Essentials of materials science and engineering si edition 3rd edition askeland solutions manual by diane.agustin567

Essentials of Materials Science and Engineering SI Edition 3rd Edition Askeland Solutions Manual Visit to download the full and correct content document: https://testbankdeal.com/dow nload/essentials-of-materials-science-and-engineering-si-edition-3rd-edition-askeland -solutions-manual/

Chapter 6 Mechanical Properties: Part One 6–3

An 3780 N force is applied to a 0.375 cm-diameter nickel wire having a yield strength of 310 MPa and a tensile strength of 379 MPa. Determine (a) whether the wire will plastically deform; and (b) whether the wire will experience necking. Solution: (a) First determine the stress acting on the wire: S = F/A0 = 3780 N / [(π/4)(0.375 cm)2] = 342 MPa Because S is greater than the yield strength of 310 MPa, the wire will plastically deform. (b) Because S is less than the tensile strength of 379 MPa, no necking will occur.

6–4

A force of 100,000 N is applied to an iron bar with a cross-sectional area of 10 mm × 20 mm and having a yield strength of 400 MPa and a tensile strength of 480 MPa. Determine (a) whether the bar will plastically deform; and (b) whether the bar will experience necking. Solution:

(a) First determine the stress acting on the wire: S = F/A0 = 100,000 N / [(10 mm)(20 mm)] = 500 N/mm2 = 500 MPa Because S is greater than the yield strength of 400 MPa, the wire will plastically deform. (b) Because S is greater than the tensile strength of 480 MPa, the wire will also neck.

6–5

Calculate the maximum force that a 0.5 cm-diameter rod of Al2O3, having a yield strength of 241 MPa, can withstand with no plastic deformation. Express your answer in N. Solution:

F = SA0 = (241 MPa)(π/4)(0.5 cm)2 = 47.4 × 102 N

6–6

A force of 20,000 N will cause a 1 cm × 1 cm bar of magnesium to stretch from 10 cm to 10.045 cm. Calculate the modulus of elasticity, in GPa. Solution: The strain e is e = (10.045 cm – 10 cm)/10 cm = 0.0045 cm/cm The stress S is S = 20,000 N/[(10 mm)(10 mm)] = 200 N/mm2 = 200 MPa E = S/e = 200 MPa / (0.0045 cm/cm) = 44,444 MPa = 44.4 GPa

6–7

A polymer bar’s dimensions are 2.5 cm × 5 cm × 37.5 cm. The polymer has a modulus of elasticity of 4137 MPa. What force is required to stretch the bar elastically from 37.5 cm to 38.13 cm? Solution: The strain e is e = (38.13 cm – 37.5 cm) / (37.5 cm) = 0.016 cm/cm The stress S is S = Ee = (4137 MPa)(0.016 cm/cm) = 66.2 MPa The force is then F = SA0 = (66.2 MPa) (2.5 cm)(5 cm) = 827.5×102 N

6–8

An aluminum plate 0.5 cm thick is to withstand a force of 50,000 N with no permanent deformation. If the aluminum has a yield strength of 125 MPa, what is the minimum width of the plate? Solution: The area is A0 = F/S = 50,000 N / (125 N/mm2) = 400 mm2 The minimum width is w = A0/t = (400 mm2)(0.1 cm/mm)2/(0.5 cm) = 8 cm

6–9

A steel cable 3.13 cm in diameter and 1500 cm long is to lift a 18,140 kg load. What is the length of the cable during lifting? The modulus of elasticity of the steel is 207 × 103 MPa. Solution: The stress is S = F/A0 = (18,140 kg)/[π/4(3.1 cm)2] = 235.7 MPa The strain is e = S/E = 235.7 MPa / (207 × 103 MPa) = 0.0011 cm/cm e = (lf – 1500 cm) / 1500 cm = 0.0011 cm/cm lf = 1502 cm

6–10 (a) Carbon nanotubes are one of the stiffest and strongest materials known to scientists and engineers. Carbon nanotubes have an elastic modulus of 1.1 TPa (1 TPa = 1012 Pa). If a carbon nanotube has a diameter of 15 nm, determine the engineering stress sustained by the nanotube when subjected to a tensile load of 4 μN (1 μN = 10–6 N) along the length of the tube. Assume that the entire crosssectional area of the nanotube is load bearing. (b) Assume that the carbon nanotube is only deformed elastically (not plastically) under the load of 4 μN. The carbon nanotube has a length of 10 µm (1 μm = 10–6 m). What is the tensile elongation (displacement) of the carbon nanotube in nanometers (1 nm = 10–9 m)? Solution: (a) The engineering stress is given by F A0

where S is the engineering stress, F is the applied load, and A0 is the initial cross–sectional area of the nanotube. A0 is equal to A0 =

π 4

where D is the nanotube diameter. Substituting into the expression for S,

π 4

F D2

π 4

4 ×10−6 N

= 22.64 × 109 Pa = 22.64 GPa

(15 ×10−9 m)2

(b) According to Hooke’s Law, S = Ee where E is the elastic modulus and the engineering strain e is given by e=

Δl l0

where Δl is the elongation and l0 is the initial nanotube length. Rearranging,

Sl0 (22.64 × 109 Pa) (10 × 10−6 m) Δl = = = 2.06 × 10−7 m = 206 nm 12 E (1.1 × 10 Pa)

6–13

Derive the expression ε = ln(1 + e), where ε is the true strain and e is the engineering strain. Note that this expression is not valid after the onset of necking. Solution: True strain ε is given by

⎛l ⎞ ⎝ l0 ⎠

ε = ln ⎜ ⎟ where l is the instantaneous length and l0 is the initial length. The instantaneous length can be written l = l0 + Δl where Δl is the change in length. Substituting,

⎛l ⎞ ⎝ l0 ⎠

⎛ l + Δl ⎞ ⎛ Δl ⎞ ⎟ = ln ⎜1 + ⎟ l0 ⎠ ⎝ l0 ⎠ ⎝

ε = ln ⎜ ⎟ = ln ⎜ 0

The engineering strain e is given by e=

Δl l0

such that

⎛

ε = ln ⎜1 + ⎝

6–14

Δl ⎞ ⎟ = ln(1 + e) l0 ⎠

The following data were collected from a standard 1.263 cm-diameter test specimen of a copper alloy (l0 = 5 cm): Load (N) 0 13,340 26,690 33,360 40,030 46,700 53,380 55,160 50,710

Δ/ (cm) 0.00000 0.00418 0.00833 0.01043 0.0225 0.1 0.65 1.25 (maximum load) 2.55 (fracture)

After fracture, the total length was 7.535 cm, and the diameter was 0.935 cm. Plot the engineering stress–strain curve and calculate (a) the 0.2% offset yield strength; (b) the tensile strength; (c) the modulus of elasticity; (d) the % elongation; 84 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly available website, in whole or in part.

(e) the % reduction in area; (f) the engineering stress at fracture; (g) the true stress at necking; and (h) the modulus of resilience. Solution:

S = F /[(π/4)(1.26 × 10–2)2] = F/1.25 × 10–4 e = Δl / 5

(a) 0.2% offset yield strength = 320.3 MPa (b) tensile strength = 441.2 MPa (c) E = (213.5 – 0) / (0.001666 – 0) = 12.8 × 104 MPa (d) % elongation =

(7.535 − 5) ×100 = 50.7% 5

(e)

(π/4)(1.26) 2 − (π/4)(0.935) 2 ×100 (π/4)(1.26) 2 = 45.3%

% reduction in area =

(f) Engineering stress at fracture = 405.7 MPa (g)

σ = S (1 + e ) = 55,160 / ⎡⎢(π / 4 ) (1.26 ×10 −2 ) ⎤⎥ (1 + 1.25 ×10 −2 / 5 ) = 534 MPa 2

⎣

⎦

(h) From the graph, yielding begins at about 266.9 MPa. Thus: ½(yield strength)(strain at yield) = ½ (266.9)(0.00208) = 0.28 MPa

6–15

The following data were collected from a 1 cm-diameter test specimen of polyvinyl chloride (l0 = 5 cm): Load (N) 0 1334 2669 4003 5338 6672 7384 7117 6316

Δ/ (cm) 0.00000 0.0187 0.0374 0.0594 0.08 0.115 0.175 (maximum load) 0.235 0.3 (fracture)

After fracture, the total length was 5.23 cm, and the diameter was 0.983 cm. Plot the engineering stress-strain curve and calculate (a) the 0.2% offset yield strength; (b) the tensile strength; (c) the modulus of elasticity; (d) the % elongation; (e) the % reduction in area; (f) the engineering stress at fracture; (g) the true stress at necking; and (h) the modulus of resilience. Solution:

S = F / [(π/4)(1 × 10–2)2] = F/0.79 × 10–2 e = (Δl – 5) / 5

(a) 0.2% offset yield strength = 80 MPa (b) tensile strength = 90.1 MPa

(5.23 − 5) ×100 = 4.6% 5

(e) % reduction in area =

(π/4)(1) 2 − (π/4)(0.983) 2 ×100 = 3.4% (π/4)(1) 2

(f) engineering stress at fracture = 80 MPa 2 (g) σ =S (1+e ) = 7384 / ⎡( π/4 )(1) ⎤ (1+0.175/5) = 94 MPa ⎣ ⎦

(h) From the figure, yielding begins near 65.9 MPa. Thus: ½(yield strength)(strain at yield) = ½ (65.9)(0.016) = 0.53 MPa 6–16

The following data were collected from a 12-mm-diameter test specimen of magnesium (l0 = 30.00 mm): Load (N) 0 5000 10,000 15,000 20,000 25,000 26,500 27,000 26,500 25,000

Δ/ (mm) 0.0000 0.0296 0.0592 0.0888 0.15 0.51 0.90 1.50 (maximum load) 2.10 2.79 (fracture)

After fracture, the total length was 32.61 mm, and the diameter was 11.74 mm. Plot the engineering stress-strain curve and calculate (a) the 0.2% offset yield strength; (b) the tensile strength; (c) the modulus of elasticity; (d) the % elongation; (e) the % reduction in area; (f) the engineering stress at fracture; (g) the true stress at necking; and (h) the modulus of resilience.

Solution::

S = F / [(π/4)(12 mm)2] = F/113.1 e = (Δl – 30.00)/30.00

(a) 0.2% offset yield strength = 186 1 MPa (b) tensile strength = 27,000/113.1 = 239 MP Pa (c) E = (115,000/113.11 – 0) / (0.08888/30 – 0) = 44,800 MP Pa E = 44.8 GPa (d) % eloongation =

(332.61 − 30) × 100 = 8.7% % 30

(e) % redduction in areaa = (π /4) (12)) − (π /4)2 (11..74) ×100 = 4.3% 4 (π /4) (12) 2

(f) enginneering stresss at fracture = 221 MPa 2 (g) σ = S (1 + e ) = 27 7,000 / ⎡⎣(π / 4 ) (12 ) ⎤⎦ (1 + 1.5 / 30.00 ) = 251 MPaa

(h) From the figure,, yielding begins b near 133 MPa. Thus: ½(yiield M strenggth)(strain att yield) = ½ (133)(0.00296) = 0.2 MPa

6–17

The follo owing data were colleccted from a 20-mm-diaameter test specimen of o a ductile caast iron (l0 = 40.00 mm): Load (N) 0 25,000 0 50,000 0 75,000 0 90,000 0 105,00 00 120,00 00 131,00 00 125,00 00

Δ/ (m mm) 0..0000 0..0185 0..0370 0..0555 0..20 0..60 1..56 4..00 (maximuum load) 7..52 (fracturee)

After fraccture, the total length waas 47.42 mm m, and the diaameter was 18.35 mm. Plot P the engineeering stresss–strain curvve and calcullate (a) the 0.2 2% offset yield strength;; (b) the ten nsile strengthh; (c) the mo odulus of elaasticity; (d) the % elongation; (e) the % reduction inn area; (f) the eng gineering strress at fractuure; (g) the tru ue stress at necking; n and (h) the mo odulus of ressilience. Solution::

S = F / [(π/4)(20 [ mm m)2] = F/3144.2

e = (Δl – 40.00)/40.000 (a) 0.2% % offset yielld strength = 274 MPa

(b) tensile strength = 131,000/314.2 = 417 MPa (c) E = (75,000/314.2 – 0) / (0.0055/40.00 – 0) = 172,000 MPa = 172 GPa (d) % elongation =

(47.42 − 40) × 100 = 18.55% 40

(e) % reduction in area =

(π /4) (20)2 − (π /4) (18.35)2 ×100 = 15.8% (π /4) (20)2

(f) engineering stress at fracture = 398 MPa

σ = S (1 + e) = 131, 000 / ⎡⎣( π / 4 ) ( 20 ) ⎤⎦ (1 + 4.00 / 40.00 ) 2

(g)

σ = 459 MPa

(h) From the figure, yielding begins near 240 MPa. Thus, ½(yield strength)(strain at yield) = ½ (240)(0.001388) = 0.17 MPa 6–18

The following data were collected from a test specimen of cold-rolled and annealed brass. The specimen had an initial gage length l0 of 35 mm and an initial crosssectional area A0 of 10.5 mm2. Load (N) 0 66 177 327 462 797 1350 1720 2220 2690 2410

Δl (mm) 0.0000 0.0112 0.0157 0.0199 0.0240 1.72 5.55 8.15 13.07 22.77 (maximum load) 25.25 (fracture)

(a) Plot the engineering stress–strain curve and the true stress–strain curve. Since the instantaneous cross-sectional area of the specimen is unknown past the point of necking, truncate the true stress–true strain data at the point that corresponds to the ultimate tensile strength. Use of a software graphing package is recommended. (b) Comment on the relative values of true stress–strain and engineering stress– strain during the elastic loading and prior to necking. (c) If the true stress–strain data were known past the point of necking, what might the curve look like? (d) Calculate the 0.2% offset yield strength. (e) Calculate the tensile strength. (f) Calculate the elastic modulus using a linear fit to the appropriate data. 90 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly available website, in whole or in part.

(a a) The tablee below conntains the vaalues for enngineering sttress and strrain S Solution: and true stress s and strrain.

Load (N N) 0 66 177 327 462 797 1350 1720 2220 2690 2410

Disp placement (mm m) 0.0000 0.0112 0.0157 0.0199 0.02440 1.72 5.55 8.15 13.07 22.77 25.25

Engineerin ng Strain 0 0.000320 0.000449 0.000569 0.000686 0.049143 0.15857 0.23286 0.37343 0.65057 0.72143

Engineeering Stress (MPa) 0 6.29 16.86 31.14 44.00 75.91 129 164 211 256 230

True Straain 0 0.0000320 0.0000448 0.0000568 0.0000685 0.0447974 0.144719 0.200933 0.31731 0.500112 0.544315

True Streess (MP Pa) 0 6.299 16.87 31.17 44.003 79.664 149 202 290 423 ******

A ploot of the engineering streess and strain and true stress and strrain is shoown below. The T true streess-strain daata is truncatted at the pooint that corresponds to t the ultimaate tensile strrength.

a (b) Durinng the elasttic loading, the values of true strress–strain and enginneering streess–strain are a nearly identical. Upon cllose inspecction, howeever, the truue stress vaalues are grreater than the enginneering stresss values, andd the true strrain values are a less than the enginneering strainn values. These T differeences becom me much more m pronoounced past the point of yieldingg. In tensioon, the crooss– sectioonal area beccomes smaller as the teest proceeds. Thus the true t stress, which is computed c using the insstantaneous cross–sectioonal area rather r than the initial cross–section c nal area, is larger than the enginneering stresss. In tensioon, the lengtth of the saample becom mes longer during thhe test. Thhus by accoounting forr instantaneous 91 © 2014 Cengage Learning. All Rights Reservved. May not be scanneed, copied, or duplicated, or posted to a publlicly available website, in whole or in part.

changges in lengthh, the true strain is sm maller than the t engineerring strainn prior to neccking. Thesee trends holdd for every tension test of o a metallic material. (c) If the true stress--strain data were w knownn past the pooint of neckiing, c wouldd not have a maximum m. The truee stress woould the curve continnually increaase. (d) The 0.2% 0 offset yield y strengthh is approxim mately 46 MPa. M (e) The ultimate u tennsile strengthh is the maaximum enggineering strress sustaiined during a tension test. t This vaalue can be read from the table or the plot as a 256 MPa. (f) The best b estimatee is obtained using a lineear fit to the data during the elasticc loading. The T elastic modulus m dettermined in this fashionn is 105 GPa. G 6 6–19

Consiider the tenssile stress–sttrain diagram ms in Figuree 6–28 labeled 1 and 2 and a answeer the follow wing questionns. These diiagrams are typical of metals. m Consiider each part p as a sepparate questiion that has no n relationshhip to previoous parts of the questiion.

Figure 6––28 Stress–sstrain curvess for Problem m 6–36. (a) Saamples 1 and 2 are idenntical exceptt for the graain size. Whhich sample has the smaller graains? How do d you know w? (b) Saamples 1 annd 2 are iddentical exceept that theyy were testted at differrent tem mperatures. Which wass tested at the lower temperature? t ? How do you y kn now? (c) Saamples 1 andd 2 are differrent materialls. Which saample is touggher? Explaiin. (d) Saamples 1 andd 2 are identtical except that t one of thhem is a purre metal and the oth her has a small s percenntage alloyinng addition. Which sam mple has been allloyed? How w do you know? (e) Giiven the streess–strain cuurves for maaterials 1 andd 2, which material m has the low wer hardnesss value on thhe Brinell haardness scalee? How do you y know? 92 © 2014 Cengage Learning. All Rights Reservved. May not be scanneed, copied, or duplicated, or posted to a publlicly available website, in whole or in part.

(f) Are the stress–strain curves shown true stress–strain or engineering stress– strain curves? How do you know? (g) Which of the two materials represented by samples 1 and 2 would exhibit a higher shear yield strength? How do you know? Solution: (a) Sample 1 has the smaller grains. Smaller grains result in more grain boundaries. Grain boundaries hinder the motion of dislocations. A higher stress must then be applied to cause dislocations to continue to move and change the shape of the material. This makes the material stronger. Sample 1 is stronger than Sample 2; therefore, all other factors being identical, Sample 1 has the smaller grains. (b) Sample 1 was tested at the lower temperature because it has the higher strength and lower ductility. As temperature increases, it becomes easier for dislocations to move leading to lower strengths and higher ductilities. Since Sample 1 is stronger and less ductile than Sample 2, it was tested at the lower temperature. (c) Sample 2 is tougher. Toughness is the ability to absorb energy prior to fracture. Toughness is a combination of strength and ductility and is represented graphically by the area under the stress–strain curve. The area under the stress–strain curve of Sample 2 is larger; therefore, Sample 2 is tougher. (d) Sample 1 has been alloyed. Introducing a solute to the metal increases the strength. Solute atoms hinder the motion of dislocations. A higher stress must then be applied to cause dislocations to continue to move and change the shape of the material. This makes the material stronger. Sample 1 is stronger than Sample 2; therefore, Sample 1 has been alloyed. (e) Sample 2 has the lower hardness value on the Brinell hardness scale. Hardness is directly proportional to strength. Sample 2 is weaker than Sample 1; therefore, it has the lower hardness. (f) The stress–strain curves shown represent true stress and true strain because they do not exhibit maxima as do tensile engineering stress and engineering strain curves. (g) Sample 1 will have the higher shear yield strength because it has the higher tensile yield strength.

6–20

A bar of Al2O3 that is 0.625 cm thick, 1.25 cm wide, and 22.5 cm long is tested in a three-point bending apparatus, with the supports located 15 cm apart. The deflection of the center of the bar is measured as a function of the applied load. The data are shown below. Determine the flexural strength and the flexural modulus. Force (N) 64.5 128.5 193 257.5 382.5

Deflection (cm) 0.0063 0.013 0.019 0.025 0.037 (fracture)

Solution: stress = 3LF /( 2wh 2 ) = (3)(15×10−2 )F/[(2)(1.25×10−2 )(0.626×10−2 ) = 46.1× 104 F

The flexural strength is the stress at fracture or 46.1×104 F = 46.1 × 104 × 382.5 = 176.3 MPa The flexural modulus can be calculated from the linear curve; picking the first point as an example:

FM = 6–21

FL3 (64.5 N)(15×10−2 )3 = 28.3×104 MPa = 4wh3δ (4)(1.25×10−2 )(0.625×10−2 )(0.0063×10−2 )

A three-point bend test is performed on a block of ZrO2 that is 20 cm long, 1.25 cm wide, and 0.625 cm thick and is resting on two supports 10 cm apart. When a force of 1780 N is applied, the specimen deflects 0.093 cm and breaks. Calculate (a) the flexural strength; and (b) the flexural modulus, assuming that no plastic deformation occurs.

Solution:

(3)(1780 N)(10×10−2 m) (a) flexuralstrength = 3FL /(2wh ) = (2)(1.25×10−2 m)(0.625×10−2 m) 2

= 546.6 MPa

(b) flexural modulus = FL3 /(4wh3δ ) =

(1780 N)(10×10−2 m)3 (4)(1.25×10 −2 m)(0.625×10−2 m)(0.093×10−2 m)

=15.7×104 MPa

6–22

A three-point bend test is performed on a block of silicon carbide that is 10 cm long, 1.5 cm wide, and 0.6 cm thick and is resting on two supports 7.5 cm apart. The sample breaks when a deflection of 0.09 mm is recorded. The flexural modulus for silicon carbide is 480 GPa. Assume that no plastic deformation occurs. Calculate (a) the force that caused the fracture; and (b) the flexural strength. Solution: (a) The force F required to produce a deflection of 0.09 mm is

F = (flexural modulus)(4wh3δ)/L3 F = (480,000 MPa)(4)(15 mm)(6 mm)3(0.09 mm) / (75 mm)3 F = 1327 N (b) Flexural strength = 3 FL/(2wh2) = (3)(1327 N)(75 mm)/[(2)(15 mm)(6 mm)2] = 276 MPa 6–23

A thermosetting polymer containing glass beads is required to deflect 0.5 mm when a force of 500 N is applied. The polymer part is 2 cm wide, 0.5 cm thick, and 10 cm long. If the flexural modulus is 6.9 GPa, determine the minimum distance between the supports. Will the polymer fracture if its flexural strength is 85 MPa? Assume that no plastic deformation occurs. Solution: The minimum distance L between the supports can be calculated from the flexural modulus. L3 = 4wh3δ(flexural modulus)/F = (4)(20 mm)(5 mm)3(0.5 mm)(6.9 GPa)(1000 MPa/GPa)/500 N

L3 = 69,000 mm3 or L = 41 mm The stress acting on the bar when a deflection of 0.5 mm is obtained is

σ = 3FL/(2wh2 ) = (3)(500 N)(41 mm)/[(2)(20 mm)(5 mm)2 ] = 61.5 MPa The applied stress is less than the flexural strength of 85 MPa; the polymer is not expected to fracture. 95 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly available website, in whole or in part.

6–24

The flexural modulus of alumina is 310.3 × 103 MPa and its flexural strength is 317 MPa. A bar of alumina 0.75 cm thick, 2.5 cm wide, and 25 cm long is placed on supports 17.5 cm apart. Determine the amount of deflection at the moment the bar breaks, assuming that no plastic deformation occurs. Solution: The force required to break the bar is

F = 2wh2(flexural strength)/(3L) F = (2)(2.5×10െ2m)(0.75×10െ2m)(317 MPa)/[(3)(17.5×10െ2 m)] = 17×102 N The deflection just prior to fracture is δ = FL3/[(4wh3)(flexural modulus)] δ = (17×10൅2 N)(17.5×10െ2m)3/[(4)(2.5×10െ2m) (0.75 ×10െ2m)3(310.3×103 MPa)] = 7×10-4m 6–26

A Brinell hardness measurement, using a 10-mm-diameter indenter and a 500-kg load, produces an indentation of 4.5 mm on an aluminum plate. Determine the Brinell hardness number (HB) of the metal. Solution:

6–27

HB =

500 kg (π /2) (10 mm) [10 − 102 − 4.52 ]

= 29.8

When a 3000-kg load is applied to a 10-mm-diameter ball in a Brinell test of a steel, an indentation of 3.1 mm is produced. Estimate the tensile strength of the steel. Solution:

HB =

3000 kg (π /2) (10 mm) [10 − 102 − 3.12 ]

= 388

Tensile strength = 3.38 HB = (3.38)(388) = 1311.44 MPa 6–29

The elastic modulus of a metallic glass is determined to be 95 GPa using nanoindentation testing with a diamond Berkovich tip. The Poisson’s ratio of the metallic glass is 0.36. The unloading stiffness as determined from the loaddisplacement data is 5.4 × 105 N/m. The maximum load is 120 mN. What is the hardness of the metallic glass at this indentation depth? Solution: The hardness of a material as determined by nanoindentation is H=

Pmax Ac

where Pmax is the maximum load imposed over the contact area. The maximum load is given in the problem statement, but the contact area must be determined. The reduced elastic modulus Er is related to the contact area by 96 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly available website, in whole or in part.

Er =

π S 2 β Ac

where S is the unloading stiffness and β is a constant for the shape of the indenter being used (β = 1.034 for a Berkovich indenter). The reduced elastic modulus is also given by

1 1 − ν 2 1 − ν i2 = + Er E Ei where E and v are the elastic modulus and Poisson’s ratio of the material being indented, respectively, and Ei and vi are the elastic modulus and Poisson’s ratio of the indenter, respectively (for diamond, Ei = 1.141 TPa and vi = 0.07). Substituting to find the reduced elastic modulus, 1 1 − 0.362 1 − 0.07 2 = + Er 95 × 109 Pa 1.141 × 1012 Pa Er = 99.66 × 109 Pa. Solving for the contact area,

π S2 π (5.4 ×105 N/m)2 Ac = 2 2 = = 2.157 ×10−11 m2 2 9 2 4β Er 4(1.034 )(99.66 ×10 Pa) Thus, the hardness is

H= 6–30

Pmax 120 ×10−3 N = = 5.6 ×109 Pa = 5.6 GPa −11 2 2.157 ×10 m Ac

The following data were obtained from a series of Charpy impact tests performed on four steels, each having a different manganese content. Plot the data and determine (a) the transition temperature of each (defined by the mean of the absorbed energies in the ductile and brittle regions); and (b) the transition temperature of each (defined as the temperature that provides 50 J of absorbed energy).

Test Temperature (°C) –100 –75 –50 –25 0 25 50 75 100

0.30% Mn 2 2 2 10 30 60 105 130 130

Impact energy (J) 0.39% 1.01% Mn Mn 5 5 5 7 12 20 25 40 55 75 100 110 125 130 135 135 135 135

1.55% Mn 15 25 45 70 110 135 140 140 140

Impact Energy (J)

150

1.55% Mn 1.01% Mn 0.39% Mn 0.30% Mn

100

50 J

0 -150

-100

-50

100

150

Temperature (ūC) Solution: (a) The transition temperatures are defined by the mean of the absorbed energies in the brittle and ductile regions:

0.30% Mn: (2 + 130) / 2 = 66 J

T = 28˚C

0.39% Mn: (5 + 135) / 2 = 70 J

T = 8˚C

1.01% Mn: (5 +135) / 2 = 70 J

T = –3˚C

1.55% Mn: (15 + 140) / 2 = 78 J

T = –20˚C

(b) The transition temperatures defined by 50 J are

0.30% Mn: T = 17˚C 0.39% Mn: T = –5˚C 1.01% Mn: T = –18˚C 1.55% Mn: T = –44˚C 98 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly available website, in whole or in part.

6–31

Plot the transition temperature versus manganese content using the data in the previous problem and discuss the effect of manganese on the toughness of steel. What is the minimum manganese allowed in the steel if a part is to be used at 0ºC? Solution:

Transition Temperature (ūC)

30 20 10

Average Absorbed Energy

0 -10

50 J Absorbed Energy

-20 -30 -40 -50

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

% Manganese Increasing the manganese increases the toughness and reduces the transition temperature; manganese is therefore a desirable alloying element for improving the impact properties of the steel. If the part is to be used at 0 ˚C, we would want at least 1.0% Mn in the steel based on the mean absorbed energy criterion or 0.39% Mn based on the 50 J criterion. 6–32

The following data were obtained from a series of Charpy impact tests performed on four ductile cast irons, each having a different silicon content. Plot the data and determine (a) the transition temperature of each (defined by the mean of the absorbed energies in the ductile and brittle regions); and (b) the transition temperature of each (defined as the temperature that provides 10 J of absorbed energy). (c) Plot the transition temperature versus silicon content and discuss the effect of silicon on the toughness of the cast iron. What is the maximum silicon allowed in the cast iron if a part is to be used at 25ºC?

Test Impact energy (J) Temperature 2.55% 2.85% 3.25% 3.63% (oC) Si Si Si Si –50 2.5 2.5 2 2 –5 3 2.5 2 2 0 6 5 3 2.5 25 13 10 7 4 50 17 14 12 8 75 19 16 16 13 100 19 16 16 16 125 19 16 16 16 Solution: 20

2.55% Si 2.85% Si 3.25% Si 3.63% Si

Impact Energy (J)

10 J

-60

-40

-20

100

120

Temperature (ūC)

Transition Temperature (ūC)

70 60 50 40

10 J Absorbed Energy

Average Absorbed Energy

20 10

2.4

2.6

2.8

3.0

3.2

3.4

3.6

% Silicon

3.8

(a) Transition temperatures defined by the mean of the absorbed energies are

2.55% Si: mean energy = (2.5 + 19)/2 = 9.8 J; 2.85% Si: mean energy = (2.5 + 16)/2 = 8.3 J; 3.25% Si: mean energy = (2 + 16)/2 = 9 J; 3.63% Si: mean energy = (2 + 16)/2 = 9 J;

T = 13oC T = 16oC T = 35oC T = 55oC

(b) Transition temperatures defined by 10 J are

2.55% Si: 2.85% Si: 3.25% Si: 3.63% Si:

T = 14oC T = 25oC T = 40oC T = 60oC

Increasing the silicon decreases the toughness and increases the transition temperature; silicon, therefore, reduces the impact properties of the cast iron. If the part is to be used at 25oC, we would want a maximum of about 2.9% Si in the cast iron. 6–33

FCC metals are often recommended for use at low temperatures, particularly when any sudden loading of the part is expected. Explain. Solution: FCC metals do not normally display a transition temperature; instead the impact energies decrease slowly with decreasing temperature and, in at least some cases (such as some aluminum alloys), the energies even increase at low temperatures. The FCC metals can obtain large ductilities, giving large areas beneath the true stress-strain curve.

6–34

A steel part can be made by powder metallurgy (compacting iron powder particles and sintering to produce a solid) or by machining from a solid steel block. Which part is expected to have the higher toughness? Explain. Solution: Parts produced by powder metallurgy often contain considerable amounts of porosity due to incomplete sintering; the porosity provides sites at which cracks might easily nucleate. Parts machined from solid steel are less likely to contain flaws that would nucleate cracks, therefore improving toughness.

6–36

A number of aluminum-silicon alloys have a structure that includes sharp-edged plates of brittle silicon in the softer, more ductile aluminum matrix. Would you expect these alloys to be notch-sensitive in an impact test? Would you expect these alloys to have good toughness? Explain your answers. Solution: The sharp-edged plates of the brittle silicon may act as stress-raisers, or notches, thus giving poor toughness to the alloy. The presence of additional notches, such as machining marks, will not have a

significannt effect, siince there are a already very largee numbers of “notches”” due to the microstructuure. Consequuently this tyype of alloy is expected to have pooor toughneess but is not n expectedd to be notcch sensitive. 6–40

A load veersus displaccement diagrram is shownn in Figure 6–29 6 for a metallic m glasss. A metallic glass g is a non-crystallin n ne (amorphoous) metal. The samplee was testedd in compression. Therefoore, even thrrough the loaad and displlacement vallues are plottted as positivve, the sampple length was w shorteneed during thhe test. Thee sample had a length in the directioon of loadinng of 6 mm and a crosss-sectional area a of 4 mm2. Numericaal values for the load andd displacemeent are givenn at the poinnts marked with w a circle an nd an X. Thhe first data point is (0, 0). Sample failure is inddicated withh an X. Answeer the follow wing questionns. (a) Calcu ulate the elastic modulus.. (b) How does d the elasstic moduluss compare too the moduluus of steel? (c) Calcu ulate the engiineering streess at the prooportional lim mit. (d) Consiider your ansswer to part (c) to be thee yield strenggth of the material. Is this a high yield y stress or a low yield stress? Support S youur answer with w an orderr of magniitude compaarison for a tyypical polyccrystalline metal. m (e) Calcu ulate the truee strain at thhe proportionnal limit. Reemember thaat the lengthh of the saample is decrreasing in coompression. (f) Calcu ulate the totall true strain at failure. (g) Calcu ulate the worrk of fracturee for this meetallic glass based b on enggineering strress and sttrain.

Figure 6– –29 Load veersus displacement for a metallic glass g tested in i compresssion for Probleem 6–40.

Solution: (a) The elastic modulus may be determined by taking the slope of an engineering stress – engineering strain curve. The slope can be based on the data points of (0, 0) and (0.100 mm, 7020 N) for the load and displacement. The engineering stress S is given by F , A0

where F is the applied force and A0 is the initial cross–sectional area. The engineering strain e is given by Δl l0

where ∆l is the displacement and l0 is the initial length. The elastic modulus E is given by E=

S e

according to Hooke’s Law. Therefore, at the load and displacement values of (0.100 mm, 7020 N), the last point in the linear region, E=

F l0 7020 N 6 mm = × = 105,300 MPa = 105.3 GPa 2 A0 Δl 4 mm 0.100 mm

(b) The elastic modulus of this metallic glass is lower than that of steel, which has an elastic modulus of 200 GPa. (c) At the proportional limit, the applied load is 7020 N. The engineering stress S is S=

F 7020 N = = 1755 MPa A0 4 mm 2

(d) This is a high stress. Pure crystalline metals have yield strengths on the order of tens of MPa or less. Alloys have yield strengths on the order of hundreds of MPa. Some tool steels have yield strengths as high as 2 GPa, but strengths as high as 1755 MPa are rare for a metal. (e) The true strain is given by

⎛

ε = ln(1 + e) = ln ⎜1 + ⎝

Δl ⎞ ⎟ l0 ⎠

Substituting, ⎛ ⎝

ε = ln ⎜1 −

0.100 mm ⎞ ⎟ = −0.01681 = −1.681% 6 mm ⎠

(f) The true strain e at failure is calculated as ⎛ ⎝

ε = ln ⎜1 −

0.105 mm ⎞ ⎟ = −0.01765 = −1.765% 6 mm ⎠

(g) The work of fracture can be approximated as the area under an engineering stress-strain curve. The area under the curve can be calculated as the area of a triangle plus the area of a rectangle. The area of the rectangle will be approximated by assuming that the stress in the plastic region has a constant value of 1755 MPa. The engineering strain at the proportional limit is e=

Δl −0.1 mm = = −0.0167 l0 6 mm

and the engineering strain at failure is e=

Δl −0.105 mm = = −0.0175 l0 6 mm

Positive values of strain will be used for the purpose of finding the area under the stress–strain curve.

Work 1 = ∫ S de = ×1755 MPa × 0.0167 + 1755 MPa × (0.0175 − 0.0167) Volume 2 Work = 16.1×106 J/m3 . Volume