Concept explainers
(a)
The critical crack length for failure of the steel.
(a)
Answer to Problem 11.11P
The critical crack length for failure of the steel is
Explanation of Solution
Given:
Number of cycles is
Stress intensity range is
The critical-stress intensity factor is
Crack growth rate is
Value of
Concept used:
Write the expression for stress intensity range.
Here,
For internal cracks the value of geometrical factor is
Calculation:
Substitute
Calculate the critical length of failure.
Conclusion:
Thus, the critical crack length for failure of the steel is
(b)
The minimum detectable crack length required for non-destructive to guarantee 4000 cycles fatigue life.
(b)
Answer to Problem 11.11P
The minimum detectable crack length required for non-destructive to guarantee 4000 cycles fatigue life is
Explanation of Solution
Concept used:
Write the expression for the number of crack propagation cycle to failure.
Here,
Write the expression for crack growth rate.
Take log on both sides.
The above expression represents a straight line with slope
Calculation:
Substitute
Simplify above expression for
Conclusion:
Thus, the minimum detectable crack length required for non-destructive to guarantee 4000 cycles fatigue life is
Want to see more full solutions like this?
Chapter 11 Solutions
Materials Science And Engineering Properties
- Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 43.1° and 47.9°, respectively, with the tensile axis. If the critical resolved shear stress is 20.7 MPa, will an applied stress of 45 MPa cause the single crystal to yield or slip? If not, what stress will be necessary?arrow_forwardSketch the stress-strain curve of the following tension test??? In the tension test of a metal fracture occurs at maximum load. The conditions at fracture were: A = 100 mm and L, = 60 mm, the initial values were: A = 150 mm and L. 40 mm.arrow_forwardFor germanium at a tensile stress of 410 MPa and a temperature of 332C , what is the primary creep mechanism? The shear modulus of germanium is 41 GPa.arrow_forward
- State the difference between cleavage and fracture of minerals, using quartz and mica as examples.arrow_forwardThe aluminum bar ACDB is rigidly connected to a steel tube in CD. The data is presented in the table. If the load P is equal to 125 kN, determine: The stress of aluminum in segment AC. Answ: 63.66 MPa The stress of aluminum in segment CD. Answ: 13.93 MPa The stress in steel in segment CD. Answ: 39.79 MPa The total deformation of the bar. Answ: 0.29273 mmarrow_forward
- Materials Science And Engineering PropertiesCivil EngineeringISBN:9781111988609Author:Charles GilmorePublisher:Cengage Learning