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Torsion of Circular shaftsTorsion of Circular shafts
In this chapter, weIn this chapter, we shall considershall consider the problem of twisting or torsion.the problem of twisting or torsion.
A slender element subjectedA slender element subjected primarily toprimarily to twist is called a shaft.twist is called a shaft.
An example is that of a lathe whereinAn example is that of a lathe wherein the torque generated by a motor isthe torque generated by a motor is transmitted to a pulley by a shaft intransmitted to a pulley by a shaft in torsion.torsion.
Torsion of Circular shaftsTorsion of Circular shafts
In the transmission of power by a shaft inIn the transmission of power by a shaft in torsion, we are interested in the stressestorsion, we are interested in the stresses generated due to the twisting moment beinggenerated due to the twisting moment being resisted by the shaft.resisted by the shaft.
Torsion of Circular shaftsTorsion of Circular shafts
We will do the analysis as follows:We will do the analysis as follows:
1.1. Examine the geometric behaviour of a twisted shaftExamine the geometric behaviour of a twisted shaft and construct a plausible deformation modeland construct a plausible deformation model
2.2. Use stress strain relationshipUse stress strain relationship
3.3. Apply the equilibrium conditionsApply the equilibrium conditions
Geometry of deformation of aGeometry of deformation of a
twisted circular shafttwisted circular shaft
We consider theWe consider the shaft of length Lshaft of length L loaded only withloaded only with twisting momentstwisting moments at the ends.at the ends. Figure (b) showsFigure (b) shows that every crossthat every cross section A issection A is subjected to thesubjected to the same twistingsame twisting momentmoment
Geometry of deformation of aGeometry of deformation of a
twisted circular shafttwisted circular shaft
The onlyThe only possible modepossible mode of deformationof deformation is one in whichis one in which the crossthe cross-- sections remainsections remain unun--deformeddeformed but rotatebut rotate relative to onerelative to one another.another.
Geometry of deformationGeometry of deformation --AssumptionsAssumptions
Plane sections perpendicular to thePlane sections perpendicular to the shaft axis remain plane.shaft axis remain plane. Straight diameters are carried intoStraight diameters are carried into straight diameters after twisting.straight diameters after twisting. There is no lengthening or shorteningThere is no lengthening or shortening of the shaftof the shaft
Geometry of deformationGeometry of deformation
Shaft behaves like a series of disks thatShaft behaves like a series of disks that rotate slightly with respect to onerotate slightly with respect to one another when the shaft is twisted.another when the shaft is twisted.
Geometry of deformationGeometry of deformation
The deformationThe deformation can be observed bycan be observed by drawing somedrawing some gridlines on a shaftgridlines on a shaft subjected to twist.subjected to twist.
Geometry of deformation of aGeometry of deformation of a
twisted circular shafttwisted circular shaft
We further make the plausibleWe further make the plausible assumption that extensional strainsassumption that extensional strains are zero.are zero.
The following are the strains duringThe following are the strains during twistingtwisting γγθθrr == γγrzrz =0=0 γγθθzz = r d= r dφφ /dz/dz εεθθ == εεrr == εεzz =0=
Stress Strain relationsStress Strain relations
From stressFrom stress--strain relations, we canstrain relations, we can obtain the following values of theobtain the following values of the stresses during twistingstresses during twisting ττθθrr == ττrzrz =0=0 ττθθzz =G r d=G r dφφ/dz/dz σσθθ == σσrr == σσzz =0=
EquilibriumEquilibrium
From equilibrium,From equilibrium, we can writewe can write
∫∫rr ττθθzz dA = MdA = Mtt
Or,Or,
∫∫ r r dr r dφφ/dz GdA/dz GdA
= M= Mtt G dG dφφ/dz I/dz Ipp = M= Mtt
Stresses and deformationStresses and deformation
We have obtained:We have obtained:
ddφφ/dz = M/dz = Mtt /(G I/(G Ipp))
Where IWhere Ipp is called the Polar Momentis called the Polar Moment of Inertia and is given by:of Inertia and is given by:
IIpp== ∫∫ rr^22 dAdA
For a shaft of length L, the totalFor a shaft of length L, the total angle of twist between the ends is:angle of twist between the ends is:
φφ == ∫∫ MMtt /(G I/(G Ipp) dz = M) dz = Mtt L/(G IL/(G Ipp))
Torsion of Hollow Circular shaftsTorsion of Hollow Circular shafts
We can hence write:We can hence write:
ddφφ/dz = M/dz = Mtt /(G I/(G Ipp))
Where IWhere Ipp == ππ(r(roo^44 - - rrii^44 )/2)/
The shear stress is:The shear stress is:
ττθθzz = r d= r dφφ/dz = M/dz = Mtt r/ Ir/ Ipp
For a shaft of length L, the totalFor a shaft of length L, the total angle of twist between the ends is:angle of twist between the ends is:
φφ == ∫∫ MMtt /(G I/(G Ipp) dz = M) dz = Mtt L/(G IL/(G Ipp))
Hollow versus circular shaftsHollow versus circular shafts
Stress Analysis in TorsionStress Analysis in Torsion
The principalThe principal stresses instresses in torsion aretorsion are equal tensionequal tension andand compressioncompression inclined at 45inclined at 45 degrees to thedegrees to the axis of theaxis of the shaft.shaft.
Mohr circle for an element in torsionMohr circle for an element in torsion
Combined stresses due to tensionCombined stresses due to tension
and torsionand torsion
Combined stresses due to tensionCombined stresses due to tension
and torsionand torsion
Strain energy per unit volume (Axial load)Strain energy per unit volume (Axial load)
Work done by force = FWork done by force = Fδδ/2= Strain energy stored/2= Strain energy stored in the infinitesimal elementin the infinitesimal element Here F=Here F= σσxx dy dz anddy dz and δδ ==εεxxdxdx Or, W= (Or, W= (σσxx dy dzdy dz εεxxdx)/2dx)/ Strain energy per unit volume= (Strain energy per unit volume= (σσxx εεxx)/2)/
Strain energy per unit volume (shear load)Strain energy per unit volume (shear load)
Work done by force = FWork done by force = Fδδ/2= Strain energy stored/2= Strain energy stored in the infinitesimal elementin the infinitesimal element Here F=Here F= ττxyxy dx dz anddx dz and δδ ==γγxyxydydy Or, W= (Or, W= (ττxyxy dx dzdx dz γγxyxydy)/2dy)/ Strain energy per unit volume= (Strain energy per unit volume= (ττxyxy γγxyxy)/2)/
Onset of yielding in torsionOnset of yielding in torsion
The principal stresses in torsion are asThe principal stresses in torsion are as follows:follows:
Onset of yielding in torsionOnset of yielding in torsion
According to Von Mises criterion,According to Von Mises criterion, yielding occurs when the effectiveyielding occurs when the effective stress reaches a value equal to yieldstress reaches a value equal to yield stress Y i.e. whenstress Y i.e. when
√3√3 ττθθzz = Y= Y
Or,Or, ττθθzz = Y/ √3 = 0.577 Y= Y/ √3 = 0.577 Y
Onset of yielding in torsionOnset of yielding in torsion
According to Tresca Criterion, theAccording to Tresca Criterion, the maximum shear stress reaches amaximum shear stress reaches a value equal to Y/2, the failurevalue equal to Y/2, the failure occurs.occurs.
For torsion, the maximum shearFor torsion, the maximum shear stress isstress is ττθθzz. Thus failure occurs. Thus failure occurs when:when:
ττθθzz = Y/2 = 0.5 Y= Y/2 = 0.5 Y
