Deflection angle for idle gear development

  When I develop the machine, I used to use some formulars in textbook which I studied at my university. I used the Castigliano's theorems so often and several cases were already solved so I just needed to apply it to my real case. But below case was not in my textbook, which I needed this case for my idle gear development so I made it by myself.

$$M_{AB} = -M' - \frac{F a c}{l} \qquad M_{BC} = -M' + F(c - l - a)$$
$$\frac{\partial M_{AB}}{\partial M'} = -1 \qquad \frac{\partial M_{BC}}{\partial M'} = -1$$

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$$\theta_A = \int_0^{l+a} M \left( \frac{\partial M}{\partial M'} \right) \frac{dc}{EI} \quad,\quad 2M' = 0$$
$$= \int_0^l -\frac{F a c}{l} (-1) \frac{dc}{EI} \;+\; \int_l^{l+a} F(x - l - a)(-1)\frac{dc}{EI}$$ $$= \frac{Fa}{EI} \int_0^l \frac{x}{l} dx \;-\; \frac{F}{EI} \int_l^{l+a} (x - l - a)\, dx$$
$$= \frac{Fa l}{2EI} \;-\; \frac{F}{EI} \left[ \frac{x^2}{2} - lx - ax \right]_{l}^{\, l+a}$$ $$= \frac{Fa l}{2EI} \;-\; \frac{F}{EI} \left[ \frac{(l+a)^2}{2} - l(l+a) - a(l+a) - \left( \frac{l^2}{2} - l^2 - al \right) \right]$$
$$= \frac{Fa l}{2EI} + \frac{Fa^2}{2EI} = \boxed{ \frac{Fa}{2EI}(l + a) } \quad (\text{단위: rad})$$

My idle gear had a cantilever structure and it has difficult design because of not enough space for idle gear. In a limited space, it received lots of force from partner gear, which was used to transfer the engine power to wheel. Shaft diameter could not be big enough and bearing selection was also limited due to the space. With those conditions, deflection and deflection angle had to be considered thoroughly. I could use the deflection formular from my textbook and for deflection angle, I used this formular. When shaft length is fixed, deflection angle multiplies shaft length would be deflection amount at the shaft end. (unit: rad x mm = mm)