---
title: "Gearbox shaft design with spur gear"
subject: "Machine Design II"
chapter: 5
key: "semester_7/machine_design_ii:gearbox-shaft-design-asme"
source: https://www.bctengineer.com/q/semester_7/machine_design_ii/gearbox-shaft-design-asme
site: BCT Engineering
---

# Gearbox shaft design with spur gear

Machine Design II · Chapter 5: System Design

- Asked in: year not given, 8 marks
- Video: [Design of Shaft With Gears | Design of Shaft for Gears | Design of Machine Elements 1 | DME](https://www.youtube.com/watch?v=MCF4E6YiXlU) — Mahesh Gadwantikar
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/machine_design_ii/gearbox-shaft-design-asme
- Whole chapter: https://www.bctengineer.com/md/semester_7/machine_design_ii/chapter-5

## Question

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.

## 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.

## Answer

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

```text
   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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the tangential force Ft found for the 28-tooth, module 6 gear transmitting 18 kW at 450 rpm?
   A. Ft = T/diameter = 2274 N
   B. Ft = T/pitch radius = 382/0.084 = 4547 N
   C. Ft = Fr tan 20 degrees = 600 N
   D. Ft = P/N = 40 N
   Answer: B. T = 382 N m and the pitch radius is mz/2 = 84 mm, so Ft = 4547 N and Fr = Ft tan 20 degrees = 1655 N. (Key point: Tangential gear force)

2. How is the bending moment at mid-span obtained for the central gear load?
   A. M = FL/4 in each plane, then combined as sqrt(MH^2 + MV^2) = 725.9 N m
   B. M = FL/2 in one plane only
   C. M = F x L, adding MH and MV directly
   D. M = T/2
   Answer: A. A central load on a simply supported shaft gives M = FL/4. The horizontal and vertical moments are combined vectorially. (Key point: Bending moment)

3. What allowable shear stress is used when the keyway reduces the 55 MPa by 25%?
   A. 55 MPa
   B. 41.25 MPa
   C. 13.75 MPa
   D. 68.75 MPa
   Answer: B. The keyway weakens the shaft, so tau = 0.75 x 55 = 41.25 MPa. (Key point: Allowable shear with keyway)

4. Which expression gives the shaft diameter, and what size is chosen?
   A. d^3 = 16/(pi tau) x (Kb M + Kt T), giving 40 mm
   B. d^3 = 16/(pi tau) x sqrt((Kb M)^2 + (Kt T)^2) = 52.6 mm, so a standard 55 mm shaft
   C. d^3 = 32 M/(pi tau), giving 45 mm
   D. d^2 = 4 T/(pi tau), giving 50 mm
   Answer: B. The ASME equation combines shock-factored moment and torque in a square root, and d = 52.6 mm rounds up to 55 mm. (Key point: ASME code equation)

## Related in Machine Design II

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- [Gear ratios of a four-speed car](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Aautomobile-gear-ratio-selection) — Machine Design II, chapter 5
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## Related in other subjects

- [Tooth forces on a spur gear pinion](https://www.bctengineer.com/question.md?key=semester_6%2Ftheory_of_mechanism_and_machine_i%3Aspur-gear-tooth-forces-numerical) — Theory of Mechanism and Machine I, chapter 8
- [Diameter of a shaft carrying a spur gear](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Agear-loaded-shaft-diameter) — Machine Design I, chapter 5
- [Lewis method module for spur gears](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Alewis-spur-gear-module-design) — Machine Design I, chapter 9
- [Compound gear train speeds, torques and tooth forces](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Acompound-gear-train-forces) — Machine Design I, chapter 9