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ASNT NDT LEVEL III NDT BASIC Question Bank Problems and Solutions MATERIALS AND PROCESSING FOR NDT TECHNOLOGY 29. A particular material has an Ultimate Tensile Stress (UTS) of 45000 psi and a yield strength of 30000 psi. An allowable design stress is l5000 psi. What is the factor of safety based on UTS? Safety factor = Ultimate Tensile Stress (UTS) / Allowable design stress = 45000/ 15000 = 3

30. A steel with 28 points of carbon contains Carbon % = Carbon points/100 = 28/100. = 0.28% carbon

72. The tensile stress in a bar of 15 in. long and 3 in. wide and 2 in. thick subjected to a tensile load, acting along the length, of 30000 pounds will be equal to Stress = Load / Initial cross -sectional area = 30000 / (3 x 2) = 5000 psi

155. Calculate the factor of safety if a material has a working stress of 15000 lbs per square inch and an Ultimate Tensile Stress (UTS) of 60000 lbs per square inch. Safety factor = Ultimate Tensile Stress (UTS) / Allowable design stress = 60000/ 15000 =4

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239. When carrying out a tensile test the maximum load to fracture is 3.2 tons and the test piece is 0 5 inches diameter. What is the tensile strength? Diameter = D = 0 5 inches Radius = r = D/2 = 0 5/2 = 0.25 inches Tensile strength = Load / Initial cross -sectional area = Load / (3.14 x r2) = 3.2 / (3.14 x 0.252) = 16.29 tons per square inch

275. Calculate the factor of safety given that a material has an Ultimate Tensile Strength (UTS) of 16000 psi. An allowable design stress is 4000 psi. Safety factor = Ultimate Tensile Stress (UTS) / Allowable design stress = 16000/ 4000 = 4

291. ASTM grain size number 4 corresponds to: n =4 N = 2 (n - 1) = 2 (4 - 1) = 8 grain in 1 sq. inch at 100X.

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OTHER NDT METHODS 74. Total atmospheric pressure at sea level is 101 kPa. Find the partial pressure of helium (He), if helium concentrations is 5 ppm (by volume). Helium concentrations = CHe-gas = 5 ppm = 0.0005% Ptotal = 101 kPa PHe = Ptotal (CHe-gas / 100) = 101 x (0.0005 /100) = 0.0005 kPa

177. An excellent radiograph is obtained under given exposure conditions with a tube current of 5 milli-amperes and an exposure time of 14 minutes. If other conditions are not changed, what exposure time would be required if the X-ray tube current could be raised to 10 milli-amperes? M1 T1 = M2 T2 5 x 14 = 5 x T2 T2 = 5 x 14 / 10 = 7 minutes

186. Express 2220 GBq in TBq 1 GBq = 109 Bq 2220 GBq = 2220 x 109 Bq = 2220 x 109/ 1012 = 2.22 TBq

191. Converting old units to SI units, 100 rads = 1 gray (Gy). How- many mGys are there in 1 rad? 1 Gy = 100 Rads 1 Rads = 1/100 Gy = (1/100) x 1000 Gy = 10 mGy

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192. What thickness of a flat type penetrameter (ASTM) would be used on a 35 mm material thickness for a 2% sensitivity? T=S X where T = thickness of a flat type penetrameter (mm) X = section thickness to be radiographed (mm) S = penetrameter sensitivity (%) T = (2 / 100 ) x 35 = 0.7 mm

205. Now the activity of Ir-192 source is 8 TBq, what will be its approximate activity after 150 days? A1 = Initial Activity (today) = 8 TBq

A2 = Activity after 150 d = ?

time gap = t = 150 d number of half-life = n = t/ T1/2 = 150/75 = 2 A2 / A1 = 1/2n = 1/22 = 1/4 A2 = A1 x (1/4) = 8 x (1/4) = 2 TBq

206. An unshielded radioactive source gives a dose rate of 1600 mR/h at 1.8 m (6 ft). What would be the unshielded dosage rate at 7.3 m (24 ft)? I2

I1 = 1600 mR/h d1 = 6 ft

?

d2 = 24 ft

I2 = I1 (d1) / (d2)

= 1600 (6) / (24)

= 100 mR/h

207. The tenth-value layer of lead for 250 kVp X-rays is 2.9 mm. What thickness of lead would be needed to reduce the exposure rate for this energy of radiation by a factor of 1000? Dose reduction factor = 1000 = 103 Required shielding =3 TVL = 3 x 2.9 = 8.7 mm 4

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208. If an exposure time of 2 minutes was necessary using a 0.6 m (2 ft) source-to-film distance for a particular exposure, what time would be necessary if a 0.3 m (1 ft) source-to-film distance is used and all other variables remain the same? t1 /(d1)2

= t2 / (d2)2 t2

= t1 (d2)2 / (d1)2 = 2 x (1)2 / (2)2

= 0.5 min. = 30 s

209. The use of 3 half-value layers of shielding will reduce the radiation intensity by a factor of: Dose reduction factor = 2n = 23 = 8

Prepared by: PRAVEEN INSTITUTE OF RADIATION TECHNOLOGY (NDT TRAINING DIVISION) CHENNAI- 600 063 INDIA. www.pirtchennai.com Email: [email protected] 6.5, 17 q

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