Gear Force Analysis The three meshing gears shown in Figure 11.13a have a module of 5 mm and a 20° pressure angle. Driving gear 1 transmits 40 kW at 2000 rpm to idler gear 2 on shaft B. Output gear 3 is mounted to shaft C, which drives a machine. Determine and show, on a free-body diagram, a. The

Question

Gear Force Analysis

The three meshing gears shown in Figure 11.13a have a module of 5 mm and a 20° pressure angle. Driving gear 1 transmits 40 kW at 2000 rpm to idler gear 2 on shaft [latex]B[/latex]. Output gear 3 is mounted to shaft [latex]C[/latex], which drives a machine. Determine and show, on a free-body diagram,

a. The tangential and radial forces acting on gear 2

b. The reaction on shaft [latex]B[/latex]

Assumptions: The idler gear and shaft transmit power from the input gear to the output gear. No idler shaft torque is applied to the idler gear. Friction losses in the bearings and gears are omitted.

Accepted Answer

The pitch diameters of gears 1 and 3, from Equation 11.4 [latex](m = \frac{d}{N})[/latex], are [latex]d_{1} = N_{1}m = 20(5) = 100 mm ext {and} d_{3} = N_{3}m = 30(5) = 150 mm.[/latex]

a. Through the use of Equation 1.15,

[latex] kW =\frac{F_{t} V}{1000}=\frac{T n}{9549} [/latex] (1.15)

[latex] T=\frac{9549 kW }{n}=\frac{9549(40)}{2000}=191 N \cdot m [/latex]

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

[latex] T=\frac{d}{2} F_n \cos \phi=\frac{d}{2} F_t [/latex] (11.21)

[latex] \begin{array}{l} F_t=F_n \cos \phi \ F_r=F_n \sin \phi=F_t an \phi \end{array} [/latex] (11.19)

[latex] F_{t, 12}=\frac{T_1}{r_1}=\frac{191}{0.05}=3.82 kN , \quad F_{r, 12}=3.82 an 20^{\circ}=1.39 kN [/latex]

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

[latex] F_{t, 32}=3.82 kN , \quad F_{r, 32}=1.39 kN [/latex]

The forces are shown in proper directions in Figure 11.13b.

b. Equilibrium of [latex]x-[/latex] and [latex]y-[/latex]directed forces acting on the idler gear gives [latex]R_{Bx} = R_{By} = 3.82 + 1.39 = 5.21 kN.[/latex] The reaction on the shaft [latex]B[/latex] is then

[latex] R_B=\sqrt{5.21^2+5.21^2}=7.37 kN [/latex]

acting as depicted in Figure 11.13b.

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.

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

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