New version page

# PSU EMCH 316 - Lab 1

Pages: 3
Documents in this Course

6 pages

4 pages

## This preview shows page 1 out of 3 pages.

View Full Document
Do you want full access? Go Premium and unlock all 3 pages.

Unformatted text preview:

CONCLUSIONLauren Wagner Section 09 Jan 27, 2022 Professor Lodhi LAB 1 Notch Sensitivity of Low Ductile Materials OBJECTIVE To see how a notch stress concentration in a low ductile material will affect the effective strength of the material. PROCEDURE • First our lab group measured the width, thickness, and ligament width (all in mm) • Then we found the net notch depth by subtracting the ligament width from the total width of the specimen • Next our lab group found the failure load (N) by using the MTS Criterion tension testing machine. To do this our group loaded the specimen into the machine and inserted the dimensions then started the test. After we collected the data from the test and found the failure load. • Lastly our group toke all the data collected and was able to find the total area of the specimen, the area of the ligament, the percent full specimen area and the effective strength of the ligament and plotted the effective strength of the ligament vs the percent full specimen area. DATA AND RESULTS • Equations: o Total area: (Width)*(Thickness) Ex: (31.66 mm)*(1.56 mm) = 49.3896 mm^2 o Alig: (ligament width)*(thickness) Ex: (28.36 mm) * ( 1.56 mm) = 44.6628 mm^2 o A% : [(Alig)/(toatal area)]*100 Ex: (44.6628 mm / 49.3896 mm) * 100 = 90.42956412 % o σeff-lig: (failure load)/(Alig) ex: (13417.976 N)/(44.6628 mm) = 300.428455 MPA Material Specimen Width, w (mm) Thickness, t (mm) Net Notch Depth, a (mm) Ligament Width, wlig (mm) Failure Load, P (mm) 6061-T6 1 31.66 1.56 3.03 28.63 13417.976 6061-T6 2 31.66 1.56 6.09 25.57 11618.354 6061-T6 3 31.66 1.56 9.38 22.28 9955.775 7075-T6 1 31.72 1.6 3.91 27.81 15896.562 7075-T6 2 31.67 1.57 5.94 25.73 12508.602 7075-T6 3 31.62 1.63 9.33 22.29 11139.643 ST1011A 31.75 1.52 2.41 29.34 12859ST1011A 31.7 1.42 4.14 27.56 11462 ST1011A 31.71 1.42 4.11 27.6 11456 ST1018 31.73 1.52 0.97 30.76 19712 ST1018 31.72 1.53 2.58 29.14 17296 ST1018 31.76 1.58 4.195 27.565 16202 Material Specimen Total Area, A Alig A% σeff-lig ( ) ( ) % ( ) 6061-T6 - - - 100 310 6061-T6 1 49.3896 44.6628 90.42956412 300.428455 6061-T6 2 49.3896 39.8892 80.76437145 291.2656559 6061-T6 3 49.3896 34.7568 70.37271004 286.4410705 7075-T6 - - - 100 572 7075-T6 1 50.752 44.496 87.67339218 357.2582255 7075-T6 2 49.7219 40.3961 81.24407957 309.6487532 7075-T6 3 51.5406 36.3327 70.49335863 306.6010233 ST1011A - - 100 340 ST1011A 48.26 44.5968 92.40944882 288.3390737 ST1011A 45.014 39.1352 86.94006309 292.8821112 ST1011A 45.0282 39.192 87.03878903 292.3045519 ST1018 - - 100 440 ST1018 48.2296 46.7552 96.94295619 421.6001643 ST1018 48.5316 44.5842 91.86633039 387.9401223 ST1018 50.1808 43.5527 86.79156171 372.0090832ANALYSIS OF DATA • Determination of specimen full area o Total area: (Width)*(Thickness) Ex: (31.66 mm)*(1.56 mm) = 49.3896 mm^2 • Derivation and determination of ligament area o Alig: (ligament width)*(thickness) Ex: (28.36 mm) * ( 1.56 mm) = 44.6628 mm^2 • Calculation of percent of full specimen area o A% : [(Alig)/(toatal area)]*100 Ex: (44.6628 mm / 49.3896 mm) * 100 = 90.42956412 % • Calculation of ligament effective ultimate strength o σeff-lig: (failure load)/(Alig) ex: (13417.976 N)/(44.6628 mm) = 300.428455 MPA DISCUSSION OF RESULTS • If an alloy is flaw or notch insensitive, the ligament effective ultimate strengths would not depend on the effective cross-sectional areas. Is alloy sensitive to the presence of geometrical discontinuities? Support your response with references to the four sets of data collected. The different alloys are sensitive to the presence of the geometrical discontinuities. This is because the tensile strength of the effective ligaments that were tested would be all the same if it mattered on the effective cross-sectional area, they are sensitive because they matter on the notch depth. • A very soft and ductile Aluminum alloy 1100 is the type commonly used in electrical conductors. Based on your experiments, what would your expectations be as to its notch sensitivity? I would expect it to have a lower notch sensitivity this is because the aluminum alloy is very ductile and looking at the data that was collected, the lower the ductile the lower the notch sensitivity was. If the material was brittle, then the notch sensitivity would be higher. • Comment on the fracture surfaces. Were they expected for this kind of test? The fracture surfaces were expected during this test. This is because they were needed to find the fracture stress, and the ultimate strength. The surfaces were brittle if the surfaces that broke were very straight across and smooth. The surfaces that were ductile were able to bend and stay intact longer than the brittle materials. CONCLUSION During the lab the difference between materials and their notch sensitivity was determined by looking at the fracture stress and ultimate strength. The data was able to be gathered from the MTS Criterion tension testing

View Full Document