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: 1.8


$$\sigma_{AC} = 1084 \space psi $$

Work Step by Step

$\underline{1) Setup:}$ Cross sectional area of AC: $A_{AC} = \frac{1}{8} * 1 = 0.125 \space in^2$ $\underline{2) Internal Forces:}$ Since link AC is a two force member the reaction at point A is known to be colinear with the link AC. This means that the x and y components of the force in the link AC are related as follows: $AC_y = tan 30^{\circ} AC_x$ $\space\space\space{\boxed{EQ 1}}$ Examining the Free Body diagram of the rectangular plate and applying Equilibrium equations: $\sum{M_B} = 0;$ $\space\space\space 120*2-120*12-AC_x * 16 + AC_y *10 = 0\space\space\space\boxed{EQ 2}$ Combining the above equations 1 and 2: $120*2-120*12-AC_x * 16 + tan 30^{\circ} AC_x*10 = 0$ combining terms and solving for AC_x: $AC_x = -117.34 \space lb$ (Compression) $boxed{EQ 1} \space \Rightarrow AC_y = -67.75 \space lb$ (Compression) Total (axial) force in link AC: $AC=\sqrt{AC_x^2+AC_y^2} = 135.5 \space lb$ (Compression) $\underline{3) Average Normal Stress:}$ $\sigma_{AC} = \frac{AC}{A_{AC}}= \frac {-135.5\space lb}{0.125 \space in^2} = \boxed{-1084.0 \space psi\space (Compression)} \space\space\leftarrow ANS$
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