The strain-hardening exponent for the metal needs to be determined. Whether the metal is FCC, BCC or HCP needs to be explained. Concept introduction: Strain hardening is called work hardening. Strengthening occurs due to plastic deformation. Strengthening occurs due to dislocation movement.

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Answer 1

Question

Chapter 8, Problem 8.10P

Interpretation Introduction

Interpretation:

The strain-hardening exponent for the metal needs to be determined. Whether the metal is FCC, BCC or HCP needs to be explained.

Concept introduction:

Strain hardening is called work hardening. Strengthening occurs due to plastic deformation. Strengthening occurs due to dislocation movement.

Expert Solution & Answer

Answer to Problem 8.10P

Strain hardening exponent for the metal 'n' is 0.12. The value of strain hardening is in BCC metals.

Explanation of Solution

Given Information:

The measurements in the plastic regions are as follows:

Force (1b)	Change in gage length (in) $(Δ1)$	Diameter (in)
27,500	0.2103	0.4800
27,000	0.4428	0.4566
25,700	0.6997	0.4343

Calculation:

Now, stress acting on the metal bar at 27500 lb

$σ1=fd1Af$ --------------- (1)

But

$Af=πFdi2=area of bar$

Where,

$Fd1=Force of metal bar=27500 lbd1=Diameter of bar=0.48 in.$

Equation (1) becomes,

$σ1=27500π4× 0.482σ1=151970Psi$

Now, stress acting at

$kd2=27000lb,d2=0.4566inσ2=fd2A2σ2=fd2π4d22σ2=27000π4× 0.45662σ2=164893Psi$

At,

$kd3=27000lb,d3=0.4343inσ3=257000π4× 0.43432σ3=173486Psi$

Calculating 'n' for force 27500 lb act on metal by using formula.

$σ1=kξ1nln σ1lnk+nlnξ$ --------------- (2)

Where,

$ξ=true strain=0.1σ1=27500lb$

Equation (2) becomes,

$ln(151970)=lnk+nln(0.1)11.9314=lnk+n(2.3026)$ --------------------- (a)

Now,the force of 25700 lb act on metal, strain hardening is given by,

$σ3=kξ3n$

Applying logarithm on both sides

$ln σ3lnk+nlnξ$

Here,

$ξ=True strain=0.3σ3=Stress acting=25700lb$

Equation (3) becomes,

$ln(173486)=lnk+nln(0.3)12.0636=lnk+n(1.2040)$ ---------------- (b)

From equation (a) and (b),

$n=0.12$

Conclusion

Thus, strain hardening exponent for metal 0.12 which is in BCC range.

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Answer 2

Question

Chapter 8, Problem 8.10P

Interpretation Introduction

Interpretation:

The strain-hardening exponent for the metal needs to be determined. Whether the metal is FCC, BCC or HCP needs to be explained.

Concept introduction:

Strain hardening is called work hardening. Strengthening occurs due to plastic deformation. Strengthening occurs due to dislocation movement.

Expert Solution & Answer

Answer to Problem 8.10P

Strain hardening exponent for the metal 'n' is 0.12. The value of strain hardening is in BCC metals.

Explanation of Solution

Given Information:

The measurements in the plastic regions are as follows:

Force (1b)	Change in gage length (in) $(Δ1)$	Diameter (in)
27,500	0.2103	0.4800
27,000	0.4428	0.4566
25,700	0.6997	0.4343

Calculation:

Now, stress acting on the metal bar at 27500 lb

$σ1=fd1Af$ --------------- (1)

But

$Af=πFdi2=area of bar$

Where,

$Fd1=Force of metal bar=27500 lbd1=Diameter of bar=0.48 in.$

Equation (1) becomes,

$σ1=27500π4× 0.482σ1=151970Psi$

Now, stress acting at

$kd2=27000lb,d2=0.4566inσ2=fd2A2σ2=fd2π4d22σ2=27000π4× 0.45662σ2=164893Psi$

At,

$kd3=27000lb,d3=0.4343inσ3=257000π4× 0.43432σ3=173486Psi$

Calculating 'n' for force 27500 lb act on metal by using formula.

$σ1=kξ1nln σ1lnk+nlnξ$ --------------- (2)

Where,

$ξ=true strain=0.1σ1=27500lb$

Equation (2) becomes,

$ln(151970)=lnk+nln(0.1)11.9314=lnk+n(2.3026)$ --------------------- (a)

Now,the force of 25700 lb act on metal, strain hardening is given by,

$σ3=kξ3n$

Applying logarithm on both sides

$ln σ3lnk+nlnξ$

Here,

$ξ=True strain=0.3σ3=Stress acting=25700lb$

Equation (3) becomes,

$ln(173486)=lnk+nln(0.3)12.0636=lnk+n(1.2040)$ ---------------- (b)

From equation (a) and (b),

$n=0.12$

Conclusion

Thus, strain hardening exponent for metal 0.12 which is in BCC range.

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Answer 3

Question

Chapter 8, Problem 8.10P

Interpretation Introduction

Interpretation:

The strain-hardening exponent for the metal needs to be determined. Whether the metal is FCC, BCC or HCP needs to be explained.

Concept introduction:

Strain hardening is called work hardening. Strengthening occurs due to plastic deformation. Strengthening occurs due to dislocation movement.

Expert Solution & Answer

Answer to Problem 8.10P

Strain hardening exponent for the metal 'n' is 0.12. The value of strain hardening is in BCC metals.

Explanation of Solution

Given Information:

The measurements in the plastic regions are as follows:

Force (1b)	Change in gage length (in) $(Δ1)$	Diameter (in)
27,500	0.2103	0.4800
27,000	0.4428	0.4566
25,700	0.6997	0.4343

Calculation:

Now, stress acting on the metal bar at 27500 lb

$σ1=fd1Af$ --------------- (1)

But

$Af=πFdi2=area of bar$

Where,

$Fd1=Force of metal bar=27500 lbd1=Diameter of bar=0.48 in.$

Equation (1) becomes,

$σ1=27500π4× 0.482σ1=151970Psi$

Now, stress acting at

$kd2=27000lb,d2=0.4566inσ2=fd2A2σ2=fd2π4d22σ2=27000π4× 0.45662σ2=164893Psi$

At,

$kd3=27000lb,d3=0.4343inσ3=257000π4× 0.43432σ3=173486Psi$

Calculating 'n' for force 27500 lb act on metal by using formula.

$σ1=kξ1nln σ1lnk+nlnξ$ --------------- (2)

Where,

$ξ=true strain=0.1σ1=27500lb$

Equation (2) becomes,

$ln(151970)=lnk+nln(0.1)11.9314=lnk+n(2.3026)$ --------------------- (a)

Now,the force of 25700 lb act on metal, strain hardening is given by,

$σ3=kξ3n$

Applying logarithm on both sides

$ln σ3lnk+nlnξ$

Here,

$ξ=True strain=0.3σ3=Stress acting=25700lb$

Equation (3) becomes,

$ln(173486)=lnk+nln(0.3)12.0636=lnk+n(1.2040)$ ---------------- (b)

From equation (a) and (b),

$n=0.12$

Conclusion

Thus, strain hardening exponent for metal 0.12 which is in BCC range.

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Answer 4

Question

Chapter 8, Problem 8.10P

Interpretation Introduction

Interpretation:

The strain-hardening exponent for the metal needs to be determined. Whether the metal is FCC, BCC or HCP needs to be explained.

Concept introduction:

Strain hardening is called work hardening. Strengthening occurs due to plastic deformation. Strengthening occurs due to dislocation movement.

Expert Solution & Answer

Answer to Problem 8.10P

Strain hardening exponent for the metal 'n' is 0.12. The value of strain hardening is in BCC metals.

Explanation of Solution

Given Information:

The measurements in the plastic regions are as follows:

Force (1b)	Change in gage length (in) $(Δ1)$	Diameter (in)
27,500	0.2103	0.4800
27,000	0.4428	0.4566
25,700	0.6997	0.4343

Calculation:

Now, stress acting on the metal bar at 27500 lb

$σ1=fd1Af$ --------------- (1)

But

$Af=πFdi2=area of bar$

Where,

$Fd1=Force of metal bar=27500 lbd1=Diameter of bar=0.48 in.$

Equation (1) becomes,

$σ1=27500π4× 0.482σ1=151970Psi$

Now, stress acting at

$kd2=27000lb,d2=0.4566inσ2=fd2A2σ2=fd2π4d22σ2=27000π4× 0.45662σ2=164893Psi$

At,

$kd3=27000lb,d3=0.4343inσ3=257000π4× 0.43432σ3=173486Psi$

Calculating 'n' for force 27500 lb act on metal by using formula.

$σ1=kξ1nln σ1lnk+nlnξ$ --------------- (2)

Where,

$ξ=true strain=0.1σ1=27500lb$

Equation (2) becomes,

$ln(151970)=lnk+nln(0.1)11.9314=lnk+n(2.3026)$ --------------------- (a)

Now,the force of 25700 lb act on metal, strain hardening is given by,

$σ3=kξ3n$

Applying logarithm on both sides

$ln σ3lnk+nlnξ$

Here,

$ξ=True strain=0.3σ3=Stress acting=25700lb$

Equation (3) becomes,

$ln(173486)=lnk+nln(0.3)12.0636=lnk+n(1.2040)$ ---------------- (b)

From equation (a) and (b),

$n=0.12$

Conclusion

Thus, strain hardening exponent for metal 0.12 which is in BCC range.

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Answer 5

Chapter 8, Problem 8.10P

Interpretation Introduction

Interpretation:

The strain-hardening exponent for the metal needs to be determined. Whether the metal is FCC, BCC or HCP needs to be explained.

Concept introduction:

Strain hardening is called work hardening. Strengthening occurs due to plastic deformation. Strengthening occurs due to dislocation movement.

Expert Solution & Answer

Answer to Problem 8.10P

Strain hardening exponent for the metal 'n' is 0.12. The value of strain hardening is in BCC metals.

Explanation of Solution

Given Information:

The measurements in the plastic regions are as follows:

Force (1b)	Change in gage length (in) $(Δ1)$	Diameter (in)
27,500	0.2103	0.4800
27,000	0.4428	0.4566
25,700	0.6997	0.4343

Calculation:

Now, stress acting on the metal bar at 27500 lb

$σ1=fd1Af$ --------------- (1)

But

$Af=πFdi2=area of bar$

Where,

$Fd1=Force of metal bar=27500 lbd1=Diameter of bar=0.48 in.$

Equation (1) becomes,

$σ1=27500π4× 0.482σ1=151970Psi$

Now, stress acting at

$kd2=27000lb,d2=0.4566inσ2=fd2A2σ2=fd2π4d22σ2=27000π4× 0.45662σ2=164893Psi$

At,

$kd3=27000lb,d3=0.4343inσ3=257000π4× 0.43432σ3=173486Psi$

Calculating 'n' for force 27500 lb act on metal by using formula.

$σ1=kξ1nln σ1lnk+nlnξ$ --------------- (2)

Where,

$ξ=true strain=0.1σ1=27500lb$

Equation (2) becomes,

$ln(151970)=lnk+nln(0.1)11.9314=lnk+n(2.3026)$ --------------------- (a)

Now,the force of 25700 lb act on metal, strain hardening is given by,

$σ3=kξ3n$

Applying logarithm on both sides

$ln σ3lnk+nlnξ$

Here,

$ξ=True strain=0.3σ3=Stress acting=25700lb$

Equation (3) becomes,

$ln(173486)=lnk+nln(0.3)12.0636=lnk+n(1.2040)$ ---------------- (b)

From equation (a) and (b),

$n=0.12$

Conclusion

Thus, strain hardening exponent for metal 0.12 which is in BCC range.

Answer 6

Chapter 8, Problem 8.10P

Interpretation Introduction

Interpretation:

The strain-hardening exponent for the metal needs to be determined. Whether the metal is FCC, BCC or HCP needs to be explained.

Concept introduction:

Strain hardening is called work hardening. Strengthening occurs due to plastic deformation. Strengthening occurs due to dislocation movement.

Expert Solution & Answer

Answer to Problem 8.10P

Strain hardening exponent for the metal 'n' is 0.12. The value of strain hardening is in BCC metals.

Explanation of Solution

Given Information:

The measurements in the plastic regions are as follows:

Force (1b)	Change in gage length (in) $(Δ1)$	Diameter (in)
27,500	0.2103	0.4800
27,000	0.4428	0.4566
25,700	0.6997	0.4343

Calculation:

Now, stress acting on the metal bar at 27500 lb

$σ1=fd1Af$ --------------- (1)

But

$Af=πFdi2=area of bar$

Where,

$Fd1=Force of metal bar=27500 lbd1=Diameter of bar=0.48 in.$

Equation (1) becomes,

$σ1=27500π4× 0.482σ1=151970Psi$

Now, stress acting at

$kd2=27000lb,d2=0.4566inσ2=fd2A2σ2=fd2π4d22σ2=27000π4× 0.45662σ2=164893Psi$

At,

$kd3=27000lb,d3=0.4343inσ3=257000π4× 0.43432σ3=173486Psi$

Calculating 'n' for force 27500 lb act on metal by using formula.

$σ1=kξ1nln σ1lnk+nlnξ$ --------------- (2)

Where,

$ξ=true strain=0.1σ1=27500lb$

Equation (2) becomes,

$ln(151970)=lnk+nln(0.1)11.9314=lnk+n(2.3026)$ --------------------- (a)

Now,the force of 25700 lb act on metal, strain hardening is given by,

$σ3=kξ3n$

Applying logarithm on both sides

$ln σ3lnk+nlnξ$

Here,

$ξ=True strain=0.3σ3=Stress acting=25700lb$

Equation (3) becomes,

$ln(173486)=lnk+nln(0.3)12.0636=lnk+n(1.2040)$ ---------------- (b)

From equation (a) and (b),

$n=0.12$

Conclusion

Thus, strain hardening exponent for metal 0.12 which is in BCC range.

Answer 7

Chapter 8, Problem 8.10P

Interpretation Introduction

Interpretation:

The strain-hardening exponent for the metal needs to be determined. Whether the metal is FCC, BCC or HCP needs to be explained.

Concept introduction:

Strain hardening is called work hardening. Strengthening occurs due to plastic deformation. Strengthening occurs due to dislocation movement.

Answer 8

Expert Solution & Answer

Answer to Problem 8.10P

Strain hardening exponent for the metal 'n' is 0.12. The value of strain hardening is in BCC metals.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given Information:

The measurements in the plastic regions are as follows:

Force (1b)	Change in gage length (in) $(Δ1)$	Diameter (in)
27,500	0.2103	0.4800
27,000	0.4428	0.4566
25,700	0.6997	0.4343

Calculation:

Now, stress acting on the metal bar at 27500 lb

$σ1=fd1Af$ --------------- (1)

But

$Af=πFdi2=area of bar$

Where,

$Fd1=Force of metal bar=27500 lbd1=Diameter of bar=0.48 in.$

Equation (1) becomes,

$σ1=27500π4× 0.482σ1=151970Psi$

Now, stress acting at

$kd2=27000lb,d2=0.4566inσ2=fd2A2σ2=fd2π4d22σ2=27000π4× 0.45662σ2=164893Psi$

At,

$kd3=27000lb,d3=0.4343inσ3=257000π4× 0.43432σ3=173486Psi$

Calculating 'n' for force 27500 lb act on metal by using formula.

$σ1=kξ1nln σ1lnk+nlnξ$ --------------- (2)

Where,

$ξ=true strain=0.1σ1=27500lb$

Equation (2) becomes,

$ln(151970)=lnk+nln(0.1)11.9314=lnk+n(2.3026)$ --------------------- (a)

Now,the force of 25700 lb act on metal, strain hardening is given by,

$σ3=kξ3n$

Applying logarithm on both sides

$ln σ3lnk+nlnξ$

Here,

$ξ=True strain=0.3σ3=Stress acting=25700lb$

Equation (3) becomes,

$ln(173486)=lnk+nln(0.3)12.0636=lnk+n(1.2040)$ ---------------- (b)

From equation (a) and (b),

$n=0.12$

Answer 12

Conclusion

Thus, strain hardening exponent for metal 0.12 which is in BCC range.

Answer 13

Ch. 8 - Prob. 8.1P Ch. 8 - Prob. 8.2P Ch. 8 - Prob. 8.3P Ch. 8 - Prob. 8.4P Ch. 8 - Prob. 8.5P Ch. 8 - Prob. 8.6P Ch. 8 - Prob. 8.7P Ch. 8 - Prob. 8.8P Ch. 8 - Prob. 8.9P Ch. 8 - Prob. 8.10P

The strain-hardening exponent for the metal needs to be determined. Whether the metal is FCC, BCC or HCP needs to be explained. Concept introduction: Strain hardening is called work hardening. Strengthening occurs due to plastic deformation. Strengthening occurs due to dislocation movement.

Answer to Problem 8.10P

Explanation of Solution

Want to see more full solutions like this?

Chapter 8 Solutions