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CHAPTER 6 MECHANICAL PROPERTIES OF METALS

6.1

What are the three main metallurgical stages that a sheet of a cold-worked metal such as aluminum or copper goes through as it is heated from room temperature to an elevated temperature just below its melting point? The three main stages are recovery, recrystallization and grain growth.

6.2

Describe the microstructure of a heavily cold-worked metal of an Al-0.8% Mg alloy as observed with an optical microscope at 100× (see Fig. 6.2a). Describe the microstructure of the same material at 20,000× (see Fig. 6.3a). At 100×, one observes a highly elongated grain structure. Whereas at 20,000×, dislocation tangles and banded cells or subgrains, produced by extensive cold work, are evident within the microstructure.

6.3

Describe what occurs microscopically when a cold-worked sheet of metal such as aluminum undergoes a recovery heat treatment. When a cold-worked sheet of metal such as aluminum is subjected to a recovery heat treatment, many dislocations are destroyed or move into lower energy configurations through the polygonization process. In this process, low-angle grain boundaries are formed and the metal ductility is significantly increased while the strength is only slightly reduced.

6.4

When a cold-worked metal is heated into the temperature range where recovery takes place, how are the following affected: (a) internal residual stresses, (b) strength, (c) ductility, and (d) hardness? (a) (b) (c) (d)

6.5

Internal stresses are greatly reduced. The metal strength is only slightly reduced. The metal ductility is usually significantly increased. The hardness of the metal is slightly reduced.

Describe what occurs microscopically when a cold-worked sheet of metal such as aluminum undergoes a recrystallization heat treatment. During a recrystallization heat treatment, new strain-free grains are nucleated, and after sufficient time, grow until the grain structure is completely recrystallized.

6.6

When a cold-worked metal is heated into the temperature range where recrystallization takes place, how are the following affected: (a) internal residual stresses, (b) strength, (c) ductility, and (d) hardness? Smith

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(a) (b) (c) (d) 6.7

Any internal stresses are relieved. The metal tensile strength is significantly reduced. The ductility of the metal is greatly increased. The hardness of the metal is substantially reduced.

Describe two principal mechanisms whereby primary recrystallization can occur. The two principal mechanisms of primary recrystallization are: the expansion of an isolated nucleus with a deformed grain; the migration of a high-angle grain boundary into a more highly deformed region of the metal.

6.8

What are five important factors that affect the recrystallization process in metals? Five important factors affecting the metal recrystallization process are: 1. 2. 3. 4. 5.

6.9

the extent of deformation of the metal prior to recrystallization; the temperature used for the recrystallization process; the length of time of the recrystallization process; the initial grain size of the metal; the composition of the metal, in terms of metal purity.

What generalizations can be made about the recrystallization temperature with respect to (a) the degree of deformation, (b) the temperature, (c) the time of heating at temperature, (d) the final grain size, and (e) the purity of the metal? (a) In order for recrystallization to occur, the metal must possess a minimum degree of deformation. The greater the extent of deformation above this required minima, the lower the temperature required for recrystallization. (b) Increasing the temperature decreases the time required for complete recrystallization. (c) Increasing the rate of heating raises the recrystallization temperature. (d) The final grain size depends primarily upon the original extent of deformation; the greater the degree of deformation, the lower the annealing temperature required for recrystallization. (e) With decreasing metal purity, the recrystallization temperature rises. Thus, solidsolution alloying additions increase the recrystallization temperature.

6.10

If it takes 115 h to 50 percent recrystallize an 1100-H18 aluminum alloy sheet at 250ºC and 10 h at 285ºC, calculate the activation energy in kilojoules per mole for this process. Assume an Arrhenius-type rate behavior. Assuming an Arrhenius-type rate behavior, t = CeQ / Rt , a system of two equations can be used to find the actuation energy: t1 = CeQ / RT1 where t1 = 115 h = 6900 min; T1 = 250
C = 523 K t2 = CeQ / RT2 where t2 = 10 h = 600 min; T2 = 285
C = 558 K

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Dividing these equations, term by term, t1  Q  1 1   = exp   −   t2  R  T1 T2     1 6900 min Q 1   = exp  −   600 min  8.314 J/mol ⋅ K  523 K 558 K  

{

}

11.5 = exp (1.443 × 10−5 )Q Q= 6.11

ln(11.5) = 169, 255 J/mol = 169.3 kJ/mol 1.443 × 10−5

If it takes 12.0 min to 50 percent recrystallize a piece of high-purity copper sheet at 140ºC and 200 min at 88ºC, how many minutes are required to recrystallize the sheet 50 percent at 100ºC? Assume an Arrhenius-type rate behavior. Since all three cases pertain to 50 percent recrystallization, the case data can be used to assess the time required for 100ºC. First, Q must be calculated for the given data: t1 = 12 min = CeQ / RT1 = CeQ /[(413 K) R ] t2 = 200 min = CeQ / RT2 = CeQ /[(361 K) R ]

Dividing,   1 12 min 1   Q = exp  −   200 min  8.314 J/mol ⋅ K  413 K 361 K   ln(0.06) =  (−4.195 × 10−5 )Q  Q=

−2.813 = 67, 056 J/mol −4.195 × 10−5 mol/J

Q may now be used to determine the time required at 100ºC (373 K): 12 min 1    67, 056 J/mol  1 = exp  −   t2  8.314 J/mol ⋅ K  413 K 373 K   t2 = 6.12

12 min = 97.4 min e−2.094

If it takes 80 h to completely recrystallize an aluminum sheet at 250ºC and 6 h at 300ºC, calculate the activation energy in kilojoules per mole for this process. Assume an Arrhenius-type rate behavior.

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Assuming Arrhenius-type behavior, t1 = 80 h = CeQ / RT1 = CeQ /[(523 K) R ] t2 = 6 h = CeQ / RT2 = CeQ /[(573 K) R ]

Dividing,   1 80 h Q 1   = exp  −   6h  8.314 J/mol ⋅ K  523 K 573 K  

{

}

ln(13.33) = (2.007 × 10−5 )Q Q=

6.13

2.59 = 129, 048 J/mol=129 kJ/mol 2.007 ×10 −5 mol/J

What are the characteristics of the surface of a ductile fracture of a metal? A ductile fracture surface has a dull, fibrous appearance and often resembles a “cup-andcone” configuration.

6.14

Describe the three stages in the ductile fracture of a metal. The three stages consist of: 1. The specimen elongates, forming a necked region in which cavities form. 2. The cavities coalesce in the neck center, forming a crack which propagates toward the specimen surface in a direction perpendicular to the applied stress. 3. As the crack approaches the surface, its growth direction shifts to 45º with respect to the tension axis. This redirection allows for the formation of the cup-and-cone configuration and facilitates fracture.

6.15

What are the characteristics of the surface of a brittle fracture of a metal? A brittle fracture surface typically appears shiny with flat facets which are created during cleavage fracture.

6.16

Describe the three stages in the brittle fracture of a metal. The three stages of brittle fracture are: 1. As a result of plastic deformation, dislocations become concentrated along slip planes; 2. Shear stresses increase in areas where dislocations are impeded from movement. As a result, microcracks are nucleated. 3. Microcracks propagate as a result of further increases in shear stress. Stored elastic strain energy can also contribute to the propagation.

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6.17

Describe the simple impact test that uses a Charpy V-notch sample. In the Charpy V-notch test, a heavy hammer attached to a pendulum is released from a known height, and during its downward swing, strikes and fractures a notched specimen. The energy absorbed by the specimen, a measure of toughness, is calculated based upon the change in potential energy of the pendulum.

6.18

How does the carbon content of a plain-carbon steel affect the ductile-brittle transition temperature range? As the carbon content of the steel increases, the ductile-brittle transition temperature range increases in terms of both the width of the range and the temperature values.

6.19

Determine the critical crack length for a through crack contained within a thick plate of 7075-T751 aluminum alloy that is under uniaxial tension. For this alloy KIC = 22.0 ksi in. and σf = 82.0 ksi. Assume Y = π . K IC = Y σ f π a 1K a =  IC π  Y σ f

2

2

 1  22.0 ksi in.   =   = 0.0073 in.  π  π (82.0 ksi) 

For an internal through crack, the critical length is ac = 2a = 0.015 in. 6.20

Determine the critical crack length for a through crack in a thick plate of 7150-T651 aluminum alloy that is in uniaxial tension. For this alloy KIC = 25.5.0 ksi in. and σf = 400.0 ksi. Assume Y = π . K IC = Y σ f π a 1K a =  IC π  Y σ f

2

2

 1  25.5 MPa m  −4  =   = 4.12 × 10 m π   π (400.0 MPa) 

For an internal through crack, the critical length is ac = 2a = 8.24 ×10-4 m = 0.824 mm. 6.21

The critical stress intensity (KIC) for a material for a component of a design is 23.0 ksi in. What is the applied stress that will cause fracture if the component contains an internal crack 0.13 in. long? Assume Y = 1. The applicable crack length is a = ½ ac = (0.13 in.)/2 = 0.065 in. Substituting,

σf = Smith

K IC Y πa

=

23.0 ksi in. = 50.9 ksi (1) π (0.065 in.)

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6.22

What is the largest size (mm) internal through crack that a thick plate of aluminum alloy 7075-T651 can support at an applied stress of (a) three-quarters of the yield strength and (b) one-half of the yield strength? Assume Y = 1. From Table 6.1, for 7075-T651, K IC = 24.2 MPa m and σ YS = 495 MPa. 3 3 (a) Given σ f = σ YS , we calculate, σ f = (495 MPa) = 371.25 MPa. Assuming Y = 1, 4 4 the crack length is, 2

2

1K  1  24.2 MPa m  −3 a =  IC  =   = 1.35 ×10 m = 1.35 mm π  Y σ f  π  (1)(371.25 MPa)  Thus, for an internal through crack, ac = 2a = 2.70 mm 1 1 (b) Given σ f = σ YS = (495 MPa)= 247.5 MPa, and assuming Y = 1, 2 2

1K a =  IC π  Y σ f

2

2

 1  24.2 MPa m  −3  =   = 3.04 ×10 m = 3.04 mm π (1)(247.5 MPa)   

Thus, for an internal through crack, ac = 2a = 6.08 mm. 6.23

A Ti-6Al-4V alloy plate contains an internal through crack of 1.90 mm. What is the highest stress (MPa) that this material can withstand without catastrophic failure? Assume Y = π . The applicable crack length is a = ½ ac = (1.90 mm)/2 = 0.95 mm. And from Table 6.1, K IC = 55.0 MPa m. Substituting,

σf = 6.24

K IC Y πa

=

55.0 MPa m

π (9.5 × 10−4 m)

= 568.0 MPa

Using the equation K IC = σ f π a , plot the fracture stress (MPa) for aluminum alloy 7075-T651 versus surface crack size a (mm) for a values from 0.2 mm to 2.0 mm. What is the minimum size surface crack that will cause catastrophic failure? In order to generate the required plot, we must first calculate the fracture stress for values of a ranging from 0.2 mm to 2.0 mm using the given relation:

σf =

Smith

K IC πa

where K IC = 24.2 MPa m based on Table 6.1. Substituting,

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σf =

24.2 MPa m 13.6534 MPa m = a πa

This relation is used to calculate and plot the following values of fracture stress versus crack size. Problem 6.24: Fracture Stress vs. Surface Crack Length

Crack Length, a (mm)

Fracture Stress, σ f

0.2 0.4 0.6 0.8 1.0 1.4 1.8 2.0

965.4 682.7 557.4 482.7 431.8 364.9 321.8 305.3

(MPa)

Fracture Stress (MPa)

1000

800

600 0.76, 495

400

200 0.0

0.5

1.0

1.5

2.0

Surface Crack Length, a (mm)

The minimum surface crack length, for catastrophic failure, would correspond to a fracture stress equal to the material’s yield stress value. Thus, from Table 6.1, σ YS = 495 MPa. Substituting, 2

2

1 K  1  24.2 MPa m  −3 a =  IC  =   = 0.76 × 10 m = 0.76 mm π  σ f  π  495.0 MPa  This point has been included in the plot generated for fracture stress vs. surface crack size. One may deduce that for crack lengths above the minima of 0.76 mm, the alloy will fracture catastrophically. Whereas, for values of a less than 0.76 mm, the alloy will yield and fracture at 495 MPa in the normal manner. 6.25

Determine the critical crack length (mm) for a through crack in a thick 2042-T6 alloy plate that has a fracture toughness KIC = 23.5 MPa m and is under a stress of 300 MPa. Assume Y = 1.

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K IC = Y σ f π a

1K a =  IC π  Y σ f

2

2

 1  23.5 MPa m  −3  =   = 1.95 ×10 m = 1.95 mm π (1)(300.0 MPa)   

For an internal through crack, the critical length is ac = 2a = 3.90 mm. 6.26

Describe a metal fatigue failure. Fatigue failure in a metal is caused by the application of repetitive or cyclic loading; the part, subjected to cyclic stressing, fails often at a stress value well below its normal limit.

6.27

What two distinct types of surface areas are usually recognized on a fatigue failure surface? The two distinct types of surface areas typically observed on a fatigue failure surface are: a smooth region of “beach” marks created by rubbing action between the crack surface areas; and a rough region which is formed through the fracture process.

6.28

Where do fatigue failures usually originate on a metal section? Fatigue failures typically originate at a point of high stress concentration, such as a sharp corner or notch.

6.29

What is a fatigue test SN curve, and how are the data for the SN curve obtained? A fatigue test SN curve is a plot of the fatigue stress to which a specimen is subjected versus the corresponding cycles, or stress reversals, up to and including the point of failure. The number of cycles is plotted on a logarithmic scale while the fatigue strength is plotted on either a linear or logarithmic scale, depending on the data. The SN data is typically obtained by repeatedly subjecting a rotating specimen to reverse or fluctuating bending while counting the cycles until destruction occurs. However, specimens may also be subjected to reversed or fluctuating axial stresses, torsional stresses, or combined stresses in testing.

6.30

How does the SN curve of a carbon steel differ from that of a high-strength aluminum alloy? In an SN curve for a carbon steel, the stress at failure levels off as the number of cycles exceeds the metal’s endurance limit of approximately 106 cycles. Whereas the SN curve for a high strength aluminum alloy continues to gradually decrease as the number of cycles is increased above 106.

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6.31

A fatigue test is made with a maximum stress of 25 ksi (172 MPa) and a minimum stress of -4.00 ksi (-27.6 MPa). Calculate (a) the stress range, (b) the stress amplitude, (c) the mean stress, and (d) the stress ratio. (a) The range of stress is σ r = σ max − σ min = 25 ksi − (−4 ksi) = 29 ksi (199.8 MPa). σ (b) The stress amplitude is σ a = r = 14.5 ksi (99.9 MPa). 2 σ max + σ min 25 ksi − 4ksi = = 10.5 ksi (72.3 MPa). (c) The mean stress is σ m = 2 2 σ −4 ksi = −0.16. (d) The stress ratio is R = min = σ max 25 ksi

6.32

A fatigue test is made with a mean stress of 17,500 psi (120 MPa) and a stress amplitude of 24,000 psi (165 MPa). Calculate (a) the maximum and minimum stresses, (b) the stress ratio, and (c) the stress range. Given σ m = 17.5 ksi (120 MPa) and σ a = 24 ksi (165 MPa), (a)

2σ m = 2(17.5 ksi) = σ max + σ min 35 ksi = σ max + σ min 2σ a = 2(24.0 ksi) = σ max − σ min 48 ksi = σ max − σ min Adding these two equations and solving,

σ max = 41.5 ksi (285.9 MPa), σ min = -6.5 ksi (44.8 MPa)

6.33

σ min −6.5 ksi = = −0.157 σ max 41.5 ksi

(b)

R=

(c)

σ r = σ max − σ min = 41.5 ksi − (−6.5 ksi) = 48 ksi (330.7 MPa)

Describe the four basic structural changes that take place when a homogeneous ductile metal is caused to fail by fatigue under cyclic stresses. For a homogeneous ductile metal subjected to fatigue, four stages of structural changes are observed: 1. Crack initiation: Plastic deformation causes the onset and early development of fatigue damage. 2. Slipband crack growth (Stage I): Slipband intrusions and extrusions are created on the surface of the metal while damage along persistent slipbands

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occurs within the sample. As a result, cracks form at or near the surface and propagate along planes subjected to high shear stresses. 3. Crack growth on planes of high tensile stress (Stage II): The slow crack growth of Stage I is replaced by rapid crack propagation as the crack direction shifts to a direction perpendicular to the direction of maximum tensile stress. During this stage, striations are formed. 4. Ultimate ductile failure: The crack achieves an area sufficient to cause the rupture of the sample by ductile fracture. 6.34

Describe four major factors that affect the fatigue strength of a metal. The four primary factors which affect the fatigue strength of a metal are: stress concentration, surface roughness, surface condition and environment.

6.35

A large flat plate is subjected to constant-amplitude uniaxial cyclic tensile and compressive stresses of 120 and 35 MPa, respectively. If before testing the largest surface crack is 1.00 mm and the plain-strain fracture toughness of the plate is 35 MPa m , estimate the fatigue life of the plate in cycles to failure. For the plate m = 3.5 and A = 5.0 × 10-12 in MPa and meter units. Assume Y = 1.3. The final crack length, from the fracture toughness equation, is: 1K a f =  IC π  Y σ f

2

2

 1  35 MPa m   =   = 0.01602 m = 16.02 mm  π  (1.3)(120 MPa) 

The fatigue life in cycles, Nf , is then, Nf =

a −f ( m / 2)+1 − a0− ( m / 2)+1

[ −(m / 2) + 1] Aσ mπ m / 2Y m

(0.0160 m)−1.75+1 − (0.001 m)−1.75+1 = [ −1.75 + 1] (5.0 ×10−12 )(120 MPa)3.5π 1.75 (1.3)3.5

= 1.18×105 cycles

6.36

Same as Prob. 6.35 with tensile stress of 70 MPa. If the initial and critical crack lengths are 1.25 mm and 12 mm in the plate and the fatigue life is 2.0 × 106 cycles, calculate the maximum tensile stress in MPa that will produce this life. Assume m = 3.0 and A = 6.0 × 10-13 in MPa and meter units. Assume Y = 1.2. The maximum tensile stress is calculated as:

σm =

Smith

a −f ( m / 2)+1 − a0− ( m / 2)+1

[ −(m / 2) + 1] Aπ m / 2Y m N f

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3 − +1

3 − +1

(0.012 m) 2 − (0.00125 m) 2 = 3.318 ×106 σ3 = 3   −13 1.5 3 6  − 2 + 1 (6.0 × 10 )(π )(1.2) (2.0 ×10 )

σ = 149 MPa 6.37

Same as Prob. 6.35. Compute the final critical surface crack length if the fatigue life must be a minimum of 7.0 × 105 cycles. Assume the initial maximum edge surface crack length of 1.80 mm and a maximum tensile stress of 160 MPa. Assume m = 1.8 and A = 7.5 × 10-13 in MPa and meter units. Assume Y = 1.25. The final critical crack length can be using the equation for fatigue life, Nf =

a −f ( m / 2)+1 − a0− ( m / 2)+1

[ −(m / 2) + 1] Aσ mπ m / 2Y m −

7 ×105 = 5

7 ×10 =

af

1.8 +1 2

− (0.0018 m)

−

1.8 +1 2

[ −(1.8 / 2) + 1] (7.5 ×10−13 )(160 MPa)1.8π 1.8 / 2 (1.25)1.8 a 0.1 f − 0.5315

2.913 × 10−9 0.5336 = a 0.1 f a f = 0.00187 m = 1.87 mm 6.38

Same as Prob. 6.35. Compute the critical surface crack length if fatigue life must be 8.0 × 106 cycles and maximum tensile stress is 21,000 psi. Assume m = 3.5 and A = 4.0 × 10-13 in ksi and inch units. Initial crack (edge) is 0.120 in. Y = 1.15. Nf =

a −f ( m / 2)+1 − a0− ( m / 2)+1

[ −(m / 2) + 1] Aσ mπ m / 2Y m −

8 × 106 = 8 × 106 =

af

3.5 +1 2

− (0.12 in. )

−

3.5 +1 2

[ −(3.5 / 2) + 1] (4.0 ×10−13 )(21 ksi)3.5π 3.5 / 2 (1.15)3.5 a −f 0.5 − 4.9047 −1.53937 × 10−7

a f = (3.67322)−1/ 0.75 = 0.176 in. 6.39

What is metal creep?

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Creep of a metal refers to the slow, progressive plastic deformation of a metal subjected to a constant load or stress. Thus creep is the time dependent strain of a metal. 6.40

For which environmental conditions is the creep of metals especially important industrially? Creep is of particular importance to industrial applications involving high stress and high temperature environments.

6.41

Draw a typical creep curve for a metal under constant load and at a relatively high temperature, and indicate on it all three stages of creep.

6.42

The following creep data were obtained for a titanium alloy at 50 ksi and 400ºC. Plot the creep strain versus time (hours) and determine the steady-state creep rate for these test conditions.

Smith

Strain (in./in.)

Time (h)

Strain (in./in.)

Time (h)

0.010 × 10-2 0.030 × 10-2 0.050 × 10-2

2 18 40

0.075 × 10-2 0.090 × 10-2 0.110 × 10-2

80 120 160

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Prob. 6.42: Rate of Creep Strain 1.2E-03

Strain (in./in.)

1.0E-03 8.0E-04 6.0E-04 4.0E-04 2.0E-04 0.0E+00 0

25

50

75

100

125

150

175

Time (h)

Creep Rate =

6.43

∆ε ∆t

=

(1.1 × 10−3 ) − (0.76 × 10−3 ) = 4.53 × 10−6 in./in./h 160 h − 85 h

Equiaxed MAR-M 247 alloy is to support a stress if 276 MPa (Fig. 6.36). Determine the time to stress rupture at 850ºC. From Fig. 6.36, for a stress of 276 MPa, the L.M. parameter value for Equiaxed MAR-M 247 is 26 × 103 K·h. Thus, P = T ( K )(20 + log tr ), T = 850
C + 273 = 1123 K 26,000 = (1123 K)(20 + log tr ) log tr = 3.152 tr = 1419 h

6.44

DS MAR-M 247 alloy (Fig. 6.36) is to support a stress of 207 MPa. At what temperature (ºC) will the stress rupture lifetime be 210 h? From Fig. 6.36, for a stress of 207 MPa, the L.M. parameter value for DS MAR-M 247 alloy is 27.5 × 103 K·h. Thus, P = T ( K )(20 + log tr )

27,500 = [ T(K)] (20 + log (210 h)) T(K) =

6.45

27,500 = 1232 K = 959o C 22.32

If DS CM 247 LC alloy (right curve of Fig. 6.36) is subjected to a temperature of 960ºC for three years, what is the maximum stress that it can support without rupturing?

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Given tr = 3 yr = 26,280 h and T = 960
C + 273 = 1233 K , P = 1233(log 26,280 + 20) = 30,109

From Fig. 6.36, for P = 30.1 × 103, the stress is approximately 96 MPa. 6.46

A Ti-6Al-2Sn-4Zr-6Mo alloy is subjected to a stress of 20,000 psi. How long can the alloy be used under this stress at 500ºC so that no more than 0.2% creep strain will occur? Use Fig. 6.37. From Fig. 6.37, for a stress of 20 ksi, the L.M. parameter for 2% stress is P ≈ 31,000. Thus, P = T (R)(20 + log t0.2 ), T = 500
C=932
F + 460 = 1392
R 31,000 = (1392
R)(20 + log t0.2 ) log t0.2 = 2.270 t0.2 = 186.2 h

6.47

Gamma titanium aluminide is subjected to a stress of 50,000 psi. If the material is to be limited to 0.2% creep strain, how long can this material be used at 1100ºF? Use Fig. 6.37. From Fig. 6.37, for a stress of 50 ksi, the L.M. parameter for 2% stress is P ≈ 36,600. Thus, P = T (R)(20 + log t0.2 ), T = 1100
F + 460 = 1560
R 36,600 = (1560
R)(20 + log t0.2 ) log t0.2 = 3.4615 t0.2 = 2894.0 h ≈ 121 days

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