Question: Assume you are recruited by a construction company in Fiji as a Trainee Engineer. Your company received a metal bar for construction work, while the construction engineer would like to know the crystal structure of the metal bar, such as, is a BCC or FCC or HCP. Neither company nor in Fiji has got any sophisticated high-cost equipment or facility to test its crystal structure. However, the company possesses simple tensile test equipment and a digital Vernier caliper. The foreman of the company already performed the tensile test, and he obtained the following results. The gauge length of the bar is 50 mm, and the diameter is 12.63 mm. -Free Course Hero Question Answer.

Question Description: Assume you are recruited by a construction company in Fiji as a Trainee Engineer. Your company received a metal bar for construction work, while the construction engineer would like to know the crystal structure of the metal bar, such as, is a BCC or FCC or HCP. Neither company nor in Fiji has got any sophisticated high-cost equipment or facility to test its crystal structure. However, the company possesses simple tensile test equipment and a digital Vernier caliper. The foreman of the company already performed the tensile test, and he obtained the following results. The gauge length of the bar is 50 mm, and the diameter is 12.63 mm.

The company has no idea of what to do; usually, they send the metal samples to Australia for these kinds of tests. It is although time and cost consuming, at this time you cannot send the metal sample due to the Covid-19 pandemic crisis as all forms of transportation are stopped. However, you got an idea that is using the available data you can predict strain-hardening constants. Indeed, anyone of the constants, say strain hardening exponent could be enough to compare against the crystal structure as what category it belongs to. Can you help them out? You can use any open resource or book to solve the question; however, it has to be presented in the solution along with the reference.

FOR THE FIRST TEST

E=Young MODULUS =10.27 GPA

For the SECOND test

E=5.26 GPA

For the third test

E=5.10 GPA

Average=10.27+5.26+5.10/3

E=20.63/3=6.87

E=6.87 GPA

This VALUE can be compared by standard value and decision can be taken

Step-by-step explanation

Image transcription text

`F = 122 K N Daimeter = 12 mm 2 Area of Cross section = 1 x 1 2 = 1/3. 112 mm2 strem = 122 x100 0 - 1078. 57 N/ mm2 113. 112 strain : 5. 258 = 0.105 50 E = Young's Modulus = 1078-57 = 10272 - 09 ripa E = 10 . 27 GPA 0. 105 2 F = 120 Dai = 11-45 A= I X1 - 45 = 102. 98 moo 8 tres = 120 x 1000 = 1165 - 27 N)mm 2 102- 98 Strain = 11.07 - 0/2214. 50 dibets / asets = 3 E = 11 65 . 27 - 5263. 19 0 . 2214 E = 5 . 2 6 GPA. 3) F = 114 Daimeter = 11- 45 Area of cross - section = 1/2x 11- 45 - 102.98 Stren= 114x1000 = 1107. 102-98 strain = 10.85 / 50 = 0 . 217 E = STren 1107 0 . 217 E = 5101. 38 mp= 5.10`