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HASHEMITE UNIVERSITY
Faculty of EngineeringMechanical Engineering Department
Student Name : Ahmed Hani Salem Al-Omari
Student Reg. No : 431900
Section No. : 4
Lab. Day : Wednesday
Lab. Date : 29/ 11 / 2006
Experiment # : 8
Experiment Title: Compression Test
Submitted to
Instructor: Dr. Ahmed Al-Shyyab
Engineer: Yousef Zakariya
Due Date: 06 / 12 / 2006
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Objective:We want from this experiment to:
o Observe the stress - strain behavior of some metals under compression load.o Determine the strength and other properties of various materials.
Theory:o In theory the compression test is just the opposite the tensile test. However, there are special
limitations on the compression test:
1- Appling a truly axial load is difficult.2- There is always a tendency for bending stresses to be set up.3-
Friction between the heads of the testing machine or bearing plates and the end surfaces of thesample.
o The strain stress diagrams in compression test have different shapes from those in thetension test.
o With ductile metals (steel, aluminum and copper), The proportional limits in compressiontest are very closed for those in tension test and the initial regions of their stress strain
diagram are very similar to tension test diagrams. When yielding begins, the behavior is
quite different.
o In a tension test, the specimen is stretched, necking may occur, and fracture ultimately takesplace. When a small specimen of ductile material is compressed, it begins to bulge
outward on the sides and become barrel shaped. With increasing load, the specimen is
flattened out, thus offering increased resistance to further shortening (which means thestress-strain curve goes upward).
o These characteristics are illustrated in Fig. 1, which shows a compression stress-straindiagram for copper.
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o With brittle materials (brass): typically have an initial linear region followed by a regionin which the shortening increases. Thus the stress strain diagram has a shape is similarto the shape in tension test.
o Brittle materials reach much higher ultimate stress in compression than tension, the brittlematerials actually fracture or break at maximum load unlike ductile material.
o As we see that we will use the same laws which we applied in tension experiment.
Equipments:Same apparatus of tension test (UTM), but the clamps of the tension test were replaced by the
compression jig parts.
Universal Testing Machine (UTM): The machine is digital type Tensile Strength Test Machine. Capable
doing the following tests:
1. Tensile test.2. Compression test.
3. 3 points bending test.
4. Direct shear test.
It uses sensor which has high accuracy of the load value. Experimenters can get well-done results.
Experimenter can save the result by mean of connecting the U.T.M. and Computer. The machine is
composed of:
Loading part:
Main body, crosshead, crosshead moving part, jig part, load cell sensor, and displacement sensor.
Measuring part:
Load display, strain display, and speed control device.
Precautions:1- Machined surfaces should be finished to 1.6m or better.2- Test specimens ends should be flat and parallel within .0005 in/in.3- Test specimens should be loaded concentrically.
Data Results & Analysis:
With: Brass material.L = 9 mm.
D = 9 mm.
Ao = * D2 / 4=63.62 mm.
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P (N) (mm) Stress (Mpa) Strain0 0 0 0
3250 0.043 51.0845646 0.004778
4790 0.114 75.29078906 0.012667
8210 0.2 129.0474693 0.022222
20870 0.4 328.0414964 0.044444
24970 0.5 392.4866394 0.05555629250 0.8 459.7610814 0.088889
30880 1 485.3819554 0.111111
32500 1.2 510.845646 0.133333
34130 1.4 536.46652 0.155556
35750 1.6 561.9302106 0.177778
40370 2 634.548884 0.222222
44470 2.4 698.994027 0.266667
50290 2.8 790.4746935 0.311111
59180 3.4 930.2106256 0.377778
62610 3.67 984.1244892 0.407778
58500 3.76 919.5221628 0.417778
13000 3.79 204.3382584 0.421111
18470 3.87 290.3175102 0.43
19500 3.95 306.5073876 0.438889
18640 3.97 292.9896259 0.441111
Stress Strain Curve
0
200
400
600
800
1000
1200
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
Strain
Stress(Mpa)
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Sample of calculations:
1- For second reading in the table (1):Stress = load / area = 3250 / 63.62 = 51.0845.
Strain = / length = 0.043 / 9 = 0.004778
2- Proportional limit = 370.37 MPa.3- Yield point = 0.0684- Yield stress for an offset 0.2% = 371.43 MPa.5-
Ultimate stress = u = 984.1244892 MPa.6- Fracture (Rupture) stress = 984.1244892 MPa.
7- Elongation = [3.97/9]*100% = 44.11 %8- Modulus of elasticity = E = slop = /
= (370.37 - 200)/ (0.05 0.03)
= 8518.5 MPa.
9- Modulus of resilience = UR= (250)2/ (2*25000) = 1.25 MPa.10-Modulus of toughness = UT = (2/3) ( max * max) = (2/3)( 984.1244892 * 0.407778 )
= 267.54 MPa.
11- for brass = 0.3412-Shear modulus of elasticity = G = (8518.5)/ [2 (1- 0.34 )]
= 6453.41 MPa.13-Bulk modulus of elasticity = K = (8518.5)/ [3 (1- 2* 0.34 )]
= 8873.44 MPa.
Discussion & Conclusion: COMPARING BETWEEN ENGINEERING & TRUE STRESS:1. Ao for specimen is used in our calculations, the resultant stress is called nominal stress,
conventional stress, or engineering stress.
2. When we obtain another exact value or accurate value of the axial stress, the last is called thetrue stress, which can be calculated by using the actual area of the bar at the cross section
where the failure occurs.
COMPARING BETWEEN COMPRESSION TEST & TENSILE TEST:1. In compression the tall of specimen is decrease, but in tensile test the tall is increase.2. In compression test the width of specimen increase, but in tensile test the width is
decrease
3. In compression test the true stress is lower than engineering stress; because the areaincreasing with time. And vise-versa for tensile test.
ADVANTAGES FOR STRESS STRAIN CURVEo Gets the modulus of elasticity, the shear modulus of elasticity, and the bulk modulus of
elasticity.
o Knowing the characteristics of the materials by the diagram and know the behaviour ofmaterials under compression load.
o This experiment gives us information to know the failure cases for the structure ortested specimen.