Question 11.5: Gear Force Analysis The three meshing gears shown in Figure ......

The pitch diameters of gears 1 and 3, from Equation (11.4), are $d_1=N_1 m=20(5)=100  mm$ and $d_3=N_3 m=30(5)=150  mm$.

$m=\frac{d}{N}$    (11.4)

a. Through the use of Equation (1.15),

$kW =\frac{F V}{1000}=\frac{T n}{9549}$      (1.15)

$T=\frac{9549 kW }{n}=\frac{9549(40)}{2000}=191  N \cdot m$

By Equation (11.21) and Equation (11.19), the tangential and radial forces of gear 1 on gear 2 are then

$T=\frac{d}{2} F_n \cos \phi=\frac{d}{2} F_t$      (11.21)

$\begin{aligned} & F_t=F_n \cos \phi \\ & F_r=F_n \sin \phi=F_t \tan \phi \end{aligned}$     (11.19)

$F_{t, 12}=\frac{T_1}{r_1}=\frac{191}{0.05}=3.82  kN \quad F_{r, 12}=3.82 \tan 20^{\circ}=1.39  kN$

Inasmuch as gear 2 is an idler, it carries no torque, so the tangential reaction of gear 3 on 2 is also equal to $F_{t, 12}$. Accordingly, we have

$F_{t, 32}=3.82  kN , \quad F_{r, 32}=1.39  kN$

The forces are shown in the proper directions in Figure 11.13(b).

b. The equilibrium of $x-$ and $y$-directed forces acting on the idler gear gives $R_{B x}=R_{B y}=3.82+1.39=5.21  kN$. The reaction on the shaft $B$ is then

$R_B=\sqrt{5.21^2+5.21^2}=7.37  kN$

acting as depicted in Figure 11.13(b).

Comments: When a combination of numerous gears is used as in a gear train, usually the shafts supporting the gears lie in different planes and the problem becomes a little more involved. For this case, the tangential and radial force components of one gear must be further resolved into components in the same plane as the components of the meshing gear. Hence, forces along two mutually perpendicular directions may be added algebraically.