KEY. FINAL EXAM Chemistry May 2012 Professor Buhro. ID Number:

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1 KEY FINAL EXAM Chemistry May 2012 Professor Buhro KEY Signature KEY Print Name Clearly ID Number: Information This is a closed-book exam; no books, notes, other students, other student exams, or any other resource materials may be consulted or examined during the exam period Calculators are permitted Partial credit will be given for partially correct reasoning in support of incorrect or correct final answers Potentially useful information, formulas, values, etc, are provided on the last pages of this exam, which you may detach for convenience THIRD MIDTERM COMPREHENSIVE FINAL 1 (10 pts) 6 (20 pts) 2 (24 pts) 7 (20 pts) 3 (20 pts) 8 (18 pts) 4 (16 pts) 9 (06 pts) 5 (10 pts) 10 (04 pts) 11 (12 pts) Subtotal (80 pts) Subtotal (80 pts) Grand Total (160 pts) Final Semester Grade 1

1 10 total pts Carbon nanotubes are individual sheets from the graphite structure rolled into covalently sealed cylinders (see below) They are believed to be the strongest materials available Some

study (Nature Nanotech 2008, 3, 626-631) measured the tensile strengths of the strongest carbon nanotubes to be in the range of 97-110 GPa Are these nanotubes exhibiting their theoretical strengths?

2 1 10 total pts Carbon nanotubes are individual sheets from the graphite structure rolled into covalently sealed cylinders (see below) They are believed to be the strongest materials available Some useful properties of carbon nanotubes are given in the table Property Value Cylindrical-end surface energy 601 J/m 2 Stiffness (E) 955 GPa C C bond length 0142 nm (a) 05 pts A recent experimental study (Nature Nanotech 2008, 3, ) measured the tensile strengths of the strongest carbon nanotubes to be in the range of GPa Are these nanotubes exhibiting their theoretical strengths? If not, what percentage of theoretical strength has been achieved? Please show your work carefully and explain your conclusions briefly 1/2 1/2 surf σ E (601 J/m 2 )( Pa) 11 c Pa 201 GPa (3 pts) d m 110 GPa % (1 pt) 201 GPa No, the nanotubes do not exhibit theoretical cleavage stress, but about half of that value (1 pt), which is still a very high value (b) 05 pts Please determine the strain value (expressed as a percentage) at the point of fracture (fracture strain) in a nanotube having a tensile strength of 110 GPa (110 GPa) (%) % (5 pts) E (955 GPa) 2

3 2 24 total pts Please consider a fiber-matrix composite having the criss-cross continuousaligned structure represented below All of the continuous fibers lie in the plane of the page, but successive layers of fibers alternate between vertical and horizontal orientations with respect to the bottom of this page The total volume fraction of the fibers is 050 Other useful values are listed in the table, including the stiffnesses of the fibers and matrix Please calculate the stiffness of the composite (E c ) when under tensile stress in the directions A (left and right), B (top and bottom), and C (in and out, perpendicular to the page) Put your work in the areas marked A, B, and C Recall that: B C E c = E m V m + KE f V f Parameter Value E m 213 GPa E f 335 GPa K * 100 K * 000 *Parallel () and perpendicular () refer to the directions parallel and perpendicular to the long axis of the fibers a 04 pts (A) A A B C E c (213 GPa)(050) (100)(335 GPa)(025) (000)(335 GPa)(025) (100)(335 GPa)(025) 84 GPa b 04 pts (B) E c (213 GPa)(050) (100)(335 GPa)(025) (000)(335 GPa)(025) (100)(335 GPa)(025) 84 GPa c 04 pts (C) E c (213 GPa)(050) (000)(335 GPa)(025) (000)(335 GPa)(025) (213 GPa)(050) 11 GPa 3

4 2 (cont) Now please consider a different fiber-matrix composite having the unidirectional continuous-aligned structure represented below All of the continuous fibers lie in the plane of the page, and all are oriented in the same direction The total volume fraction of the fibers is 050 The parameters given in the table on the previous page remain valid for this composite also Please calculate the stiffness of the composite (E c ) when under tensile stress in the directions A (left and right), and B (top and bottom) Put your work in the areas marked A, and B B A A d 04 pts (A) B E c (213 GPa)(050) (000)(335 GPa)(050) (213 GPa)(050) 11 GPa e 04 pts (B) E c (213 GPa)(050) (100)(335 GPa)(050) (100)(335 GPa)(050) 170 GPa f 04 pts Please briefly describe the usage conditions for which the criss-cross and unidirectional composites would each be most appropriate, and briefly justify your answer If the intended use stresses the composite in only one direction, then the unidirectional composite will provide the greatest stiffness (170 GPa) However, if the intended use stresses the material in various in-plane directions, then the criss-cross composite provides the greatest overall stiffness (84 GPa) 4

5 3 20 total pts Please provide brief answers to the following questions (a) 05 pts Can a brittle solid resist fracture at low applied stresses? If so, please explain why If not, please explain why Yes; if the applied stress is sufficiently low, the critical (Griffith) crack length will be longer than the lengths of any surface cracks, and the cracks will not propagate (b) 05 pts Why is the work of fracture (W) always greater than the surface energy ( surf ) of the newly created fracture surfaces? W is always greater because fracture induces structural changes (rearrangements, dispruptions) several layers below the crack surfaces (4 pts), which consumes additional (strain) energy (1 pt) (c) 05 pts Why do metals have very large works of fracture (Ws)? The large Ws of metals are the result of dislocation generation and motion (4 pts), which consumes large amounts of (strain) energy (1 pt) (d) 05 pts Why are structural materials in which the tensile strengths exceed 1-2% of E not generally useful? Larger strains induce deformations in structures that can generally not be tolerated (or make the structure fail, 5 pts) Or, larger strains and the attendant deformations result in flutter and/or Euler collapse of structures (5 pts) 5

6 4 16 total pts This question concerns the stress-concentration factor K for surface cracks in conventional silica glass Please calculate K for the various scenarios given below (a) 05 pts Please first consider the most-extreme condition when a crack is closed at the tip by the spacing of just one bond Assume that the radius of curvature at the crack tip is equal to a single Si O bond distance of 0161 nm, and that the crack is 100 m in length Calculate K, and show your work K m m cracktip 9 (b) 05 pts We know that the stress concentration will decrease as the radius of curvature at the crack tip becomes larger Please now assume that the 100-m length crack is closed at the tip by two Si O bond distances, such that the radius of curvature is doubled in comparison to part (a) above Calculate K, and show your work What is the percentage decrease in the K factor? K (3 pts) m m cracktip % (2 pts) 159 (c) 06 pts To demonstrate the severity of the stress-concentration problem, please consider what would be necessary to reduce the K factor for the 100-m length crack to 200, which should be sufficient to prevent brittle fracture at fairly high stress levels Please calculate the radius of curvature at the crack tip that is required to reduce K to a value of 200 How many Si O bond distances does this radius correspond to? Show your work m x m x m m m 6 x m m (3 pts) y bond distances m m 24,800 (3 pts) 6

7 5 10 total pts The following exercises concern the weak-interface mechanism for crack stopping (a) 05 pts Please explain how the crack-induced stress parallel to the direction of crack propagation ( ) works together with a weak interface to stop a crack The parallel stress pulls back on the crack tip, and in the opposite direction that the crack tip is advancing towards Significantly, this stress is maximized a short distance in front of the advancing crack tip Consequently, it arrives at the weak interface before the crack does, and pulls the interface open, creating a large radius of curvature at the crack tip (or a tee-shaped crack stopper), consuming strain energy and stopping the crack [The answer must indicate (1) that the parallel stress is centered in advance of the crack tip, and (2) that it pulls the interface open stopping the crack 3 pts for either one, and 2 pts for the other] (b) 05 pts Please consider a ceramic material having a tensile strength of 200 MPa Weak interfaces have been engineered into the ceramic as shown below The interface strength is 100 MPa On the figure provided (below on the right), please complete the sketch of the material after the crack has encountered the interface, and briefly explain your sketch weak interface [3 pts] The interface strength must be less than or equal to 20% (1/5) of the general tensile strength of the material so that the parallel stress may pull it open In this example, the weak interface is not weak enough, and the crack passes through unimpeded [2 pts for noting that the interface strength is too large] 7

8 6 20 total pts These questions concern the interpretation and use of XRD data (a) 05 pts The XRD pattern of an MX compound having a conventional cubic crystal structure contains a 100 reflection at o 2 Please identify the structure exhibited by the MX compound, and provide a brief justification for your answer The 100 reflection establishes a primitive crystal structure (2 pts) The only conventional cubic structure for MX compounds that is primitive is the CsCl structure (3 pts), which is thus indicated here (b) 05 pts Please determine the lattice parameter a, and show your work d Å a 429 Å (5 pts) 2sin 2sin( 2071 ) 2 (c) 10 pts The XRD pattern below was obtained from a metallic specimen having a conventional crystal structure Please identify the structure, and determine the values of the lattice parameter or parameters Show your work in support of your answers The structure is hcp (2 pts) 1542 Å d 279Å 100 2sin 2sin( 321 ) 2 (2 pts) 1/ Å = a 3a 2 2 a = 322 Å (2 pts) d 002 = 1542 Å 2sin( 342 ) Å (2 pts) 262 Å = c c 2 c = 524 Å (2 pts) 8

9 7 (20 total pts) Unit cells from several common crystal structures are given below In each case, the larger, gray spheres are anions, and the smaller, black spheres cations Please assign (write) a common structure name below each of the unit cells Please also provide the stoichiometry of each structure, as in M 2 X 3, for example (M = cation; X = anion) (a) (b) (c) (a = b c; = = 90, = 120 ) (a = b = c; = = = 90 ) (a = b = c; = = = 90 ) Wurtzite, MX Rock salt or NaCl, MX Fluorite, MX 2 (d) (e) (f) (a = b c; = = 90, = 120 ) (a = b c; = = = 90 ) (a = b = c; = = = 90 ) NiAs, MX Rutile, MX 2 Perovskite, M A M B X 3 (g) (h) (i) (a = b c; = = 90, = 120 ) (a = b = c; = = = 90 ) (a = b = c; = = = 90 ) (j) Wurtzite, MX Perovskite, M A M B X 3 CsCl, MX Zinc blende, MX [1 pt each correct structure name; 1 pt each correct stoichiometry] (a = b = c; = = = 90 ) 9

10 8 18 total pts The following exercises concern the Fe-B phase diagram shown below C B A (a) 04 pts Please identify each eutectic point by listing the temperature and weight-percent boron at which it occurs 1174 C, 4 wt% B (2 pts) 1500 C, 256 wt% B (2 pts) (b) 02 pts Please identify each congruent melting point (excluding those of the elements Fe and B) by listing the temperature and weight-percent boron at which it occurs 1650 C, 16 wt% B (2 pts) (c) 02 pts Please identify each incongruent melting point by listing the temperature and weightpercent boron at which it occurs 1389 C, 85 wt% B (2 pts) (d) 04 pts Please identify each line compound by its chemical formula FeB (2 pts) Fe 2 B (2 pts) (e) 06 pts Please list the phase(s) present in the regions labeled A, B, and C A: FeB + B ss (2 pts) B: B ss (2 pts) C: L + B ss (2 pts) 10

11 9 06 total pts Will dissolution of small amounts of the following compounds into NaCl increase its ionic conductivity, decrease its ionic conductivity, or have little or no effect? (a) LiBr (b) CaBr 2 (c) Na 2 S Little or no effect Increase Decrease total pts Will dissolution of small amounts of the following compounds into AgCl increase its ionic conductivity, decrease its ionic conductivity, or have little or no effect? (a) CdCl 2 (b) Ag 2 S Decrease Increase total pts (1 pt each) These true-false and fill-in-the-blank exercises concern structural materials (a) Materials are generally weaker than their chemical bonds because of stress concentration (b) Please translate the latin sentence, ut tensio sic uis, into English As the extension, so the force (c) Please translate the latin sentence, ut tensio sic uis, into a mathematical expression = E (f) The primary function of a structural material is to generate opposing force (g) True or False (circle one) If a crack tip is sharp enough, the crack will propagate (h) True or False (circle one) If a crack is long enough, the crack will propagate (i) The stiffness of a material is determined by the stiffness of its chemical bonds (j) The theoretical strength of a material is estimated by what quantity? (theoretical) cleavage stress or (theoretical) shearing stress (k) True or False (circle one) For each material, the critical crack length (l g ) is a constant (l) True or False (circle one) Brittle fracture is the only mechanism for cleaving (breaking) a solid (m) Plastic deformation occurs by slip or shearing (n) Structures are generally weaker than the materials they are constructed from because of stress concentration 11

FINAL EXAM KEY. Professor Buhro. ID Number:

FINAL EXAM KEY. Professor Buhro. ID Number: FINAL EXAM Chemistry 465 KEY 10 May 011 Professor Buhro KEY Signature KEY Print Name Clearly ID Number: Information. This is a closed-book exam; no books, notes, other students, other student exams, or