TECHNICAL AND STUDY SKILLS FOR ENGINEERS: CE01201-7-2013-1-2013-1

LABORATARY REPORTONANALYSIS ON MECHANICAL PROPERTIES AND BEHAVIOUR OF MATERIALS: TENSILE TEST BYAditya Narayan Singh Rajput ( 11033091 ) Submitted to:Prof. DAVID CHESHIREMSc MECHANICAL ENGINEERINGFACULTY OF COMPUTING, ENGINEERING AND TECHNOLOGYSubmission date 25thOCTOBER 2013

CONTENTS

1. AIMS AND OBJECTIVES2. THEORYa. STRESSb. STRAINc. MODULUS OF ELASTICITYd. YIELD STRESSe. ULTIMATE TENSILE STRESS3. METHOD4. RESULTS5. SOFTWARE SIMULATION ANSYS WORK BENCH 14.56. RESULTS7. REFLECTIONS ON RESULTS AND DISCUSSION.8. BIBLIOGRAPHY.

AIMS AND OBJECTIVES1. To obtain the stress-strain relationship and material property for the test sample provided. [Laboratory Report Assignment, Prof. D Cheshire]

2. To verify the results by the use of ANSYS SOFTWARE.

THEORYThe Tensile test plays a very vital role in determining the property of material by providing number of data of the specimen. By this we can test the properties and then the material can be used for different roles in daily life keeping in mind its mechanical property like tension, compression, and torsion and bending. [http://ocw.metu.edu.tr/pluginfile.php/1216/mod_resource/content/0/Tension_Test_-_ekme_Deneyi.PDF]STRESS: Stress is the internal resistance, or counterforce, of a material to the distorting effects of an external force or load. These counter forces tend to return the atoms to their normal positions. The total resistance developed is equal to the external load. This resistance is known as stress. [http://me.yeditepe.edu.tr/courses/me402/lab%20manuals/tension%20test.pdf] Here by the definition it means that the stress is a resistive force which acts opposite to the applied force with the same magnitude.STRESS = FORCE/AREA.STRAIN: Strain is defined as the ratio of (Final Length Actual Length) to the actual length.STRAIN = dL/LMODULUS OF ELASTICITY : It is a description of the material to deform elastically when a certain amount of force is applied to it. The ratio between stress and strain is termed as Youngs Modulus of Elasticity. E = STRESS/STRAIN.

FIGURE-1

https://www.google.co.uk/search?q=modulus+of+elasticity&source=lnms&tbm=isch&sa=X&ei=HY9qUqLFKseyhAeVpoCICw&sqi=2&ved=0CAcQ_AUoAQ&biw=1920&bih=979#q=ELASTIC+RANGE&tbm=isch&imgdii=_The graph illustrates the relationship between stress and strain for a ductile metal.YEILD STRESS: It is defined as that point from where the material does not retain to its original shape also after releasing it from the applied force. From here it enters the plastic region. In the graph above after the point B plastic limit starts.ULTIMATE TENSILE STRENGHT: This is the highest value of stress which a material is able to withstand before breaking. http://en.wikipedia.org/wiki/Ultimate_tensile_strength

METHODFIGURE-2

https://blackboard.staffs.ac.uk/bbcswebdav/pid-1123280-dt-content-rid-1550182_1/courses/BB0000010931/BB0000005236_ImportedContent_20130118120412/BB0000005237_ImportedContent_20120904010158/Module%20Study%20Space/Laboratory%20Skills%20Laboratory%20Practice%20-%20Mechanical%20Automotive%20Tensile%20Test%20Laboratory/Tensile%20Test%20Labwork.pdfLLOYD LR30K- TENSOMETER.In this test:Material used: CARBON STEEL.CODE REFERENCE: CHEAT TREATMENT: NORMALISED AT 8600 CELCIUS.1. Measured the dimensions of the given specimen that is its diameter and its gauge length. D=4.98mm L=27.25mmFIGURE-3Diagram of the Specimen: https://www.google.co.uk/search?q=tensile+testing&source=lnms&tbm=isch&sa=X&ei=9rNqUr63IJGrhQfb9oHwCA&sqi=2&ved=0CAcQ_AUoAQ&biw=1440&bih=799#facrc=_&imgdii=_&imgrc=gR9bAuMClWytcM%3A%3B-k78z6_4-01TNM%3Bhttp%253A%252F%252Fwww.hsc.csu.edu.au%252Fengineering_studies%252Fapplication%252Flift%252F3210%252Fimage009.png%3Bhttp%253A%252F%252Fwww.hsc.csu.edu.au%252Fengineering_studies%252Fapplication%252Flift%252F3210%252Findex.html%3B305%3B3572. Now take the slack in the machine and then zero the machine.3. Now run the NEXGEN software and select the Simple Tensile.bch4. Now enter the values of diameter and the gauge length which we measured from the specimen in the Batch Document.5. Now run the batch document, suddenly steadily increasing load will start applying on the specimen and it will measure the elongation at each load step. At certain point the specimen will fail.6. Now save the data recorded throughout the test using 200 data points.

Results:Using the theory and the formulas then plotted the graph between Stress v/s Strain in the excel sheet using 200 data points.FIGURE-4

Graph from the excel: Stress v/s StrainFinal Gauge length = 35.510mmDifference in gauge length = 35.510-27.25 = 8.26 mmStrain = dL/L = 8.26/27.25 = 0.30311Upper Yield Stress = 342.14 MPaLower Yield Stress = 339.77 MPaUltimate Tensile Stress = 526.78 MPa

FIGURE-5Graph which the Machine has generated with the given values of gauge length and diameter.Now from the recorded data by the machine the values are:Youngs Modulus = 27616 MPaUpper Yield Stress = 350.59 MPaLower Yield Stress = 334.09 MPaUltimate Tensile Stress = 527.59 MPa

SOFTWARE STIMULATION Figure 6

This is the entering door to the software. The figure 6 is generated by clicking on STATIC STRUCTURE. Here first choose the Engineering Data to put the values of Yield strength and Ultimate tensile stress.

Figure 7

In this page values of stress are given according to the material property, which is in this case is tensile yield strength = 350 MPa and tensile ultimate strength = 527 MPa.After this return to the project.

Figure 8

Select the Geometry for drawing the basic structure in figure 8.

Figure 9In figure 9 now select the millimetre as the unit of the dimensions.

Figure 10

In figure 10 choose the plane on which the base of the element is to be drawn here XZ plane, so that it can be extruded in Y direction.

Figure 11

In figure 11 the basic geometry is a circle because the base of a cylinder is a circle, keeping the diameter = 4.98mm as of specimen. Then by clicking the command extrude and generate and go to model section.Figure 12

This figure contains a geometry with D = 4.98mm and Length = 27.25mm in the model section. Click on the Static structure for applying support and load.

Figure 13In figure 13 frictionless support has been provided on the base of the specimen.

Figure 14

Figure 14 depicts the surface where the force is applied and in +Y direction with the magnitude of 169.6566 N. The value was taken from one of the 200 data which appeared from the testCALCULATION:Stress = 8.7145 MPa.Elongation = 0.1052 mm. Stress = Force/ AreaForce (F) = Stress* AreaArea= 3.14*r^2 = 19.46831 mm^2So, F = 169.6566 N

Figure 15

Click on solutions for applying stress (equivalent non mises), strain (equivalent non mises), and deformation (total and directional). Then click solve to get the stress, strain and the deformation.

Figure 16

It shows the stress on the specimen.

Figure 17

It shows the elastic strain on the specimen.

Figure 18

This figure shows the total deformation on the specimen.

Figure 19

This figure depicts the directional deformation in +Y direction.Results:Stress = 8.7136 MPaStrain = 0.435Total Deformation = 1.1875 * 10^-6 mDirectional Deformation = 3.2547* 10^-8 m

Reflections on Results & Discussion:1. On comparing the results which are obtained from the laboratory results and which are obtained from ANSYS software we can say that there is a very negligible amount of difference in both the stresses.2. The strain in both the cases has a slight difference.3. But there is a bit difference in elongations in both the processes.4. This difference happened because the Extensometer is not used in the Laboratory work so might be the machine has not noted the correct deformation, it may have read the other deformations also which had happened during the test.5. Another reason about difference in the results may be because of the temperature difference in both the processes.6. Also sometimes the machine noise can be a reason for having the different results.7. Now we can see in the figure that after attaining of the ultimate tensile stress the value of the Stress decreases with increase in the amount of strain this is because after attaining the yield stress the specimen start necking so before it the stress is calculated with the actual area but now the area so no more the same that is why this happens.

Bibliography1. Wikipedia2. Google3. https://www.google.co.uk/webhp?source=search_app&gws_rd=cr&ei=JaJqUuf4BYKLhQfj0IGADw#q=why+does+there+is+a+fall+of+strain+after+the+ultimate+tensile+stress4. [http://ocw.metu.edu.tr/pluginfile.php/1216/mod_resource/content/0/Tension_Test_-_ekme_Deneyi.PDF