Log in
Log in Simulate Now
Log in Simulate Now
Help > OnScale Solve Validation Cases > Validation Case: Torque in a Stepped Shaft

Validation Case: Torque in a Stepped Shaft


This article is part of the series of validation cases performed using OnScale Solve.

In this case, we have a stepped shaft that is subject to three torque loads. The torques have magnitudes of 3000 \text{N} \cdot \text{m}, 2000 \text{N} \cdot \text{m}, and 800 \text{N} \cdot \text{m}. The aim of the analysis is to determine the maximum shear stress on this shaft.


The shaft geometry is avalable as an Onshape document.

Figure 1. Stepped Shaft



  • Young’s Modulus (E) = 208 \cdot 10^{9}~\text{Pa}
  • Poisson’s ratio (\nu) = 0.3


  • Fixed restraint on the face at point A, as shown in Fig. 2.
  • Load at the end of each step
    • Torque 1 = 3000~\text{N} \cdot \text{m}
    • Torque 2 = 2000~\text{N} \cdot \text{m}
    • Torque 3 = 800~\text{N} \cdot \text{m}

Figure 2. Freebody Diagram


A second-order tetrahedral mesh is used for this study.

Reference Solution

The solution is derived using methods in [1]. We solve for the moment in each of the three sections T_{AB}, T_{BC}, and T_{CD} using the free body diagram in Fig. 2 as a reference, where T_{R} represents the reaction torque.

This produces the set of equilibrium equations:

T_{R} = T_1 + T_2 + T_3 \\ T_{CD} = T_R - T_1 - T_2 \\ T_{BC} = T_R - T_1 \\ T_{AB} = T_R

We solve the system of equations with 4 unknowns to obtain the moment in each section and the moment reaction at the fixed end.

T_{R} = 5800 \ \text{N} \cdot \text{m} \\ T_{CD} = 800 \ \text{N} \cdot \text{m} \\ T_{BC} = 2800 \ \text{N} \cdot \text{m} \\ T_{AB} = 5800 \ \text{N} \cdot \text{m}

The maximum shear stress on the outer surface of the circular shaft is computed from the corresponding torque.

\tau_{\text{max}} = \frac{T\cdot r}{I_{p}}


  • \tau_{\text{max}} = Maximum shear stress
  • T = Torque
  • r = Radius of cross section
  • I_{p} = Polar moment of inertia


\begin{align*} \tau_{AB} = \frac{T_{AB} \cdot r_{AB}}{I_{p,~AB}} = \frac{(5800~\text{N}\cdot\text{m})(0.04~\text{m})}{\frac{\pi(0.04~\text{m})^4}{2}} = 57.7~\text{MPa} \end{align*}

\tau_{BC} = \frac{T_{BC} \cdot r_{BC}}{I_{p,~BC}} = \frac{(2800~\text{N}\cdot\text{m})(0.03~\text{m})}{\frac{\pi(0.03~\text{m})^4}{2}} = 66.0~\text{MPa} = \tau_{\text{max}}

\tau_{CD} = \frac{T_{CD} \cdot r_{CD}}{I_{p,~CD}} = \frac{(500~\text{N}\cdot\text{m})(0.02~\text{m})}{\frac{\pi(0.02~\text{m})^4}{2}} = 63.7~\text{MPa}

Results Comparison

The table below compares the values obtained using the analytical equation described above and the analysis performed using OnScale Solve. In addition, a mesh convergence study has been conducted.

# of VerticesAnalytical, \tau_{\text{max}} (MPa)OnScale Solve, \tau_{\text{max}} (MPa)Difference (%)

A graphical representation as been further presented in Fig. 3.

Figure 3. Mesh Convergence Study


B. J. Goodno, Mechanics of materials, 9th ed. Cengage Learning, 2021.