Non-destructive Evaluation（无损检测）NDE（PPT讲稿）

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 40MPam. 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

Tensile strength(Catastrophic failure area =10 mmx5 mm=50x 10-6 m2 max load (tensile strength x area) )= safety factor =(900MPaX50X106m2)÷1.6 =28.1kN (tensile fracture at load>28. 1 kN) 202128 & Dr Jonathan c.y. Chung: NDE

Tensile strength (Catastrophic failure) area = 10 mm x 5 mm = 50 x 10-6 m2 max. load = (tensile strength x area)  safety factor = (900MPa x 50 x 10-6 m2 )  1.6 = 28.1 kN (tensile fracture at load > 28.1 kN) ❖2021/2/8 ❖Dr. Jonathan C.Y. Chung: NDE ❖6

Fracture ToughneSS (require information of crack length) MC=ao vaa Assume geometric correction factor, a=1 max=Kc/V(πa) Max load oXA÷( safety factor) KIc /v(a)x A-(safety factor =40 MPam/(31416Xa)×50×106m2÷ safety factor When a=2mm, max load=(2000÷0.07927)/1.5=16.8kN When a=1mm, max load=(2000÷0.05605)15=236kN When a=0.1mm, max load=(2000÷0.01772)1.5=752kN 202128 & Dr Jonathan c.y. Chung: NDE

Fracture Toughness (require information of crack length) KIC =   (a) Assume geometric correction factor,  = 1 max = KIC /(a) Max load =  x A  (safety factor) = KIC /(a) x A  (safety factor) = 40MPam /(3.1416 x a) x 50 x 10-6 m2  (safety factor) When a = 2 mm, max load = (2000  0.07927)/1.5 = 16.8 kN When a = 1 mm, max load = (2000  0.05605)/1.5 = 23.6 kN When a = 0.1 mm, max load = (2000  0.01772)/1.5 = 75.2 kN ❖2021/2/8 ❖Dr. Jonathan C.Y. Chung: NDE ❖7

NDE methods for location of defects Surface defects Internal defects detection Magnetic particle method detection Radiographic methods Visual inspection 8 Electromagnetic methods 冷 Eddy current method o liquid penetrant test &o Barkhausen Noise Inspection ☆ Magnetic particle Principle method Material defects(grinding damage re-tempering burn, Re- g burn residue stresses 令 Acoustic methods 202128 & Dr Jonathan c.y. Chung: NDE

NDE methods for location of defects Surface defects detection ❖Visual inspection ❖Liquid penetrant test ❖Magnetic particle method ❖2021/2/8 ❖Dr. Jonathan C.Y. Chung: NDE ❖8 Internal defects detection ❖ Magnetic particle method ❖ Radiographic methods ❖ Electromagnetic methods ❖ Eddy current method ❖ Barkhausen Noise Inspection Principle Material defects (grinding damage, re-tempering burn, Re￾hardening burn, residue stresses ❖ Acoustic methods

Visual inspection It should never be omitted Use low-power magnifying glass or microscopes (remember to take permanent photographic record ■ Surface roughness Touch inspection using finger along the surface(2-3 cm/s Light reflection method ☆ No-parallex method ■ Penetrant test 202128 & Dr Jonathan c.y. Chung: NDE ◆9

Visual inspection ◼ It should never be omitted. ◼ Use low-power magnifying glass or microscopes (remember to take permanent photographic record) ◼ Surface roughness: ❖Touch inspection using finger along the surface (2-3 cm/s.) ❖Light reflection method ❖No-parallex method ◼ Penetrant test ❖2021/2/8 ❖Dr. Jonathan C.Y. Chung: NDE ❖9

Penetrant test Suitable for locating surface discontinuities, such as cracks seams, laps, laminations in non-porous materials Applicable to in-process final, and maintenance inspection ■ ASTME165 General procedure: Thoroughly clean the surface Apply penetrant on the surface Liquid penetrant enter small openings by capillary action Remove liquid completely and apply developer ( dry or wet uThe penetant bleed out onto the surface showing the location of the surface defect 202128 & Dr Jonathan c.Y. Chung: NDE 今10

Penetrant test ◼ Suitable for locating surface discontinuities, such as cracks, seams, laps, laminations in non-porous materials. ◼ Applicable to in-process, final, and maintenance inspection. ◼ ASTM E 165 General procedure: ◼ Thoroughly clean the surface ◼ Apply penetrant on the surface ◼ Liquid penetrant enter small openings by capillary action ◼ Remove liquid completely and apply developer (dry or wet) ◼ The penetant bleed out onto the surface showing the location of the surface defect ❖2021/2/8 ❖Dr. Jonathan C.Y. Chung: NDE ❖10