3. For steels, the endurance limit is approximately half of the tensile strength, and the fatigue strength at 10' cycles is approximately 90% of the tensile strength. The S-N curves can be approximated by a straight lines between 10' and 10* cycles when plotted as log(S ) versus log(N). Beyond 10 cycles the curves are horizontal. a) Using the information described above for steels, sketch the salient features of the S-N curve and write a mathematical expression for S as a function of N for the sloping part of the S-N curve. b) A steel part fails in 12,000 cycles. Use the above expression to find what % decrease of applied (cyclic) stress would be necessary to increase in the life of the part by a factor of 2.5 (to 30,000 cycles). c) Alternatively, what % increase in tensile strength would achieve the same increase in life without decreasing the stress?

3. For steels, the endurance limit is approximately half of the tensile strength, and the fatigue strength at 10' cycles is approximately 90% of the tensile strength. The S-N curves can be approximated by a straight lines between 10' and 10* cycles when plotted as log(S ) versus log(N). Beyond 10 cycles the curves are horizontal. a) Using the information described above for steels, sketch the salient features of the S-N curve and write a mathematical expression for S as a function of N for the sloping part of the S-N curve. b) A steel part fails in 12,000 cycles. Use the above expression to find what % decrease of applied (cyclic) stress would be necessary to increase in the life of the part by a factor of 2.5 (to 30,000 cycles). c) Alternatively, what % increase in tensile strength would achieve the same increase in life without decreasing the stress?

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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6. The following engineering stress-strain data were obtained for 0.2% C plain carbon steel. (a) Plot the engineering stress-strain curve (b) Determine the ultimate tensile strength for the alloy (c) Determine the percent elongation at fracture (d) Plot the true stress-strain curve Engineering strain, in./in. Engineering stress, ksi 30 0.001 55 0.002 60 0.005 68 0.01 72 0.02 74 0.04 75 0.06 76 0.08 75 0.1 73 0.12 69 0.14 65 0.16 56 0.18 51 0.19(fracture)
Cyclic Stresses (Fatigue); The S-N Curve 4. The fatigue data for a ductile cast iron are given as follows: Cycles to Failure Stress Amplitude [MPa (ksi)] 248 (36.0) 236 (34.2) 224 (32.5) 213 (30.9) 201 (29.1) 193 (28.0) 193 (28.0) 193 (28.0) 1 × 105 3 x 105 1 x 10⁰ 3 x 106 1 x 107 3 x 107 1 108 3 × 108 (a) Make an S-N plot (stress amplitude versus logarithm cycles to failure) using these data. (b) What is the fatigue limit for this alloy? (c) Determine fatigue lifetimes at stress amplitudes of 230 MPa (33,500 psi) and 175 MPa (25,000 psi). (d) Estimate fatigue strengths at 2 × 105 and 6 × 106 cycles.
a) Explain the term Thermal Fatigue. In which situations it is very prevalent. Write down the mechanical properties of the metal which are most affected by thermal fatigue. b) If, for carbon steel, operating at 140 MPa, the Larson Miller parameter (LMP ) is 24000. Calculate the time of rupture of this material at 800 °C. Where as, LMP = T(20+logt.); t = time to rupture in hour and Tis in deg Kelvin
The engineering stress-strain curve below was obtained for a precipitation hardened Aluminum alloy. What is the approximate Yield Strength for this alloy in psi? Engineering Stress Based on Original Area (psi) 50,000 45,000 40,000 35,000 30,000 25,000 20,000 15,000 10,000 5,000 0 O 0.02 0.04 0.06 Aluminum 6061-T6 0.08 0.1 0.12 Engineering Strain (in/in) 0.14 X 0.16 0.18
70% + | 8 0 4. An application requires ultimate tensile strength and yield strength of a steel at 110 ksi and 91 ksi, respectively. A data table is attached in the back of the test. Answer the following 4 questions: 4.1. Can SAE 1040 steel be selected for this application? 4.2. If "no" is the answer in Part I, the following Part II, III, and IV can be ignored. If "yes" is the answer in Part I, which condition of SAE 1040 should be selected? 4.3. Why is that steel with the condition in part II selected? 4.4. Is the selected steel brittle or ductile? and Why? Page 4 of 6
An S-N plot from fatigue testing of a steel is provided below. Use the plot to determine the following values for this steel. 1. Fatigue limit: 2. Fatigue lifetime at a stress amplitude of: 415 MPa: 275 MPa: 3. Fatigue strength at: 5x104 cycles: 5x105 cycles 500 400 300 200 100 0 1.E+04 Stress Amplitude (MPa) 1.E+05 1.E+06 Cycles to Failure 1.E+07 1.E+08
a . Sketch stress strain curve if the result shown in table represent the force and extension happened in steel, and show Mechanical properties that we get from tensile test on curve? (10p)Note : Lo=80 mm , Do=10 mm , use excel to plot the curve ExtensionLoad(mm) (N)0 0.900.83 4694.341.67 4831.412.50 4781.083.33 4918.834.17 4926.585.00 5257.075.83 5437.016.66 5575.888.33 5775.189.16 5847.5210.83 5965.4111.67 6010.5312.50 6042.5713.33 6072.2614.16 6092.9315.00 6113.2416.67 6140.3617.50 6146.3718.33 6148.1419.16 6149.1725.00 5940.2125.83 5675.3326.67 4725.52b. What is meant by modulus of rigidity? if it increases what does happen to material? (2p)
4. Based on Figure 1 shows, for a carbon steel, the partial tensile engineering stress-strain curve. Determine; (i) Yield strength (o) (ii) Ultimate Tensile Strength (oTs) (ii) Young Modulus (E) 300 2000 10° psi 300 MPa 2000 200 200 1000 1000 100 100 0.000 0.005 0.010 Strain 0.015 0.000 0.020 0.040 0.060 0.080 Strain Figure 1: Engineering stress-strain curve for alloy steel Stress (MPa) Stress Stress (10° psi)
(b) Figure 1 shows, the tensile engineering stress-strain curve for an aluminum alloy. Based on Figure 1, determine: (1) Yield strength (a) (ii) Ultimate Tensile Strength (ors) (ii) Young Modulus (E) | Stress vs Strain 350 300 250 200 150 100 50 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 Strain (mm/mm) Figure 1: Engineering stress-strain curve for aluminum alloy Stress (N/mm2)
Q2c) Listed in the table below is the tensile stress-strain data for different grades of steels. Utilizing the data given answer the three queries given below. Tensile Strain at Fracture Strength Strength Fracture Strength Modulus Material Yield Elastic (MPa) (MPa) (MPa) (GPa) А 410 1440 0.63 265 410 200 220 0.40 105 250 C 815 950 0.25 500 610 800 650 0.14 720 210 E Fractures before yielding 650 550 1) Which will experience the greatest percent reduction in area? Why? 2) Which is the strongest? Why? 3) Which is the stiffest? Why? B.
3. A cylindrical specimen of cold-worked copper has a ductility, %EL. If its cold-worked radius is d, what was its radius before deformation? Use %EL=25%; d = 10mm Ductility (%EL) 70 60 50 40 30 20 Brass 1040 Steel 10 Copper 0 0 10 110 20 30 40 50 60 70 Percent cold work
b) If, for carbon steel, operating at 140 MPa, the Larson Miller parameter (LMP ) is 24000. Calculate the time of rupture of this material at 800 °c. Where as, LMP = T(20+logt.); t time to rupture in hour and Tis in deg Kelvin