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

Stellen Sie sich einen Fall vor, in dem das auf die kreisförmige Welle ausgeübte Drehmoment innerhalb der Grenze des Hookeschen Gesetzes liegt, sodass es zu keiner bleibenden Verformung kommt.
Erinnern Sie sich nun an den Ausdruck für die Scherdehnung. Multipliziert man ihn mit dem Steifigkeitsmodul und verwendet man das Hookesche Gesetz für Schubspannung und Dehnung, so kann ein Ausdruck für die Schubspannung in einer Welle bestimmt werden.
Es sei daran erinnert, dass die Summe der Momente der Elementarkräfte, die auf einen beliebigen Querschnitt der Welle ausgeübt werden, gleich der Größe des auf die Welle ausgeübten Drehmoments sein muss.
Das Ersetzen der Schubspannung und das Neuanordnen der Terme ergibt einen Ausdruck mit einem integralen Term, der das polare Trägheitsmoment des Querschnitts in Bezug auf seinen Mittelpunkt darstellt.
Weitere Umlagerungen und Substitutionen für maximale Schubspannung ergeben die elastische Torsionsformel für die Schubspannung in einer starren, gleichmäßigen Kreiswelle.
Bei einer Hohlwelle mit r1 und r2 als Innen- und Außenradien wird das polare Trägheitsmoment jedoch als Differenz in der vierten Potenz zweier Radien ausgedrückt.

kreisförmige welle – spannung im linearen bereich

Circular Shaft - Stresses in Linear Range

Stellen Sie sich ein Szenario vor, in dem eine kreisförmige Welle einem Drehmoment ausgesetzt ist, das innerhalb der Grenzen des Hookeschen Gesetzes bleibt und eine bleibende Verformung vermeidet. Daher wird die Formel für die Scherdehnung überarbeitet. Diese Formel wird mit dem Steifigkeitsmodul multipliziert und dann das Hookesche Gesetz für die Scherspannung und -dehnung angewendet. Daraus lässt sich die Gleichung für die Scherspannung in einer Welle ableiten.

<img alt="Equation 1" src="/files/ftp_upload/15614/15614_Equation_1.svg" style="text-align: center; height: 28px;">

Darüber hinaus ist es wichtig zu bedenken, dass die Summe der Momente der Elementarkräfte, die auf einen beliebigen Querschnitt der Welle wirken, mit dem auf diese Welle ausgeübten Drehmoment identisch sein muss. Ein Integralterm entsteht, wenn die Gleichung angepasst wird, um die Scherspannung zu ersetzen. Dieser Begriff bezeichnet das polare Trägheitsmoment des Querschnitts bezüglich seines Mittelpunkts. Nach weiteren Anpassungen und Ersetzungen für die maximale Scherspannung kann die elastische Torsionsformel für die Scherspannung in einer gleichmäßig starren kreisförmigen Welle abgeleitet werden.

<img alt="Equation 2" src="/files/ftp_upload/15614/15614_Equation_2.svg" style="text-align: center; height:40px;">

Anders verhält es sich jedoch bei einer Hohlwelle, bei der r_1 und r_2 als Innen- und Außenradius dargestellt werden. In diesem Fall wird das polare Trägheitsmoment als Differenz der vierten Potenz beider Radien ausgedrückt.

Kreisförmige Welle – Spannung im linearen Bereich

Stellen Sie sich einen Fall vor, in dem das auf die kreisförmige Welle ausgeübte Drehmoment innerhalb der Grenze des Hookeschen Gesetzes liegt, sodass es zu keiner bleibenden Verformung kommt.
Erinnern Sie sich nun an die ...

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Torsion

KEY TERMS AND CONCEPTS

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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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