Mechanics of Materials, 7th Edition

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

Chapter 1 - Problems - Page 22: 1.11

Answer

$$\sigma_{BE}=25.55\space ksi$$

Work Step by Step

1)Support Reactions Moment Equilibrium: $\sum M_{A} = 0$ $H_y * 36 - 80*9 - 80*18-80*27 = 0$ $\rightarrow H_y = 120 \space kip$ By symmetry of loading and gemoetry, $A_y = H_y= 120 \space kip$ 2)By Inspection, elements AC and CE are zero force members. 3)Equilibrium of forces at joint A $\space\space\space\bullet$ Y-forces Equilibrium: $\sum F_y = 0.8 F_{AB} + 120 = 0$ $\rightarrow F_{AB} = -150\space kip $ (Compression) $\space\space\space\bullet$ X-forces Equilibrium: $\sum F_x = A_x-150*0.6=0$ $\rightarrow A_x=90 \space kip$ (Horizontal Rxn at A) 4)Make a cut through elements BD, BE, CE. $\space\space\space\bullet$ Y-forces Equilibrium: $\sum F_y=0$ $ 120 - F_{BE} * 0.8 = 0 $ $\Rightarrow F_{BE} = 150\space kip$ (Tension) 5)Axial Stress in element BE $\sigma_{BE} = \frac{F_{BE}}{A_{BE}} = \frac{150\space kip}{5.87\space in^2} = \boxed{25.55 \space ksi}\space\space \leftarrow ANS$
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