New version page

NCSU MSE 200 - Exam#2 Review Notes

Upgrade to remove ads

This preview shows page 1-2-3 out of 9 pages.

Save
View Full Document
Premium Document
Do you want full access? Go Premium and unlock all 9 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 9 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 9 pages.
Access to all documents
Download any document
Ad free experience

Upgrade to remove ads
Unformatted text preview:

MSE 200 Exam#2 Review NotesChapter 4: Imperfections in solidsChapter 6: Mechanical properties of metalsMSE 200 Practice Exam #2 Spring,2020Multiple choice questions (3 points each): Carefully circle or write only one answer.% Cold workConstants and Equations for Exam #2Equations:©Copyright. 2020. A. Singh. North Carolina State University. All right reserved. MSE 200 Exam#2 Review Notes Chapter 4: Imperfections in solids • # of defects as a function of temperature • Point defects (vacancies, interstitials, self-interstitials, substitutional impurities) • Solubility rules for substitutional and interstitial alloys • Line defects ( dislocations [edge, screw], Burger’s vector direction relative to dislocation line for each) • Area defects ( Grain boundaries, surfaces); single crystal versus polycrystal • Chapter 6: Mechanical properties of metals • Tension tests (dog bone samples) • Engineering stress vs. Engineering strain (elastic modulus, Yield strength, tensile strength) • What is engineering stress? • What is engineering strain? • How do they change as a function of temperature? • Poisson’s ratio • Necking • Toughness, Ductility, Resilience, Hardness Chapter 7: Dislocations and strengthening mechanisms • How do dislocations participate in plastic damage? • What type of stress is needed? (tensile or shear) • How do dislocations interact with one another? • How does crystallinity play a role in plastic yielding? • Slip systems • Resolved shear stress and critical resolved shear stress • Strengthening mechanisms - Grain boundaries (smaller grains stronger, less ductile) - Alloying or solid solution strengthening (increased alloying results in stronger, less ductile) - Work or strain hardening (more % CW, increased strength, less ductility, reversible) - Precipitation hardening©Copyright. 2020. A. Singh. North Carolina State University. All right reserved. MSE 200 Practice Exam #2 Spring,2020 Name (print) (Singh) I have neither given nor received unauthorized aid on this test. X (signature) No notes, books, or information stored in calculator memories may be used. Cheating will be punished severely. All of your work must be written on these pages and turned in. To receive full or partial credit on numerical problems, you must show your calculations in step-by-step fashion. Units must be shown when applicable and all plots must have labeled axes. Be sure that you read and answer all parts of each question. Multiple choice questions (3 points each): Carefully circle or write only one answer. 1. Which of the following processes makes a material more ductile? a. decreasing the temperature b. adding an alloying impurity c. recrystallization after cold work d. none of the above 2. HCP materials are more brittle than FCC metals because of a. lower number of slip systems b. larger number of slip systems c. higher density d. larger volume of a unit cell 3. The critical resolved shear stress is: a. the minimum shear stress required to initiate slip b. the shear stress projected onto a plane of atoms c. the shear stress when ɸ and λ are equal 45° d. none of the above 4. As the percent cold work increases: a. yield strength increases b. tensile strength increases c. ductility decreases d. all of the above 5. Toughness depends on: a. strain rate b. temperature c. microstructure d. all of the above©Copyright. 2020. A. Singh. North Carolina State University. All right reserved. 6. Consider an extra half plane got inserted in top half of a crystal resilting in an edge dislocation; the region just above this line defect will experience a. tension b. compression c. neither tension or compression d. both tension and compression 7 Which statement is trure regarding intersitial impurity in metal a. interstitial impurities are point defects and will always make host metal weak b. interstitial impurities tend to make metal more ductile c. interstitial impurities does not affect mechanical properties of host metal because they diffuse much faster d. interstitial impurities increase packing density©Copyright. 2020. A. Singh. North Carolina State University. All right reserved. 8. A Cu-Zn alloy has Hall-Petch coefficients of stress are 25 MPa and 12.5 MPa(mm1/2). Predict the change in yield strength of this alloy when average grain size diameter of 1x10-3 mm is reduced to half©Copyright. 2020. A. Singh. North Carolina State University. All right reserved. 9. Two previously undeformed specimens of the same metal are to be plastically deformed by reducing their cross-sectional area. One has a circular cross-section, and the other is rectangular. During the deformation the circular cross-section went down from 15.2 to 11.4 mm diameter, while the rectangular was 125x175 mm before deformation and became 75x200 mm after deformation. Which of these will be harder? Why? Show you calculations.©Copyright. 2020. A. Singh. North Carolina State University. All right reserved. 10. A cylindrical specimen of a metal alloy that has a diameter of 5.0 mm. A tensile force of 1000 N applied longitudinal direction produces an elastic reduction in diameter of 2.8x10-4 mm . Calculate the modulus of elasticity for this alloy that has a Poisson's ratio of 0.30 11. Plot on the axis below curves showing the stress strain relationship of a metal as solid line and using dashed line show a modified stress starin relationship if the same metal underwent 50% cold work. Label your axes and legends©Copyright. 2020. A. Singh. North Carolina State University. All right reserved. 12. The variation in tensile strength with % cold work for copper is shown in the graph. A copper wire having an initial diameter of 3.0 mm must be uniaxially drawn to obtain a tensile strength of 450 MPa. Determine the final diameter of the wire in mm. 650 600 550 500 450 400 350 300 0 20 40 60 80 100 % Cold work Tensile strength co -worke copper d ld of Tensile strength (MPa)©Copyright. 2020. A. Singh. North Carolina State University. All right reserved. y Constants and Equations for Exam #2 Constants and Conversions: 1 amu = 1 g/mol 1 GPa = 109 Pa 1 MPa = 106 Pa K = oC + 273 109 nm = 1 m 102 cm = 1 m NA = 6.02 x 1023 atoms/mole k = 8.62 x 10-5 eV/atom-K Equations: Nv= Ne−Qv/ kT ε = (li–l0)/l0 = ∆l/l0 σ = F/A0 σ=Eε τ = F/A0 ν = –εx/ εz = –εy/ εz %EL = [(lf – l0)/l0] x


View Full Document
Download Exam#2 Review Notes
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Exam#2 Review Notes and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Exam#2 Review Notes 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?