chapter 5 torsion

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Chapter

5TORSION:

5.1 Introduction5.2 Deformation of a Circular Shaft5.3 The Torsion Formula5.4 Axial and Transverse Shear Stresses5.5 Stresses on Inclined Planes5.6 Angle of Twist5.7 Statically Indeterminate Shafts5.8 Design of Circular Shafts5.9 Stress Concentrations

5.1 Introduction:= , = units of T is N-m, lb-ft

Figure 5.1

Figure 5.2( a) &(b)

In many engineering applications, members are required to transmitting atorque from one plane to a parallel plane and to carry these torsional loads.The simplest members for accomplished this function is called a shafts.Shafts are commonly used to connect an engine or motor to a pump,compressor, axle, or similar members. Shafts connecting gears and pulleysare a common application involving torsion members.

Torque is moment that tends to twist a member about the longitudinal axis.Figure 5.1 illustrated produced couples T1 and T2 that are called torques,twisting couples or twisting moments.

Consider a long, circular shaft (circular cross section is an efficient shapefor resisting torsional loads), of length L and radius c that is fixed at oneend A, as shown in Figure 6.2a.

When an external torque T is applied to the free end of the shaft at B, theshaft is deforms as shown in Figure 6.2b. All cross sections of the shaftare subjected to the same torque T, therefore, the shaft is said to be inpure torsion.

Longitudinal lines in Figure 6.2b are twisted into helixes as the free endof the shaft rotates through an angle . The angle of rotation is called asthe angle of twist or angular displacement. The angle of twist changesalong the length L of the shaft. For prismatic shaft, will vary linearlybetween the ends of the shafts.

The twisting deformation does not distort cross sections of the shaft inany way and the overall length remains constants. The followingassumptions can be applied to torsion of shafts that have circular-eithersolid or hollow- cross-sections:

51

Chapter

5TORSION:



Figure 5.1

Figure 5.2( a) &(b)







51

Chapter

5TORSION:



Figure 5.1

Figure 5.2( a) &(b)







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1) The plane sections before twisting remain plane after twisting (pointsin a given plane remain in that plane after twisting). In other words,circular cross sections do not warp as they twist.2) Cross sections rotate about and remain perpendicular to thelongitudinal axis of the shaft.3) Furthermore, each cross section remains undistorted as it rotatesrelative to neighboring cross sections. In other words, the cross sectionsremains circular and the expansion or contraction of a cross section doesnot occur. Thus all normal strains are zero and redial lines remain straightand redial as the cross section rotates.4) The distances between cross sections remain constant during thetwisting deformation. In other words, no axial strain occurs in a round shaftas it twists.

5) The material is homogeneous and isotropic.

In this chapter, we will develop formulas for the stresses anddeformations produced in circular bars subjected to torsion, such asdrive shafts, thin-walled members. This chapter covers severaladditional topics related to torsion: statically indeterminate members,strain energy, thin-walled tube of noncircular section, stressconcentration, and nonlinear behavior.

5.2 Torsional Shear Strain ():

Figure 5.3a

To investigate the deformations that occur during twisting, a short segment of the shaft shown in Figure 6.2 is isolated in Figure 6.3a. The shaft radius C ;however, for more generally, an interior cylindrical portion for an arbitrary radial

distance (where < ) from the center of the shaft will be examined asshown in Figure 5.3b .

As the shaft twist, the two cross sections of the segment rotate about the x-axis, and the line element CD on the undeformed shaft is twisted into helix C'D'

The angular difference between the rotation of the two cross sections is equalto , which creates a shear strain in the shaft. The shear strain is equal to the angle between line elements C'D' and C'D'',as shown in Figure 5.3b, which is given by:

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Figure 5.3a





52





Figure 5.3a





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Figure 5.3b

tan = = The distance can also be expressed by the arc length = ,

which gives:

tan = = = =

As the length of the shaft segment decrease to zero, the shear strainbecomes:

= (5.1) The quantityd , represents the rate of change of the angle of twist, is

the angle of twist per unit length. Note that Eq. (5.1) show that the shearstrain in a circular shaft varies linearly with respect to the redialcoordinate ; therefore, the shear strain at the shaft centerline (i.e., = 0) is zero while the largest shear strain occurs for the largest value of (i.e., = c ), which occurs on the outermost surface of the shaft. Then themaximum shear strain at the outer radius c is given by:

= (5.2) Eqs. (5.1) and (5.2) can be combined to express at any radial coordinate

in terms of as:

= (5.3) The above equations are valid for a circular bar of any material, elasticor inelastic, linear or nonlinear.

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Figure 5.3b


which gives:

tan = = = =





in terms of as:


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Figure 5.3b


which gives:

tan = = = =





in terms of as:


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5.2 Deformation of a Circular Shaft (another procedure):Deformation of a Circular Shaft: Consider a bar or shaft of circular cross section twisted by acouple T, assume the left-hand end is fixed and the right-hand end will rotate a small angle,called angle of twist.

By observation, if angle of rotation is small, length of shaft and its radius remain unchanged.

If every cross section has the same radius and subjected to the same torque, the angle (x)will vary linearly between ends under twisting deformation.

Important assumptions:1. The plane cross sections perpendicular to the axis of the bar remain plane after theapplication of a torque: points in a given plane remain in that plane after twisting.

2. Furthermore, expansion or contraction of a cross section does not occur, nor does ashortening or lengthening of the bar. Thus all normal strains are zero.

3. The material is homogeneous and isotropic.4. If is small neither the length L nor its radius will change. The torque T twist the bar (Figure 5.3a), the longitudinal line AB deform into the helix

AB', and the angle BOB' represent the total angle of twist . The turning of the right end through a small angle of twist d relative to the left end

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54








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(that is m = m') of an element of length dx in such a shaft (Figure 5.3b). Note that the straight line mn, initially parallel to the axis of the bar, distorts onto linemn', which is approximately straight.

The magnitude of is given by angle nmn'. Consider an element of the bar dx, on its outer surface we choose an small elementmmnn, during twisting the element rotate a small angle d, the element is in a state ofpure shear, and deformed into mmnn', observe that the length of nn' is:= . ( ) And the shear strain may be taken:tan = . ( )= = ( ) , represents the rate of change of the angle of twist, denote as the angle of twist per

unit length or the rate of twist. Then the maximum shear strain at the outer radius c isgiven by: = (5.1)

Additionally, the shear strain inside the bar or for an arbitrary radial distance fromthe center of the bar is: = (5.2)

Eq. (5.2), show that the shear strain in a circular shaft varies linearly with the distancefrom the axis of the shaft.

Eliminating, from eqs. (5.1) and (5.2), we have:= (5.3)5.3 The Torsion Formula:

Material Behaviour: Shear stress in the bar of a linear elastic materialis: = = . . (5.4) For solid shaft, shear stress varies from zero at shafts longitudinalaxis to maximum value at its outer surface.

and in circular bar vary linearly with the radial distance fromthe center, the maximum values max and max occur at the outersurface..

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Due to proportionality of triangles, or using Hookes law and Eq.5.4, yields: = (5.5) Consider a bar subjected to pure torsion, the shear force acting onan element area dA is dA, the moment of this force about the axisof bar is ( dA). The applied torque is:= ( ) = ,= . . ( )

Hear the integral is called the polar moment of inertia J of the entirecross section of a bar. For solid circular and hollow cross sections, J isgiven by the formulas:= = . . . (5.6), where c is the radius of the bar

In terms of the bar diameter, = = . . ( . )= ( ) . . (5.8) , where c and b are the inner and outerradii of the bar.In terms of the diameters,= (d d ) = 0.03125(do4 di4) . . (5.9)For thin-walled circular members (i.e., r/t 10), an approximateformula for J is given by: = 2 . (5.10)where r is the mean radius and t is the thickness of the tube,

421 cJ

414221 ccJ

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respectively. The maximum shear stress on the outer surface is: = . (5.11) Shear stress at an arbitrary radial distance from the center of the

bar is: = . (5.12) Because Hooks law was used in the derivation of Eqs.(5.4),(5.11), and (5.12), these formulas are valid if the shear stresses do notexceed the proportional limit of the material shear. The aboverelationships are referred to as the torsion formula and based only upongeometric concepts; they are valid if shaft is circular, its materialhomogenous, and it behaves in a linear-elastic manner.

Furthermore, The shear stress acting on the plane of the cross section areaccompanied by shear stresses of the same magnitude acting on longitudinalplane of the bar as depicted in Figure 5.6. If the material is weaker in shear on longitudinal plane than on cross-sectional planes, as in the case of a circular bar made of wood, the first crackdue to twisting will appear on the surface in longitudinal direction will besubjected to tensile and compressive stresses.

Sign Convention: T is positive, by right-hand rule, is directedoutward from the shaft.

5.4 Stresses on inclined planes:The elastic strain formula eq. (5.12) can be used to calculate the absolute maximum torsional stress ona transverse section in a circular shaft subjected to pure torsion. Thus we need to find location whereratio / is maximum, and to draw a torque diagram (internal torque T vs. X along shaft). Onceinternal torque throughout shaft is determined, maximum ratio of / can be identified. For this investigation, the stresses at a point in the shaft of Figure 5.7a will be analyzed. If the equations of equilibrium are applied to the free body diagram of Figure 5.7b, the following resultsare obtained: = ( cos ) cos + ( sin ) sin = 0

From which:

Figure 5.5

Figure 5.6

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= (cos sin ) = cos 2 (5.13)and : = ( cos ) sin ( sin ) cos = 0From which: = 2 cos sin = sin 2 . . (5.14)These results are shown in the graph of Figure 5.8. from which, we observe = = , and= 0, when = 45 , and = = ,

and = 0, when = 135 . thus theThe state of pure shear stress is

equivalent to equal tensile and compressive stresses onan element rotation through an angle of 45o.

If a twisted bar is made of material that isweaker in tension than in shear, failure will occur in

tension along a helix inclined at 45o, such as chalk.

Example 5.1(P (5.1) ):Given: A hollow steel shaft of outer radius c is fixed at one end and subjected to a torque T at theother end. If c = 30 mm , T = 2 kN m, all = 80 MPa.Find: If the shearing stress is not to exceed all , what is the required inner radius b?SOLUTION:

4 4( )2

allTc

c b

or

c b.

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36

4 42(2 10 )(0.03)80(10 ) [(0.03) ]b

or

6 6 480(10 )[0.81(10 ) ] 38.197b Solving,

0.024 24b m mm Example 5.2

Given: A hollow steel shaft with an outside diameter of 100 mm and a wall thickness of 10 mm issubjected to a pure torque of T = 5,500 N-m.Find:(a) The maximum shear stress in the hollow shaft.(b) The minimum diameter of a solid steel shaft for which the maximum shear stress is the same as inpart (a) for the same torque T.SOLUTION:

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Example 5.3Given: A solid 0.75-in.-diameter shaft issubjected to the torques shown in Fig. P6.8. Thebearings shown allow the shaft to turn freely.

Find:

(a) Plot a torque diagram showing the internaltorque in segments (1), (2), and (3) of the shaft.

Use the sign convention presented.(b) The maximum shear stress magnitude in the shaft.SOLUTION: by equilibrium:

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Example 5.4:Given: A compound shaft (Fig. P6.7) consists of brass segment (1) and aluminumsegment (2). Segment (1) is a solid brass shaft with an allowable shear stress of 60MPa. Segment (2) is a solid aluminum shaft with an allowable shear stress of 90 MPa.The torque of T = 23,000 N-m is applied at C.Find: the minimum required diameter of (a) the brass shaft and (b) the aluminumshaft.

SOLUTION:

Example 5.5(P-5.4):

Given: The circular shaft is subjected to the torques shown in Fig. P5.4.Find: What is the largest shearing stress in the member and where does it occur?SOLUTION: Apply the method of sections between the changes of load points:

50EFT N m 30DET N m 120CD BCT T N m 80ABT N m

We have4 4 6 4

2 [(0.025) (0.015) ] 0.534 10hJ m Thus,

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max :Tc J

3 32 2(80) 3.26(0.025)ABT MPac

3 62(120) 120(0.025)4.89 , 5.62(0.025) 0.534 10BC CDMPa MPa

Example 5.6(P-5.12):Given: A hollow shaft is made by rolling a plate of thickness t into a cylindrical shape and

welding the edges along the helical seams oriented at an angle to the axis of the member (Fig.P5.12). The permissible tensile and shear stresses in the weld are all = 100 MPa, all = 55MPa, respectively, and D = 120 mm, t = 5 mm, = 60o.Find: The maximum torque that can be applied to the shaft?

SOLUTION:

60 90 150o 4 4 6 4[(0.06) (0.055) ] 5.984 10

2J m

From Eq. (5.14a) at 150o :'

sin 300 0.866ox 6

6(0.06)100 10 0.866 0.866

5.984 10Tc TJ

or 11.52T kN m Similarly, Eq. (5.14b):

' 'cos300 0.5ox y

gives

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(0.06)55 10 0.5 , 10.975.986 10

T T kN m Thus,

10.97allT kN m

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5.5 Torsional Deformation (Angle of twist ):i. Prismatic Shafts: let us consider the deformation of a bar of

length L, of uniform circular cross section, fixed at one end andsubjected to torque T at the free end as shown in Figure 5.13.The rotation of the cross section at the free end of the shaft,called the angle of twist , is obtained as:

Geometry of deformation:= (5.1)Equilibrium conditions: = . (5.11)Material Behaviour ( Hook's law): = = = . . ( )

Equations (5.1) and (a) can then be combined to obtain the following expression of the angle of twist of theshaft:

= (5.15)Where is measured in radians. The direction of the angle of twist is the same as that of the torque T.Torsional spring stiffness: = = (5.16)ii. Multiple Prismatic Shafts (Stepped): in this case of shaft made of several segments with various

cross sections subjected to torques at a number of locationsas shown in Figure 5.14, Eq. (5.15) may applied to each partseparately.

For this purpose, internal reaction torque Tiand the angle of twist i for each part must be calculated as,

= . ( ) In general, the total angle of twist of the

shaft is then obtained by the algabric sum as:

= = (5.17)

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Here, the subscript i is an index identifing the various parts of the shaft and n is the total number of segments,each having constant values for Gi, Ji,Li, and Ti. In the U.S. Customary system, the consistent units are G[ psi ], T [ lbin ] , and L [ in.], and J [ in4] in the SI system, the consistent units are G [ Pa ], T [Nm ], L [ m ], and J [ m4]. The unit of in Eq. (3.15) is radians, regardless of

which system of unit is used in the computation. Sign convention: Represent torques as vectors using the right-hand rule, the same sign convention appliesto both T & which are taken as + if a right hand screw advances with the direction of the vector whentwisted in the direction indicated by the torque as illustrated in Figure 5.15.iii. Non-prismatic Shafts:

In this case, either the torque or the section or both changes continuously along the axis of the shaft,such as depicated in Figure 5.16a. In this situation, the torque Tx of a cross-section located

at a distance x from the end of the shaft may be obtained fromthe conditions of equilibrium (Figure 5.16b). Hence assuming the taper is small and gradual, the

maximum shearing stress at a cross section can be detrmine bythe application of the tortional formula.

Equation (5.15) may be applied to disk element of thickness . The differential angle of rotationfor the element equals: =

Integration of this expression results in the total angle of twist for a nonprismatic shaft as,= . (5.18)5.6 Power transmission:

Shafts are used to transmit power as shown in Figure 5.17. The power P is transmited by a torque Trotating at the angular speed is given by = , where is measured in radians per unit time.If the shaft is rotating with a frequency of revolutions per unit time, then = 2 , which gives= (2 ). Therefore, the torque can be expressed as:= (2 ) . . (5.19)

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In SI units, P usually measured in watts (1 = 1 . / ) and in hertz (1 = 1 / ); Eq.(5.19) then determines the torque T in NmIn U.S. Customary units with P in lbin./s and f in hertz, Eq. (5.19) calculates the torque T in lbin.

Because power in U.S. Customary units is oftenexpressed in horsepower (1 = 550 / = 396 10 ./ ), a convenient form of Eq.(5.19) is:( . ) = ( )2 396 10 ( . / )1( )

which simplifies to :( . ) = 63.0 10 ( )(1 = 745.7 = 745.7 . ).

5.7 Statically indeterminate problems:Note: Analogous to the case of axially loaded members. Needed torques cannot be determined using staticequilibrium equations alone. Additional compatibility conditions are needed.1. Draw the required free-body diagrams and write the equations of equilibrium.2. Derive the compatibility equations from the restrictions imposed on the angles of twist. Satisfies structuralintegrity (or geometric continuity) written in terms of displacements at a point in a given direction.3. Use the torque- twist relationships in Eq.(5.15) to express the angles of twist in the compatibility equationsin terms of the torques. Angle of twist at a given X-section is the same (i.e., unique) regardless of whichsegment joined at X-section is used to determine the angle of twist.4. Solve the equations of equilibrium and compatibility for the torques.

Example 5.7Given: A compound shaft supports several pulleys asshown in Figure E7. Segments (1) and (4) are solid25-mm-diameter steel [G = 80 GPa] shafts. Segments(2) and (3) are solid 50-mm-diameter steel shafts. Thebearings shown allow the shaft to turn freely.

Find:

(a) The maximum shear stress in the compound shaft.(b) The rotation angle of pulley D with respect to pulley B.(c) The rotation angle of pulley E with respect to pulley A.SOLUTION:

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Example 5.8Given: Figure (a) shows a steel shaft (G =80GPa) of length L = 1.5 m and diameter d = 25 mm thatcarries a distributed toque of intensity (torque per unit length) = ( / ), where tB = 200N m/m.

Find:

(1) the maximum shear stress in the shaft; and(2) The angle of twist.SOLUTION:

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Example 5.9Given: The shaft in Figure (a) consists of a 3-in diameteraluminum segment (G = 4106 psi) that is rigidly joined to a2-in diameter steel segment (G = 12106 psi). The ends ofthe shaft are attached to rigid supports, the torque =10 . is applied.Find: The maximum shear stress developed in eachsegment?

SOLUTION:by equilibrium: = 0, (10 10 ) = 0, This problemis statically indeterminate.

Example 5.10Given: The four rigid gears, loaded as shown in Figure (a), areattached to a 2-in.-diameter steel shaft.Compute: The angle of rotation of gear A relative to gear D. Use

G = 12106 psi for the shaft.

SOLUTION: It is convenient to represent the torques as vectors

(using the right-hand rule) on the FBDs in Fig. (b).

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Example 5.11(P5.48):Given: Two transmission shafts-one a hollow tube with an outer diameter of 100 mm and aninner diameter of 60 mm, the other solid with a diameter of 100 mm are each to transmit 200kW. If both operate at f = 6 Hz, compute the highest shearing stresses in each.

SOLUTION:

5.8 stress concentrations:

The derivation of the torsion formula,J

Tcmax , assumed a circular shaft with uniform cross-

section loaded through rigid end plates.The use of flange couplings, gears and pulleys attached to shafts by keys in keyways, andcross-section discontinuities can cause stress concentrations.Experimental or numerically determined concentration factors are applied as,

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JTcKK nommax .

Here the nominal or average shear stress is obtained for the smaller diameter shaft.

5.9 Torsion of Noncircular Members:

Previous torsion formulas are valid foraxisymmetric or circular shafts.

Planar cross-sections of noncircular shafts

do not remain planar and stress and straindistribution do not vary linearly For uniform rectangular cross-

section(table 5.1),

GabTL

and,abT

32max

Example 5.12:Given: A stepped shaft with a major diameter of D = 20 mm and a minor diameter of d = 16 mmis subjected to a torque of 25 N-m. A full quarter-circular fillet having a radius of r = 2 mm isused to transition from the major diameter to the minor diameter.Find: The maximum shear stress in the shaft?

SOLUTION:

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Example 5.13:

Given: A 40-mm-diameter shaft contains a 10-mm-deep U-shaped groove that has a 6-mm radiusat the bottom of the groove. The maximum shear stress in the shaft must be limited to 60 MPa.Find: If the shaft rotates at a constant angular speed of 22 Hz, determine the maximum powerthat may be delivered by the shaft.

SOLUTION:

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Example 5.14:Given: A torque of magnitude T = 1.5 kip-in. isapplied to each of the bars shown in Fig below.The allowable shear stress is specified as= 8 .Find: The minimum required dimension b foreach bar.

SOLUTION:

SOLUTION:SOLUTION: Apply the method of sections between the changes of load points: Find: The maximum torque that can be applied to the shaft? SOLUTION:SOLUTION:

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