Materials Science And Engineering Properties
1st Edition
ISBN: 9781111988609
Author: Charles Gilmore
Publisher: Cengage Learning
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Chapter 11, Problem 9ETSQ
To determine
The procedure not to be expected to increase the fatigue crack initiation life of plain low carbon steel.
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A high-strength steel has a yield strength of 1380 MPa and fracture toughness of 91 MPavm. A
surface crack of 2.5 mm is found at the surface. At what max. applied stress level will catastrophic
failure occur? (Y = 1.00 for internal crack, and Y=1.12 for surface crack)
Chapter 11 Solutions
Materials Science And Engineering Properties
Ch. 11 - Prob. 1CQCh. 11 - Prob. 2CQCh. 11 - Prob. 3CQCh. 11 - Prob. 4CQCh. 11 - Prob. 5CQCh. 11 - Prob. 6CQCh. 11 - Prob. 7CQCh. 11 - Prob. 8CQCh. 11 - Prob. 9CQCh. 11 - Prob. 10CQ
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- An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 34 MPaym. It has been determined that fracture results at a stress of 221 MPa when the maximum (or critical) internal crack length is 2.94 mm. a) Determine the value of Yona for this same component and alloy at a stress level of 287 MPa when the maximum internal crack length is 1.47 mm. MPaymarrow_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_forwardAfter an inspection it is found that a structural ceramic part has no flaws greater than 100 micrometers in size, calculate the maximum service stress (in MPa) available with SiC. Assume that Y= 1arrow_forward
- 2. Please estimate the number of cycles to failure of a steel specimen under tensile fatigue loading with the following parameters. The R ratio is 3, mean stress 200 MPa, yield strength 450 MPa, ultimate tensile strength 560 MPa, Young’s modulus 200 GPa, KIC = 140 MPa . Assume the initial crack length is 0.1 mm.arrow_forward(b) An aluminium plate (100 mm width x 300 mm height x 4 mm thick) with a centre-crack is subjected to a mode-l service load of 40 kN. Assume that the fracture toughness of the material is 24 MPa Vm and Y-1.0. (ii) What is the safety factor on crack length? Assume that the yield strength of aluminium is 425 MPa.arrow_forwardA structural component in the shape of a flat plate 24.3 mm thick is to be fabricated from a metal alloy for which the yield strength and plane strain fracture toughness values are 533 MPa and 22.0 MPa-m1/2, respectively. For this particular geometry, the value of Y is 1.3. Assuming a design stress of 0.4 times the yield strength, calculate the critical length of a surface flaw. What formulas do i use ? And how do i use them?arrow_forward
- 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_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_forward2- What is the largest size (mm) internal through crack that a thick plate of aluminium alloy 7075-T651 can support at an applied stress of (a) three-quarters of the yield strength and (b) one-half of the yield strength? Assume Y = 1. for 7075-T651, KỊC = 24.2 MPa ym and oYS = 495 MPa.arrow_forward
- d) No increase Which of the following is not a property of steel materials? C a) Homogeneous and isotropic b) Linearly elastic stress-strain behavior c) Recyclable d) Fire-resistantarrow_forwardGiven your understanding of what initiates and controls failure in materials, which of the following will increase the failure strength or lifetime of a test piece or component and why? a. Decreasing the difference between the maximum and minimum stress values, as this effects the stress concentration factor b. Decreasing the temperature below the brittle-ductile transition temperature, to make it harder C. Polishing to reduce surface defects Od. Increasing its volume, to give a larger cross sectional area Oe. Increasing the grain size so there are less grain boundaries to initiate failurearrow_forwardQ3: A cylindrical specimen of steel having an original diameter of (12.8mm) is tensile tested to fracture and found to have engineering fracture strength of (450MP ). If its cross-sectional diameter at fracture is (10.7mm), determine: (1) the ductility in terms of percent reduction in area and (2) the true stress at fracture.arrow_forward
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