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

Answer

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

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) $\underline{3) Average Normal Stress:}$ $\sigma_{AC} = \frac{AC}{A_{AC}}= \frac {-135.5\space lb}{0.125 \space in^2} = \boxed{-1084.0 \space psi\space (Compression)} \space\space\leftarrow ANS$
Small 1486857185
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.