---
title: "Principal stresses in a shaft under combined loads"
subject: "Strength of Materials"
chapter: 6
key: "semester_4/strength_of_materials_mechanical:shaft-combined-bending-torsion-stresses"
source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/shaft-combined-bending-torsion-stresses
site: BCT Engineering
---

# Principal stresses in a shaft under combined loads

Strength of Materials · Chapter 6: Transformation of stress and strain

- Asked in: year not given, 6 marks
- Video: [Mechanics of Material: Lesson 44 - Combined Loading Introduction Problem](https://www.youtube.com/watch?v=9_X8Ir7LLnk) — Jeff Hanson
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/shaft-combined-bending-torsion-stresses
- Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-6

## Question

A solid circular shaft 60 mm in diameter is subjected to a bending moment of 1.5 kN m and a torque of 2.0 kN m at a section. For a point on the surface at the top of the shaft where the bending stress is tensile, find the principal stresses, the maximum shear stress and the inclination of the principal planes to the shaft axis.

## Short answer

Bending stress σx = 32M/(πd³) = 70.7 MPa, shear stress τ = 16T/(πd³) = 47.2 MPa. Principal stresses: σ1 = 94.3 MPa, σ2 = –23.6 MPa. Maximum shear stress = 58.9 MPa. Principal plane normal inclined at 26.6° to the shaft axis.

## Answer

#### Stresses at the surface point
$$
\sigma_x = \frac{32M}{\pi d^3} = \frac{32\times1.5\times10^6}{\pi\times60^3} = 70.7\ \text{MPa (tensile)}
$$

$$
\tau_{xy} = \frac{16T}{\pi d^3} = \frac{16\times2.0\times10^6}{\pi\times60^3} = 47.2\ \text{MPa}
$$

and $\sigma_y = 0$ (no stress transverse to the axis).

#### Principal stresses
$$
\frac{\sigma_x}{2} = 35.4\ \text{MPa}, \qquad R = \sqrt{35.4^2 + 47.2^2} = 58.9\ \text{MPa}
$$

$$
\sigma_1 = 35.4 + 58.9 = 94.3\ \text{MPa (tension)}
$$

$$
\sigma_2 = 35.4 - 58.9 = -23.6\ \text{MPa (compression)}
$$

#### Maximum shear stress
$$
\tau_{max} = R = 58.9\ \text{MPa}
$$

(The same value follows from $\frac{16}{\pi d^3}\sqrt{M^2+T^2} = \frac{16\times2.5\times10^6}{\pi\times60^3} = 58.9$ MPa, which is the basis of the equivalent torque $T_e = \sqrt{M^2+T^2} = 2.5$ kN m.)

#### Principal planes
$$
\tan2\theta_p = \frac{2\tau_{xy}}{\sigma_x} = \frac{2\times47.2}{70.7} = 1.335 \;\Rightarrow\; 2\theta_p = 53.1^\circ,\ \theta_p = 26.6^\circ
$$

The plane carrying $\sigma_1$ has its normal at $26.6^\circ$ to the shaft axis, so the plane itself is inclined at $26.6^\circ$ to the cross-section, i.e. at $63.4^\circ$ to the shaft axis. The plane carrying $\sigma_2$ is $90^\circ$ from it: its normal is at $116.6^\circ$ to the axis, so the plane makes $26.6^\circ$ with the shaft axis.

**Answer: $\sigma_1 = 94.3$ MPa, $\sigma_2 = -23.6$ MPa, $\tau_{max} = 58.9$ MPa; $\theta_p = 26.6^\circ$ and $116.6^\circ$ (angles of the plane normals to the shaft axis).**

## Check yourself

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

1. Which stresses act at the top surface point of the shaft?
   A. σx = 32M/πd³ = 70.7 MPa, τ = 16T/πd³ = 47.2 MPa, σy = 0
   B. σx = 16M/πd³, τ = 32T/πd³, σy = 0
   C. σx = 32M/πd³ = 70.7 MPa, τ = 16T/πd³, σy = 70.7 MPa
   D. σx = 32T/πd³, τ = 16M/πd³, σy = 0
   Answer: A. Bending gives the axial stress 32M/πd³ and torsion gives surface shear 16T/πd³; there is no transverse stress. (Key point: Stresses at the surface point)

2. What are the principal stresses?
   A. 70.7 MPa and 0
   B. 94.3 MPa tension and -23.6 MPa compression
   C. 35.4 MPa and -35.4 MPa
   D. 58.9 MPa and -58.9 MPa
   Answer: B. σ = 35.4 ± R with R = sqrt(35.4^2 + 47.2^2) = 58.9, giving 94.3 and -23.6 MPa. (Key point: Principal stresses)

3. How is τmax found directly from M and T?
   A. τmax = 16(M + T)/πd³, using 3.5 kN m
   B. τmax = 16 sqrt(M² - T²)/πd³
   C. τmax = 16 sqrt(M² + T²)/πd³, using Te = 2.5 kN m
   D. τmax = 32 sqrt(M² + T²)/πd³
   Answer: C. R = (16/πd³)sqrt(M²+T²) with Te = sqrt(1.5²+2²) = 2.5 kN m gives 58.9 MPa. (Key point: Equivalent torque)

4. How is the inclination of the principal planes found?
   A. tan2θp = σx/2τxy, giving θp = 36.6°
   B. tan2θp = τxy/σx, giving θp = 16.6°
   C. tan2θp = 2τxy/σx, giving θp = 53.1°
   D. tan2θp = 2τxy/σx = 1.335, giving θp = 26.6°
   Answer: D. 2θp = 53.1°, so θp = 26.6° for the plane carrying σ1. (Key point: Principal plane inclination)

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