Materials Science and Engineering: An Introduction

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

Chapter 15 - Characteristics, Applications, and Processing of Polymers - Questions and Problems - Page 631: 15.17

Answer

$TS = 112.5 MPa$

Work Step by Step

Required: Tensile strength at a number-average molecular weight of 40,000 g/mol Solution: Using Equation 15.3 and substituting the given values, two expressions with two unknowns will be expressed: $TS = TS_{∞} - \frac{A}{M_{n}}$ $50 MPa = TS_{∞} - \frac{A}{30,000 g/mol}$ $150 MPa = TS_{∞} - \frac{A}{50,000 g/mol}$ Computing the two simultaneously, we find: $TS_{∞} = 300 MPa, A= 7.50 \times 10^{6} MPa-g/mol$ Substituting the computed values into Equation 15.3 for $M_{n} = 40,000 g/mol$: $TS = TS_{∞} - \frac{A}{M_{n}} = (300 MPa) - \frac{(7.50 \times 10^{6} MPa-g/mol)}{(40,000 g/mol)} = 112.5 MPa$
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