270 GOVERNING EQUATIONS FOR A DEFORMABLE SOLID CHAP. 10 This variational statement is called Reissner's principle (see Ref. 8).(a) Transform IR to Tp, by requiring the stresses to satisfy the stressdisplacement relations. Hint: Note (10-101). k(b) Transform FIR to -IH by restricting the geometry to be linear (a= aand eij = (ui, j + uj, i)/2) and requiring the stresses to satisfy the stressequilibrium equations and stress boundary conditions on DQ. Hint: Integrate uijeij by parts, using (10-81).10-29. Interpret (10-90) as St. Venant Theory ofdQ -1c where PQ is a force applied at Q in the direction of the displacement measure, dQ. Torsion-Feexure of Prismatic Members 11-1. INTRODUCTION AND NOTATION A body whose cross-sectional dimensions are small in comparison with itsaxial dimension is called a member. If the centroidal axis is straight and theshape and orientation of the normal cross section are constant,-'-the member is said to be prismatic. We define the member geometry with respect to a global reference frame (X1, X2, X3), as shown in Fig. 11-1. The X1 axis istaken to coincide with the centroidal axis and X2, X are taken as the principalinertia directions. We employ the following notation for the cross-sectionalproperties: A = f dx, dx, = fS dA I2 = j'(x 3)2 dA (11-1) I3 =-(X)2 dA Since X2, X3 pass through the centroid and are principal inertia directions,the centroidal coordinates and product of inertia vanish: Ax2 = -x dJXxdAz = 0 3 0 (11-2) 123 = (fX2 dA 0X3 One can work with an arbitrary orientation of the reference axes, but this willcomplicate the derivation. St. Venant's theory of torsion-flexure is restricted to linearbehavior. It is an exact linear formulation for a prismatic member subjected to a prescribed f The case where the cross-sectional shape is constant but the orientation varies along thecentroidal axis is treated in Chapter 15. 271272 I I '" TORSION-FLEXURE OF PRISMATIC MEMBERS CHAP. 11 distribution of surface forces applied on the end cross sections. Later, in Chapter 13, we modify the St. Venant theory to account for displacement restraint at the ends and for geometric nonlinearity. X2 X1 Fig. 11-1. Notation for prismatic member. The distribution of surface forces on a cross section is specified in terms of its statically equivalent force system at the centroid. Figure 11-1 shows the stress components on a positive face. We define F+, M+ as the force and moment vectors acting at the centroid which are statically equivalent to the distribution of stresses over the section. The components of F, M+ are called stress resultants and stress couples, respectively, and their definition equations are F = J, 11 dA F = ffU1 2 dA F3 = fOr 3 dA M, = f(x261 -3c 12)dA (11-3) M2 = fJJX 3a 1 1 dA M3 = -fx 2 all dA The internal force and moment vectors acting on the negative face are denoted by F_, M_ Since F_ = -+ M_ = -M+ (11-4) it follows that the positive sense of the stress resultants and couples for the negative face is opposite to that shown in Fig. 11-1. We discuss next the pure-torsion case, i.e., where the end forces are statically equivalent to only M1. We then extend the formulation to account for flexure SEC. 11-2. THE PURE-TORSION PROBLEM 273 and treat torsional-flexural coupling. Finally, we describe an approximate procedure for determining the flexural shear stress distribution in thin-walled sections. 11-2. THE PURE-TORSION PROBLEM Consider the prismatic member shown in Fig. 11-2. There are no boundary forces acting on the cylindrical surface. The boundary forces acting on the end cross sections are statically equivalent to just a twisting moment M1. Also, there is no restraint with respect to axial (out-of-plane) displacement at the ends. The analysis of this member presents the pure-torsion problem. In what follows, we establish the governing equations for pure torsion, using the approach originally suggested by St. Venant. X2 ill1 Ailli ---*-i--Fig. 11-2. Prismatic member in pure torsion. Rather than attempt to solve the three-dimensional problem directly, we impose the following conditions on the behavior and then determine what problem these conditions correspond to. 1. Each cross section is rigid with respect to deformation in its plane, i.e., e2 = 83 = 723 = 0. 2. Each cross section experiences a rotation w1 about the X1 axist and an out-of-plane displacement u1. These conditions lead to the following expansions for the in-place displace-ments: 112 = -1X3 113 = +0 1X2 The corresponding linear strains are 82 = 83 = Y23 = 0 C1 =' ll. 1 (11-6) 1 2 = U1, 2 + 12. 1 = tl, 2 -X3 01, 1 713 = U1, 3 + u3,1 ==0U1,3 ± X2 ,11 t Problem 11-treats the general case where the cross section rotates about an arbitrary point.274 TORSION-FLEXURE OF PRISMATIC MEMBERS CHAP. 11 Now, the strains must be independent of xl since each cross section is subjected to the same moment. This requires col 1 = const -kl (11-7) U1 = lt1 (X2, X3 ) We consider the left end to be fixed with respect to rotation and express cot, Ul as (11-8) = kl(qtU1 where d = 0t(x2, x 3) defines the out-of-plane displacement (warping) of a cross section. The strains and stresses corresponding to this postulated displacement behavior are E = E2 = 3 = Y23 = 0 Y12 = k (t, 2 -3) (11-9) T13 = 1(¢t, 3 + X2) and al0' = 2 2 = = 01 1 3 3= a'2 3 a012 = Gyl2 = Gkl(t 2 - x3) = 51 2(x2, x3) (11-10) U13 = G1 3 = Gki(4t 3 + x2) = a13(x2, x3) We are assuming that the material is isotropict and there are no initial strains. One step remains, namely, to satisfy the stress-equilibrium equations and stress boundary conditions on the cylindrical surface. The complete system of linear stress-equilibrium equations, (10-49), reduces to a 2 1,2 + a3 1,3 = 0 (11-11) Substituting for the shearing stresses and noting that Gkl is constant lead to the differential equation (U.4+ -a--)J =V24, == 0 (11-12) which must be satisfied at all points in the cross section. The exterior normal n for the cylindrical surface is perpendicular to the Xl direction. Then a,, = 0, and the stress boundary conditions, (10-49), reduce to Pn = {n26021 + xn3a 3 1 = 0 (11-13) Using (11-10), the boundary condition for , is Xn2(¢, 2 - X3) + (Z3(t,. 3 + 2) = 0 (11-14) a0n -- ,,2X 3 -n3X2 (on S) t Problem 11-3 treats the orthotropic case. SEC. 11-2. THE PURE-TORSION PROBLEM 275 The pure-torsion problem involves solving V2 , = 0 subject to (11-14). Once A0 is known, we determine the distribution of transverse shearing stresses from