For every 300 mm width perpendicular to the drawing
$F_a = \left( \frac{1}{2}K_a \gamma H^2 \right)(0.3)$
$F_a = \left[ \frac{1}{2}(\frac{1}{3})(17.3)(2.4^2) \right](0.3)$
$F_a = 4.9824 ~ \text{kN}$
$V_{max} = R = F_a$
$V_{max} = 4.9824 ~ \text{kN}$
$M_{max} = M = 0.8F_a$
$M_{max} = 0.8(4.9824)$
$M_{max} = 3.985\,92 ~ \text{kN}\cdot\text{m}$
Maximum Bending Stress
$f_b = \dfrac{6M}{bd^2}$
$f_b = \dfrac{6(3.985\,92)(1000^2)}{300(75^2)}$
$f_b = 14.17 ~ \text{MPa}$ ← [ B ] answer for part 1
Maximum Shear Stress
$f_v = \dfrac{3V}{2bd}$
$f_v = \dfrac{3(4.9824)(1000)}{2(300)(75)}$
$f_v = 0.332 ~ \text{MPa}$ ← [ B ] answer for part 2
Minimum thickness of plank to prevent failure
From part (1)
(fb = 14.17 MPa) > (Fb = 10.4 MPa) ← failed in bending
From part (2)
(fv = 0.332 MPa) < (Fv = 0.8 MPa) ← passed in shear
To prevent failure, increase the thickness of the plank according to bending
$F_b = \dfrac{6M}{bd^2}$
$10.4 = \dfrac{6(3.985\,92)(1000^2)}{300d^2}$
$d = 87.55 ~ \text{mm}$
Use d = 90 mm ← [ A ] answer for part 3
