Met-2023: Concepts of Materials Science I Sample Questions & Answers,(2009) ( Met, PR, FC, MP, CNC, McE )
|
|
- Elfreda Atkinson
- 7 years ago
- Views:
Transcription
1 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 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 cm. (at. Wt. for Fe g/mol ) Q-4.(a) The planar density of the (112) plane in BCC iron is 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 Aº ) (ii) FCC silver ( a Aº ) (iii) FCC copper ( a 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 Aº ) Met-223
2 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 g/cm 3 and its lattice parameter is 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 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 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 H atoms/ cm 3 are in equilibrium with the hot side of the foil, and 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 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 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 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 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 The stress-number of cycles to failure (S-N) curves for a tool steel and an aluminum alloy. Met-223
6 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 where, a = the lattice parameter or lattice constant h,k,l = Miller indices of adjacent planes of atoms Met-223
7 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 nm. The lattice parameter of FCC copper is nm. The number of copper atoms, per cm 3 is 4 atoms / cell ( cm) 22 = copper atoms/cm At room temperature, T = = 298 K: n v = ( ) ep [(- 2,)/ (1.987) (298)] = vacancies/cm 3 We wish to produce 1 times this number, or n v = vacancies/cm 3. We could do this by heating the copper to a temperature at which this number of vacancies forms: n v = = ( ) ep (-2,/1.987T) ep (- 2,/1.987T ) = /( ) = T = 2, / (1.987)(26.87) = 375 K = 12 C Met-223
8 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 cm. (at. wt. for Fe = 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 g / mole density = = g / cm ( 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) ( g / mole) density = = g / cm ( cm ) (6.21 atoms/ mol) g / cm ( g / mole) atoms / unit cell = = atoms ( cm ) (6.21 atoms/ mol) 3 no. of vacancies per unit cell = =.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 = vacancies/cm 3 Q-4.(a)The planar density of the (112) plane in BCC iron is 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 9 (a) (b) Calculate the length of the Burgers vector in the following materials: (i) BCC niobium ( a Aº ) (ii) FCC silver ( a Aº ) (iii) FCC copper ( a Aº ) (i) For BCC niobium, a = 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 = Aº The directions of the Burgers vector, are in [l 1] for FCC metal Met-223
10 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 = 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 = ( nm ) = 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 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 MPa applied stress direction - [111] direction. the critical resolved shear stress ι =? τ = σ CosφCosλ From the Fig: Met-223
12 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 τ = 3.51 = MPa 3 the critical resolved shear stress, τ = 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 Aº ) (a) copper metal, FCC structure, 4 atoms/ unit cell, a A cm Q cal mol R cal mol K 8 = , = , = 2, /, = /. 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 = = = atom / cm ( a ) ( cm) Q 2, nv = n = RT = vacancies / cm 24 ep( ) 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 , v 1, =? T = C+ = K = Q n Q nv = nep RT Met-223
13 13 nv Q = ep n RT Q ep( ) = Q= 128 cal/ mol Q-7. The density of a sample of FCC palladium is g/cm 3 and its lattice parameter is 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 = g/cm 3, 4 atoms/ unit cell At. wt. of Pd g/mol a = A= 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 = = g/ cm cm atom mol ( ) (6.2 1 / ) 3 (a) (b) For theoretical density g/cm 3, 4 atoms/ unit cell For the given density g/cm 3,? atoms/ unit cell For the given density g/cm 3, atoms/ unit cell no. of vacancies per unit cell (n v ) = =.95 n v =.95 n = no. of atom per unit cell = 4 n v n.95 = = nv no. of vacancies per unit cell = = = vacancies / cm cm volume of unit cell ( cm) 2 3 Q-8.(a) Define the rate of Diffusion of Fick's First Law. Met-223
14 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 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 = 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 ( ) = jump / sec Q-9.(a) Define (i) diffusion (ii) diffusion coefficient. Met-223
15 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 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 / ( cm ) 3 = W atoms/cm 3 In the tungsten -1 atom % thorium alloy, the number of thorium atoms is: C Th = (.1 ) ( ) = Th atoms/cm 3 In the pure tungsten, the number of thorium atoms is zero. Thus, the concentration gradient is: ΔC / Δ = /.1 cm = Tho atom / cm 3.cm T = 2 C K = 2273 K (a) for volume diffusion, D = 1.ep(-12,/RT) cm 2 / s J = - D.Δc/Δ = - 1.ep(-12,/ ) ( ) = Th atoms/cm 2.s (b) for grain boundary, D =.74 ep(-9,/rt) cm 2 / s J = - D.Δc/Δ = -.74 ep(-9,/ ) ( ) = 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/ ) ( ) = 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 17 Q-1.(b) The diffusion coefficient for Cr in Cr 2 O 3 is 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 = cm 2 /s at T 1 = 727 C K = 1 K D = cm 2 /s at T 2 = 14 C K = 1673 K (i) the activation energy Q =? ( ii ) the constant D o. =? ep ( Q D = D ) RT Q = eq. (1) D ep( ) RT1 Q = eq. (2) 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( ) Q= 5923 cal/ mol (ii) = Q D ep( ) RT = 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 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 =.68 = erf [ ] Dt From Table 2-3, we find that, Dt = or.1.71 Dt =.1 Dt = = ( ).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 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 K = 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 = 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
20 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 H atoms/ cm 3 are in equilibrium with the hot side of the foil, and 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 = = 923 K C initial = H atoms/ cm 3 C final = 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 =? (i) the concentration gradient (Δc/Δ) = (.1) (2.54) = 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, / ) = J = - D.Δc/Δ = -.12 ep( 36, / ) ( = 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 21 By Fick's second law Cs C = erf [ C C 2 Dt s ] =.667 = erf [ ] Dt From Table 2-3, we find that, Dt = Dt = (.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 / T ) D.t =.133, D =.133/ t, D =.133/ 72 D =.23ep( 32,9 / T ) and D =.133/ / 72 =.23ep( 32,9 / T ) T = 118 K = = 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 ] =.733 = erf [ ] Dt From Table 2-3, we find that, Dt = Met-223
22 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 / ) = cm / s D.t =.45, t =.45 / D =.45 / = 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 / ) = cm 2 /s.6 erf [ ] = erf [.144] From Table 2-3, we find that, Dt =, [ erf 2 ] = Dt.161 Cs.45 =.161 C.2 s C s =.53 C % Met-223
23 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 ] =.1667 = erf [ ] 2 Dt erf [ ] =.1667, From Table 2-3, we find that, 2 Dt 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 / ) = cm 2 /s Dt. = =.149, = Dt =.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 24.8 C = erf [ ] 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 / ) = cm 2 /s D.t = =.342,.25 erf [ ] = erf [ ] 2 Dt =.3655, erf [ ] = Dt 2 Dt.8 C = erf [ ] = 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 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 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, = , 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 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 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 = 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 / psi = in/in E ε l l l f f = == = l lf = in Met-223
28 28 ε ( lateral) Poisson's ratio: μ = =.3 ε ( longitudional ε(lateral) = - (μ ) ε(longitudinal) = - (.3 ) (.24868) = in/in Df D Df.4 = = in / in D.4 Df =.39997in the length of the bar = in the diameter of the bar = 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 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 = 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, / =.235 ε l l D D ε =.235, Engineering strain = ε = = l D D = 2 in., D =? D D =, D = in the diameter of the opening die = 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 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
30 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 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 days. 24. hr..6.min Number of cycle / year = N = 1.min..1. year..1. day..1. hr. N = cycles/year From Fig. 3-13, S-N curve for tool steel, N = ,,, applied stress σ = 72, Psi or 72 Ksi. applied stress σ must be less than 72, Psi or 72 Ksi. Met-223
32 L. F By the equation, σ = 3 d , = 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 , = 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
33 33 Met-223
34 34 Met-223
Mechanical Properties of Metals Mechanical Properties refers to the behavior of material when external forces are applied
Mechanical Properties of Metals Mechanical Properties refers to the behavior of material when external forces are applied Stress and strain fracture or engineering point of view: allows to predict the
More informationChapter 5: Diffusion. 5.1 Steady-State Diffusion
: Diffusion Diffusion: the movement of particles in a solid from an area of high concentration to an area of low concentration, resulting in the uniform distribution of the substance Diffusion is process
More informationCHAPTER 7 DISLOCATIONS AND STRENGTHENING MECHANISMS PROBLEM SOLUTIONS
7-1 CHAPTER 7 DISLOCATIONS AND STRENGTHENING MECHANISMS PROBLEM SOLUTIONS Basic Concepts of Dislocations Characteristics of Dislocations 7.1 The dislocation density is just the total dislocation length
More informationChapter Outline Dislocations and Strengthening Mechanisms
Chapter Outline Dislocations and Strengthening Mechanisms What is happening in material during plastic deformation? Dislocations and Plastic Deformation Motion of dislocations in response to stress Slip
More informationThe atomic packing factor is defined as the ratio of sphere volume to the total unit cell volume, or APF = V S V C. = 2(sphere volume) = 2 = V C = 4R
3.5 Show that the atomic packing factor for BCC is 0.68. The atomic packing factor is defined as the ratio of sphere volume to the total unit cell volume, or APF = V S V C Since there are two spheres associated
More informationSolution for Homework #1
Solution for Homework #1 Chapter 2: Multiple Choice Questions (2.5, 2.6, 2.8, 2.11) 2.5 Which of the following bond types are classified as primary bonds (more than one)? (a) covalent bonding, (b) hydrogen
More informationConcepts of Stress and Strain
CHAPTER 6 MECHANICAL PROPERTIES OF METALS PROBLEM SOLUTIONS Concepts of Stress and Strain 6.4 A cylindrical specimen of a titanium alloy having an elastic modulus of 107 GPa (15.5 10 6 psi) and an original
More informationChapter Outline Dislocations and Strengthening Mechanisms
Chapter Outline Dislocations and Strengthening Mechanisms What is happening in material during plastic deformation? Dislocations and Plastic Deformation Motion of dislocations in response to stress Slip
More informationChapter Outline. Mechanical Properties of Metals How do metals respond to external loads?
Mechanical Properties of Metals How do metals respond to external loads? Stress and Strain Tension Compression Shear Torsion Elastic deformation Plastic Deformation Yield Strength Tensile Strength Ductility
More informationLösungen Übung Verformung
Lösungen Übung Verformung 1. (a) What is the meaning of T G? (b) To which materials does it apply? (c) What effect does it have on the toughness and on the stress- strain diagram? 2. Name the four main
More information14:635:407:02 Homework III Solutions
14:635:407:0 Homework III Solutions 4.1 Calculate the fraction of atom sites that are vacant for lead at its melting temperature of 37 C (600 K). Assume an energy for vacancy formation of 0.55 ev/atom.
More informationDefects Introduction. Bonding + Structure + Defects. Properties
Defects Introduction Bonding + Structure + Defects Properties The processing determines the defects Composition Bonding type Structure of Crystalline Processing factors Defects Microstructure Types of
More informationChapter Outline. Diffusion - how do atoms move through solids?
Chapter Outline iffusion - how do atoms move through solids? iffusion mechanisms Vacancy diffusion Interstitial diffusion Impurities The mathematics of diffusion Steady-state diffusion (Fick s first law)
More informationStress Strain Relationships
Stress Strain Relationships Tensile Testing One basic ingredient in the study of the mechanics of deformable bodies is the resistive properties of materials. These properties relate the stresses to the
More informationExperiment: Crystal Structure Analysis in Engineering Materials
Experiment: Crystal Structure Analysis in Engineering Materials Objective The purpose of this experiment is to introduce students to the use of X-ray diffraction techniques for investigating various types
More informationLECTURE SUMMARY September 30th 2009
LECTURE SUMMARY September 30 th 2009 Key Lecture Topics Crystal Structures in Relation to Slip Systems Resolved Shear Stress Using a Stereographic Projection to Determine the Active Slip System Slip Planes
More informationLecture 14. Chapter 8-1
Lecture 14 Fatigue & Creep in Engineering Materials (Chapter 8) Chapter 8-1 Fatigue Fatigue = failure under applied cyclic stress. specimen compression on top bearing bearing motor counter flex coupling
More informationObjectives. Experimentally determine the yield strength, tensile strength, and modules of elasticity and ductility of given materials.
Lab 3 Tension Test Objectives Concepts Background Experimental Procedure Report Requirements Discussion Objectives Experimentally determine the yield strength, tensile strength, and modules of elasticity
More informationMechanical Properties - Stresses & Strains
Mechanical Properties - Stresses & Strains Types of Deformation : Elasic Plastic Anelastic Elastic deformation is defined as instantaneous recoverable deformation Hooke's law : For tensile loading, σ =
More informationIn order to solve this problem it is first necessary to use Equation 5.5: x 2 Dt. = 1 erf. = 1.30, and x = 2 mm = 2 10-3 m. Thus,
5.3 (a) Compare interstitial and vacancy atomic mechanisms for diffusion. (b) Cite two reasons why interstitial diffusion is normally more rapid than vacancy diffusion. Solution (a) With vacancy diffusion,
More informationCh. 4: Imperfections in Solids Part 1. Dr. Feras Fraige
Ch. 4: Imperfections in Solids Part 1 Dr. Feras Fraige Outline Defects in Solids 0D, Point defects vacancies Interstitials impurities, weight and atomic composition 1D, Dislocations edge screw 2D, Grain
More informationME 612 Metal Forming and Theory of Plasticity. 1. Introduction
Metal Forming and Theory of Plasticity Yrd.Doç. e mail: azsenalp@gyte.edu.tr Makine Mühendisliği Bölümü Gebze Yüksek Teknoloji Enstitüsü In general, it is possible to evaluate metal forming operations
More informationPROPERTIES OF MATERIALS
1 PROPERTIES OF MATERIALS 1.1 PROPERTIES OF MATERIALS Different materials possess different properties in varying degree and therefore behave in different ways under given conditions. These properties
More informationCH 6: Fatigue Failure Resulting from Variable Loading
CH 6: Fatigue Failure Resulting from Variable Loading Some machine elements are subjected to static loads and for such elements static failure theories are used to predict failure (yielding or fracture).
More informationσ y ( ε f, σ f ) ( ε f
Typical stress-strain curves for mild steel and aluminum alloy from tensile tests L L( 1 + ε) A = --- A u u 0 1 E l mild steel fracture u ( ε f, f ) ( ε f, f ) ε 0 ε 0.2 = 0.002 aluminum alloy fracture
More informationThe mechanical properties of metal affected by heat treatment are:
Training Objective After watching this video and reviewing the printed material, the student/trainee will learn the basic concepts of the heat treating processes as they pertain to carbon and alloy steels.
More informationDIFFUSION IN SOLIDS. Materials often heat treated to improve properties. Atomic diffusion occurs during heat treatment
DIFFUSION IN SOLIDS WHY STUDY DIFFUSION? Materials often heat treated to improve properties Atomic diffusion occurs during heat treatment Depending on situation higher or lower diffusion rates desired
More informationSolid Mechanics. Stress. What you ll learn: Motivation
Solid Mechanics Stress What you ll learn: What is stress? Why stress is important? What are normal and shear stresses? What is strain? Hooke s law (relationship between stress and strain) Stress strain
More informationIntroduction To Materials Science FOR ENGINEERS, Ch. 5. Diffusion. MSE 201 Callister Chapter 5
Diffusion MSE 21 Callister Chapter 5 1 Goals: Diffusion - how do atoms move through solids? Fundamental concepts and language Diffusion mechanisms Vacancy diffusion Interstitial diffusion Impurities Diffusion
More informationUnit 6: EXTRUSION. Difficult to form metals like stainless steels, nickel based alloys and high temperature metals can also be extruded.
1 Unit 6: EXTRUSION Introduction: Extrusion is a metal working process in which cross section of metal is reduced by forcing the metal through a die orifice under high pressure. It is used to produce cylindrical
More informationCrystal Defects p. 2. Point Defects: Vacancies. Department of Materials Science and Engineering University of Virginia. Lecturer: Leonid V.
Crystal Defects p. 1 A two-dimensional representation of a perfect single crystal with regular arrangement of atoms. But nothing is perfect, and structures of real materials can be better represented by
More informationLABORATORY EXPERIMENTS TESTING OF MATERIALS
LABORATORY EXPERIMENTS TESTING OF MATERIALS 1. TENSION TEST: INTRODUCTION & THEORY The tension test is the most commonly used method to evaluate the mechanical properties of metals. Its main objective
More informationLecture 18 Strain Hardening And Recrystallization
-138- Lecture 18 Strain Hardening And Recrystallization Strain Hardening We have previously seen that the flow stress (the stress necessary to produce a certain plastic strain rate) increases with increasing
More informationMaterials Issues in Fatigue and Fracture
Materials Issues in Fatigue and Fracture 5.1 Fundamental Concepts 5.2 Ensuring Infinite Life 5.3 Finite Life 5.4 Summary FCP 1 5.1 Fundamental Concepts Structural metals Process of fatigue A simple view
More informationEDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES OUTCOME 2 ENGINEERING COMPONENTS TUTORIAL 1 STRUCTURAL MEMBERS
ENGINEERING COMPONENTS EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES OUTCOME ENGINEERING COMPONENTS TUTORIAL 1 STRUCTURAL MEMBERS Structural members: struts and ties; direct stress and strain,
More informationTorsion Tests. Subjects of interest
Chapter 10 Torsion Tests Subjects of interest Introduction/Objectives Mechanical properties in torsion Torsional stresses for large plastic strains Type of torsion failures Torsion test vs.tension test
More informationStrengthening. Mechanisms of strengthening in single-phase metals: grain-size reduction solid-solution alloying strain hardening
Strengthening The ability of a metal to deform depends on the ability of dislocations to move Restricting dislocation motion makes the material stronger Mechanisms of strengthening in single-phase metals:
More informationTIE-31: Mechanical and thermal properties of optical glass
PAGE 1/10 1 Density The density of optical glass varies from 239 for N-BK10 to 603 for SF66 In most cases glasses with higher densities also have higher refractive indices (eg SF type glasses) The density
More informationEach grain is a single crystal with a specific orientation. Imperfections
Crystal Structure / Imperfections Almost all materials crystallize when they solidify; i.e., the atoms are arranged in an ordered, repeating, 3-dimensional pattern. These structures are called crystals
More informationProperties of Materials
CHAPTER 1 Properties of Materials INTRODUCTION Materials are the driving force behind the technological revolutions and are the key ingredients for manufacturing. Materials are everywhere around us, and
More informationMaterial Deformations. Academic Resource Center
Material Deformations Academic Resource Center Agenda Origin of deformations Deformations & dislocations Dislocation motion Slip systems Stresses involved with deformation Deformation by twinning Origin
More informationMECHANICS OF MATERIALS
T dition CHTR MCHNICS OF MTRIS Ferdinand. Beer. Russell Johnston, Jr. John T. DeWolf ecture Notes: J. Walt Oler Texas Tech University Stress and Strain xial oading - Contents Stress & Strain: xial oading
More informationGENERAL PROPERTIES //////////////////////////////////////////////////////
ALLOY 625 DATA SHEET //// Alloy 625 (UNS designation N06625) is a nickel-chromium-molybdenum alloy possessing excellent resistance to oxidation and corrosion over a broad range of corrosive conditions,
More informationMETU DEPARTMENT OF METALLURGICAL AND MATERIALS ENGINEERING
METU DEPARTMENT OF METALLURGICAL AND MATERIALS ENGINEERING Met E 206 MATERIALS LABORATORY EXPERIMENT 1 Prof. Dr. Rıza GÜRBÜZ Res. Assist. Gül ÇEVİK (Room: B-306) INTRODUCTION TENSION TEST Mechanical testing
More informationCHAPTER 6: DIFFUSION IN SOLIDS. Inter-diffusion. Simple Diffusion. Diffusion- Steady and Non-Steady State ISSUES TO ADDRESS...
CHAPTER 6: DIFFUSION IN SOLIDS Diffusion- Steady and Non-Steady State ISSUES TO ADDRESS... Gear from case-hardened steel (C diffusion) Diffusion - Mass transport by atomic motion How does diffusion occur?
More informationUniaxial Tension and Compression Testing of Materials. Nikita Khlystov Daniel Lizardo Keisuke Matsushita Jennie Zheng
Uniaxial Tension and Compression Testing of Materials Nikita Khlystov Daniel Lizardo Keisuke Matsushita Jennie Zheng 3.032 Lab Report September 25, 2013 I. Introduction Understanding material mechanics
More informationFigure 1: Typical S-N Curves
Stress-Life Diagram (S-N Diagram) The basis of the Stress-Life method is the Wohler S-N diagram, shown schematically for two materials in Figure 1. The S-N diagram plots nominal stress amplitude S versus
More informationFatigue. 3. Final fracture (rough zone) 1. Fatigue origin. 2. Beach marks (velvety zone)
Fatigue Term fatigue introduced by Poncelet (France) 1839 progressive fracture is more descriptive 1. Minute crack at critical area of high local stress (geometric stress raiser, flaws, preexisting cracks)
More informationENGINEERING COUNCIL CERTIFICATE LEVEL
ENGINEERING COUNCIL CERTIICATE LEVEL ENGINEERING SCIENCE C103 TUTORIAL - BASIC STUDIES O STRESS AND STRAIN You should judge your progress by completing the self assessment exercises. These may be sent
More informationMECHANICAL PRINCIPLES HNC/D PRELIMINARY LEVEL TUTORIAL 1 BASIC STUDIES OF STRESS AND STRAIN
MECHANICAL PRINCIPLES HNC/D PRELIMINARY LEVEL TUTORIAL 1 BASIC STUDIES O STRESS AND STRAIN This tutorial is essential for anyone studying the group of tutorials on beams. Essential pre-requisite knowledge
More informationLecture 12. Physical Vapor Deposition: Evaporation and Sputtering Reading: Chapter 12. ECE 6450 - Dr. Alan Doolittle
Lecture 12 Physical Vapor Deposition: Evaporation and Sputtering Reading: Chapter 12 Evaporation and Sputtering (Metalization) Evaporation For all devices, there is a need to go from semiconductor to metal.
More informationUltrasonic Technique and Device for Residual Stress Measurement
Ultrasonic Technique and Device for Residual Stress Measurement Y. Kudryavtsev, J. Kleiman Integrity Testing Laboratory Inc. 80 Esna Park Drive, Units 7-9, Markham, Ontario, L3R 2R7 Canada ykudryavtsev@itlinc.com
More informationAC 2008-2887: MATERIAL SELECTION FOR A PRESSURE VESSEL
AC 2008-2887: MATERIAL SELECTION FOR A PRESSURE VESSEL Somnath Chattopadhyay, Pennsylvania State University American Society for Engineering Education, 2008 Page 13.869.1 Material Selection for a Pressure
More information4.461: Building Technology 1 CONSTRUCTION AND MATERIALS FALL TERM 2004 SCHOOL OF ARCHITECTURE AND PLANNING: MIT
4.461: Building Technology 1 CONSTRUCTION AND MATERIALS Professor John E. Fernandez FALL TERM 2004 SCHOOL OF ARCHITECTURE AND PLANNING: MIT Concrete and Composites Stadelhofen Station Zurich Santiago Calatrava
More informationTechnology of EHIS (stamping) applied to the automotive parts production
Laboratory of Applied Mathematics and Mechanics Technology of EHIS (stamping) applied to the automotive parts production Churilova Maria, Saint-Petersburg State Polytechnical University Department of Applied
More informationSurface Treatments. Corrosion Protective coatings for harsh environments (catalytic converters, electrochemical cells )
Surface Treatments Applications Biomedical (biocompatible coatings on implants, drug coatings for sustained release ) Mechanical Tribological friction and wear (tool steels, implants ) Fatigue minimize
More informationLecture slides on rolling By: Dr H N Dhakal Lecturer in Mechanical and Marine Engineering, School of Engineering, University of Plymouth
Lecture slides on rolling By: Dr H N Dhakal Lecturer in Mechanical and Marine Engineering, School of Engineering, University of Plymouth Bulk deformation forming (rolling) Rolling is the process of reducing
More informationMSE 528 - PRECIPITATION HARDENING IN 7075 ALUMINUM ALLOY
MSE 528 - PRECIPITATION HARDENING IN 7075 ALUMINUM ALLOY Objective To study the time and temperature variations in the hardness and electrical conductivity of Al-Zn-Mg-Cu high strength alloy on isothermal
More informationCrystal Structure of Aluminum, Zinc, and their Alloys By: Omar Fajardo Sebastian Henao Devin Baines ENGR45, F2014, SRJC
Crystal Structure of Aluminum, Zinc, and their Alloys By: Omar Fajardo Sebastian Henao Devin Baines ENGR45, F2014, SRJC Purpose The purpose of this experiment was to examine and observe the microstructure
More informationBending, Forming and Flexing Printed Circuits
Bending, Forming and Flexing Printed Circuits John Coonrod Rogers Corporation Introduction: In the printed circuit board industry there are generally two main types of circuit boards; there are rigid printed
More informationEDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQF LEVEL 3 OUTCOME 1 - LOADING SYSTEMS TUTORIAL 3 LOADED COMPONENTS
EDEXCEL NATIONAL CERTIICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQ LEVEL 3 OUTCOME 1 - LOADING SYSTEMS TUTORIAL 3 LOADED COMPONENTS 1. Be able to determine the effects of loading in static engineering
More informationLecture 9, Thermal Notes, 3.054
Lecture 9, Thermal Notes, 3.054 Thermal Properties of Foams Closed cell foams widely used for thermal insulation Only materials with lower conductivity are aerogels (tend to be brittle and weak) and vacuum
More informationAppendice Caratteristiche Dettagliate dei Materiali Utilizzati
Appendice Caratteristiche Dettagliate dei Materiali Utilizzati A.1 Materiale AISI 9840 UNI 38NiCrMo4 AISI 9840 Steel, 650 C (1200 F) temper, 25 mm (1 in.) round Material Notes: Quenched, 540 C temper,
More informationMassachusetts Institute of Technology Department of Mechanical Engineering Cambridge, MA 02139
Massachusetts Institute of Technology Department of Mechanical Engineering Cambridge, MA 02139 2.002 Mechanics and Materials II Spring 2004 Laboratory Module No. 5 Heat Treatment of Plain Carbon and Low
More informationWJM Technologies excellence in material joining
Girish P. Kelkar, Ph.D. (562) 743-7576 girish@welding-consultant.com www.welding-consultant.com Weld Cracks An Engineer s Worst Nightmare There are a variety of physical defects such as undercut, insufficient
More informationMartensite in Steels
Materials Science & Metallurgy http://www.msm.cam.ac.uk/phase-trans/2002/martensite.html H. K. D. H. Bhadeshia Martensite in Steels The name martensite is after the German scientist Martens. It was used
More informationFatigue of Metals Copper Alloys. Samuli Heikkinen 26.6.2003
Fatigue of Metals Copper Alloys Samuli Heikkinen 26.6.2003 T 70 C Temperature Profile of HDS Structure Stress amplitude 220 MPa Stress Profile of HDS Structure CLIC Number of Cycles f = 100 Hz 24 hours
More informationChapter Outline. How do atoms arrange themselves to form solids?
Chapter Outline How do atoms arrange themselves to form solids? Fundamental concepts and language Unit cells Crystal structures Simple cubic Face-centered cubic Body-centered cubic Hexagonal close-packed
More informationALLOY 2205 DATA SHEET
ALLOY 2205 DATA SHEET UNS S32205, EN 1.4462 / UNS S31803 GENERAL PROPERTIES ////////////////////////////////////////////////////// //// 2205 (UNS designations S32205 / S31803) is a 22 % chromium, 3 % molybdenum,
More informationTensile Testing of Steel
C 265 Lab No. 2: Tensile Testing of Steel See web for typical report format including: TITL PAG, ABSTRACT, TABL OF CONTNTS, LIST OF TABL, LIST OF FIGURS 1.0 - INTRODUCTION See General Lab Report Format
More informationDIN 17172-78 STEEL PIPES FOR PIPE LINES FOR THE TRANSPORT OF COMBUSTIBLE FLUIDS AND GASES
DIN 17172-78 STEEL PIPES FOR PIPE LINES FOR THE TRANSPORT OF COMBUSTIBLE FLUIDS AND GASES For connection with the International Draft Standards 3183 and 3845 published by the International Organization
More informationMATERIALS AND MECHANICS OF BENDING
HAPTER Reinforced oncrete Design Fifth Edition MATERIALS AND MEHANIS OF BENDING A. J. lark School of Engineering Department of ivil and Environmental Engineering Part I oncrete Design and Analysis b FALL
More informationx100 A o Percent cold work = %CW = A o A d Yield Stress Work Hardening Why? Cell Structures Pattern Formation
Work Hardening Dislocations interact with each other and assume configurations that restrict the movement of other dislocations. As the dislocation density increases there is an increase in the flow stress
More informationORIENTATION CHARACTERISTICS OF THE MICROSTRUCTURE OF MATERIALS
ORIENTATION CHARACTERISTICS OF THE MICROSTRUCTURE OF MATERIALS K. Sztwiertnia Polish Academy of Sciences, Institute of Metallurgy and Materials Science, 25 Reymonta St., 30-059 Krakow, Poland MMN 2009
More informationStructural welding is a process by which the parts that are to be connected are heated and
CHAPTER 6. WELDED CONNECTIONS 6.1 INTRODUCTORY CONCEPTS Structural welding is a process by which the parts that are to be connected are heated and fused, with supplementary molten metal at the joint. A
More informationNorth American Stainless
North American Stainless Flat Products Stainless Steel Grade Sheet 310S (S31008)/ EN 1.4845 Introduction: SS310 is a highly alloyed austenitic stainless steel designed for elevated-temperature service.
More informationRelevant Reading for this Lecture... Pages 83-87.
LECTURE #06 Chapter 3: X-ray Diffraction and Crystal Structure Determination Learning Objectives To describe crystals in terms of the stacking of planes. How to use a dot product to solve for the angles
More informationA Comparison of FC-0208 to a 0.3% Molybdenum Prealloyed Low-Alloy Powder with 0.8% Graphite
A Comparison of FC-0208 to a 0.3% Molybdenum Prealloyed Low-Alloy Powder with 0.8% Graphite Francis Hanejko Manager, Customer Applications Hoeganaes Corporation Cinnaminson, NJ 08077 Abstract Iron copper
More informationSheet metal operations - Bending and related processes
Sheet metal operations - Bending and related processes R. Chandramouli Associate Dean-Research SASTRA University, Thanjavur-613 401 Table of Contents 1.Quiz-Key... Error! Bookmark not defined. 1.Bending
More informationAN EXPLANATION OF JOINT DIAGRAMS
AN EXPLANATION OF JOINT DIAGRAMS When bolted joints are subjected to external tensile loads, what forces and elastic deformation really exist? The majority of engineers in both the fastener manufacturing
More information9. TIME DEPENDENT BEHAVIOUR: CYCLIC FATIGUE
9. TIME DEPENDENT BEHAVIOUR: CYCLIC FATIGUE A machine part or structure will, if improperly designed and subjected to a repeated reversal or removal of an applied load, fail at a stress much lower than
More informationM n = (DP)m = (25,000)(104.14 g/mol) = 2.60! 10 6 g/mol
14.4 (a) Compute the repeat unit molecular weight of polystyrene. (b) Compute the number-average molecular weight for a polystyrene for which the degree of polymerization is 25,000. (a) The repeat unit
More informationSTRESS AND DEFORMATION ANALYSIS OF LINEAR ELASTIC BARS IN TENSION
Chapter 11 STRESS AND DEFORMATION ANALYSIS OF LINEAR ELASTIC BARS IN TENSION Figure 11.1: In Chapter10, the equilibrium, kinematic and constitutive equations for a general three-dimensional solid deformable
More informationMechanical Properties and Fracture Analysis of Glass. David Dutt Chromaglass, Inc.
Mechanical Properties and Fracture Analysis of Glass David Dutt Chromaglass, Inc. IES ALC Williamsburg 2006 2 IES ALC Williamsburg 2006 3 Outline The Ideal The Practical The Reality IES ALC Williamsburg
More informationChapter Outline: Phase Transformations in Metals
Chapter Outline: Phase Transformations in Metals Heat Treatment (time and temperature) Microstructure Mechanical Properties Kinetics of phase transformations Multiphase Transformations Phase transformations
More informationMaterial data sheet. EOS CobaltChrome MP1. Description
EOS CobaltChrome MP1 EOS CobaltChrome MP1 is a cobalt-chrome-molybdenum-based superalloy powder which has been optimized especially for processing on EOSINT M systems. This document provides information
More informationTensile Testing Laboratory
Tensile Testing Laboratory By Stephan Favilla 0723668 ME 354 AC Date of Lab Report Submission: February 11 th 2010 Date of Lab Exercise: January 28 th 2010 1 Executive Summary Tensile tests are fundamental
More informationMCE380: Measurements and Instrumentation Lab. Chapter 9: Force, Torque and Strain Measurements
MCE380: Measurements and Instrumentation Lab Chapter 9: Force, Torque and Strain Measurements Topics: Elastic Elements for Force Measurement Dynamometers and Brakes Resistance Strain Gages Holman, Ch.
More informationALLOY C276 DATA SHEET
ALLOY C276 DATA SHEET //// Alloy C276 (UNS designation N10276) is a nickel-molybdenum-chromium-iron-tungsten alloy known for its corrosion resistance in a wide range of aggressive media. It is one of the
More informationFluids and Solids: Fundamentals
Fluids and Solids: Fundamentals We normally recognize three states of matter: solid; liquid and gas. However, liquid and gas are both fluids: in contrast to solids they lack the ability to resist deformation.
More informationActivity 2.3b Engineering Problem Solving Answer Key
Activity.3b Engineering roblem Solving Answer Key 1. A force of 00 lbs pushes against a rectangular plate that is 1 ft. by ft. Determine the lb lb pressure in and that the plate exerts on the ground due
More informationFluid Mechanics: Static s Kinematics Dynamics Fluid
Fluid Mechanics: Fluid mechanics may be defined as that branch of engineering science that deals with the behavior of fluid under the condition of rest and motion Fluid mechanics may be divided into three
More informationdifferent levels, also called repeated, alternating, or fluctuating stresses.
Fatigue and Dynamic Loading 1 Fti Fatigue fil failure: 2 Static ti conditions : loads are applied gradually, to give sufficient i time for the strain to fully develop. Variable conditions : stresses vary
More informationModule #17. Work/Strain Hardening. READING LIST DIETER: Ch. 4, pp. 138-143; Ch. 6 in Dieter
Module #17 Work/Strain Hardening READING LIST DIETER: Ch. 4, pp. 138-143; Ch. 6 in Dieter D. Kuhlmann-Wilsdorf, Trans. AIME, v. 224 (1962) pp. 1047-1061 Work Hardening RECALL: During plastic deformation,
More informationIonic and Metallic Bonding
Ionic and Metallic Bonding BNDING AND INTERACTINS 71 Ions For students using the Foundation edition, assign problems 1, 3 5, 7 12, 14, 15, 18 20 Essential Understanding Ions form when atoms gain or lose
More informationSAND CAST CHILL CAST LM4 - TF
1 This alloy conforms with British Standards 1490 and is similar to the obsolete specifications BS.L79 and D.T.D 424A. Castings may be in the cast (M) of fully heat treated (TF) conditions. CHEMICAL COMPOSITION
More informationWeld Cracking. An Excerpt from The Fabricators' and Erectors' Guide to Welded Steel Construction. The James F. Lincoln Arc Welding Foundation
Weld Cracking An Excerpt from The Fabricators' and Erectors' Guide to Welded Steel Construction The James F. Lincoln Arc Welding Foundation Weld Cracking Several types of discontinuities may occur in welds
More informationThe Mechanical Properties of Glass
The Mechanical Properties of Glass Theoretical strength, practical strength, fatigue, flaws, toughness, chemical processes Glass Engineering 150:312 Professor Richard Lehman Department of Ceramics and
More informationChapter 12 - Liquids and Solids
Chapter 12 - Liquids and Solids 12-1 Liquids I. Properties of Liquids and the Kinetic Molecular Theory A. Fluids 1. Substances that can flow and therefore take the shape of their container B. Relative
More informationCERAMICS: Properties 2
CERAMICS: Properties 2 (Brittle Fracture Analysis) S.C. BAYNE, 1 J.Y. Thompson 2 1 University of Michigan School of Dentistry, Ann Arbor, MI 48109-1078 sbayne@umich.edu 2 Nova Southeastern College of Dental
More information