Consider a cylindrical specimen of a steel alloy (see figure below) 15.0 mm in diameter and 75 mm long that is pulled in tension. Determine its elongation when a load of 20,000 N is applied. 600 500 400 500 400 300 300 200 200 100 100 0.000 0.004 0.002 Strain 0.006 0.00 0.04 0.08 0.12 0.16 0.20 Strain Stress (MPa) Stress (MPa)

Consider a cylindrical specimen of a steel alloy (see figure below) 15.0 mm in diameter and 75 mm long that is pulled in tension. Determine its elongation when a load of 20,000 N is applied. 600 500 400 500 400 300 300 200 200 100 100 0.000 0.004 0.002 Strain 0.006 0.00 0.04 0.08 0.12 0.16 0.20 Strain Stress (MPa) Stress (MPa)

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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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.)
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
Example: Convert the change in length data in Table 3-2 to engineering stress and strain and plot a stress-strain curve Homework- help Table 3-2 The results of a tensile test of a 0.505 in. diameter aluminum alloy test bar, initial length (1o) = 2 in. Calculated LTO Load (Ib) Change in Length (in.) Stress (psi) Strain (in./in.) 0.000 1000 0.001 0.0005 4,993 14,978 24,963 34,948 37,445 39,442 39,941 39,691 37,944 3000 0.003 0.0015 5000 0.005 0.0025 7000 0.007 0.0035 7500 0.030 0.0150 7900 0.080 0.0400 8000 (maximum load) 0.120 0.0600 7950 0.160 0.0800 7600 (fracture) 0.205 0.1025
Stress Strain Diagram The Data shown in the table have been obtained from a tensile test conducted on a high-strength steel. The test specimen had a diameter of 0.505 inch and a gage length of 2.00 inch. Using software. plot the Stress-Strain Diagram for this steel and determine its: A= TTdT(050s A %3D 1. Proportional Limit, 2. Modulus of Elasticity, 3. Yield Strength (SY) at 0.2% Offset, 4. Ultimate Strength (Su), 5. Percent Elongation in 2.00 inch, 6. Percent Reduction in Area, 7. Present the results (for Steps 1-6) in a highly organized table. e Altac ie sheet (as problelle 4 A = 0.2.002 BEOINNING of the effort Elongation (in) Elongation (In) Load Load #: #3 (Ib) (Ib) 1 0.0170 15 12,300 0.0004 1,500 16 12,200 0.0200 0.0010 3. 3,100 17 12,000 0.0275 0.0016 4,700 18 13,000 0.0335 5. 6,300 0.0022 19 15,000 0.0400 0.0026 6. 8,000 20 16,200 0.055 0.0032 9,500 21 17,500 0.0680 0.0035 8. 11,000 22 18,800 0.1080 0.0041 11,800 23 19,600 0.1515 0.0051 24 20,100 0.2010 10 12,300 0.0071 25…
Problem F2 stress (ksi) 130 120 110 100 90 80 70 60 50 40 30 20 10 0 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 strain (in/in) Above you will find the experimental stress-strain diagram of 1045 steel. Calculate the permanent set if a cylindrical specimen with a diameter of 2 in is loaded to the ultimate stress and then unloaded. Provide your answer in units of in/in with 3 significant figures after the decimal. The elastic modulus of 1045 steel is 29,000 ksi.
. From the tensile stress-strain behaviour for a casted iron as shown in the following curve. Please answer the following questions (1) Estimate the modulus of elasticity, (2) Calculate the change in length of a specimen with original length of 100mm that is subjected to a tensile stress of 300MPa. (Hint: Using the Hooke's law to get the strain.) 600 500 400 - 300 - 200 - 100 - 2 4 6. 8 10 12 14 16 Strain Strength (MPa)
Problem No. 2 The above load-elongation data were obtained for a specimen with Lo = 0.1 m and ro 7 mm. Plot the stress-strain data and determine the E, G, and of Load (KN) 0.0 12.8 36.7 43.5 47.0 50.1 61.4 71.6 74.2 71.6 67.6 Elongation (mm) 0 0.0201 0.0690 0.1173 0.1898 0.2864 1.1684 3.5052 7.3025 10.2235 13.7287
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An engine parts is being tested with a load of 60000 lb. The allowable tensile stress is 10000 psi, modulus of elasticity of 40x106 psi. If the original length of specimen is 42 inches with elongation not exceeding 0.0015 in, what diameter of the specimen is rejected? a. 4.2 in b. 3.0 in c. 2.5 in d. 5.17 in
The following data (in given figure) were obtained from the tensile test of Aluminum alloy. The initial diameter of test specimen was 0.505 inch and gauge length was 2.0 inch. Plot the stress strain diagram and determine(a) Proportional Limit (b) Modulus of Elasticity (c) Yield Stress at 0.2% offset (d) Ultimate Stress and(e) Nominal Rupture Stress.
An eight-inch-long steel has a cross section area of 0.25 in2. After a load of 5000 lbs has been applied, the length is found to be 8.0065 in. What is the unit deformation induced? Select one: a. 0.00081 b. 0.00041 c. 0.0081 d. 0.004
Example A cylindrical specimen of aluminum having a diameter of 0.505 in. (12.8 mm) and a gauge length of 2.000 in. (50.800 mm) is pulled in tension. Use the load- elongation characteristics tabulated below to complete parts (a) through (). Load Length Ib in. (a) Plot the data as engineering stress 50.800 2.000 7,330 1,650 50.851 2.002 versus engineering strain. 15,100 3,400 50.902 2.004 (b) Compute the modulus of elasticity. 23,100 5,200 50.952 2.006 30,400 6,850 51.003 2.008 (c) Determine the yield strength at a strain 34,400 7,750 51.054 2.010 offset of 0.002. 38,400 8,650 51.308 2.020 41,300 9.300 51.816 2.040 (d) Determine the tensile strength of this 44,800 10,100 52.832 2.080 46,200 10,400 53.848 2.120 alloy. 47,300 10,650 54.864 2.160 (e) What is the approximate ductility, in 47,500 10,700 55.880 2.200 46,100 10,400 percent elongation? 56.896 2.240 44,800 10,100 57.658 2.270 ) Compute the modulus of resilience. WILEY Chapter 8- 51 42,600 9,600 58.420 2.300 36,400 8,200 59.182…