ME Tensile Strength, Percentage of Elongation and the Reduction

     ME – 241                              Introductionto Solid Mechanics       TensileTest   Prepared by:  Name: Ulvi Umut  Surname: YAVUZLARStudent ID: 38971533166 Student E-Mail: [email protected]

tr  Due Date: December 12th,2017 INDEX?INTRODUCTION…….……………………3 ?THEORY OF EXPERIMENT……………..4 ?EXPERIMENT PROCEDURE……………7 ?GRAPHS AND CALCULATIONS ……….8 ?RESULTS AND DISCUSSION ..….

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.……12 ?ASTM STANDARDS…………

……………13 ?REFERENCES…………………………….14 ?CONCLUSION………………………….

…15                                 ?Introduction TensileTest  The tensile test, as known as the”tension test”,is one of a most fundamental type of mechanical test that used on a material.

Tensile test is really inexpensive which makes this test more preferable. From tensile test, how thematerial will react to forces being applied in tension can be determined. Asthe material is pulled by machine, material’s strength can be found along itselongation.From the stress-strain curveof the tensile test, the values which can be found are Modulus of Elasticity,Yield Strength, the Tensile Strength, Percentage of Elongation and theReduction in Area.

Also the Toughness, Resilience, Poisson’s ratio can be foundby using the tensile test. In the report, these valueswill be found by doing the calculations and in the Results and Discussion partof the report, these calculations will be explained and discussed. Thisexperiment is made with two different specimens.(one is made from aluminum andthe other is made from steel)      Specimen’sraw material = Aluminum             Specimen’s raw material = SteelDiameter ofspecimen = 12 mm                        Diameter of specimen = 15 mmGage length= 59 mm                                        Gage length = 84 mmFinal Gagelength = 69 mm                              Final Gagelength = 95 mm                 After fracture :                             Max.load =39.56276(kN)                                Max.

load = 86.94378 (kN)Max.stress = 349.

8033 (MPa)                           Max stress = 492(MPa)       ?Theory of Experiment Formulas That Will Be Used in Lab Report Engineering Stress ?= P/A0 Engineering stress can be found by dividing the AppliedForce to Cross Sectional area of the specimen before the deformation has takenplace. True Stress?T =P/A                                     True stress can be found by dividing the AppliedForce to Cross Sectional area of the specimen at the point which the load is applied.                                                             Engineering Strain                               ? = ? / L0                                              Engineering strain can be found by dividing theTotal Elongation to Original value of the gage length. TrueStrain         L                     ?T = ?(1/L)dL = ln(L/L0)                            L0                                   True strain can be found by taking the integral of(1/(Changed value of gage length)) which is equal to ln((Changed value of gagelength)/(Original value of the gage length)). Hooke’s Law? = Normal Stress? = ? * E                 ?  = NormalStrain                                      E = Modulus of Elasticity(Young’sModulus)  Yield Point In order to find the yield point, take the load at the point wherestrain is (0.2%) divided by the cross-sectional area.     Ductility Ductility is ability ofmaterial that undergoes permanent deformation (through the reduction in crosssection area) by flexing or twisting at room temperature without fracturing.     Gage LengthDistance along the specimenthat the calculations of extension are made is called ”gage length”.

Sometimes,distance between the grips are taken as gage length.Difference Between Engineeringand True Stress/Strain True stress and strain are often not required. When the yieldstrength is surpassed, the material deforms.

The component hasfailed because it doesn’t have the original intended shape anymore.Furthermore, a significant difference develops between the two curves only when necking begins.But when the necking begins, the component is extremely deformed and no longersupplies its expected use. In the graph, true stress continues  increasing after necking,although the load requireddecreases, the area decreases even and even more.

 ToolsUsed in Experiment           Aluminum                                                                 SteelSpecimen        Specimen                             Caliper                                                           Instron Load?Procedure of Experiment            To determine the gage lengthand the diameter of the cross section of aluminum specimen, the aluminumspecimen was measured with the caliper. The diameter of the aluminum specimenwas determined (12.00 mm),the gage length of it was determined (59.00 mm) and scribed into the specimen inorder to measure the distance between the two marks after the tensile test wascompleted. After that, same measurements are made for the steel specimen andthe gage length was determined (84.00 mm), the diameter was determined (15.

00 mm).         To space the specimen equally between the twoclamps, the specimen got loaded to Instron load frame’s jaws. The extensometerswere fitted to the reduced gage section of the specimen, providing that theaxial extensometer was set correctly when attaching it to the gage and that thetransverse extensometer was attached to complete the diameter of specimen. The Instron load frame was loadedby using the scroll wheel to provide that the specimen was accurately loaded intothe frame and ensured that it wasn’t slipping in the jaws.

After that, by usingthe software, load was released so that the extensometers were zeroed. The testgot started and the specimen was loaded, resulting in a mensureable strain.This increase in the rate of strain may caused some error but this increasesped up the test.             ?Graphs and Calculations ?Graphsof Aluminum Specimen                                   ?Calculationsof Aluminum Specimen Hooke’s Law??=?*E290MPa=E*(0.029)                             so experimental value of E = 10000 MPa                                     ———————————————————————————Elongation%?( lfinal-l0)/(l0)            (l0= 59 mm, lfinal = 69 mm) = ((69mm-59mm)/(59mm))*100%     so Elongation% = 16.

949%              ———————————————————————————Cross-Sectional Area? ?*(((diameter)2)/4)= ?*(((15mm)2)/4)so the Cross-Sectional Area=176.715 mm2———————————————————————————     Area of Reduction(RA%)r0 = (12mm)/2 = 6 mmV = ?*r2*h = ?*(59mm)*(6mm)2 = ?*(69mm)*(r12)so r1 = 5.13 mm  ; fracture area =  ?*(r1)2so RA% = (((?*r02)-(?*r12))/(?*r02))*100%=26.8975%                                    ———————————————————————————Ultimate Tensile Strength?(Maximum Load)/(Cross-Sectional Area)=(39562.76 N)/(113.097 mm2)  UltimateTensile Strength is 349.

8033 MPa———————————————————————————           Note : There is a r1value because few seconds before fracture, the specimen’s gage lengthincreased and the specimen lost its default shape and we got a new radiusvalue(r1= 5.13 mm).?Graphs of Steel Specimen                 ?Calculationsof Steel Specimen Hooke’s Law??=?*E410MPa=E*(0.025)                             so experimental value of E = 16400 MPa                                    ———————————————————————————Elongation%?( lfinal-l0)/(l0)            (l0= 84 mm, lfinal = 95 mm) = ((95mm-84mm)/(84mm))*100%     so Elongation% = 13.095%              ———————————————————————————Cross-Sectional Area? ?*(((diameter)2)/4)= ?*(((15mm)2)/4)so the Cross-Sectional Area=176.715 mm2———————————————————————————     Area of Reduction(RA%)r0 = (15mm)/2 = 7.

5 mmV = ?*r2*hso ?*(84mm)*(7.5mm)2= ?*(95mm)*r12so r1 = 7.05 mm  ; fracture area =  ?*(r1)2so RA% = (((?*r02)-(?*r12))/(?*r02))*100%=11.640%                                    ———————————————————————————Ultimate Tensile Strength?(Maximum Load)/(Cross-Sectional Area)=(39562.76 N)/(176.715 mm2)  UltimateTensile Strength is 349.8033 MPa———————————————————————————      Note : There is a r1value because few seconds before fracture, the specimen’s gage length increasedand the specimen lost its default shape and we got a new radius value(r1= 7.05 mm).

   ?Results and Discussion  Values Aluminum Steel Diameter (mm) 12 mm 15 mm Cross-Sectional Area (mm2) 113.097 mm2 176.715 mm2 Gage Length (mm)? Final Gage Length (mm) 59 mm ? 69mm 84 mm ? 95mm Young’s Modulus (Gpa) 10 Gpa 16.4 GPa Load at Yield Poin t(N) 30.22 * 103 N 72.

45 *103 N Yield Strength (MPa) 290 Mpa 410 Mpa Maximum Load (N) 39562.76 N 86943.78 N Ultimate Tensile Strength (MPa) 349.8033 Mpa 492 Mpa Elongation% 16.

949% 13.095% Area of Reduction% (RA%) 26,8975% 11.640% Fracture Strain 0.138 0.146  2.Discuss about the yield strength and the different methods that are used tofind them. TheYield Strength is found by taking the load at the point where strain is (0.

2%)divided by the cross-sectional area.   3.Compare the stress-strain diagrams for Aluminum and steel and discuss theparameters in the tables. In comparison, the stress-strain diagrams have similarities but they arenot the same because the different raw materials have used in specimens. Thefracture strains are close to each other but the other values are not the sameor similar values to each other. 4.Discuss the different fracture types for aluminum and steel.Aluminum specimen broke from near to the middle of the specimen but thesteel specimen broke further from the middle of the specimen.

Steel specimenbroke cup&cone style.    ?ASTM STANDARDS                        10 to 38°C 50 to 100°F must be theroom temperature for the test unless it has special conditions. The valuesspecified in the SI units must be accepted separately from the inch / poundunits. The valuesspecified in each system are not exact equivalents; therefore, each system mustbe used independently of each other. Combining values in two systems can causeinconsistency with the standard.

      Strength and Ductility values of materials under uniaxial tensilestresses can be found by applying the tensile test to materials. Tensile testsgive information about the ductility and strength of materials under uniaxialtensile stresses. This information might be useful in comparisons of materials,quality control, alloy development.

While doing the calculations, large radiusfillets(the fillets at the ends of the gage length) must be used in brittlematerials.During a tensile test, the limits of crosshead speed must be in (mm/min)or (in./min) form. The limits of the crosshead speed must be farther quali?edby representing different limits for specimens which have different types, raw materialsand sizes.                      ?Conclusion         Inthe report, aluminum specimen and steel specimen’s Young’s Modulus,Elongation%, Area of Reduction%, Fracture Strain and other values have beenfound by doing some calculations and looking at the Graphs and measured values.         The comparison of thediameters, cross sectional areas, gage lengths, Young’s Modulus, load at yieldpoints, yield strengths, maximum loads, ultimate tensile strengths, elongationpercentages, area of reduction percentages, fracture strains of the aluminumspecimen and steel specimen has done and a table of these values are made.

          ASTM Standards arementioned and these standards are summarized. For the tensile test, the ASTMStandards are the valid standards.                          ?References ?AmericanAssociation State Highway and Transportation Officials Standard/AASHTO No.: T68- An American National Standard ? https://stephanfavilla.files.wordpress.com/2011/03/tensile-testing-laboratory.pdf ?http://eng.sut.ac.th/metal/images/stories/pdf/Lab_3Tensile_Eng.pdf ? Mechanics of MaterialsHibbeler 9th Edition ?http://www.engineeringarchives.com/les_mom_truestresstruestrainengstressengstrain.html ?https://www.astm.org/Standards/A285.htm ?http://www.instron.us/en-us/our-company/library/test-types/tensile-test ?http://www.academia.edu/14171023/How_to_measure_Yield_strength