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