Materiaalkunde NMC113 Materials Science Probleme vir Hfstk. 6 / Problems for Chap. 6

Transcription

1 Materiaalkunde NMC113 Materials Science Probleme vir Hfstk. 6 / Problems for Chap /03/30 STRESS AND STRAIN (See answers on p S2 of Callister) 1. Callister prob Callister prob Callister prob Callister prob Callister prob Callister prob Callister prob Callister prob Callister prob Callister prob Callister prob Callister prob Callister prob Callister prob Callister prob Callister prob HARDNESS 1. Callister prob Callister prob TOUGHNESS AND FRACTURE TOUGHNESS Example 1. A component is made from a Ti alloy with fracture toughness of 55 MPa m, yield strength of 910 MPa, elastic modulus E of 114 MPa and y = 1,5 a) Calculate the critical internal flaw length in the component when subjected to a stress equal to ½ the yield strength. b) Calculate the specific surface energy γ of the alloy. Solution a) K 1c = yσ c πc c = (55 / 1,5x455) 2 /π = 0,00207 m = 2,07 mm Flaw length = 2x2,07 = 4,14 mm b) K 1c = 2Eγ γ = K 1c 2 /2E = (55x10 6 ) 2 / 2x 114x10 9 = 13x10 3 [J/m 2 ] PROBLEMS 1. Prob. E8.1 E8.3 in Ashby (Theory) 2. Prob. E8.4 in Ashby (9,769 MPa m, 477 J/m 2 ) 3. Determine the maximum surface crack length in a MgO component when the component must not fracture if a tensile stress of 13,5 MPa is applied to the component. Let the specific surface energy γ for MgO be 1,0 J/m 2 and E 220 GPa. (0,786 mm) 4. A 4340 steel component with fracture toughness of 54,8 MPa m is exposed to a tensile stress of 1030 MPa. Will the component experience fracture if the largest internal crack has a total length of 1 mm long. Assume y = 1 (No.; σ < σ f ) 5. An Al alloy component has fracture toughness of 40 MPa m and fractures at a stress of 300 MPa when the maximum or critical internal total crack length is 4 mm. Will fracture occur of the same component when it has a surface crack of length of 3 mm under a stress of 260 MPa. 1

2 y 1. (Yes) 6. A structural component made from steel has K 1c = 98,9 MPa m and yield strength σ y = 860 MPa. The internal flaw size resolution of a NDT apparatus is 3 mm. Will the critical internal flaw size for the steel be detected by NDT apparatus if the design stress is half the yield strength, i.e. a safety factor of 2, and y = 1 (Yes) 7. A steel plate of unit thickness with a surface crack of 1 mm length is subjected to a tensile stress of 200 MPa. The yield strength of the steel is 320 MPa. and y = 1,5. a) Calculate the theoretical size r y of the process zone or plastic region in the steel plate. ( 0,439 mm) b) Calculate the theoretical size r y of the process zone under the same conditions as above but with a yield strength of 350 MPa. ( 0,367 mm) c) What can be deduced from the difference in the answers to questions a) and c) above. (Higher σ y, lower r y or plastic zone material less ductile ) 8. Consider the impact test results shown in the figure below. a) At what temperature will material A fracture completely brittle and what temp. plastic. Give a brief explanation for your answer. ( - 20 o C, > 35 o C) b) What is the 20 J BTT for materials A and B respectively ( -4 & 22 o C resp.) c) Which one of the given materials will be the most suitable to be used at 10 o C. Give a brief explanation for your answer. ( A, BTT < 10 o C) d) Which one of the given materials will have the lowest fracture toughness at 10 o C. ( B, brittle fracture) e) Which one of the given materials will have the largest process zone at 10 o C and why. (A, plastic fracture) Fig. Impact test results for two hypothetical materials 9. Problems of your own selection in Callister FATIGUE Examples 1. The high cycle fatigue life of a material is given by Basquin s law: σ N f 0,1 = 400 a) Calculate the cycles to fracture N f for the material which has a tensile strength of 300 MPa when subjected to a cyclic stress with an amplitude of 55 MPa about a mean stress of 75 MPa. b) Calculate the max. and min. stresses applied Solution: a) Apply Goodman s rule: ( to find σ with σ m = 0) σ σm = σ σo [ 1 - σ m /σ ts ] σ σm = 2 σ a = 110 σ σo = 110/ [ 1 75/300] 2

3 = 146,67 MPa Now apply Basquin s law: N f = 0,1 400/ σ = 0,1 400/146,67 = 0,1 2,727 = cycles b) σ a = (σ max - σ min )/ σ m = (σ max + σ min )/ From 9.1 σ max = 2σ a + σ min a) subst. in 9.2 2σ m = 2σ a + 2σ min σ min = σ m - σ a = = 20 MPa subst. in a) σ max = 2x = 130 MPa 2. Assume Basquin s law for a material is given by : σ N f 0,1 = A component of the material is subjected to variable stress cycles with σ m = 0 in an application until it fractures. The measured stress cycles were: 1) cycles at a stress range of 60 MPa for which N f = (calculated) 2) cycles at a stress range of 140 MPa for which N f = cycles (calculated) 3) cycles at an unknown stress range Determine the unknown stress range Solution. Apply Miner s rule: Σ i=1 n N i /N fi = 1 = N 60 /N f60 + N 140 /N f140 + N x /N fx = / / / N fx = 0,5 + 0, / N fx / N fx = 0,3 N fx = Now apply Basquin s law: σ = 420 / = 420 / 5,25 = 80 MPa. 3. A 7 mm dia. rod of 2014-T 6 Al alloy is subjected in a test to a reversed tension-compression cycling load along its axis a) If the maximum and minimum loads are and 5773 N respectively, determine the fatigue life of the rod. Use Fig below) b) If the cyclic stress was applied at a frequency of 5 cycles per sec., how many hours did the test last. Solution a) Calc. stress amplitude. A = π d 2 /4 = π.49/4 = 38,49 mm 2 σ a = F/A = 5773 / 38,49 = 150 MPa From Fig. 8.34: fatigue life at 150 MPa = 10 x cycles = 10 7,3 cycles. ( Determine x by interpolating between the powers 7 and 8) b) Hours = cycles / cycles per second x 3600 = 10 7,3 /5x 3600 = 1108,48 hrs. 3

4 Fig Impact test results for some materials (Taken from Callister, W.D Matl. Sc & Eng. An Introduction. New York: John Wiley) PROBLEMS 1. Probl. E9.2 in Ashby (Theory) 2. Probl. E9.3 in Ashby ( If σ ts = 900 MPa, σ e = 300 MPa Fig. 9.8) 3. Probl. E9.4 in Ashby ( b) No, c) Yes, d) Apply Goodmans rule. Yes) 4. Probl. E9.7 in Ashby (96 MPa) 5. Probl. E9.8 in Ashby ( a) R = 0, b) σ m = 600 MPa, c) σ a = 900 MPa, N = cycl.) 6. Probl. E9.9 in Ashby ( No failure, K calc. = 1,98 MPa m) 7. The following condensed data was obtained from analysing the stress cycles (σ m = 0) on a component for which Basquin s law is given by: σ N 0,09 f = 380 Stress amplitude (MPa) No. of cycles N f (Calc. fatigue life) After how many remaining cycles will the component fracture when subjected to a stress amplitude of 40 MPa. ( cycles) 8. A cylindrical steel bar is subjected to repeated compression-tension stress cycling along its axis. If the load amplitude is N, calculate the minimum allowable diameter of the bar to ensure that fatigue failure does not occur. Use Fig above. ( For σ e = 313 MPa, d min = 16,46 mm) 9. A bar of 2014-T4 Al alloy with cross sectional dimensions of 12 x 12 mm is subjected to repeated tension-compression loading along its axis. Determine the (use Fig above): a) stress amplitude to yield a fatigue life of 10 6,1 cycles ( 200 MPa) b) maximum and minimum stress applied to the bar when the mean stress was 35 MPa (σ max = 235 MPa, σ min = -165 MPa) c) maximum and minimum loads applied to the bar (F max = N, F min = N) 10. Consider the S-N curve for red brass given in Fig Data was obtained by a bendingrotating test. Determine the: a) Fatigue strength at 4 x 10 6 cycles (± 100 MPa) b) Assume the test apparatus was rotating at 1200 r.p.m. What is the stress amplitude applied when the sample fractured after hrs. (147 MPa) 4

5 11. Three identical fatigue samples (A,B and C) are made from a non-ferrous alloy. The samples are subjected to cyclic stresses as given in the table below. Rank the specimens in order of highest to lowest expected fatigue life. ( Based on σ a and σ m : B>A>C ) Specimen σ max (MPa) σ min (MPa) A B C Problems of your own selection in Callister CREEP 1. Let the steady state creep rate ε ss for a stainless steel at 800 o C be given by the equation: ε ss = 1,53x10-10 σ 8 with σ in MPa a) Calculate the ε ss at 800 o C when the applied stress is 57 MPa (1,7x10-4 /s) 2. Refer to the deformation mechanism map for stainless steel on page 306 in Ashby see below. a) If the ε ss of the steel must not exceed 10-8 /s at 600 o C, what is the largest stress that could be applied. (ca. 49 MPa) b) What type of creep occurs in the stainless steel at the conditions stated in a) (power law creep) c) Describe the mechanisms by which the creep occurs in b) (Theory) d) Determine the constants B [MPa -1 s -1 ] and n in the equation ε ss = Bσ n at 600 o C if is observed from the map that : ε ss = 10-8 /s at σ = 49 MPa and ε ss = 10-6 /s at σ = 88 MPa. (B = 4,931x10-22, n = 7,874) d) Determine the constant C in the equation: ε ss = C.exp (-Q/RT) at a stress of 100 MPa for the stainless steel. The activation energy is equal to 341,1 kj/mol, R = 8,314 J/mol.K and it is observed from the map that ε ss = 10-8 /s at 495 o C and ε ss = 10-4 /s at 655 o C ( C = 15,85x10 14 /s) 3. Problems of your own selection in Callister 5