Essentials of Materials Science and Engineering, SI Edition
Essentials of Materials Science and Engineering, SI Edition
4th Edition
ISBN: 9781337672078
Author: ASKELAND, Donald R., WRIGHT, Wendelin J.
Publisher: Cengage Learning
Question
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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)2n/2(ai)2n/2](2n)(fnΔσnπn/2)

Expert Solution & Answer
Check Mark

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 constantn=3.1andc=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)2n/2(ai)2n/2](2n)(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=3.1and1 .8×10 -10 respectivlyN=number of cycles=350,000F=1.0

Equation (2) becomes.

350,000= 2[ a 1 (23.1)/2 (0.010) (23.1)/2 (23.1)(1.8× 10 10 ) (1.0) 3.1 (100) 3.1 π 3.1/2

350,000= 2[ a 1 0.55 562.3413] (23.1)(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 =323.806562.3413

=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)2n/2(ai)2n/2](2n)(fnΔσnπn/2)

Here,

Under compression

Δσ=250MPaa1=0.0476mmac=3.18mmΔσ=250MPaf=1.0n=3.1c=1.8×1010

Therefore equation (2) becomes

N=2[ (3.18) (23.1)/2 (0.0476) (23.1)/2(23.1)(1.8× 10 10) (1.0) 3.1 (250) 3.1π 3.1/2=429.5280.03168N=13,560cycles

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 13,560cycles.

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