E Mech 316DATA AND RESULTSANALYSIS OF DATACONCLUSIONS18 Name: Lauren Wagner Section: 09 Date: 2/10/22 Instructor: Lodhi E Mech 316 LAB #2 Tensile Behavior of a Ductile Metal OBJECTIVE • To measure and determine the Tensile behavior of the ductile metal aluminum 6061-T6 through a dog bone specimen. PROCEDURE • First our group toke the measurement of the diameter of the specimen at the reduced section in the middle of the dog bone specimen. We also toke an initial measurement of the gauge-length of the sample. • Next the group toke the specimen and loaded it into the MTS Criterion testing machine so that ¾ of the gripping surface was in contact with the specimen. • Next the extensometer was attached, and the zeroing pin was put into place on the sample. • • Run the test according to the instructions provided, making sure to remove the zeroing pin when prompted without disturbing the extensometer’s position. • Once the sample has surpassed the proportional limit, our group replaced the zeroing pin and removed the extensometer. • Lastly our group removed the fractured specimen from the machine, measured the final diameter and then sketched the failed specimen DATA AND RESULTS Table 1. Dog bone dimensions.19 QUANTITY DIMENSION ( mm ) Initial diameter, di 6.4 mm Sample Gauge length, l0 77.15 mm Extensometer Gauge length, le 25.4 mm Final diameter, df 2.64 mm Table 2. Load, Elongation, Stress, and Strain data. Load Crosshead Elongation, ΔLCH Extensometer Elongation, ΔLEX Stress Crosshead Strain, εCH Extensometer Strain, εEX ( N ) ( ) ( ) ( mpa) ( mm ) ( mm ) Total length 86.47mm Figure 1. Fracture surface. ANALYSIS OF DATA • Determination of specimen Area: A= (pi/4)(d)2 A=(pi/4)(6.4)220 A=32.17 mm • Sample calculation of Stress & Strain: Stress = F/A = (37.30)/(32.15) = 1.16 mpa Crosshead strain = F/a = (0.00404)/(32.15) = 0.00125 mm Extensometer strain = extensometer / Extensometer Gauge length = (0.000373)/25.4) = 1.46*10^-5 mm • Determination of Modulus of Elasticity, E: E = (stress/strain) = (122.029)/(0.001989016) = 61014 MPa • Determination of 0.2% Offset Yield Strength, σy: (σy) * 0.002 = 1.159496*0.002 = 0.002% • Determination of Ultimate Tensile Strength, σu: The highest max value = 231.7034 MPa • Determination of Percent Elongation, ε%: = (( final gauge length – initial gauge length) / (initial gauge length) )*100 = ((86.47-25.4)/(86.47))*100 = 70.62% • Determination of Modulus of Resilience, Ur (estimate σP first): Area under the graph= (0.5)*(198.4)*(0.003)= 297600 Pa • Determination of Modulus of Toughness, UT: =((Sigma yield + sigma UTS)/2)(strain at fracture) =((231.003+205.3)/2)*(0.1366706) = 29.814 MPa • Determination of Toughness, T:21 Area under the curve = 297600 + 600000 = 897600 Pa • Determination of Percent Reduction in Area: ((2.64-6.4)/6.4)*100 = -58.75 Figure 2. Full data plot (Crosshead) Figure 3. Yield strength plot (Extensometer)22 DISCUSSION OF RESULTS 1. Was the deformation of the specimen entirely uniform? Explain. The deformation of the specimen was uniform until it hit the UTS and started necking. This caused the specimen to not be uniform. 2. Was ratio of the gauge length to diameter ratio appropriate by ASTM standards? (at least 4:1 ratio) The ratio was calculated to be 12:1 which is appropriate because it is greater than 4:1 3. Does Al 6061-T6 work harden? If so, to what degree. Explain in detail. All materials that undergoes plastic deformation undergoes work hardening. This also shows that Al 6061 T6 can undergo work hardening. This is computed by subtracting ultimate tensile strength by yield strength and then dividing the that by the yield strength.23 4. How did the Modulus of Toughness compare with the Toughness? In what type of material is the Toughness exactly equal to the Modulus of Toughness? The toughness is greater than the modulus of toughness for this material. The brittle material will be exactly equal to the modulus of toughness. This is because there is no plastic region. 5. Describe the fracture surface of the specimen? Is this expected? Be sure to refer to Figure 1. The specimen had necking occur. This is seen because there is a divot into the specimen showing that it was necking. As well as the specimen’s diameter was smaller after the necking than the original diameter before the load was applied. CONCLUSIONS Al 6061-T6 when having a load applied will have necking occur before