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
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Chapter 11 Solutions
Materials Science And Engineering Properties
- For a specimen of a steel alloy with a plane strain fracture toughness of 80 MPa√m, fracture results at a stress of 510 MPa when the maximum (or critical) internal crack length is 6 mm. For the same alloy, will fracture occur at a stress level of 380 MPa when the maximum internal crack is 9.0 mm? Why or why not? Select the most appropriate answer based on your calculation. Select one: a. It will not fracture b. Not enough information c. It will fracturearrow_forwardThe steel has yield strength of 550 MPa and a fracture toughness of 40 MPa m^1/2.What will be the limiting design stress if the maximum tolerable crack is 3.0 mm in length and no plastic deformation is permitted?arrow_forwardA ceramic part is used under a complete reverse cyclic stress with a stress amplitude (S) of 250 MPa. The yield strength and fracture toughness of materials is 550 MPa and 12.5 MPa*sqrt(m), respectively. Y is 1.4. What is the critical surface crack length?arrow_forward
- State the difference between cleavage and fracture of minerals, using quartz and mica as examples.arrow_forwardThe data shown in the table below were obtained from a tensile test of high-strength steel. The test specimen had a diameter of 13mm and a gage length of 50mm. At the fracture, the elongation between the gage marks was 3.0mm and the minimum diameter was 10.7mm. Plot the conventional stress-strain curve for the steel and determine the proportional limit, modulus of elasticity (i.e the slope of initial part of the stress-strain curve), yield stress at 0.1% offset, ultimate stress, percent elongation in 50mm, and percentage reduction in area Tensile-Test Data Load(kN) Elongation (mm) 5 0.005 10 0.015 30 0.048 50 0.084 60 0.099 64.5 0.109 67.0 0.119 68.0 0.137 69.0 0.160 70.0 0.229 72.0 0.259 76.0 0.330 84.0 0.584 92.0 0.853 100.0 1.288 112.0 2.814 113.0 Fracturearrow_forwardA steel specimen is tested in tension. The specimen is 25 mm wide by 12.5 mm thick in the test region. By monitoring the load dial of the testing machine, it was found that the specimen yielded at a load of 160 kN and fractured at 214 kN. a. Determine the tensile stress at yield and at fracture. b. If the original gauge length was 100 mm, estimate the gauge length when the specimen is stressed to 1/2 the yield stress.arrow_forward
- Materials Science And Engineering PropertiesCivil EngineeringISBN:9781111988609Author:Charles GilmorePublisher:Cengage Learning