Question

Consider a long circular cylinder of elastic-perfectly plastic material that is subjected to a torque \(T\) at its outer surface \(r=a\). The state of stress in the cylinder can be characterized by an azimuthal shear stress \(\tau\). Determine the torque for which an elastic core of radius \(c\) remains. Assume that the yield stress in shear is \(\sigma_{0}\). In the elastic region the shear stress is proportional to the distance from the axis of the cylinder \(r\). What is the torque for the onset of plastic yielding? What is the maximum torque that can be sustained by the cylinder?

Short answer

The torque for onset of plastic yielding is \(T_y = \sigma_0 \frac{\pi a^3}{2}\), and the maximum torque is \(T_{max} = \frac{2}{3}\pi \sigma_0 a^3\).

Step-by-step solution

Understand Shear Stress Distribution

For a cylinder under torsion, the shear stress \(\tau\) is proportional to the radial distance \(r\) from the axis. This is expressed as \(\tau = \frac{T}{J}r\), where \(J\) is the polar moment of inertia. In the elastic region, \(\tau = \frac{T}{J}r\) until it reaches the yield stress \(\sigma_0\) at \(r=c\).

Elastic Core Torque Calculation

The critical condition for the elastic-plastic interface is where the shear stress reaches the yield stress \(\sigma_0\). Set \(\sigma_0 = \frac{T}{J}c\) to find the torque for which an elastic core of radius \(c\) remains. Therefore, \(T = \sigma_0 J / c\).

Define Polar Moment of Inertia

The polar moment of inertia \(J\) for a solid cylinder is given by \(\pi a^4/2\). This is used to relate torque, shear stress, and radius.

Solve for Onset of Plastic Yielding Torque

Substitute \(J = \pi a^4/2\) into the expression for \(T\) to determine the torque \(T_y\) for which plastic yielding begins at the surface. \[T_y = \sigma_0 \frac{\pi a^4}{2c}\].

Maximum Sustain Torque Calculation

When the entire cross-section yields plastically, the maximum torque occurs. This is calculated by integrating the shear stress \(\tau = \sigma_0\) across the entire radius from 0 to \(a\). This gives \(T_{max} = \frac{2}{3}\pi \sigma_0 a^3\).

Key concepts

Shear Stress Distribution

Shear stress distribution in a cylinder under torsion is an essential concept in understanding material stress responses. When a cylinder is subjected to a torque, the shear stress

• increases linearly with radial distance from the center of the cylinder.
• is zero at the center and reaches a maximum on the outer surface.

This linear relationship in the elastic region can be represented mathematically by:\[\tau = \frac{T}{J}r\]where:

• \(\tau\) is the shear stress,
• \(T\) is the applied torque,
• \(J\) is the polar moment of inertia,
• \(r\) is the radial distance from the cylinder's axis.

By applying a torque to the cylinder, the stress gradually spreads from the axis to the surface, creating a distinct distribution based on the material's elastic properties. As the torque and resulting stress increase, the material approaches the critical yield stress \(\sigma_0\). This marks the boundary between elastic and plastic deformation regions in the material.

Polar Moment of Inertia

The polar moment of inertia \(J\) is a measure of an object's ability to resist torsion or twisting. For a solid circular cylinder, it is defined as:\[J = \frac{\pi a^4}{2}\]where:

• \(a\) is the outer radius of the cylinder.

This calculation plays a crucial role in determining how the material withstands twisting forces. It factors into the relationship between torque, shear stress, and radial distance. Understanding the polar moment of inertia is vital because:

• It indicates the resistance level of a cylinder against rotational deformation.
• Larger polar moments correspond to greater resistance against torsional stress.

In practical applications, knowing \(J\) helps predict when a material will reach the yield point and how it behaves under various loads, ensuring structural integrity.

Plastic Yielding

Plastic yielding signifies the point when a material undergoes irreversible deformation. In a cylinder under torsion, plastic yielding begins at the outer surface once shear stress meets the yield shear stress \(\sigma_0\). Before this point, the material behaves elastically, returning to its original form upon removing the stress. When the shear stress distribution extends past the elastic limit to \(\sigma_0\), the material at and beyond the point of yielding enters a plastic state:

• The stress is no longer proportional to the radial distance.
• Permanent deformation occurs at these points.

Monitoring this transition between elastic and plastic states is crucial in engineering to prevent structural failures. It helps design systems capable of withstanding specific torques without permanent damage.

Torque Calculation

Calculating torque is essential for predicting when a material will transition from elastic behavior to plastic deformation. It involves assessing the torque needed at different stages:- **Onset of Plastic Yielding**: Torque at which yielding begins on the outer surface is given by substituting the polar moment of inertia into the torque equation:\[T_y = \sigma_0 \frac{\pi a^4}{2c}\]Here, \(c\) represents the radius of the elastic core, signifying the point inward from the surface where yielding has not yet occurred.- **Maximum Sustained Torque**: When the entire cylinder yields plastically, the maximum torque that can be withstood is determined by integrating the yield shear stress across the cylinder’s cross-section:\[T_{max} = \frac{2}{3}\pi \sigma_0 a^3\]This calculation is critical for safely determining the maximum operational limits of cylindrical structures. It ensures that applied forces do not exceed the material's yield capacity, preventing structural failures.