Question
A shearing force of 57.3 Newtons is applied to an aluminum rod (S=2.5010 10 N/m 2 ). If the rod is 14.2 m long and
A shearing force of 57.3 Newtons is applied to an aluminum rod (S=2.501010 N/m2). If the rod is 14.2 m long and has a cross-sectional area of 1.0010-5 m2, what is the displacement of the end of the rod, in millimeters?
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Fundamentals of Heat and Mass Transfer
Authors: Incropera, Dewitt, Bergman, Lavine
6th Edition
978-0470055540, 471457280, 470881453, 470055545, 978-0470881453, 978-0471457282
