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Compound gear train speeds, torques and tooth forces

Short answer

Output speed = 160 rpm, output torque = 660 N.m, output power = 11.06 kW. First pair: Ft = 1989 N, Fr = 724 N; second pair: Ft = 4775 N, Fr = 1738 N. Forces derived from pitch diameters d=mz and torque after efficiency multipliers.

A 12 kW motor running at 1440 rpm drives a compound reduction gear train of two spur gear pairs, all of module 4 mm and 20 degree pressure angle. Pinion 1 has 20 teeth and meshes with gear 2 of 60 teeth. Gear 3 of 24 teeth is mounted on the same shaft as gear 2 and meshes with gear 4 of 72 teeth. Take 96 percent efficiency for each pair of gears. Find the output speed, the output torque and power, and the tangential and radial tooth forces on the first and the second pair.

 Motor ->[1]---[2]  shaft 2
         20T   60T  (2 and 3 fixed together)
               [3]---[4] -> output
               24T   72T

Speed ratio and output speed

Ratio=z2z1×z4z3=6020×7224=3×3=9\text{Ratio} = \frac{z_2}{z_1}\times\frac{z_4}{z_3} = \frac{60}{20}\times\frac{72}{24} = 3 \times 3 = 9 n2=14403=480 rpm,n4=4803=160 rpmn_2 = \frac{1440}{3} = 480\ \text{rpm}, \qquad n_4 = \frac{480}{3} = 160\ \text{rpm}

Torques and power

T1=60P2πn1=60×12 0002π×1440=79.58 N.mT_1 = \frac{60P}{2\pi n_1} = \frac{60 \times 12\,000}{2\pi \times 1440} = 79.58\ \text{N.m}

Each pair multiplies the torque by its ratio and by the efficiency 0.96:

T2=T1×3×0.96=229.2 N.mT4=T2×3×0.96=660.0 N.m\begin{aligned} T_2 &= T_1 \times 3 \times 0.96 = 229.2\ \text{N.m} \\ T_4 &= T_2 \times 3 \times 0.96 = 660.0\ \text{N.m} \end{aligned} Pout=12×0.962=11.06 kW(check: 660.0×2π×160/60=11.06 kW)P_{out} = 12 \times 0.96^2 = 11.06\ \text{kW} \quad(\text{check: } 660.0 \times 2\pi \times 160/60 = 11.06\ \text{kW})

Tooth forces (pitch diameters d=mzd = mz)

First pair (pinion 1: d1=4×20=80d_1 = 4 \times 20 = 80 mm):

Ft=T1d1/2=79.580.04=1989 NFr=Fttan⁡20∘=1989×0.364=724 NFn=Ftcos⁡20∘=2117 N\begin{aligned} F_t &= \frac{T_1}{d_1/2} = \frac{79.58}{0.04} = 1989\ \text{N} \\ F_r &= F_t \tan 20^\circ = 1989 \times 0.364 = 724\ \text{N} \\ F_n &= \frac{F_t}{\cos 20^\circ} = 2117\ \text{N} \end{aligned}

Second pair (pinion 3: d3=4×24=96d_3 = 4 \times 24 = 96 mm, torque T2=229.2T_2 = 229.2 N.m):

Ft=229.20.048=4775 NFr=4775×0.364=1738 NFn=4775cos⁡20∘=5081 N\begin{aligned} F_t &= \frac{229.2}{0.048} = 4775\ \text{N} \\ F_r &= 4775 \times 0.364 = 1738\ \text{N} \\ F_n &= \frac{4775}{\cos 20^\circ} = 5081\ \text{N} \end{aligned}

The tangential force produces the torque; the radial force pushes the gears apart and loads the shaft bearings.

Answer: Output 160 rpm; output torque 660 N.m; output power 11.06 kW; first pair Ft = 1989 N, Fr = 724 N; second pair Ft = 4775 N, Fr = 1738 N.

✓ Answer checked by Claude Opus 5.5 (high effort)

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