Question

(GPa) A cylindrical machine part 250 mm long with a diameter of 10 mm is subjected to an axial load of 28,500 N. Using the list of materials shown in the table below, answer questions 1 through 2. Yield Strength Young's Modulus Material [MPa] Density (g/cm] Aluminum alloy 255 70 2.75 Steel alloy 400 207 7.85 Titanium alloy 800 110 4.50 Copper alloy 250 110 8.50 Brass alloy 340 100 8.40 1. If the elongation is not to exceed 0.6 mm, which material is suitable (assume elastic deformation)? a. copper b. titanium c. steel d. aluminum e. brass 2. If the applied load produces an elongation of 0.4 mm and Poisson's ratio is v = 0.3, the reduction in diameter is most near to a. 0.0065 mm b. 0.0056 mm c. 0.0048 mm d. 0.0032 mm e. 0.0016 mm 1920 FA MEE235 Midterm 2 Quiz 11/19/2019

Answer 1

Given data elongation Sel - 0.4mm, delomm Poisson's Ratio V: 0:3 We know that Poisson's Ratio 2= lateral strain Jongitudinal Stain - Sdil = d.se d Sel = 250 0.3 - 80 x 250 TO X04 Sel - 0.3 x 10 x0.4 s 0.0048 mm So elongationes diameter = 0.0048 mm Any Hence option (C) i correet,

Given data Fox Aluminium A length 'b's 250 mm diameter d's 10mm load'p' = 28,500N mium Alloy from table, E= 70 GPO - 70x103 MPa By Hook's law Stress a Strain Opel where it û Young modules 4Pe TIJZE : Sb-Elongation - RE - 112E AS = 4x28500x250 1.296 mm 3.14x10)2 x 70 x 10 For Steel Alloy Given data. Vield Strength = 400 MPa 3 me E=207 Gla=2078 10 MPa 8=7.85 g/cm3 - Se = 4x28 500x250 2 = 0.238 mm 3.14x(1072 x 207x 103 For Titanium Alloy Yield Strength = 800 mpa, E = 110 G Pacllox1:3MPa 8=4.50g/cm3 81 - 4x28500x250 iz = 0.825 mm 3.14 x10)2 x 110x13 For copper Alloy Yield strength = 25ompa E=110 Gla= 110x13MPa - 2=8.5og/cm3 ses 4x28500X250 2=0.825 mm 3:14x11072x110x133 For Brass Alloy Vield strength = 340MPa, E = 100 Gira = loox 103 MPa - 8 = 8.491cm3 soux 28500X250 0.9076 3.14x(10)2 x 100X103 -0.908 mm 0 So from above calculation, it is clear that the elongation (El) is not exceed 0.6mm for steel alloy. Hence option (C) is correct.