Report analisi multi field problems

A)

B) Fig.174 Point 1 different material, coupling A) A-A B) W-A

A) 56

B) Fig.175 Point2 different material, coupling A) A-A B) W-A

A)

B) Fig.176 Point 3

57

A)

B) Fig.177 Point 4

The behavior is very different for the materials considered. While for isotropic and orthotropic

material the first peak is more or less at the same frequencies, for sandwich plate the first peak is at

lower frequencies, there are more resonance peaks and finally the pressure value is generally higher

than the other two materials.

6. CONCLUSIONS

With this exercise we have seen how a plate interacts when placed between two fluids.

In general, if we have a sandwich plate the trend is more complex: pressure with higher values and

with more peaks.

The trend is also different depending on the fluids used: if used a Water-Air (W-A) coupling in

general the pressure will have higher values is greater presence of peaks, on the contrary an Air-Air

coupling (A-A) returns us lower values and less peaks; the first peak in the latter case you have at

frequency values more others than the W-A coupling

58

Cap.7 THERMO-MECHANICAL ANALYSIS

Let's see the thermo-mechanical problem: this is always derived from PVD, considering an extended

form of it, called the Hamilton principle: {̅

{} } {}

({}

= ∫ − − ∙ ) =

⏟

{ }−{ }

{} []{} []{ } []{ } [][]{} []{}

= = − = −

⏟

{}

{ } []{}

= = mechanical deformation

{ } {}

= =thermal deformation

{} = vector of the coefficient of thermal expansion

= − = temperature increase

0

{̅} []

{ }

= ; ; = = temperature gradient

[] = [ ; ; ] = partial derivative vector

[]̅

{} = { , , } = = Heat flux

0 0

11

[] 0 0

= [ ] =tensor of thermal conduction coefficients (in the material

22

0 0

33 0

11 12

0

→ [ ] =

reference system) tensor in the global reference

12 22

0 0

33

= variation of entropy in the structure

{̅ }{} = weak form of the Fourier equation, takes into account the change in

temperature.

− ∙ = term considered in the case of dissipative plastic deformation: this term is often

overlooked.

We need to consider a set of Constitutive equations that tie variables to unknowns and consider fem

+ kinematic model approximations along the thickness:

Constitutive equations: {} []{} []{ } []{ } [][]{} {}

= = − = −

[]̅

{} [][]

= =

{} {} {} []{}

= + ∙ = + ∙

59

=

0

= material-specific heat per unit mass

= density

= reference temperature

0

Fem+Kinematic model along the thickness:

{} () (, } {} () (,

= ){ = ){ }

() (, () (,

= ) = )

By substituting and expanding on the subscripts we can lead back to the following algebraic system:

[ ] [ ] {} {}

[ ]{ } = { }

[ ] [ ] {} {0}

We can write the final system in a contracted form by assuming an equivalent stiffness matrix and an

equivalent primary variable vector. This vector can be divided into two vectors that contain the known

quantities given by boundary conditions and a vector of the unknown quantities to be calculated,

respectively.

{} { }

= ;

= unknow quantities

= assigned quantities given by BC

Boundary conditions are no longer assigned by giving zero value to some displacement components,

but we can remove rows and columns in the stiffness matrix associated with null degrees of freedom.

However, if the components are not null, it is easier to assign conditions to the boundary as above,

with the known value. 0

{}

{ }

[ ] + [ ] { } =

0

0 {}

[ ] { } = =

{} {}

{ }

[ ] = +

0

In solving the exercises, we will often use this formulation.

If the temperature profile is linear, we expect, given the relationships, that the displacement is

parabolic, this is because the field of deformations must be consistent (the deformations are

proportional to the temperature and are the derivatives of displacements). If the temperature has a

generic distribution that can be parabolic/cubic/etc... we expect the displacements to be of a higher

polynomial order. This makes us understand that in thermo-mechanical problems it is important to

use higher order theories, classical models are not suitable in this case. This is a problem in most

finite model software that only implement classic models.

In the exercises we will compare the temperature distribution for different plate configurations, both

material and lamination. We will see the distribution of the displacements, in particular the transversal

composition as for previous exercises, and we will evaluate the stress to check the distribution of

these stress, in particular in the multilayer plate. 60

In code, what is implemented is the mating problem. Applying both the kinematic model and FEM,

we derived the final algebraic system, in which we can consider both contributions of the Hamilton

principle or, for example, we can overlook the entropy variation and consider only the thermal energy

associated with the heat flow, and in this case we have an equivalent stiffness matrix that includes the

classical stiffness matrix , the coupling matrix that includes the thermal stress (the matrix depends on

the technical expansion coefficients) and another matrix in which the temperature profile is calculated

in accordance with the thermal conductivity coefficients of the material.

7.1 Thin plate, orthotropic material, temperature/displacement

As a first exercise we perform a static analysis of a square plate one meter long subject to a

temperature distribution not considering other mechanical loads. The load vector will not be null

because we remember that we assume as external loads the vector given by the conditions at the

contour.

7.1 INPUT FILE

Input files are like vibro-acoustic problems. In this case we consider two meshes:

15x15 Q9 elements;

- 10x10 Q9 elements;

-

The results obtained for the same analysis are the same, so we can directly use the mesh with 100 Q9

elements.

The code for thermo mechanical analysis is 111. You have to change and enter this code even if the

analysis is static, because there is another procedure implemented only for thermal problems. Thanks

to this input, the code recognizes that it must also calculate the thermal matrix involved in the final

algebraic system.

Regarding the conditions at the contour, we have a plate simply resting.

Fig.178 Boundary Condition, sine from the distribution of

As for the edges parallel to the x-direction, the displacement components perpendicular to the edge

itself are free, while the other two components are null. For edges parallel to the y-direction, and

components are null. The temperature increase is imposed as a boundary condition on the top and

bottom surface of the plate and represented by the last columns of the T-PLATE rows: we have two

values on the lower and upper surface; all other values are intermediate between the plate. The first

four columns always refer to the plane to which we are applying the condition to the contour. The

= −0.005 = 0.005

last coefficient will be for the upper surface and to the lower surface: the

value of this coefficient is the inverse of the z-coordinate of the surface of the plate in question. The

temperature distribution is made sine so that the values can be compared with the analytical results.

0+

+

= sin ( ) sin ( ) =

−

−

= sin ( ) sin ( ) =

0

61

We could also consider a linear distribution along the thickness, assigning temperature values along

the thickness.

Fig.179 Boundary Condition, linear distribution of

We could first consider an isotropic material, because even in this simple case we could see in the

results that the distribution of displacements will not be linear but parabolic. This is the case with the

theory because the displacements are proportional to the temperature profile. For this exercise,

however, we consider an orthotropic material since the analytical solution has been calculated with

this material. Fig.180 MATERIAL

The mechanical properties of the material used are unusual, for example, the Young E module has a

very low value. This is because we are analyzing an academic case without any real physical

application. In fact, the goal is just to validate the model. Young's modulus and G-cutting module are

however given so that they have different values in the direction of the fibers than the other directions.

Unlike static analyses, we must also provide thermal properties. To insert these properties, you add

two lines to the file:

the first, T-EXP, provides the six coefficients of thermal expansion of the material; the last

- three coefficients are usually null even for orthotropic materials, but there are cases when

they must be considered;

the second, T-CON, provides thermal conduction coefficients; are organized into groups of

- three: the first three represent the first line of the tensor, the second three the second, and

finally the last three the third line of the tensor.

Please note that all these coefficients are provided in the material reference system.

the ID used for these two rows must be the same as for the material (second column).

As far as lamination is concerned, we are considering a plate with three layers of equal thickness with

lamination 0°-90°-0°. Fig.181 LAMINATION

62

The model used for analysis is the LE model. We use a linear expansion with 3 elements and 4 total

nodes along th