Question

Problem 1 (6 points): A cylindrical rod of Ni 200 alloy has the following properties: E = 204 GPa, v = 0.31. It is loaded elastically in compression at 12.5 kN. If the original rod length and diameter are 20 mm and 15 mm, respectively, determine the rod length and diameter under load.

Answer 1

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Given . E 204 Gpa Rod ve 0.31 Fr 125KN (compression) 12.5kN compression L= 20mm da 15mm Since the there Hod SO id in will be decrease compression in original diameter length and increase in in we need length To find change (stress) 6- f A & cstrain ) ble E 2 Alto HE 2. Az 02. Ix (15 x 103) 2 ų ų Os 176.625% 10-6 TA 1.766% 104 m2 & estrain ) = - 1215 x 103 (ve -re sign as of compression Putting all values 1.766710-4 x 2048 109 LE -3.47810-4 caxial strain) strain

change in length 2 AL LE e (stain) change in lengre a 20 x (-3.47010-4) Ale – 6.948 10 3 original mm length Therefore rod length after compression is Le 2 L & AL 6.94x10-3 2 L- 19.9931 mm change in To find poisson's ratio diameter require axial strain haver to radial strain relation. E; (strain) 2-VE 2-0.31*(-3.478 10-4) L Ei 2 1.0757 x 10-4. Ad a I dx Ee (change diameter) 15% 1.0757 810-4 1.6140 10 3 Therefore diameter after compression will 1 mm be do 2 dt Ad dea 15 + 1.614610-3 de a 15.00 16 mm