Non-destructive Evaluation NDE Dept of Physics and Materials Science City University of Hong Kong H E. Davis, G.E. Troxell, in chapter 16 of"The Testing of Engineering Materials".1982 J.S. Ceurter et al. "Advanced Materials Processes"(April 2002), p 29-31 T. Adams, "Advanced Materials Processes"(April 2002), p 32-34 202128 & Dr Jonathan c.Y. Chung: NDE
Non-destructive Evaluation NDE Dept. of Physics and Materials Science City University of Hong Kong References: 1. H.E. Davis, G.E. Troxell, in chapter 16 of “The Testing of Engineering Materials”, 1982. 2. J.S. Ceurter et al., “Advanced Materials Processes” (April 2002), p.29-31. 3. T. Adams, “Advanced Materials Processes” (April 2002), p.32-34. ❖2021/2/8 ❖Dr. Jonathan C.Y. Chung: NDE ❖1
Various purposes Locate defects( Why ? Determine dimension, physical, or mechanical characteristics Determine residue StreSs XRD 202128 & Dr Jonathan c.Y. Chung: NDE 今2
Various Purposes • Locate defects (Why ?) • Determine dimension, physical, or mechanical characteristics • Determine Residue Stress (XRD) ❖2021/2/8 ❖Dr. Jonathan C.Y. Chung: NDE ❖2
Advantage of knowing the defects i Defects are usually stress raiser Stress raiser can cause pre-mature failure Over design to overcome pre-mature failure> Bulky/heavy design Catastrophic/sudden/unpredicted failure loss of lives and money Quality control Better design 202128 & Dr Jonathan c.y. Chung: NDE
Advantage of Knowing the defects • Defects are usually stress raiser • Stress raiser can cause pre-mature failure→ Over design to overcome pre-mature failure→ Bulky/heavy design • Catastrophic/sudden/unpredicted failure→ loss of lives and money • Quality control • Better design ❖2021/2/8 ❖Dr. Jonathan C.Y. Chung: NDE ❖3
Better design example) Consider a rectangular bar 10mm x 5 mm which will be used to support some load. the steel chosen had yield strength, tensile strength and fracture toughness being 600MPa, 900MPa and 40MPavm If the corresponding design safety factors are 1.2, 1.6 and 1.5 respectively. What is the allowable load? (a)yielding failure(25 kN) (tEnsile fracture (>28. 1 kN) (c)Fracture toughness(crack size dependant)2 mm 168kN;1mm:23.6kN;0.1mm:75.2kN 202128 & Dr Jonathan c.y. Chung: NDE
Better design (example) Consider a rectangular bar 10mm x 5 mm which will be used to support some load. The steel chosen had yield strength, tensile strength and fracture toughness being 600MPa, 900MPa and 40MPam. If the corresponding design safety factors are 1.2, 1.6 and 1.5 respectively. What is the allowable load? (a)Yielding failure (>25 kN) (b)Tensile fracture (>28.1 kN) (c)Fracture toughness (crack size dependant)2 mm: 16.8kN; 1mm: 23.6kN; 0.1mm: 75.2kN ❖2021/2/8 ❖Dr. Jonathan C.Y. Chung: NDE ❖4
Yield strength(plastic deformation) area =10 mmx5 mm=50x 10-6 m2 max load yield strength x area): safety factor =(600MPaX50X106m2)÷1.2 =25kN (plastic deformation at load 25 kN) 202128 & Dr Jonathan c.y. Chung: NDE
Yield strength (plastic deformation) area = 10 mm x 5 mm = 50 x 10-6 m2 max. load = (yield strength x area) safety factor = (600MPa x 50 x 10-6 m2 ) 1.2 = 25 kN (plastic deformation at load > 25 kN) ❖2021/2/8 ❖Dr. Jonathan C.Y. Chung: NDE ❖5