---
title: "Diameter of a shaft carrying a spur gear"
subject: "Machine Design I"
chapter: 5
key: "semester_6/machine_design_i:gear-loaded-shaft-diameter"
source: https://www.bctengineer.com/q/semester_6/machine_design_i/gear-loaded-shaft-diameter
site: BCT Engineering
---

# Diameter of a shaft carrying a spur gear

Machine Design I · Chapter 5: Design of shafts

- 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_6/machine_design_i/gear-loaded-shaft-diameter
- Whole chapter: https://www.bctengineer.com/md/semester_6/machine_design_i/chapter-5

## Question

A solid shaft of length 800 mm rests on two bearings A and B at its ends. A spur gear of pitch radius 100 mm is mounted at C, 300 mm from bearing A. The gear transmits a tangential force of 4 kN acting vertically downward and a radial force of 1.456 kN acting horizontally (20 degree pressure angle). The torque is delivered to the shaft through a coupling at B end. Find the reactions in both planes, the resultant bending moment at C and the shaft diameter using the maximum shear stress theory with allowable shear stress 40 MPa, Kb = 1.5 and Kt = 1.0.

## Short answer

Torque =400 N·m. Vertical reactions:2.5,1.5 kN; Mv=750 N·m. Horizontal:0.91,0.546 kN; Mh=273 N·m. Resultant moment M=798.1 N·m. Te=√((1.5×798.1)²+400²)=1262.3 N·m. d³=16Te/(π×40)=1.607×10⁵ mm³ → d=54.4 mm →55 mm.

## Answer

```text
        4 kN (down)     Fr=1.456 kN (horizontal)
            |
  A---------C--------------B----> coupling (torque)
  |<-300->|<-----500----->|
  |<----------800--------->|
```

#### Torque

$$
T = F_t \times r = 4000 \times 0.1 = 400\ \text{N.m}
$$

#### Vertical plane (load 4 kN at C)

$$
\begin{aligned} R_{Av} &= 4 \times \frac{500}{800} = 2.5\ \text{kN}, \qquad R_{Bv} = 4 - 2.5 = 1.5\ \text{kN} \\ M_v &= R_{Av} \times 0.3 = 2.5 \times 0.3 = 0.75\ \text{kN.m} = 750\ \text{N.m} \end{aligned}
$$

#### Horizontal plane (load 1.456 kN at C)

$$
\begin{aligned} R_{Ah} &= 1.456 \times \frac{500}{800} = 0.91\ \text{kN}, \qquad R_{Bh} = 1.456 - 0.91 = 0.546\ \text{kN} \\ M_h &= 0.91 \times 0.3 = 0.273\ \text{kN.m} = 273\ \text{N.m} \end{aligned}
$$

Check: in each plane $R_A + R_B$ equals the applied load, and moments about A balance: vertical $4 \times 0.3 = 1.2 = 1.5 \times 0.8$ kN.m; horizontal $1.456 \times 0.3 = 0.437 = 0.546 \times 0.8$ kN.m.

#### Resultant bending moment at C

$$
M = \sqrt{750^2 + 273^2} = 798.1\ \text{N.m}
$$

#### Diameter

$$
\begin{aligned} T_e &= \sqrt{(K_b M)^2 + (K_t T)^2} = \sqrt{(1.5 \times 798.1)^2 + (1.0 \times 400)^2} \\ &= \sqrt{1197.2^2 + 400^2} = 1262.3\ \text{N.m} \end{aligned}
$$

$$
d^3 = \frac{16 \times 1262.3 \times 10^3}{\pi \times 40} = 1.607\times10^5\ \text{mm}^3
$$

$$
d = 54.4\ \text{mm}
$$

Use the standard size 55 mm.

**Answer: Reactions (vertical) 2.5 kN and 1.5 kN, (horizontal) 0.91 kN and 0.546 kN; M at C = 798 N.m; d = 54.4 mm, adopt 55 mm.**

## Check yourself

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

1. What is the torque on the shaft?
   A. 4000 N.m
   B. 40 N.m
   C. 400 N.m (Ft x r = 4000 x 0.1)
   D. 800 N.m
   Answer: C. Torque equals tangential force times pitch radius. (Key point: Torque)

2. What is RAv for 4 kN at 300 mm from A on an 800 mm shaft?
   A. 1.5 kN
   B. 2.0 kN
   C. 4 kN
   D. 2.5 kN
   Answer: D. RAv = 4 x 500/800 = 2.5 kN. (Key point: Vertical reaction)

3. How is the bending moment at C found?
   A. Square root of the sum of squares of vertical and horizontal moments: 798 N.m
   B. Arithmetic sum: 1023 N.m
   C. Only the vertical moment: 750 N.m
   D. Only the horizontal moment: 273 N.m
   Answer: A. Moments in perpendicular planes combine vectorially: sqrt(750^2+273^2). (Key point: Resultant moment)

4. The diameter obtained is
   A. 42 mm
   B. 54.4 mm, adopt 55 mm
   C. 47.6 mm
   D. 75.9 mm
   Answer: B. Te = 1262.3 N.m, d^3 = 16Te/(pi x 40). (Key point: Diameter)

## Related in Machine Design I

- [Spur gear terminology and standard tooth system](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Aspur-gear-terminology) — Machine Design I, chapter 9
- [Hollow shaft against solid shaft](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Ahollow-versus-solid-shaft) — Machine Design I, chapter 5
- [Shaft diameter for power and bending](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Astatic-shaft-power-bending-diameter) — Machine Design I, chapter 5
- [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
- [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
- [ASME code equation for shaft diameter](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Aasme-code-shaft-diameter-derivation) — Machine Design I, chapter 5

## Related in other subjects

- [Gearbox shaft design with spur gear](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Agearbox-shaft-design-asme) — Machine Design II, chapter 5
- [Greatest bolt force, 100 kN at 500 mm](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abolt-force-shaft-100kn-500mm) — Design of Steel and Timber Structure, chapter 4
- [Pitch, bearing contact and composite gear tests](https://www.bctengineer.com/question.md?key=semester_3%2Fmetrology%3Agear-testing-methods-notes) — Metrology, chapter 11
- [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