CHAPTER 8 TORSION Summary For a solid or hollow shaft of uniform circular cross-section throughout its length,the theory of pure torsion states that Tt G0 万=R=D where T is the applied external torque,constant over length L; J is the polar second moment of area of shaft cross-section .πD4 32 for a solid shaft and (dfor a hollow shat; 32 D is the outside diameter;R is the outside radius; d is the inside diameter; t is the shear stress at radius R and is the maximum value for both solid and hollow shafts; G is the modulus of rigidity(shear modulus);and 6 is the angle of twist in radians on a length L. For very thin-walled hollow shafts J=2nr3t,where r is the mean radius of the shaft wall and t is the thickness. Shear stress and shear strain are related to the angle of twist thus: t=- -R=Gy Strain energy in torsion is given by T2L GJ82 U= 2G= 2L 4Gx volume for solid shafis For a circular shaft subjected to combined bending and torsion the equivalent bending moment is M.=[M+√/(M2+T)] and the equivalent torque is T。=√/(M2+T2) where M and T are the applied bending moment and torque respectively. The power transmitted by a shaft carrying torque T at o rad/s Tw. 176
CHAPTER 8 TORSION Sommary For a solid or hollow shft of uniform circular cross-section throughout its length, the theory of pure torsion states that T T GO J R=L -=- where Tis the applied external torque, constant over length L; J is the polar second moment of area of shaft cross-section x(D4 - d 4, for a hollow shaft; xD4 32 32 = - for a solid shaft and D is the outside diameter; R is the outside radius; d is the inside diameter; T is the shear stress at radius R and is the maximum value for both solid and hollow shafts; G is the modulus of rigidity (shear modulus); and 8 is the angle of twist in radians on a length L. For very thin-walled hollow shafts J = 2nr3t, where T is the mean radius of the shaft wall and t is the thickness. Shear stress and shear strain are related to the angle of twist thus: GB L T=-R=G~ Strain energy in torsion is given by U=- x volume for solid shafis 2GJ 2L For a circular shaft subjected to combined bending and torsion the equivalent bending moment is Me = i[M + J(Mz +T ')I and the equivalent torque is where M and T are the applied bending moment and torque respectively. The pa~er transmitted by a shaft carrying torque Tat o rad/s = To. T, = +J( M +T 2, 176
$8.1 Torsion 177 8.1.Simple torsion theory When a uniform circular shaft is subjected to a torque it can be shown that every section of the shaft is subjected to a state of pure shear(Fig.8.1),the moment of resistance developed by the shear stresses being everywhere equal to the magnitude,and opposite in sense,to the applied torque.For the purposes of deriving a simple theory to describe the behaviour of shafts subjected to torque it is necessary to make the following basic assumptions: (1)The material is homogeneous,i.e.of uniform elastic properties throughout. (2)The material is elastic,following Hooke's law with shear stress proportional to shear strain. (3)The stress does not exceed the elastic limit or limit of proportionality. (4)Circular sections remain circular. (5)Cross-sections remain plane.(This is certainly not the case with the torsion of non- circular sections.) (6)Cross-sections rotate as if rigid,i.e.every diameter rotates through the same angle. Resisting Applied shear Complementory shear Applied torque T Fig.8.1.Shear system set up on an element in the surface of a shaft subjected to torsion. Practical tests carried out on circular shafts have shown that the theory developed below on the basis of these assumptions shows excellent correlation with experimental results. (a)Angle of twist Consider now the solid circular shaft of radius R subjected to a torque T at one end,the other end being fixed(Fig.8.2).Under the action of this torque a radial line at the free end of the shaft twists through an angle 0,point A moves to B,and AB subtends an angle y at the fixed end.This is then the angle of distortion of the shaft,i.e.the shear strain. Since angle in radians arc radius arc AB=R0=Ly Y=R0/L (8.1) From the definition of rigidity modulus shear stress t G= shear strain y
§8.1 Torsion 177 8.1. Simple torsion theory When a uniform circular shaft is subjected to a torque it can be ShOWn that every sectiOn of the shaft is subjected to a state of pure shear (Fig. 8.1 ), the moment of resistance developed by the shear stresses being everywhere equal to the magnitude, and opposite in sense, to the applied torque. For the purposes of deriving a simple theory to describe the behaviour of shafts subjected to torque it is necessary to make the following basic assumptionS: (1) The material is homogeneous, i.e. of uniform elastic properties throughout. (2) The material is elastic, following Hooke's law with shear stress proportional to shear strain. (3) The stress does nOt exceed the elastic limit or limit of proportionality. (4) Circular SectiOnS remain circular. (5) Cross-sectioDS remain plane. (This is certainly nOt the case with the torsion of DODcircular SectiOnS.) (6) Cross-sectioDS rotate as if rigid, i.e. every diameter rotates through the same angle. Fig. 8.1. Shear system set up on an elem-ent in thesufface of a shaft subjected to torsion. Practical tests carried out on circular shafts have shown that the theory developed below on the basis of these assumptions shows excellent correlation with experimental results. (a) Angle of twist Consider now the solid circular shaft of radius R subjected to a torque T at one end, the other end being fixed (Fig. 8.2). Under the action of this torque a radial line at the free end of the shaft twists through an angle 9, point A moves to B, and AB subtends an angle y at the fixed end. This is then the angle of distortion of the shaft, i.e. the shear strain. SinCe angle in radians = arc + radius arc AB = R8 = Ly y = R8/ L (8.1) From the definition of rigidity modulus shear stress T shear strain y G=
178 Mechanics of Materials §8.1 Fig.8.2. (8.2) where t is the shear stress set up at radius R. Therefore equating eqns.(8.1)and(8.2), RO T L-G a-2(-) (8.3) where t'is the shear stress at any other radius r. (b)Stresses Let the cross-section of the shaft be considered as divided into elements of radius r and thickness dr as shown in Fig.8.3 each subjected to a shear stress t'. Fig.8.3.Shaft cross-section. The force set up on each element stress x area =t'x 2nr dr (approximately)
178 Mechanics of Materials $8.1 T .. Fig. 8.2. T y=- G where T is the shear stress set up at radius R. Therefore equating eqns. (8.1) and (8.2), where T' is the shear stress at any other radius r. (b) Stresses Let the cross-section of the shaft be considered as divided into elements of radius r and thickness dr as shown in Fig. 8.3 each subjected to a shear stress z'. Fig. 8.3. Shaft cross-section. The force set up on each element = stress x area = 2' x 2nr dr (approximately)
$8.2 Torsion 179 This force will produce a moment about the centre axis of the shaft,providing a contribution to the torque =(t'X2πrdr)Xr =2πx'r2dr The total torque on the section T will then be the sum of all such contributions across the section, i.e. T= 2nt'r2 dr 0 Now the shear stress t'will vary with the radius r and must therefore be replaced in terms of r before the integral is evaluated. From eqn.(8.3) GO t'= L T= 2 L R 2nr3dr L 0 The integral 2dr is called the polar second moment ofarea J,and may be evaluated asa standard form for solid and hollow shafts as shown in $8.2 below. T GO or 万= (8.4) Combining egns.(8.3)and(8.4)produces the so-called simple theory of torsion: Tt GO 了=R=L (8.5) 8.2.Polar second moment of area As stated above the polar second moment of area J is defined as 2nr3dr
$8.2 Torsion 179 This force will produce a moment about the centre axis of the shaft, providing a contribution to the torque = (7' x 2nrdr) x r = 2n7'r2 dr The total torque on the section T will then be the sum of all such contributions across the section, i.e. T = 2nz'r2dr J i 0 Now the shear stress z' will vary with the radius rand must therefore be replaced in terms of r before the integral is evaluated. From eqn. (8.3) R .. 0 = E L jkr3 dr 0 The integral 5 0" 2nr3 dr is called the polar second moment of area J, and may be evaluated as a standard form for solid and hollow shafts as shown in $8.2 below. GO L .. T=-J or T GO JL - =- Combining eqns. (8.3) and (8.4) produces the so-called simple theory of torsion: T z G8 -- - J-R-L 8.2. Polar second moment of area As stated above the polar second moment of area J is defined as J = 2nr3dr 0 i
180 Mechanics of Materials $8.3 For a solid shaft, -[ 2πR4 元D4 or 4 32 (8.6) For a hollow shaft of internal radius r, =R-ror五0-的 (8.7) For thin-walled hollow shafts the values of D and d may be nearly equal,and in such cases there can be considerable errors in using the above equation involving the difference of two large quantities of similar value.It is therefore convenient to obtain an alternative form of expression for the polar moment of area. Now 2πr3dr=Σ(2πrdr)r2 =2Ar2 where A(=2xr dr)is the area of each small element of Fig.8.3,i.e.J is the sum of the Ar2 terms for all elements. If a thin hollow cylinder is therefore considered as just one of these small elements with its wall thickness t=dr,then J=Ar2 =(2nrt)r2 =2r3t(approximately) (8.8) 8.3.Shear stress and shear strain in shafts The shear stresses which are developed in a shaft subjected to pure torsion are indicated in Fig.8.1,their values being given by the simple torsion theory as Ge -TR Now from the definition of the shear or rigidity modulus G, t=Gy It therefore follows that the two equations may be combined to relate the shear stress and strain in the shaft to the angle of twist per unit length,thus G0 t =Gy (8.9)
180 For a solid shafi, Mechanics of Materials nD* or - 4 32 2n~4 =- For a hollow shaft of internal radius r, J=2n r3dr=2n - i [:I: x x = -(R4-r4) or -(D4-d*) 2 32 $8.3 For thin-walled hollow shafis the values of D and d may be nearly equal, and in such cases there can be considerable errors in using the above equation involving the difference of two large quantities of similar value. It is therefore convenient to obtain an alternative form of expression for the polar moment of area. Now J = 2nr3dr = C(2nrdr)r’ = AY’ 0 i where A ( = 2nr dr) is the area of each small element of Fig. 8.3, i.e. J is the sum of the Ar2 terms for all elements. If a thin hollow cylinder is therefore considered as just one of these small elements with its wall thickness t = dr, then J = Ar’ = (2nrt)r’ = 2xr3t (approximately) (8.8) 8.3. Shear stress and shear strain in shafts The shear stresses which are developed in a shaft subjected to pure torsion are indicated in Fig. 8.1, their values being given by the simple torsion theory as GO L 7=-R Now from the definition of the shear or rigidity modulus G, r = Gy It therefore follows that the two equations may be combined to relate the shear stress and strain in the shaft to the angle of twist per unit length, thus
$8.4 Torsion 181 or,in terms of some internal radius r, G0 -LI-GY (8.10) These equations indicate that the shear stress and shear strain vary linearly with radius and have their maximum value at the outside radius(Fig.8.4).The applied shear stresses in the plane of the cross-section are accompanied by complementary stresses of equal value on longitudinal planes as indicated in Figs.8.1 and 8.4.The significance of these longitudinal shears to material failure is discussed further in $8.10. Complementary longitudinal shears Fig.8.4.Complementary longitudinal shear stress in a shaft subjected to torsion. 8.4.Section modulus It is sometimes convenient to re-write part of the torsion theory formula to obtain the maximum shear stress in shafts as follows: Tt 万=R TR With R the outside radius of the shaft the above equation yields the greatest value possible for t (Fig.8.4), TR i.e. Tmax= tmu-Z (8.11) where Z =J/R is termed the polar section modulus.It will be seen from the preceding section that: πD3 for solid shafts, Z= (8.12) 16 and for hollow shafts, z=0*-d4 (8.13) 16D
$8.4 Torsion 181 or, in terms of some internal radius r, (8.10) These equations indicate that the shear stress and shear strain vary linearly with radius and have their maximum value at the outside radius (Fig. 8.4). The applied shear stresses in the plane of the cross-section are accompanied by complementary stresses of equal value on longitudinal planes as indicated in Figs. 8.1 and 8.4. The significance of these longitudinal shears to material failure is discussed further in 88.10. Fig. 8.4. Complementary longitudinal shear stress in a shaft subjected to torsion. 8.4. Section modulus It is sometimes convenient to re-write part of the torsion theory formula to obtain the maximum shear stress in shafts as follows: TT JR -=- With R the outside radius of the shaft the above equation yields the greatest value possible for T (Fig. 8.4), i.e. TR 7-= - J T Z .. T-=- (8.11) where 2 = J/R is termed the polar section modulus. It will be seen from the preceding section that: nD3 16 for solid shafts, Z=- (8.12) and for hollow shafts, ~(D1-d~) 160 Z- (8.13)
182 Mechanics of Materials §8.5 8.5.Torsional rigidity The angle of twist per unit length of shafts is given by the torsion theory as 6 T L"G可 The quantity GJ is termed the torsional rigidity of the shaft and is thus given by GJ=T 0/L (8.14) i.e.the torsional rigidity is the torque divided by the angle of twist(in radians)per unit length. 8.6.Torsion of hollow shafts It has been shown above that the maximum shear stress in a solid shaft is developed in the outer surface,values at other radii decreasing linearly to zero at the centre.It is clear, therefore,that if there is to be some limit set on the maximum allowable working stress in the shaft material then only the outer surface of the shaft will reach this limit.The material within the shaft will work at a lower stress and,particularly near the centre,will not contribute as much to the torque-carrying capacity of the shaft.In applications where weight reduction is of prime importance as in the aerospace industry,for instance,it is often found advisable to use hollow shafts. The relevant formulae for hollow shafts have been introduced in $8.2 and will not be repeated here.As an example of the increased torque-to-weight ratio possible with hollow shafts,however,it should be noted for a hollow shaft with an inside diameter half the outside diameter that the maximum stress increases by 6%over that for a solid shaft of the same outside diameter whilst the weight reduction achieved is approximately 25 8.7.Torsion of thin-walled tubes The torsion of thin-walled tubes of circular and non-circular cross-section is treated fully in Mechanics of Materials 2. 8.8.Composite shafts-series connection If two or more shafts of different material,diameter or basic form are connected together in such a way that each carries the same torque,then the shafts are said to be connected in series and the composite shaft so produced is therefore termed series-connected (Fig.8.5)(see Example 8.3).In such cases the composite shaft strength is treated by considering each component shaft separately,applying the torsion theory to each in turn;the composite shaft will therefore be as weak as its weakest component.If relative dimensions of the various parts are required then a solution is usually effected by equating the torques in each shaft,e.g.for two shafts in series T=C1J8=G2J202 (8.15) L2 1E.J.Hearn,Mechanics of Materials 2,3rd edition(Butterworth-Heinemann,Oxford,1997)
182 Mechanics of Materials $8.5 8.5. Torsional rigidity The angle of twist per unit length of shafts is given by the torsion theory as e~ L=GJ - The quantity GJ is termed the torsional rigidity of the shaft and is thus given by T GJ =- 91~ (8.14) i.e. the torsional rigidity is the torque divided by the angle of twist (in radians) per unit length. 8.6. Torsion of hollow shafts It has been shown above that the maximum shear stress in a solid shaft is developed in the outer surface, values at other radii decreasing linearly to zero at the centre. It is clear, therefore, that if there is to be some limit set on the maximum allowable working stress in the shaft material then only the outer surface of the shaft will reach this limit. The material within the shaft will work at a lower stress and, particularly near the centre, will not contribute as much to the torque-carrying capacity of the shaft. In applications where weight reduction is of prime importance as in the aerospace industry, for instance, it is often found advisable to use hollow shafts. The relevant formulae for hollow shafts have been introduced in $8.2 and will not be repeated here. As an example of the increased torque-to-weight ratio possible with hollow shafts, however, it should be noted for a hollow shaft with an inside diameter half the outside diameter that the maximum stress increases by 6 % over that for a solid shaft of the same outside diameter whilst the weight reduction achieved is approximately 25 %. 8.7. Torsion of thin-walled tubes The torsion of thin-walled tubes of circular and non-circular cross-section is treated fully in Mechanics of Materials 2. t 8.8. Composite shafts - series connection If two or more shafts of different material, diameter or basic form are connected together in such a way that each carries the same torque, then the shafts are said to be connected in series and the composite shaft so produced is therefore termed series-connected (Fig. 8.5) (see Example 8.3). In such cases the composite shaft strength is treated by considering each component shaft separately, applying the torsion theory to each in turn; the composite shaft will therefore be as weak as its weakest component. If relative dimensions of the various parts are required then a solution is usually effected by equating the torques in each shaft, e.g. for two shafts in series T=----- GlJlO1 =- Ll L2 t E. J. Hearn, Mechanics of Materials 2, 3rd edition (Butterworth-Heinemann, Oxford, 1997). (8.15)
$8.9 Torsion 183 Fig.8.5."Series-connected"shaft-common torque. In some applications it is convenient to ensure that the angles of twist in each shaft are equal,i.e.0 =02,so that for similar materials in each shaft J1=J2 Ly L2 or =4 LJ (8.16 8.9.Composite shafts-parallel connection If two or more materials are rigidly fixed together such that the applied torque is shared between them then the composite shaft so formed is said to be connected in parallel (Fig.8.6). Fixed end end Torque T Fig.8.6."Parallel-connected"shaft-shared torque. For parallel connection, total torque T=T1+T2 (8.17) In this case the angles of twist of each portion are equal and TiL:T2L2 三 (8.18) G1J1G2J2
58.9 Torsion T 183 Fig. 8.5. “Series<onnected shaft - common torque. In some applications it is convenient to ensure that the angles of twist in each shaft are equal, i.e. 8, = f12, so that for similar materials in each shaft Jl Jz Ll L2 LZ J2 --_ - - Ll =- J1 or (8.16) 8.9. Composite shafts - parallel connection If two or more materials are rigidly fixed together such that the applied torque is shared between them then the composite shaft so formed is said to be connected in parallel (Fig. 8.6). Torque T Fig. 8.6. “Parallelconnected” shaft - shared torque. For parallel connection, total torque T = TI +Tz In this case the angles of twist of each portion are equal and (8.17) (8.18)
184 Mechanics of Materials §8.10 i.e.for equal lengths (as is normally the case for parallel shafts) - T2 G2J2 (8.19) Thus two equations are obtained in terms of the torques in each part of the composite shaft and these torques can therefore be determined. The maximum stresses in each part can then be found from TiRi 、T2R2 t1= J and t2= J2 8.10.Principal stresses It will be shown in $13.2 that a state of pure shear as produced by the torsion of shafts is equivalent to a system of biaxial direct stresses,one stress tensile,one compressive,of equal value and at 45 to the shaft axis as shown in Fig.8.7;these are then the principal stresses. 459 Fig.8.7.Shear and principal stresses in a shaft subjected to torsion. Thus shafts which are constructed from brittle materials which are notably weaker under direct stress than in shear(cast-iron,for example)will fail by cracking along a helix inclined at 45 to the shaft axis.This can be demonstrated very simply by twisting a piece of chalk to failure (Fig.8.8a).Ductile materials,however,which are weaker in shear,fail on the shear planes at right angles to the shaft axis(Fig.8.8b).In some cases,e.g.timber,failure occurs by cracking along the shear planes parallel to the shaft axis owing to the nature of the material with fibres generally parallel to the axis producing a weakness in shear longitudinally rather than transversely.The complementary shears of Fig.8.4 then assume greater significance. 8.11.Strain energy in torsion It will be shown in $11.4 that the strain energy stored in a solid circular bar or shaft subjected to a torque T is given by the alternative expressions T2L GJ02 2 U -Te-2GJ2L=4Gx volume (8.20)
184 Mechanics of Materiols 88.10 i.e. for equal lengths (as is normally the case for parallel shafts) (8.19) Thus two equations are obtained in terms of the torques in cach part of the composite shaft and these torques can therefore be determined. The maximum stresses in each part can then be found from T2 R2 TI =- and T~ =- Tl Rl Jl J2 8.10. Principal stresses It will be shown in 813.2 that a state of pure shear as produced by the torsion of shafts is equivalent to a system of biaxial direct stresses, one stress tensile, one compressive, of equal value and at 45" to the shaft axis as shown in Fig. 8.7; these are then the pMcipal stresses. Fig. 8.7. Shear and principal stresses in a shaft subjected to torsion. Thus shafts which are constructed from brittle materials which are notably weaker under direct stress than in shear (cast-iron, for example) will fail by cracking along a helix inclined at 45" to the shaft axis. This can be demonstrated very simply by twisting a piece of chalk to failure (Fig. 8.8a). Ductile materials, however, which are weaker in shear, fail on the shear planes at right angles to the shaft axis (Fig. 8.8b). In some cases, e.g. timber, failure occurs by cracking along the shear planes parallel to the shaft axis owing to the nature of the material with fibres generally parallel to the axis producing a weakness in shear longitudinally rather than transversely. The complementary shears of Fig. 8.4 then assume greater significance. 8.11. Strain energy in torsion It will be shown in 511.4 that the strain energy stored in a solid circular bar or shaft subjected to a torque Tis given by the alternative expressions T2L GJg2 72 2GJ 2L 4G u = = - - - --_ - xvolume (8.20)
§8.11 Torsion 185 Fig.8.8a.Typical failure of a brittle material (chalk)in torsion.Failure occurs on a 45 helix owing to the action of the direct tensile stresses produced at 45 by the applied torque. Fig.8.8b.(Foreground)Failure of a ductile steel in torsion on a plane perpendicular to the specimen longitudinal axis.Scribed lines on the surface of the specimen which were parallel to the longitudinal axis before torque application indicate the degree of twist applied to the specimen.(Background) Equivalent failure of a more brittle,higher carbon steel in torsion.Failure again occurs on 45 planes but in this case,as often occurs in practice,a clean fracture into two pieces did not take place
§8.11 Torsion 185 Fig. 8.8a. Typical failure of a brittle material (chalk) in torsion. Failure occurs on a 45° helix owing to the action of the direct tensile stresses produced at 45° by the applied torque. Fig. 8.8b. (Foreground) Failure of a ductile steel in torsion on a plane perpendicular to the specimen longitudinal axis. Scribed lines on the surface of the specimen which were parallel to the longitudinal axis before torque application indicate the degree of twist applied to the specimen. (Background) Equivalent failure of a more brittle, higher carbon steel in torsion. Failure again occurs on 450 planes but in this case, as often occurs in practice, a clean fracture into two pieces did not take place
