**Application of Cellulose Microfibrils in Polymer**

Application of Cellulose Microfibrils in Polymer

(Parte **1** de 2)

6 Mechanical Properties and Behavior

6–24A850-lb force is applied to a 0.15-in. diameter nickel wire having a yield strength of 45,0 psi and a tensile strength of 5,0 psi. 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/A=850 lb / (π/4)(0.15 in.)2=48,100 psi

Because sis greater than the yield strength of 45,0 psi, the wire will plastically deform.

(b) Because sis less than the tensile strength of 5,0 psi, no necking will occur.

6–25Aforce of 100,0 N is applied to a 10 m ×20 m iron bar 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/A=100,0 N / (10 m)(20 m) =500 N/m2=500 MPa

Because sis greater than the yield strength of 400 MPa, the wire will plastically deform.

(b) Because sis greater than the tensile strength of 480 MPa, the wire will also neck.

6–25(c)Calculate the maximum force that a 0.2-in. diameter rod of Al2O3, having a yield strength of 35,0 psi, can withstand with no plastic deformation. Express your answer in pounds and newtons.

6–26Aforce of 20,0 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, both in GPa and psi.

Solution:The strain eis e=(10.045 cm −10 cm)/10 cm =0.0045 cm/cm

6–27Apolymer bar’s dimensions are 1 in. ×2 in. ×15 in. The polymer has a modulus of elasticity of 600,0 psi. What force is required to stretch the bar elastically to 15.25 in.?

Solution:The strain eis e=(15.25 in. −15 in.) / (15 in.) =0.01667 in./in.

The stress sis s=Ee=(600,0 psi)(0.01667 in./in.) =10,0 psi The force is then F=sA=(10,0 psi)(1 in.)(2 in.) =20,0 lb

6–28An aluminum plate 0.5 cm thick is to withstand a force of 50,0 N with no permanent deformation. If the aluminum has a yield strength of 125 MPa, what is the minimum width of the plate?

6–29A3-in.-diameter rod of copper is to be reduced to a 2-in.-diameter rod by being pushed through an opening. To account for the elastic strain, what should be the diameter of the opening? The modulus of elasticity for the copper is 17 ×106psi and the yield strength is 40,0 psi.

The strain is also e=(2 in. −do ) / do

=0.00235 in./in.

2 −do =0.00235 d o

The opening in the die must be smaller than the final diameter.

6–30Asteel cable 1.25 in. in diameter and 50 ft long is to lift a 20 ton load. What is the length of the cable during lifting? The modulus of elasticity of the steel is 30 ×106 psi.

6–33The following data were collected from a standard 0.505-in.-diameter test specimen of a copper alloy (initial length (lo ) =2.0 in.):

56The Science and Engineering of MaterialsInstructor’s Solution Manual

Load Gage Length Stress Strain (lb) (in.) (psi) (in./in.)

After fracture, the gage length is 3.014 in. and the diameter is 0.374 in. Plot the data 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 fracture, and (h) the modulus of resilience.

(a) 0.2% offset yield strength =45,0 psi (b) tensile strength =62,0 psi (c) E=(30,0 −0) / (0.001665 −0) =18 ×106psi

50Stress (ksi) yielding 0.2% offset

CHAPTER 6Mechanical Properties and Behavior57

6–34The following data were collected from a 0.4-in. diameter test specimen of polyvinyl chloride (lo =2.0 in.):

Load Gage Length Stress Strain (lb) (in.) (psi) (in./in.)

After fracture, the gage length is 2.09 in. and the diameter is 0.393 in. Plot the data 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 fracture, and (h) the modulus of resilience.

(a) 0.2% offset yield strength =1,600 psi (b) tensile strength =12,729 psi (c) E=(7160 −0) / (0.01187 −0) =603,0 psi

Stress (ksi)

Yielding 0.2% offset

58The Science and Engineering of MaterialsInstructor’s Solution Manual

6–35The following data were collected from a 12-m-diameter test specimen of magnesium (lo =30.0 m):

LoadGage LengthStress Strain (N)(m)(MPa) (m/m)

After fracture, the gage length is 32.61 m and the diameter is 1.74 m. Plot the data 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 fracture, and (h) the modulus of resilience.

(a) 0.2% offset yield strength =186 MPa (b) tensile strength =238.7 MPa (c) E=(132.6 −0) / (0.00296 −0) =4,800 MPa =4.8 GPa

Stress (Mpa) 0.2% offset

Yielding

CHAPTER 6Mechanical Properties and Behavior59

6–36The following data were collected from a 20 m diameter test specimen of a ductile cast iron (lo =40.0 m):

LoadGage LengthStress Strain (N)(m)(MPa) (m/m)

After fracture, the gage length is 47.42 m and the diameter is 18.35 m. Plot the data 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 fracture, and (h) the modulus of resilience.

(a) 0.2% offset yield strength =274 MPa (b) tensile strength =417 MPa (c) E=(238.7 −0) / (0.001388 −0) =172,0 MPa =172 GPa

0.2% offset Yielding

Strain (m/m)

60The Science and Engineering of MaterialsInstructor’s Solution Manual

6–39Abar of Al2O3that is 0.25 in. thick, 0.5 in. wide, and 9 in. long is tested in a threepoint bending apparatus, with the supports located 6 in. 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 Deflection Stress (lb) (in.)(psi)

The flexural strength is the stress at fracture, or 24,768 psi.

The flexural modulus can be calculated from the linear curve; picking the first point as an example:

FM = FL3 =

6–40(a)A0.4-in. diameter, 12-in. long titanium bar has a yield strength of 50,0 psi, a modulus of elasticity of 16 ×106psi, and Poisson’s ratio of 0.30. Determine the length and diameter of the bar when a 500-lb load is applied.

Solution:The stress is σ=F/A=500 lb/(π/4)(0.4 in.)2=3,979 psi

Stress (ksi)

CHAPTER 6Mechanical Properties and Behavior61

The applied stress is much less than the yield strength; therefore Hooke’s law can be used.

lf − l o lo 12 in.

lf=12.00298 in.

From Poisson’s ratio, m=−elat/ elong=0.3 elat =− (0.3)(0.00024868) =− 0.0000746 in./in.

df − d o

= df−0.4 in. =−0.0000746 in./in.

df=0.39997 in.

6–40(b)When a tensile load is applied to a 1.5-cm diameter copper bar, the diameter is reduced to 1.498-cm diameter. Determine the applied load, using the data in Table 6–3.

Solution:From Table 6–3, m=−elat/ elong=0.36

6–41Athree-point bend test is performed on a block of ZrO2that is 8 in. long, 0.50 in. wide, and 0.25 in. thick and is resting on two supports 4 in. apart. When a force of

400 lb is applied, the specimen deflects 0.037 in. and breaks. Calculate (a) the flexural strength and (b) the flexural modulus, assuming that no plastic deformation occurs.

Solution:(a) flexural strength =3FL/2wh2=(3)(400 lb)(4 in.) =76,800 psi(2)(0.5 in.)(0.25 in.)2

(4)(0.5 in.)(0.25 in.)3(0.037 in.)

6–42Athree-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 m is recorded. Calculate (a) the force that caused the fracture and (b) the flexural strength. The flexural modulus for silicon carbide is 480 GPa. Assume that no plastic deformation occurs.

Solution:(a) The force F required to produce a deflection of 0.09 m is

62The Science and Engineering of MaterialsInstructor’s Solution Manual

6–43(a)Athermosetting polymer containing glass beads is required to deflect 0.5 m 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 Lbetween the supports can be calculated from the flexural modulus.

L3=69,0 m3or | L=41 m |

The applied stress is less than the flexural strength of 85 MPa; the polymer is not expected to fracture.

6–43(b)The flexural modulus of alumina is 45 ×106psi and its flexural strength is 46,0 psi. Abar of alumina 0.3 in. thick, 1.0 in. wide, and 10 in. long is placed on supports 7 in. 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

6–52ABrinell hardness measurement, using a 10-m-diameter indenter and a 500 kg load, produces an indentation of 4.5 m on an aluminum plate. Determine the Brinell hardness number HB of the metal.

Solution:

6–53When a 3000 kg load is applied to a 10-m-diameter ball in a Brinell test of a steel, an indentation of 3.1 m is produced. Estimate the tensile strength of the steel.

Solution:

Tensile strength =500 HB =(500)(388) =194,0 psi

6–55The data below 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 (defined by the mean of the absorbed energies in the

CHAPTER 6Mechanical Properties and Behavior63 ductile and brittle regions) and (b) the transition temperature (defined as the temperature that provides 50 J absorbed energy). Plot the transition temperature versus manganese content and discuss the effect of manganese on the toughness of steel. What would be the minimum manganese allowed in the steel if a part is to be used at 0oC?

Solution:

Test temperatureImpact energy (J) oC0.30% Mn0.39% Mn1.01% Mn1.5% Mn

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

0.30% Mn: mean energy =2 +(130 +2)/2 =68 J; | T=27oC |

0.39% Mn: mean energy =5 +(135 +5)/2 =75 J; | T=10oC |

1.01% Mn: mean energy =5 +(135 +5)/2 =75 J; | T=0oC |

50 J Average

T r ansition T e mper ature

64The Science and Engineering of MaterialsInstructor’s Solution Manual

Increasing the manganese increases the toughness and reduces the tran sition temperature; manganese is therefore a desirable alloying element for improving the impact properties of the steel.

If the part is to be used at 25oC, we would want at least 1.0% Mn in the steel based on the mean absorbed energy criterion or 0.36% Mn based on the 50 J criterion.

6–57The 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 (defined by the mean of the absorbed energies in the ductile and brittle regions) and (b) the transition temperature (defined as the temperature that provides 10 J absorbed energy). Plot the transition temperature versus silicon content and discuss the effect of silicon on the toughness of the cast iron. What would be the maximum silicon allowed in the cast iron if a part is to be used at 25oC?

Solution:

Test temperatureImpact energy (J) oC 2.5% Si 2.85% Si3.25% Si 3.63% Si

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

3.25% Si: mean energy =2 +(16 +2)/2 =1 J; | T=45oC |

3.63% Si: mean energy =2 +(16 +2)/2 =1 J; | T=65oC |

Average20

T r ansition T e mper ature

C)Impact energy (J)

CHAPTER 6Mechanical Properties and Behavior65

(b) Transition temperatures defined by 10 J are: 2.5% Si: T=15oC 2.85% Si: T=25oC 3.25% Si: T=38oC 3.63% Si: T=56oC

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.

6–58FCC 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.

6–59Asteel 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.

6–62Anumber 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 significant effect, since there are already very large numbers of “notches” due to the microstructure. Consequently this type of alloy is expected to have poor toughness but is not expected to be notch sensitive.

within the alumina. Would doing this affect the toughness of the ceramic matrix composite? Explain. (These materials are discussed in later chapters.)

Solution:The SiC fibers may improve the toughness of the alumina matrix. The fibers may do so by several mechanisms. By introducing an interface (between the fibers and the matrix), a crack may be blocked; to continue growing, the crack may have to pass around the fiber, thus increasing the total energy of the crack and thus the energy that can be absorbed by the material. Or extra energy may be required to force the crack through the

66The Science and Engineering of MaterialsInstructor’s Solution Manual interface in an effort to continue propagating. In addition, the fibers may begin to pull out of the matrix, particularly if bonding is poor; the fiber pull-out requires energy, thus improving toughness. Finally, the fibers may bridge across the crack, helping to hold the material together and requiring more energy to propagate the crack.

6–68Aceramic matrix composite contains internal flaws as large as 0.001 cm in length.

The plane strain fracture toughness of the composite is 45 and the tensile strength is 550 MPa. Will the flaw cause the composite to fail before the tensile strength is reached? Assume that f=1.

Solution:Since the crack is internal, 2a =0.001 cm =0.00001 m. Therefore a=0.000005 m

The applied stress required for the crack to cause failure is much larger than the tensile strength of 550 MPa. Any failure of the ceramic should be expected due to the massive overload, not because of the presence of the flaws.

(Parte **1** de 2)