EGR 355: ENGINEERING MATERIALS

1. Give electron configurations for the Fe3+ and S2- ions.

When the atom is an ion, it means that the number of protons does not equal the number of electrons. The charge of the atom will them be displayed at the top right (usually) corner of the chemical symbol. So, an antimony atom with charge +2 has an electron configuration of 1s22s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p1. Notice that the 5p3 changed into a 5p1. Be careful when the configuration of an uncharged atom ends in anything but an s and p orbital set. When you take away electrons, you can only take them away from the valence orbitals (the s and p orbitals). So, if a configuration ends in 4s2 3d7, and the atom gains a charge of +2, then the configuration would change to end with 4s0 3d7. Notice that 3d7 does not change, instead, the s orbital electrons are lost.

2. Why are the atomic weights of the elements generally not integers? Cite two reasons.

Atomic weights are a function of both the number of protons and the number of neutrons. The identity of each element is defined by the number of protons it has. However, the same element can have multiple isotopes, each of which has a different number of neutrons. So the mass of a specific atom of one element does not weigh exactly the same as all other.The atomic mass listed in the periodic is found by taking the average of all the different isotopes of a given element found in nature, weighted for their natural abundance. 

Note that the atomic mass is not the same as the atomic weight. The atomic mass is the weight of one specific isotope of one atom, and is expressed in "atomic mass units" or amu. The atomic weight is expressed in grams per mole (g/mol) and is the weighted average of all the isotopes weighted by their abundance. 

3. Offer an explanation as to why covalently bonded materials are generally less dense than ionically or metallically bonded ones.

First, it is not strictly true, but it may be because covalent bonds are between two non-metals and ionic bonds are between a non-metal and a metal. Most non-metals are in the top right of the periodic table, and so have lighter atomic weights, whereas metals take up most of the periodic table and so in general are heavier than non-metals (with obvious exceptions such as the very light metal

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lithium for instance). Also ionic compounds tend to have a long range bonding structure and because of the nature of ionic bonds tend to be tightly packed. Covalent molecules, while the atoms are tightly "packed" in the molecule, the molecules are not so tightly packed against one another, leaving more empty space, and thus reducing the density. 

Also, what may cause the added space between the covalently bonded molecules is that fact that the molecular bonds are highly directional (which in many instances causes assymetric molecules) where ionic and metallic bonds are non-directional. This can reduce intermolecular attractions to only weak van der waals forces.

Covalent bonds, unlike both metallic and ionic bonds, are highly directional. The directionality of the bonds makes it less easy to form close-packed and dense structures in covalently bonded materials because the bond angle requirements force more open volume in the crystal structure. More open spaces translate into lower density.

4. What is the difference between crystal structure and crystal system?

A crystal structure is described by both the geometry of, and atomic arrangements within, the unit cell, whereas a crystal system is described only in terms of the unitcell geometry. For example, face-centered cubic and body-centered cubic are crystal structures that belong to the cubic crystal system.

A crystal structure is described by both the geometry of, and atomic arrangements within, the unit cell, whereas a crystal system is described only in terms of the unit cell geometry. For example, face-centered cubic and body-centered cubic are crystal structures that belong to the cubic crystal system.

5. For cubic crystals, as values of the planar indices h, k, and l increase, does the distance between adjacent and parallel planes (i.e., the interplanar spacing) increase or decrease? Why?

The distances decrease. Mathmatically the bottom of the equation gets

bigger resulting in a smaller overall value. Conceptually The interplanar spacing between adjacent and parallel planes decreases as the values of h, k, and l increase. As values of the planar indices increase, the magnitude of the denominator in Equation 3.15 increases, with the result that the interplanar spacing (dhkl) decreases.

6. Do noncrystalline materials display the phenomenon of allotropy (or polymorphism)? Why or why not?

Noncrystalline materials do not display the phenomenon of allotropy; since a noncrystalline material does not have a defined crystal structure, it cannot have more than one crystal structure, which is the definition of allotropy.

7. Cite the primary differences between elastic, elastic, and plastic deformation behaviors.

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Elastic deformation returns to it's orginal shape after a strain is applied. 

Plastic deformation returns to a deformed shape after a strain is applied. The material's molecular bonds are strained to the point of fracture, making it not possible to return to the same state.

Elastic deformation is time-independent and nonpermanent, anelastic deformation is time-dependent and nonpermanent, while plastic deformation is permanent.

8. Questions: Of those metals listed in Table below, (a) Which will experience the greatest percent reduction in area? Why?(b) Which is the strongest? Why?(c) Which is the stiffest? Why?

Material Yield Strength (MPa)

Tensile strength(MPa)

Strain at Fracture

Fracture strength (MPa)

Elastic Modulus(GPa)

A 310 340 0.23 265 210B 100 120 0.40 105 150C 415 550 0.15 500 310D 700 850 0.14 720 210E Fractures before yielding 650 350

(a) Material B will experience the greatest percent area reduction since it has the highest strain at fracture, and, therefore is most ductile. (b) Material D is the strongest because it has the highest yield and tensile strengths. (c) Material E is the stiffest because it has the highest elastic modulus.

Difference between strong and stiffIn materials science, the strength of a material is its ability to withstand an applied stress without failure. Yield strength refers to the point on the engineering stress-strain curve (as opposed to true stress-strain curve) beyond which the material begins deformation that cannot be reversed upon removal of the loading. Thus strength is related to plastic deformation.Stiffness is the resistance of an elastic body to deformation by an applied force (tensile, shear, torsional, etc.). For the axial tension or compression, Young’s modulus is the measure of the stiffness of a material. The stiffness of a structure is of principal importance in many engineering applications, so the modulus of elasticity is often one of the primary properties considered when selecting a material. A high modulus of elasticity is sought when deflection is undesirable, while a low modulus of elasticity is required when flexibility is needed.

Ultimate Yield Strength: Is the maximum stress developed in a material during a tensile test.A good indicator of the presence of defects in the crystal structure of the material.

It is not used too much in design due to the amount of plastic deformation that has taken place

before reaching this point

9. Make a schematic plot showing the tensile engineering stress–strain behavior for a typical

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metal alloy to the point of fracture. Now superimpose on this plot a schematic compressive engineering stress-strain curve for the same alloy. Explain any differences between the two curves.

The schematic stress-strain graph on which is plotted the two curves is shown below. The initial linear (elastic) portions of both curves will be the same. Otherwise, there are three differences between the two curves which are as follows: (1) Beyond the elastic region, the tension curve lies below the compression one. The reason for this is that, during compression, the cross-sectional area of the specimen is increasing—that is, for two specimens that have the same initial cross-sectional area (A0), at some specific strain value the instantaneous cross-sectional area in compression will be greater than in tension. Consequently, the applied force necessary to continue deformation will be greater for compression than for tension; and, since stress is defined according to Equation 7.1 as σ = FA0

10. Which of the following is the slip system for the simple cubic crystal structure? Why?

{100}<110>{110}<110>{100}<010>{110}<111>

The slip system for some crystal structure corresponds to the most densely packed crystallographic plane, and, in that plane, the most closely packed crystallographic direction. For simple cubic, the most densely packed atomic plane is the {100} type plane; the most densely packed direction within this plane is an <010> type direction. Therefore, the slip system for simple cubic is {100} <010>.

11. Explain the difference between resolved shear stress and critical resolved shear stress .

Resolved shear stress is the shear component of an applied tensile (or compressive) stress resolved along a slip plane that is other than perpendicular or parallel to the stress axis. The critical resolved shear stress is the value of resolved shear stress at which yielding begins; it is a property of the material.

12. Would you expect a crystalline ceramic material to strain harden at room temperature? Why or why not?

No, it would not be expected. In order for a material to strain harden it must be plastically deformed; since ceramic materials are brittle at room temperature, they will fracture before any plastic deformation takes place.

13. Briefly explain why some metals (e.g., lead and tin) do not strain harden when deformed at room temperature.

Metals such as lead and tin do not strain harden at room temperature because their recrystallization temperatures lie below room temperature (Table 7.2).

14. Would you expect it to be possible for ceramic materials to experience recrystallization? Why or why not?

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No, recrystallization is not expected in ceramic materials. In order to experience recrystallization, a material must first be plastically deformed, and ceramic materials are too brittle to be plastically deformed.

15. What is the difference between the states of phase equilibrium and metastability?

For the condition of phase equilibrium the free energy is a minimum, the system is completely stable meaning that over time the phase characteristics are constant. For metastability, the system is not at equilibrium, and there are very slight (and often imperceptible) changes of the phase characteristics with time.

16. A copper-nickel alloy of composition 70 wt% Ni-30 wt% Cu is slowly heated from a temperature of 1300 ° C . (a) At what temperature does the first liquid phase form?(b) What is the composition of this liquid phase?(c) At what temperature does complete melting of the alloy occur?(d) What is the composition of the last solid remaining prior to complete melting?

Upon heating a copper-nickel alloy of composition 70 wt% Ni-30 wt% Cu from 1300°C and utilizing Figure 9.3a:(a) The first liquid forms at the temperature at which a vertical line at this composition intersects the α-(α + L) phase boundary—i.e., about 1350°C;(b) The composition of this liquid phase corresponds to the intersection with the (α + L)-L phase boundary, of a tie line constructed across the α + L phase region at 1350°C—i.e., 59 wt% Ni;(c) Complete melting of the alloy occurs at the intersection of this same vertical line at 70 wt% Ni with the (α + L)-L phase boundary—i.e., about 1380°C;(d) The composition of the last solid remaining prior to complete melting corresponds to the intersection with α-(α + L) phase boundary, of the tie line constructed across the α + L phase region at 1380°C—i.e., about 78 wt% Ni.

17. Is it possible to have a copper-nickel alloy that, at equilibrium, consists of an α phase of composition 37 wt% Ni-63 wt% Cu, and also a liquid phase of composition 20 wt%Ag-80 wt% Cu? If so, what will be the approximate temperature of the alloy? If this is not possible, explain why?

It is not possible to have a Cu-Ni alloy which at equilibrium consists of a liquid alloy phase of composition 20 wt% Ni-80 wt% Cu and an α phase of composition 37 wt% Ni-63 wt% Cu. A single tie line does not exist within the α + L region that intersects the phase boundaries at the given compositions. At 20 wt% Ni, the L-(α + L) phase boundary is at about 1200°C, whereas at 37 wt% Ni the (L + α)-α phase boundary is at about 1225

18. The figure below is hafnium-vanadium phase diagram, for which only single-phase region are labeled. Specify temperature-composition points at which all eutectics eutectoids, peritectics and congruent phase transformations occur. Also for each write the reaction upon cooling.

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There are two eutectics on this phase diagram. One exists at 18 wt% V-82 wt% Hf and 1455°C. The reaction upon cooling is L → βHf + HfV2

The other eutectic exists at 39 wt% V-61 wt% Hf and 1520°C. This reaction upon cooling is

L → HfV2 + V(solid solution)

There is one eutectoid at 6 wt% V-94 wt% Hf and 1190°C. Its reaction upon cooling is

βHf → αHf + HfV2

There is one congruent melting point at 36 wt% V-64 wt% Hf and 1550°C. The reaction upon cooling is

L → HfV2

No peritectics are present

1. If the attractive force between a pair of Sr 2+ and O 2- ions is 1.29 10 -8 N and the ionic radius of the O 2- ion is 0.132 nm, calculate the ionic radius of the Sr 2+ ion in nanometers.

2. What are the indices of the directions shown in the unit cubes of Fig. P3.34?

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3. Determine the tensile stress that must be applied to the axis of a high-purity copper single crystal to cause slip on the system. The resolved shear stress for the crystal is 0.85 MPa.

4. Equiaxed MAR-M 247 alloy is to support a stress of 276 MPa (Fig. 7.31). Determine the time to stress rupture at 850 C.

3.64753x10^22 hours?

5. As the alloy is cooled below the eutectic temperature, the tin content in the alpha phase and the lead content in the beta phase are further reduced. However, at room temperature (20ºC), equilibrium is not achieved because the diffusion rate is so slow.

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Referring to Fig. 8.13, if the solvus line is extrapolated to 20ºC, the approximate composition of alpha and beta are 2.0% and 100.0 %, respectively. Thus,

6. A hypereutectoid plain carbon steel contains 10.7 wt % eutectoid Fe 3C. What is its average carbon content in weight percent?

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