From the FBD of Joint
C
$\Sigma F_V = 0$
$2\left( \frac{5}{\sqrt{89}}F \right) = 20$
$F = 2\sqrt{89} ~ \text{kN}$
Axial Stress on Member AC
$\sigma_{AC} = \dfrac{2\sqrt{89}(1000)}{150(100)}$
$\sigma_{AC} = 1.258 ~ \text{MPa}$ ← [ A ] answer for part 3
From the FBD of Joint A
$\Sigma F_H = 0$
$F_H = \frac{8}{\sqrt{89}}F$
$F_H = \frac{8}{\sqrt{89}}\left( 2 \sqrt{89} \right)$
$F_H = 16 ~ \text{kN}$
For Minimum Length of x
The failure plane is section M-M. Mode of failure is shearing parallel to grain of the bottom chord.
FH = Allowable Shear Parallel to Grain × Area of Section M-M
$F_H = \tau A$
$16(1000) = 1(100x)$
$x = 160 ~ \text{mm}$ ← [ C ] answer for part 1
For Minimum Length of y
The failure plane is section N-N. Mode of failure is bearing. For the bottom chord, the bearing is parallel to grain and for the top chord, the bearing is at an angle θ to the grain. The critical bearing is the one that is at an angle to grain, hence, top chord.
For allowable bearing strength at surface N-N, use Hankinson's Formula:
$F_n = \dfrac{F_c \, F_{c\bot}}{F_c \sin^2 \theta + F_{c\bot}\cos^2 \theta}$
$F_n = \dfrac{11(5)}{11\left( \frac{5}{\sqrt{89}} \right)^2 + 5\left( \frac{8}{\sqrt{89}} \right)^2}$
$F_n = \frac{979}{119} ~ \text{MPa}$
FH = Allowable Bearing Stress × Area of Section N-N
$F_H = F_n A$
$16(1000) = \frac{979}{119}(100y)$
$y = 19.45 ~ \text{mm}$ ← [ D ] answer for part 2
