Mechanics of Materials, 7th Edition

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

Chapter 1 - Problems: 1.42

Answer

$$Area_{AB} = 0.2687 \space in^2$$

Work Step by Step

1) Free Body Diagram of $\overline{BCDE}$ $\space\space\space\bullet$ Moment Equillibrium: $\sum M_{B} = 0$ $5 * 1.4 + 600* 4.2 *2.1- D_y * 2.8 = 0$ $\Rightarrow D_y = 4.39\space kip \uparrow$ $\space\space\space\bullet$ y-Force Equillibrium: $\sum F_y = 0$ $AB_{y}-5-600*4.2+4.39 = 0$ $\Rightarrow AB_y=3.13 \space kip \uparrow$ $\space\space\space\bullet$ Total force in AB: $AB = \frac{AB_y}{sin35} = 5.457\space kip$ 2)Allowable axial stress: $\sigma_{all} = \frac{\sigma_U}{F.S.} = \frac{65}{3.2} = 20.313\space ksi$ 3)Design Area of AB: $\sigma_{all} = \frac{AB_y}{A_{AB}}\space\Rightarrow A_{AB} = \frac{AB_y}{\sigma_{all}} = \frac{5.457}{20.313} = \boxed{0.2687 \space in^2} \space\space\leftarrow\space ANS$
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