Strength of Material- Young Modulus Experiment

Experiment 1 : Tensile Test

Objective

i. To estimate the stress () and strain () of the material.ii. To study the stress and strain behavior of the material using - diagram.

iii. To study the various regions of the - diagram.iv. To determine the Young’s modulus or the Modulus of Elasticity.v. To study the ductile behavior of the material.

Abstracts

Tensile are fundamental for understanding properties of different material, and how they will behave under load. This experiment tested four different material, including steel, copper, aluminum, and copper alloy (brass). Each material was tested by using a pull tester and data was recorded. The data from each test was used to determine valuable material properties such as ultimate tensile strength, modulus of elasticity, and yield strength. Other calculated properties included true fracture strength, percent reduction of area, and percent elongation. These material properties were used for comparing the material to each other, and to define the material as brittle or ductile.

The results of the tensile tests showed that the steel was the strongest material. It had the highest ultimate tensile strength (699 MPa), the greatest yield strength (684 MPa), and the largest true fracture strength (554.2 MPa). The copper had a higher yield (310 MPa) than the aluminum (292.1 MPa), a higher ultimate tensile strength (315 MPa), but a lower true fracture strength (165 MPa). All of the materials besides the brass proved to be ductile, especially the copper, which had a percent elongation of 9.6%. The brass sample averaged a percent elongation of only 3.1%.

Methodology

Stress and strain relationship

When a specimen is subjected to an external tensile loading, the metal will undergo elastic and plastic deformation. Initially, the metal will elastically deform giving a linear relationship of load and extension. These two parameters are then used for the calculation of the engineering stress and engineering strain to give a relationship as follow

σ= PAo

ε=Lf−Lo

Lo= δ

L0

Where, is the stress

is the strain

P is the external axial tensile load

Ao is the original cross-sectional area of the specimen

is the deformation of the specimen

Lo is the original length of the specimen

Lf is the final length of the specimen

Young’s modulus, E

During elastic deformation, the engineering stress-strain relationship follows the Hook's Law and the slope of the curve indicates the Young's modulus (E)

E=σε

Young's modulus is of importance where deflection of materials is critical for the required engineering applications. This is for examples: deflection in structural beams is considered to be crucial for the design in engineering components or structures such as bridges, building, ships, and others. The applications of tennis racket and golf club also require specific values of spring constants or Young's modulus values.

Figure 1: Stress-strain relationship under uniaxial tensile loading

Yield strength, y

By considering the stress-strain curve beyond the elastic portion, if the tensile loading continues, yielding occurs at the beginning of plastic deformation. The yield stress, σy, can be obtained by dividing the load at yielding (Py ) by the original cross-sectional area of the specimen (Ao ) as shown

σ y=P y

Ao

The yield point can be observed directly from stress-strain curve of the metals. At the yield point, the specimen continues to extend without a significant change in the stress level. Load increment is then followed with increasing strain.

Ultimate Tensile Strength

Beyond yielding, continuous loading leads to an increase in the stress required to permanently deform the specimen as shown in the stress-strain curve. At this stage, the specimen is strain hardened or work hardened if the load is continuously applied, the stress-strain curve will reach the maximum point, which is the ultimate tensile strength. At this point, the specimen can withstand the highest stress before necking takes place. This can be observed by a local reduction in the cross sectional area of the specimen.

Rupture Strength, rupture

After necking, plastic deformation is not uniform and the stress decreases accordingly until fracture. The rupture strength (σrupture) can be calculated from the load at rupture divided by the original cross-sectional area, Ao, as expressed

σ rupture=Prupture

Ao

Tensile ductility

Tensile ductility of the specimen can be represented as percentage of elongation or percentage of reduction in area as expressed in the equations given below

Percentage of Elongation=|Lf −Lo

Lo|x100 %

Percentage of Reduction∈Area=|A f −Ao

Ao|x 100 %

Where Af is the cross-sectional area of the specimen at fracture.

Fracture characteristics

Metals with good ductility normally exhibit a so-called cup and cone fracture characteristic whereas metal with low ductility appear a flat fracture surface as shown in figure below.

Necking starts when the stress-strain curve has passed the maximum point where plastic deformation is no longer uniform. This type of fracture surface signifies high energy absorption during the fracture process due to large amount of plastic deformation taking place, also indicating good tensile ductility. Metals such as aluminum and copper normally exhibit ductile fracture. For brittle metals, the fracture surfaces usually consist of flat.

Figure 2: Fracture characteristics

Apparatus

1. Universal Testing Machine2. Sample specimens of metal3. Vernier caliper

Figure 3: Universal Testing Machine

Figure 3a: Force display

Figure 3b: Elongation display

Figure 4: Sample specimens of metal

Procedure

1. The original length and diameter of four materials was measured by vernier caliper and readings were recorded.

2. The load frame of the upper part was adjusted to fit the test specimen inside the dial gauge.

3. The dial gauge was loosened and lowered it on the column and a test specimen was chosen and screwed tightly into the jaws of the machine.

4. The force display and elongation display of pull tester was adjusted back to zero reading.

5. The hand wheel was turn slowly to increment of 0.01mm for each elongation and the test specimen was observed.

6. The force applied and elongation was recorded base on the elongation display and force display from time by time until the test specimen reached its necking region and then fracture.

7. The true fracture force and final elongation were recorded.

8. The final length and diameter of four materials was measured by vernier caliper and readings were recorded.

9. Step 2 to step 8 was repeated by replacing each different material test specimen.

Results

Based on the observation, we can conclude that steel, copper and aluminum have good ductility and behave as ductile material as all of the sample specimen undergo necking process and exhibit a cup and cone fracture at the fracture surface. For brass, it behaves as brittle surface and appear a smooth fracture at the fracture surface.

Table 1: The original length, diameter and area for four materials

Materials Length, lo (m) Diameter, do (m) Area, A0 = π do4

4 (m2¿/x10−6

Steel 0.025 0.00505 20.03

Copper 0.025 0.00505 20.03

Aluminum 0.025 0.00505 20.03

Brass 0.025 0.00505 20.03

Table 2: The final length, diameter and area for four materials

Materials Length, lf (m) Diameter, d f (m) Area, A f = π d f2

4 (m2¿/x10−6

Steel 0.02642 0.00315 7.79

Copper 0.02739 0.00256 5.15

Aluminum 0.02705 0.00288 6.54

Brass 0.02578 0.00382 11.46

Table 3: The percent reduction of area, and the percent elongation for four materials

Materials Percent of elongation (%) Percent of reduction of area (%)

Steel 5.7 61.1Copper 9.6 74.3

Aluminum 8.2 67.4

Brass 3.1 42.8

Table 4: The yield strength, ultimate tensile strength, true rupture strength, and modulus of elasticity for four materials

Materials Steel Copper Aluminum Brass

Yield strength, σ y (MPa)

684 310 292.1 -

Ultimate tensile strength, σ u (MPa) 699 315 302 434.3

True rupture strength, σ r (MPa) 554.2 165 290 434.3

Modulus of elasticity, E (GPa) 63.1 40 150 37.4

Table 5: The percent error of modulus of elasticity for four materials

Materials Theoretical modulus of elasticity, ET (GPa)

Experimental modulus of elasticity, EE (GPa)

Percentage Error (%)

Steel 200 63.1 68.5Copper 117 40 65.8

Aluminum 69 150 117.4

Brass 114 37.4 67.2

Percentage error=¿

*The approximate theoretical values of Young Modulus are taken from Wikipedia.

Website: https://en.wikipedia.org/wiki/Young%27s_modulus

Steel

Elongation (mm) Force (kN) Strain (mm/mm) Stress (kPa)0 0 0 00.01 1.3 0.0004 64895.47370.02 2.1 0.0008 104831.14980.03 2.7 0.0012 134782.90690.04 3.2 0.0016 159742.70440.05 3.8 0.0020 189694.46150.06 4.4 0.0024 219646.21860.07 4.9 0.0028 244606.01610.08 5.4 0.0032 269565.81370.09 5.7 0.0036 284541.69230.10 6.2 0.0040 309501.48980.11 6.7 0.0044 334461.28740.12 7.1 0.0048 354429.12540.13 7.5 0.0052 374396.96350.14 7.9 0.0056 394364.80150.15 8.2 0.0060 409340.68010.16 8.5 0.0064 424316.55860.17 9.0 0.0068 449276.35620.18 9.3 0.0072 464252.23470.19 9.6 0.0076 479228.11330.20 10.0 0.0080 499195.95130.25 11.3 0.0100 564091.42500.30 12.5 0.0120 623994.93920.35 13.2 0.0140 658938.65570.40 13.6 0.0160 678906.49380.42 13.7 0.0168 683898.45330.46 13.9 0.0184 693882.37230.50 14.0 0.0200 698874.33180.56 14.0 0.0224 698874.33180.60 13.7 0.0240 683898.45330.70 13.7 0.0280 683898.45330.80 13.3 0.0320 663930.61530.90 13.3 0.0360 663930.61531.00 12.8 0.0400 638970.81771.10 12.5 0.0440 623994.93921.20 12.3 0.0480 614011.02011.30 12.0 0.0520 599035.14161.40 11.3 0.0560 564091.42501.42 11.1 0.0568 554107.5060

0 0.01 0.02 0.03 0.04 0.05 0.060

100000

200000

300000

400000

500000

600000

700000

800000

Proportional Limit(0.0076, 479228.11)

Yield Point(0.0168, 683898.45)

Ultimate Strength(0.0224, 698874.33)

Rupture Strength(0.0568, 554107.50)

Stress-Strain Curve of Steel

Stress,

Strain, (mm/mm)

0 0.01 0.02 0.03 0.04 0.05 0.060

100000

200000

300000

400000

500000

600000

700000

800000


Yield Point(0.0168, 683898.45)



Stress-Strain Curve of Steel

Stress,

Strain, (mm/mm)

0.0000 0.0010 0.0020 0.0030 0.0040 0.0050 0.0060 0.0070 0.00800

100000

200000

300000

400000

500000

600000

(0, 0)

(0.0076, 479228.1133)

Best Line Fit of Stress-Strain Curve of Steel Within Proportional Limit

Stress, (kPa)

Strain, (mm/mm)

Young modulus of Steel, E s = slope of stress-strain curve within proportional limit

E=y2− y1

x2− x1

E s=479228.1133 x103−0

0.0076−0=63.0563 x109≈ 63.1 GPa

At yield point,

δ=0.42 mm, P y = 13.70 kPa

Yield Strength, σ y=P y

Ao= 13.7 x103

20.03 x10−6 ≈ 684 MPa

At fracture point,

δ=1.42 mm, Pr = 11.10 kPa

True Rupture Strength,σ r=P r

Ao= 11.1 x103

20.03 x 10−6 ≈ 554.2 MPa

Copper

0.0000 0.0010 0.0020 0.0030 0.0040 0.0050 0.0060 0.0070 0.00800

100000

200000

300000

400000

500000

600000

(0, 0)

(0.0076, 479228.1133)

Best Line Fit of Stress-Strain Curve of Steel Within Proportional Limit

Stress, (kPa)

Strain, (mm/mm)

Elongation (mm) Force (kN) Strain (mm/mm) Stress (kPa)

0 0 0 00.01 0.6 0.0004 29951.75710.02 1.0 0.0008 49919.59510.03 1.4 0.0012 69887.43320.04 1.7 0.0016 84863.31170.05 2.0 0.0020 99839.19030.06 2.4 0.0024 119807.02830.07 2.7 0.0028 134782.90690.08 3.0 0.0032 149758.78540.09 3.3 0.0036 164734.66390.10 3.5 0.0040 174718.58300.11 3.9 0.0044 194686.42100.12 4.1 0.0048 204670.34000.13 4.5 0.0052 224638.17810.14 4.7 0.0056 234622.09710.15 4.9 0.0060 244606.01610.16 5.1 0.0064 254589.93520.18 5.5 0.0072 274557.77320.20 5.9 0.0080 294525.61130.22 6.1 0.0088 302013.55050.24 6.2 0.0096 307005.51010.26 6.2 0.0104 309501.48980.30 6.3 0.0120 314493.44930.55 6.3 0.0220 314493.44930.60 6.1 0.0240 304509.53030.70 6.0 0.0280 299517.57080.80 6.0 0.0320 299517.57080.90 6.0 0.0360 299517.57081.10 5.8 0.0440 289533.65181.30 5.6 0.0520 279549.73271.50 5.4 0.0600 269565.81371.70 5.1 0.0680 254589.93521.90 4.8 0.0760 239614.05662.10 4.3 0.0840 214654.25912.30 3.7 0.0920 184702.50202.39 3.3 0.0956 164734.6639

0.00 0.02 0.04 0.06 0.08 0.10 0.120

50000

100000

150000

200000

250000

300000

350000


Yield Point(0.0104, 309501.49) Ultimate Strength

(0.0220, 314493.45)


Stress-Strain Curve of Copper

Stress, (kPa)

Strain, (mm/mm)

0.00 0.02 0.04 0.06 0.08 0.10 0.120

50000

100000

150000

200000

250000

300000

350000


Yield Point(0.0104, 309501.49) Ultimate Strength

(0.0220, 314493.45)


Stress-Strain Curve of Copper

Stress, (kPa)

Strain, (mm/mm)

0.0000 0.0010 0.0020 0.0030 0.0040 0.0050 0.0060 0.00700

50000

100000

150000

200000

250000

300000

(0, 0)

(0.0064, 254589.9352)

Best Fit Line of Stress-Strain Curve of Copper Within Proportional Limit

Stress, (kPa)

Strain, (mm/mm)

Young modulus of Copper, Ec = slope of stress-strain curve within proportional limit

E=y2− y1

x2− x1

Ec=254589.9352 x 103−0

0.0064−0=39.7797 x109 ≈ 40GPa

At yield point,

δ=0.26 mm, P y = 6.20 kPa

Yield Strength, σ y=P y

Ao= 6.20 x103

20.03 x10−6 ≈ 310 MPa

At fracture point,

δ=2.39 mm, Pr = 3.30 kPa

True Rupture Strength , σr=P r

Ao= 3.30 x 103

20.03 x 10−6 =165 MPa

Aluminum

Elongation (mm) Force (kPa) Strain (mm/mm) Stress (kPa)0 0 0 0

0.0000 0.0010 0.0020 0.0030 0.0040 0.0050 0.0060 0.00700

50000

100000

150000

200000

250000

300000

(0, 0)

(0.0064, 254589.9352)

Best Fit Line of Stress-Strain Curve of Copper Within Proportional Limit

Stress, (kPa)

Strain, (mm/mm)

0.02 2.80 0.0008 139774.86640.04 4.80 0.0016 239614.05660.06 5.30 0.0024 264573.85420.08 5.40 0.0032 269565.81370.10 5.50 0.0040 274557.77320.12 5.55 0.0048 277053.75300.14 5.60 0.0056 279549.73270.16 5.60 0.0064 279549.73270.18 5.70 0.0072 284541.69230.20 5.75 0.0080 287037.67200.22 5.75 0.0088 287037.67200.24 5.75 0.0096 287037.67200.26 5.80 0.0104 289533.65180.28 5.85 0.0112 292029.63150.30 5.85 0.0120 292029.63150.35 5.90 0.0140 294525.61130.40 5.90 0.0160 294525.61130.45 5.90 0.0180 294525.61130.50 5.90 0.0200 294525.61130.55 5.90 0.0220 294525.61130.60 5.90 0.0240 294525.61130.65 5.95 0.0260 297021.59100.70 5.95 0.0280 297021.59100.75 5.95 0.0300 297021.59100.85 6.00 0.0340 299517.57080.95 6.00 0.0380 299517.57081.05 6.00 0.0420 299517.57081.15 6.00 0.0460 299517.57081.25 6.00 0.0500 299517.57081.35 6.00 0.0540 299517.57081.45 6.00 0.0580 299517.57081.55 6.00 0.0620 299517.57081.65 6.05 0.0660 302013.55051.75 6.00 0.0700 299517.57081.85 6.00 0.0740 299517.57081.95 5.90 0.0780 294525.61132.05 5.80 0.0820 289533.6518

0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.090

50000

100000

150000

200000

250000

300000

350000


Yield Point(0.0120, 292029.63)



Stress-Strain Curve of Aluminum

Stress, (kPa)

Strain, (mm/mm)

0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.090

50000

100000

150000

200000

250000

300000

350000


Yield Point(0.0120, 292029.63)



Stress-Strain Curve of Aluminum

Stress, (kPa)

Strain, (mm/mm)

0.0000 0.0002 0.0004 0.0006 0.0008 0.0010 0.0012 0.0014 0.0016 0.00180

50000

100000

150000

200000

250000

300000

(0, 0)

(0.0016, 239614.0566)

Best Fit Line of Stress-Strain Curve of AluminumWithin Proportional Limit

Stress, (kPa)

Strain, (mm/mm)

Young modulus of Aluminum, EA = slope of stress-strain curve within proportional limit

E=y2− y1

x2− x1

EA=239614.0566 x 103−0

0.0016−0=149.7588 x 109 ≈150 GPa

At yield point,

δ=0.30 mm, P y = 5.85 kPa

Yield Strength , σ y=P y

Ao= 5.85 x103

20.03 x10−6 ≈ 292.1 MPa

At fracture point,

δ=2.05 mm, Pr = 5.80 kPa

True Rupture Strength,σ r=P r

Ao= 5.8 x103

20.03 x 10−6 ≈ 290 MPa

Brass

Elongation (mm) Force (kN) Strain (mm/mm) Stress (kPa)0 0 0 00.01 0.0 0.0004 00.02 0.3 0.0008 12479.89880.03 0.7 0.0012 34943.71660.04 1.0 0.0016 49919.59510.05 1.4 0.0020 69887.43320.06 1.8 0.0024 87359.29150.07 2.1 0.0028 104831.14980.08 2.5 0.0032 124798.98780.09 2.8 0.0036 137278.88660.10 3.1 0.0040 154750.74490.11 3.5 0.0044 174718.58300.12 3.8 0.0048 189694.46150.13 4.1 0.0052 204670.34000.14 4.5 0.0056 224638.17810.15 4.7 0.0060 234622.09710.16 5.0 0.0064 249597.97570.17 5.1 0.0068 254589.93520.18 5.4 0.0072 269565.81370.19 5.5 0.0076 274557.77320.20 6.0 0.0080 299517.57080.22 6.3 0.0088 314493.44930.24 6.6 0.0096 329469.32790.26 7.0 0.0104 349437.16590.28 7.2 0.0112 359421.08500.30 7.4 0.0120 369405.00400.34 7.8 0.0136 389372.84200.38 8.2 0.0152 409340.68010.42 8.2 0.0168 409340.68010.46 8.4 0.0184 419324.59910.50 8.4 0.0200 419324.59910.54 8.4 0.0216 419324.59910.60 8.5 0.0240 424316.55860.65 8.5 0.0260 424316.55860.70 8.5 0.0280 424316.55860.75 8.5 0.0300 424316.55860.78 8.7 0.0312 434300.4776

0.0000 0.0050 0.0100 0.0150 0.0200 0.0250 0.0300 0.03500

50000

100000

150000

200000

250000

300000

350000

400000

450000

500000



Stress-Strain Curve of Brass

Stress, (kPa)

Strain, (mm/mm)

0.0000 0.0050 0.0100 0.0150 0.0200 0.0250 0.0300 0.03500

50000

100000

150000

200000

250000

300000

350000

400000

450000

500000



Stress-Strain Curve of Brass

Stress, (kPa)

Strain, (mm/mm)

0.0000 0.0010 0.0020 0.0030 0.0040 0.0050 0.0060 0.0070 0.00800

50000

100000

150000

200000

250000

300000

(0, 0)

(0.0072, 269565.8137)

Best Fit Line of Stress-Strain Curve of BrassWithin Proportional Limit

Stress, (kPa)

Strain, (mm/mm)

Young modulus of Brass, Eb = slope of stress-strain curve within proportional limit

E=y2− y1

x2− x1

Eb=269565.8137 x 103−0

0.0072−0=37.4397 x 109≈ 37.4 GPa

At fracture point,

δ=0.78 mm, Pr = 8.70 kPa

True Rupture Strength, σ r=P r

Ao= 8.7 x 103

20.03 x 10−6 ≈ 434.3 MPa

Discussion

1. After carrying out the experiment, we recorded the results. These results were then used to plot graphs of stress versus strain so that are able to see the relationships between each material.

2. As a general idea and based on the results obtained, we assume that copper and aluminum are ductile and brass as brittle. As we did not know the exact composition of the steel alloy, we took steel as a ductile material based on the shape of its curve in the stress strain graph. Harder steel alloys are usually more brittle than softer ones. However, although we take steel as a ductile material, it is still much stronger than other ductile materials (copper and aluminum).

3. From Table 4, which sums up most of the results of the experiment, we can see that Steel is the strongest material as it has the highest ultimate strength followed by brass, copper then aluminum. Steel also has the highest rupture strength. This is closely followed by brass aluminum then copper. Steel is the strongest because it is an alloy of iron. The carbon in steel provides a doping agent which increases strength in the crystal structure of the iron. This allows steel to become stronger and lighter than pure iron.

4. From the graphs, we see that copper has the highest ductility. As ductility increases with the maximums strain a material can handle, copper proves to be the most ductile material of the four. Ductile material can be stretch easily into wire due to their necking and yielding regions of their stress strain graph. On the other hand, Brass is the most brittle material as it does not take much elongation to rupture it.

5. Problems occurred when comparing Young’s Modulus. Between materials, aluminum was found to have the highest Young’s Modulus, higher than that of steel. This is theoretically impossible as Young’s Modulus is essentially a degree of stiffness. As Young’s Modulus increases, the material is stiffer.

6. As we can see from Table 5, we encountered extremely high percentage errors when comparing the theoretical and experimental Young’s Modulus. Errors are highly likely to have occurred here. Some of the possible error and problems faced are discussed below.

7. An important aspect of the experiment that was overlooked was the specimen alignment. A deviated alignment will significantly influence the results. If the alignment is off center, the rupture strength will be lowered considerably. This is because the experiment setup is no longer uniaxial. Unwanted side loading and

bending moments will cause the specimen to rupture with lower force. This affects brittle materials more than ductile materials. To counter this problem, we should always check the machine’s alignment before starting the experiment.

8. During the experiment, we found out that the hydraulics in machine were faulty. When increasing the force, we sometimes experienced a lost of control over the amount of the force we were adding. The wheel would lose its grip and rotate forward a great deal. The hydraulic system of the machine should be constantly checked and maintained in good condition.

9. Worn machine component also contributed to the error. For examples, worn out thread would lose their grip. The specimen may slip or break inside the gripped area. A dirty experimental setup will also affect the grip of the threads. The setup should be cleaned to prevent this error.

10. When carrying out the experiment, we did not take into account the decrease in diameter when the specimen is subjected to the load. We did not measure the gradual change in diameter so the stress is calculated using the same surface area. To accurately measure this decrease in diameter, we should change to an electronic setup.

11. To carry out the experiment, we used a shaped specimen that will concentrate the stress within the gage length. If the specimen is incorrectly machined, fracture could occur outside the gage length and result in strain errors. This is important because we want to avoid having a break or fracture within the area being gripped. We should always check the conditions of the specimen before the experiment.

12. The display of the force was also a problem. As the smallest division could only measure 0.5kN, we did not have the accuracy to control the force well. To solve this, we should use a computerized system to exert and measure the force.

13. Inaccurate reading of the specimen dimensions will also cause errors. Worn micrometers or calipers should be replaced and care should be taken when recording specimen dimensions. To prevent this error, a computer based test systems should be used to read the micrometer or caliper directly, thus eliminating data entry errors.

14. Parallax errors and human error also affected the results of the experiment. We should take greater care when taking the readings and increasing the force.

Conclusion

1. To estimate the stress and strain on a material, we plot a stress-strain graph. We do this so that we can then analyze the behaviors of the materials under different stresses and strain more accurately.

2) & 3) Brittle material such as brass do not have a yield, strain hardening and necking regions to their graph. They rupture as soon as the ultimate strength is reached. Ductile materials have yield, strain hardening and necking regions to their graph. They will not rupture when the ultimate strength is reached. They undergo strain hardening and necking before breaking at rupture strength.

4) Young’s Modulus can be found by calculating E=σ/ε. Young’s Modulus is also equivalent to the gradient of the region before reaching the proportional limit in a stress-strain graph. As Young’s Modulus increases, the stiffness of the material increases. We could not determine the Young’s modulus of a material accurately due to errors. The experimental Young’s Modulus for all the materials are:

1. Steel, 63.1 Gpa2. Copper, 40 Gpa3. Aluminum, 150 Gpa4. Brass, 37.4 Gpa

5) Ductile materials are different from brittle materials as they have yield, strain hardening and necking regions to their graph. They undergo strain hardening and necking before breaking at rupture strength.

As we could achieve four out of the five objectives, the experiment was a success.

Strength of Material- Young Modulus Experiment

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