Material Tensile Testing

Material Original Diameter Original Length Length After Test

Aluminium Alloy 4.97mm 26.98mm 36.8mm

Steel 5.04mm 27.15mm 30.6mm

Brass 5.0mm 27.19mm 30.8mm

From these results I can find the ultimate tensile strength, Young’s modulus, percentage elongation and from the graph the yield point and proof stress of the material.

Aluminium sample: TS1030-2011-T3

UTS =(MAX LOAD)/(ORIGINAL CROSS SECTIONAL AREA)

Area=πr^2, r = half of diameter, D = 4.97 mm, r = 2.485 mm A=π'〖'2.485'〗'^2 = 19.40 mm²

Maximum Load is approximately 7200 N as shown on the graph in green.

UTS= 7200/19.4 = 371.134 Nmm²

Percentage elongation = (CHANGE IN LENGTH)/(ORIGINAL LENGTH) X 100

Percentage elongation = (9.82 mm )/(26.98 mm) X 100 = 36.39 %

The yield point as shown on the graph in aluminium is demonstrated as proof stress and is 0.2% of the strain as shown. Where the material is no longer in its elastic region and will plastically deform this is shown on the graph in RED with the force applied at approximately 6400 N and at 1 mm of elongation.

Proof Stress = (Load at 0.2% proof )/(Original Cross Sectional Area) = 6400/19.4 = 329.896 Nmm²

Young’s modulus = STRESS/STRAIN

Using the excel spread sheet data the stress and strain can be calculated.

Stress = FORCE/(CROSS SECTION AREA) = 5952.667N/19.4mm²

Strain = (CHANGE IN LENGTH)/(ORIGINAL LENGTH) = 0.8mm/26.98mm

To calculate young’s modulus, I calculate the stress over strain at 0.8 mm as this is where the yield point is and the relationship is no longer linear. It is as follows.

Young’s modulus = 306.8385052/0.029651594 = 10348.12859 Nmm², = 10348.12859 MPa

For this although I have gain results, I need to verify these results to see if they are in the expected region. If they are not in an ideal situation I would perform the test again and compare the results further.

Source Young’s Modulus Figure

My Results 10348.12859 MPa = 10.381 GPa

Matweb (2016) 70 GPa

Engineering Materials 1- An introduction to Properties, Applications and Design. Ashby and Jones (2005) 69-79 GN m⁻²

As it clearly shows in the table the result that I have obtained is not the expected result from recognised sources. There may be many reasons for this, the obvious reason may be that my calculations are incorrect and there may be a mistake. Another reason could be the type of aluminium data that is used and different alloying elements alter the properties and therefore alter the young’s modulus. Another factor could be the validity of the test, whether there was an error in performing the tensile test itself and another factor could be a defect in the material that was used as the test piece. In all circumstances the test would need to be carried out again to eliminate the factor of making a mistake when performing the test. And then recalculated and the results compared against each other.

Steel tensile test

UTS =(MAX LOAD)/(ORIGINAL CROSS SECTIONAL AREA)

Area=πr^2, r = half of diameter, D = 5.04 mm, r = 2.52 mm A=π'〖'2.52'〗'^2 = 19.95 mm²

Maximum Load is approximately 12400 N as shown on the graph in green.

UTS= 12400/19.95 = 621.553 Nmm²

Percentage elongation = (CHANGE IN LENGTH)/(ORIGINAL LENGTH) X 100

Percentage elongation = (3.45 mm )/(27.15 mm) X 100 = 12.7 %

The yield point as shown on the graph is where the material is no longer in its elastic region and will plastically deform this is shown on the graph in RED and is approximately at 11600 N and at 1.3 mm of elongation.

Yield Stress = (Load at Yield )/(Original Cross Sectional Area) = 11600/19.95 = 581.453 Nmm²

Young’s modulus = STRESS/STRAIN

Using the excel spread sheet data the stress and strain can be calculated.

Stress = FORCE/(CROSS SECTION AREA) = 11534.67N/19.95mm² = 578.1789474

Strain = (CHANGE IN LENGTH)/(ORIGINAL LENGTH) = 1.3mm/27.15mm = 0.047882136

To calculate young’s modulus, I calculate the stress over strain at 1.3 mm as this is where the yield point is. It is as follows.

Young’s modulus = 578.1789474/0.047882136 = 12075.04494 Nmm², = 12075.04494 MPa

For this again although I have gain results, I need to verify these results to see if they are in the expected region just as before. If they are not in an ideal situation I would perform the test again and compare the results further.

Source Young’s Modulus Figure

My Results 12075.04494 MPa = 12.075 GPa

Matweb (2016) 205 GPa

Engineering Materials 1- An introduction to Properties, Applications and Design. Ashby and Jones (2005) 200 GN m ⁻²

Again just like the aluminium results there seems to be a trend that my results are significantly lower than that of the expected results and as before there could be many reasons for this. However I will continue and analyse the brass tensile test further.

Brass tensile test

UTS =(MAX LOAD)/(ORIGINAL CROSS SECTIONAL AREA)

Area=πr^2, r = half of diameter, D = 5.0 mm, r = 2.5 mm A=π'〖'2.5'〗'^2 = 19.634 mm²

Maximum Load is approximately 10800 N as shown on the graph in green.

UTS= 10800/19.634 = 550.066 Nmm²

Percentage elongation = (CHANGE IN LENGTH)/(ORIGINAL LENGTH) X 100

Percentage elongation = (3.61 mm )/(27.19 mm) X 100 = 13.276 %

The yield point as shown on the graph is demonstrated as proof stress with an offset of 0.2% of the strain where the material is no longer in its elastic region and will plastically deform this is shown on the graph in RED with the force applied at approximately 8600 N and at 1.2mm of elongation.

Proof Stress = (Load at 0.2% proof)/(Original Cross Sectional Area) = 8600/19.634 = 438.015 Nmm²

Young’s modulus = STRESS/STRAIN

Using the excel spread sheet data the stress and strain can be calculated.

Stress = FORCE/(CROSS SECTION AREA) = 8460.667N/(19.634 mm²) = 430.9191708

Strain = (CHANGE IN LENGTH)/(ORIGINAL LENGTH) = 1mm/27.19mm = 0.036778227

To calculate young’s modulus, I calculate the stress over strain at 1 mm as this is where the yield point is and the relationship between the 2 is no longer linear. It is as follows.

Young’s modulus = 430.9191708/0.036778227 = 11716.69225 Nmm², = 11716.69225 MPa

For this again although I have gain results, I need to verify these results to see if they are in the expected region just as before. If they are not in an ideal situation I would perform the test again and compare the results further.

Source Young’s Modulus Figure

My Results 11716.69225 MPa = 11.716 GPa

Matweb (2016) 82- 117 GPa

Engineering Materials 1- An introduction to Properties, Applications and Design. Ashby and Jones (2005) 103- 124 GN m ⁻²

As with the previous results my calculated results do not cohere to what is expected from other sources and I think there may be an error in my calculations or the tests were not carried out correctly. It is unlikely to be a material flaw or imperfection in all 3 test samples. Therefore the obtain the correct figures the tests would need to be ran again to be able to obtain and gather data that is within the expected region, if the test results were similar to the original tests than the fault may be within the machine itself and therefore not reliable in testing the material.

One thing that can be found from the data is the characteristics of the materials themselves. To start with the aluminium sample shows ductile properties as this has the highest percentage elongation but with less force to fracture the sample. The steel however had the lowest amount of elongation and a lot more force had to be applied to fracture the sample. This would indicate the material has higher tensile strength and is a stronger metal, in which compared to aluminium and brass this is true. The brass data indicates a less ductile metal than aluminium but more pliable than steel. With a similar percentage elongation to steel but with a lower amount of force this indicates a metal with good strength and some ductility which again is true to the metal properties.

Hardness Testing

The next test was to determine the hardness of the material with respect to a scale. In this case the scale was Brinell hardness scale. Each sample was tested 5 times to get an average of the hardness. The results are as follows;

Material Brinell Hardness Average Hardness Expected value Matweb (2016) Expected value

Brass 110, 112, 112, 102, 108 108.8 55-73 53F (Engineering handbook, 2016)

ABS 469, 414, 431, 417, 421 430.4 191-473 103-112 Rockwell R (Test standard 2016)

High Impact Polystyrene 403, 419, 365, 413, 404 400.8 42-84 Rockwell scale L 55 Rockwell scale L (kutz, 2002)

Steel 211, 208, 214, 212, 210 211 105 105 (Pivotpins, 2016)

As the results show there are many different figures and values available in terms of hardness and with different scales. The polymer materials were the hardest material to find any data on as the tests that are completed tend to be using Rockwell machines opposed to brinell and therefore the data is not directly comparable. However it does show in comparison to other materials they are not as hard even if the values do not translate the same figure. The metal samples were in brinell scales and therefore can be compared, the brass that I tested seems to have a higher value than that of the data that I have found. Meaning the brass was not as hard as the available data from other tests. There may be many reasons for this and the most likely is that the brass compositions are not the same and therefore the final material will have different hardness properties. The steel value followed a similar trend whereby the values I found were of a harder grade of steel but from both sources the values were the same so it appears the independent tests were of a similar, if not of the same material.

Izod impact testing.

The next test was to carry out an impact test, this one being an Izod test. The data is as follows.

Material Width Depth Start Energy Break Value Value Expected value Expected value

ABS (1) 1.58 mm 12.770 mm 25.046 J 1.34095 J 66.4615 KJ/M² 848.713 J/M 10-20 KJ/M² (BPF, 2016) 0.1- 0.5

KJ/M

(Kutz,2002, 338)

ABS (2) 1.7 mm 13.13 mm 25.046 J 1.2406 J 55.5815 KJ/M² 729.785 J/M 10-20 KJ/M² (BPF, 2016) 0.1- 0.5

KJ/M

(Kutz,2002, 338)

Acrylic 3.89 mm 12.86 mm 25.046 J 1.0580 J 21.15510 KJ/M² 272.003 J/M 0.118 - 1.47 J/cm (Matweb, 2016) 21.6 J/M

(Akrylic)

As the table shows just like the other tests the values vary significantly from one source to the next and to my values. This again I think is because of the composition of the materials and the fact that they may not be the same. However the properties can be investigated from the results that are present, it appears from the table that the acrylic took less energy to break and therefore is less impact resistant than ABS however this may also be a factor of the condition of the notch and therefore may not be an accurate reading for the impact resistance.

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