19.8
Plastic Deformation in Circular Shafts
When materials are subjected to forces that surpass their yield strength, they undergo a process known as plastic deformation. This results in a perma…
When materials are subjected to stresses exceeding their yield strength, plastic deformation occurs, causing a permanent strain. In the case of circular shafts, plastic deformation changes its configuration.
An accurate assessment of plastic deformation necessitates the determination of stress distribution within the circular shaft.
If the maximum shearing stress in the material is known, then plotting a shearing-stress-strain diagram gives the corresponding maximum shearing strain.
Recall that the shearing strain varies linearly with the distance from the axis of the shaft. The relationship between shearing strain and radial distance is established by substituting the maximum shearing strain value. Further, the relationship between shearing stress and radial distance is obtained.
Using the integral relation and substituting for the elemental area and the polar moment of inertia in terms of shaft radius, the ultimate torque that causes shaft failure can be determined by maximizing the value of the material's ultimate shearing stress.
The corresponding fictitious stress is called the modulus of rupture in the torsion of the given material.
View the full transcript and gain access to JoVE Core videos
Q1: What happens to a circular shaft when plastic deformation occurs?
Plastic deformation causes permanent strain and changes the shaft's configuration. When materials experience stresses exceeding their yield strength, they undergo this irreversible alteration in shape. Understanding the resulting stress distribution within the circular shaft is essential for predicting failure and designing safe components.
Q2: How does shearing strain relate to distance from the shaft axis?
Shearing strain varies linearly with radial distance from the shaft axis. By substituting the maximum shearing strain value into this relationship, engineers can establish how strain changes across the shaft's cross-section. This linear relationship is fundamental to determining stress distribution and predicting failure under torsional loading.
Q3: What is the modulus of rupture in torsion?
The modulus of rupture in torsion is a fictitious stress value representing the maximum stress a material can endure before failing under torsional loading. It is derived by maximizing the material's ultimate shearing stress using integral relations and the polar moment of inertia. This value helps engineers determine the ultimate torque that causes shaft failure.
Q4: How is the ultimate torque causing shaft failure determined?
The ultimate torque is calculated using integral relations that incorporate the elemental area and polar moment of inertia in terms of shaft radius. By maximizing the material's ultimate shearing stress value through this mathematical relationship, engineers can predict the torque at which the shaft will fail under plastic deformation conditions.
Q5: Why is a shearing-stress-strain diagram important for plastic deformation analysis?
A shearing-stress-strain diagram reveals the maximum shearing strain corresponding to the maximum shearing stress in the material. This graphical relationship enables engineers to visualize how the material responds to torsional loading and accurately assess plastic deformation behavior, which is critical for predicting failure and designing reliable shafts.
Q6: What role does stress distribution play in assessing plastic deformation in shafts?
Accurate assessment of plastic deformation requires determining the stress distribution within the circular shaft. Understanding how shearing stress varies with radial distance allows engineers to identify critical failure zones and predict the ultimate torque capacity. This knowledge is essential for safe shaft design and failure prevention.
Q7: How does the relationship between shearing stress and radial distance affect shaft design?
The relationship between shearing stress and radial distance is derived from the linear strain relationship and material properties. By establishing this stress distribution, engineers can optimize shaft geometry and material selection to withstand applied torques while minimizing weight and cost, ensuring reliable performance under torsional loading conditions.