Mechanics of Materials, 7th Edition

Published by McGraw-Hill Education
ISBN 10: 0073398233
ISBN 13: 978-0-07339-823-5

Chapter 1 - Problems: 1.15

Answer

$$D_{max} = 43.4 \space mm $$

Work Step by Step

Solution: 1)Area of Shear Force Distribution: Equal to the surface area on the inside of the punched hole: $A = \pi * D * t$ , where D is hole punch Diameter and t is thickness of polysterene Sheet 2)Shear Stress required for material failure: $\tau_{FAIL} = \frac {F_{MAX}}{A} = 55 \space MPa$ $\tau_{FAIL} = \frac{45 * 10^{3}}{ \pi * D_{MAX} * 6*10^{-3}} \geq 55 *10^6$ $\Rightarrow \space\space D_{MAX}\leq \frac{45*10^{3}}{\pi * 6*10^{-3} * 55 * 10^{6}} = 0.0434 \space m$ $\boxed{D_{MAX} = 43.4 \space mm } \space\space\leftarrow\space ANS$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.