Diameter of a shaft carrying a spur gear

        4 kN (down)     Fr=1.456 kN (horizontal)
            |
  A---------C--------------B----> coupling (torque)
  |<-300->|<-----500----->|
  |<----------800--------->|

Torque

$$T = F_t \times r = 4000 \times 0.1 = 400\ \text{N.m}$$

Vertical plane (load 4 kN at C)

$$\begin{aligned} R_{Av} &= 4 \times \frac{500}{800} = 2.5\ \text{kN}, \qquad R_{Bv} = 4 - 2.5 = 1.5\ \text{kN} \\ M_v &= R_{Av} \times 0.3 = 2.5 \times 0.3 = 0.75\ \text{kN.m} = 750\ \text{N.m} \end{aligned}$$

Horizontal plane (load 1.456 kN at C)

$$\begin{aligned} R_{Ah} &= 1.456 \times \frac{500}{800} = 0.91\ \text{kN}, \qquad R_{Bh} = 1.456 - 0.91 = 0.546\ \text{kN} \\ M_h &= 0.91 \times 0.3 = 0.273\ \text{kN.m} = 273\ \text{N.m} \end{aligned}$$

Check: in each plane $R_A + R_B$ equals the applied load, and moments about A balance: vertical $4 \times 0.3 = 1.2 = 1.5 \times 0.8$ kN.m; horizontal $1.456 \times 0.3 = 0.437 = 0.546 \times 0.8$ kN.m.

Resultant bending moment at C

$$M = \sqrt{750^2 + 273^2} = 798.1\ \text{N.m}$$

Diameter

$$\begin{aligned} T_e &= \sqrt{(K_b M)^2 + (K_t T)^2} = \sqrt{(1.5 \times 798.1)^2 + (1.0 \times 400)^2} \\ &= \sqrt{1197.2^2 + 400^2} = 1262.3\ \text{N.m} \end{aligned}$$ $$d^3 = \frac{16 \times 1262.3 \times 10^3}{\pi \times 40} = 1.607\times10^5\ \text{mm}^3$$ $$d = 54.4\ \text{mm}$$

Use the standard size 55 mm.

Answer: Reactions (vertical) 2.5 kN and 1.5 kN, (horizontal) 0.91 kN and 0.546 kN; M at C = 798 N.m; d = 54.4 mm, adopt 55 mm.