breve intro a e.f

Chapter 1

Introduction to Finite Element

Analysis

1.1 Introduction

Finite element analysis is a numerical method is a numerical method that is used

in the analysis of complex mechanical and structural problems. It is used to

obtain numerical solutions and can be applied over a wide range of objective

functions and over a wide range of loading conditions. The method can be used

for analysis of static problems and dynamic problems and can even be used to

analyse linear systems and non-linear systems. A linear system is a type of

system where a linear relation-ship exists between the force and deflection and

these systems do not take plastic deformation. A non-linear system is a type of

system that takes plastic deformation into account and can allow testing all the

way up to the point(s) of fracture.

Finite element analysis is not restricted to mechanical and structural analysis.

In the methods most basic form, it can be applied to problems that involve heat

loss, fluid flow and even electric potential. Due to this wide range of application,

the method is very popular. The method is used in the analysis of many complex

two-dimensional problems, but can also be applied for problems involving three-

dimensions. As problems become more complex with finite element analysis, they

can take longer to solve. Thankfully, computers can be used to solve complex

1

1. Introduction to Finite Element Analysis

finite element problems and a wide range of finite element analysis software exists

today (LUSAS, ABAQUS, ANSYS).

Figure 1.1: Computer finite element analysis of an I-Beam

The method works by taking a problem and modelling it into smaller elements

through the use of nodes and elements. Physical and geometric properties are

assigned to the elements and loads and displacements are applied to the nodes.

From that, a finite element analysis is carried out in an attempt to earn numerical

values. The best way to see finite element analysis in action, is to see how it is

applied and used to solve simple basic problems.

1.2 Worked Example 1

The following worked example will be used to develop a one degree of freedom

finite element. This simple problem will be solved and then the results will

be compared to a theoretical solution. Consider a cantilever L. It is axially

restrained at one end and subjected to a axial load W at the other end. The

member has a continuous cross-sectional area of A and a Youngs Modulus of E.

The theoretical stress of the beam, σ, can be obtained using

σ =Force

Area(1.1)

2


Figure 1.2: Axially loaded member

The load applied is W and the cross-sectional area is A. Therefore the equa-

tion becomes

σ =W

A(1.2)

The theoretical strain of the beam, ǫ, can be calculated using

E =σ

ǫ(1.3)

Rearranging

ǫ =σ

E(1.4)

Substituting the theoretical stress obtained previously, equation 1.1, into

equation 1.4 to give

ǫ =W

EA(1.5)

With the theoretical results obtained, the finite element method can now be

applied to the cantilever beam. The results obtained from the finite element

method can then be compared to the theoretical results obtained for stress and

strain.

1.2.1 Step 1 - Creating an element

In order to apply the finite element method, the problem shown in Figure 1.1

must be modeled as a finite element. As the beam is a simple cantilever, with

a uniform cross sectional area, we can model this as one single element. This

is achieved by first applying a node to the fixed end of the beam and a node

3


at the free end of the beam. An element can be created between the nodes, as

demonstrated in Figure 1.3.

Figure 1.3: Cantilever beam as an element

The element has a length of L and has two nodes, nodes 1 and 2, at either

end of the beam. Each node has one degree of freedom, since the original problem

is such that any point on the member has one degree of freedom, u. The finite

element model is then used to obtain a solution.

1.2.2 Step 2 - Nodal displacements

Now that a finite element model has been established, the nodal displacements

are now taken into consideration, however, the restraints of the cantilever beam

are ignored for the timed being to create and understand the general relationship

for displacement along the cantilever beam.

Figure 1.4: Axially loaded member

If a horizontal load is applied to the element, node 1 will displace to 1 ′ and

node 2 will displace to 2 ′. The displacement of node 1 to 1 ′ can be considered

as u1 and the displacement of node 2 to 2 ′ can be considered as u2. This is

illustrated in Figure 1.4. These displacements can be written as the displacement

vector, u.

u =

[

u1

u2

]

(1.6)

4


The displacement along the member is required. Since there are two nodal

displacements, a function with two unknowns can be used (usually, but not al-

ways, a complete polynomial is used). Thus, the displacement function can be

written as.

u = a1 + a2x (1.7)

Where u is the displacement at x along element 1 2, a1 and a2 are unknown

constants and x is a co-ordinate along the beam. Using the displacement function,

we can determine a1 and a2 in terms of the displacements u1 and u2.

At node 1, x = 0. Substituting into the displacement function, 1.7,

u1 = a1 + a2(0) (1.8)

Therefore

u1 = a1 (1.9)

Now consider the other node. At node 2, x = L. Substituting into the

displacement function,

u2 = a1 + a2L (1.10)

Previously, the coefficient a1 was determined. Therefore,

u2 = u1 + a2L (1.11)

Rearranging it terms of the constant a2,

L =u2 − u1

L(1.12)

With both a1 and a2 determined, they can be substituted back into the dis-

placement function, to give

u = u1 +u2 − u1

Lx (1.13)

5


1.2.3 Step 3 - Strain

With the displacement function obtained, the strain can be determined along the

member. The strain in the member, ǫ, can be calculated using

ǫ =Change in Length

Original Length(1.14)

Figure 1.5 illustrates the strain along the member

Figure 1.5: Strain along the member

From Figure 1.5, the original length of the beam is denoted as dx. Node 1 is

displaced by u and node 2 is displaced by u+ ∂u

∂xdx. Therefore,

Change in Length = u+∂u

∂xdx− u (1.15)

This simplifies to

Change in Length =∂u

∂xdx (1.16)

With expressions for the change in length and the original length, an expres-

sion for the strain can be derived.

ǫ =Change in Length

Original Length=

∂u

∂xdx

dx(1.17)

This simplifies to

ǫ =∂u

∂x(1.18)

Previously, an expression for displacement, u, was derived to give equation

1.13. By differentiating this with respect to x, we obtain

6


∂u

∂x= (0) +

u2 − u1

L(1) (1.19)

∂u

∂x=

u2 − u1

L(1.20)

Therefore,

ǫ =u2 − u1

L(1.21)

This expression can be expressed in matrix format

ǫ =[

−1L

1L

]

[

u1

u2

]

(1.22)

Consider the the matrix

[

−1L

1L

]

This is the strain-displacement matrix and can be written as [B]. Now con-

sider the vector

[

u1

u2

]

This is the displacement vector and can be written as u. Therefore, the

expression for strain can be written as

ǫ = [B] u (1.23)

In this worked example, we are dealing with a one dimensional problem, there

is only one strain. Later we will be looking at two dimensional and three di-

mensional situations where there can be 3 or 6 strains at any point in a body,

respectively.

7


1.2.4 Step 4 - Stress

With an expression for the strain, the stress in the element can now be determined.

The expression to obtain stress is given as

σ = [D] ǫ (1.24)

[D] is the elasticity matrix and the notation is used by convention. As we

are dealing with a one dimensional problem, only one stress will be obtained.

However, σ, can contain a number of stresses depending on the problem. In

this case, as there is only one stress, the matrix [D] only contains one term the

modulus of elasticity E.

σ = E ǫ (1.25)

Previously, an expression for strain was derived. Substituting that into ex-

pression 1.25 yields,

σ = E [B] u (1.26)

1.2.5 Step 5 - Nodal Loads

The nodal loads now need to be related to the internal stresses.

Figure 1.6: Nodal loads

This yields the nodal force vector

P =

[

P1

P2

]

(1.27)

However, rather than apply just to the particular loading shown above a

general relationship will be derived. This can be done with the aid of virtual

work.

8


1.2.6 Step 6 - Application of Virtual Work

Virtual work can be used to relate the external forces acting on the beam with

the internal forces acting from within the beam.

External Work Done = Internal Work Done (1.28)

Given a set of external applied forces, P , we can apply a set of virtual dis-

placements, u. Therefore, the external work applied can be expressed as

External Work Done = uT P (1.29)

As for the internal work, the internal stresses, σ, equilibrate P and the internal

strains, ǫ, are compatible with u. From this, the internal work can be calculated

using

Internal Work Done = ǫT σ Per Unit Vol (1.30)

This becomes

Internal Work Done =

∫

V

ǫT σ dV (1.31)

Now, consider the expression for strain that was derived previously.

ǫ = [B] u (1.32)

By taking the transpose of the strain, we obtain

ǫT = [B]T uT (1.33)

This can then be substituted into expression 1.31 to give


∫

V

uT [B]T σ dV (1.34)

This expression can be simplified further by using the expression for strain,

expression 1.26, that was derived previously. This gives

9



∫

V

uT [B]T E [B] u dV (1.35)

We can now apply equilibrium.

External Work Done = Internal Work Done (1.36)

uT P =

∫

V

uT [B]T E [B] u dV (1.37)

uT is not a function of the geometry of the member. Therefore, this can be

rearranged to obtain

uT P = uT

∫

V

[B]T E [B] dV u (1.38)

Let∫

V

[B]T E [B] dV = [k] (1.39)

This is the stiffness matrix. Therefore

uT P = uT [k] u (1.40)

This can be simplified further to obtain the final expression

P = [k] u (1.41)

Where P is the nodal force vector, [k] is the element stiffness matrix and u

is the nodal dispalcement vector. As mentioned previously, the important point

to note is that the derivation is completely general. No assumptions have been

made regarding the loads, element type, material properties, etc.

1.2.7 Step 7 - Stiffness Matrix

In order to solve the problem, we mus first derive the stiffness matrix. For our 2

noded element, consider the stiffness matrix expression

10


Let

[k] =

∫

V

[B]T E [B] dV (1.42)

Rearranging

[k] = [B]TE[B]

∫

V

dV (1.43)

Now, consider

∫

V

dV (1.44)

This is the equivalent of

∫ ∫ ∫

dxdydz (1.45)

By integrating with respect to y and z, we obtain

∫

L

O

dxyz (1.46)

yz is the equivalent of the area, A. Therefore

∫

L

O

dxA = A

∫

L

O

dx = AL (1.47)

Substituting

[k] = [B]TE[B]AL (1.48)

Now consider the strain-displacement matrix [B]. From previous, it was found

[B] =[

−1L

1L

]

By taking the transpose of the strain-displacement matrix, we obtain

[B]T =

[

−1L

1L

]

Substituting

11


[k] =

[

−1L

1L

]

E[

−1L

1L

]

AL (1.49)

Multiplying through, we obtain

[k] =

[

EA

L−

EA

L

−EA

L

EA

L

]

(1.50)

This can then be used to solve the original problem.

1.2.8 Step 8 - Solution

Using what has been derived in the previous steps, we can now return to the

original problem.

Figure 1.7: Original Problem

Figure 1.8: Finite Element

This problem can be solved using

P = [k]u (1.51)

From previous, the nodal load vector is

P =

[

P1

P2

]

(1.52)

12


The displacement vector is

u =

[

u1

u2

]

(1.53)

The stiffness matrix is

[k] =

[

EA

L−

EA

L

−EA

L

EA

L

]

(1.54)

Substituting these into equation 1.51, we obtain

[

P1

P2

]

=

[

EA

L−

EA

L

−EA

L

EA

L

][

u1

u2

]

(1.55)

We can now apply the beams properties, known displacements and known

loads to expression 1.55. In this case

u1 = 0 (Node 1 is restrained and cannot be displaced)

u2 = Unknown

P1 = Unknown

P2 = W

E = E

A = A

L = L

13


Therefore

[

P1

W

]

=

[

EA

L−

EA

L

−EA

L

EA

L

][

0

u2

]

(1.56)

Multiplying the matrices, we obtain

P1 = −EA

Lu2 (1.57)

W =EA

Lu2 (1.58)

Rearranging expression 1.58, we obtain an expression for u2

u2 =WL

EA(1.59)

This can then be used with expression 1.57 to find the unknown nodal force.

P1 = −EA

L

(

WL

EA

)

= −W (1.60)

u2 can also be used to obtain the strain within the beam. Using expression

1.2.3

ǫ =[

−1L

1L

]

[

0WL

EAu2

]

(1.61)

Multiplying through, the expression for strain becomes

ǫ =W

EA(1.62)

Using the strain, the stress can be obtained.

σ = [D]W

EA(1.63)

As mentioned previously, [D] contains only one value, E, as we are dealing

with a one dimensional problem. Therefore, the stress becomes

σ =W

A(1.64)

14


Using basic finite element method theory, it is clear to see that the results

obtained through the finite element method agree with the theoretical results

that were calculated earlier.

15



Now consider the following problem.

Figure 1.9: Axially loaded member with two loads applied

We can use finite element analysis to obtain the stresses and strains within

this beam. However, this beam now has two applied loads at different lengths of

the beam - one at the end of the beam and one half way across the length of the

beam. In worked example 1, we were able to solve the problem by modelling the

entire beam with a single two-noded element. However, if we were to do that for

this case, we would neglect the load applied half way across the beam. In order

to solve this, we must divide the structure with two two-noded elements. Each

element has a length of L

2.

16


Figure 1.10: The original problem and the equivalent finite element model

From this we can establish a relationship between the loads and displacements

using

P = [k] u (1.65)

However, we cannot apply this directly to the entire member. We must first

apply the relationship to the individual elements and then use them to create an

overall relationship for the beam.

1.3.1 Element 1

Consider element 1 and apply the unique nodal loads to the element.

We can obtain the the nodal force vector, the displacement vector and the

stiffness matrix to form a relationship using 1.65. The nodal vector for element

1 is given as

P (1) =

[

P(1)1

P(1)2

]

(1.66)

The displacement vector for element 1 is

17


Figure 1.11: Element 1

u(1) =

[

u(1)1

u(1)2

]

(1.67)

Considering the element length is L

2, the cross sectional area is A and Young’s

modulus is E, the stiffness matrix is for element 1 is

[k(1)] =

[

2EA

L−

2EA

L

−2EA

L

2EA

L

]

(1.68)

Therefore, using 1.65, the relationship for element 1 is

[

P(1)1

P(1)2

]

=

[

2EA

L−

2EA

L

−2EA

L

2EA

L

] [

u(1)1

u(1)2

]

(1.69)

1.3.2 Element 2

Now consider element 2 and apply the unique nodal loads to the element.

The nodal vector for element 2 is given as

P (2) =

[

P(2)2

P(2)3

]

(1.70)


18


Figure 1.12: Element 2

u(2) =

[

u(2)2

u(2)3

]

(1.71)


2, the cross sectional area is A

2and Young’s


[k(2)] =

[

EA

L−

EA

L

−EA

L

EA

L

]

(1.72)

The relationship for element 2 is

[

P(2)2

P(2)3

]

=

[

EA

L−

EA

L

−EA

L

EA

L

] [

u(2)2

u(2)3

]

(1.73)

1.3.3 Overall Relationship

We can now create an overall relationship for the member. However, before we

can do this, we must first apply compatibility. We must first obtain the structure

load vector, structure displacement vector and structure stiffness matrix, which

will then be used to obtain an overall relationship for the member.

F = [K] δ (1.74)

Where

19


F = Structure load vector

[K] = Structure stiffness matrix

δ = Structure displacement vector

20


Figure 1.13: Compatibility

Consider the forces. We can apply compatibility to create relationships for

the forces that affect each node. Therefore,

F1 = P(1)1

F2 = P(1)2 + P

(2)2

F3 = P(2)3

Therefore, the structure load vector becomes

F =

F1

F2

F3

(1.75)

Considering the displacements and applying compatibility, we obtain

δ1 = u(1)1

21


δ2 = u(1)2 = P

(2)2

δ3 = u(2)3

Therefore, the structure displacement vector becomes

δ =

δ1

δ2

δ3

(1.76)

Now consider the stiffness matrix. The stiffness matrix from element 1 can be

combined with the stiffness matrix from element 2 to give the structure stiffness

matrix.

[K] =

2EA

L−

2EA

L0

−2EA

L

(

2EA

L+ EA

L

)

−EA

L

0 −EA

L

EA

L

(1.77)

This simplifies to give

[K] =EA

L

2 −2 0

−2 3 −1

0 −1 1

(1.78)

Therefore, the overall relationship for the member is

F1

F2

F3

=

EA

L

2 −2 0

−2 3 −1

0 −1 1

δ1

δ2

δ3

(1.79)

22


1.3.4 Solution

As a relationship between the loads and the displacements has been derived for

the member, we can now solve the problem. We can apply the known loads and

displacements that act on the member. By inspection, the forces are

F1 = Unknown

F2 = 2W

F3 = W

By inspection, the displacements are

δ1 = 0

δ2 = Unknown

δ3 = Unknown

Using these, expression 1.79 becomes

F1

2W

W

=

EA

L

2 −2 0

−2 3 −1

0 −1 1

0

δ2

δ3

(1.80)

Multiplying through, we obtain

F1 = −2EAδ2

L(1.81)

2W =3EAδ2

L−

2EAδ3

L(1.82)

23


W = −EAδ2

L+

EAδ3

L(1.83)

Solving gives gives the unknown force and the unknown displacements

F1 = −3W (1.84)

δ2 =3WL

2EA(1.85)

δ3 =5WL

2EA(1.86)

These values can then be used to obtain expressions for stress and strains for

the two elements.

For element 1, the strain obtained is

ǫ(1) =3W

EA(1.87)

And the stress for element 1 is

σ(1) =3W

A(1.88)

For element 2, the strain obtained is

ǫ(2) =2W

EA(1.89)

And the stress for element 2 is

σ(2) =2W

A(1.90)

24



So far, we have dealt with problems where point loads have been applied to

beams. Consider the following problem.

Figure 1.14: Member with UDL applied

We now have a cantilever beam of length L where a load of w per unit length

has been applied. Young’s modulus, E, and the cross-sectional area, A, are

uniform through out the beam. We can still use finite element analysis to calculate

the stresses and strains within the member.

1.4.1 Theoretical results

We can first use basic theory to obtain results.

The theoretical stress of the beam, σ, can be obtained using

σ =Force

Area(1.91)

The load applied is w over length L and the cross-sectional area isA. Therefore

σ =wL

A(1.92)

The theoretical strain of the beam, ǫ, is calculated using

ǫ =σ

E(1.93)

25


This becomes

ǫ =wL

EA(1.94)

Figure 1.15: Theoretical stress along the member

If we plot the theoretical stress along the member, as shown in figure 1.15, the

maximum stress occurs at the fixed end of the beam. Moving down the beam, the

stress decreases linearly. At the end of the beam, no stress exists on the beam.

Finite element analysis will then be used to obtain the stresses and strains

and the results obtained will be compared to the theoretical results.

1.4.2 Modelling with a single two-noded element

This problem is similar to worked example 1. Therefore, we will model this using

one two-noded element.

As this is modeled using a single two-noded element, it is clear to see that the

element force vector, element stiffness matrix and element displacement vector

are the same as the structure load vector, structure stiffness matrix and structure

26


Figure 1.16: Two-noded element

displacement matrix respectively. Therefore, we can easily create a relationship

between the forces and displacements that will occur on the member.

[

F1

F2

]

=EA

L

[

1 −1

−1 1

] [

δ1

δ2

]

(1.95)

We can then apply the known displacements to the relationship. By inspec-

tion, the displacements are

δ1 = 0

δ2 = Unknown

We can also apply the known forces to the relationship. However, as we are

dealing with a distributed load, we must take greater care. First consider node

2. Whilst the distributed load acts on node 2, not all of it does. Therefore, the

force acting on node 2 is

27


F2 =wL

2

Now consider node 1. As with node 2, not all of the distributed load acts

on this node. However, we must also take into consideration that there is an

unknown reaction at node 1. Therefore, the forces that act on node 1 can be

expressed as

F1 = R1 +wL

2

Where R1 is the unknown reaction. Using these, 1.95 becomes

[

F1

wL

2

]

=EA

L

[

1 −1

−1 1

] [

0

δ2

]

(1.96)

Solving gives

F1 = −EA

Lδ2 (1.97)

δ2 =wL2

2EA(1.98)

Substituting 1.98 into 1.97 gives

F1 = −wL

2(1.99)

This can then be used to obtain the unknown reaction, R1

F1 = R1 +wL

2(1.100)

Therefore

R1 = −wL (1.101)

The stress and strain for the member can now be obtained. Using expression

1.98, the strain is given as

28


ǫ =wL

2EA(1.102)

And the stress is given as

σ =wL

2A(1.103)

Figure 1.17 illustrates the results obtained, plotted against the theoretical

results.

Figure 1.17: Single two noded element stress vs. theoretical stress

Whilst the unknown reaction obtained through finite element analysis agrees

with the theoretical result, the results for the stress and strain obtained through

finite element do not agree with the theoretical stress and strain. We can see that

the results obtained through theory show maximum stress occurring at the fixed

end of the beam and the stress decreases linearly through the beam. The results

obtained through finite element analysis show that the expressions for stress and

strain are uniform throughout the beam, which is not the case, although the finite

element results do accurately predict the stress and strain at the midpoints of

the beam. However, these are artifacts of the analysis and may not necessarily

be true.

Whilst the results obtained do not agree with theory, it does not mean that

29


we cannot use finite element analysis with problems where a distributed load

is applied. It means that we must model our problem much more carefully to

improve the accuracy of the analysis. Now, we will try to model the problem

using two, two-noded elements.

1.4.3 Modelling with two, two-noded elements

We will now model the problem using two two-noded elements, each with lengthL

2.

Figure 1.18: Two two-noded finite element model

As in worked example 2, we must first determine the element load vector,

element displacement vector and element stiffness matrix and create the load-

30


displacement relationship for each element before establishing an overall load-

displacement relationship for the structure. For element 1, the element load

vector is

P (1) =

[

P(1)1

P(1)2

]

(1.104)


u(1) =

[

u(1)1

u(1)2

]

(1.105)




[k(1)] =

[

2EA

L−

2EA

L

−2EA

L

2EA

L

]

(1.106)

Therefore, the load-displacement relationship for element 1 is

[

P(1)1

P(1)2

]

=

[

2EA

L−

2EA

L

−2EA

L

2EA

L

] [

u(1)1

u(1)2

]

(1.107)

As for element 2, the nodal vector for element 2 is

P (2) =

[

P(2)2

P(2)3

]

(1.108)


u(2) =

[

u(2)2

u(2)3

]

(1.109)




[k(2)] =

[

2EA

L−

2EA

L

−2EA

L

2EA

L

]

(1.110)

31


The relationship for element 2 is

[

P(2)2

P(2)3

]

=

[

2EA

L−

2EA

L

−2EA

L

2EA

L

] [

u(2)2

u(2)3

]

(1.111)

In order to create an overall load-displacement relationship, we must first

apply compatibility.

Figure 1.19: Compatibility

Compatibility for the forces

F1 = P(1)1

F2 = P(1)2 + P

(2)2

F3 = P(2)3

The structure load vector becomes

32


F =

F1

F2

F3

(1.112)

Displacement compatibility

δ1 = u(1)1

δ2 = u(1)2 = P

(2)2

δ3 = u(2)3

Structure displacement vector

δ =

δ1

δ2

δ3

(1.113)

Now consider the stiffness matrix. The stiffness matrix from element 1 can be

combined with the stiffness matrix from element 2 to give the structure stiffness

matrix.

[K] =

2EA

L−

2EA

L0

−2EA

L

(

2EA

L+ 2EA

L

)

−2EA

L

0 −2EA

L

2EA

L

(1.114)

This gives

[K] =2EA

L

1 −1 0

−1 2 −1

0 −1 1

(1.115)

Therefore, the overall relationship for the structure is

33


F1

F2

F3

=

2EA

L

1 −1 0

−1 2 −1

0 −1 1

δ1

δ2

δ3

(1.116)

We can then solve this by applying known loads and forces. By inspection,

the known applied loads at each node are

F1 = R1 +wL

4

F2 =wL

2

F3 =wL

4

The known displacements are

δ1 = 0

δ2 = Unknown

δ3 = Unknown

Therefore

F1

wL

2

FwL4

=

2EA

L

1 −1 0

−1 2 −1

0 −1 1

0

δ2

δ3

(1.117)

Solving this gives

F1 = −6wL

8(1.118)

34


δ2 =3wL2

8EA(1.119)

δ3 =wL2

2EA(1.120)

These are then used to obtain the unknown reaction, R1.

R1 = −wL (1.121)

The strains and stresses can then be calculated for each element. For element

1, the strain is calculated as

ǫ(1) =3wL

4EA(1.122)

And the stress in element 1 is

σ(1) =3wL

4A(1.123)

For element 2, the strain is calculated as

ǫ(2) =wL

4EA(1.124)

And the stress is

σ(2) =wL

4A(1.125)

Now, withe reference to figure 1.20, we can compare the results obtained with

the two two-noded elements, to the theoretical results and the single two-noded

element.

35


Figure 1.20: Two two-noded finite element model

By dividing the structure into two elements, the accuracy of the solution

has improved. However, it still does not entirely match the theoretical results.

Therefore it can be said that by splitting the model up with more elements,

the solution approaches the theoretical solution. However, by doing so, this will

also increase the size of the structure stiffness matrix. Increasing the size of the

stiffness matrix also increases the complexity of the problem and will take more

time to evaluate, hence the need for a computer.

36

breve intro a e.f

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