1d Finite Elements - Theory

STRUCTURAL ENGINEERING CV3403

Dynamics

Plates

Finite Element Analysis

Stability

1

Dr P. Mergos

Few element stiffness matrices can be formulated by the direct method as the case for bars and beams.

Now we will use a more general method to formulate element stiffness matrices. This method is called the formal stiffness matrix approach.

Applying the formal stiffness matrix approach, we can define stiffness matrices for more types of elements like for example the triangular or rectangular element subjected to plane stress.

We will deal with displacement-based elements that are based on pre-determined displacement fields. The displacement fields cannot always be known (like for prismatic bars and beams without distributed loads). So they need to be approximated.

Finite Element Analysis (FEA)

Introduction

Structural Engineering (CV3403)

One dimensional elements

2

The basic steps for the formal stiffness matrix approach are:

1. Identify the problem

2. Select a suitable element displacement field function

3. Relate element displacement field to nodal displacements

4. Relate element strain field to nodal displacements

5. Relate element stress field to nodal displacements

6. Relate element end forces to nodal displacements

We will follow these steps for all finite element types


Formal stiffness matrix approach



3

Here, we define an appropriate co-ordinate and node numbering system for the element.

After defining the degrees of freedom of the element, the nodal displacement (d.o.f.) vector {d} and nodal forces vector {r} can be established.

The stiffness matrix of the element [k] (we assume for simplicity here that local and global co-ordinate systems coincide) is defined by the following equation

Hence, in order to determine [k] we have to relate element end forces {r} to element nodal displacements {d}.


1. Identify the problem


rdk ][


4


1. Identify the problem bar element


We assume a prismatic bar element of length L elastic modulus E and cross section area A

Nodes are allowed to displace only in the axial direction u1 and u2.

Corresponding element end forces are F1 and F2

The nodal d.o.f. and force matrices are:

2

1}{

F

Fr

2

1}{

u

ud


5


1. Identify the problem beam element


2

2

1

1

}{

M

F

M

F

r

2

2

1

1

}{

v

v

d

Description

We assume a 2D prismatic beam element along axis X of length L elastic modulus E, cross section area A and moment of inertia about the centroidal axis parallel to Z axis I.

The 2D beam element has one node at each end with 2 nodal d.o.f which are lateral translation v and rotation about Z-axis . In total, the beam element has 4 nodal d.o.f.

Assumptions

We assume the Euler-Bernoulli beam theory. Hence, we ignore shear deformations (Timoshenko beam theory) and we consider only bending deformations.



v

v1 v2

1

2

v

EI

EI

6

Here we select a continuous function that approximates the element displacement field and satisfies boundary conditions at the nodal d.o.f.

This procedure is called interpolation

In almost all cases we use polynomials which are continuous and single-valued.

Hence, if {u(x)} is the displacement field that we want to interpolate over the element and x is the distance, we have:

Where:

n=1 linear interpolation, n=2 quadratic interpolation, n=3 cubic interpolation

Since we want to define coefficients i (i=0n), the number of these coefficients should be equal to the element d.o.f. (boundary conditions).


2. Select element displacement field function



u(x) = o + 1 x + 2x2 ++ nx

n = i xin

i=0 {u x } = X {} (1)

X = 1 x x2xn and = o 1 2nT (2)

7

Interpolation in finite elements is piecewise. This means that displacement field is smooth inside the element but may not be smooth between two adjacent elements.

Cm represents the level of continuity of piecewise interpolations. It means that the m-th derivative of the displacement field is inter-element continuous.

It is required that derivatives of u of order m are included as d.o.f. if we want the field to be Cm

C0 elements ensure only displacement continuity. They are used to model plane bodies.

C1 elements ensure displacement and slope continuity. They are used to model beams, plates and shells.


2. Select element displacement field function


displacement

x

Elements boundary

C1

C0

Element 1 Element 2

slope

x

Elements boundary

C0

C1

Element 1 Element 2

C0


8

The bar element has two d.o.f.

For bar elements, the level of continuity is C0

Hence, we can determine two polynomial coefficients and the displacement field should be of first order (linear interpolation)

The displacement field {u(x)} is the axial displacement u(x) over the element. We write

Where:


2. Select element displacement field function bar element



u(x) = u x = o + 1 x = i xi1

i=0 {u x } = X {} (3)

X = 1 x and = o 1T (4)

9

The beam element has four d.o.f.

For beam elements, the level of continuity is C1

Hence, we can determine four polynomial coefficients and the displacement field should be of third order (cubic interpolation)

The displacement field {u(x)} is the transverse displacement v(x) over the element

Where:


2. Select element displacement field function beam element



u x = v x = o + 1 x + 2 x2 + 3 x

3 = i xi3

i=0 {u x } = X {} (5)

X = 1 x x2 x3 and = o 1 2 3T (6)

10

Here we want to correlate displacement field {u(x)} with nodal d.o.f. vector {d}

{d} {u(x)}

By applying boundary conditions for every d.o.f. we get the following equation

Where [A] has as many rows as the element d.o.f. Combining Eqs. (3) and (7) we obtain:

In Eq. (8), [N] is the interpolation matrix and each element of it Ni is called shape function. Ni is the displacement field {u(x)} when nodal d.o.f. di=1 and all other d.o.f. dj=0 (ji).

In FEA, {d} is typically calculated by solving the structure system of equations [K]{D}={R}. Then, it can be used to define displacement field {u(x)}.


3. Relate displacement field to nodal displacements



= = 1 {} (7)

() = [] 1 = {} where = X 1 (8)

11

The bar element is a case of C0 linear interpolation It is:

Hence the matrix [A] is:

Hence, N1=(L-x)/L and N2=x/L as shown in the figure



x

u2

= 1 = 1 , = 1 = + 1 ,

= 0 1 = + 1 0 , = 2 = + 1

In matrix form, it is written as: 12

=1 01

1

[] =1 01

[]1=1

01 1

The interpolation matrix N=[N1 N2] becomes

x=0 x=L

u1

u(x)

x x=0 x=L

1

N1(x)

x x=0 x=L

N2(x) 1

= X 1 = 1 1 0

1

1

= [

] (9)


3. Relate displacement field to nodal displacements bar element

12

The bar element shape functions derived previously can be regarded as special cases of the Lagranges interpolation shape functions given by the following general function:



x

u2

=1 2 [ ] ( )

1 2 [ ] ( )

In this formula, the terms in the brackets are omitted to obtain Nk and kn. For linear interpolation (previous bar element) n=2. For quadratic interpolation n=3 and so on. Lagranges interpolation is a C0 interpolation scheme. Hence, it is appropriate for bar elements and plane problems. On the other hand, if it is applied to beams it may drive to discontinuous slopes between adjacent elements. Substituting Lagranges formula for the previous bar element (n=2, x1=0, x2=L) we obtain

1 =2

2 1=

0= 1 /

2 =1

1 2=0

0 = /

x=0 x=L

u1

u(x)

x x=0 x=L

1

N1(x)

x x=0 x=L

N2(x) 1


3. Relate displacement field to nodal displacements bar element

13

In this case, we have for the lateral displacement (x):

Hence the matrix [A] is:

For this interpolation we used both displacements and slopes. This is called Hermitian interpolation and it is C1 interpolation.



n = 3 X = 1 x x2 x3 , = o 1 2 3 T

v x = o + 1x + 2x

2 + 3x3 v x = 1 + 22x + 33x

2 (10) x = 0 v1 = o (11) x = 0 1 = 1 (12) x = L v2 = o + 1 L + 2L

2 + 3L3 (13)

x = L 2 = 1 + 22L + 33L2 (14)

In matrix form, it is written as: v11v22

=

1 0 1 0

01L1

0 0 L2

2L

0 0 L3

3L2

123

A =

1 0 1 0

01L1

0 0 L2

2L

0 0 L3

3L2

A 1 =

1 0

3

L2

2

L3

01

2

L1

L2

0 0

3

L

2

2

L3

0 0

1

L

1

L2


3. Relate displacement field to nodal displacements beam element

14

The equations and shapes of the shape functions Ni are shown in the figures.

To avoid the inversion of [A] one can use the definition of shape function Ni=the displacement field when d.o.f. di=1 and all other d.o.f. are zero. By Eqs. (11-14) and for N1, we obtain:

= 0 1 = = 1

= 0 1 = 1 = 0

= 2 = 1 + 22 + 3

3 = 0

= 2 = 22 + 332 = 0

Solving the system of the last two equations we obtain that

2 = 3

2 3 = 2/

3

Finally, we substitute in Eq. (10) to get the shape function N1=v. We can do the same for all Ni

1 = 1 32/2 + 23/3



The interpolation matrix becomes = X 1 = 1 2 3

1 0

3/2

2/3

01

2/L

1/2

0 0

3/2

2/3

0 0

1/

1/2

= [1 2 3 4]


3. Relate displacement field to nodal displacements beam element

1=v

2=v

3=v

4=v

1

v1=1

v2=1

2

v

v

v

v

15

Cook 2002

Here we want to define strain field {(x)} from the displacement field {u(x)}

{d} {u(x)} {(x)}

The strains {(x)} at any point within an element can be obtained from the displacement field by differentiation. The exact form of differentiation depends on the type of the problem being examined. For example, for the bar element strains correspond to first order derivatives of the displacements, while for the beam element the strain field is associated with curvatures which correspond to second order derivatives of the lateral displacement field v(x)

If is the matrix correlating {(x)} with {u(x)} we can write

() = () (15)

Combining Eqs. (8) and (15) we obtain

() = = (16)

Where [B]=[] is the so called strain matrix relating directly element strain field {(x)} to nodal displacement vector {d}


4. Relate strain field to nodal displacements



16

For the bar element, the strain displacement relationship is

() =

= () (17)

Where

=

(18)

Hence, matrix [B] becomes

[B]=[] =1

2

=

1

1

(19)


4. Relate strain field to nodal displacements bar element



17

For the beam element, the strain that we examine is the curvature

(x) =d2v

dx2= v(x) (20)

Where

=2

2 (21)


B = [] =21

2 22

2 23

2 24

2=

6

2+12

3

4

+6

2

6

212

3

2

+6

2 (22)


4. Relate strain field to nodal displacements beam element



18

Here we relate internal stresses {(x)} to the element strain field {(x)}

{d} {u(x)} {(x)} {(x)}

The stress {(x)} at any point within an element can be obtained from the strain {(x)} at the same point by applying the theory of elasticity.

If [D] is the constitutive matrix then we obtain in the general case

(x) = D ( (x) - (x) ) = D (x) + (x) (23)

Where (x) = { x } and {(x)} is the initial strain vector.

The previous equation can be written also as

(x) = D 1 (x) + (x) (24)

Combining Eq. (16) and (23) we get

(x) = D B {} + (x) (25)


5. Relate stress field to nodal displacements



19

For bar elements, we obtain by applying Hookes law (= ) that [D]=E

where E is the modulus of elasticity.

Hence, we can write

(x) = E B {} + (x)

(x) =

+ (x) (26)


5. Relate stress field to nodal displacements bar elements



20

For beams, we obtain by Euler-Bernoulli beam theory that M=I . Hence [D]=EI

where E is the modulus of elasticity and I is the moment of inertia about the centroidal axis parallel to Z axis

Hence, we can write

(x) = B {} + (x)

(x) = 6

2+12

3

4

+6

2

6

212

3

2

+6

2 {} + (x) ) (27)


5. Relate stress field to nodal displacements beam elements



21

In this section we will relate nodal loads to nodal displacements. To serve this goal we will apply the principle of virtual work. This principle is equivalent to equilibrium.

A virtual displacement is an imaginary and very small change in the configuration of the system.

We assume that a virtual displacement takes place relative to the equilibrium configuration when all loads have been fully applied.

During this small displacement it is assumed that loads and stresses remain constant.

According to this principle, during every virtual displacement imposed on the element, the total work done by external forces should be equal to the total internal work done by the stresses.

We designate as {d} the arbitrary virtual nodal displacement vector that causes the virtual displacement field {u(x)} and corresponding strain field {(x)}.

Furthermore, we assume that external body forces {F} act in the volume of the element V and surface tractions {} on the surface of the element S and concentrated nodal forces {P} at the nodes of the element.


6. Relate end loads to nodal displacements



22

The principle of virtual work gives, Eq. (27)

= (x) = + + (27)

If {} are initial strains, then it holds {}=[D]({}-{}).

By substituting {(x)} and {(x)} from Eq. (16) and (25) respectively and taking into

account that {d} is not a function of the coordinates, we obtain that

[] x x = + + (28)

Since {d} is arbitrary we can eliminate it from both sides and we finally obtain

( []) {d} = + + + [] x (29)

Eq. (29) is equivalent to the node equilibrium equation:

P + {re} r = 0 r = + = + (30)

Where {P} are the external loads applied directly to the nodes, {re} are the equivalent loads applied to the nodes due to the external loads distributed over the element and {r} are the loads applied to the member ends due to element deformation.





23

According to this observation, we obtain:

r = = []) d = [] (31)

re = + + [] x (32)

With Eq. (31) we can determine the stiffness matrix of the element and with Eq. (32) the equivalent nodal loads {re} due to body forces {F} (e.g. gravity loads), surface tractions {} and due to initial element strains {o}.

It is worth noting that Eq. (32) provides consistent equivalent nodal loads, which means that these loads are determined by the use of the same shape functions as the ones used for the element stiffness matrix.

It is also noted that in the case of more than one elements Eqs. (27)-(30) are valid when a summation symbol for all elements is applied before these equations.





24

In the following figure the complete procedure applied to determine the element stiffness matrix is presented.


Element stiffness matrix determination - overview



{d} {r}

{} {}

kinematic

constitutive

equilibrium

stiffness

= D {}

= {}

= {}

= ( []) d

25

For the prismatic bar element with area A, Eq. (30) becomes:

= ( 1

1

1

1

)

= (

1

1

1

1

)

0 =

1/ 1/1/ 1/

(33)

More analytically, for example, for the stiffness matrix coefficient k12

12 = 1

1

=

0 /2 = /

0


Element stiffness matrix bar element



26

For the prismatic beam element with flexural rigidity EI, Eq. (30) becomes:

= ( )

0 = (34)

Where B is given by Eq. (22)

More analytically, for example, for the stiffness matrix coefficient k34

34 = 6

212

3

2

+6

2 =

12

3+60

4 722

5 = 6

0/2

0


Element stiffness matrix beam element



L

EI

L

EI

L

EI

L

EIL

EI

L

EI

L

EI

L

EIL

EI

L

EI

L

EI

L

EIL

EI

L

EI

L

EI

L

EI

4626

612612

2646

612612

22

2323

22

2323

v1

1

v2

2

27

For the prismatic bar element with area A and elastic modulus E, Eq. (32) becomes:

For the load P acting at a distance L/3 from point 1:

re = =

=3=

3

3

=2/3/3

For the distributed load q(kN/m):

re = = 1

() =

+1

( + 1)( + 2)

+1

( + 2)

For an initial imperfection L:

re L = [] x =

1

1

/ =

1

1

=

11

For uniform temperature change T (=thermal expansion coefficient, e.g. for concrete 10-5 per degree Celsius)

re T = [] x =

1

1

= 1

1

= 11

For all of the above: = re P + re + re L + re T


Element equivalent nodal loads bar element



1 2

u2 u1 L/3 2L/3

P q=xb

28

For the prismatic beam element with rigidity EI, Eq. (32) becomes:

For the uniform distributed load q(kN/m):

re = = () =

/2

2/12/2

2/12

For example: re(4) = q 2

+3

2

0=

3

3+4

42 0= 2/12

For a temperature difference between the bottom and top surface of a section with height h:

re T = [] x =

=

0101

For all of the above:

= re + re T


Element equivalent nodal loads beam element



1 2

v2 v1

L

q

1 2

29


The 3-node bar element - introduction


The 2-node bar element developed previously displays constant axial strain along its length. Hence, it cannot be accurate when distributed forces act inside the element length.

In the following, we examine again a prismatic bar element of length L elastic modulus E and cross section area A.

However, the bar element consists of 3 nodes with displacements in the axial direction u1, u2.and u3.

For simplicity, we assume that the intermediate node is at the middle point of the bar. The formulation can easily be modified for other locations of the intermediate node.

Corresponding element end forces are F1, F2 and F3


3

2

1

}{

F

F

F

r

3

2

1

}{

u

u

u

d


u1 u3

u2

F1 F2 F3

x

1 2

3

L/2 L/2

30


The 3-node bar element displacement field



This bar element has three d.o.f. This means that we can determine three polynomial coefficients i

The displacement field {u(x)} is the axial displacement u(x) over the element. We write

Where:

u(x) = u x = o + 1 x + 2 2 = i x

i2i=0 {u x } = X {}

X = 1 x x2 and = o 1 2T

The boundary conditions are:

= 0 1= + 1 0 + 2 0, = /2 2 = + 1 /2 + 2

2/4 = 3 = + 1 + 2

2

In matrix form, it is written as: 123

=1 0 01 /2 2/4

1 2

12

[] =1 0 01 /2 2/4

1 2

u1 u3

u2

F1 F2 F3

x

1 2

3

L/2 L/2

31


The 3-node bar element displacement field



The interpolation matrix is then calculated by:

Where:

1 = 2

2

3

+ 1 , 2 = 4

2

+ 4

, 3= 2

2

,

= 1 2 3 = X 1

The interpolation functions can also be calculated by their

definition. For example for N1 we can solve the following

system of equations to obtain and o, 1, and 2. = 0 1 = + 1 0 + 2 0, = 0 = + 1 + 2

2 = /2 0 = + 1 /2 + 3

2/4

The interpolation functions are presented in the figures: They can also be calculated from Lagranges interpolation function (e.g. for N2):

2 =x1 x x3 x

(1 2)(3 2)=0 x x

(0 2)(

2)= 4

2

+ 4

[N1]

1 2 3

[N3]

1 2 3

[N2]

1 2

3

1

1

1

32


The 3-node bar element strain field



For the bar element, the strain displacement relationship is

() =

= ()

Where:

=


[B]=[] =1

2

3

Where: 1

=4x

L23

L

2

=

8x

L2+4

L

3

=4x

L21

L

33


The 3-node bar element element stiffness matrix



The stress field is then determined by applying Hookes law. Hence, if we ignore initial stresses, we can write

(x) = D B x = []

We observe that both strains and stresses can vary linearly along the element. We recall that for the 2-node bar element strains and stresses are constant within the element.

For the prismatic 3-node bar element with area A, the stiffness matrix is 3x3 and is calculated as:

= ( [] ) =EA ( )

0 =

3

7 8 18 16 81 8 7

34

1d Finite Elements - Theory

Documents

Transcript of 1d Finite Elements - Theory