CHAPTER 5 TORSION OF NON-CIRCULAR AND THIN-WALLED SECTIONS Summary For torsion of rectangular sections the maximum shear stress tmax and angle of twist e are given by T Tmax kidb2 L=kadbG ki and k2 being two constants,their values depending on the ratio d/b and being given in Table 5.1. For narrow rectangular sections,k=k2=. Thin-walled open sections may be considered as combinations of narrow rectangular sections so that T 3T Tmax kidb2= ∑db2 T 3T The relevant formulae for other non-rectangular,non-tubular solid shafts are given in Table 5.2. For thin-walled closed sections the stress at any point is given by T T= 2At where A is the area enclosed by the median line or mean perimeter and t is the thickness. The maximum stress occurs at the point where t is a minimum. The angle of twist is then given by TL 0= which,for tubes of constant thickness,reduces to ATs ts L=442G=2AG where s is the length or perimeter of the median line. 141
CHAPTER 5 TORSION OF NON-CIRCULAR AND THIN-WALLED SECTIONS Summary For torsion of rectangular sections the maximum shear stress tmax and angle of twist 0 are given by T tmax = ~ kldb2 T - e L k2db3G kl and k2 being two constants, their values depending on the ratio dlb and being given in Table 5.1. For narrow rectangular sections, kl = k2 = i. Thin-walled open sections may be considered as combinations of narrow rectangular sections so that 3T ___- - T Ckldb2 Cdb2 rmax = 3T - - T - - 0 - L Xk2db’G GCdb’ The relevant formulae for other non-rectangular, non-tubular solid shafts are given in For thin-walled closed sections the stress at any point is given by Table 5.2. T 2At r=- where A is the area enclosed by the median line or mean perimeter and t is the thickness. The maximum stress occurs at the point where t is a minimum. The angle of twist is then given by - e=----/ds TL 4A2G t which, for tubes of constant thickness, reduces to Ts rs - - e L 4A2Gt 2AG where s is the length or perimeter of the median line. 141
142 Mechanics of Materials 2 $5.1 Thin-walled cellular sections may be solved using the concept of constant shear flow g(=rt),bearing in mind that the angles of twist of all cells or constituent parts are assumed equal. 5.1.Rectangular sections Detailed analysis of the torsion of non-circular sections which includes the warping of cross-sections is beyond the scope of this text.For rectangular shafis,however,with longer side d and shorter side b,it can be shown by experiment that the maximum shearing stress occurs at the centre of the longer side and is given by T Tmax kidb2 (5.1) where k is a constant depending on the ratio d/b and given in Table 5.1 below. Table 5.1.Table of k and k2 values for rectangular sections in torsiont dib 】.0 15 1.75 2.0 2.5 3.0 4.0 6.0 8.0 10.0 00 ky 0.208 0.231 0.239 0.246 0.258 0.267 0.282 0.299 0.307 0.313 0.333 k2 0.141 0.196 0.2140.229 0.249 0.263 0.281 0.299 0.307 0.313 0.333 S.Timoshenko,Strength of Materials.Part I.Elementary Theory and Problems.Van Nostrand.New York. The essential difference between the shear stress distributions in circular and rectangular members is illustrated in Fig.5.1,where the shear stress distribution along the major and minor axes of a rectangular section together with that along a"radial"line to the corner of the section are indicated.The maximum shear stress is shown at the centre of the longer side,as noted above,and the stress at the corner is zero. Fig.5.1.Shear stress distribution in a solid rectangular shaft. The angle of twist per unit length is given by 0、T L=kadbG (5.2) kz being another constant depending on the ratio d/b and also given in Table 5.1
142 Mechanics of Materials 2 $5.1 Thin-walled cellular sections may be solved using the concept of constant shear flow q(= ~t), bearing in mind that the angles of twist of all cells or constituent parts are assumed equal. 5.1. Rectangular sections Detailed analysis of the torsion of non-circular sections which includes the warping of cross-sections is beyond the scope of this text. For rectangular shcrfrs, however, with longer side d and shorter side 6, it can be shown by experiment that the maximum shearing stress occurs at the centre of the longer side and is given by ’F I where kl is a constant depending on the ratio dlb and given in Table 5.1 below. Table 5.1. Table of kl and k2 values for rectangular sections in torsion‘“’. dlb 1.0 1.5 1.75 2.0 2.5 3.0 4.0 6 .O 8.0 10.0 00 kl 0.208 0.231 0.239 0.246 0.258 0.267 0.282 0.299 0.307 0.313 0.333 k2 0.141 0.196 0.214 0.229 0.249 0.263 0.281 0.299 0.307 0.313 0.333 “” S. Timoshenko, Strength of Materials. Part I, Elemeiiruri Theor) aid Problenrs, Van Nostrand. New York. The essential difference between the shear stress distributions in circular and rectangular members is illustrated in Fig. 5.1, where the shear stress distribution along the major and minor axes of a rectangular section together with that along a “radial” line to the corner of the section are indicated. The maximum shear stress is shown at the centre of the longer side, as noted above, and the stress at the comer is zero. Fig. 5.1. Shear stress distribution in a solid rectangular shaft. The angle of twist per unit length is given by T - 8 L kzdb3G k2 being another constant depending on the ratio dlb and also given in Table 5.1
§5.2 Torsion of Non-circular and Thin-walled Sections 143 In the absence of Table 5.1,however,it is possible to reduce the above equations to the following approximate forms: T db2 B+1 db3d+1.8b1 (5.3) 42TLJ 42TLI and 0= GA+Gd4b4 (5.4) where A is the cross-sectional area of the section (bd)and J=(bd/12)(b2+d2). 5.2.Narrow rectangular sections From Table 5.I it is evident that as the ratio d/b increases,i.e.the rectangular section becomes longer and thinner,the values of constants k and k2 approach 0.333.Thus,for narrow rectangular sections in which d/b>10 both k and k2 are assumed to be 1/3 and eqns.(5.1)and (5.2)reduce to 3T Tmax= db2 (5.5) 03T L=dbG (5.6) 5.3.Thin-walled open sections There are many cases,particularly in civil engineering applications,where rolled steel or extruded alloy sections are used where some element of torsion is involved.In most cases the sections consist of a combination of rectangles,and the relationships given in eqns.(5.1) and(5.2)can be adapted with reasonable accuracy provided that: (a)the sections are "open",i.e.angles,channels.T-sections,etc.,as shown in Fig.5.2; (b)the sections are thin compared with the other dimensions. Fig.5.2.Typical thin-walled open sections
85.2 Torsion of Non-circular and Thin-walled Sections 143 In the absence of Table 5.1, however, it is possible to reduce the above equations to the following approximate forms: T T rmax = [3 + 1.83 = -[3d db3+ 1.8bI (5.3) and 42TW 42TW GA4 Gd4b4 (5.4) (j=- - - where A is the cross-sectional area of the section (= bd) and J = (bd/12)(b2 + d2). 52. Narrow rectangular sections From Table 5.1 it is evident that as the ratio d/b increases, i.e. the rectangular section becomes longer and thinner, the values of constants k, and k2 approach 0.333. Thus, for narrow rectangular sections in which dlb > IO both kl and k2 are assumed to be 113 and eqns. (5.1) and (5.2) reduce to 3T db2 hax = - e 3~ - L db3G (5.5) (5.6) 53. Thin-walled open sections There are many cases, particularly in civil engineering applications, where rolled steel or extruded alloy sections are used where some element of torsion is involved. In most cases the sections consist of a combination of rectangles, and the relationships given in eqns. (5.1) and (5.2) can be adapted with reasonable accuracy provided that: (a) the sections are “open”, i.e. angles, channels. T-sections, etc., as shown in Fig. 5.2; (b) the sections are thin compared with the other dimensions. -F I Fig. i 5.2. Typical thin-walled open sections
144 Mechanics of Materials 2 §5.3 For such sections egns.(5.1)and(5.2)may be re-written in the form TT tmax kidb=Zi (5.7) T and T L=kadbG JegG (5.8) where Z'is the torsion section modulus =Z'web +Z'flanges kidib+kid2b2+...etc. =∑k1db2 and Jeg is the "effective"polar moment of area or"equivalent J"(see $5.7) J eq web Jeg fianges k2dib+k2d2b+...etc. =∑k2db3 T i.e. tmax三 ∑k1db2 (5.9) 0 T and L=G∑kdb (5.10) and for d/b ratios in excess of 10,k=k2=.so that 3T Tmax= ∑db2 (5.11) 8 3T L=G∑ab (5.12) To take account of the stress concentrations at the fillets of such sections,however,Timo- shenko and Young'suggest that the maximum shear stress as calculated above is multiplied by the factor 6 Aa (Figure 5.3).This has been shown to be fairly reliable over the range 0<a/b<0.5.In the event of sections containing limbs of different thicknesses the largest value of b should be used. b Fig.5.3. S.Timoshenko and A.D.Young.Strength of Materials.Van Nostrand.New York.1968 edition
144 Mechanics of Materials 2 $5.3 For such sections eqns. (5.1) and (5.2) may be re-written in the form T kldb2 Z’ - T Tmax = and T ~- - - T - I9 L k2db3G J,,G _- where Z’ is the torsion section modulus = Z’ web + Z’ flanges = kldlbt + kld2b; + . . . etc. = Ckldb2 and J,, is the “effective” polar moment of area or “equivalent J” (see $5.7) = J,, web + J,, flanges = k2dl b: + k2d2b: + . . . etc. = Ck2db3 T kldb2 i.e. Tmax = and l and for d/b ratios in excess of 10, kl = k:! = 3, so that 3T Tmax = ~ db2 3T - e - L GCdb3 (5.9) (5.10) (5.11) (5.12) To take account of the stress concentrations at the fillets of such sections, however, Timoshenko and Young? suggest that the maximum shear stress as calculated above is multiplied bv the factor (Figure 5.3). This has been shown to be fairly reliable over the range 0 < a/b < 0.5. In the event of sections containing limbs of different thicknesses the largest value of b should be used. Fig. 5.3 ‘S. Timoshenko and AD. Young, Strength offuteritrls, Van Nostrand. New York. 1968 edition
§5.4 Torsion of Non-circular and Thin-walled Sections 145 5.4.Thin-walled split tube The thin-walled split tube shown in Fig.5.4 is considered to be a special case of the thin-walled open type of section considered in $5.3.It is therefore treated as an equivalent rectangle with a longer side d equal to the circumference (less the gap),and a width b equal to the thickness. T Then Tmax kdb2 9 T and I= kdb3G d.meon circumference .2r Fig.5.4.Thin tube with longitudinal split. where k and k2 for thin-walled tubes are usually equal to. It should be noted here that the presence of even a very small cut or gap in a thin-walled tube produces a torsional stiffness(torque per unit angle of twist)very much smaller than that for a complete tube of the same dimensions. 5.5.Other solid(non-tubular)shafts Table 5.2 (see p.146)indicates the relevant formulae for maximum shear stress and angle of twist of other standard non-circular sections which may be encountered in practice. Approximate angles of twist for other solid cross-sections may be obtained by the substi- tution of an elliptical cross-section of the same area A and the same polar second moment of area J.The relevant equation for the elliptical section in Table 5.2 may then be applied. Alternatively,a very powerful procedure which applies for all solid sections,however irregular in shape,utilises a so-called"inscribed circle"procedure described in detail by RoarkT.The procedure is equally applicable to thick-walled standard T,I and channel sections and is outlined briefly below: Inscribed circle procedure Roark shows that the maximum shear stress which is set up when any solid section is subjected to torque occurs at,or very near to,one of the points where the largest circle which R.J.Roark and W.C.Young.Formulas for Stress Strain,5th edn.McGraw-Hill.Kogakusha
95.4 Torsion of Non-circular and Thin-walled Sections 145 5.4. Thin-walled split tube The thin-walled split tube shown in Fig. 5.4 is considered to be a special case of the thin-walled open type of section considered in 65.3. It is therefore treated as an equivalent rectangle with a longer side d equal to the circumference (less the gap), and a width b equal to the thickness. - Then and I T - e _-- L k2db3G demeon ctrcumference : 2rrr Fig. 5.4. Thin tube with longitudinal split. where kl and kf for thin-walled tubes are usually equal to f. It should be noted here that the presence of even a very small cut or gap in a thin-walled tube produces a torsional stiffness (torque per unit angle of twist) very much smaller than that for a complete tube of the same dimensions. 5.5. Other solid (non-tubular) shafts Table 5.2 (see p. 146) indicates the relevant formulae for maximum shear stress and angle of twist of other standard non-circular sections which may be encountered in practice. Approximate angles of twist for other solid cross-sections may be obtained by the substitution of an elliptical cross-section of the same area A and the same polar second moment of area J. The relevant equation for the elliptical section in Table 5.2 may then be applied. Alternatively, a very powerful procedure which applies for all solid sections, however irregular in shape, utilises a so-called “inscribed circle” procedure described in detail by Roarkt . The procedure is equally applicable to thick-walled standard T, I and channel sections and is outlined briefly below: Inscribed circle procedure Roark shows that the maximum shear stress which is set up when any solid section is subjected to torque occurs at, or very near to, one of the points where the largest circle which ’ R.J. Roark and W.C. Young, Formulas for Sfress & Strain, 5th edn. McCraw-Hill, Kogakusha
146 Mechanics of Materials 2 §5.5 Table 5.2ta) Cross-section Maximum shear stress Angle of twist per unit length Elliptic 16T 4x2T/ πb2h A4G at end of minor axis XX wherehand A is the area of cross-section =h/4 Equiloteral triangle 20T 46.2T b3 b4G at the middle of each side Regular hexogon 入 T 0.217Ad 0.133Ad2 where d is the diameter of inscribed circle and A is the cross-sectional area From S.Timoshenko.Strength of Materials.Part ll,Advanced Theory and Problems.Van Nostrand,New York,p.235. Approximate angles of twist for other solid cross-sections may be obtained by the substitution of an equivalent elliptical cross. section of the same area A and the same polar second moment of area /The relevant equation for the elliptical section in Table 5.2 may then be applied. can be constructed within the cross-section touches the section boundary-see Fig.5.5. Normally it occurs at the point where the curvature of the boundary is algebraically the least, convex curvatures being taken as positive and concave or re-entrant curvatures negative. The maximum shear stress is then obtained from either: Tmax r o =()c where,for positive curvatures (i.e.straight or convex boundaries), C= D +o15(-2月 1+ π2D4 16A2 with D=diameter of the largest inscribed circle, r=radius of curvature of boundary at selected position(positive), A cross-sectional area of section
146 Mechanics of Materials 2 $5.5 Table 5.2'") Cross-section Maximum shear stress Angle of twist per unit length EI liptic I 16T 4n2TJ rb2h AJG Equilateral triangle 20 T b' at the middle of each side - 46.2T b4 G ~ Regu lor hem T 0.217 Ad T 0.133 Ad2G where d is the diameter of inscribed circle and A is the cross-sectional area "') From S. Timoshenko. Strength of Materials. Part 11, Adwnced Theory nml Problems. Van Nostrand, New York, p. 235. Approximate angles of twist for other solid cross-sections may be obtained by the substitution of an equivalent elliptical crosssection of the same area A and the same polar second moment of area J. The relevant equation for the elliptical section in Table 5.2 may then be applied. can be constructed within the cross-section touches the section boundary - see Fig. 5.5. Normally it occurs at the point where the curvature of the boundary is algebraically the least, convex curvatures being taken as positive and concave or re-entrant curvatures negative. The maximum shear stress is then obtained from either: where, for positive curvatures (i.e. straight or convex boundaries), with D = diameter of the largest inscribed circle, r = radius of curvature of boundary at selected position (positive), A = cross-sectional area of section
§5.6 Torsion of Non-circular and Thin-walled Sections 147 Largest inscribed circle stress Fig.5.5.Inscribed circle stress evaluation procedure. or,for negative curvatures (concave or re-entrant boundaries): D 2 C=- π2D tanh 1+ 16A2 with =angle through which a tangent to the boundary rotates in travelling around the re-entrant position (radians)and r being taken as negative. For standard thick-walled open sections such as T,1,Z,angle and channel sections Roark also introduces formulae for angles of twist based upon the same inscribed circle proce- dure parameters. 5.6.Thin-walled closed tubes of non-circular section (Bredt-Batho theory) Consider the thin-walled closed tube shown in Fig.5.6 subjected to a torque T about the Z axis,i.e.in a transverse plane.Both the cross-section and the wall thickness around the periphery may be irregular as shown,but for the purposes of this simplified treatment it must be assumed that the thickness does not vary along the length of the tube.Then,if t is the shear stress at B and r'is the shear stress at C (where the thickness has increased to t')then,from the equilibrium of the complementary shears on the sides AB and CD of the element shown,it follows that udz =r't'dz t t'r i.e.the product of the shear stress and the thickness is constant at all points on the periphery of the tube.This constant is termed the shear flow and denoted by the symbol g (shear force per unit length). Thus g=rt constant (5.13) The quantity q is termed the shear flow because if one imagines the inner and outer boundaries of the tube section to be those of a channel carrying a flow of water,then. provided that the total quantity of water in the system remains constant,the quantity flowing past any given point is also constant
$5.6 Torsion of Non-circular and Thin-walled Sections 147 Largest inscribed circle / stress position Fig. 5.5. Inscribed circle stress evaluation procedure. or, for negative curvatures (concave or re-entrant boundaries): D 1 + O.1181oge I - - -0.238- tanhC= n2D4 [ { ( :) z} ?] with 4 = angle through which a tangent to the boundary rotates in travelling around the re-entrant position (radians) and r being taken as negative. For standard thick-walled open sections such as T, I, Z, angle and channel sections Roark also introduces formulae for angles of twist based upon the same inscribed circle procedure parameters. 5.6. Thin-walled closed tubes of non-circular section (Bredt-Batho theory) Consider the thin-walled closed tube shown in Fig. 5.6 subjected to a torque T about the Z axis, i.e. in a transverse plane. Both the cross-section and the wall thickness around the periphery may be irregular as shown, but for the purposes of this simplified treatment it must be assumed that the thickness does not vary along the length of the tube. Then, if r is the shear stress at B and r’ is the shear stress at C (where the thickness has increased to t’) then, from the equilibrium of the complementary shears on the sides AB and CD of the element shown, it follows that rt dz = r‘t’ dz tt = r’t’ i.e. the product of the shear stress and the thickness is constant at all points on the periphery of the tube. This constant is termed the shearjow and denoted by the symbol q (shear force per unit length). Thus q = tt = constant (5.13) The quantity q is termed the shear flow because if one imagines the inner and outer boundaries of the tube section to be those of a channel carrying a flow of water, then, provided that the total quantity of water in the system remains constant, the quantity flowing past any given point is also constant
148 Mechanics of Materials 2 §5.6 0 Fig.5.6.Thin-walled closed section subjected to axial torque. At any point,then,the shear force on an element of length ds is =rtds =gds and the shear stress is g/t. Consider now,therefore,the element BC subjected to the shear force o=gds rt ds. The moment of this force about O =dT=Op where p is the perpendicular distance from O to the force O. dT=gds p Therefore the moment,or torque,for the whole section = apds-afpds But the area COB=base x height =ipds ie. dA=pds or 2dA pds torque T=2q dA T=2qA (5.14) where A is the area enclosed within the median line of the wall thickness. Now,since 9=t it follows that T=2tA T or t=2At (5.15) where t is the thickness at the point in question
148 Mechanics of Materials 2 $5.6 Fig. 5.6. Thin-walled closed section subjected to axial torque. At any point, then, the shear force Q on an element of length ds is Q = rt ds = q ds and Consider now, therefore, the element BC subjected to the shear force Q = qds = ttds. The moment of this force about 0 the shear stress is q/t. =dT=Qp where p is the perpendicular distance from 0 to the force Q. .. dT = qdsp Therefore the moment, or torque, for the whole section But the area COB = 4 base x height = ipds i.e. dA= ipds or 2dA=pds torque T = 2q dA .. s T = 2qA where A is the area enclosed within the median line of the wall thickness. Now, since q = rt it follows that or T = 2ttA - T t= - 2At (5.14) (5.15) where t is the thickness at the point in question
$5.7 Torsion of Non-circular and Thin-walled Sections 149 It is evident,therefore,that the maximum shear stress in such cases occurs at the point of minimum thickness. Consider now an axial strip of the tube,of length L,along which the thickness and hence the shear stress is constant.The shear strain energy per unit volume is given by U-7 Thus,with thickness t,width ds and hence V=tLds T2L fds 842G/7 But the energy stored equals the work done =T0. 1 T9=8 T2L fds A2GJ1 The angle of twist of the tube is therefore given by TL ds 4A2Gt For tubes of constant thickness this reduces to TLs S 0= 442G=2AG (5.16) where s is the perimeter of the median line. The above equations must be used with care and do not apply to cases where there are abrupt changes in thickness or re-entrant corners. For closed sections which have constant thickness over specified lengths but varying from one part of the perimeter to another: [+2+3+…etc. 元=4A2G+26 5.7.Use of "equivalent J"for torsion of non-circular sections The simple torsion theory for circular sections can be written in the form: 0T L-GI and,as stated on page 143,it is often convenient to express the twist of non-circular sections in similar form:
$5.7 Torsion of Non-circular and Thin-walled Sections 149 It is evident, therefore, that the maximum shear stress in such cases occurs at the point of Consider now an axial strip of the tube, of length L, along which the thickness and hence minimum thickness. the shear stress is constant. The shear strain energy per unit volume is given by Thus, with thickness t, width ds and hence V = tLds - 1 (’>’ & ds 2At But the energy stored equals the work done = ;TO. The angle of twist of the tube is therefore given by @=-I- TL ds 4A2G t For tubes of constant thickness this reduces to tLs 4A2Gt 2AG - TLs @=--- (5.16) where s is the perimeter of the median line. abrupt changes in thickness or re-entrant comers. one part of the perimeter to another: The above equations must be used with care and do not apply to cases where there are For closed sections which have constant thickness over specified lengths but varying from 1 T si s2 s3 - + - + - +... etc. -=-[ e L 4A2G ti t2 t3 5.7. Use of “equivalent J” for torsion of non-circular sections The simple torsion theory for circular sections can be written in the form: 8T L GJ and, as stated on page 143, it is often convenient to express the twist of non-circular sections in similar form: -- - -
150 Mechanics of Materials 2 $5.8 ie. L GJea where Jeg is the "equivalent J'or "effective polar moment of area"for the section in question. Thus,for open sections: T T L=EkadbG GJeg with Jeg Ek2db3 (=Edb3 for d/b>10). Similarly,for square tubes of closed section: 0 TLs T T L4A2GI G[4A2t/s]GJeq and Jeg 4A2t/s. The torsional stiffness of any section,i.e.the ratio of torque divided by angle of twist per unit length,is then directly given by the value of GJ or GJeq i.e. T Stitness==GJ (or) 5.8.Thin-walled cellular sections The Bredt-Batho theory developed in the previous section may be applied to the solution of problems involving cellular sections of the type shown in Fig.5.7. R Fig.5.7.Thin-walled cellular section. Assume the length RSMN is of constant thickness ti and subjected therefore to a constant shear stress t.Similarly,NOPR is of thickness t2 and stress t2 with NR of thickness t3 and stress t3. Considering the equilibrium of complementary shear stresses on a longitudinal section at N,it follows that T1h=2h+t33 (5.17)
150 Mechanics of Materials 2 $5.8 i.e. where J,, is the “equivalent J’ or “effective polar moment of area” for the section in question. Thus, for open sections: T -- T - 0 L Ck2db3G GJeq - with J,, = Ck2db3 (= iCdb3 for dlb > 10). Similarly, for square tubes of closed section: 6’ TLs T -- T - - L 4A2Gt G[4A2t/s] GJeq and J,, = 4A2t/s. unit length, is then directly given by the value of GJ or GJ,, Le. The torsional stiffness of any section, i.e. the ratio of torque divided by angle of twist per T Stiffness = ~ = GJ (or GJes). @/L 5.8. Thin-walled cellular sections The Bredt-Batho theory developed in the previous section may be applied to the solution of problems involving cellular sections of the type shown in Fig. 5.7. S R P Fig. 5.7. Thin-walled cellular section. Assume the length RSMN is of constant thickness tl and subjected therefore to a constant shear stress rl . Similarly, NOPR is of thickness t2 and stress t2 with NR of thickness tg and stress r3. Considering the equilibrium of complementary shear stresses on a longitudinal section at N, it follows that Tltl = t2t2 + T3t3 (5.17)