Tensile TEST Amzar Syakir BIN Asmadi PDF

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LABORATORY ASSESSMENT FORM MEC 291 (MECHANICS AND MATERIALS LAB) EXPERIMENT: VENUE

TENSILE TEST :

MATERIAL STRENGTH LAB 1

PREPARED BY: NAME SIGNATURE

1.

GROUP

AIMAN DANIAL BIN MOHD GHAZALI

B1

UiTM ID NO.

2020488314

Aiman

2. AMZAR SYAKIR BIN ASMADI

B1

2020457294

Amzar

3. AIMAN RIZQHAN BIN ROSZI

B1

2020892978

Rizqhan

DATE PERFORMED

:

8/11/2021

DATE SUBMITTED

:

14/11/2021

LECTURER`S NAME

:

MUHAMMAD FARIS ABD MANAP

ASSISTANT LECTURER`S NAME: ABDUL HALIM BIN SAAD LAB TECHNICIAN`S NAME :

For office use only ASSESSMENT

No 1 2 3 4 5

Items

Marks (%)

Conducting experiment following the procedure. (CO1:PO5) Able to use laboratory equipment precisely to obtain experiment data and results (CO1:PO5) Results (data, graph, calculation and etc. (CO1:PO5) Discussion. (CO2:PO4)

10

Conclusions.(CO2:PO4)

10

6

Safety .(CO3:PO9)

7

Interaction and Participation.(CO3:PO9)

8

Q&A (CO3:PO9) Total score

10 20 20

Weightage

Score (%) 1

2

3

1x scale(……….) = 1x scale(……….) = 2x scale(……….) = 2x scale(……….) = 1x scale(……….) =

10

1xscale(…..…..)

10

1xscale(…..…..)

10

1xscale(….…..)

100 Oct2021

1.0

Objective To obtain a general understanding of how different materials and cross sections behave under uniaxial tensile loading. ii. To determine the stress-strain relationship and compare mechanical/material properties of various materials and cross section. iii. To obtain the mechanical properties: the modulus of elasticity, the yield stress, the ultimate stress, the fracture stress and the ductility ratio. i.

2.0

Apparatus

Universal testing machine, vernier caliper, steel ruler, steel specimen.

Universal testing machine

Figure 4: A typical tensile test specimen 3.0

Theory

Mechanical testing play an important role in evaluating fundamental properties of engineering materials (i.e: modulus of elasticity, Poisson’s ratio, ultimate strength, yield strength, fracture strength, resilience, toughness, % reduction in area and % elongation) as well as in developing new materials and in controlling the quality of materials for use in design and construction. Most of these engineering values are found by graphing the stress and strain values from testing. A number of experimental techniques are developed for mechanical testing of engineering materials subjected to tension, compression, bending and torsion loading. Ductile materials will neck down through the plastic range before rupture (Figure 1a). Brittle materials do not neck down significantly (Figure 1b). Instead they fail sharply and abruptly at the maximum load because brittle materials do not exhibit much plasticity.

Figure 1: Typical of failure of materials

When a specimen is loaded so that the resultant force passes through the centroid of the specimen cross section, the loading is called as axial and can be either tensile or compressive. The test measures force and change of length of the specimen which are used to calculate nominal stress and nominal strain. The term nominal is used to indicate that the stress is based on the original test specimen cross section area and the strain is based on the original gage length as shown in Figure 4. Stress is a measure of the intensity of an internal force. Stress is defined as the force P per unit area A: P

Stress, σ= (SI unit; N/m2) A

Strain is a measure of the deformation that has occurred in a material. In the case where the magnitude of deformation is the same over the entire length of a body, strain is defined as: Strain, ε =

Lf−Lo

(m/m-i.e. dimensionless)

Lo

where: Lo = the initial length Lf = final length A typical stress-strain diagram from a tensile test for structural steel is shown in Figure 2. The particular properties are designated on the Figure 2 and are described as below: 1. Young’s Modulus (Modulus of elasticity), E Young’s Modulus is the ratio of stress to strain for the initial straight line portion of the stress-strain curve (slope of the straight line). Determined by: E=

σp εp

where:

σp = proportional limit stress εp = proportional limit strain

2. Proportional limit Proportional limit is the value of engineering stress (the load is divided by the initial cross-sectional area) at the point where the straight-line portion of the stress-strain curve ends. 3. Yield point Yield point is a point on the stress-strain curve, after which there is a significant increase in strain with little or no increase in stress. The corresponding stress is called the Yield strength/Stress of the material. For materials that do not possess well-defined yield point, “offset method” is used to determine it.

4. Elastic limit Elastic limit is the value of stress on the stress-strain curve after which the material deforms plastically (maximum stress for which stress will be directly proportional to strain). 5. Ultimate strength Ultimate strength is the highest value pf apparent stress on the stress-strain curve. It is also known as the tensile (or compressive) strength. 6. Fracture strength Fracture strength is the value of stress at the point of final fracture on the stress-strain curve. 7. Percent elongation Percent elongation is the measure of the deformation at the point of final fracture. Determined by: %𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 =

𝐿𝑓−𝐿𝑜

× 100

𝐿𝑜

8. Percent reduction of area Percent reduction of area is the measure of the fracture ductility. Determined by: %𝑅𝐴 =

𝐴𝑓−𝐴 𝑜

× 100

𝐴𝑜

where; Ar = the final cross − sectional area at the location of fracture Ao = the initial cross − sectional area

9. Ductility Ductility is the characteristic of a material where the material can undergo large plastic deformations before fracture, especially in tension. Ductility of materials is measured by ductility ratio; 𝑑𝑢𝑐𝑡𝑖𝑙𝑖𝑡𝑦, 𝜇 =

𝜀𝑢 𝜀𝑦

where ;

𝜀 𝑢 = 𝑡 ℎ𝑒 𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑠𝑡𝑟𝑎𝑖𝑛 𝜀𝑦 = 𝑡ℎ𝑒 𝑦𝑖𝑒𝑙𝑑 𝑠𝑡𝑟𝑎𝑖𝑛

Figure 2: A typical stress-strain diagram for ductile material

4.0

Brief Procedure 1. Measure the dimensions of the each test specimen before and after test and fill in the table 1. Mark the gauge length on the test specimen. 2. Switch on the machine. 3. Mount the test specimen in the grips of the machine. 4. Apply and record load and the corresponding deformation. 5. Repeat steps (1) to (4) for various type of the test specimen.

5.0

Result

Table 1: Dimension of the appropriate tested specimen Initial (unit: mm) Material Steel

Final (unit: mm)

Lo

Ao (mm2)

bo

ho

Lf

Af (mm2)

bf

hf

80.00

46.02

17.70

2.60

129.98

16.47

12.20

1.35

Table 2: Determine the following observation for the tested specimen No

Force (N)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

2700 4500 6800 9550 12750 17750 21700 21450 22950 23700 24350 24700 24250 22600 19300

Elongation (mm) 0.75 1.13 1.47 1.79 2.10 2.47 3.61 6.21 15.07 18.68 24.64 33.49 45.63 48.43 49.98

Stress (Pa)

Strain

58670.14 97783.57 147761.84 207518.47 277053.46 385701.87 471534.12 466101.70 498696.22 514993.48 529117.77 536723.16 526944.81 491090.83 419382.88

0.0094 0.0141 0.0184 0.0224 0.0263 0.0309 0.0453 0.0776 0.1884 0.2335 0.3080 0.4186 0.5704 0.6054 0.6248

Table 3: Determine the following observation load for the tested specimen

Material

Load at Elastic Limit (N)

Load at Upper Yield Point (N)

Load at Lower Yield Point (N)

Steel

20600

21700

21450

Ultimate Breaking Load Load (N) (N) 24700 19300

Table 4: Determine the following properties for the tested specimen

Proportional Nominal Material Limit Stress Fracture (Pa) Stress (Pa) Steel

450000

10671.25

Actual Fracture Stress (Pa)

% Reduction in Area

Strain

491090.83

64.21

0.6248

% Elongation Ductility

182.44

9.24

Table 5: Determine the % error of the tested specimen

Properties

Modulus of Elasticity

Experimental Reference % Difference

11.84MPa 195GPa 100

Material Steel

6.0

Calculation 1. Stress Based on the data from the Table 1 and Table 2, σ=

P A

=

2700 46.02×10 −3

= 58670.14 Pa

2. Strain ε=

Lf −Lo Lo

= 0.75 80

= 0.0094

0.2% offset Yield Stress 471.53Pa 471.53Pa 0

Yield Stress

Ultimate Stress

471.53kPa 260MPa 99.8

536.72kPa 655MPa 99.9

3. Young’s Modulus E= =

σp σp

450000 0.0380

= 11842105.26Pa

4. Percent elongation %elongation =

Lf−Lo L0

=

129.98−46.02 46.02

× 100

= 182.44%

5. Percent reduction of area %RA =

Ao −Af

× 100

Ao

=

46.02−16.47 46.02

= 64.21%

6. Ductility 𝜇= =

𝜀𝑢 𝜀𝑦

0.4186 0.0453

= 9.24

× 100

× 100

FAKULTI KEJURUTERAAN MEKANIKAL UNIVERSITI TEKNOLOGI MARA PULAU PINANG

7. Percentage difference i. Modulus of elasticity =

195×109−1182110 195×10 9

× 100

= 100% ii. Yield stress =

260×106 −471530 260×106

× 100

= 99.8% iii. Ultimate stress =

655×10 6−536720 655×10 6

× 100

= 99.9% 7.0

Discussion Based on the data from the Table 4, the percentage reduction of area obtained is 64.21%. However, the reference value is 56.24%, which is lower than experimental value. Next, percentage of elongation from the table is 182.44% that has a large difference from the reference value, which is 23.16%. In my opinion, the cause of discrepancies is the size of tested specimen that affect the results. The size of the specimen may influence the calculation and result. This experiment only includes one specimen which is mild steel. By referring the Table 5, the properties of the tested specimen for Modulus of Elasticity obtained is 11.84MPa. In addition, the obtained yield stress is 471.53kPa and the ultimate stress is 536.72kPa. From the stress-strain graph, the yield point is 466.1kPa which is slightly lower than the yield strength, 471.53kPa. In my opinion, yield strength is the best indication for material’s fitness in a particular tensile application. This is because yield strength is the maximum stress needed to change the shape of the material. The proportional limit obtained for the tested specimen is 450kPa, plotted from the stress-strain graph while the elastic limit is 466.1kPa. Therefore, as long as we already have the data for yield strength, the elastic limit is more important indicator for material’s mechanical behavior. This is because elastic limit is defined as the point at which there is no permanent deformation in the structure of tested specimen. The advantages of stress-strain graph are easier to determine the material’s mechanical properties such as ultimate stress, fracture stress and yield stress. Other than that, we also can obtain the material’s behavior and the process for example, elastic region, yielding, strain hardening and necking.

8.0

Conclusion In conclusion, from this experiment, we have obtained the general understanding of how the specimen behave under uniaxial tensile loading. Besides, we also have determined the stressstrain relationship and mechanical properties of the specimen. Lastly, we have obtained these mechanical properties; the modulus of elasticity, the yield stress, the ultimate stress, the fracture

FAKULTI KEJURUTERAAN MEKANIKAL UNIVERSITI TEKNOLOGI MARA PULAU PINANG

stress and the ductility ratio. There are several ways to improve the results of this experiment, which we must ensure our eye is perpendicular to the scale reading of vernier caliper to avoid parallax error. Secondly, we have to make sure that the tested specimen is fitted correctly to the universal testing machine to obtain more accurate results....