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# Torque and its Application in Simulation

By Tristan Akalumhe 15 March 2022

When discussing rotational motion and power transmission, engineers need the right boundary conditions to simulate their design scenarios. In this blog post, we will dive deeper into the application of torque in CAE simulation.

##### Fig. 1 Wrench and Bolt

Torque is a moment due to a force that tends to cause rotation of a body. Torque can be written as the cross product of a force vector and the position vector from the axis of rotation.

\vec{\tau}=\vec{r} \times \vec{F}

Or in component form:

\vec{\tau}=\left(F_{z} r_{y}-F_{y} r_{z}\right) \hat{i}-\left(F_{z} r_{x}-F_{x} r_{z}\right) \hat{j} + \left(F_{y} r_{x}-F_{x} r_{y}\right) \hat{k}

A practical example of torque is a bolt and wrench, as shown in Fig 1. Application of a force at the end of the wrench arm creates a torque about the axis of the bolt which will tighten or loosen it, depending on the direction of the force.

The SI unit for a torque is the Newton-Meter (Nm).

## Why is torque important?

The availability of torque as a boundary condition in simulation software creates a means for engineers to model the torsional forces in their designs effectively. The torque load in OnScale Solve allows us to apply a pure moment to a body without applying a translational force component. A motor shaft is a typical example of a torsional member that requires analysis because it is subject to heavy torque loads while transmitting mechanical power to, for example, the propeller of an airplane. We want to analyze whether the shaft will fail in shear due to the applied torque.

##### Fig. 2 Shear Stress on Circular Cross Section

We can look at the relationship between the shear stress and torque for a solid bar of circular cross section to understand the expected results. We start by considering an element of area dA located at a radial distance \rho from the axis of the circular bar, while using Fig 2 as a guide. The shear force acting on this element is equal to \tau \cdot dA , where \tau is the shear stress at radius \rho . Thus, the torque about the axis is simply the shear force multiplied by the distance from the center. Hence, the increment of torque at radius \rho is:

dT = (\tau d A) \cdot \rho=\frac{\tau_{\max }}{r} \rho^{2} dA

in which the shear stress at radius \rho is related to the max shear stress at the outer surface r as:

\tau=\frac{\rho}{r} \tau_{\max }

This relationship is only applicable to linearly elastic materials. We can further determine the torque over the entire cross-sectional area using:

T=\int_{A} d T=\frac{\tau_{\max }}{r} \int_{A} \rho^{2} d A=\frac{\tau_{\max }}{r} I_{p}

in which, I_{p} is the polar moment of inertia of the circular cross section.

I_{p}=\int_{A} \rho^{2} dA

Thus, making it possible to express the maximum shear stress in terms of the torque using the equation below:

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

## Now That We Have Covered the Theory, Let’s Dive into an OnScale Solve Example

We see applications of shafts in many systems for the purpose of power transmission. Engineers are often faced with the design problem of determining the ideal size of a shaft to transmit power at a specified rotational speed, without exceeding allowable stresses.

##### Fig. 3 Torque Load on a Shaft in OnScale Solve

For example, you could be tasked with designing a rotating circular shaft that transmits 40 kW to a gear. Here, we demonstrate a simple cylindrical shaft as shown in Fig 3. The CAD model is available as an Onshape document for download. This shaft operates at 667 RPM and your design objective is to ensure that the diameter of 0.041 m is adequately sized so that the shaft will not exceed an allowable shear stress of 42 MPa during operation. The torque can be determined using an equation that relates the torque, power, and number of revolutions per minute:

T=\frac{60 \cdot P}{2 \cdot \pi \cdot n}=\frac{60(400000\:\mathrm{Nm})}{2 \pi\left(667\:\mathrm{rpm}\right)}=573 \:\mathrm{Nm}

The torque is applied to a sectioned end of the shaft as depicted in Fig 3 with the axis of rotation aligned with the axis of the shaft. The opposite end of the shaft is fully restrained.

##### Fig. 5 Third Principal Stress

Results from the simulation can be viewed in the Results view in OnScale Solve with reference to Fig 4 and Fig 5. We are interested in the maximum shear stress in the shaft, which can be computed using the formula below. \sigma_{1} represents the first principal stress and \sigma_{3} represents the third principal stress.

\tau_{\max } =0.5 \cdot(\sigma_{1}-\sigma_{3})
\tau_{\max }=0.5 \cdot(44.05 - (-44.10)) = 44.075\:MPa

We can use this equation because the principal stresses are uniform around the shaft in this case. The maximum shear stress will soon be available as an output in OnScale Solve. Most importantly, this result informs us that we are not within the allowable stress of 42 MPa, so we would have to redesign the shaft by making the diameter bigger.

In conclusion, we briefly discussed the theory behind torque, its importance, and covered a practical example of its usage.

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