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.

*F*_{H} = 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:

$F_n = \dfrac{F_c \, F_{c\bot}}{F_c \sin^2 \theta + F_{c\bot}\cos^2 \theta}$

$F_n = \dfrac{11(5)}{11\left( \frac{5}{\sqrt{89}} \right)^2 + 5\left( \frac{8}{\sqrt{89}} \right)^2}$

$F_n = \frac{979}{119} ~ \text{MPa}$

*F*_{H} = 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*