1 Met-223: Concepts of Materials Science I Sample Questions & Answers,(29) ( Met, PR, FC, MP, CNC, McE ) Q-1.Define the following. (i) Point Defects (ii) Burgers Vector (iii) Slip and Slip system (iv) Interplanar spacing (v) Frenkel Defect (vi) Schottky defect Q-2.Design a heat treatment that will provide 1 times more vacancies in copper than are normally present at room temperature. About 2, cal/mol are required to produce a vacancy in copper.the lattice parameter of FCC copper is.36151 nm. Q-3.Determine the number of vacancies needed for a BCC iron lattice to have a density of 7.87 g/cm. The lattice parameter of the iron is 2.866 1-8 cm. (at. Wt. for Fe- 55.847 g/mol ) Q-4.(a) The planar density of the (112) plane in BCC iron is 9.94 1 14 atoms/cm 2. Calculate (i) the planar density of the (11) plane and (ii) the interplanar spacings for both the (112) and (11) planes. On which plane would slip normally occur? (b) Calculate the length of the Burgers vector in the following materials: (i) BCC niobium ( a 3.294 Aº ) (ii) FCC silver ( a 4.862 Aº ) (iii) FCC copper ( a 3.6151 Aº ) Q-5.(a) Define Schmid's law. (b) An aluminum crystal slips on the (111) plane and in the [11] direction with a 3.5 MPa stress applied in the [111] direction. What is the critical resolved shear stress? Q-6.(a) Calculate the number of vacancies per cm 3 epected in copper at 185 C (just below the melting temperature). The energy for vacancy formation is 2, cal/mol. ( a for Cu - 3.6151 Aº ) Met-223
2 (b) The fraction of lattice points occupied by vacancies in solid aluminum at 66 c C is 1-3. What is the energy required to create vacancies in aluminum? Q-7. The density of a sample of FCC palladium is 11.98 g/cm 3 and its lattice parameter is 3.892 A. Atomic mass of Pd is 16.4 g/mol. Calculate (a) the fraction of the lattice points that contain vacancies and (b) the total number of vacancies in a cubic centimeter of Pd. Q-8.(a) Define the rate of Diffusion of Fick's First Law. (b) Atoms are found to move from one lattice position to another at the rate of 5 1O 5 jumps per second at 4 c C when the activation energy for their movement is 3, cal/mol. Calculate the jump rate at 75 C. Q-9.(a) Define (ii) diffusion and (ii) diffusion coefficient. (b) Consider a diffusion couple set up between pure tungsten and a tungsten-1 at % thorium alloy. After several minutes of eposure at 2 C, a transition zone of.1 cm thickness is established. What is the flu of thorium atoms at this time if diffusion is due to (a) volume diffusion, (b) grain boundary diffusion, and (c) surface diffusion? The lattice parameter of BCC tungsten is 3.165A. Diffusion Coefficient for Thorium in Tungsten Surface Grain boundary Volume.47 ep(-66,4/rt).74 ep(-9,/rt) 1.ep(-12,/RT) Q-1.(a) Define Activation energy. (b) The diffusion coefficient for Cr in Cr 2 O 3 is 6 1-15 cm 2 /s at 727 C and is 11 9 cm 2 /s at 14 C. Calculate (i) the activation energy and ( ii ) the constant D o. Q-11.The surface of a.1% C steel is to be strengthened by carburizing. In carburizing, the steel is placed in an atmosphere that provides 1.2% C at the surface of the steel at a high temperature. Carbon then diffuses from the surface into the steel. For optimum properties, the steel must contain.45% C at a depth of.2 cm below the surface. Design a carburizing heat Met-223
3 treatment that will produce these optimum properties. Assume that the temperature is high enough (at least 9 C) so that the iron has the FCC structure. Q-12.(a) Define Fick's Second Law. (b) We find that 1 h are required to successfully carburize a batch of 5 steel gears at 9 C, where the iron has the FCC structure. We find that it costs $ 1 per hour to operate the carburizing furnace at 9 C and $ 15 per hour to operate the furnace at 1 C. Is it economical to increase the carburizing temperature to 1 C? Q-13.A.1 in. BCC iron foil is used to separate a high hydrogen gas from a low hydrogen gas at 65 C. 5 1 8 H atoms/ cm 3 are in equilibrium with the hot side of the foil, and 2 1 3 H atoms/cm 3 are in equilibrium with the cold side. Determine (a) the concentration gradient of hydrogen and (b) the flu of hydrogen through the foil. Q-14.What temperature is required to obtain.5% C at a distance of.5 mm beneath the surface of a.2% C steel in 2 h, when 1.1% C is present at the surface? Assume that the iron is FCC. Q-15.A.15% C steel is to be carburized at 11 C, giving.35% C at a distance of 1mm beneath the surface. If the surface composition is maintained at.9% C, what time is required? Q-16.A.2% C steel is to be carburized at 12 C in 4 h, with a point.6 mm beneath the surface reaching.45% C. Calculate the carbon content required at the surface of the steel. Q-17.A 1.2% C tool steel held at 115 C is eposed to oygen for 48 h. The carbon content at the steel surface is zero. To what depth will the steel be decarburized to less than.2% C? Q-18.A BCC steel containing.1% N is nitrided at 55 C for 5 h. If the nitrogen content at the steel surface is.8%, determine the nitrogen content at.25 mm from the surface. Met-223
4 Q-19.Define the following. (i) % Elongation (ii) Endurance limit (iii) Endurance Ratio (iv) Fatigue life (v) Elastic deformation Q-2.(a) Define Engineering Stress, True stress and Engineering Strain, True Strain. (b) A force of 1, N is applied to a 1 mm 2 mm iron bar having a yield strength of 4 MPa and a tensile strength of 48 MPa. Determine (i) whether the bar will plastically deform and (ii) whether the bar will eperience necking. Q-21.An aluminum alloy that has a plane strain fracture toughness of 25, psi- in.fails when a stress of 42, psi is applied. Observation of the fracture surface indicates that fracture began at the surface of the part. Estimate the size of the flaw that initiated fracture. Assume that f = 1.1. Q-22.Define Hooke's law and Poisson's ratio. Q-23.A.4-in. diameter, 12-in. long titanium bar has a yield strength of 5, psi, a modulus of elasticity of 16 1 6 psi, and Poisson's ratio of.3. Determine the length and diameter of the bar when a 5-Ib load is applied. Q-24.A 3-in.diameter rod of copper is to be reduced to a 2-in. diameter rod by being pushed through an opening. To account for the elastic strain, what should be the diameter of the opening? The modulus of elasticity for the copper is 17 1 6 psi and the yield strength is 4, psi. Q-25.Which factors does depend on the ability of a material to resist the growth of a crack? Q-26.Define the following. (i) Tensile strength (ii) Yield strength (iii) Ductility Q-27.A ceramic part for a jet engine has a yield strength of 75, psi and a plane strain fracture toughness of 5, psi - in.. To be sure that the part does not fail, we plan to Met-223
5 assure that the maimum applied stress is only one-third the yield strength. We use a nondestructive test that will detect any internal flaws greater than.5 in. long. Assuming that f = 1.4, does our nondestructive test have the required sensitivity? Eplain. Q-28.A large steel plate used in a nuclear reactor has a plane strain fracture toughness of 8, psi- in and is eposed to a stress of 45, psi during service. Design a testing or inspection procedure capable of detecting a crack at the surface of the plate before the crack is likely to grow at a catastrophic rate. Q-29.A 85-lb force is applied to a.15-in. diameter nickel wire having a yield strength of 45, psi and a tensile strength of 55, psi. Determine (a) whether the wire will plastically deform and (b) whether the wire will eperience necking. Q-3.A solid shaft for a cement kiln produced from the tool steel must be 96 inches long and must survive continuous operation for one year with an applied load of 12,5 lb. The shaft makes one revolution per minute during operation. Design a shaft that will satisfy these requirements. NOTES;(1) For question no. 11,12,13,14,15,16,17,18, need to provide error function Table 2-3 and Diffusion Coefficient Table 2-1 (2) For question no. 3, need to provide Figure 3-13. The stress-number of cycles to failure (S-N) curves for a tool steel and an aluminum alloy. Met-223
6 Met-223 Concepts of Materials Science I Sample Questions and Answers, (29) Q-1.Define the following. (i) Point Defects (ii) Burgers Vector (iii) Slip and Slip system (iv) Interplanar spacing (v) Frenkel Defect (vi) Schottky defect (i) Point Defects Point defects are localized disruptions of the lattice involving one or, possibly, several atoms. These imperfections, shown in Figure, may be introduced by movement of the atoms when they gain energy by heating, during processing of the material, by introduction of impurities, or intentionally through alloying. Point defects are (1) vacancy, (2) interstitial atom, (3) substitutional atom, (4) Frenkel defect, and (5) Schottky defect. All of these defects disrupt the perfect arrangement of the surrounding atoms. (ii) Burgers vector ( b ) The direction and the distance that a dislocation moves in each step. The length of Burgers vector (b) is equal to the repeat distance. (iii) Slip and slip system The process by which a dislocation moves and causes a material to deform is called slip. The direction, in which the dislocation moves is the slip direction and the plane in which the dislocation moves is the slip plane. The combination of slip direction and slip plane is the slip system. (iv) Interplanar spacing The distance between the two adjacent planes of atoms with the same Miller indices is called the interplanar spacing d (hkl) and the general equation, d ( hkl) = a h + k + l 2 2 2 where, a = the lattice parameter or lattice constant h,k,l = Miller indices of adjacent planes of atoms Met-223
7 (v) Frenkel Defects A Frenkel defect is a vacancy-interstitial pair formed when an ion jumps from a normal lattice point to an interstitial site, leaving behind a vacancy. (vi) Schottky Defects A Schottky defect is a pair of vacancies in an ionically bonded material; both an anion and a cation must be missing from the lattice if electrical neutrality is to be preserved in the crystal. These are common in ceramic materials with the ionic bond. Q-2. Design a heat treatment that will provide 1 times more vacancies in copper than are normally present at room temperature. About 2, cal/mol are required to produce a vacancy in copper. The lattice parameter of FCC copper is.36151 nm. The lattice parameter of FCC copper is.36151 nm. The number of copper atoms, per cm 3 is 4 atoms / cell (3.61511 cm) 22 = 8.47 1 copper atoms/cm 3 8 3 At room temperature, T = 25 + 273 = 298 K: n v = (8.47 1 22 ) ep [(- 2,)/ (1.987) (298)] = 1.815 1 8 vacancies/cm 3 We wish to produce 1 times this number, or n v = 1.815 1 11 vacancies/cm 3. We could do this by heating the copper to a temperature at which this number of vacancies forms: n v = 1.815 1 11 = (8.47 1 22 ) ep (-2,/1.987T) ep (- 2,/1.987T ) = 1.851 11 /(8.47 1 22 ) =.2141-11 T = 2, / (1.987)(26.87) = 375 K = 12 C Met-223
8 By heating the copper slightly above 1 C, then rapidly cooling the copper back to room temperature, the number of vacancies in the structure may be one thousand times greater than the equilibrium number of vacancies. Q-3. Determine the number of vacancies needed for a BCC iron lattice to have a density of 7.87 g/cm. The lattice parameter of the iron is 2.866 1-8 cm. (at. wt. for Fe = 55.847 g/mol ) Theoretical density of iron can be calculated from the lattice parameter and the atomic mass. Since the iron is BCC, two iron atoms are present in each unit cell. theoretical density of iron ; 2 atoms / unit cell 55.847 g / mole density = = 7.8814 g / cm -8 3 23 (2.8661 cm ) (6.21 atoms/ mol) Let's calculate the number of iron atoms and vacancies that would be present in each unit cell for the required density of 7.87 g/cm 3. 3 ( atoms / unit cell) (55.847 g / mole) density = = 7.8814 g / cm -8 3 23 (2.8661 cm ) (6.21 atoms/ mol) 3 7.8814 g / cm (55.847 g / mole) atoms / unit cell = = 1.997 atoms -8 3 23 (2.8661 cm ) (6.21 atoms/ mol) 3 no. of vacancies per unit cell = 2 1.9971 =.29 Or, there should be.29 vacancies per unit cell. The number of vacancies per cm 3 is Vacancies/cm 3 =.29 vacancies per unit cell / (2.866l -8 cm) 3 = 1.231 2 vacancies/cm 3 Q-4.(a)The planar density of the (112) plane in BCC iron is 9.94 1 14 atoms/cm 2. Calculate (i) the planar density of the (11) plane and (ii) the interplanar spacings for both the (112) and (11) planes. On which plane would slip normally occur? Met-223
9 (a) (b) Calculate the length of the Burgers vector in the following materials: (i) BCC niobium ( a 3.294 Aº ) (ii) FCC silver ( a 4.862 Aº ) (iii) FCC copper ( a 3.6151 Aº ) (i) For BCC niobium, a = 3.294 Aº The directions of the Burgers vector, are in [l 11] for BCC metal The repeat distance is along the [ll1] directions and is equal to one-half of the body diagonal, since lattice points are located at corners and centers of body. Body diagonal distance = 3 a = 3 (.3294 nm ) =.575nm The length of the Burgers vector, or the repeat distance, is: 1 b = (.575 nm ) =.2852 nm 2 (ii) For FCC silver, a = 4.862 Aº The directions of the Burgers vector, are in [l 1] for FCC metal Met-223
1 The repeat distance is along the [ll] directions and is equal to one-half of the face diagonal, since lattice points are located at corners and center of the face. Face diagonal distance = 2 a = 2 (.4862 nm) =.5778 nm The length of the Burgers vector, or the repeat distance, is: 1 b = (.5778 nm ) =.2889 nm 2 (iii) Copper is FCC, a =.36151 nm The directions of the Burgers vector, are of the form(l 1). The repeat distance along the (ll) directions is one-half the face diagonal, since lattice points are located at corners and centers of faces. Face diagonal distance = 2 a= ( 2) (.36151) =.51125nm The length of the Burgers vector, or the repeat distance, is: 1 b = (.51125 nm ) =.25563 nm 2 Q-5.(a) Define Schmid's law. Schmid's law The relationship between shear stress, the applied stress, and the orientation of the slip system and the resolved shear stress ι in the slip direction is F τ = Cos φ Cos λ, τ = σ CosφCosλ A where, τ = the resolved shear stress in the slip direction σ = the applied stress φ = the angle between the direction of the force and the normal to the slip plane, λ = the angle between the direction of force and the slip direction. Met-223
11 Q-5.(b) An aluminum crystal slips on the (111) plane and in the [11] direction with a 3.5 MPa stress applied in the [111] direction. What is the critical resolved shear stress? (111) slip plane [111] applied stress direction and normal to slip plane [11] slip direction Aluminum crystal slips plane - (111) plane slip direction - [11] direction stress applied σ - 3.5 MPa applied stress direction - [111] direction. the critical resolved shear stress ι =? τ = σ CosφCosλ From the Fig: Met-223
12 the same, the applied stress direction and the direction of the normal to the slip plane are the angle φ =, Cos = 1 2 a Cos λ = = 3 a 2.8165 3 = 2 τ = 3.51 = 2.875 MPa 3 the critical resolved shear stress, τ = 2.875 MPa Q-6.(a) Calculate the number of vacancies per cm 3 epected in copper at 185 C (just below the melting temperature). The energy for vacancy formation is 2, cal/mol. ( a = for Cu - 3.6151 Aº ) (a) copper metal, FCC structure, 4 atoms/ unit cell, a A cm Q cal mol R cal mol K 8 = 3.6151, = 3.6151 1, = 2, /, = 1.987 /. T = 185 C+ 273= 1358 K, n v =? Q nv = nep RT no. of atom per unit cell n = volumeof unit cell 4 atom per unit cell 4 n = = = 1.16 1 atom / cm 3 8 3 ( a ) (3.61511 cm) Q 2, nv = n = RT 1.9871358 19 3 = 5.12 1 vacancies / cm 24 ep( ) 1.16 1 ep( ) 24 3 Q-6.(b) The fraction of lattice points occupied by vacancies in solid aluminum at 66 c C is 1-3.What is the energy required to create vacancies in aluminum? Aluminium metal is FCC structure, 4 atoms/ unit cell, n 3 66 273 933, v 1, =? T = C+ = K = Q n Q nv = nep RT Met-223
13 nv Q = ep n RT Q 1.987 933 3 1 ep( ) = Q= 128 cal/ mol Q-7. The density of a sample of FCC palladium is 11.98 g/cm 3 and its lattice parameter is 3.892 A. Atomic mass of Pd is 16.4 g/mol. Calculate (a) the fraction of the lattice points that contain vacancies and (b) the total number of vacancies in a cubic centimeter of Pd. FCC Pd density = 11.98 g/cm 3, 4 atoms/ unit cell At. wt. of Pd - 16.4 g/mol a = 3.892 A= 3.8921 8 cm nv nv ( a) =?, ( b) =? 3 n cm mass of unit cell theoretical density = volumeof unit cell 4 atoms per unit cell 16.4 g / mol ρ = = 12.85 g/ cm cm atom mol 8 3 23 (3.892 1 ) (6.2 1 / ) 3 (a) (b) For theoretical density 12.85 g/cm 3, 4 atoms/ unit cell For the given density 11.98 g/cm 3,? atoms/ unit cell For the given density 11.98 g/cm 3, 3.995 atoms/ unit cell no. of vacancies per unit cell (n v ) = 4 3.995 =.95 n v =.95 n = no. of atom per unit cell = 4 n v n.95 = = 2.375 1 4 3 3 nv no. of vacancies per unit cell 2.375 1 = = = 1.6 1 vacancies / cm 3 8 3 cm volume of unit cell (3.892 1 cm) 2 3 Q-8.(a) Define the rate of Diffusion of Fick's First Law. Met-223
14 The rate at which atoms diffuse in a material can be measured by the flu J, which is defined as the number of atoms passing through a plane of unit area per unit time. Fick's first law eplains the net flu of atoms: c J = D. where, J is the flu (atoms/cm 2.s), D is the diffusivity or diffusion coefficient (cm 2 /s), and Δc/Δ is the concentration gradient (atoms/cm 3.cm). Q-8.(b) Atoms are found to move from one lattice position to another at the rate of 5 1 5 jumps per second at 4 c C when the activation energy for their movement is 3, cal/mol. Calculate the jump rate at 75 C. the rate of diffusion = 5 1 5 jumps per second at 4 c C the activation energy Q = 3, cal/mol. the jump rate at 75 C =? T1 = 4 C+ 273 = 673 T2 = 75 C+ 273 = 123 By the equation, Rate of diffusion =.ep( Q c ) RT K K Q = ----- eq. (1) 5 51 c ep( ) RT1 Q = c ep ( ) ----- eq. (2) RT eq. (2) eq.(1) 2 Q Q = c ep ( ) / 5 1 = c ep ( ) 5 RT2 RT1 3, 5 3, = cep ( ) / 5 1 = cep ( ) 1.987 673 1.987 123 = jump 9 1.8 1 / sec Q-9.(a) Define (i) diffusion (ii) diffusion coefficient. Met-223
15 (a) (i) Diffusion Diffusion is the movement of atoms within a material. The rate of diffusion is governed by the Arrhenius relationship that is, the rate increases eponentially with temperature. Rateof diffusion = c Q ep ( ) RT where, Q is the activation energy (cal/mol) R is the gas constant (1.987 cal/mol.k) T is the absolute temperature in (K) C o is a constant for a given diffusion system 1 (ii) Diffusion coefficient (D) The diffusion coefficient D is related to temperature by an Arrhenius equation. The diffusion coefficient depends on temperature and activation energy. ep ( Q D = D ) RT where, D is the diffusion coefficient ( cm 2 / s ) Q is the activation energy (cal/mol) R is the gas constant (1.987 cal/mol.k) T is the absolute temperature in (K) D o is a constant for a given diffusion system Q-9.(b) Consider a diffusion couple set up between pure tungsten and a tungsten-1 at % thorium alloy. After several minutes of eposure at 2 C, a transition zone of.1 cm thickness is established. What is the flu of thorium atoms at this time if diffusion is due to (a) volume diffusion, (b) grain boundary diffusion, and (c) surface diffusion? The lattice parameter of BCC tungsten is 3.165A. Diffusion Coefficient for Thorium in Tungsten Surface Grain boundary Volume.47 ep(-66,4/rt).74 ep(-9,/rt) 1.ep(-12,/RT) Met-223
16 The lattice parameter of BCC tungsten is 3.165A. the number of tungsten atoms/cm 3 is: no. of W atoms per unit cell / vol. of unit cell = 2 atom per cell / (a ) 3 = 2 / (3.165 1-8 cm ) 3 = 6.3 1 22 W atoms/cm 3 In the tungsten -1 atom % thorium alloy, the number of thorium atoms is: C Th = (.1 ) ( 6.3 1 22 ) = 6.3 1 2 Th atoms/cm 3 In the pure tungsten, the number of thorium atoms is zero. Thus, the concentration gradient is: ΔC / Δ = - 6.3 1 2 /.1 cm = - 6.3 1 22 Tho atom / cm 3.cm T = 2 C + 273 K = 2273 K (a) for volume diffusion, D = 1.ep(-12,/RT) cm 2 / s J = - D.Δc/Δ = - 1.ep(-12,/1.987 2273 ) (- 6.3 1 22 ) = 1.82 1 1 Th atoms/cm 2.s (b) for grain boundary, D =.74 ep(-9,/rt) cm 2 / s J = - D.Δc/Δ = -.74 ep(-9,/1.987 2273 ) (- 6.3 1 22 ) = 1.3 1 13 Th atoms/cm 2.s (c) for surface diffusion, D =.47 ep(-66,4/rt) cm 2 / s J = - D.Δc/Δ = -.47 ep(-66,4/1.987 2273 ) (- 6.3 1 22 ) = 12.2 1 15 Th atoms/cm 2.s Q-1.(a) Define Activation energy. Activation Energy (Q) The atom is originally in a low-energy, relatively stable state. In order to move to a new location, the atom must overcome an energy barrier. This energy barrier is the activation energy Q. This energy is gained by heat supply. In diffusion, the activation energy is related to the energy required to move an atom from one lattice site to another. The activation energy Q is epressed in (cal/mol). Met-223
17 Q-1.(b) The diffusion coefficient for Cr in Cr 2 O 3 is 6 1-15 cm 2 /s at 727 C and is 11 9 cm 2 /s at 14 C. Calculate (i) the activation energy and ( ii ) the constant D o. The diffusion coefficient for Cr in Cr 2 O 3 D = 6 1-15 cm 2 /s at T 1 = 727 C + 273 K = 1 K D = 1 1 9 cm 2 /s at T 2 = 14 C + 273 K = 1673 K (i) the activation energy Q =? ( ii ) the constant D o. =? ep ( Q D = D ) RT Q = ----- eq. (1) 15 6 1 D ep( ) RT1 Q = ----- eq. (2) 9 1 1 D ep( ) RT2 eq. (1) eq.(2) 6 1 ep ( Q Q = D ) /1 1 = D ep ( ) 15 9 RT1 RT2 15 Q 9 Q 6 1 = Dep( )/ 11 = Dep( ) 1.987 1 1.987 1673 Q= 5923 cal/ mol (ii) = Q 9 1 1 D ep( ) RT2 5923 1.987 1673 9 1 1 = D ep( ) D 2 =.55 cm / sec Q-11.The surface of a.1% C steel is to be strengthened by carburizing. In carburizing, the steel is placed in an atmosphere that provides 1.2% C at the surface of the steel at a high temperature. Carbon then diffuses from the surface into the steel. For optimum properties, the steel must contain.45% C at a depth of.2 cm below the surface. Design a carburizing heat treatment that will produce these optimum properties. Assume that the temperature is high enough (at least 9 C) so that the iron has the FCC structure. Met-223
18 Cs = 1.2% C, C =.45% C, C =.1% C, =.2cm Given: t=?, T =? By Fick's second law, Cs C = erf [ ] C C 2 Dt s 1.2.45.2 =.68 = erf [ ] 1.2.1 2 Dt From Table 2-3, we find that,.2.71 2 Dt = or.1.71 Dt =.1 Dt = =.71 2. ( ).198 Any combination of D and t whose product is.198 will work. For carbon diffusing in FCC iron, the diffusion coefficient is related to temperature by: ep ( Q D = D ) RT From the table 2-1, D =.23, Q = 32,9 cal/mol 329 D =.23 ep( ) 1.987T Therefore, the temperature and time of the heat treatment are related by:.198 t = D.198 t =.23ep 16558/ ( T ) Some typical combinations of temperatures and times are: If T = 9 C = 1173 K, then, t = 1 16,174 s = 32.3 hr If T= 1 C = 1273 K, then, t = 36,36 s = 1.7 hr If T = 11 C = 1373 K, then, t = 14,88 s = 4.13 hr If T = 12 C = 1473K, then, t = 6,56 s = 1.82 hr Q-12.(a) Define Fick's Second Law. Composition Profile (Fick's Second Law) Met-223
19 Fick's second law describes the dynamic or non-steady state diffusion of atoms by the differential equation dc/dt = D (d c/d 2 ), whose solution depends on the boundary conditions for a particular situation. One solution is Cs C = erf [ ] C C 2 Dt s where, C s - concentration of the diffusing atoms at the surface of the material C - the concentration of the diffusing atom at location below the surface after time t. C - the initial uniform concentration of the diffusing atoms in the material t - the diffusion time in (s) the depth from the surface of the material ( cm ) D - the diffusion coefficient Q-12.(b) We find that 1 h are required to successfully carburize a batch of 5 steel gears at 9 C, where the iron has the FCC structure. We find that it costs $ 1 per hour to operate the carburizing furnace at 9 C and $ 15 per hour to operate the furnace at 1 C. Is it economical to increase the carburizing temperature to 1 C? T = 9 C + 273 K = 1 173 K, t 1173 = 1 hour T = 1 C = 273 K = 1273 K., t 1273 =? For carbon diffusing in FCC iron, from Table, the activation energy Q = 32,9 cal/mol. for the same carburizing treatment at 1 C as at 9 C: D 1273.t 1273 = D 1173.t 1173.t 1273 = D 1173.t 1173 / D 1273 t 1273 (1h)ep( 32,9) /(1.987).(1173) = ep( 32,9) /(1.987).(1273) t 1273 = 3.299 hour At 9 C, the cost per part is ($ 1/h) (1h)/5 parts = $ 2/part At l C, the cost per part is ($15/h) (3.299h)/5 parts=$ 9.9/part Considering only the cost of operating the furnace, increasing the temperature reduces the heat-treating cost of the gears and increases the production rate. Met-223
2 Q-13. A.1 in. BCC iron foil is used to separate a high hydrogen gas from a low hydrogen gas at 65 C. 5 1 8 H atoms/ cm 3 are in equilibrium with the hot side of the foil, and 2 1 3 H atoms/cm 3 are in equilibrium with the cold side. Determine (i)the concentration gradient of hydrogen and (ii)the flu of hydrogen through the foil. In BCC iron foil, the flu of hydrogen through the foil, from hot side to cold side. at 65 C., T = 65 + 273 = 923 K C initial = 5 1 8 H atoms/ cm 3 C final = 2 1 3 H atoms/cm 3 thickness - Δ =.1 in. =.1 in 2.54 cm (i) the concentration gradient (Δc/Δ) =? (ii) the flu of hydrogen through the foil =? 3 8 21 51 (i) the concentration gradient (Δc/Δ) = (.1) (2.54) = - 19691 8 H atoms/ cm 3 (ii) the flu of hydrogen through the foil, J = - D.Δc/Δ ep ( Q D = D ) RT From Table 2-1, H in BCC iron, D =.12, Q = 36 D =.12 ep( 36, /1.987.923) = J = - D.Δc/Δ = -.12 ep( 36, /1.987.923) (- 19691 =.33 1 8 H atoms/ cm 2.s 8 ) Q-14. What temperature is required to obtain.5% C at a distance of.5 mm beneath the surface of a.2% C steel in 2 h, when 1.1% C is present at the surface? Assume that the iron is FCC. Given: C = 1.1% C, C =.5% C, C =.2% C, =.5 mm=.5cm s t= 2hr36= 72sec, T =? Met-223
21 By Fick's second law Cs C = erf [ C C 2 Dt s ] 1.1.5.5 =.667 = erf [ ] 1.1.2 2 Dt From Table 2-3, we find that,.5.685 2 Dt = Dt.5 2 2. = (.685) =.133 ep ( Q D = D ) RT From the table 2-1, C in FCC iron, D =.23, Q = 32,9 cal/mol D =.23ep( 32,9 /1.987. T ) D.t =.133, D =.133/ t, D =.133/ 72 D =.23ep( 32,9 /1.987. T ) and D =.133/ 72.133/ 72 =.23ep( 32,9 /1.987. T ) T = 118 K = 118 273 = 97 C Q-15. A.15% C steel is to be carburized at 11 C, giving.35% C at a distance of 1mm beneath the surface. If the surface composition is maintained at.9% C, what time is required? Given; C =.9% C, C =.35% C, C =.15% C, = 1 mm=.1cm s T = 11 C + 273= 1373 K, time( t) =? By Fick's second law Cs C = erf [ C C 2 Dt s ].9.35.1 =.733 = erf [ ].9.15 2 Dt From Table 2-3, we find that,.1.786 2 Dt = Met-223
22 D.t = [.1/2.786 ] 2 =.45 ep ( Q D = D ) RT From the table 2-1, C in FCC iron, D =.23, Q = 32,9 cal/mol D -6 2 =.23ep(-32,9 /1.987.1373) = 1.332 1 cm / s D.t =.45, t =.45 / D =.45 / 1.332 1-6 = 34 s time ( t ) = 34 sec = 51 min Q-16. A.2% C steel is to be carburized at 12 C in 4 h, with a point.6 mm beneath the surface reaching.45% C. Calculate the carbon content required at the surface of the steel. Given; C =?, C =.45% C, C =.2% C, =.6 mm=.6cm s T = 12 C + 273= 1473 K, time( t) = 4hr = 4 36 = 44sec By Fick's second law Cs C = erf [ ] C C 2 Dt s Cs.45.6 = erf [ ] C.2 2 Dt s.6 erf [ 2 Dt ] =? ep ( Q D = D ) RT From the table 2-1, C in FCC iron, D =.23, Q = 32,9 cal/mol -6 D =.23ep( 32,9 /1.987.1473) = 3.19 1 cm 2 /s.6 erf [ ] = erf [.144] 6 2 3.191.4. 36. From Table 2-3, we find that,.144 2 Dt =, [ erf 2 ] = Dt.161 Cs.45 =.161 C.2 s C s =.53 C % Met-223
23 Q-17. A 1.2% C tool steel held at 115 C is eposed to oygen for 48 h. The carbon content at the steel surface is zero. To what depth will the steel be decarburized to less than.2% C? Given; C =.% C, C =.2% C, C = 1.2% C, ( cm) =? s T = 115 C + 273= 1423 K, time( t) = 4 hr = 4 36sec = 44sec By Fick's second law Cs C = erf [ C C 2 Dt s ].2 1.2 =.1667 = erf [ ] 2 Dt erf [ ] =.1667, From Table 2-3, we find that, 2 Dt.149 2 Dt =, ep ( Q D = D ) RT From the table 2-1, C in FCC iron, D =.23, Q = 32,9 cal/mol -6 D =.23ep( 32,9 /1.987.1423) = 2.34 1 cm 2 /s Dt. = 6 2.341.48..36 =.149, =. 149 2 Dt 2.5929 =.5929,, =.177 cm Q-18. A BCC steel containing.1% N is nitrided at 55 C for 5 h. If the nitrogen content at the steel surface is.8%, determine the nitrogen content at.25 mm from the surface. Given; C =.8 N%, C =? N%, C =.1 N%, =.25mm=.25cm s T = 55 C + 273= 823 K, time( t) = 5hr = 5 36sec = 8sec By Fick's second law C C s s C C = erf [ 2 ] Dt Met-223
24.8 C = erf [ ].8.1 2 Dt ep ( Q D = D ) RT From the table 2-1, N in BCC iron, D =.47, Q = 18,3 cal/mol -8 D =.47 ep( 18,3 /1.987..823) = 6.488 1 cm 2 /s D.t = 8 6.488..1..5..36 =.342,.25 erf [ ] = erf [ ] 2 Dt 2.342 =.3655, erf [ ] =.394 2 Dt 2 Dt.8 C = erf [ ] =.394.8.1 2 Dt C =.49 N% Q-19. Define the following. (i) % Elongation (ii) Endurance limit (iii) Endurance Ratio (iv) Fatigue life (v) Elastic deformation (i) % Elongation - The total percentage increases in the length of a specimen during a tensile test. f % Elongation = 1% l l where l f final length of the specimen l initial length of the specimen l (ii) Endurance limit - The stress below which a material will not fail in a fatigue test. (iii) Endurance Ratio - The endurance limit divided by the tensile strength of the material. The ratio is about.5 for many ferrous metals. Endurance limit Endurance ratio =.5 trnsile strength The endurance ratio allows us to estimate fatigue properties from the tensile test. Met-223
25 (iv) Fatigue life - The number of cycles permitted at a particular stress before a material fails by fatigue. Fatigue life tells us how long a component survives at a particular stress. (v) Elastic deformation - Deformation of the material that is recovered when the applied load is removed. Q-2. (a) Define Engineering Stress, True stress and Engineering Strain, True Strain. (b) A force of 1, N is applied to a 1 mm 2 mm iron bar having a yield strength of 4 MPa and a tensile strength of 48 MPa. Determine (i) whether the bar will plastically deform and (ii) whether the bar will eperience necking. (a) Engineering Stress - The applied load, or force, divided by the original cross sectional area of the material. force F Engineering stress = = σ = initial. crosssec tional. area A True stress - The load divided by the actual cross-sectional area of the specimen at that load. True stress = σ 1 = F1 A 1, σ 2 = F A 2 2 Engineering Strain - The amount that a material deforms per unit length in a tensile test. l l Engineering strain = ε = l where l final length of the specimen l initial length of the specimen True Strain - The strain, given by ε t = ln (l/l ), produced in a material. l True Strain ε t = ln l where l final length of the specimen l initial length of the specimen Met-223
26 Q-2.(b) Applied force = 1, N (i) area of iron bar = 1 mm 2 mm = 2 mm 2 yield strength - 4 MPa tensile strength - 48 MPa. force F Engineering stress = = σ = initial. crosssec tional. area A σ = 1, N / 2 mm 2 = 5 N/mm 2 1 MPa = 1 N/mm 2, 5 MPa applied stress σ is greater than yield strength ( 5 MPa > 4 MPa ),therefore, the iron bar will plastically deform. (ii) applied stress σ is greater than tensile strength ( 5 MPa > 48 MPa ),therefore, the bar will occur necking. Q-21. An aluminum alloy that has a plane strain fracture toughness of 25, psi- in.fails when a stress of 42, psi is applied. Observation of the fracture surface indicates that fracture began at the surface of the part. Estimate the size of the flaw that initiated fracture. Assume that f = 1.1. plane strain fracture toughness K c = 25, psi- in. stress σ = 42, psi is applied. the fracture began at the surface of the part. the size of the flaw =?. Assume that f = 1.1. K IC = f.σ. π. a 25, = 1.1 42, π. a a =.93 in the initial flaw size on the surface.93 in Q-22. Define Hooke's law and Poisson's ratio. Hooke's law The relationship between stress and strain in the elastic portion of the Met-223
27 stress-strain curve. The modulus of elasticity, or Young's modulus, E, is the slope of the stress-strain curve in the elastic region. This relationship is Hooke's law : σ E = ε where E - Young's modulus σ - engineering stress ε - engineering strain Poisson's ratio The ratio between the lateral and longitudinal strains in the elastic region. Poisson's ratio, μ, relates the longitudinal elastic deformation produced by a simple tensile or compressive stress to the lateral deformation that occurs simultaneously: ε ( lateral) μ =, ( μ is about.3) ε ( longitudional Q-23. A.4-in. diameter,12-in. long titanium bar has a yield strength of 5, psi, a modulus of elasticity of 16 1 6 psi, and Poisson's ratio of.3. Determine the length and diameter of the bar when a 5-Ib load is applied. a titanium bar, diameter =.4-in., length = 12-in. yield strength = 5, psi, modulus of elasticity E = 16 1 6 psi, Poisson's ratio μ =.3. the length =? and the diameter of the bar =? when a 5-Ib load is applied. F = 5 lb the stress σ = F/A = 5 lb / (π/4 )(.4 in.) 2 = 3979 psi By Hook's Law, E the strain Engineering strain = σ ε = σ ε = = 3979 psi / 16 1 6 psi =.24868 in/in E ε l l l f f = == = l 12 12.24868 lf = 12.298 in Met-223
28 ε ( lateral) Poisson's ratio: μ = =.3 ε ( longitudional ε(lateral) = - (μ ) ε(longitudinal) = - (.3 ) (.24868) = -.746 in/in Df D Df.4 = = -.746 in / in D.4 Df =.39997in the length of the bar = 12.298 in the diameter of the bar =.39997 in Q-24. A 3-in.diameter rod of copper is to be reduced to a 2-in. diameter rod by being pushed through an opening. To account for the elastic strain, what should be the diameter of the opening? The modulus of elasticity for the copper is 17 1 6 psi and the yield strength is 4, psi. A copper rod is to be reduced in diameter, D = 3-in., D 1 = 2-in. dia, the diameter of the opening? modulus of elasticity for the copper, E = 17 1 6 psi yield strength σ = 4, psi. To get 2 in diameter bar, the diameter of the opening must be smaller than the final dia. E = σ, ε = σ / E = 4, / 17 1 6 =.235 ε l l D D ε =.235, Engineering strain = ε = = l D D = 2 in., D =?.235 2 D D =, D = 1.995 in the diameter of the opening die = 1.995 in. Q-25.Which factors does depend on the ability of a material to resist the growth of a crack? The ability of a material to resist the growth of a crack depends on a large number of factors: Met-223
29 1). the larger the flaws size, the smaller the permitted stress. 2). Increasing the strength of a given metal usually decreases ductility and gives a lower fracture toughness. 3).Thicker, more rigid materials have lower fracture toughness than thin materials. 4). Increasing the rate of application of the load, such as in an impact test, typically reduces the fracture toughness of the material. 5). Increasing the temperature normally increases the fracture toughness, just as in the impact test. 6). A small grain size normally improves fracture toughness, more point defects and dislocations reduce fracture toughness. Q-26.Define the following. (i) Tensile strength (ii) Yield strength (iii) Ductility (i) Tensile strength The stress obtained at the highest applied force is the tensile strength, which is the maimum stress on the engineering stress-strain curve. (ii) Yield strength deformation. The stress applied to a material that just causes permanent plastic (iii) Ductility The ability of a material to be permanently deformed without breaking when a force is applied. Ductility measures the amount of deformation that a material can withstand without breaking. Q-27. A ceramic part for a jet engine has a yield strength of 75, psi and a plane strain fracture toughness of 5, psi - in.. To be sure that the part does not fail, we plan to assure that the maimum applied stress is only one-third the yield strength. We use a nondestructive test that will detect any internal flaws greater than.5 in. long. Assuming that f = 1.4, does our nondestructive test have the required sensitivity? Eplain. Met-223
3 A ceramic part for a jet engine, yield strength σ = 75, psi plane strain fracture toughness K IC = 5, psi - in.. the maimum applied stress = one-third the yield strength. = 1/3 75, psi internal flaw size = 2.a, f = 1.4, K IC = f.σ. π. a 5 = 1.4 1/3 75,. π. a a =.65 in the length of internal flaw = 2.a = 2.65 in =.13 in Non- destructive test can detect the flaw size of length =.5 in. Now, the flaw size is.13 in. and therefore, our nondestructive test have no the required sensitivity. Q-28. A large steel plate used in a nuclear reactor has a plane strain fracture toughness of 8, psi- in and is eposed to a stress of 45, psi during service.design a testing or inspection procedure capable of detecting a crack at the surface of the plate before the crack is likely to grow at a catastrophic rate. steel plate used in a nuclear reactor, applied stress σ = 45, psi plane strain fracture toughness K IC = 8, psi - in. surface crack = a =?, f = 1, Under these condition, to determine the minimum size of crack. K IC = K c = f.σ. π. a 8, = 1 45, π. a a = 1 in. This minimum crack size ( 1 in.) on the surface can be observed visually. If the growth rate of the crack is slow, the inspection is performed by regular method. Q-29. A 85-lb force is applied to a.15-in. diameter nickel wire having a yield strength of 45, psi and a tensile strength of 55, psi. Determine Met-223
31 (a) whether the wire will plastically deform and (b) whether the wire will eperience necking. nickel wire, yield strength = 45, psi, tensile strength = 55, psi. applied force = 85-lb, initial diameter D =.15-in. (i)whether the wire will plastically deform and (ii)whether the wire will eperience necking. force F Engineering stress on the wire = = σ = initial. cross sec tional. area A σ = 85 lb / (π/4) (d ) 2 = 48,1 Psi (i) Applied stress σ is greater than yield strength ( 48,1 Psi > 45, Psi ), therefore, the nickel wire will plastically deform. (ii)applied stress σ is less than tensile strength ( 48,1 Psi < 55, Psi ), therefore, the nickel wire will no necking occur. Q-3. A solid shaft for a cement kiln produced from the tool steel must be 96 inches long and must survive continuous operation for one year with an applied load of 12,5 lb. The shaft makes one revolution per minute during operation. Design a shaft that will satisfy these requirements. a tool steel solid shaft for a cement kiln, length L = 96 in., applied load ( F ) = 12,5 lb. continuous operation for one year, and one revolution per minute during operation. Design a shaft that will satisfy these requirements. It means that minimum diameter of the shaft (d ) =? 1 cycle..365. days. 24. hr..6.min Number of cycle / year = N = 1.min..1. year..1. day..1. hr. N = 5.265 1 5 cycles/year From Fig. 3-13, S-N curve for tool steel, N = 5.265 1 5,,, applied stress σ = 72, Psi or 72 Ksi. applied stress σ must be less than 72, Psi or 72 Ksi. Met-223
32 1.18. L. F By the equation, σ = 3 d 1.18..96..125 72, = 3 d d = 5.54 in For these conditions, the diameter of the shaft 5.54 in. will operate for one year. But, safety is required in the design without failure. In Fig. endurance limit 6, Psi, < 72, Psi., this condition is minimum diameter required to prevent failure. 1.18..96..125 6, = 3 d d = 5.88 in The condition that will operate for more than one year. without failure, the minimum diameter of the shaft (d ) is 5.88 in * * * * * * * * * * * * *END* * * * * * * * * * * * Met-223
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