All of the constants in the equation fir tensile yield strength as function of solid solution composition.

Question

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Answer 1

Question

Chapter 7, Problem 7.6P

(a)

To determine

All of the constants in the equation fir tensile yield strength as function of solid solution composition.

(a)

Expert Solution

Answer to Problem 7.6P

The constants in the equation fir tensile yield strength as function of solid solution composition is $8×109 N/m2⋅af1/2$ .

Explanation of Solution

Concept used:

Draw the diagram for yield strength of iron as a function of the atom fraction of carbon to the one-half power as shown below:

Materials Science and Engineering Properties, SI Edition, Chapter 7, Problem 7.6P

Write the expression for tensile strength of solid solution strengthened alloy.

$σy=σ0+kstcs1/2$ …… (1)

Here, $σy$ is the tensile strength of solid solution strengthened alloy, $σ0$ is the tensile yield stress at zero concentration of solid-solution atoms, $cs$ is the solute chemical composition in terms of atom fraction and $kst$ is constant.

Calculation:

Refer to figure 1; the value of $σ0$ at zero concentration of atom fraction of carbon is $7×107 N/m2$ .

Refer to figure 1; at atom fraction of carbon as $10×10−3 af1/2$ the corresponding yield stress is $15×107 N/m2$ .

Substitute $15×107 N/m2$ for $σy$ , $7×107 N/m2$ for $σ0$ and $10×10−3 af1/2$ for $cs1/2$ in equation (1).

$15×107 N/m2=7×107 N/m2+kst(10× 10 −3 af 1/2)kst=8× 107 N/m2( 10× 10 −3 af 1/2 )kst=8×109 N/m2⋅af1/2$

Conclusion:

Thus, the constants in the equation fir tensile yield strength as function of solid solution composition is $8×109 N/m2⋅af1/2$ .

(b)

To determine

The yield strength of iron-carbon alloy.

(b)

Expert Solution

Answer to Problem 7.6P

The yield strength of iron-carbon alloyis $33.53×107 N/m2$ .

Explanation of Solution

Substitute $8×109 N/m2⋅af1/2$ for $kst$ , $7×107 N/m2$ for $σ0$ and $0.0011$ for $cs$ in equation (1).

$σy=7×107 N/m2+(8× 109 N/m2⋅ af 1/2)(0.0011 af)1/2=7×107 N/m2+26.53×107 N/m2=33.53×107 N/m2$

Conclusion:

Thus, the yield strength of iron-carbon alloyis $33.53×107 N/m2$ .

(c)

To determine

Comment if fraction of carbon increased thenthe strength to be given by an extension of the line in the figure or not.

(c)

Expert Solution

Explanation of Solution

If the concentration of solid solution atoms becomes large then the precipitation of atoms starts to form. As mentioned if concentration is increased from $0.0011$ to $0.0015$ af of carbon the precipitates of iron carbides starts to form in alloy.

The strength dependence of precipitates is different than the strength dependence of solid solution. Therefore, it is not possible to give extension of line in figure.

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Answer 2

Question

Chapter 7, Problem 7.6P

(a)

To determine

All of the constants in the equation fir tensile yield strength as function of solid solution composition.

(a)

Expert Solution

Answer to Problem 7.6P

The constants in the equation fir tensile yield strength as function of solid solution composition is $8×109 N/m2⋅af1/2$ .

Explanation of Solution

Concept used:

Draw the diagram for yield strength of iron as a function of the atom fraction of carbon to the one-half power as shown below:

Materials Science and Engineering Properties, SI Edition, Chapter 7, Problem 7.6P

Write the expression for tensile strength of solid solution strengthened alloy.

$σy=σ0+kstcs1/2$ …… (1)

Here, $σy$ is the tensile strength of solid solution strengthened alloy, $σ0$ is the tensile yield stress at zero concentration of solid-solution atoms, $cs$ is the solute chemical composition in terms of atom fraction and $kst$ is constant.

Calculation:

Refer to figure 1; the value of $σ0$ at zero concentration of atom fraction of carbon is $7×107 N/m2$ .

Refer to figure 1; at atom fraction of carbon as $10×10−3 af1/2$ the corresponding yield stress is $15×107 N/m2$ .

Substitute $15×107 N/m2$ for $σy$ , $7×107 N/m2$ for $σ0$ and $10×10−3 af1/2$ for $cs1/2$ in equation (1).

$15×107 N/m2=7×107 N/m2+kst(10× 10 −3 af 1/2)kst=8× 107 N/m2( 10× 10 −3 af 1/2 )kst=8×109 N/m2⋅af1/2$

Conclusion:

Thus, the constants in the equation fir tensile yield strength as function of solid solution composition is $8×109 N/m2⋅af1/2$ .

(b)

To determine

The yield strength of iron-carbon alloy.

(b)

Expert Solution

Answer to Problem 7.6P

The yield strength of iron-carbon alloyis $33.53×107 N/m2$ .

Explanation of Solution

Substitute $8×109 N/m2⋅af1/2$ for $kst$ , $7×107 N/m2$ for $σ0$ and $0.0011$ for $cs$ in equation (1).

$σy=7×107 N/m2+(8× 109 N/m2⋅ af 1/2)(0.0011 af)1/2=7×107 N/m2+26.53×107 N/m2=33.53×107 N/m2$

Conclusion:

Thus, the yield strength of iron-carbon alloyis $33.53×107 N/m2$ .

(c)

To determine

Comment if fraction of carbon increased thenthe strength to be given by an extension of the line in the figure or not.

(c)

Expert Solution

Explanation of Solution

If the concentration of solid solution atoms becomes large then the precipitation of atoms starts to form. As mentioned if concentration is increased from $0.0011$ to $0.0015$ af of carbon the precipitates of iron carbides starts to form in alloy.

The strength dependence of precipitates is different than the strength dependence of solid solution. Therefore, it is not possible to give extension of line in figure.

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Answer 3

Question

Chapter 7, Problem 7.6P

(a)

To determine

All of the constants in the equation fir tensile yield strength as function of solid solution composition.

(a)

Expert Solution

Answer to Problem 7.6P

The constants in the equation fir tensile yield strength as function of solid solution composition is $8×109 N/m2⋅af1/2$ .

Explanation of Solution

Concept used:

Draw the diagram for yield strength of iron as a function of the atom fraction of carbon to the one-half power as shown below:

Materials Science and Engineering Properties, SI Edition, Chapter 7, Problem 7.6P

Write the expression for tensile strength of solid solution strengthened alloy.

$σy=σ0+kstcs1/2$ …… (1)

Here, $σy$ is the tensile strength of solid solution strengthened alloy, $σ0$ is the tensile yield stress at zero concentration of solid-solution atoms, $cs$ is the solute chemical composition in terms of atom fraction and $kst$ is constant.

Calculation:

Refer to figure 1; the value of $σ0$ at zero concentration of atom fraction of carbon is $7×107 N/m2$ .

Refer to figure 1; at atom fraction of carbon as $10×10−3 af1/2$ the corresponding yield stress is $15×107 N/m2$ .

Substitute $15×107 N/m2$ for $σy$ , $7×107 N/m2$ for $σ0$ and $10×10−3 af1/2$ for $cs1/2$ in equation (1).

$15×107 N/m2=7×107 N/m2+kst(10× 10 −3 af 1/2)kst=8× 107 N/m2( 10× 10 −3 af 1/2 )kst=8×109 N/m2⋅af1/2$

Conclusion:

Thus, the constants in the equation fir tensile yield strength as function of solid solution composition is $8×109 N/m2⋅af1/2$ .

(b)

To determine

The yield strength of iron-carbon alloy.

(b)

Expert Solution

Answer to Problem 7.6P

The yield strength of iron-carbon alloyis $33.53×107 N/m2$ .

Explanation of Solution

Substitute $8×109 N/m2⋅af1/2$ for $kst$ , $7×107 N/m2$ for $σ0$ and $0.0011$ for $cs$ in equation (1).

$σy=7×107 N/m2+(8× 109 N/m2⋅ af 1/2)(0.0011 af)1/2=7×107 N/m2+26.53×107 N/m2=33.53×107 N/m2$

Conclusion:

Thus, the yield strength of iron-carbon alloyis $33.53×107 N/m2$ .

(c)

To determine

Comment if fraction of carbon increased thenthe strength to be given by an extension of the line in the figure or not.

(c)

Expert Solution

Explanation of Solution

If the concentration of solid solution atoms becomes large then the precipitation of atoms starts to form. As mentioned if concentration is increased from $0.0011$ to $0.0015$ af of carbon the precipitates of iron carbides starts to form in alloy.

The strength dependence of precipitates is different than the strength dependence of solid solution. Therefore, it is not possible to give extension of line in figure.

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Answer 4

Question

Chapter 7, Problem 7.6P

(a)

To determine

All of the constants in the equation fir tensile yield strength as function of solid solution composition.

(a)

Expert Solution

Answer to Problem 7.6P

The constants in the equation fir tensile yield strength as function of solid solution composition is $8×109 N/m2⋅af1/2$ .

Explanation of Solution

Concept used:

Draw the diagram for yield strength of iron as a function of the atom fraction of carbon to the one-half power as shown below:

Materials Science and Engineering Properties, SI Edition, Chapter 7, Problem 7.6P

Write the expression for tensile strength of solid solution strengthened alloy.

$σy=σ0+kstcs1/2$ …… (1)

Here, $σy$ is the tensile strength of solid solution strengthened alloy, $σ0$ is the tensile yield stress at zero concentration of solid-solution atoms, $cs$ is the solute chemical composition in terms of atom fraction and $kst$ is constant.

Calculation:

Refer to figure 1; the value of $σ0$ at zero concentration of atom fraction of carbon is $7×107 N/m2$ .

Refer to figure 1; at atom fraction of carbon as $10×10−3 af1/2$ the corresponding yield stress is $15×107 N/m2$ .

Substitute $15×107 N/m2$ for $σy$ , $7×107 N/m2$ for $σ0$ and $10×10−3 af1/2$ for $cs1/2$ in equation (1).

$15×107 N/m2=7×107 N/m2+kst(10× 10 −3 af 1/2)kst=8× 107 N/m2( 10× 10 −3 af 1/2 )kst=8×109 N/m2⋅af1/2$

Conclusion:

Thus, the constants in the equation fir tensile yield strength as function of solid solution composition is $8×109 N/m2⋅af1/2$ .

(b)

To determine

The yield strength of iron-carbon alloy.

(b)

Expert Solution

Answer to Problem 7.6P

The yield strength of iron-carbon alloyis $33.53×107 N/m2$ .

Explanation of Solution

Substitute $8×109 N/m2⋅af1/2$ for $kst$ , $7×107 N/m2$ for $σ0$ and $0.0011$ for $cs$ in equation (1).

$σy=7×107 N/m2+(8× 109 N/m2⋅ af 1/2)(0.0011 af)1/2=7×107 N/m2+26.53×107 N/m2=33.53×107 N/m2$

Conclusion:

Thus, the yield strength of iron-carbon alloyis $33.53×107 N/m2$ .

(c)

To determine

Comment if fraction of carbon increased thenthe strength to be given by an extension of the line in the figure or not.

(c)

Expert Solution

Explanation of Solution

If the concentration of solid solution atoms becomes large then the precipitation of atoms starts to form. As mentioned if concentration is increased from $0.0011$ to $0.0015$ af of carbon the precipitates of iron carbides starts to form in alloy.

The strength dependence of precipitates is different than the strength dependence of solid solution. Therefore, it is not possible to give extension of line in figure.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 7, Problem 7.6P

(a)

To determine

All of the constants in the equation fir tensile yield strength as function of solid solution composition.

(a)

Expert Solution

Answer to Problem 7.6P

The constants in the equation fir tensile yield strength as function of solid solution composition is $8×109 N/m2⋅af1/2$ .

Explanation of Solution

Concept used:

Draw the diagram for yield strength of iron as a function of the atom fraction of carbon to the one-half power as shown below:

Materials Science and Engineering Properties, SI Edition, Chapter 7, Problem 7.6P

Write the expression for tensile strength of solid solution strengthened alloy.

$σy=σ0+kstcs1/2$ …… (1)

Here, $σy$ is the tensile strength of solid solution strengthened alloy, $σ0$ is the tensile yield stress at zero concentration of solid-solution atoms, $cs$ is the solute chemical composition in terms of atom fraction and $kst$ is constant.

Calculation:

Refer to figure 1; the value of $σ0$ at zero concentration of atom fraction of carbon is $7×107 N/m2$ .

Refer to figure 1; at atom fraction of carbon as $10×10−3 af1/2$ the corresponding yield stress is $15×107 N/m2$ .

Substitute $15×107 N/m2$ for $σy$ , $7×107 N/m2$ for $σ0$ and $10×10−3 af1/2$ for $cs1/2$ in equation (1).

$15×107 N/m2=7×107 N/m2+kst(10× 10 −3 af 1/2)kst=8× 107 N/m2( 10× 10 −3 af 1/2 )kst=8×109 N/m2⋅af1/2$

Conclusion:

Thus, the constants in the equation fir tensile yield strength as function of solid solution composition is $8×109 N/m2⋅af1/2$ .

(b)

To determine

The yield strength of iron-carbon alloy.

(b)

Expert Solution

Answer to Problem 7.6P

The yield strength of iron-carbon alloyis $33.53×107 N/m2$ .

Explanation of Solution

Substitute $8×109 N/m2⋅af1/2$ for $kst$ , $7×107 N/m2$ for $σ0$ and $0.0011$ for $cs$ in equation (1).

$σy=7×107 N/m2+(8× 109 N/m2⋅ af 1/2)(0.0011 af)1/2=7×107 N/m2+26.53×107 N/m2=33.53×107 N/m2$

Conclusion:

Thus, the yield strength of iron-carbon alloyis $33.53×107 N/m2$ .

(c)

To determine

Comment if fraction of carbon increased thenthe strength to be given by an extension of the line in the figure or not.

(c)

Expert Solution

Explanation of Solution

If the concentration of solid solution atoms becomes large then the precipitation of atoms starts to form. As mentioned if concentration is increased from $0.0011$ to $0.0015$ af of carbon the precipitates of iron carbides starts to form in alloy.

The strength dependence of precipitates is different than the strength dependence of solid solution. Therefore, it is not possible to give extension of line in figure.

Answer 6

Chapter 7, Problem 7.6P

(a)

To determine

All of the constants in the equation fir tensile yield strength as function of solid solution composition.

(a)

Expert Solution

Answer to Problem 7.6P

The constants in the equation fir tensile yield strength as function of solid solution composition is $8×109 N/m2⋅af1/2$ .

Explanation of Solution

Concept used:

Draw the diagram for yield strength of iron as a function of the atom fraction of carbon to the one-half power as shown below:

Materials Science and Engineering Properties, SI Edition, Chapter 7, Problem 7.6P

Write the expression for tensile strength of solid solution strengthened alloy.

$σy=σ0+kstcs1/2$ …… (1)

Here, $σy$ is the tensile strength of solid solution strengthened alloy, $σ0$ is the tensile yield stress at zero concentration of solid-solution atoms, $cs$ is the solute chemical composition in terms of atom fraction and $kst$ is constant.

Calculation:

Refer to figure 1; the value of $σ0$ at zero concentration of atom fraction of carbon is $7×107 N/m2$ .

Refer to figure 1; at atom fraction of carbon as $10×10−3 af1/2$ the corresponding yield stress is $15×107 N/m2$ .

Substitute $15×107 N/m2$ for $σy$ , $7×107 N/m2$ for $σ0$ and $10×10−3 af1/2$ for $cs1/2$ in equation (1).

$15×107 N/m2=7×107 N/m2+kst(10× 10 −3 af 1/2)kst=8× 107 N/m2( 10× 10 −3 af 1/2 )kst=8×109 N/m2⋅af1/2$

Conclusion:

Thus, the constants in the equation fir tensile yield strength as function of solid solution composition is $8×109 N/m2⋅af1/2$ .

Answer 7

Chapter 7, Problem 7.6P

(a)

To determine

All of the constants in the equation fir tensile yield strength as function of solid solution composition.

Answer 8

(a)

Expert Solution

Answer to Problem 7.6P

The constants in the equation fir tensile yield strength as function of solid solution composition is $8×109 N/m2⋅af1/2$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Concept used:

Draw the diagram for yield strength of iron as a function of the atom fraction of carbon to the one-half power as shown below:

Materials Science and Engineering Properties, SI Edition, Chapter 7, Problem 7.6P

Write the expression for tensile strength of solid solution strengthened alloy.

$σy=σ0+kstcs1/2$ …… (1)

Here, $σy$ is the tensile strength of solid solution strengthened alloy, $σ0$ is the tensile yield stress at zero concentration of solid-solution atoms, $cs$ is the solute chemical composition in terms of atom fraction and $kst$ is constant.

Calculation:

Refer to figure 1; the value of $σ0$ at zero concentration of atom fraction of carbon is $7×107 N/m2$ .

Refer to figure 1; at atom fraction of carbon as $10×10−3 af1/2$ the corresponding yield stress is $15×107 N/m2$ .

Substitute $15×107 N/m2$ for $σy$ , $7×107 N/m2$ for $σ0$ and $10×10−3 af1/2$ for $cs1/2$ in equation (1).

$15×107 N/m2=7×107 N/m2+kst(10× 10 −3 af 1/2)kst=8× 107 N/m2( 10× 10 −3 af 1/2 )kst=8×109 N/m2⋅af1/2$

Conclusion:

Thus, the constants in the equation fir tensile yield strength as function of solid solution composition is $8×109 N/m2⋅af1/2$ .

Answer 13

(b)

To determine

The yield strength of iron-carbon alloy.

(b)

Expert Solution

Answer to Problem 7.6P

The yield strength of iron-carbon alloyis $33.53×107 N/m2$ .

Explanation of Solution

Substitute $8×109 N/m2⋅af1/2$ for $kst$ , $7×107 N/m2$ for $σ0$ and $0.0011$ for $cs$ in equation (1).

$σy=7×107 N/m2+(8× 109 N/m2⋅ af 1/2)(0.0011 af)1/2=7×107 N/m2+26.53×107 N/m2=33.53×107 N/m2$

Conclusion:

Thus, the yield strength of iron-carbon alloyis $33.53×107 N/m2$ .

Answer 14

(b)

To determine

The yield strength of iron-carbon alloy.

Answer 15

(b)

Expert Solution

Answer to Problem 7.6P

The yield strength of iron-carbon alloyis $33.53×107 N/m2$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Substitute $8×109 N/m2⋅af1/2$ for $kst$ , $7×107 N/m2$ for $σ0$ and $0.0011$ for $cs$ in equation (1).

$σy=7×107 N/m2+(8× 109 N/m2⋅ af 1/2)(0.0011 af)1/2=7×107 N/m2+26.53×107 N/m2=33.53×107 N/m2$

Conclusion:

Thus, the yield strength of iron-carbon alloyis $33.53×107 N/m2$ .

Answer 20

(c)

To determine

Comment if fraction of carbon increased thenthe strength to be given by an extension of the line in the figure or not.

(c)

Expert Solution

Explanation of Solution

If the concentration of solid solution atoms becomes large then the precipitation of atoms starts to form. As mentioned if concentration is increased from $0.0011$ to $0.0015$ af of carbon the precipitates of iron carbides starts to form in alloy.

The strength dependence of precipitates is different than the strength dependence of solid solution. Therefore, it is not possible to give extension of line in figure.

Answer 21

(c)

To determine

Comment if fraction of carbon increased thenthe strength to be given by an extension of the line in the figure or not.

Answer 22

(c)

Expert Solution

Explanation of Solution

If the concentration of solid solution atoms becomes large then the precipitation of atoms starts to form. As mentioned if concentration is increased from $0.0011$ to $0.0015$ af of carbon the precipitates of iron carbides starts to form in alloy.

The strength dependence of precipitates is different than the strength dependence of solid solution. Therefore, it is not possible to give extension of line in figure.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Ch. 7 - Prob. 1CQ Ch. 7 - Prob. 2CQ Ch. 7 - Prob. 3CQ Ch. 7 - Prob. 4CQ Ch. 7 - Prob. 5CQ Ch. 7 - Prob. 6CQ Ch. 7 - Prob. 7CQ Ch. 7 - Prob. 8CQ Ch. 7 - Prob. 9CQ Ch. 7 - Prob. 10CQ

All of the constants in the equation fir tensile yield strength as function of solid solution composition.

All of the constants in the equation fir tensile yield strength as function of solid solution composition.

Answer to Problem 7.6P

Explanation of Solution

The yield strength of iron-carbon alloy.

Answer to Problem 7.6P

Explanation of Solution

Comment if fraction of carbon increased thenthe strength to be given by an extension of the line in the figure or not.

Explanation of Solution

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Chapter 7 Solutions