Question

700 600 T-5A-2.55n titanium alloy 4340 500 400 Mamuntres SMP 1045 300 Ductile contien 200 700 30 brass 2014-T6 Alloy 100 EG21A-T6 Mga 10 10 10 10 10 Cycles totallur, #3. A ceramic part for a jet engine has a yield strength of 75,000 psi and a plane strain fracture toughness, i.e., Kc, of 5000 psivin. To be sure that the part does not fail, we plan to assure that the maximum applied stress is only one-third the yield strength. We use a nondestructive test that will detect any internal flaws greater than 0.05-in. long. Assuming that the geometry factor is Y - 1.4, does our nondestructive test have the required sensitivity? Explain your answer briefly. Answer: #4. We want a copper bar to have a yield strength of at least 65,000 psi and a final diameter of 0.375-in. What is the minimum diameter of the original bar? Use the figure below to solve the problem Answer:

Answer 1

soln # Given = kic- [50oopsitin), f = Y = 1.4 know that the strain fracture toughness of ceramic part is given by Kics folita Kic Jita Squaring both side fo we get Kic 2 Kic a = fo Ta fo where f is geometry factor T. *o is applied stress a is flaw size. So the maximum applied Stress is equal to the one third of the yield strength .ie. 1 lyield strength ) given in ques 1 (75000 psi) 3 25000 psi Now putting all the value in about formula me get sooo psico 1.4x2500o psi 0 65 in

Also the length of internal flaw is za je 2 a 2x0.6065 = 0.0013 in Ans Since the value is less than 0.05 it will not get detected and hence will cause a failure therefore this test is not good.