Name: torsion formula for circular shafts
Uploaded: 2024-05-23T12:59:40.000Z
Duration: 1 min 13 s
Channel: JoVE

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.

arbre circulaire – contrainte dans une plage linéaire

Circular Shaft - Stresses in Linear Range

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 déformation permanente. Ainsi, la formule de déformation en cisaillement est revisitée. Cette formule est multipliée par le module de rigidité, puis la loi de Hooke pour la contrainte et la déformation de cisaillement est appliquée. En conséquence, l’équation de la contrainte de cisaillement dans un arbre peut être dérivée.

<img alt="Equation 1" src="/files/ftp_upload/15614/15614_Equation_1.svg" style="text-align: center; height: 28px;">

De plus, il est essentiel de rappeler que la somme des moments des forces élémentaires agissant sur n'importe quelle section transversale de l'arbre doit être identique au couple appliqué sur cet arbre. Un terme intégral apparaît lorsque l'équation est ajustée pour remplacer la contrainte de cisaillement. Ce terme désigne le moment d'inertie polaire de la section transversale par rapport à son centre. Après d'autres ajustements et substitutions pour la contrainte de cisaillement maximale, la formule de torsion élastique peut être dérivée pour la contrainte de cisaillement dans un arbre circulaire uniformément rigide.

<img alt="Equation 2" src="/files/ftp_upload/15614/15614_Equation_2.svg" style="text-align: center; height:40px;">

Cependant, le scénario diffère pour un arbre creux où r_1 et r_2 sont représentés par les rayons intérieur et extérieur. Dans ce cas, le moment d'inertie polaire est exprimé comme la différence de puissance quatrième des deux rayons.

Arbre circulaire – Contrainte dans une plage linéaire

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 le ...

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torsion formula for circular shafts

Torsion in a circular shaft is analyzed when the applied torque stays within Hooke's law. In this linear range, the shaft does not undergo permanent deformation. The discussion uses shearing strain, shearing stress, and the modulus of rigidity to build the stress equation.

The derivation also requires the moments of the small forces on a cross-section to equal the applied torque. When the shearing stress expression is rewritten, an integral term appears. This term represents the polar moment of inertia, which measures how the cross-section resists twisting about its center.

After the maximum shearing stress is substituted, the elastic torsion formula is obtained for a uniformly rigid circular shaft. The same idea is then applied to a hollow shaft. For that case, the polar moment of inertia depends on the inner radius r1 and outer radius r2, and it is given by the difference of their fourth powers.

Torsion Formula for Circular Shafts

stresses in a shaft

Torque and Shear in a Twisted Shaft

A shaft under torque develops internal shear forces as it twists. In this setup, equal and opposite torques act on the shaft PQ. To study the effect, a section is cut perpendicular to the shaft axis at an arbitrary point R. The free-body diagram of the QR segment shows the shearing forces applied by the PR portion. These forces act as the shaft resists twisting. Using equilibrium on the QR segment links the internal shear forces in the cut section to the internal torque. Here, r is the...

deformation in a circular shaft

Shear Strain in a Circular Shaft

Shear strain in a circular shaft describes how the shaft’s surface changes shape under torsion. A circular shaft keeps each cross-section flat and undistorted as it twists. Each section behaves like a rigid slab that rotates in place. To study the strain inside the shaft, consider a cylindrical section with length L and radius R. One end of this section is fixed. The radius at a point inside the section is labeled r. Before the load is applied, a small square element lies on the surface of...

circular shaft - stresses in linear range

Torsion in a circular shaft is analyzed when the applied torque stays within Hooke's law. In this linear range, the shaft does not undergo permanent deformation. The discussion uses shearing strain, shearing stress, and the modulus of rigidity to build the stress equation. The derivation also requires the moments of the small forces on a cross-section to equal the applied torque. When the shearing stress expression is rewritten, an integral term appears. This term represents the polar moment...

angle of twist - elastic range

Elastic Twist in a Circular Shaft

Elastic twist in a circular shaft describes how a cylindrical shaft rotates when torque is applied. The shaft has a length L and a uniform radius r. A torque at the free end creates shearing strain, which is greatest at points farther from the shaft axis. When the shaft stays within the elastic range, the shearing strain can be written in terms of the applied torque, radial distance, polar moment of inertia, and modulus of rigidity. The polar moment of inertia is a geometric measure of how the...

angle of twist: problem solving

Shaft Torsion Between Pulleys B and C

Shaft torsion between pulleys B and C is found by analyzing how torque twists an aluminum shaft. An electric motor applies 700 N·m of torque and produces stable rotation. The shaft carries pulleys B and C, which experience torques of 300 N·m and 400 N·m. The calculation starts with a section cut between pulleys B and C. A free-body diagram is used to study the cut cross-section. Because pulley B has an anticlockwise torque, the torque at the cut must balance it in the opposite direction. The...

design of transmission shafts

Sizing a Transmission Shaft by Torque

A transmission shaft is sized from the power it must carry and the speed it rotates at. These two design limits guide the choice of shaft material and cross-sectional size. The goal is to keep the maximum shearing stress within the elastic limit while still transmitting the needed power. Power is directly linked to torque, which is the twisting force applied to the shaft. By rearranging the power relation, the torque can be found from the required power and the shaft speed. That torque value...

stress concentrations in circular shafts

Why Shafts Develop Local Stress Peaks

Circular shafts can develop local stress peaks when torque is applied through real machine parts. The elastic torsion formula works best for a circular shaft with a uniform cross-section and rigid plates fixed to its ends. In practice, torque is often passed through flange couplings or gears, and these parts use keys in keyways to connect to the shaft. That loading method changes the stress pattern near the point where torque enters the shaft. The stress distribution no longer matches the...

plastic deformation in circular shafts

Torsional Failure and Plastic Strain in Shafts

Plastic deformation in circular shafts happens when torsional force pushes the material past its yield strength. At that point, the shaft does not return to its original shape. Instead, it keeps a permanent strain. To measure this change, the stress pattern inside the shaft must be understood. The maximum shearing stress is calculated first. Then a shearing-stress-strain diagram can be drawn to find the maximum shearing strain. Shearing strain changes linearly with the distance from the shaft...

circular shafts - elastoplastic materials

Torsion Limits in Circular Shafts

Circular shafts under torsion show a clear change from elastic behavior to plastic deformation as torque increases. In the elastic limit, stress rises in direct proportion to the distance from the shaft’s center. The outer material carries the highest stress first, and the shaft can still return to its original shape. At a critical stress level, the shaft begins to yield. This marks the maximum torque the shaft can carry without permanent deformation. Beyond this point, a plastic region forms...

residual stresses in circular shafts

Residual Stress After Shaft Torsion

Residual stress can remain in a circular shaft after torque causes plastic deformation. This happens in elastoplastic materials, which show both elastic and plastic behavior. The plastic change can come from high shearing stress or large strain. Residual stress is the internal stress that stays in the material after the external force is removed. In a shaft under torque, the angle of twist does not return fully to its original value. That lasting change shows that residual stress is still...

torsion of noncircular members

Torsion in Square and Rectangular Bars

Torsion in square and rectangular bars shows how shape changes the way a material twists. Circular shafts keep their cross-sections nearly unchanged because they are axis-symmetric. That symmetry helps the stress spread evenly, so the shaft can resist torsion without distorting. Square bars behave differently because they do not have the same axial symmetry. When twisted, their cross-sections distort, although the diagonals and the lines connecting the midpoints stay as exceptions. This means...

thin-walled hollow shafts

Shear Flow in Thin-Walled Shafts

Shear flow in a thin-walled hollow shaft describes how torsional force is carried through the wall. When a small segment of width dx is isolated, it is still in equilibrium, but it experiences torsional shearing forces at its ends. These forces come from the longitudinal shearing stress on the segment's minor surface multiplied by that surface area. The result is a uniform shear flow through the structure. To study the wall more closely, a small section with length ds is considered. The force...

key terms and concepts

Learn how torque, shearing stress, and polar moment of inertia lead to the elastic torsion formula for solid and hollow circular shafts in the linear range.

Torsion Formula for Circular Shafts | JoVE K12

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Torsion Formula for Circular Shafts

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