Essentials Of Materials Science And Engineering, Si Edition
Essentials Of Materials Science And Engineering, Si Edition
4th Edition
ISBN: 9781337629157
Author: Donald R. Askeland, Wendelin J. Wright
Publisher: Cengage Learning
Question
Book Icon
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%.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
How can construction project managers find a balance between speeding up schedules and the risks of making more mistakes and needing rework, especially when using methods like fast tracking?
Derive the equation below ah ap ax 12μ ax, +( ah ap ay 12μ ay Where P P (x, y) is the oil film pressure. 1..ah 2 ax
Word File Edit View Insert Format Tools Table Window Help PEPSI AutoSave 5-Marca, Christopher Read-Only Compatibility Mode - Saved to my Mac Insert Draw Design Layout References Mailings Review View Picture Format X Aptos (Body) ▾ 10 A A Aa Po BIU x, x A DA EvE E1INT AaBbCcl Pleading 2 Lalacoder AaBbCcl AalibGDdEe AaBbCcDr Abde ABC Normal Title No Spacing Heading 3 Heading 3 Subtitie Suble B d Only To save a copy of this document, click Duplicate. Problem 10) For the Boolean.expression, which is the correct MSOP expression? I Y = wxyz+xy+w Expression A. I Y = wxyz+y 663 words Expression B. English (United States) Accessibility: Investigate A APR LO M O stv I
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
MATLAB: An Introduction with Applications
Engineering
ISBN:9781119256830
Author:Amos Gilat
Publisher:John Wiley & Sons Inc
Text book image
Essentials Of Materials Science And Engineering
Engineering
ISBN:9781337385497
Author:WRIGHT, Wendelin J.
Publisher:Cengage,
Text book image
Industrial Motor Control
Engineering
ISBN:9781133691808
Author:Stephen Herman
Publisher:Cengage Learning
Text book image
Basics Of Engineering Economy
Engineering
ISBN:9780073376356
Author:Leland Blank, Anthony Tarquin
Publisher:MCGRAW-HILL HIGHER EDUCATION
Text book image
Structural Steel Design (6th Edition)
Engineering
ISBN:9780134589657
Author:Jack C. McCormac, Stephen F. Csernak
Publisher:PEARSON
Text book image
Fundamentals of Materials Science and Engineering...
Engineering
ISBN:9781119175483
Author:William D. Callister Jr., David G. Rethwisch
Publisher:WILEY