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Unformatted text preview: Experiment I a. Procedure Tension test is commonly used to determine the properties of the material. That helps to show the material’s behaviors under tensile or bending conditions. To do the experiment firstly we mark the specimen to measure the gage length and diameter of the specimen to do the calculations after the experiment done. Secondly, we do the experiment with tension test machine. When we were doing the experiment, the recorder measured the load varying against time until it fractures. After the testing, we plot the load versus elongation graph and the engineering stress strain graph. Results b. The graph was drawn by hand and it is on the back side of the report. c. The graph was drawn by hand and it is on the back side of the report. d. I. With the 0.2% offset method I found σ y = 613 MPa. II. From elastic region’s slope (Hooke’s Law σ= Eε) E = 10 GPa. III. UTS is biggest stress value on engineering stress strain diagram. UTS= 674.5 MPa IV. Fracture stress σ f = 471 MPa e. I. εy = 6.3% II. ε UTS = 9.3% III. ε f = 16.5% f. I. %Δl= [(l fl )/l ]*100 %Δl= [(86.2077.02)/77.02]*100= 11.92% II. %RA=[(AA f )/A ]*100 %RA= [(121.1556.21)/121.15]*100= 53.60% g. I. U r =1/2 σ y ε y = 0.5*613*0.063= 19.31MPa= 19.31* 106 J/ mm³ . II. I found the area under the curve UT= 96.86* 106 J/ mm³ . h. In the elastic deformation region we assume volume constant A *l = A y *l y (6.21)*(6.21)*π*77.02= Ay*77.02*1.063 dy= 12.05 mm ν=  (Δd/d )/ (Δl/l ) ν=  (0.37/12.42)/ (4.85/77.02)= 0.473 i. Brinell hardness of the specimen [2] UTS(MPa)= 3.45*HB 674.5= 3.45*HB => HB= 195.5 MPa j. σ T = F/ A i (true stress) [2] ε T = ln (L i /L o ) (true strain) [2] Assume volume constant before UTS A *l = A f *l f σ t = σ (1+ε) ε t = ln (1+ε) Onset neck area we find the data before UTS and try to find the K and n values by logs of true stress and strain. y = 0.1873x + 3.0661 2.852 2.854 2.856 2.858 2.86 2.862 2.864 2.8661.141.131.121.111.11.091.081.07 log σ T versus log ε T Excel founds that the slope is 0.1873 that gives us the n value. We can find K from that log K= 3.0661 K=1165 MPa Also from the equation σ T = K ε n 732.5=K (0.084) (0.1873) K = 1165 MPa k. Results Experimental Theoretical[2] Yield strength σ y 613Mpa 290490 MPa Young’ modulus E 10Gpa 207 GPa Ultimate tensile strength σ UTS 674.5Mpa 520590 MPa Fracture Stress σ f 471Mpa 260MPa Yield strain ε y 6.3% UTS strain ε UTS 9.3% Fracture strain ε F 16.5% Percent elongation % ∆l 11.92% 1228 % Percent reduction in area % RA 53.60% Resilience U R 19.31*106 J/mm³ Toughness U T 96.86*106 J/mm³ Poisson’s ratio ( ν ) 0.473 0.3 Brinell hardness 195.5 Mpa True ultimate stress 737.2 Mpa True fracture stress 1015.1 Mpa True ultimate strain 0.089 True yield strain 0.056 N 0.1873 K 1165 MPa Discussion The materials mechanical properties are important for using field. If we want spring we want higher elastic modulus value with high yield strength value. This specimen has we want higher elastic modulus value with high yield strength value....
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This note was uploaded on 12/27/2011 for the course MATH 286 taught by Professor Adad during the Spring '11 term at Middle East Technical University.
 Spring '11
 adad

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