.How does the concept of strain hardening affect the ductility and toughness of materials in structural integrity assessments?
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This occurs due to the rearrangement and alignment of dislocations in the material's microstructure. Show more…
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Explain what the meaning of strain hardening is, how it happens in the materials. Do you think strain hardening is useful? Explain your answer
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A novel steel composite is being tested to determine its toughness. Compact tension samples are fatigue cracked and then loaded to failure. Three samples have been tested as shown in Table 1. Table 1 Data for Question 4 Sample W (mm) B (mm) a (mm) Load at failure P (kN) 1 50 5 24.9 6.2 2 50 20 25.1 28.1 3 50 30 24.6 42.5 a. i. Use the ‘Compact tension specimen’ geometry in the ‘K-calculator’ spreadsheet to calculate the value of K at failure for the three samples. ii. In order for a valid fracture toughness value to be obtained, both the sample thickness B and the final crack length, a, should satisfy B, a ≥ 2.5(KQ / σyield)² where KQ is the estimated toughness value of K at failure, and σyield is the yield strength of the material, which in this case is 700 MPa. Calculate whether the samples are sufficiently thick (i.e. whether B is large enough) to meet this criterion. Are the final crack lengths acceptable? b. A cylindrical pressure vessel of 2 m radius and 0.05 m wall thickness is manufactured from a steel with σyield of 700 MPa and fracture toughness of 90 MPa ∙∙m. The internal pressure of the vessel during operation is expected to reach 0.75 MPa. Before the component is installed, a non-destructive testing (NDT) contractor reveals a flaw of depth 17 mm. i. Assess if the component is safe to be used. Use the ‘Plate in tension – through-thickness edge crack’ geometry in the K-calculator. You will need to calculate the hoop stress in the wall of the pressure vessel, assuming it to be a thin-walled cylinder. ii. A safety-conscious engineer orders the flaw to be repaired. After repairing, NDT reveals that welding has introduced a new flaw, 6 mm in depth. It is assumed that as a result of welding, the residual stress in the region of the flaw is close to the yield strength of the steel. Will the vessel be safe to use in this situation? Explain how you have come to your conclusion. iii. Careful weld simulation reveals that in the region of the flaw, the residual stress is actually only 25% of the yield stress. Is the flaw safe under these conditions? iv. The safety engineer wants to purchase some new NDT equipment to detect flaws. Considering the stress conditions described in part (iii), if a flaw must be detected before it grows to half the critical length, which of the devices given in Table 2 would be suitable? Explain your answer. Table 2 List of possible NDT devices Device Minimum detectable flaw (mm) 1 1 2 5 3 10 4 16
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