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: -- - -