---
title: "Shaft diameter for power and bending"
subject: "Machine Design I"
chapter: 5
key: "semester_6/machine_design_i:static-shaft-power-bending-diameter"
source: https://www.bctengineer.com/q/semester_6/machine_design_i/static-shaft-power-bending-diameter
site: BCT Engineering
---

# Shaft diameter for power and bending

Machine Design I · Chapter 5: Design of shafts

- Asked in: year not given, 6 marks
- Video: [Design of Shafts - Part 2 | Numerical Problems & Full Solution | Machine Design](https://www.youtube.com/watch?v=xgIo9yMKjuI) — CADxMACHINES
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/machine_design_i/static-shaft-power-bending-diameter
- Whole chapter: https://www.bctengineer.com/md/semester_6/machine_design_i/chapter-5

## Question

A solid shaft transmits 25 kW at 360 rpm. The bending moment at the critical section is 450 N.m. The shaft material has ultimate tensile strength 560 MPa and yield strength 330 MPa, and the shaft has a keyway. The load is applied with minor shocks, so take Kb = 1.5 and Kt = 1.2. Using the ASME code, find the diameter of the shaft.

## Short answer

Torque T =60P/(2πn) =663.1 N·m. Allowable shear: min(0.30×330,0.18×560)=99 MPa, reduced by25% for keyway →74.25 MPa. Equivalent twisting moment Te =√((1.5×450)²+(1.2×663.1)²) =1043.5 N·m. Then d³ = (16 Te)/(π τ_allow) gives d=41.5 mm, use42 mm standard.

## Answer

Use the ASME code equation $d^3 = \dfrac{16}{\pi \tau_{allow}}\sqrt{(K_b M)^2 + (K_t T)^2}$.

#### Torque

$$
T = \frac{60P}{2\pi n} = \frac{60 \times 25\,000}{2\pi \times 360} = 663.1\ \text{N.m}
$$

#### Allowable shear stress

$$
\begin{aligned} 0.30\,S_{yt} &= 0.30 \times 330 = 99\ \text{MPa} \\ 0.18\,S_{ut} &= 0.18 \times 560 = 100.8\ \text{MPa} \end{aligned}
$$

The smaller value is 99 MPa. With a keyway the stress is reduced by 25 percent:

$$
\tau_{allow} = 0.75 \times 99 = 74.25\ \text{MPa}
$$

#### Equivalent moment

$$
\begin{aligned} K_b M &= 1.5 \times 450 = 675\ \text{N.m} \\ K_t T &= 1.2 \times 663.1 = 795.8\ \text{N.m} \\ T_e &= \sqrt{675^2 + 795.8^2} = 1043.5\ \text{N.m} \end{aligned}
$$

#### Diameter

$$
d^3 = \frac{16 \times 1043.5 \times 10^3}{\pi \times 74.25} = 71\,572\ \text{mm}^3
$$

$$
d = 41.5\ \text{mm}
$$

Adopt the next standard size, 42 mm.

**Answer: d = 41.5 mm, so use a 42 mm shaft.**

## Check yourself

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

1. What is the torque for 25 kW at 360 rpm?
   A. 69.4 N.m
   B. 331.5 N.m
   C. 1326 N.m
   D. 663.1 N.m
   Answer: D. T = 60P/(2 pi n) = 60 x 25000/(2 pi x 360) = 663.1 N.m. (Key point: Torque)

2. What allowable shear stress is used (Syt = 330, Sut = 560 MPa, keyway)?
   A. 74.25 MPa
   B. 99 MPa
   C. 100.8 MPa
   D. 75.6 MPa
   Answer: A. Smaller of 99 and 100.8 is 99, then 0.75 x 99 = 74.25 MPa. (Key point: Allowable stress)

3. What is Te = sqrt((KbM)^2 + (KtT)^2) here?
   A. About 1237 N.m
   B. About 1043.5 N.m
   C. About 795.8 N.m
   D. About 675 N.m
   Answer: B. KbM = 675, KtT = 795.8, so Te = 1043.5 N.m. (Key point: Equivalent moment)

## Related in Machine Design I

- [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
- [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
- [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
- [Soderberg design with reversed bending and steady torque](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Asoderberg-bending-torsion-shaft) — Machine Design I, chapter 5
- [Bending stress and safety factor of a spur pinion](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Aagma-bending-stress-safety-factor) — Machine Design I, chapter 9

## Related in other subjects

- [Solid shaft 300 kW, 120 rpm with twist limit](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Asolid-shaft-300kw-120rpm-twist-limit) — Strength of Materials, chapter 6
- [Solid shaft 300 kW; hollow shaft saving](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Asolid-shaft-300kw-percent-saving) — Strength of Materials, chapter 6
- [Hollow shaft 1670 kW at 4500 rpm: size](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ahollow-shaft-1670kw-4500rpm) — Strength of Materials, chapter 6
- [Hollow versus solid shaft for power](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Ahollow-versus-solid-shaft-power) — Strength of Materials, chapter 4