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