Problem: A tensile test specimen of stainless steel alloy having a with a diameter of 12.6 mmand a 50 mm gage length was tested to fracture. Load and deformation data obtained during thetest are given:Determine:(a) the modulus of elasticity.(b) the proportional limit.Change in length(mm)0.000.000.05Load(c) the ultimate strength.(kN)0.000.0025.79(d) the yield strength (0.20% offset).(e) the fracture stress.51.590.10(f) the true fracture stress if the final diameter66.550.20of the specimen at the location of the fracture73.360.3080.180.45was 8.89 mm87.770.630.811.011.1895.35101.42105.98112.05118.13123.461.411.661.942.27128.032.602.82131.85133.39134.95135.733.153.353.683.934.26136.53136.56135.84135.14133.664.664.945.225.475.705.82130.67128.44124.69120.92117.91116.525.95fracture

Problem: A tensile test specimen of stainless steel alloy having a with a diameter of 12.6 mm and a 50 mm gage length was tested to fracture. Load and deformation data obtained during the test are given: Determine: (a) the modulus of elasticity. (b) the proportional limit. Change in length (mm) 0.00 0.00 0.05 0.10 Load (c) the ultimate strength. (kN) 0.00 0.00 (d) the yield strength (0.20% offset). (e) the fracture stress. 25.79 51.59 66.55 73.36 (f) the true fracture stress if the final diameter 0.20 0.30 0.45 of the specimen at the location of the fracture 80.18 was 8.89 mm 87.77 95.35 0.63 0.81

Problem: A tensile test specimen of stainless steel alloy having a with a diameter of 12.6 mm and a 50 mm gage length was tested to fracture. Load and deformation data obtained during the test are given: Determine: (a) the modulus of elasticity. (b) the proportional limit. Change in length (mm) 0.00 0.00 0.05 0.10 Load (c) the ultimate strength. (kN) 0.00 0.00 (d) the yield strength (0.20% offset). (e) the fracture stress. 25.79 51.59 66.55 73.36 (f) the true fracture stress if the final diameter 0.20 0.30 0.45 of the specimen at the location of the fracture 80.18 was 8.89 mm 87.77 95.35 0.63 0.81

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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hings The following results were obtained during a tensile test to destruction on a mild steel test piece of diameter 15.96mm and 80mm gauge length Extension for a load of 40KN = 0.08mm Maximum load applied during test = 93kN Diameter at fracture = 12.85mm %3D Final length between gauge points 106mm From these results, determine: i) The modulus of elasticity for mild steel ii) The tensile strength iii) Percentage of elongation iv) Percentage of reduction in area
1. Plot the engineering stress & strain diagram of an alloy having a tensile test result found in Table 1. The tensile test specimen has a diameter of 12.5mm and a gage length of 50.0mm. The given alloy is used to make a 30.0mm diameter cylinder, which is placed inside a hardened circular steel casement with a 30.01mm inner diameter. Table 1: Tensile test results of an alloy Change In Length (mm) Change In Diameter (mm) Load (kN) 0.000 0.0000 0.0000 4.364 0.0254 0.0019 -0.0057 13.092 0.0762 21.819 0.1270 0.0095 30.547 32.729 0.1778 0.7620 34.911 3.0480 30.01 mm O F Rigid Plate Cylindrical Alloy -Steel casement Figure k Section view of the steel casement encapsulating the cylindrical alloy 2. Determine the required minimum value of F such that the cylindrical alloy would touch the walls of the steel casement.
1. Plot the engineering stress & strain diagram of an alloy having a tensile test result found in Table 1. The tensile test specimen has a diameter of 12.5mm and a gage length of 50.0mm. The given alloy is used to make a 30.0mm diameter cylinder, which is placed inside a hardened circular steel casement with a 30.01mm inner diameter. Table 1: Tensile test results of an alloy Change In Length (mm) Change In Diameter (mm) Load (kN) 0.000 0.0000 0.0000 4.364 0.0254 -0.0019 13.092 0.0762 -0.0057 21.819 0.1270 -0.0095 30.547 0.1778 32.729 0.7620 34.911 3.0480 30.01 mm Ø F Rigid Plate Cylindrical Alloy - Steel casement Figure 1: Section view of the steel casement encapsulating the cylindrical alloy 2. Determine the required minimum value of F such that the cylindrical alloy would touch the walls of the steel casement.
The table below shows the deformation data that resulted from applying a pure tensile load to a brass alloy rod with an initial length of 30 mm and a diameter of 10 mm. This rod was subjected to necking and beyond necking deformation. The data at fracture is shown in the last row. Applied Load (N) Length (mm) Diameter (mm) 75,100 33.8 68, 250 34.9 50,200 35.7 6.7 Solve for the TRUE STRESS/ES and TRUE STRAINS. 8.3 7.8
A bronze alloy specimen having a diameter of 12.8 mm and a gage length of 50 mm was tested to fracture. Stress and strain data obtained during the test are shown in Figure. Determine: 400 - Upper scale a. the modulus of elasticity. 320 b. the proportional limit. Lower scale c. the ultimate strength. 240 d. the yield strength (0.20% offset). 160 e. the modulus of resilience. f. the true fracture stress if the 80 final diameter of the specimen at the location of 0.100 0.005 0.300 0.015 0.200 0.400 the fracture was 10.5 mm. 0.010 0.020 g. The Poisson's ratio. Strain (mm/mm) Stress (MPa)
A tensile test specimen of stainless steel alloy having a diameter of 0.495 in. and a gage length of 2.00 in. was tested to fracture. Stress and strain data obtained during the test are shown. Determine the true fracture stress if the final diameter of the specimen at the location of the fracture was 0.350 in. Stress (ksi) 120 Upper scale Lower scale H 0.0 0.020 0.040 0.060 0.080 0.100 0.120 0.0 0.002 0.004 0.006 0.008 0.010 0.012 Strain (in./in.) 90 60 30 0- 300 ksi O 126 ksi EA O 237 ksi O 202 ksi O 261 ksi
In tensile test a plain carbon steel specimen has a (40mm) gauge length and the Final area (A final) of specimen after tensile test was 264.327. The load which caused fracture was (122.5 KN). After fracture, the final length was 47.516mm The % reduction in area is * O 15.819% 18.819 O 17.819 O 20.819
A tensile test specimen of 1045 hot-rolled steel having a diameter of 0.505 in. and a gage length of 2.00 in. was tested to fracture. Stress and strain data obtained during the test are shown. Determine the ultimate strength. Stress (ksi) 60 50- 40- 30- 20 10- 0 0 0 Upper scale O 66 ksi 28 ksi O 72 ksi O 53 ksi O 62 ksi Lower scale. 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.002 0.004 0.006 0.008 0.010 0.012 0.014 Strain (in./in.)
Force P and length change AL data are given in table below for the initial portion of a tension test on 7075-T651 Al alloy. The diameter before testing was 9.07 mm., and the gage length Linitial for t length change measurement was 50.8 mm. What tension force is required to cause yielding in a bar of the same material but with a diameter of 20 mm? P, kN AL mm 7.22 0.0839 14.34 0.1636 21.06 26.8 31.7 34.1 35.0 0.241 0.308 0.380 0.484 0.614 36.0 0.924 36.5 1.279 36.9 37.2 1.622 1.994
In tensile test a plain carbon steel specimen has a (40mm) gauge length and the Final area (A final) of specimen after tensile test was 264.327. The load which caused fracture was (122.5 KN). After fracture, the final length was 47.516mm True stress at fracture is. * O 463.441 N/mm2, O 525.441 N/mm2, O 254.441 N/mm2, 498.441 N/mm2,
A tensile test for a copper specimen has been performed and the following data are obtained. - Percentage of Elongation = 69 % - Percentage of Reduction in Area = 39 % - Final length after fracture = 35.5 mm - Final Diameter after fracture = 3.6 mm & - Ultimate stress = 396 MPa SOLUTION: i) Initial Length (in mm) = ii) Final Area (in mm2) = iii) Initial Area (in mm2) =