Homework SourseA solid round bar of aluminium 6061T6 with a diameter of 159
A solid round bar of aluminium (6061-T6) with a diameter of 15.95mm and length of 600mm is, loaded in torsion. Take the material properties of 6061-T6 aluminium to be E=70GPa and G=26GPa.
Question 3: Strain analysis A solid round bar of aluminium (6061-T6) with a diameter of 15.95 mm and length of 600 mm is loaded in torsion. Take the material properties of 6061-T6 aluminum to be E = 70 GPa and G = 26 GPa. A total of for strain gauges are cemented on the bar as illustrated in the annotated photograph below The orientations of each gauge relative to the axis of the ba r are presented in the table below to rotating end end (degrees) 21.8 50.0 22.9 48.1 gauge 4 Measurements from gauges 2, 3 and 4 will be analysed to detemine the strains £x, Ey, and yxy You will use the measured strain from gauge 1 as a cross-check on the analysis you have performed. The strain registered on a strain gauge can be calculated from the change of resistance of the gauge (AR) provided the gauge factor (GF) and unstrained resistance of the gauge (R) are known. GF R The manufacturer of the strain gauges used in this demonstration provides the following value GF-2.10 (dimensionless) Further specifications of the strain gauges used in this demonstration can be found in the file strain-gauge-datasheet.pdf. The resistances of the four strain gauges in the unloaded state (T= 0) and in the loaded state (T= 40 N.m) are presented in the table below R (2) 120.12 120.20 120.19 120.17 T=40N·m Ri (2) 119.98 119.97 120.36 120.40 gaugeSolution
a) Strain epsilon x = 1 / G.F * Delta R / R
= 1 / 2.1 * (119.97 - 120.20 ) / 120.20 = -9.1118 * 10^-4
Strain epsilon y = 1 / G.F * Delta R / R
= 1 / 2.1 * (120.36 - 120.19 ) / 120.19 = 6.73537 * 10^-4
Strain epsilon xy = 1 / G.F * Delta R / R
= 1 / 2.1 * (120.40 - 120.17 ) / 120.17 = 9.11407 * 10^-4
Theoretical value of strain
epsilon xy = T * l / G * J = 40 * 0.6 / (26 * 10^9 * pi * 0.01595^4 / 32)
= 0.145276
