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

Validation Case: Torque in a Stepped Shaft

Overview

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.

Geometry

The shaft geometry is avalable as an Onshape document.

Figure 1. Stepped Shaft

Material

Steel

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

Physics

  • 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

Meshing

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}}

where:

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

Thus,

\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 Vertices Analytical, \tau_{\text{max}} (MPa) OnScale Solve, \tau_{\text{max}} (MPa) Difference (%)
5978 66.00 66.87 1.32
34410 66.00 66.35 0.53
74681 66.00 66.19 0.29
154473 66.00 66.10 0.15
322691 66.00 66.07 0.11

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

Figure 3. Mesh Convergence Study

References

[1]
B. J. Goodno, Mechanics of materials, 9th ed. Cengage Learning, 2021.