19.3
Arbre circulaire – Contrainte dans une plage linéaire
Considérons un scénario dans lequel un arbre circulaire est soumis à un couple qui reste dans les limites de la loi de Hooke, évitant ainsi toute défo…
Considérons un cas où le couple appliqué à l'arbre circulaire est dans la limite de la loi de Hooke, il n'y a donc pas de déformation permanente.
Maintenant, rappelez-vous l’expression pour la déformation de cisaillement. En la multipliant par le module de rigidité et en utilisant la loi de Hooke pour la contrainte de cisaillement et la déformation, une expression de la contrainte de cisaillement dans un arbre peut être déterminée.
Rappelons que la somme des moments des forces élémentaires exercées sur n’importe quelle section de l’arbre doit être égale à l’intensité du couple exercé sur l’arbre.
En substituant la contrainte de cisaillement et en réarrangeant les termes, on obtient une expression avec un terme intégral, qui représente le moment polaire d’inertie de la section transversale par rapport à son centre.
D’autres réarrangements et substitutions pour la contrainte de cisaillement maximale donnent la formule de torsion élastique pour la contrainte de cisaillement dans un arbre circulaire uniforme et rigide.
Cependant, pour un arbre creux dont les rayons intérieur et extérieur sont r1 et r2, le moment d’inertie polaire s’exprime comme une différence de puissance quatre de deux rayons.
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Q1: What is the elastic torsion formula for shearing stress in a circular shaft?
The elastic torsion formula determines shearing stress in a rigid uniform circular shaft by combining Hooke's Law with the polar moment of inertia. It derives from multiplying shearing strain by the modulus of rigidity, then ensuring the sum of moments on any cross-section equals the applied torque. This formula applies when torque remains within Hooke's law limit, preventing permanent deformation.
Q2: How does the polar moment of inertia differ between solid and hollow circular shafts?
For a solid circular shaft, the polar moment of inertia is calculated from the fourth power of the radius. For a hollow shaft with inner radius r1 and outer radius r2, the polar moment of inertia is expressed as the difference in the fourth power of both radii. This difference accounts for the removed material in the hollow center.
Q3: Why is the modulus of rigidity important in deriving shaft stress equations?
The modulus of rigidity relates shearing stress to shearing strain through Hooke's Law. When multiplied by the shearing strain expression, it enables derivation of the shearing stress equation for the shaft. This material property is essential for connecting elastic deformation behavior to the applied torque.
Q4: What role does the sum of moments play in the elastic torsion formula?
The sum of moments of elementary forces exerted on any cross-section of the shaft must equal the magnitude of the applied torque. This equilibrium condition is fundamental to deriving the elastic torsion formula. Substituting this relationship into the stress equation produces the integral term representing polar moment of inertia.
Q5: When can the elastic torsion formula be applied to a circular shaft?
The elastic torsion formula applies when torque remains within Hooke's law limit, ensuring no permanent deformation occurs. Under these conditions, the shaft exhibits linear elastic behavior, and shearing stress is directly proportional to shearing strain. This linear range assumption is critical for formula validity.
Q6: How does maximum shearing stress relate to the polar moment of inertia in torsion?
Maximum shearing stress is inversely proportional to the polar moment of inertia. After substituting for maximum shearing stress in the equilibrium equation, the polar moment of inertia emerges as a key geometric property. Larger polar moments of inertia reduce maximum stress for the same applied torque, which is why hollow shafts are often preferred in design of transmission shafts.
Q7: What assumptions must be satisfied for the elastic torsion formula to hold?
The shaft must be rigid and uniform with constant cross-section. Torque must remain within Hooke's law limit to prevent plastic deformation. The formula assumes linear elastic material behavior where shearing stress is proportional to shearing strain. These conditions ensure the derived relationships between torque, stress, and geometric properties remain valid.