Materials Science and Engineering: An Introduction

Published by Wiley
ISBN 10: 1118324579
ISBN 13: 978-1-11832-457-8

Chapter 6 - Mechanical Properties of Metals - Questions and Problems - Page 211: 6.37

Answer

%RA = 59.66% %EL = 46%

Work Step by Step

Given: cylindrical metal specimen with original diameter of 12.8 mm and gauge length of 50.80 mm diameter at point of fracture is 8.13 mm fractured gauge length is 74.17 mm Required: ductility in terms of percent reduction in area and in terms of elongation Solution: Using Equation 6.12, it follows: %$RA = \frac{π (\frac{d_{0}}{2})^{2} - π (\frac{d_{f}}{2})^{2} }{π (\frac{d_{0}}{2})^{2}} \times 100 = \frac{π (\frac{12.8 mm}{2})^{2} - π (\frac{8.13 mm}{2})^{2} }{π (\frac{12.8 mm}{2})^{2}} \times 100 $= 59.66 % Using Equation 6.11: %$EL = (\frac{l_{f}-l_{0}}{l_{0}}) \times 100 = \frac{(74.17 mm - 50.80 mm)}{50.80 mm} \times 100 $ = 46%
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