The number of large stress amplitude cycles that can be sustained before failure need sto be determined, if the material has a crack of the initial length of 0.010 mm and then subjected to calculating tensile structures of 25 and 125 MPa for 350,000 cycles. Concept introduction: Equation of strain fracture toughness of the material. k k = f σ π a Material subjected to cycle string is given by, N = 2 [ ( a c ) 2 − n / 2 − ( a i ) 2 − n / 2 ] ( 2 − n ) ( f n Δ σ n π n / 2 )

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Answer 1

Question

Chapter 7, Problem 7.38P

Interpretation Introduction

Interpretation:

The number of large stress amplitude cycles that can be sustained before failure need sto be determined, if the material has a crack of the initial length of 0.010 mm and then subjected to calculating tensile structures of 25 and 125 MPa for 350,000 cycles.

Concept introduction:

Equation of strain fracture toughness of the material.

$kk=fσπa$

Material subjected to cycle string is given by,

$N=2[(ac)2−n/2−(ai)2−n/2](2−n)(fnΔσnπn/2)$

Expert Solution & Answer

Answer to Problem 7.38P

The cycles that sustained before failure are given by 13,560 cycles.

Explanation of Solution

Given Information:

$Initial length=0.010mkk=Fracture toughness=25MPam Materials constant n=3.1 and c=1 .8×10 -10 f=1.0 σ=Compressive stress=250MPa cycles=350,000$

Calculation:

Stress fracture toughness $kk$ is given by:

$kk=fσπa$ ------------------(1)

Here,

$F=geometry factor=1.0 σ=Applied stress=250MPa a=Crack lengthkk=Fracture toughness=25MPam$

Rearranging equation (1),

$a=1π(kkfσ)a=1π(2.5(1.0)×250)a=0.00318ma=0.00318×103mma=3.19mm$

Therefore, critical crack length when 250 MPa stress is applied is given by 3.18mm

Let, a₁ is the initial flaw size when material subjected to stress in cyclic testing.

Now, the number of cycles (N) is given to find out crack (a₁)

$N=2[(ac)2−n/2−(ai)2−n/2](2−n)(fnΔσnπn/2)$ --------------------(2)

Here,

$a1=initial flow size =0.010mmac=flow needed for fractureΔσ=difference between maximum and minimum cyclical stress =125MPa-25MPa =100MPa n and c=material constant=3.1 and 1 .8×10 -10 respectivlyN=number of cycles=350,000F=1.0$

Equation (2) becomes.

$350,000= 2[ a 1 (2−3.1)/2 − (0.010) (2−3.1)/2 (2−3.1)(1.8× 10 −10 ) (1.0) 3.1 (100) 3.1 π 3.1/2$

$350,000= 2[ a 1 −0.55 −562.3413] (2−3.1)(1.8× 10 −10 ) (1.0) 3.1 (100) 3.1 π 3.1/2$

$350,000=−1,080.89( a 1 −0.55 −562.3413)$

$− 350,000 1,080.89 = ( a 1 ) −0.55 −562.3413$

$−323.806= ( a 1 ) −0.55 −562.3413$

$( a 1 ) −0.55 =−323.806−562.3413$

$=238.534$

$a 1 =4.76× 10 −5 m$

$=4.76× 10 −5 ×1000mm$

$a 1 =0.0476mm$

Hence the length of the crack $a1=0.0476 mm$

Now a number of cycles (N) for the crack length of $0.0476 mm$ is given by,

$N=2[(ac)2−n/2−(ai)2−n/2](2−n)(fnΔσnπn/2)$

Here,

Under compression

$Δσ=250MPaa1=0.0476mmac=3.18mmΔσ=250MPaf=1.0n=3.1c=1.8×10−10$

Therefore equation (2) becomes

$N=2[ (3.18) (2−3.1)/2− (0.0476) (2−3.1)/2(2−3.1)(1.8× 10 −10) (1.0) 3.1 (250) 3.1π 3.1/2=−429.5280.03168N=13,560 cycles$

Required number of cycles $13,560 cycles$ .

Conclusion

Hence by using strain fracture toughness of material equation and also a number of cycles equation, the number of cycles (N) before failure is $13,560 cycles$ .

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Answer 2

Question

Chapter 7, Problem 7.38P

Interpretation Introduction

Interpretation:

The number of large stress amplitude cycles that can be sustained before failure need sto be determined, if the material has a crack of the initial length of 0.010 mm and then subjected to calculating tensile structures of 25 and 125 MPa for 350,000 cycles.

Concept introduction:

Equation of strain fracture toughness of the material.

$kk=fσπa$

Material subjected to cycle string is given by,

$N=2[(ac)2−n/2−(ai)2−n/2](2−n)(fnΔσnπn/2)$

Expert Solution & Answer

Answer to Problem 7.38P

The cycles that sustained before failure are given by 13,560 cycles.

Explanation of Solution

Given Information:

$Initial length=0.010mkk=Fracture toughness=25MPam Materials constant n=3.1 and c=1 .8×10 -10 f=1.0 σ=Compressive stress=250MPa cycles=350,000$

Calculation:

Stress fracture toughness $kk$ is given by:

$kk=fσπa$ ------------------(1)

Here,

$F=geometry factor=1.0 σ=Applied stress=250MPa a=Crack lengthkk=Fracture toughness=25MPam$

Rearranging equation (1),

$a=1π(kkfσ)a=1π(2.5(1.0)×250)a=0.00318ma=0.00318×103mma=3.19mm$

Therefore, critical crack length when 250 MPa stress is applied is given by 3.18mm

Let, a₁ is the initial flaw size when material subjected to stress in cyclic testing.

Now, the number of cycles (N) is given to find out crack (a₁)

$N=2[(ac)2−n/2−(ai)2−n/2](2−n)(fnΔσnπn/2)$ --------------------(2)

Here,

$a1=initial flow size =0.010mmac=flow needed for fractureΔσ=difference between maximum and minimum cyclical stress =125MPa-25MPa =100MPa n and c=material constant=3.1 and 1 .8×10 -10 respectivlyN=number of cycles=350,000F=1.0$

Equation (2) becomes.

$350,000= 2[ a 1 (2−3.1)/2 − (0.010) (2−3.1)/2 (2−3.1)(1.8× 10 −10 ) (1.0) 3.1 (100) 3.1 π 3.1/2$

$350,000= 2[ a 1 −0.55 −562.3413] (2−3.1)(1.8× 10 −10 ) (1.0) 3.1 (100) 3.1 π 3.1/2$

$350,000=−1,080.89( a 1 −0.55 −562.3413)$

$− 350,000 1,080.89 = ( a 1 ) −0.55 −562.3413$

$−323.806= ( a 1 ) −0.55 −562.3413$

$( a 1 ) −0.55 =−323.806−562.3413$

$=238.534$

$a 1 =4.76× 10 −5 m$

$=4.76× 10 −5 ×1000mm$

$a 1 =0.0476mm$

Hence the length of the crack $a1=0.0476 mm$

Now a number of cycles (N) for the crack length of $0.0476 mm$ is given by,

$N=2[(ac)2−n/2−(ai)2−n/2](2−n)(fnΔσnπn/2)$

Here,

Under compression

$Δσ=250MPaa1=0.0476mmac=3.18mmΔσ=250MPaf=1.0n=3.1c=1.8×10−10$

Therefore equation (2) becomes

$N=2[ (3.18) (2−3.1)/2− (0.0476) (2−3.1)/2(2−3.1)(1.8× 10 −10) (1.0) 3.1 (250) 3.1π 3.1/2=−429.5280.03168N=13,560 cycles$

Required number of cycles $13,560 cycles$ .

Conclusion

Hence by using strain fracture toughness of material equation and also a number of cycles equation, the number of cycles (N) before failure is $13,560 cycles$ .

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Answer 3

Question

Chapter 7, Problem 7.38P

Interpretation Introduction

Interpretation:

The number of large stress amplitude cycles that can be sustained before failure need sto be determined, if the material has a crack of the initial length of 0.010 mm and then subjected to calculating tensile structures of 25 and 125 MPa for 350,000 cycles.

Concept introduction:

Equation of strain fracture toughness of the material.

$kk=fσπa$

Material subjected to cycle string is given by,

$N=2[(ac)2−n/2−(ai)2−n/2](2−n)(fnΔσnπn/2)$

Expert Solution & Answer

Answer to Problem 7.38P

The cycles that sustained before failure are given by 13,560 cycles.

Explanation of Solution

Given Information:

$Initial length=0.010mkk=Fracture toughness=25MPam Materials constant n=3.1 and c=1 .8×10 -10 f=1.0 σ=Compressive stress=250MPa cycles=350,000$

Calculation:

Stress fracture toughness $kk$ is given by:

$kk=fσπa$ ------------------(1)

Here,

$F=geometry factor=1.0 σ=Applied stress=250MPa a=Crack lengthkk=Fracture toughness=25MPam$

Rearranging equation (1),

$a=1π(kkfσ)a=1π(2.5(1.0)×250)a=0.00318ma=0.00318×103mma=3.19mm$

Therefore, critical crack length when 250 MPa stress is applied is given by 3.18mm

Let, a₁ is the initial flaw size when material subjected to stress in cyclic testing.

Now, the number of cycles (N) is given to find out crack (a₁)

$N=2[(ac)2−n/2−(ai)2−n/2](2−n)(fnΔσnπn/2)$ --------------------(2)

Here,

$a1=initial flow size =0.010mmac=flow needed for fractureΔσ=difference between maximum and minimum cyclical stress =125MPa-25MPa =100MPa n and c=material constant=3.1 and 1 .8×10 -10 respectivlyN=number of cycles=350,000F=1.0$

Equation (2) becomes.

$350,000= 2[ a 1 (2−3.1)/2 − (0.010) (2−3.1)/2 (2−3.1)(1.8× 10 −10 ) (1.0) 3.1 (100) 3.1 π 3.1/2$

$350,000= 2[ a 1 −0.55 −562.3413] (2−3.1)(1.8× 10 −10 ) (1.0) 3.1 (100) 3.1 π 3.1/2$

$350,000=−1,080.89( a 1 −0.55 −562.3413)$

$− 350,000 1,080.89 = ( a 1 ) −0.55 −562.3413$

$−323.806= ( a 1 ) −0.55 −562.3413$

$( a 1 ) −0.55 =−323.806−562.3413$

$=238.534$

$a 1 =4.76× 10 −5 m$

$=4.76× 10 −5 ×1000mm$

$a 1 =0.0476mm$

Hence the length of the crack $a1=0.0476 mm$

Now a number of cycles (N) for the crack length of $0.0476 mm$ is given by,

$N=2[(ac)2−n/2−(ai)2−n/2](2−n)(fnΔσnπn/2)$

Here,

Under compression

$Δσ=250MPaa1=0.0476mmac=3.18mmΔσ=250MPaf=1.0n=3.1c=1.8×10−10$

Therefore equation (2) becomes

$N=2[ (3.18) (2−3.1)/2− (0.0476) (2−3.1)/2(2−3.1)(1.8× 10 −10) (1.0) 3.1 (250) 3.1π 3.1/2=−429.5280.03168N=13,560 cycles$

Required number of cycles $13,560 cycles$ .

Conclusion

Hence by using strain fracture toughness of material equation and also a number of cycles equation, the number of cycles (N) before failure is $13,560 cycles$ .

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Answer 4

Question

Chapter 7, Problem 7.38P

Interpretation Introduction

Interpretation:

The number of large stress amplitude cycles that can be sustained before failure need sto be determined, if the material has a crack of the initial length of 0.010 mm and then subjected to calculating tensile structures of 25 and 125 MPa for 350,000 cycles.

Concept introduction:

Equation of strain fracture toughness of the material.

$kk=fσπa$

Material subjected to cycle string is given by,

$N=2[(ac)2−n/2−(ai)2−n/2](2−n)(fnΔσnπn/2)$

Expert Solution & Answer

Answer to Problem 7.38P

The cycles that sustained before failure are given by 13,560 cycles.

Explanation of Solution

Given Information:

$Initial length=0.010mkk=Fracture toughness=25MPam Materials constant n=3.1 and c=1 .8×10 -10 f=1.0 σ=Compressive stress=250MPa cycles=350,000$

Calculation:

Stress fracture toughness $kk$ is given by:

$kk=fσπa$ ------------------(1)

Here,

$F=geometry factor=1.0 σ=Applied stress=250MPa a=Crack lengthkk=Fracture toughness=25MPam$

Rearranging equation (1),

$a=1π(kkfσ)a=1π(2.5(1.0)×250)a=0.00318ma=0.00318×103mma=3.19mm$

Therefore, critical crack length when 250 MPa stress is applied is given by 3.18mm

Let, a₁ is the initial flaw size when material subjected to stress in cyclic testing.

Now, the number of cycles (N) is given to find out crack (a₁)

$N=2[(ac)2−n/2−(ai)2−n/2](2−n)(fnΔσnπn/2)$ --------------------(2)

Here,

$a1=initial flow size =0.010mmac=flow needed for fractureΔσ=difference between maximum and minimum cyclical stress =125MPa-25MPa =100MPa n and c=material constant=3.1 and 1 .8×10 -10 respectivlyN=number of cycles=350,000F=1.0$

Equation (2) becomes.

$350,000= 2[ a 1 (2−3.1)/2 − (0.010) (2−3.1)/2 (2−3.1)(1.8× 10 −10 ) (1.0) 3.1 (100) 3.1 π 3.1/2$

$350,000= 2[ a 1 −0.55 −562.3413] (2−3.1)(1.8× 10 −10 ) (1.0) 3.1 (100) 3.1 π 3.1/2$

$350,000=−1,080.89( a 1 −0.55 −562.3413)$

$− 350,000 1,080.89 = ( a 1 ) −0.55 −562.3413$

$−323.806= ( a 1 ) −0.55 −562.3413$

$( a 1 ) −0.55 =−323.806−562.3413$

$=238.534$

$a 1 =4.76× 10 −5 m$

$=4.76× 10 −5 ×1000mm$

$a 1 =0.0476mm$

Hence the length of the crack $a1=0.0476 mm$

Now a number of cycles (N) for the crack length of $0.0476 mm$ is given by,

$N=2[(ac)2−n/2−(ai)2−n/2](2−n)(fnΔσnπn/2)$

Here,

Under compression

$Δσ=250MPaa1=0.0476mmac=3.18mmΔσ=250MPaf=1.0n=3.1c=1.8×10−10$

Therefore equation (2) becomes

$N=2[ (3.18) (2−3.1)/2− (0.0476) (2−3.1)/2(2−3.1)(1.8× 10 −10) (1.0) 3.1 (250) 3.1π 3.1/2=−429.5280.03168N=13,560 cycles$

Required number of cycles $13,560 cycles$ .

Conclusion

Hence by using strain fracture toughness of material equation and also a number of cycles equation, the number of cycles (N) before failure is $13,560 cycles$ .

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Answer 5

Chapter 7, Problem 7.38P

Interpretation Introduction

Interpretation:

The number of large stress amplitude cycles that can be sustained before failure need sto be determined, if the material has a crack of the initial length of 0.010 mm and then subjected to calculating tensile structures of 25 and 125 MPa for 350,000 cycles.

Concept introduction:

Equation of strain fracture toughness of the material.

$kk=fσπa$

Material subjected to cycle string is given by,

$N=2[(ac)2−n/2−(ai)2−n/2](2−n)(fnΔσnπn/2)$

Expert Solution & Answer

Answer to Problem 7.38P

The cycles that sustained before failure are given by 13,560 cycles.

Explanation of Solution

Given Information:

$Initial length=0.010mkk=Fracture toughness=25MPam Materials constant n=3.1 and c=1 .8×10 -10 f=1.0 σ=Compressive stress=250MPa cycles=350,000$

Calculation:

Stress fracture toughness $kk$ is given by:

$kk=fσπa$ ------------------(1)

Here,

$F=geometry factor=1.0 σ=Applied stress=250MPa a=Crack lengthkk=Fracture toughness=25MPam$

Rearranging equation (1),

$a=1π(kkfσ)a=1π(2.5(1.0)×250)a=0.00318ma=0.00318×103mma=3.19mm$

Therefore, critical crack length when 250 MPa stress is applied is given by 3.18mm

Let, a₁ is the initial flaw size when material subjected to stress in cyclic testing.

Now, the number of cycles (N) is given to find out crack (a₁)

$N=2[(ac)2−n/2−(ai)2−n/2](2−n)(fnΔσnπn/2)$ --------------------(2)

Here,

$a1=initial flow size =0.010mmac=flow needed for fractureΔσ=difference between maximum and minimum cyclical stress =125MPa-25MPa =100MPa n and c=material constant=3.1 and 1 .8×10 -10 respectivlyN=number of cycles=350,000F=1.0$

Equation (2) becomes.

$350,000= 2[ a 1 (2−3.1)/2 − (0.010) (2−3.1)/2 (2−3.1)(1.8× 10 −10 ) (1.0) 3.1 (100) 3.1 π 3.1/2$

$350,000= 2[ a 1 −0.55 −562.3413] (2−3.1)(1.8× 10 −10 ) (1.0) 3.1 (100) 3.1 π 3.1/2$

$350,000=−1,080.89( a 1 −0.55 −562.3413)$

$− 350,000 1,080.89 = ( a 1 ) −0.55 −562.3413$

$−323.806= ( a 1 ) −0.55 −562.3413$

$( a 1 ) −0.55 =−323.806−562.3413$

$=238.534$

$a 1 =4.76× 10 −5 m$

$=4.76× 10 −5 ×1000mm$

$a 1 =0.0476mm$

Hence the length of the crack $a1=0.0476 mm$

Now a number of cycles (N) for the crack length of $0.0476 mm$ is given by,

$N=2[(ac)2−n/2−(ai)2−n/2](2−n)(fnΔσnπn/2)$

Here,

Under compression

$Δσ=250MPaa1=0.0476mmac=3.18mmΔσ=250MPaf=1.0n=3.1c=1.8×10−10$

Therefore equation (2) becomes

$N=2[ (3.18) (2−3.1)/2− (0.0476) (2−3.1)/2(2−3.1)(1.8× 10 −10) (1.0) 3.1 (250) 3.1π 3.1/2=−429.5280.03168N=13,560 cycles$

Required number of cycles $13,560 cycles$ .

Conclusion

Hence by using strain fracture toughness of material equation and also a number of cycles equation, the number of cycles (N) before failure is $13,560 cycles$ .

Answer 6

Chapter 7, Problem 7.38P

Interpretation Introduction

Interpretation:

The number of large stress amplitude cycles that can be sustained before failure need sto be determined, if the material has a crack of the initial length of 0.010 mm and then subjected to calculating tensile structures of 25 and 125 MPa for 350,000 cycles.

Concept introduction:

Equation of strain fracture toughness of the material.

$kk=fσπa$

Material subjected to cycle string is given by,

$N=2[(ac)2−n/2−(ai)2−n/2](2−n)(fnΔσnπn/2)$

Expert Solution & Answer

Answer to Problem 7.38P

The cycles that sustained before failure are given by 13,560 cycles.

Explanation of Solution

Given Information:

$Initial length=0.010mkk=Fracture toughness=25MPam Materials constant n=3.1 and c=1 .8×10 -10 f=1.0 σ=Compressive stress=250MPa cycles=350,000$

Calculation:

Stress fracture toughness $kk$ is given by:

$kk=fσπa$ ------------------(1)

Here,

$F=geometry factor=1.0 σ=Applied stress=250MPa a=Crack lengthkk=Fracture toughness=25MPam$

Rearranging equation (1),

$a=1π(kkfσ)a=1π(2.5(1.0)×250)a=0.00318ma=0.00318×103mma=3.19mm$

Therefore, critical crack length when 250 MPa stress is applied is given by 3.18mm

Let, a₁ is the initial flaw size when material subjected to stress in cyclic testing.

Now, the number of cycles (N) is given to find out crack (a₁)

$N=2[(ac)2−n/2−(ai)2−n/2](2−n)(fnΔσnπn/2)$ --------------------(2)

Here,

$a1=initial flow size =0.010mmac=flow needed for fractureΔσ=difference between maximum and minimum cyclical stress =125MPa-25MPa =100MPa n and c=material constant=3.1 and 1 .8×10 -10 respectivlyN=number of cycles=350,000F=1.0$

Equation (2) becomes.

$350,000= 2[ a 1 (2−3.1)/2 − (0.010) (2−3.1)/2 (2−3.1)(1.8× 10 −10 ) (1.0) 3.1 (100) 3.1 π 3.1/2$

$350,000= 2[ a 1 −0.55 −562.3413] (2−3.1)(1.8× 10 −10 ) (1.0) 3.1 (100) 3.1 π 3.1/2$

$350,000=−1,080.89( a 1 −0.55 −562.3413)$

$− 350,000 1,080.89 = ( a 1 ) −0.55 −562.3413$

$−323.806= ( a 1 ) −0.55 −562.3413$

$( a 1 ) −0.55 =−323.806−562.3413$

$=238.534$

$a 1 =4.76× 10 −5 m$

$=4.76× 10 −5 ×1000mm$

$a 1 =0.0476mm$

Hence the length of the crack $a1=0.0476 mm$

Now a number of cycles (N) for the crack length of $0.0476 mm$ is given by,

$N=2[(ac)2−n/2−(ai)2−n/2](2−n)(fnΔσnπn/2)$

Here,

Under compression

$Δσ=250MPaa1=0.0476mmac=3.18mmΔσ=250MPaf=1.0n=3.1c=1.8×10−10$

Therefore equation (2) becomes

$N=2[ (3.18) (2−3.1)/2− (0.0476) (2−3.1)/2(2−3.1)(1.8× 10 −10) (1.0) 3.1 (250) 3.1π 3.1/2=−429.5280.03168N=13,560 cycles$

Required number of cycles $13,560 cycles$ .

Conclusion

Hence by using strain fracture toughness of material equation and also a number of cycles equation, the number of cycles (N) before failure is $13,560 cycles$ .

Answer 7

Chapter 7, Problem 7.38P

Interpretation Introduction

Interpretation:

The number of large stress amplitude cycles that can be sustained before failure need sto be determined, if the material has a crack of the initial length of 0.010 mm and then subjected to calculating tensile structures of 25 and 125 MPa for 350,000 cycles.

Concept introduction:

Equation of strain fracture toughness of the material.

$kk=fσπa$

Material subjected to cycle string is given by,

$N=2[(ac)2−n/2−(ai)2−n/2](2−n)(fnΔσnπn/2)$

Answer 8

Expert Solution & Answer

Answer to Problem 7.38P

The cycles that sustained before failure are given by 13,560 cycles.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given Information:

$Initial length=0.010mkk=Fracture toughness=25MPam Materials constant n=3.1 and c=1 .8×10 -10 f=1.0 σ=Compressive stress=250MPa cycles=350,000$

Calculation:

Stress fracture toughness $kk$ is given by:

$kk=fσπa$ ------------------(1)

Here,

$F=geometry factor=1.0 σ=Applied stress=250MPa a=Crack lengthkk=Fracture toughness=25MPam$

Rearranging equation (1),

$a=1π(kkfσ)a=1π(2.5(1.0)×250)a=0.00318ma=0.00318×103mma=3.19mm$

Therefore, critical crack length when 250 MPa stress is applied is given by 3.18mm

Let, a₁ is the initial flaw size when material subjected to stress in cyclic testing.

Now, the number of cycles (N) is given to find out crack (a₁)

$N=2[(ac)2−n/2−(ai)2−n/2](2−n)(fnΔσnπn/2)$ --------------------(2)

Here,

$a1=initial flow size =0.010mmac=flow needed for fractureΔσ=difference between maximum and minimum cyclical stress =125MPa-25MPa =100MPa n and c=material constant=3.1 and 1 .8×10 -10 respectivlyN=number of cycles=350,000F=1.0$

Equation (2) becomes.

$350,000= 2[ a 1 (2−3.1)/2 − (0.010) (2−3.1)/2 (2−3.1)(1.8× 10 −10 ) (1.0) 3.1 (100) 3.1 π 3.1/2$

$350,000= 2[ a 1 −0.55 −562.3413] (2−3.1)(1.8× 10 −10 ) (1.0) 3.1 (100) 3.1 π 3.1/2$

$350,000=−1,080.89( a 1 −0.55 −562.3413)$

$− 350,000 1,080.89 = ( a 1 ) −0.55 −562.3413$

$−323.806= ( a 1 ) −0.55 −562.3413$

$( a 1 ) −0.55 =−323.806−562.3413$

$=238.534$

$a 1 =4.76× 10 −5 m$

$=4.76× 10 −5 ×1000mm$

$a 1 =0.0476mm$

Hence the length of the crack $a1=0.0476 mm$

Now a number of cycles (N) for the crack length of $0.0476 mm$ is given by,

$N=2[(ac)2−n/2−(ai)2−n/2](2−n)(fnΔσnπn/2)$

Here,

Under compression

$Δσ=250MPaa1=0.0476mmac=3.18mmΔσ=250MPaf=1.0n=3.1c=1.8×10−10$

Therefore equation (2) becomes

$N=2[ (3.18) (2−3.1)/2− (0.0476) (2−3.1)/2(2−3.1)(1.8× 10 −10) (1.0) 3.1 (250) 3.1π 3.1/2=−429.5280.03168N=13,560 cycles$

Required number of cycles $13,560 cycles$ .

Answer 12

Conclusion

Hence by using strain fracture toughness of material equation and also a number of cycles equation, the number of cycles (N) before failure is $13,560 cycles$ .

Answer 13

Ch. 7 - Prob. 7.1P Ch. 7 - Prob. 7.2P Ch. 7 - Prob. 7.3P Ch. 7 - Prob. 7.4P Ch. 7 - Prob. 7.5P Ch. 7 - Prob. 7.6P Ch. 7 - Prob. 7.7P Ch. 7 - Prob. 7.8P Ch. 7 - Prob. 7.9P Ch. 7 - Prob. 7.10P

Answer to Problem 7.38P

Explanation of Solution

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