Objectives

PART 1: To show that a tensile test within the elastic limits leaves no residual elongation in the material and to show tensile test within the plastic limits leaves residual elongation in the materials.

PART 2: To derive the modulus of elasticity, E for steel, aluminium, brass and copper.

PART 3: To complete a tensile test on steel, aluminium, brass and copper.

Procedure

Equipment: TENSILE TEST MACHINE MT 3017

Figure 5.1Tensile Test Machine Mt 3017

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Figure 5.2 Test piece (before and after)

PROCEDURE:

Part 1

1. The dial gauge is loosened and it is being lowered on the column to allow the test piece to be screwed into the jaws.

2. The pressure cylinder is wind back to zero by turning anti- clockwise.3. The measurement Lo and do is checked as shown in Diagram (1b) of mild steel test

piece and then it is being screw in to the test piece.4. The max pointer of the gauge is set to zero.5. The dial gauge is set to 6 and the outer ring of the meter is turned on so that the outer

scales 0 lines up with the large pointer. Each revolution is equal to 0.1mm.6. The pressure is crank up, carefully; to 1KN and the dial gauge reading is recorded.7. The pressure is increased by turning clockwise, carefully to 2KN, 3KN, 4KN, and 5

KN and 0KN. All the dial gauge reading in table (a) part 1 is recorded.8. The pressure is reduced by carefully winding anti- clockwise to 4KN, 3KN, 4KN, and

5 KN and 0KN. All the dial gauge reading in table (a) part 1 is recorded.9. Procedure 1 to 5 is repeated.10. The pressure is crank up, carefully to 2KN and the dial gauge reading in table (b), part

1 is recorded.11. The pressure is increased by turning clockwise, carefully to 4KN, 6KN and 8KN, and

the dial gauge reading in table (b) part 1 is recorded. 12. The pressure is increased by turning clockwise, carefully to 6KN, 4KN, 2KN and

0KN, and the dial gauge reading in table (b) part 1 is recorded. 13. The pressure is increased by turning clockwise, carefully to 4KN, 6KN and 8KN, and

the dial gauge reading in table (b) part 1 is recorded.

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Part 2

1. The dial gauge is loosened and the column is lowered to allow the test piece to be screwed into jaws.

2. The pressure cylinder is wind back to zero by turning anti- clockwise.3. The measurement L0 and d0 as shown in Diagram (1b) of mild steel test piece and

the table (a) part 2 is recorded. Then the test jaws are screw into it.4. The max pointer of the pressure gauge is set to zero.5. The dial gauge is set to 6 and the outer ring on the meter is turn so that the outer

scales likes up with the larger pointer, each revolution is equal to 0.1mm.6. The pressure is crank up, carefully to 1KN and the dial gauge reading is recorded.7. The pressure is increased, by turning clockwise, carefully to 2KN, 3KN, 4KN, and

5KN, and all the dial gauge reading in table (a) part 2 is recorded.8. The pressure load is removed to zero.9. The mild steel test piece is changed to brass test piece.10. Procedure 1 to 8 is repeated.11. The brass test piece is changed to copper test piece.12. The copper test piece is changed to aluminum test piece.13. The procedure 1 to 8 is repeated.

Part 3

1. The dial gauge is loosened and the column is lowered to allow the test piece to be screwed into jaws.

2. The pressure cylinder is wind back to zero by turning anti- clockwise.3. The measurement L0 and d0 as shown in Diagram (1b) of mild steel test piece and

the table (a) part 3 is recorded. Then the test jaws are screw into it.4. The max pointer of the pressure gauge is set to zero.5. The dial gauge is set to 6 and the outer ring on the meter is turned so that the outer

scales0 lines up with the larger pointer.6. The pressure is cranked, carefully until the force elongation is equal to 0.2mm and

the pressure force is recorded.7. The pressure is increased by turning clockwise, carefully following force

elongation as shown in table (a) part 3 until the test piece breaks. All the pressure forces are recorded. The ultimate length, Lu and ultimate diameter, du of the test piece is measured as shown in Diagram (1b).

8. The pressure load is removed to zero.9. The mild steel test piece is changed to brass test piece.10. The procedure 1 to 8 is repeated.11. The brass test piece is changed to copper test piece.12. The procedure 1 to 8 is repeated.13. The copper test piece is changed to aluminum test piece.14. The procedure 1 to 8 is repeated.

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Notations

Tensile stress, σ = F/ A (N / mm2)

Cross sectional, A = π d2 / 4 (mm2)

Tensile strain, ϵ = ∆L / L

Robert Hooke’s law, σ = E. ϵ (N/mm2)

ϵa = ( Lu−L0 ) x100

L0

(%)

ϵ1 = ( A0−Au ) x 100

A0

(%)

Metal , alloy Typical modulus of elasticity, E(N/mm2)

Iron , Nickel 200-220Copper 125Brass 80-100

Aluminum 72

E = Modulus of Elasticity (N/mm2)σ = Axial Stress (N/mm2)F = Force (N)d0 = original Diameter of Test piece (mm)Du = Ultimate Diameter of Test Piece (mm)∆L= Extension of the test piece (mm)L0 = Original Length of Test piece (mm)Lu = Ultimate Length of Test Piece (mm) ϵa = Axial Strain of Test Piece when it’s breaking (%)ϵ1 = Lateral strain of test piece when it’s contracting (%)δ = Elongation (mm)

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Theory

Strength refers to the ability of a structure to resist loads without failure because of excessive stress or deformation. Typical points of interest when testing a material include: ultimate tensile strength (UTS) or peak stress; offset yield strength (OYS) which represents a point just beyond the onset of permanent deformation; and the rupture (R) or fracture point where the specimen separates into pieces.

Figure 5.1 Stress-strain Curve

A graphical description of the amount of deflection under load for a given material is the stress-strain curve. Engineering stress (S) is obtained by dividing the load (P) at any given time by the original cross sectional area (Ao) of the specimen.

S = P/Ao

(1)

Engineering strain (e) is obtained by dividing the elongation of the gage length of the specimen (∆l) by the original gauge length (lo).

e = ∆l/lo = (l - lo)/lo

(2)

Figure (5.1) depicts a typical stress-strain curve. The shape and magnitude of the curve is dependent on the type of metal being tested. Point A represents the proportional limit of a material. A material loaded in tension beyond point A when unloaded will exhibit permanent deformation. The proportional limit is often difficult to calculate, therefore, two practical

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measurements, offset yield strength (OYS) and yield by extension under load (EUL) were developed to approximate the proportional limit. The initial portion of the curve below point A represents the elastic region and is approximated by a straight line. The slope (E) of the curve in the elastic region is defined as Young’s Modulus of Elasticity and is a measure of material stiffness.

E = ∆S /∆e = (S2-S1)/(e2-e1)

(3)

Point B represents the offset yield strength and is found by constructing a line X-B parallel to the curve in the elastic region. Line X-B is offset a strain amount O-X that is typically 0.2% of the gage length. Point C represents the yield strength by extension under load (EUL) and is found by constructing a vertical line Y-C. Line Y-C is offset a strain amount O-Y that is typically 0.5% of gage length. The ultimate tensile strength, or peak stress, is represented by point D. Total elongation, which includes both elastic and plastic deformation, is the amount of uniaxial strain at fracture and is depicted as strain at point Z. Percent elongation at break is determined by removing the fractured specimen from the grips; fitting the broken ends together and measuring the distance between gage marks. Percent elongation at break reports the amount of plastic deformation only. The gage length used for measurement is reported with the result.

elongation at break(%) = ez = 100*(lz-lo)/lo

(4)

Reduction of area, like elongation at break, is a measure of ductility and is expressed in percent. Reduction of area is calculated by measuring the cross sectional area at the fracture point (Az).

reduction of area(%) = (Ao-Az)/Ao

(5)

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Results

Part 1:

a) The Elastic Range

F (kN) Dial Gauge

Reading (mm)

Calculated

σ= FA0

Calculated

∆ L (mm )

Calculated

∈=∆ LL0

0 0.000 0.000 0.000 0.00000

1 0.004 0.051 0.004 0.00011

2 0.008 0.102 0.008 0.00022

3 0.012 0.153 0.012 0.00032

4 0.016 0.204 0.016 0.00043

5 0.019 0.255 0.019 0.00051

4 0.017 0.204 0.017 0.00046

3 0.015 0.153 0.015 0.00041

2 0.012 0.102 0.012 0.00032

1 0.008 0.051 0.008 0.00022

0 0.003 0.000 0.003 0.00008

Table 5.1

b) Residual Elongation

F (kN) Dial Gauge

Reading (mm)

Calculated

σ= FA0

Calculated

∆ L (mm )

Calculated

∈=∆ LL0

0 0.000 0.000 0.000 0.00000

2 0.006 0.102 0.006 0.00016

4 0.014 0.204 0.014 0.00038

6 0.020 0.306 0.020 0.00054

8 0.026 0.408 0.026 0.00070

6 0.023 0.306 0.023 0.00062

4 0.017 0.204 0.017 0.00046

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2 0.011 0.102 0.011 0.00030

0 0.001 0.000 0.001 0.00027

2 0.007 0.102 0.007 0.00019

4 0.014 0.204 0.014 0.00038

6 0.010 0.306 0.010 0.00027

8 0.027 0.408 0.027 0.00073

Table 5.2

Part 2:

a) To Derive the Modulus Elasticity, E for Mild Steel, Brass, Copper, and Aluminum.

Mild Steel

L0=37 mm

d0=5mm

F (kN) Dial Gauge

Reading (mm)

Calculated

σ= FA0

Calculated

∆ L (mm )

Calculated

∈=∆ LL0

0 0.000 0.000 0.000 0.00000

1 0.003 0.051 0.003 0.00008

2 0.006 0.102 0.006 0.00016

3 0.010 0.153 0.010 0.00027

4 0.014 0.204 0.014 0.00038

5 0.016 0.255 0.016 0.00043

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0 0.05 0.1 0.15 0.2 0.25 0.30

0.00005

0.0001

0.00015

0.0002

0.00025

0.0003

0.00035

0.0004

0.00045

0.0005

Axial stress vs Tensile strain (Mild Steel)

Brass

L0=37 mm

d0=5mm

F (kN) Dial Gauge

Reading (mm)

Calculated

σ= FA0

Calculated

∆ L (mm )

Calculated

∈=∆ LL0

0 0.000 0.051 0.000 0.00000

1 0.001 0.102 0.001 0.00003

2 0.004 0.153 0.004 0.00011

3 0.007 0.204 0.007 0.00019

4 0.013 0.255 0.013 0.00035

5 0.018 0.051

0.018 0.00049

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0 0.05 0.1 0.15 0.2 0.25 0.30

0.0001

0.0002

0.0003

0.0004

0.0005

0.0006

Axial stress vs Tensile strain (Brass)

Copper

L0=37 mm

d0=5mm

F (kN) Dial Gauge

Reading (mm)

Calculated

σ= FA0

Calculated

∆ L (mm )

Calculated

∈=∆ LL0

0 0.000 0.051 0.000 0.00000

1 0.002 0.102 0.002 0.00005

2 0.004 0.153 0.004 0.00011

3 0.007 0.204 0.007 0.00019

4 0.012 0.255 0.012 0.00032

5 0.016 0.051 0.016 0.00043

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0 0.05 0.1 0.15 0.2 0.25 0.30

0.0001

0.0002

0.0003

0.0004

0.0005

0.0006

Axial stress vs Tensile strain (Brass)

Aluminum

L0=37 mm

d0=5mm

F (kN) Dial Gauge

Reading (mm)

Calculated

σ= FA0

Calculated

∆ L (mm )

Calculated

∈=∆ LL0

0 0.000 0.051 0.000 0.00000

1 0.007 0.102 0.007 0.00019

2 0.013 0.153 0.013 0.00035

3 0.021 0.204 0.021 0.00057

4 0.026 0.255 0.026 0.00070

5 0.034 0.051 0.034 0.00092

b) Modulus of Elasticity

Material Mild Steel Brass Copper Aluminum

Modulus of 210 80 – 100 125 72

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Elasticity

E (N

mm2)

Part 3:

a) To Complete a Tensile Test on Mil Steel, Brass, Copper, and Aluminum.

Force

Elongation,

σ (mm)

Steel

F(kN)

Aluminum

F(kN)

Brass

F(kN)

Copper

F(kN)

0.2 0 0 0 4

0.4 0 0 0 0

Material Original

Diameter, d0

(mm)

Original

Length, L0

(mm)

Ultimate

Diameter, du

(mm)

Ultimate

Length, Lu

(mm)

Mild Steel 5 37 3.1 37.8

Brass 5 37 4.8 37.6

Copper 5 37 3.9 38.7

Aluminum 5 37 4.2 39.1

Sample of calculations and formulas used in the results:

F: Force (kN)

σ : Axial Stress( N

m m2)

∈:Tensile Strain

∆ L=¿Dial Gauge Reading

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Axial Stress (σ )= FA0

Where:

A0=π d0

2

4

¿ π ×52

4

¿19.63 m m2

Then,

(σ )= FA0

¿1

19.63

¿0.051

Tensile Strain (∈)= ∆ LL0

¿ 0.00437

¿0.00011

So we apply the same formulas and the same procedure in calculating these values on the rest

of the results obtained.

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Discussion

Part 1

From the graph plotted of axial stress versus tensile strain, it is seen that axial stress increases

as the tensile strain increases. The tensile strain can be attributed to loading by the force

applied. The graph also depicts the residual elongation. From the results of the residual

elongation test, the data was analyzed and the curves compared. It was found that slight

residual elongation occurred.

Part 2

From the results of part 2, it was found that the modulus of elasticity for aluminum was the

lowest, followed by brass, copper, and steel. These findings are in accordance with the

theoretical values for the given materials.

Part 3

Tensile test (part 3) yielded incomplete data because 3 of the specimens broke before the

required force was applied. Copper, however, broke at 4kN and had the force elongation of

0.2mm. The premature breakage of the specimens could have been caused by faulty

equipment and gauges. Even multiple runs yielded the same results.

However, the broken specimens were analyzed and their final diameters and lengths were

obtained.

a) A0, , L and in table 5.1 and 5.2 were calculated. The meter reading was at 6 at the

start and the value decreased.

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b)

0 0.05 0.1 0.15 0.2 0.25 0.30

0.0001

0.0002

0.0003

0.0004

0.0005

0.0006

Axial stress vs Tensile strain (loading)

The Stress-Strain graph shows that the Copper sample experienced more plastic

deformation that the Steel sample and this is reflected by the higher percentage elongation

(fig.3). After it had fractured, the surface of the Copper was rough and irregular. The 2

half’s of the fractured sample showed a “cup” and a “cone” shape with an inclination of

approximately 45° on their fracture surfaces. In a uniaxial tensile test, this orientation

represents the angle of principle shear stress and the surface demonstrates this principle

Shear Stress caused the crystalline boundaries to slip over each other before failure (3).

Both of these observations are characteristics of ductile materials, which is a commonly

stated property of Copper. The Copper sample also displayed a higher Toughness than the

sample, which is represented by the larger area beneath the stress strain graph.

Despite being smaller than the Copper sample, the plastic region of the Steel sample is

significantly large enough to be considered to have some ductile properties. The surface

of the fracture also possessed a cup and cone geometry at a lesser extent than the copper

sample. The steel sample had a larger necking region than the Copper sample, which

explains the greater reduction in cross sectional area at the point of fracture, but as shown

in fig.5, the Steel sample showed a very rapid transition between the decreased area and

the rest of its length, whereas the Copper showed a gradual transition. Necking is a

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property of a ductile material. Referring to Engineering Materials the Yield Stress’s for

Copper is 60MPa, compared to the 200MPa value that was obtained experimentally.

The differences between these results suggest that;

a) The yield stress for copper that was predicted using a proof stress may of given an

inaccurate answer that is higher than the real value

b) The stress-stain results that were read from the machine were inaccurate. One

inaccuracy is that the experiment used the ‘nominal stress’ of the sample rather than the

‘true stress’.

However, the difference between the two is very small, particularly in the elastic region

of the test, and could not cause such a large difference between the experimental and

theoretical value of yield stress. This would mean the difference is more likely to be

caused by

(a) and that very little confidence can be placed on determining the yield stress with one

run of an experiment and by determining the yield stress using the graph. The

experimental value for the Yield Stress of Steel is within the theoretical range of value

which is between 260MPa and 1300MPa

(b). Because this value was more clearly defined on the graph than it was for copper and

it was not derived using a proof stress, it would be expected to be more accurate and

could have a high confidence placed on it. The values for the Modulus of Elasticity

obtained experimentally are around one order of magnitude smaller than values stated in

Engineering Materials which quotes it to be 200GPa for mild Steel and 124GPa for

Copper.

Determining the Modulus of a material using a uni-axial tensile Stress experiment is

generally regarded as being inaccurate and is instead commonly determined by measuring

the natural frequency of a sample using an oscillation test

(1). The reasons for this are; Recording small displacements of the sample is imprecise

due to the measuring equipment

(2). Factors such as creep can contribute to the strain

(3). When exerting large forces the equipment can begin to flex, and the displacement of

the machine is mistakenly read as a displacement of the sample The ultimate tensile

stresses recorded are very close to the theoretical values, which are 400Mpa and 500-

1880MPa for copper and Carbon Steel Alloy.

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(4)The difference between the experimental and theoretical values for the Modulus

suggests that in this case, very little confidence could be made with the results. In

conclusion, copper can be regarded as a more Ductile material than steel with a higher

Toughness, and Steel can be considered to have a higher Yield and Tensile Strength with

an equal elastic Modulus.

Conclusion

A tensile test was conducted within the elastic limits to show that it left no

residual elongation in the material. The graph of the results agreed with

the theory about residual elongation. Also, a second tensile test was

conducted within the plastic limits. This test showed that residual

elongation formed.

The modulus of elasticity, E, was derived for steel, aluminum, brass, and

copper using Hooke’s Law. It was found that mild steel had the highest

modulus of elasticity among the materials tested.

A tensile test was conducted on steel, aluminum, brass and copper.

Copper was found to be the most tensile by withstanding a 4kN force.

However certain aspects of the experiment could not be carried out due to

faulty apparatus.

Reference

1. http://www.admet.com/assets/Tensile_Testing_Basics_Quality_Magazine.pdf

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