Edgar Snchez Prados Aleix Matabacas Enebral Stephen Rhoton Melndez

Material properties exercise report

1. Objective and procedure

The aim of this exercise is to characterize the mechanical behavior of starch/barley straw composites, doing tests to four different materials composed by a matrix of starch and different concentrations of barley straw.

The tests for this characterization of the materials are related to density measurements,

tensile tests at 0'5%-1%/min strain rates for some of the samples, and creep tests only for the pure starch sample but at three different temperatures.

2. Density measurements Here we present the table of the different densities for each one of the materials:

Material Density (g/cm3)

Standard deviation (g/cm3)

Fraction of fiber

Fraction of matrix

0% of fiber 1,3684 0,0547 0 1

5% of fiber 1,3643 0,0508 0,0487 0,9513

10% of fiber 1,3597 0,0317 0,1027 0,8973

20% of fiber 1,3508 0,0793 0,2082 0,7918

As we can see in the results, the fractions that we obtained with the calculations of

percentage of each one of the materials are very close to what the labels says. The one with 20% of fiber has the highest difference, but is still within a reasonable range.

3. Fracture behavior On the videos we can observe the same test carried out in three composite materials and

one with only the matrix compound. The test consist in deforming the material at 10%/min. In the material composed only by starch (matrix material) we can see that it has the major

ductility; this material deforms more time than the others before it breaks. As we increase the barley straw concentration (%wt) in the other materials, we can appreciate that they present a lesser ductility and become more brittle. In other words, the materials deform less and break in less time as the concentration of barley straw increases.

4. Analysis of the tensile tests The stress-strain curves at 0'5%/min are the following:

0

5000

10000

15000

20000

25000

30000

35000

40000

0 0,2 0,4 0,6 0,8

Stre

ss (

kPa)

Strain

Stress-strain for 0% of fiber material

0

10000

20000

30000

40000

50000

60000

70000

0 0,2 0,4 0,6 0,8 1 1,2

Stre

ss (

kPa)

Strain

Stress-strain for 5% of fiber material

The Young's modulus at 2 and at 5 minutes for each material (and both at 0'5-1%/min strain rates) are the following:

Young's modulus (kPa)

Material At 2

minutes At 5

minutes

0'5%/min 0% of

fiber

353259 376199

1%/min 451681 331853

0'5%/min 5% of

fiber

349230 371909

1%/min 1007542 659554

0'5%/min 20% of

fiber

13008168 8140842

1%/min 9465310 ---

Note that for the material with 20% of fiber and at 1%/min strain rate, we don't have any

information after 3'85 minutes, so we can't really say what Young's modulus it would have at

minute 5 of the strain test.

Next step is to calculate the hypothetical Youngs modulus at 50%wt of fiber material. To do the calculations, first well need to find the average Youngs modulus of each material sample at the different strain rates as shown on the next table:

E (kPa)

x (%wt) 0,5%/min 1%/min

0 96355,52445 180167,218

5 151238,373 358885,224

20 4436463,599 7967349,03

0

50000

100000

150000

200000

250000

300000

0 0,05 0,1 0,15 0,2

Stre

ss (

kPa)

Strain

Stress-strain for 20% of fiber material

Once we have this data we can plot the Youngs modulus over the concentration and then we can estimate E at 50%wt. If we take the three values at a strain rate of 0,5%/min well get the following plot:

As we can see, theres some strange behavior. Instead of using the three values, well just use the ones for x=5%wt and x=20%wt. Well delete the value at x=0%wt. This way, well have linear plot that will allow us to calculate the needed Youngs modulus. At a 50%wt we consider E1 as the E at 0,5%/min rate and E2 as the E at 1%/min rate.

From which we obtain:

1 = 285682 50 1 106 =

-1000000

0

1000000

2000000

3000000

4000000

5000000

0 5 10 15 20 25

E [k

Pa]

x [%wt]

y = 285682x - 1E+06 R = 1

0

500000

1000000

1500000

2000000

2500000

3000000

3500000

4000000

4500000

5000000

0 5 10 15 20 25

E [

kP

a]

x [%wt]

E at 50%wt and 0,5%/min rate

From which we obtain:

2 = 507231 50 2 106 = Once we know this data, we can use both the loading along and the loading across to calculate the Youngs modulus of the hypothesis. Well use the following expressions: Loading along:

= + ( ) = 13284100 0,5 + 23361550 (1 0,5) = Loading across:

=

+

=1

0,513284100 +

1 0,523361550

= ,

And the results obtained are the ones on this table:

Loading along E = 18322825 kPa

Loading across E = 16937189,9 kPa

5. Analysis of the tensile tests of pure starch at different temperatures

In order to know the stress value when stable creep flow is observed for each given temperature, we can look at the respective graphics. Note that, in this case, the strain-rate for all the temperature was 0'5%/min:

y = 507231x - 2E+06 R = 1

0

1000000

2000000

3000000

4000000

5000000

6000000

7000000

8000000

9000000

0 5 10 15 20 25

E [

kP

a]

x [%wt]

E at 50%wt and 1%/min rate

We can see a stable creep flow between 30-40% of strain. The tension value in this interval is around 93431,27 kPa.

We can observe that the stable creep flow is around the same strain, but the stress applied

is a lot lower. In this case, is that of 46011,8 kPa.

0

10000

20000

30000

40000

50000

60000

70000

80000

90000

100000

0 0,2 0,4 0,6 0,8

Ten

sio

n (

kPa)

Strain

Stress-strain at 15C

0

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

0 0,2 0,4 0,6 0,8

Stre

ss (

kPa)

Strain

Stress-strain at 25C

So, we can observe again, the stable creep flow around the same strain, and the stress applied is lower. In this case, is that of 35637,48 kPa.

Here we have the stable creep flow for the same interval of strain, but even lower stress is needed than in the previous test. For a temperature of 35C, the stress applied during the stable creep flow is 23766,23 kPa.

The following thing to obtain is a graphic of the nepierian logarithm of the strain rate to stress ratio versus the inverse of temperature at K and, from there, determine the activation energy for creep flow.

-5000

0

5000

10000

15000

20000

25000

30000

35000

40000

0 0,2 0,4 0,6 0,8

Ten

sio

n (

kPa)

Strain

Stress-strain at 29C

0

5000

10000

15000

20000

25000

30000

0 0,2 0,4 0,6 0,8

Stre

ss (

kPa)

Strain

Stress-strain at 35C

With the equation given by the plot, we can calculate now the activation energy for creep flow, which is 61399,35 J/mol. In fact, were considering that the slope (m) follows the following equation:

=

Where Q is the activation energy on J/mol and R is the gas constant taking a value of R=8,134

J/molK.

Knowing the activation energy and the model for the creep flow, we can predict the stress

necessary to generate a strain rate of 2%/min at a temperature of 10C, after some operations

which is 580034,96 kPa.

To get to this final result we first need to calculate the logarithm of the stress at 10C from the

data we obtained on the plot and from there we must find first the paramater A before we can

calculate the stress at 10C and a strain rate of 2%/min. Thus, parting from the data of the plot

we determine the stress at 10C and 0,5%/min strain rate and then the parameter A we need.

The procedure is as follows:

ln() = 6178,8 (1

283,15) 9,9371 = 11,88

stress = 11,88 = 145008,74

=

= 103,41 Where is the strain rate.

Once we know the A parameter we can calculate the strain or stress at 10C and a strain rate of

2%/min. It will be:

=

=

0,02min1

103,41

6178,8 /

8,314

283,15 = ,

y = 6178,8x - 9,9371 R = 0,9966

9,8

10

10,2

10,4

10,6

10,8

11

11,2

11,4

11,6

0,0032 0,00325 0,0033 0,00335 0,0034 0,00345 0,0035

ln(S

tres

s)

1/T [K-1]

ln(Stress) vs 1/T

LN stress

Linear (LN stress)