Materials Science And Engineering Properties
Materials Science And Engineering Properties
1st Edition
ISBN: 9781111988609
Author: Charles Gilmore
Publisher: Cengage Learning
bartleby

Concept explainers

Question
Book Icon
Chapter 11, Problem 11.11P

(a)

To determine

The critical crack length for failure of the steel.

(a)

Expert Solution
Check Mark

Answer to Problem 11.11P

The critical crack length for failure of the steel is 0.102m .

Explanation of Solution

Given:

Number of cycles is 4000 .

Stress intensity range is 0 MPa to 500MPa .

The critical-stress intensity factor is 200MPam1/2 .

Crack growth rate is dadN=A(ΔK)6 .

Value of A is 2.5×1014 .

Concept used:

Write the expression for stress intensity range.

  Kmax=Yσmax(πaf)1/2 …… (1)

Here, Kmax is the max stress intensity range, af is the crack length at fracture, Y is the geometrical factor and σmax is the maximum stress..

For internal cracks the value of geometrical factor is 1.0 .

Calculation:

Substitute 1.0 for Y , 500MPa for σmax and 200MPam1/2 for Kmax in equation (1).

  200MPam1/2=(1.0)(500MPa)(π a f)1/2(π a f)1/2=0.4 m1/2πaf=0.16maf=0.051 m

Calculate the critical length of failure.

  2a=2(0.051 m)=0.102 m

Conclusion:

Thus, the critical crack length for failure of the steel is 0.102m .

(b)

To determine

The minimum detectable crack length required for non-destructive to guarantee 4000 cycles fatigue life.

(b)

Expert Solution
Check Mark

Answer to Problem 11.11P

The minimum detectable crack length required for non-destructive to guarantee 4000 cycles fatigue life is 1.0206×104 m .

Explanation of Solution

Concept used:

Write the expression for the number of crack propagation cycle to failure.

  Nf=1(n2+1)(1AΔσnπ n/2)(af( n/2)+1ai( n/2)+1) …… (2)

Here, Nf is the number of crack propagation cycle to failure, n is the slope of plot between log(dadN) versus log(ΔK) , Δσ is the stress range, A is the constant and ai is the initial detectable crack length.

Write the expression for crack growth rate.

  dadN=A(ΔK)6

Take log on both sides.

  log(dadN)=logA+6log(ΔK)

The above expression represents a straight line with slope 6 . Thus, the slope of of plot between log(dadN) versus log(ΔK) is 6 .

Calculation:

Substitute 4000 for Nf , 6 for n , 500MPa for Δσ , 0.051 m for af and 2.5×1014 MPam for A in equation (2).

  4000=1( 6 2 +1)(1 ( 2.5× 10 14 ) ( 500MPa ) 6 π 6/2 )( ( 0.051 ) ( 6/2 )+1ai ( 6/2 )+1)4000=13+1(1 12× 10 3 )[( 0.051)2ai2]4000=124× 103[( 0.051)2ai2]

Simplify above expression for ai .

  (0.051)2ai2=4000(24× 103)ai2=(0.051)2+4000(24× 103)1ai2=96000384.47ai=1.0206×104 m

Conclusion:

Thus, the minimum detectable crack length required for non-destructive to guarantee 4000 cycles fatigue life is 1.0206×104 m .

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
A very large, steel plate of yield stress 200 MPa has a crack at the centre of length 18 mm, orientated along the x-axis. If the plate is subjected to far field tensile loading of magnitude 179 MPa and is assumed to be in a state of plane strain, determine the extent of the plastic region at the crack tip, along the x-axis. You may assume Poisson's ratio is (1/3). Express your answer as an integer value of mm.
A very large, steel plate of yield stress 200 MPa and Poisson's ratio 0.3 has a crack at the centre of length 34 mm, orientated along the x-axis. If the plate is subjected to far field tensile loading of magnitude 177 MPa and is assumed to be in a state of plane stress, determine the extent of the plastic region at the crack tip, along the x-axis. You may assume Poisson's ratio is (1/3). Express your answer as an integer value of mm.
A structural component in the shape of a flat plate 20.8 mm thick is to be made from a metal with yield strength of 533 MPa and a critical fracture toughness of 22.0 MPa-m!2. Assume a crack forms in the plate, and the geometry of the crack yields a Y value of 1.5. If the plate is designed to a design stress equivalent to 0.3 times the yield strength, what is the critical crack length? Equations: 1/2 () Om К = 2 σο K. = YocTa %3D Pt
Knowledge Booster
Background pattern image
Civil Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Materials Science And Engineering Properties
Civil Engineering
ISBN:9781111988609
Author:Charles Gilmore
Publisher:Cengage Learning