How is the tangential force Ft found for the 28-tooth, module 6 gear transmitting 18 kW at 450 rpm?

Ft = T/pitch radius = 382/0.084 = 4547 N

How is the bending moment at mid-span obtained for the central gear load?

M = FL/4 in each plane, then combined as sqrt(MH^2 + MV^2) = 725.9 N m

M = F x L, adding MH and MV directly

Which expression gives the shaft diameter, and what size is chosen?

d^3 = 16/(pi tau) x sqrt((Kb M)^2 + (Kt T)^2) = 52.6 mm, so a standard 55 mm shaft

d^3 = 16/(pi tau) x (Kb M + Kt T), giving 40 mm

d^3 = 32 M/(pi tau), giving 45 mm

d^2 = 4 T/(pi tau), giving 50 mm

Practice · 8 marks

Gearbox shaft design with spur gear

Short answer

Using ASME code with K_b=1.5, K_t=1.2, equivalent moment √((1.5×725.9)²+(1.2×382)²)=1181 N·m and allowable shear 41.25 MPa gives d = 52.6 mm. Select standard 55 mm shaft.

A gearbox shaft carries a 28-tooth spur gear of module 6 mm and 20 degree full-depth teeth, which transmits 18 kW at 450 rpm. The gear is mounted midway between two bearings 600 mm apart (treat as simply supported). The shaft has a keyway. The allowable shear stress for the shaft material is 55 MPa without keyway, and the keyway reduces this by 25%. Use shock and fatigue factors K_b = 1.5 and K_t = 1.2. Find the shaft diameter using the ASME code equation and select a standard size.

Data

$P = 18$ kW, $N = 450$ rpm, $z = 28$ , $m = 6$ mm, $\phi = 20^\circ$ , $L = 600$ mm.

Torque and gear loads

T = \frac{P}{\omega} = \frac{18000}{2\pi\times450/60} = \frac{18000}{47.12} = 382.0\ \text{N m}

Pitch circle diameter $d_p = mz = 6\times28 = 168$ mm, so the pitch radius is 0.084 m.

F_t = \frac{T}{r} = \frac{382.0}{0.084} = 4547\ \text{N}, \qquad F_r = F_t\tan20^\circ = 1655\ \text{N}

Bending moments (load at mid-span, $M = FL/4$ )

\begin{aligned} M_H &= \frac{4547\times0.6}{4} = 682.1\ \text{N m} \\ M_V &= \frac{1655\times0.6}{4} = 248.3\ \text{N m} \\ M &= \sqrt{682.1^2 + 248.3^2} = 725.9\ \text{N m} \end{aligned}

   Ft, Fr (down/side)
        |
   A ---+--- B        L = 600 mm
   R_A  300  R_B      M = FL/4 at mid-span

Allowable shear stress

\tau_{all} = 55\times0.75 = 41.25\ \text{MPa}

ASME code equation

d^3 = \frac{16}{\pi\,\tau_{all}}\sqrt{(K_bM)^2 + (K_tT)^2}

\sqrt{(1.5\times725.9)^2 + (1.2\times382.0)^2} = \sqrt{1088.9^2 + 458.4^2} = 1181.4\ \text{N m}

d^3 = \frac{16\times1181.4}{\pi\times41.25\times10^{6}} = 1.4587\times10^{-4}\ \text{m}^3

d = 0.0526\ \text{m} = 52.6\ \text{mm}

Standard size: 55 mm.

Answer: $d = 52.6$ mm, so use a 55 mm shaft.

✓ Answer checked by Claude Opus 5.5 (high effort)

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