Notations and Cartesian Tensor Notation

The main notation used here is the Cartesian tensor notation. This is used to simplify and writeequations in compact form when writing long expressions and is also used in more elaborated

form in general tensor analysis and differential calculus. Tensor notation makes use of subscript

indices (1,2,3 for three dimensional problems ) to represent the Cartesian directions (x, y , z) andrenders summation symbol unnecessary when the same letter subscript occurs twice in a term.Hence , in three dimensions . = + + = + + The derivatives of a function with components are represented as = while the products of two matrices with components and are represented in the form = .Other symbols which will be used in the text are the alternating tensor and the Kroneckerdelta

defined as follows :

= And = The divergence of a vectoru is defined as . u = = and results in a scalar and x u = gives a vector.Some more examples :

2.2 Discret Systems:

The following are the examples of discrete systems :

1)


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A one degree of freedom spring mass and damper system

2)

The equations of motion are:


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3) A two degree of freedom system

Equations of motion are :

2. 3 CONTINUOUS SYSTEMS :

In contrst we show for a continous system in 2D the differential equations which are clearlycoupled partial differential equations:

The equations are for a plain strain problem shown below:


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3. Displacements and DeformationsIn this chapter and in subsequent chapters we will use both conventional notations and indicial

notations to facilitate the learning of the indicial notations.

All bodies are deformable. When they deform under load, each point displaces. If thedisplacements of each point are same we say that the body is undergoing a rigid body translation.

RIGID BODY TRANSLATION ----THE COMPONENTS OF DISPLACEMENTS OF ALL POINTS

ARE EQUAL

If a deformable body of arbitrary geometry is loaded with different kinds of arbitrary loads, the

displacements will be, in general, different at different points. As a consequence of the

variarion of displacement fields the gradients of displacements are not zero , so the stresses willbe build up in the structure due to Hookes law or other material laws.

In figure 2.1 we show the deformation of a body after loading. So the displacements of different

points can be obtained if we know coordinates of points before and after deformation.

Obviously the coordinates of any arbitrary point of the body after deformation is given by the

vector Ox (red vector) with components :

,

, and

and before loading the vectorOX

(pink vector) gives the coordinates of the same point :

and

.

The red vector Ox gives the new position of the old position (vector) of any arbitrary point .Its old position is given by the vectorOX (pink). The displacement vector u of the point is

shown in green colour.

As at different points the displacements will be, in general, different ( except in the case of rigid

body translation) , they will be functions of and .


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From above it follows that : = - (1) = - (2)

=

-

(3)

In indicial notations : = - (4) = + , i=1,..,3 (6)and in conventional notation :

x =X + u (X,Y,Z) (7)

y= Y + v(X,Y,Z) (8)

z = Z + w(X,Y,Y,Z) (9)

Therefore we have x = X + u ( Bold letters designate vectors ) ( or = + ).To simplify the exposition we proceed in conventional notations.

Derivation of Large deformation Green Lagrange Strain Tensor:

The above figure shows a body before and after deformation due to application of loads. Let the

vectordX be a line element in the undeformed body. We can write its components in indicialnotation : , i = 1,..3. After deformation the vector line element takes a new position and let it bedenoted by dx . In index notation we can write its components : , i = 1,..3.Let the square of the lengths of the deformed and undeformed line elements be given by

and

respectively with

= dx . dx =

and

=

For a very small line element we have : dx = d ( X + u) = dX + du.With equations 7, 8 and 9 the X, Y, Z components of the above expression can be written as :

dx = dX + du(X,Y,Z)

dy = dY + dv(X,Y,z)

dz = dZ+ dw(X,Y,Z)


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As all the three above expressions are functions of X,Y and Z , following expressions hold :

dx = dX +

dy = dX + (10 a, b ,c)

dz =

dX +

With equations 7, 8 and 9 and noting that X, Y and Z are independent variables we get :

dx =dX + dX + dy =dY +

dX + (11 a, b, c)dz =dZ +

dX + These equations we can put in matrix vector form :

=

.

(12)

=

. (13)Or , = . . where (14)

=

(15)

=

(16) = (17)and is the identity matrix.Therefore

.

= =

.

. .

.

=

(18)

and . - . = . = . . = . . (19)With equation 17 the above expression becomes : . - . . = .


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(20)

Now let us construct a tensor quantity E called the Green Lagrange large deformation tensoror

in index notation defined by the following expression : - = . . = = = ; i =1, 3; j =1, ..3; with ds =

and d

=

.

(21)

We note that in the third expression i.e in i is a dummy index because itis repeated. So there are summations involved in that expression. In the fourth expression both

i and j are repeated indices. So there are summations over both i and j ; i =1,..3 and j=1,..3;We remind you that as both i and j are dummy indices, we can replace them by any other

indices for example p, q or l . Because all the indices are summation indices , the last two

expressions ( 4th

and 5th

expressions) are scalar quantities as they should be because the

difference of the square of vectors is a scalar.From equations 19 , 20 and 21 we get :

= (

. (22)

Using equation 17 we get :

(23)Or, in index notation = . ) = . ( + + . ) (24)And with the definition of = as given in equation 16 we have : = ( + ) (25)

_________ ____

Linear Part Non linear Part

Here we -have introduced the notations : = (26)The tensor E is called the Green Lagrange large deformation tensor.

When the deformations are small the second term of the Green Lagrange strain tensor whichcontains products of deformations is small compared to the first term and we get the expressions

for the linear strains. = ( + ) ; i , j = 1,.,3. (27)The above expression of Green Lagrange large deformation tensor can also be derived using

index notation in the following manner : - = = ; = 1,..3; m= 1,3;

=(

+

)(

+

)=

+

+

+ Therefore - = + + + = + +


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(But as i is a dummy index in the first and second term we can replace them by any other

index , for example in the first expression we replace i by m and in the second term we replace

i by n.)

= + + ) = 2 where =1/2 ( + + ) or =1/2 ( + + )[Another way to come to the same result is by contracting the indices. + += + +

=

)

+

)

+

( with = = and ) = 2 where =1/2 ( + + ) or =1/2 ( + + )[ Explanation of = The above expression shows the multiplication of the unit matrix

with

i.e.

.

The multiplication process of can be described in the following way:Multiplication of two matricescolumn index: i column index : n

In the multiplication process a summation of the products of the elements of a row of the first

matrix with that of the elements of the columns of the second matrix is involved . The dummyindex here in this case which carries out such a summation is i and is repeated.

Therefore this multiplication will give rise to a matrix with m rows and n columns.

The matrix obtained after multiplication


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Column index : n

= . ]Therefore = . This explains contraction of indices. ]NOTE : similarly = More on large deformation Green Lagrange tensor:

The components of are obtained by simply using the equation (23) : = . ( + + . ) or ( Note: . = in index notation is given by where k, j = 1,.,3 and k is the indexof the rows and j is the index of the columns. In the expression we have

.

i.e . we have a

summation over k and this is equivalent to where is defined in equation 16 : = ;

and =

.The large deformation strain tensor contains two parts : linear part and non linear part.

Thus = + with = = Thus we get the components of the linear part of the large deformation strain tensor : = ==

+

)

+ )= + )And by multiplying with we get the nonlinear part of the large deformation straintensor: = ( + +


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= ( + + ) = ( + + )

+

+

)

+ + ) = + + )By adding the linar and nonlinear parts we get the components of large deformation tensorin conventional notation : = + ( + + + ( + + ) = + ( + + )

=

+

+

+

+

)

= + + + + ) = + + + + )Thus the components of are obtained by simply using the equation (23) : = . ( + + . ) or with thedefinition of given in equation 16 :

=

An alternative way of deriving the large deformation tensors :

Expression for large deformation in one dimension

Let us consider one dimensional case first . In this case dx = dX + dX as there is only the

displacement u in X direction and u is a function of only X i.e u = u(X).

Therefore dx.dxdX.dX = ( dX+ dX ). (dX + dX )dX .dX = dX(2 )dX + dX

. dX

= dX ( 2

+

) dX

= dX 2E dX ( in this case E is a scalar and can be denoted by .Therefore E = = 1/2 ( 2 + ) = + The large deformation expression for the two dimensional case

In two dimensional case we have the displacements u(X,Y) and v(X,Y) and = dx + dy ; = dX+ dY.


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Therefore - = . . = + - - = (dX + dX + ) + (dY + dX + . (dY + dX + ) - -

= + dX. 2 ( + + + dY. 2 ( + - - = dX. 2 (

+ + dY. 2 ( + = dX. 2

+ dX. 2 + dX + 2 + dY + dY. 2 + dX + 2 + dY= dX (2

+ dX + + dY. + dX + )+ dY (dX. + + + + + + dY)= dX ( (2

+ ) dX + )+ dY ( (

+ + ) dX + 2 + + )dY)( We observe that the above scalar is equal to the following product of a vector . Matrix and avector which is also a scalar)

= ,- .

. ,-

The matrix is the large deformation Green Lagrange tensor for the two dimensional case. Wecan derive in the same way, the three dimensional Green Lagrange tensor , but obviously muchmore laborious than the method which we have given at the beginning. This shows also the

usefulness of mathe matics.

The large deformation expresson for the general three dimensional case

In 3 dimension we are not going to derive that same way because it is cumbersome, but if you

follow the same procedure you will get the following equalities if you multiply out all the termsof the following equation

-

=

.

.

=

+

-

-

-

=

(dX + dX + + )} + {(dY + dX + . (dY + dX + )} + {(dZ + dX + . (dZ+

dX + )} - - -


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= dX { (2 + ) dX + +

+ dY {( + + ) dX + 2 + + +

)dY + (

+

+

) }dZ

+dZ { + + + + )dY + 2 + + + )dZ }The above scalar expression is equal to

[

. .

= .2.

where [E] is the large deformation Green Lagrange tensor. Thus we can derive the tensor

without defining the deformation tensor [F] .

3. 2 . TRANSFORMATION OF STRESSES AND FORCES (TENSORS AND VECTORS)

IN DI FF ERENT COORDINATE SYSTEMS.

WHY WE NEED THAT?

1. TRANSFORMATION OF VECTORS IN 2- & 3-D SPACES

Def. An entity is called a vector or a tensor of first order if it has three components , i=1,2,3 inprimed coordinate system and , m=1,2,3 in unprimed system x, y, z (x1, x2 and x3) coordinatesystem.


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PreliminariesGeometrical methodof deriving the direction cosine matrix in the case of 2dimensions:

By rotating a coordinate system along axis by an angle we get anothercoordinate system. We derive the classical formulae of the relations between components of a

vector in the two coordinate systems geometrically [see Fig. 2.1 above] . Let be a vector.Its components in primed system and unprimed systems are ( , ) and ( , , )respectively.

Therefore Therefore

and we know that Hence putting (1), (2) and (3) together we have the following system of equations,


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This can be written in matrix as follows:

Thus for the transformations of the vectors in the plane, the classical direction cosine matrix isgiven as:* +.or in terms of dot products of unit vectors of the two systems :

Let us derive the relation between the components of a vector in the primed and unprimed

coordinate systems by a general method. We will apply then the same method for the relations

between the components of tensors in primed and unprimed coordinate systems.

We know that a vector can be described in both the primed and unprimed systems in thefollowing form:

= Expanding this equation we get: = Dot product with on both sides of equation (5) gives: = From which we get:= The right hand side needs some explanation. As dot product of means projection of on which is equal to the cosine of angle between the two primed and unprimed axes we have : projtion o on projtion o on projtion o on

(Note: The vector doesnt change, but its components are different in different coordinate

systems)

(Note also: The left hand side becomes as = 1 but = 0; = 0.)


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To proceed further let us specialize for the case where the is obtained bygiving a rotation of around third xis that means coincides with . From the figure 2.3we see that axis makes an angle of with and the angle between and is therefore the projection of on is . As the angle between and is

therefore the projection of

on

is

os .

Therefore the equation (7) becomes:= ossinos= os sin Similarly taking the equation (5) and carrying dot product with we get the following: = As in the previous case we obtain the following equation:= rojtion o on rojtion o on rojtion o on From the figure 2.3 we see that X2 axis makes an angle of with X1 and the anglebetween

and

is

Therefore the projection of

on

is

. The angle between

and

is therefore the projection of on i.e. is os .Therefore the equation (10) becomes:= sin os As rotation was carried out around X3 the axis X3 and X3 coincides therefore, Hence we can write in vector matrix form as follows:

os sin sin os

(11)or =

So we get the same equations what we have derived geometrically.In indicial notations the above equation is written as: or equivalently,

Note that the summation index k in the above equation 12a is the component index of unprimed

system.


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Propert ies of matr ix P:

The equation (5) can be written in the following way: = (Note:The above equation is obtained by interchanging the right hand and left hand sides of

equation (5).)Dot product with on both sides of equation (13) gives: = From which we get as in previous cases with projtion o on os, projtion o on os sin and projtion o on os = os sin Similarly dot product with on both sides of equation (13) gives: =

=

sin os

Similarly dot product with on both sides of equation (13) gives: = or Hence we can write the above equations in vector matrix form as follows:

os sin sin os

But we see by examining equation (14) and (11) that

On the other hand the equation (12b) reads . If we left multiply both sides of thisequation by then we get: Therefore Summarizing the above derivations we get:

or

and or where and


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In the case of two dimensions we have: = and

=

Transformation of Tensors in Different Coordinate Systems:

In the previous paragraph we have derived the relations between the components of vectors of

different systems (where one system is rotated with respect to the other) with the help of matricesof P and P

T. Here we derive the relations between the components of tensors of 2

ndorders when

one coordinate system is rotated with respect to the other.

As we represent a vector v by multiplying with unit base vectors of the coordinate system itscomponents (eg. v v ) so we can also represent a general tensor of 2nd order by multiplyingits components with unit base vectors of the coordinate system.

[Definition: An entity is called as a 2nd

order tensor if it has nine components ij, i=1,2,3 and

j=1,2,3 in the unprimed frame and nine components ij in the primed frame and if thesecomponents are related by the characteristic lawij = PikPjlkl (16)]

To prove the equation (16) we proceed as in the cases of transformation of vectors in different

coordinate systems as we have done in the previous paragraph.If we write the same tensor of rank 2 in the two different bases, we obtain: Dot product with on both sides of equation (17) we get: which gives with

=

The first term of the above equation can be contracted giving: Dot product with on both sides of equation (18) we get: From where we obtain: Finally contracting the left hand side we get: or

The inverse transformation is given as: Right multiplying both the sides of equation by P we get: As we get )Now left multiplying both sides of equation (21) by P

Twe get


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As Hence

or Similarly for tensors we have,

or

Example Problems

1) Suppose the components of a vector is given is the old system . We wish to find itscomponent in the new system if the old coordinate system is rotated by around the z(X3 coordinate) coordinate. Find the P matrix and the components of the vector in the

new system.

2) A stress tensor is given in the old system

* +. The coordinate system is

rotated by around the Z or X3 system. Find the components of the stress in the newsystem.3) (Example of Stress Transformation) Stresses are given in old coordinate system. The

coordinate system is rotated by an angle . Find the general expression for the components

of stresses in the primed coordinate system as a function of the component of stresses of

the unprimed coordinate system.

a) Give the components of the General Orthotropic Material Matrix in the primed

coordinate system if its values are given in the unprimed coordinate system.

4) Composite materials :


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X.Y The equations of equilibrium in Cartesian

Coordinates :

The stress vector and its use in the case of balance of forces

One of the most important equation of mechanics is the equation of the stress vector

which relates the stress tensor and the normal vector to the surface element. One of the

major difficulties in introducing advanced notions in mechanics to junior students (at

B.Tech B.S level ) is that the importance of this equation is not emphasized at this level.

We will show here how its introduction simplifies and formalizes many important

concepts of mechanics.

This equation is

=

for i,j = 1,2,3. Here

, j=1,2,3 are the components of the normal vector on

the surface element on which we are trying to find the stress vector components. We

remind you here that stress at a point has six components and if the surface normal is

known at any surface, then only the stress vector shows the direction of the force onthat surface whose components are . We have to say something on thedimension of this equation. This equation must be always applied on a surface with

some area. Then obviously as normal vector components have no dimensions, the left

hand side will have the dimension of a force only when the right hand members are

multiplied by an area. Thats why in some books this equation is always written in the

form

=

, where dS is a surface element.

By expanding we get : + + + Because of its importance let us write also in conventional notations: +

+

+

a body on its exterior surfaces or in the interioron a cut surface.We give some examples and apply this equation for deriving one of the fundamental

equations that is the equilibrium equations.

Let us consider a solid infinitesimal volume of dimension dx dy dz as shown in the

following figure.


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Then the stress vectors on each surface can be written as dS. Now we can definethe surface integral dS which means we are integrating on all the sixsurfaces. By applying now the Gauss formula we get

dS =

dV. But

dS is the sum of all the components of the stress vector . Hence dS = dS. For static equilibrium dS = 0 . Hence dS = dV = 0giving

= 0 . This is the static equation of equilibrium . If you have dynamic bodyforces per unit volume then the surface forces must be balanced by giving

dS -

= 0 giving by applying Gauss law on the first term we

get dV - = 0 which gives = .It is clear that we have derived the equilibrium equations without drawing a figure. But

in most of the books you will find a figure as shown below. Let us explain that figure and

particularly explain the directions of arrows on different surfaces and the balance of

forces. You need the Taylor series development of all the stresses for deriving the

equilibrium equations.


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Let us consider an infinitesimal solid volume of sides dx dy dz as shown in the figure

above.

On the face 1-4-5-8 we have stresses

,

. On the opposite face 2-3-7-6

we have the same stresses but with increments + , + and + . The increments in stresses are due to the position of this face which is at adistance dy from the opposite face 1-4-5-8. On the face 4-3-7-8 ( the stresses are not

shown) we have the stresses , . On the opposite of the face 4-3-7-8 at adistance dx is the face 1-2-6-5. On the face 1-2-6-5 we have the stresses + , + and + . Here also the increments in stresses aredue to the position of this face which is at a distance dx from the opposite face 4-3-7-8.

Then we have the face 1 -2-3-4 where the stresses acting are

and

. Finally

We have on the opposite face 5-6-7-8 which is at a distance dz we have the stresses + , + and + . Here also you must have remarked that theincrement is due to derivatives of the stresses with respect to z multiplied by the

distance of this face from the opposite face i.e. dz.

A remark on the `arrows of the stresses on the faces and the indices of the stresses:

Actually the stresses on the faces which are shown are in reality components of the

stress vectors on the surface concerned. If the normal vector on the surface is in the

direction of the axis the stresses ( both normal and shear stresses) or more correctly

the stress vector components directions are also to the positive axis direction. If the

normal vector is in the direction of the negative axis direction, the stresses( the stress

vector components ) on that surface will be also in the negative direction. This is also

true for plates, any cut surfaces or external surfaces.

X.Z CONSTITUTIVE EQUATIONS

Whenever strains are developed in a body, because of material laws stresses will be

also developed in the body. For small deformation Hookes Law gives the different

components of stresses. In 3 dimension following relations hold good: = . This is the most general case for a material. The material in general containssymmetries. The coefficients. But as the stress and strain tensors aresymmetrical,we have = and = = and as = Therefore the = and = .


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We note also that for isotropic materials must be also an isotropic tensor. This means that thisfourth order tensor can be constructed from other isotropic tensors. Since is isotropic is alsoisotropic. Other fourth order isotropic tensors are : + and - . ThereforeWe can assume that :

=

+

+

+

(

-

. But because of

=

and

=

we have = + + . For the proof of this assertion we send the reader to Fung ( ).Putting this relation in = = + + we have = + + = + or = + . Therefore = + . = + . From this we have = ; Putting this in = + we have + = + . From this we get ) = .This is more commonly written as = - + = - where E = and = . Here we remind you that the shear modulus G and G= E/(2(1+ .We get from = - the different components of the strain tensors : = - = - + + ) = - + ) = - + ) = - + ) = - + ) = = =

The relation between

and E and

are the following :

=

and

=

.

We derive now the bulk modulus of compressibility :

= - . (a) Suppose we have hydrostatic compression (tension). = = - p = - + ) = p ( -2 ) =- = - + ) = - = - + ) =- Therefore = -p 3( ). Thus p =- ( ). =- .The coefficient k is called bulk modulus of elasticity.

Multiplying (a) by we have : = - = - 3 = . Therefore = . Putting this in (a)we get = - = - or + = or + = or +


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For isotropic material the elasticity tensoris the same in all coordinate frames and therefore thestructure of the Hookes law remains the same in any orthogonal coordinate system.

Therefore for the cylindrical coordinate we have

= [

]

with G =

.

For the spherical coordinate we have

=

X.Y1 LINEAR CONSTI TUTI VE EQUATI ONS FOR 3D , PLANE STRESS AND PLANE

STRESS HYPOTHESIS

3 Dimensional Constitutive Equation :

{

} =[

]


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The inverted relation is :

{ }

=

[

]

PLAIN STRAIN:

If we take the above relation and neglect the stresses , and , neglect the third equationinclude the 4

thequation and invert the matrix we get the plain stress-strain relations:

= . But 0 = - PLAIN STRESS :Similarly if we consider the first two equations and the 4

thequation of the 3D stress strain

relation above ( eqn. )and neglect and we get the plain strain stress reation :

[

]

ut 0 = .Discussion on one dimensional stress, plane stress and plane strain:One dimensional deformation:

We know from our experience that when we load a rod of circular or square cross section in

axial direction, its diameter or side contracts or increases. It is evident from the conservation

of mass or from the presence of Poissons ratio that if the rod elongates its radius or side mustbe reduced. But in one dimensional theory we neglect them because they are small. Another

observation that we wish to make is that for a rod all the points in the cross section will be

having the same displacements deformations and stresses. So what is the use of keeping all thosepoints in the cross section ? This is the reason why at undergraduate level we replace the cross

section by the central point and the rod becomes a line obtained by joining all the mid points of

the sections.


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Plane strain:

Classically plane strain is introduced in the following manner:

Consider an infinitely long cylindrical body shown in the figure below.

If the body forces and the tractions on the lateral boundaries are independent of z, then the

displacements can be taken approximately in the form u(x,y), v(x,y) and w=0. We emphasize

here the word approximately because as it will be clear soon does not satisfy the tractionfree boundary conditions at left and right ends. Because of Hookes law . In theliterature , (in this case a text book by Sod from which I have taken the figure ) admits that in the body . But the two surfaces at the end are stress free hence This means must go tozero there from its constant value inside the body. This means near the end surfaces in the body

This means according to the third equation of equilibrium

0 or

0. But as

the order of magnitudes of n 1st and the 2ndequation equilibrium both must be non zero at the two ends so that the in planestresses n must be modified to accommodate the stresses to satisfy the equation of equilibriums. In the interior of the body if , i. e. if does not vary with z then the third equation is satisfied if So in reality there is 3D a state of stress and strain at the two ends eventhough in the interior the situation may correspond to plane strain condition so long they satisfythe other two equation of equilibriums in strong or weak sense. So you have a boundary layer

here and matched asymptotic solution can be obtained by putting the boundary condition of the

inner i.e interior part of the body. We give below the following information from Sods book

which is correct for the interior part of the body:


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Plane Stress:The figure above is a thin structure bounded by 2 planes z = . Its thickness is small,therefore no appreciable variation of stresses will be in the thickness direction. The theory

further assumes that these planes are stress free i.e. = = = 0

The plane stress problem implies that are in general functions of x and y and as

is obtained through the constitutive equation ( Hookes law) at every point of the plate, so

w must be a function of x,y and z i.e. w = w(x, y, z) as x y . ( Integrating this equation we get w= w(x, y, z) ) . Therefore arenot zero. But there is no reason why

will be equal to or will be equal to . ( Rightnow we have

and because of assumptions = 0 and = 0; ). But as we see thatthese assumptions are not correct. But then = ; = = .Therefore and are linear in z. Therefore = G 0 , and = G 0 .Therfore we can put these quantities in the first two equilibrium equations and get a better

solution of u and v . Then taking the third equation of equilibrium, as

it ontins and

we get a solution of by integrating in z. This solution will bequadratic in z as and are linear functions of z. Then taking the three dimensionalelasticity constitutive equation we see that , and as well as , and on z . This is exactly what is given by Timoshenko andGoddier ( P 276).


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We give below the approximate equations for the problem as given by Saad :

X.Y SLENDER STRUCTURES : Their Geometries, Knematics and other

considerwations :

Structural elements with one dimensional stress , plain stressses and plain strains

(The following material is taken from Bathes Finite Element Procedure. The author is

thankful to him. This portion will be reworked and put in the proper context )


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( Figures above and below are from Bathe, F.E.PROCEDURES)


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X. Y STI FFNESS MATRIX CONSTRUCTION OF DI FF ERENT STRUCTURAL FI NI TE

ELEMENTS

WHAT STRESSES AND STRAINS SHOULD BE INCLUDED AND WHAT TYPE OF

DISPLACEMENT FUNCTIONS SHOULD BE USED IN THE DIFFERENT DISPLACEMENTBASED FINITE ELEMENTS ? WHY NOT STRAIN, STRESS AND DISPLACEMENT BASED

FINITE ELEMENTS?

Exact solution of a beam under uniform pressure and the exact solution of a thin plate underuniform pressure give us the order of magnitude of different stresses in these slender bodies.

Exact solution of thick and moderately thick plates and beams confirm these results.

We first give exact solutions of thin beams and plates. Then based on the order of magnitude of

stresses of slender bodies we give in the following table the stresses and strains retained inslender structural Finite Elements. We justify also the displacement functions used in the

construction of the different finite elements.


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5.Exact solutions of some impor tant slender structures

5. 1 Exact Solution of thin beam under uni form pressure loading

Following are the main results of the exact solution of a plane 2D beam ( figure ) under

uniform pressure loading. This problem corresponds to a beam under uniform pressure loading (eq.3 ) and is supported at the ends such that the integral of the stresses i.e. forces in the axial

direction on the end surfaces at the support ends are zero. The moments there are also zero(eqns. 4 and 5 ). While solving stress function is used. A polynomial of 5th

order satisfies allthe boundary and loading conditions. The beam is in the x-z plane. On the top surface we have a

distributed pressure w. On the top and bottom surface there are no force in the x direction

(eq.1). As a result these two surfaces are shear stress free. At the lower surface of the beamthere is no normal force i.e. pressure force. (eq. 2). ( Check these by applying the equation of

the stress vector = on these surfaces. Find the three components of the normal vectoron the top and the bottom surfaces and apply the equation for the stress vector. On the top

surface only the pressure load is acting. There is no force in the x direction. At the bottom

surface no forces are acting in x or z direction.) The bottom surface is force free. This problemcorresponds to an one dimensional beam simply supported at its ends that you might have

already learned in your strength of material course . Let the stress function be given by :

By satisfying the following equations below the free constants of the stress function will be

found. The boundary conditions on stresses at the top and the bottom surfaces are (first threeequations ) : (x)=0 (1)(x)=0 (2)(x)= p (3) (4)

(5) (6)The integrals on the stresses gives the forces at the supported edges (the equation 4 above ).

Integral of stresses multiplied by y gives the moments. The 5th

equation above shows that the


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moments at both ends are zero. The 6th

equation above shows that the forces at the support ends

can be calculated from the reaction forces and this must be equal to the integral of shear stresses

at both support ends.

Deriving the stress function twice with respect to z we get

, by deriving twice with respect to x we get and by deriving first w.r.t. x and then w.r.t.z we get still contain the unknown coefficients which will bedetermined by applying the boundary conditions: = 6z + 6 (2/3 = 2 + 2 + 2 = -2x- 6 Applying the boundary conditions we get the unknown coefficients we determine the unknown

coefficients :

= - p/4, = 3p/(4h); = -p/(); = - (-(2/5) )= ( - )Therefore the stresses are given by :

= - (z-

=

{ ( 1/3)

-

+ 2/3

)

= -( p/2I) ( ) xwhere I = These equations show that p. / p .If we put p = 1N then we have simply :

In our notations this gives and


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where L and h are respectively the length and the height of the beam.

Exercise( for teachers) or advanced students going for research:

Q1. What are the displacements u and v for the above problem.

5.2Exact solution of thin Kirchhoffs plates under uniform pressure loading

For completeness we derive the equilibrium equations of a thin plate. For this we take the

equilibrium equations already derived in section II.1. The equilibrium equations are given in

the absence of any volume forces by =0. In conventional notations we have:

+

+

=0

+ + =0 + + =0The coordinate system and the plate is shown in the following figure. The above equilibrium

equations are valid everywhere in the plate. In the past there was a tendency to see everything in

integrated form that is to establish the equations of equilibrium of plates in terms of moments.

This was particularly true before the digital age. But we need not to sacrifice anythingparticularly the local behavior of the solution of the stresses and particularly all the stresses we

want to compute and assess the order of magnitudes of different stresses and their distribution

in the thickness. Our intention is not to present you what the researchers have published for theirpersonal publications but how the subject need to be treated in to-days perspective taking into

account fully the advantage of the digital computation.

Let us derive the equilibrium equations in terms of moments as you will find in other texts. From

the above 3 equations we can derive the equilibrium equations in terms of moments. As in thecase of beams here also because of the slenderness of the body and the direction of the load

vector ( transverse to the plate plane) the stresses develop in such a manner that the resultant

forces in the thickness of the plate in x and y direction are zero on all the sides i.e. for exampleon the four sides of the plate we have

But thestresses in the thickness in the x and y direction above and below the mid surface particularly

at the top and bottom of the plate are much higher than the stress that is the pressure on the top

surface which is generating all other stresses. In fact the pressure stress which is generating all

the stresses is much smaller than the stresses , or at the top or bottom surfaces.But if you consider the plate globally ( integrally) then obviously as the external forces are in z

direction and therefore the reactions at the points where the plate is held fixed at the boundary


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or any other point where the displacements are restrained must be also only in the z direction

and there will be no forces( we are talking of resultant forces on the lateral surfaces and not

stresses ) in the x and y direction, so they must be zero. This is caused by the slenderness of the

body and this is the reason why the plate theories are a bit difficult to understand.

If we integrate the first two equations in z and satisfy the boundary conditions on

and

, then we get zero forces in the x and y directions. This is correct because in the x and ydirection there are no external forces. So every where we get zero forces in x and y direction.This is what we have explained above.Now we will multiply the first two equilibrium equations by z and then integrate in z .

Noting that for the thin plates we have

=

and=

=-[

]

and =

=[D]. ( ).where [D]=

[

]

.

The first two equilibrium equations multiplied by z and integrated in the thickness gives : + + = 0 + + = 0The third term can be integrated using the formula of products of functions using = - .Let us set u= z and

dz = dv. Therefore

=( z

dz. But

the first term is zero because at +h/2 andh/2 is zero and second term is the shear force .Hence + + = +-= 0 ( )Similarly if we do the same operations with the second equation we get : +- = 0 .( )


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Now let us take the third equation of equilibrium : + + =0.````Integrating this equation in z we get :

+

+

=0.

The above equation gives :

+ + - =0. But as and = -p we

have with = and = : + - .Putting equations for and in the last equation we get : + 2+ = pWe can replace here

by the curvatures ( equation ( ) ) giving finally :

We note that =- D(1- ; = -D ; = -D ;Therefore + 2 + - p= -D -D -2 D(1- -p= D(- - -2 (1- p= D( + p =0Therefore D ( + p. We remark here that the sign of the force onthe right hand side is due to the direction of the pressure which is in the negative z direction.

The simply supported thin plate under uni form pressure loading

Approximate solution ( strong solution):

Let us consider a simply supported square plate of side a and thickness h under uniformly

distributed load p on the top surface.


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This problem can be solved by developing the solution in Fourier series (Ref. 4.

Chandrashekharan.)

For the details of the derivation we send the reader to Ref. 4. Here we give only the main results.

The maximum which is found at the center is : = .287 p Here we see that as a = L )But Therefore ) = )Similarly ) = )This equation shows that

= p. L/h

On the otherhand for we have :

with

and D = we have

= (6p/ - )At z =- h/2 we have = pAssessment of order of magnitudes of stresses:From the above analysis we find that as in the case of beam here also we have if we put p =1N/ .With p =1N/ this gives and

Based on the above analysis of beams and plates we summarize in the following table thestresses and strains retained in the construction of different finite elements. The stresses which

are very small or are small are neglected.


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5.3 Stresses and strains retained in the fini te element modeling of slender and other 3D bodies

Definition : Slender Bodies Bodies whose one or two dimensions are smaller thanothers/other. Ex : Thin Plates ;Thick Plates; Thick Beams ; Thin Beams ; Rods, Membraneelements etc.

Following table gives the stresses and strains retained and neglected as well as the displacementfunctions used in different structural elements :

Structural

Element

s

Stresses and Strainsretained

Stressesand

Strains

neglected

Displacements Loadsvalidity

Validity

Rods Remainin

g stressesand

strains

u=u(x);

Displacement onlyin the direction of

the axis of the rod

In the

axialdirection

of the rod

In cross

sectionsno

gradients

allowed.

ThinBeams = = -zRemaining stressesand

strains

u= -z; ww(z); but w=w(x); = 0 = u= -z

In thetransversal

direction

of thebeam;

very thinbeam

ThickBeams

=

= z

=

Remaining stresses

and

strains

;w w(z); butw=w(x);

0 ;Therefore u = z.

In thetransvers

al

directionof thebeam.

thick tomoderatel

y thick

beams

ThinPlates

=

Remaining stresses

and

strains

;Therefore w w(z);but w=w(x,y); = 0 =

u(x,y,z)= -z; = 0 = v(x,y,z)= -

In thetransvers

al

directionof the

plate;

very thinplates


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zThick

plates

,

,

Remainin

g stresses

andstrains

;

Therefore w

w(z);

but w=w(x,y) ; 0 ;Therefore u = z. 0 ;Therefore v = z.

In the

transvers

aldirection

of theplate;

thick to

moderatel

y plates

SolidElements , , Allstressesand

strains

are

considered

u(x,y,z), v(x,y,z)and w(x,y,z) or inisoparametric

formulation

u(), v( )and w( )

In all thedirections

Allstructuralmembers

apart

from

thosementione

d above

6.1 GENERAL DISPLACEMENT BASED FINITE ELEMENT FORMULATIONFOR STATIC PROBLEMS

Most common methodologies of the displacement based finite elements are based on theprinciples of virtual displacement or work method and the principle of minimum of potential

energy method. But obviously both give same system of equations. Let us give both the

formulations:

A) Principle of virtual work postulates that the following expression holds good:

= + , - + ..(A)where

represent the stresses in equilibrium with applied loads ,

the virtual

strain and

, V is the volume of the body and

is the surface where surface forces are applied, are the virtual displacements atconcentrated loads .


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6.2. GENERAL DISPLACEMENT BASED FINITE ELEMENT FORMULATION

FOR DYNAMIC PROBLEMS

INTRODUCTION OF MASS MATRICES

If the loads are applied rapidely, inertial forces need to be considered; i.e. a truly dynamic

problem needs to be solved. Using dAlemberts principle, we can simply include the elementinertia forces as a part of the body forces. Assuming that the element accelerations areapproximated in the same way as the element displacements the total load due to inertial forces

can be constructed easily.

Using dAlemberts force as a body force in the above equation and finding the virtual workdone by this body force (which is in this case and integrating in the volume of theelement after introducing the interpolation functions (or= ) we get : - dV = - As this is an external force it is on the right hand side. Bringing this term on th right hand side

we get along with the stiffness matrix the following equation for the energy: (t) + = where F(t) is the time dependent load vector.The matrix M is the mass matrix and it depends on the displacement interpolation functions or

the rotation functions chosen for the finite element construction of the rods, membrane

elements , thin and thick beams and plates or solid.

6.3 Weak solutions or strong solutions? What do they mean?

Let us first show that solving the above equation by the above energy formulation is equivalent to solving the partial

differential equations which are obtained from the equilibrium equations with the given boundary conditions of the

problem.Obviously once more we will be using Gausss theorem to transform the problem at hand. We show that solving the

coupled partial differential equations are same as solving by the energy principles. Let us suppose that we have

surface forces acting on the body on a boundary part and there is a body force acting on the body. We know that the

equilibrium equation that must be satisfied in the body is : + (***)where is a body force. On the surface a surface traction is acting . The essential

displacement boundary conditions are : on . Here S= . We note that for this case =0 and are the components of the unit normal on the surface S of the body. Consider any continuousdisplacement field

= 0 on

. Let us form the product(

+

. Obviously (

+

= 0. We can call displacemenst. We know that ( = + and from Gauss theorem ( dV = dS.Therefore = = =


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Therefore dS = Therefore = dS - As is zero on that part of the surface where the displacements are imposed i.e.

dS =

dS=

dS= dS. Therefore = dS - .Now also considering the external forces per unit volume we get finally + = dS - + = 0 . Therefore dS + = dS+ . The term on the left hand side can be shown to be equal to . That is = . This follows from the definition of the strain and = ( + )andfrom the fact that In that expression both i and j are dummy indices. This is the equation we havegiven in (A) above ( if we remove the concentrated forces).

Therefore finally we get = dS + = dS+ . . (**)Reader who have been attentive must have remarked that the above energy correspond to small deformation

formulation as the volume considered is the initial volume before deformation and expression of the deformations

correspond to small deformation expressions. We will show latter in the chapter of large deformations after

introducing the 2nd Piola Kirchhoff and Cauchy stresses what form the equilibrium equations take if large

deformation is considered.

You have probably noted one remarkable aspect of this integral transformation using Gausss theorem. The

expression (**) contains only first derivatives of i.e which are contained in as well as in as components can be expressed in coefficients of Hookes matrix and the components of

.

Although the products of the terms and are involved, as is arbitrary finally after discretisation ( say afterintroducing interpolation functions for then differentiating the displacements and integrating in thewhole volume ) we get easily a system of equation for the unknown nodal variables which are in this casedisplacements. But the key point is that in this energy formulation which is called also a weak formulation only

first derivatives are concerned while the original 2nd order coupled partial differential equations we hve the second

derivatives of the displacements. Direct solution of these coupled 2nd order PDEs with boundary conditions are

called strong solutions. From mathematical stand point strong solutions of PDEs is a very active area of research. In

the early days many strong solutions had been obtained by difference schemes where the 2 nd and first derivatives

were replace by the difference schemes and thus the solution was obtained by solving a system of equations.

However if the boundary is curved implementation of boundary conditions are not straight forward. And also from

mathematical side the convergence properties were not available. The convergence properties of the weak solutions

procedures had been on the other hand investigated in the frame work of applying modern mathematics and the

convergence and other properties were obtained which contributes to the robustness of the method. As the boundary

conditions and the implementation of the theory are relatively easy to program many commercial codes came outand the method became very popular in the industries where complex problems are solved routinely thus reducing

the design time.

6.4 ISOPARAMETRIC ELEMENTS

The isoparametric element construction technique is one of the most successful general

techniques in finite element analysis. Several elements can be constructed using this technique.For example the stiffness matrices of quadrilateral elements with nonrectangular shapes can be

easily constructed using this technique. Same comment holds good for solid 8 noded brick


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element. This is a versatile technique and can be easily programmed. Many plate elements and

solid elements which are implemented in different codes are isoparametric elements. This is the

reason why in the following short chapter we will give a brief description of the basic steps usedin this technique.

The principle idea of the isoparametric element construction technique is to achieve the

relationship between the element displacements at any point and the element nodal pointdisplacements directly through the use of interpolatyion functions ( also called the shapefunctions). We have already used the technique while constructing the plane strain quadrilateral

elements (Q4 elements). Here we will reconsider them oncemore in a more systematic way and

will construct the thick plate elements, hexahedral elements etc. We reintroduce the rod elementconstruction which shows all the basic steps used in this technique.

To keep the presentation simple we will not put the dependence of the time variable in the

construction of the stiffness matrices . But whenever they will be used in the context of

dynamics we must take into consideration the time dependence and this will be for linear casesmanifested in the nodal variables. They will be functions of time as the decompositions

(or =

) is

then valid.

6.4.1 SLENDER BODIES

6.4.1.1 ISOPARAMETRIC CONSTRUCTION OF THE ROD ELEMENT:

Stresses and strains retained and neglected in rod elements

Step One : Stresses and strains RetainedThe general 3- dimensional components of virtual strains and stresses are given by :

and the six components of stresses are :

.

A ro is slnr boy hving two imnsions i its with n hight or its imtrris muh smllr thn th xil lngth n it is lo only in its xil irtionuniformly. There is no gradient of axial displacements in the directions y and z. The

loading in the axial directions do not have any gradient in the y and z direction. There is no

other displacement other than u which is a function of x. This is the reason why except

and the virtual strain all other stress and strain components areneglected in the virtual energy term of a rod.


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This means in rods the components

and

are neglected as it is an one

dimensional object kinematically although it has a 3D geometry.

Step Two: Kinematic relation and the constitutive equationWith the above approximations we have (x) = u(x) .Small deformation theory gives = n thn with pproprit Hooks lw w hv = E .Step three : Geometry

One dimensional geometry: As u is only a function of x and no other displacements areinvolv its D gomtry rus to lin or ll kinmti sriptions This is th rsonwhy in Strength of Material Courses we generally introduce this object as a line.Transformation of the geometry from x space to space : The domain of the element istransformed to space by the transformation x = ( ; i =1..n; n=2 ( ( . Here and are the coordinates of the two end nods of the rodand () and are the two interpolation functions with ( (

Figure 1a The two nod element is transformed from x space to space.Step Four

The Hookes law for this one dimensional case is given by : = E , where E is theelasticity modulus.

Step Five

The virtual Energy in the element is given by :


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For this one dimensional case we have the internal virtual energy in the bar = = = where is the length of theelement and A its cross section.

Step Six : Construction of B matrix

We know that = ; Introducing ( ( ( we have = = : = = = = + ) I * + = * + =

with

=

*

+

=

*

+ *

+

and

;

Determination of : As x = ( ; i =1..n; n=2; x = ( ( ;Therefore = ( ( ) = =( = /2 ; There fore 2/Therefore = * + .Evaluation of B matrix : On the otherhand as = = -; and = = ; wehave finally

=

2/

=

Step Six : Construction of K matrix : The virtual energy of deformation is : A x . By introducing = we have A x = A x A x = A x A = A = EA x = EA * + * + x = EA * + * + x = x

= where


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= x . After integrating in x the stiffness matrix for therod element is :

= * +and therefore the virtual energy A x = A x is : * + [ Derivation with isoparametric technique: As is not a function of we have thevirtual energy = A

= /2 = A

= * + = * + Therefore as the above virtual energy must be for all virtual displacements , thestiffness matrix for the bar element or rod element is :

* + ]Step 8 . THE MASS MATRIX OF THE ROD ELEMENT :

We have already given the procedure for the construction of mass matrices. The work

done by the inertial forces ( dAlemberts principle) is :

dV . ( ( ( (t)we get : dV = t t ) t t ) d the following expression for thew A * +

= [M] with [M] =

A * +


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6.4.1.2. Thick Beam Element (MINDLINS Thick Beam/ Barnoulli /Euler Beam)

Deformation of thick beams:

Step one : This also is an example from slender structures. Here the beam thickness is smaller

than the other dimension that is its length. Here we will consider a beam only in bending.

Once more the order of magnitudes of different stresses and strains shows that areneglected. are not present as in one dimensional beam theory all thederivatives in y are zero, nothing depends on y and the load does not vary in y. However Later we will show that if wewish to develop the 3 D beam , we have to take them into consideration. At present we neglect

them . the strains and stresses retained in this structural element are:

{

}


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.

Step two : Kinematics = = 0 , particularly = 0 , imposes on the displacementfield w a special and important condition. Applying the definition of we get = = 0and this gives the following fundamental nature of the displacement field w :

W= w(x); So the transversal displacement must be only a function of x only and not of z. That

is w does not depend on z, for if we derive w with respect to z we have i.e. = w(x)= 0 .

Here we assume that the displacements u for bending is : u = z

(x). This corresponds to the

rotations of the normals to the mid-surface with respect to y axis by

For thick beam

problem the dependence of displacement u on z thus becomes explicit and does not depend onw.

The Constitutive Equation :For thick beam bending we take the following constitutive equation for the relations between the and , and the shear stress and the shear strain :

with G = Third Step : The Geometry

The beam element is shown above figure.

Fourth step :

The virtual Energy in the element is given by :

with


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and the two components of stresses are :

.

We have w = w(x) , u = z (x)The virtual energy can be written in the form taking into consideration the above

splitting of strains and strains in bending and shear : = + Introducing w =

iW

iN () , u = iN () we have :

therefore the virtual bending energy :

E

dV

As we can integrate the term the virtual bending energy the followingexpression : x = EIb x with I = /12 and b thewidth of the beam.

Now observing the above expression we conclude that it is of the same type as the rod

energy except that we must replace area EA by inertia EI b and the degrees of

freedom by .Therefore the thick beam bending energy is given by : * + with = .Therefore the thick beam bending stiffness matrix is given by : = * +On the other hand the virtual shear energy is given by :

G ) ) dV = G ) dV =G + G dv + .


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is of the same type as the rod virtual internal energy and hence wecan directly use the expression found for it by replacing by =

*+

and E by G as the height, width and the length of this element are same as the

rod element . * + = * + *+Similarly the second integral G dv is of the same type as the work done bydAlemberts force of the rod element. Hence we replace the acceleration vector by

, and

by

in th bov xprssion fordAlemberts force we

get the following expression

A * + for G dv .Remaining terms will be obtained from integrating + . = G * + v =

*

+

v

= x

As =-1/L, = 1/L , = 1-x/L and = x/L we have

x

= x


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54/76

=

[

]

where [ ] =GA

[

]

= [ ]

By adding the bending stiffness matrix and the shear stiffness matrix we get the

stiffness matrix for the thick beam element. Thus assembling the bending part we getfor the thick beam stiffness matrix :

=

We would request the reader to refer Bathe, F.E. Procedures , Page where also this

stiffness matrix derived.

THI CK BEAM MASS MATRIX:

The mass matrix construction is straight forward. Like the rod element mass matrix we constructthe Mass matrix from the work done due to dAlemberts force:


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dV = wt t [M] =

and [M] = with [] = dx and[] = I dx.( See Bathe p 402)

6.4.1.3. Mindlins Thick Plate ElementDeformation of the Mindlins thick plate :

A thick beam supported at points A, B, C, D is shown in the figure below. The plate is

loaded for example with uniform load on the upper surface.

-

Deformation of the thick plate `in xz plane :


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Deformation of the thick plate in yz plane :

Step one : This also is an example from slender structures. Here the plate thickness is smallerthan the two other dimensions that is its width and length. Here we will consider a plate only in

bending. Once more the order of magnitudes of different stresses and strains shows that only

are neglected. The strains and stresses retained in this structural element are:


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{

}.

Step two : Kinematics = = 0 , particularly = 0 , imposes on the displacementfield w a special and important condition. Applying the definition of we get = = 0and this gives the following fundamental nature of the displacement field w :

W= w(x,y); So the transversal displacement must be only a function of x and y only and not of

z. That is w does not depend on z, for if we derive w with respect to z we have i.e. =

(w(x,y)) = 0 .

Here we assume that the displacements u and v for bending are : u = z v = z . Thiscorresponds to the rotations of the normals to the mid-surface with respect to y and x axis by and For thick plate problem the dependence of displacements on z thusbecomes explicit.

The Constitutive Equation :

Beacause of the constraint = = 0 the constitutive equation will be modified. For platebending we take the following constitutive equation for the relations between the components and , =1,..n ; n=2; and for transverse shear stresses and strains ( between

and

as well as

and

) :

,- * + ,- with G =


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Third Step : The GeometryBecause of the above kinematic considerations w = w(x,y) and if we take for the displacements

u and v for bending as u = z (x,y); v = z which corresponds to the rotations ofthe normals to the mid-surface with respect to y and x axis by and , forthick plate problem the dependence of displacements u and v on z becomes explicit. This is

definitely a great advantage, because in the expression for the virtual energy where a volumeintegration must be carried out ,we can then integrate in z direction analytically. As a result only

mid surface in the x-y need to be represented. Such a surface in x-y plane representing an

element is shown below. In the Isoparametric technique we will be using the same interpolation

functions for transforming the element from x-y space to space and to introduce thefunctions u( and v( in the space. We remind you that the mapping x-y space to is invertible. This means that x = x ( (x,y) . This means also that we can find u ( = u ( and particularly , , and by using chain rules.Derivation of interpolation functions :The Interpolation Functions . What they are ? How they are derived ?

Consider the mapping which transforms the quadrilateral from x-y plane to a square in the

space. (See the above figure).

Let us introduce x = 1a + + y = ;Taking the equation for x we derive the interpolation functions in the following manner. Weapply them the also for y.

We have x = 1a + + . Satisfying the , coordinates at the corner nodes weget the equations for determining 1a , 2a , 3a and . Introducing these values of thecoefficients in the equation for x and after rearranging we can write


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x = 1X . 1N ( ), + 2X . 2N ( ), + 3X . 3N ( ), + 4X . 4N ( ), .

To prove the above assertion we proceed in the following way:

For example the coordinates of the nod 1 in the

space are (-1,-1) (see the figure above).

Putting these values in the above equation for the four nods we get :

= 1a -1 + 3a . -1+ .1 = - = 1a .1 + 3a -1+ .-1 = + = 1a .1 +. 1+.1 = + = 1a .-1 +. 1+ -1 = - These equations we can write in vector matrix form :

=

; i=1..,4; j = 1,..4; with

= ; and = The matrix be easily inverted because it is orthogonal ( the scalar product of itsdifferent columns are zero). .

Its inverse is given by :

=1/4 ;Therefore ( + ) ( + ) ( + )

(

+

)

With this expressions of, , we havex =

( + ) + ( + ) + ( + ) + ( + ) = ( 1- + ( 1- + + ( 1- + ( 1- + = iX iN (, ) ; i = 1,.. n ; n=4 ; with iN (, ) = (1+ )(1+ (no sum on i on the right hand side) where (,


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nod. Similarly we have y = iY iN (, ) ; i = 1,.., n ; n = 4 ;with

iN (, ) = (1+ )(1+

Thus we have derived the interpolation functions for this 4 nod quadrilateral.

For the reason described above, the geometry of the plate degenerates to a midsurfaceand its thickness.

Therefore for a four nod quadrilateral thick plate element of thickness h we have for

the plate midsurface :

x =i

Xi

N (, ) and y =i

Yi

N (, ) ; i = 1,.., 4 where ( 1X , 1Y ), ( 2X , 2Y ), ( 3X , 3Y ) and (

4X , 4Y ) are the coordinates of the nodal points of the plate and 1N (, ) , 2N (, ) , 3N

(, ) and 4N (, ) are the bilinear interpolation functions. x = 1X . 1N ( ), + 2X . 1N (

), + 3X . 3N ( ), + 4X . 4N ( ), ;

y = 1Y . 1N ( ), + 2Y . 1N ( ), + 3Y . 3N ( ), + 4Y . 4N ( ),

Any point of the plate can be described uniquely by assigning a point on the mid-

surface ( x, y) and a height z from this point z .Let us also Introduce the inplane displacements with the help of the same interpolation

functions u =i

Ui

N (, ) and v =i

Vi

N (, ) ; i =1..n; n=4;

u = 1U . 1N ( ), + 2U . 1N ( ), + 3U . 3N ( ), + 4U . 4N ( ),

v =1

V . 1N ( ), + 2V . 1N ( ), + 3V . 3N ( ), + 4V . 4N ( ),

Here {U1 U2 U3 U4}T

and V1 V2 V3 V4}T

are the nodal displacement vectors in x and y

directions of the element.

Consequences of the geometric transformation and the Jacobian matrix :

The above geometric transformation transforms the quadrilateral to a square whose

each side is of 2 unit length.


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As x =i

Xi

N (, ) and y =i

Yi

N (, ) ; i = 1,.., 4;, we can construct the following partial

derivatives :

=

i

Xi

N (, ) =1

X

1N (, ) +

2X

2N (, ) +

3

X

3N (, ) +

4X

4N (, )

= iX iN (, ) = 1X 1N (, ) + 2X 2N (, ) + 3X 3N (, ) + 4X 4N (, ) = iY iN (, ) = 1Y 1N (, ) + 2Y 2N (, ) + 3Y 3N (, ) + 4Y 4N (, ) = iX iN (, ) = 1Y 1N (, ) + 2Y 2N (, ) + 3Y 3N (, ) + 4Y 4N (, )Definition of the Jacobian Matrix and its inverse for the general 3D cases:

Applying the chain rule of derivation we get : + + + This can be written in the following form :

=

or, with

.(a)

Similarly as


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+ + + This can be written in the following form :

=

or,

(b)

If we put (a) in (b) we get ; Therefore = Identity matrix.Therefore = . We call matrix of the transformation and is the inverse of the jacobian transformation.Definition of the Jacobian Matrix and its inverse for the 2D case:

Therefore the Jacobian matrix can be computed very easily for the quadrilateral elements as the

iN (, ) s are the polynomials in and . ). If the elements of the jacobian matrix are not

constants and are functions of (, , ), its inverse can be computed numerically by invertingthe2x2 in the cases of 2D problems or 3x3 matrix in the 3D cases.

In 2 dimensional cases we have :

= and j = ;

Fourth step :

The vir tual Energy in the element is given by :


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.

Hookes law can be conveni ently wri tten in thi s case :

[ ]

,

- * + ,

-with G =

With w = w(x,y) , u = z and v = z We have =

; Let ,-

= ,-

= * +The vir tual Energy in the element is then given by :

The virtual energy can be written in the form taking into consideration the above

splitting of strains and strains :

=

+

Introducing w =i

Wi

N (, ) , u = iN (, ) and v = z iN (, ) ; i =1..n;n=4; we have :


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=

=

{

}

=

iN iN iN iN = z with

[

]

and=

To show the true character of the above integrals for determining the stiffness matrix

elements let us develop analytically the bending stiffness matrix elements of a square

4 nod thick Mindlin plate.


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The above is the expression for the bending stiffness matrix for a squre of side two

unit of length ( Beacause it is a square det J = 1 and off diagonal terms of both the

Jacobian and its inverse are zero and both are unit matrices . This simplifies a lot and

allows us to find the different elements of the matrix

.

[K] = . For the time being let ussimplify the notatations. Only for this example of square thick plate ue remove theindex bending from B and let the 3x3 matrix .If we introduce the indices we have = .What does it mean ?

The index i and j takes the values from { 1,2,8} and the indices k and l takes the

values from { 1,2,3}. The indices I and j are the free indices so that you get finally a

8x8 matrix and the indices k and l are the summation indices because they are

repeated. The matrix has 8 rows and 3 columns . The matrix is 3x3. Both kand l takes values from { 1,2,3} . Both are the summation indices. So only twosummations are involved in evaluating any term of . For any i and any j let ustake the product . For the matrix j is the column index and thisfixed because this is a free index.

Example


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Evaluation of of this bending thick Mindlin plate element :Let us imagine we are evaluating the term . So i = 2 and j =3 . Let us consider theproduct in the above integral. If we put the value of j in the product

we get

. In this expression k is a free index l is a

summation index and 3 is a fixed number hence means it is the third column ofthe matrix B because the second index is the column index. So actually is a vectorwhich is nothing but the third column. Hence the term means as l is asummation index and k is a free index ( both k and l takes values from { 1,2,3}) that

the 3x3 matrix column of B matrix. The resultwill be obviously a vector whose free index is k = 1,2,3. That is the result will be vector

with three components as the free index is k . Let us store this vector with 3

components somewhere. We will use this vector write now. Let us call it EBc3(k), k =

1,2,3. Now let us come back to

=

. We

replace here our partial result for by the vector EBc3(k), k= 1,2,3. Weremand you that the components of this vector may contain derivatives of functionswith respect to as the elements of the matrix B contain such derivatives.Now using our partial result we get = k . Herewe see that k is a summation index and see that is to be multiplied by k. We note that k takes values from 1,2,3. We note also that is the 2nd row of the matrix and has only 3 components and k is the summation index. So we have totake the 2

ndrow

and multiply it with the

vector k which has also 3 components . We note that the three componentsof the 2nd row may contain derivatives of functions with respect to asthe elements of the matrix B and hence contain such derivatives . FinallyWe have to integrate this summation k in .Hence = k .Let us summarize what we have done with a concrete example where

Let us first evaluate the matrix . This is a 3x3 matrix symmetric and having only 4terms non zero. Let = . We can then immediately evaluatek . As the element is a square we have = and =


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The , i=1,..4 are given by = (1+ ; i= 1,..4 as this is a 4 nod squareelement.

Then the vector kis sily obtin by multiplying with the third thethird column of B matrix. We get . (As the 3

rd

column of B is

we have :)=

=

=

. On the other hand

as the second row of = , we have . k = { } =

+ .Finally as = k we have

= ) . are bilinear functions of n given above , so therederivation and integration in n poses no problem. Thus we can find any elementof the 8x8 matrix. As an exercise we recommend to calculate the element , etc. of this matrix. in detail :[K] = = for the square thick plate of side two units :


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WE know that = and = , i=1..4. Therefore the above integral becomes :

= ) .Let us simplify the notations: Let in this sub chapter = = This means if you want to compute proceed in 2 steps : First take the E matrix.Multiply it with the jth column of the B matrix that is multiply with E matrix the 7

th

column of the matrix B. Store this vector. Then take the i th row that is the 4th

row of

the matrix

. This is transpose of a vector. In our case it has 3 components. Execute

scalar multiplication of this vector with the previously obtained vector which you have

stored. Mind that both the i th row of the matrix as well as the j th column of thematrix B are functions of derivatives of with respect to and . In this special case asthe element is a square or rectangle the off diagonal terms of the inverse of the

Jacobian matrix are zero. After scalar multiplication of those two vectors integrate each


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term in and within the limits of +1 and -1 as it is shown for the case of in theequation ( ).

Similarly we have

,- iW iN iN iW iN iN .

with

=

contains etc..By applying chain rules : . Thus as and are known from the inverse of the jacobianmatrix ( which must be inverted numerically in most of the cases), all the terms like

and ; i=1,..n; n=4; can be computed and thus the matricescan be easily evaluated at the Gauss points .Finally for the virtual energy we get : = + = + =

= and h )


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he integral in

the virtual energy expression by a Gauss numerical integration scheme we get : = ( )( ) ( ) + ( )( ) ( ) = and = h

Therefore ultimately for the virtual energy due to bending and shear we get

: = where = +

Note : The matrices and are in general functions of So theintegrations in numerically because they contain also elementsof the inverses of the jacobian which may be functions of

. Mostly Gausss integration

schemes are used..In essence, the integration is replaced by a summation of the terms containingthe weights multiplied by the values of the functions at selected points.

Remarks on the Mindlins thick plate element:

1) You might have observed that the transverse shear strains and the transverse shear

stresses are not functions of z . This means it is constant in thickness. But we have

seen in chapter eq. and eq. that for the thick and thin beams and plates the

transverse shear strains and stresses are quadratic functions of z and are zero at the

lower and upper surfaces if the top and the bottom surfaces are free from tangential

forces. This is not the case with Mindlins plate theories. This means Mindlins

theory violates the stress free conditions at the stress free surfaces.

If one considers that the transverse shear stresses and strains found by the

Mindlins theory are the mean stresses and strains then also we need a correction

factor in the energy expression. The situation became s critical for general

composite plates. We show below some typical distribution of shear stresses for

composite plates and conclude that if shear st

zzzz -- fem coursematerials mtech cad cam 1st sem1

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