Essentials Of Materials Science And Engineering
Essentials Of Materials Science And Engineering
4th Edition
ISBN: 9781337670845
Author: ASKELAND
Publisher: Cengage
Question
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Chapter 7, Problem 7.37P
Interpretation Introduction

(a)

Interpretation:

The critical crack length required to cause a fracture needs to be determined.

Concept Introduction:

Relation of critical fracture toughness is as follows:

kk=fσπac

Here,

kk=Critical fracture toughnessσ=Continued cycled stressf=Geometry factor

Expert Solution
Check Mark

Answer to Problem 7.37P

Critical crack length to cause fracture is given by 0.00853 m.

Explanation of Solution

Given Information:

Critical fracture toughness=kk=55 MPamContinued cycled stress=σ=300MPaGeometry factorf=1.12Initial crack sizea1=0.02mmConstantcandn=2× 10 11and3.4respectively

Critical fracture toughness is given by

kk=fσπac

Here,

kk=Critical fracture toughness=55 MPamσ=Continued cycled stress=300MPaf=Geometry factor=1.12

Squaring on both sides

kk2=(f2σ2πac) a c = 1 π [ k k fσ ] 2 -------------------------(1)

Equation (1) becomes

a c = 1 π [ 55 1.12×300 ] 2 =0.0268πac=0.00853m

The critical length that results to fractured is 0.00853 m.

Interpretation Introduction

(b)

Interpretation:

The cycles that cause product failure needs to be determined, if critical length is 0.00853 m.

Concept Introduction:

Equation for the number of cycles that causes failure is as follows:

N=2[(ac)2n/2(ai)2n/2](2n)(fnΔσnπn/2)

Expert Solution
Check Mark

Answer to Problem 7.37P

Number cycle that will cause product failure is 50366 cycles.

Explanation of Solution

Given Information:

Critical fracture toughness=kk=55 MPamContinued cycled stress=σ=300MPaGeometry factorf=1.12Initial crack sizea1=0.02mmConstantcand n=2×10 -11and 3.4respectively

ac=Critical length causeFracture=0.00853m

Since, equation for number of cycles that causes failure is as follows:

N=2[(ac)2n/2(ai)2n/2](2n)(fnΔσnπn/2)

Here,

a1=initial flow size =0.02mm

Constantcandn=2×10-11and3.4respectively

ac=Length causes fracture=0.00853m

Putting the values,

N=2[ (0.00853) (23.4)/2 (0.02× 10 3 ) (23.4)/2(23.4)×(2× 10 11)× (1.12) 3.4 (300) 3.4π 3.4/2N=2[28.076021946.61023](1.4)(2× 10 11)(1.4701)(2.646× 108)7=50365.7cyclesN=50366cycles

Hence a number of the cycle causes product failure is 50366 cycles.

Now, the critical length that causes a fracture, if the product is removed when a crack reaches 15% of critical crack length

( a i ) n =15%×( a c )Here,ac=0.00853 m ( a i ) n =15(0.00853)=0.00128 m

Now, the number of cycles required when the product fails can be calculated as follows:

Nf=2[ ( a c ) (2n)/2 ( a i )n (2n)/2(2n)(fnΔσnπ n/2) ( a i ) n =0.00128mc=2×1011n=3.4ac=0.00853m

Equation (1) becomes,

Nf=2×[ (0.00853) (23.4)/2 (0.00128) (23.4)/2(23.4)×(2× 10 11)× (1.12) 3.4 (300) 3.4π 3.4/2Nf=2043 cycles

Cycles cause product failure is 2043 cycles.

Interpretation Introduction

(c)

Interpretation:

If cycle causes failure is given by 2043 cycles, the percentage of the useful life of the product if it is removed from service needs to be determined.

Concept Introduction:

The percentage of the useful life of the product is given by the following relation:

NfN×100

Expert Solution
Check Mark

Answer to Problem 7.37P

Percentage of the useful life of product remains is 4%.

Explanation of Solution

Given Information:

N=No. of cycle that cause fails=50366cyclesNf=Cycle causes product when removes and failures2043cycle

The percentage that of useful life of the product remains is as follows:

NfN×100=204350366×100=0.04×100=4%

Hence the percentage of useful life is 4%.

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