Modeling of mechanical behavior of materials - Reports

S

L T LT s

L

The value provided as result in Abaqus is the evaluation of this formula, its second power

must be compared with the verge value, 1, to establish the correctness of the design.

4.2. Modeling of composite vessel

The object of the work is a composite overwrapped pressure vessel, which works under very

stressful condition; the cylinder shape is popular for its light weight and high strength,

but it provides some criticalities due to the high stiffness of the fibers, that can hardly be

bent. Here are reported the dimensions of the vessel:

• L = 1830mm

• D = 889mm

26 4| COPV

• P = 1M P a

The load is applied onto the internal surface.

Figure 4.1: Vessel’s geometry

The element has been modeled with shell element, for an eight of the entire element and

the x-y-z symmetry has been exploited. The composite is made by 4 plies of graphite

fibers absorbed in an epoxy matrix generates the following elastic properties: E =

1

and

131000M P a, E = 10300M P a, ν = 0.22, G = G = 6900M P a G = 4032M P a.

2 12 12 21 23

The four plies are oriented with 45, - 45, 90, 0 degrees with respect to the x-axis, and

each of them is thick.

0.127mm Figure 4.2: Laminate layup

4.3. FEM Analysis

All the element has been meshed with an average size of and generates over

20mm 7700

nodes. The analysis has been performed with the quadratic formulation with 5 degrees

of freedom per each node, avoiding the internal torsion. 27

4| COPV Figure 4.3: Von Mises stress

As can be seen, knowing that there is a scale factor, the highest stress is on the

164x

lateral surface, where expected. But as previously said, in composites Von Mises stress

doesn’t have the same importance as the Tsai-Hill evaluation, so it is also reported.

Figure 4.4: Tsai-Hill values

In this case, the value is so it is acceptable, but this picture shows only the values in

≤ 1

the external layer. In order to better appreciate the global resistance the plot of Tsai-Hill

criterion is made as a function of the thickness:

28 4| COPV

Figure 4.5: Tsai-Hill over thickness

How can be seen, the highest value is in the middle layer, it overcomes of the ex-

0.34

ternal layer, but still remaining under 1. By considering these results, it’s clear that an

optimization process can be applied to this structure in order to minimize the weight,

a fundamental requirement due to the fact those elements are widely used in aerospace

sector.

4.4. Optimization

The first kind of optimization is the one about the weight, the solutions are simply an

epoxy-graphite or epoxy-glass, or a mixture of the two. The whole mass depends on the

density of each ply and the volume, so the number of plies. In order to minimize the

weight, without looking at the cost, the best solution is using only graphite which has the

lowest density between the possibilities: against . The second step

3 3

1800kg/m 2500kg/m

is finding the correct number of plies, which should be the lowest possible, and the correct

combination of angles in order to have the minimum Tsai-Hill value. It’s required a safety

coefficient equal to 2, so the pressure has been doubled since the very beginning. In this

case the load is only perpendicular to the surface, but two different situations are present:

on the lateral surface there should be a mojority of fibers aligned with the cylinder’s axis

so 90°; on the top and bottom surfaces the load is still perpendicular to them and so it

requires a different fibers’ orientation, 0°. The solution has been found by considering only

those two angles and by changing the number of plies and the combination of orientations:

• 10 symmetric plies, repetition of 3 sequences of 3;

Layers:

• following sequence 90, 90, 0;

Angles: 29

4| COPV Figure 4.6: Tsai-Hill MatLab Graphite

The graphs present the results of the failure criteria computed in MATLAB. The obtained

value remains below 1, as required, indicating that this combination achieves the lowest

possible mass due to the limited number of layers.

Considering the unit cost, the density and the volume of the new composite layup the

overall cost of the compost will be around assuming the density of the ply equal

12850$,

to the one of the graphite. The second optimization should be done on the cost: it is

possible to compute the number of plies required using only glass fiber, less expansive than

graphite, to obtain the same cost. The results is 30 ply, for a symmetric configuration 15

for each side. Figure 4.7: Tsai-Hill MatLab Glass Fiber

30 4| COPV

This result was obtained by slightly adjusting the orientation of the plies to minimize the

Tsai-Hill parameter, considering the smaller difference between and in glass fiber

E1 E2

compared to graphite. However, the value remains above the acceptable limit, leading to

the selection of graphite to ensure compliance while also minimizing costs.

The final optimization considers both cost and mass. However, as previously described,

the initial solution using full graphite meets all the required criteria.

As a confirmation, the same composite layup was implemented in Abaqus, and the cor-

responding solution is reported below.

Figure 4.8: Tsai-Hill Abaqus

As can be seen, the value are a little bit different, there a error from MatLab to

10%

Abaqus. This can be caused by numerical roundings and other problems strongly related

to the computations. However, considering the safety factor, the real value is around 0.51

which is acceptable. 31

5| Tensile Test

5.1. Introduction

The characterization of a material is typically conducted based on theoretical principles;

however, it is also essential to practically validate the theory and determine whether the

designed material’s properties meet the specified requirements.

The objective of the current experiment is to establish the mechanical properties of a

composite material. This involves starting with the pure polymeric matrix, Polyamide 6

(PA6), and progressively analyzing composites with varying weight fractions of reinforcing

fibers—in this case, fiberglass—ranging from to

15% 50%.

To achieve this, the following elements are required:

• a tensile test machine and specimens of composites;

• extensometer and MEMS system to collect data.

5.2. Specimen

Specimens prepared for standard tensile testing must adhere to specific guidelines outlined

by ISO regulations. Firstly, the dimensions of the specimens must be highly precise, with

tolerances in the range of hundreds of microns. In addition,the extensometer must be

attached according to precise specifications to ensure accurate sstrain measurement.

Figure 5.1: Specimen dimensions

32 5| Tensile Test

This particular specimen’s shape is perfect for the fiber reinforced composite material,

thanks to the long gauge length which allows to assume that fibers are disposed along the

Different test

main dimension. Here, are long fibers with a length radius ratio of L = 30.

D

are performed:

• with all the three composites and also the alone matrix;

5mm/min

• with all the three composites and also the alone matrix;

50mm/min

• with all the three composites and also the alone matrix;

500mm/min

this choice is made to ensure that in three different situations, all the peculiarities of the

material can be appreciated, like an intense, or less intense, viscoelastic behavior (strongly

time-dependent).

5.3. Results

Firstly is reported a comparison between the same specimen with different speed, starting

from the pure matrix and increasing the weight fraction.

Figure 5.2: PA6

In this case, due to the material’s ductility, the tensile test conducted at the lower strain

rate was interrupted to save time. Under such conditions, the material behavior closely

resembles that of a simple polymer, characterized by a prominent plateau in the plastic

region before stress increases sharply at failure. 33

5| Tensile Test

As observed, increasing the strain rate leads to a more distinct yield strength, while the

viscoelastic response becomes less pronounced. As expected, the stress values, expressed

in MPa, remain relatively low; however, the strain is substantial, reaching values of several

hundred mm/mm.

Focusing on the elastic region, it is evident that the curves share the same slope, indicating

a consistent elastic modulus. This modulus has been calculated and will be discussed in

detail later. Figure 5.3: FV15

In this graphic are presented the same curves but using the FV15 specimen, as expected,

by looking the magnitude of the strain and stresses values, is clear that the composite

is much more stiff than the simple polymer. Also in this case the increasing of the

speed generates a more appreciable brittle behavior and the amount of energy absorbed

decreases.

34 5| Tensile Test

Figure 5.4: FV30

By looking the curves, it seems that the composite with and increased amount of fibers

is less stiff than the previous one, FV15, but the values of the stress are higher and the

strain’s ones are lower; the computation of the elastic modulus will confirm the higher

stiffness. Figure 5.5: FV50 35

5| Tensile Test

FV50 is the stiffest material among the composites of this experiment, the strain at the

failure is less than 5% and the UTS can reach, in the fastest case, values around 200M P a.

Now, the effective numbers of elastic modulus, yield strength, UTS and at failure can

ε

be calculated. The elastic modulus has been computed by interpolating the data in the

elastic region between these two values: and (they come from ISO

ε = 0.0005 ε = 0.0025

regulation). The following numbers refer to the , the longitudinal stress.

E

1

ElasticM odulus(E)[Gpa] 5mm/min 50mm/min 500mm/min M assF raction

−

P A6 1 1.38 2.12

F V 15 3.57 3.65 3.95 0.15

F V 30 6.17 6.55 7.22 0.3

F V 50 11.34 12.28 13 0.5

As can be seen, the stiffness increases with ah higher amount of fibers, but it’s also a little

bit time dependent. These values have been calculated with a least square approxima-

tion with a first grade polynomial. These values can be plotted on the same line which

represents the rule of mixture, the way to calculate the elastic modulus as a function of

volume fraction. Firstly is required the calculation of the volume fraction, thanks to the

density the math is pretty simple:

• 15% = 0.0718;

• 30% = 0.1582;

• 50% = 0.3048. Figure 5.6: Longitudinal Elastic Modulus

36 5| Tensile Test

This figure demonstrates how the experimental data fit the points; Halpin-Tsai correction

was used in order to obtain a better fitting by considering the heterogeneous fiber.

2·l f

ε = r f

(experimental value which best fits the points)

η = 2 1−ηεV

f

E =

1 1−ηV

f

Following, yield st