The above figure demonstrates the impact of grain structures on fracture toughness for sheet 7000 series alloys. The fracture toughness magnitudes for the sheet alloys quoted are very large. In some cases >150MPavm Why are the values so large? OA Demonstrates the benefit of processing to promote fine grained microstructures. OB. Demonstrates the impact of retaining a deformation microstructure OC. These are plane stress fracture toughness values (Ke) so will be significantly greater than the normally quoted plane strain Kic values. D. Illustrates the benefit of transition metal grain refiners like Zr over Mn, and Cr which are more likely to be used in 2000 series alloys.

The above figure demonstrates the impact of grain structures on fracture toughness for sheet 7000 series alloys. The fracture toughness magnitudes for the sheet alloys quoted are very large. In some cases >150MPavm Why are the values so large? OA Demonstrates the benefit of processing to promote fine grained microstructures. OB. Demonstrates the impact of retaining a deformation microstructure OC. These are plane stress fracture toughness values (Ke) so will be significantly greater than the normally quoted plane strain Kic values. D. Illustrates the benefit of transition metal grain refiners like Zr over Mn, and Cr which are more likely to be used in 2000 series alloys.

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

See similar textbooks

Similar questions

Lsing the isothemal transtomation diagram for a 0.45 wes C stoel alloy (Fqure 10.40). dotemine the final microstructure in terms of just the microconstituents present) ofa small specimen that has boen subjected to the following timo-demperature traatments. In each case assume that the specimen begins at 845C (155OF), and that it has been held at this temperature long enough to have achieved a complote and homogeneous austenitic structure. • Rapidly cool to 250C (480F), hold for 10s, then quench to room temperature. • Rapidly cool to 70OC (1290F), hold for 30s, then quench to room temperature • Rapidly cool to 400C (750F), hold tor 500s, then quench to room temperature. • Rapidly cod to6C(12DFI, hadat this temparature br 35, rapidly cod to 4000 (7OFI hod tor 10 s, then quench to room bemperature. • Rapidly cool to 450C (840F), hold for 10 s, then quench to room temperature. 900 1600 800 A+F 1400 700 1200 AP 600 1000 E 500 A 800 50% 400 Mistart) 600 300 M190%) 200 400 100 200 0.1 10 10…
evwVmpreun atatus: QUESTION 34 Consider a single crystal of RENE NS, a second generation Nbased superaloy. This alloy is used in the hottest part of gas turbine engines for aerospace. The matrix phase has a face-centered-cubic (FCC) crystal structure. This single crystal has been subjected to a tensile stress that is applied along a (314) crystal direction (aka, the axis f sip occurs on a (1 1 1) plane in a (101) direction, please cakulate the stress in MPa at which ylelding will occur. The critical resolved shear stress for this alloy is 800 MPa. cose Fo cos dcos (CRSS) (1SJos dcose 3184.3 O 41,600.0 O800.1 O 6368.7 579.0 400.1 Save A A es nit to sove ond subent. Cck Sae All A es to save dll ansers 3.
A cold worked brass specimen of average grain size 0.02 mm has a yield strengh of 160 MPa. Estimate the sield strenath of this alloy affer it hasbeen heat teated and the value of ky is 12.0 MPa-mm* 2
Televio olpn Example moleda Sblaiy Ieteyo elpnie erli lliW (B Q.: A stress of 75 MPa is applied in the [001] direction on an fcc single crvstel Calculate: T. (a) the resolved shear stress acting on the (111) [-101] slip system and, (b) the resolved shear stress acting on the (111) [-110] slip system.
(d) Given that the grain size in a solid solution is determined according to the following expression: = 22-1 where Ny is the average number of grains per square inch at a magnification of M, and n is the ASTM grain size number. Carry out the following: i. Determine the ASTM grain size if 30 grains per square inch are measured at a magnificatio 100X. ii. Calculate the number of grains per square inch at a magnification of 75X.
Using the values found in the previous two questions, use the Hall-Petch equation and the graph below for a 70 Cu-30 Zn cartridge brass to predict the yield strength (in MPa) of this alloy when the average grain diameter is 1.0 x 10³ mm. Enter only the numerical value, do not include the unit. Grain size, d (mm) 10- 10-2 5 x 103 200 150 100 50 4 8. 12 16 d1/2 (mm-1/2) Yield strength (MPa)
As the head of R&D in your company which deals with aluminum, you are faced with aluminum alloy 2014 which was solution heat-treated. The alloy needs to be hardened by precipitation hardening to have a minimum yield strength of 345 MPa and ductility of at least 12% EL. What practical precipitation heat treatment with respect to temperature and time will you recommend to give these mechanical properties? Explain your answer Usc - A Z Type here to search @ 2 W S alt TRINIT # X 3 aps E D $ 4 C R Atomic percentage, magnesium O ene F % 5 T D V G 6 Y B & 7 H U N 8 J 9 11 M K Ľ L P 73 prt sc ? Ľ 1 backspace pause 40) Turn 8:59 AM 9/13/2022 11 enter 7 home A 28. 1 5 end 9
The flow stress of a coarse-grained dilute copper alloy increased from 2 to 55 MPa when the dislocation density was increased from a low value of 10 cm via cold working to a modest value of 100 cm2. Calculate the flow stress for this alloy when heavy cold working introduces a dislocation density of 10¹2 cm². An equation similar to the Hall-Petch equation has been proposed for dislocations, and is: Tflow To ka Paisl where Tflow is the flow stress (i.e., the force per unit area necessary to get plastic deformation), pdisi is the dislocation density (the dislocation line length per unit volume), and To and ka are constants for a given material. The easiest way to solve this problem is to put values into this equation twice, subtract one expression from the other, and solve for ka. Then enter your value of ka into either original equation to determine To. Keep track of units, and then solve the problem stated above.
The isothermal transformation diagram for and iron- carbon alloy of eutectoid composition is given in figure. Specify the nature of the final microstructure (in terms of microconstituents present and approximate percentages of each) of a small specimen that has been subjected to the following time-temperature treatments. In each case, assume that the specimen begins at 760°C. Show your work on the diagram. a. Cool rapidly to 300°C, hold for 5 s. b. Cool rapidly to 600°C hold for 100 s, then quench to room temperature. C. Cool rapidly to 500°C, hold for 7 s, and then quench to room temperature. Temperature (°C) 800 700 600 500 400 300 200 100 0 -1 10 6 A A A M(start) M(50%) M(90%) 1 10 M-A Eutectoid temperature P 10² Time (s) B 50% 10³ 104 105
Q-6) Which type of elements interstitial or substitutional elements can have greater influence on yield strength. Also from the table below by introducing which elements a maximum increase in yield strength of iron can be obtained? Explain Element Atomic size (pm) Size difference (%) Iron Carbon (C) Nitrogen (N) Phosphorus (P) Silicon (Si) Manganese (Mn) Molybdenum (Mo) Aluminium (AI) 126 77 -39 65 -48 107 -15 111 -12 127 +1 139 +10 143 +9 Copper (Cu) Nickel (Ni) 128 +2 124 -2
0-10) Compare the equilibrium spacing of two partial dislocations between an Al alloy (G=28GPa bie 0.165nm, stacking fault energy 200MJM ) and Gold (G=38GPA, bpartial3D0.166nm, SFE- 9mJm).
1. There are three slip systems on a FCC octahedral plane, i.e. one of the {111} planes. Assume a2 MPa tensile stress is applied along the [100] direction of a gold single crystal, whose criticalresolved shear stress is 0.91 MPa at room temperature. Demonstrate quantitatively thatmeasurable slip will not occur on any of the three slip systems containing the (111) plane as aresult of this applied tensile stress. (Hint: Schmid’s law) 2. Identify the Burgers vector (using vector notation) of a screw dislocation that can cross-slipbetween (111) and (111) planes of an FCC crystal (Hint: Use Weiss zone law). 3. Why are dislocations not considered thermodynamic equilibrium defects like vacancies?Explain by describing the relative interplay between entropy and enthalpy affecting the Gibbsfree energy of the dislocations-containing material system.