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 stress displacement relations Hint Note 10 101 b Transform FIR to IH by restricting the geometry to be linear ak a and eij ui j uj i 2 and requiring the stresses to satisfy the stress equilibrium equations and stress boundary conditions on DQ Hint Integrate uijeij by parts using 10 81 10 29 Interpret 10 90 as dQ 1c where PQ is a force applied at Q in the direction of the displacement measure dQ St Venant Theory of Torsion Feexure of Prismatic Members 11 1 INTRODUCTION AND NOTATION A body whose cross sectional dimensions are small in comparison with its axial dimension is called a member If the centroidal axis is straight and the shape 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 X axis is 1 taken to coincide with the centroidal axis and X2 X are taken as the principal inertia directions We employ the following notation for the cross sectional properties 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 x2 x dJXxdAz A 123 fX2 X3 dA 0 3 0 11 2 0 One can work with an arbitrary orientation of the reference axes but this will complicate 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 the centroidal axis is treated in Chapter 15 271 I 272 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 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 M 1 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 X1 X2 Ailli ill 1 i Fig 11 2 Prismatic member in pure torsion 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 F3 fOr 3 dA F ffU1 2 dA F J 11 dA M f x 2 61 M2 M3 fJJX 3a fx 2 3c 12 dA 11 3 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 dA 112 1X 3 all dA 113 0 1 X2 11 The internal force and moment vectors acting on the negative face are denoted by F M Since F 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 M M The corresponding linear strains are 82 83 Y23 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 C1 12 713 U1 2 U1 3 0 ll 1 12 1 11 6 tl 2 u3 1 0 U1 3 X 3 01 1 X2 1 1 t Problem 11 treats the general case where the cross section rotates about an arbitrary point I 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 U1 11 7 SEC 11 2 THE PURE TORSION PROBLEM The pure torsion problem involves solving V 0 subject to 11 14 Once is known we determine the distribution of transverse shearing stresses from 11 10 Note that t depends only on the shape of the cross section The shearing stress distribution must lead to no shearing stress resultants A0 lt1 X2 X3 We consider the left end to be fixed with respect to rotation and express cot Ul as 11 8 F Y12 E2 J k t T13 and al0 11 a01 2 Gyl2 U1 3 G 13 2 2 2 1 t 2 Gki 4t 3 11 9 3 X2 3 33 Gkl t a 2 3 51 2 x 2 x 3 a13 x2 x2 a3 1 3 j dA O0 X 2 b 3 f 11 15 dA f S exj which is just a special case of 10 81 Applying 11 15 to V2 q dA leads to Green s theorem V2 11 10 dA n2jq 5 O 3 1n3 q3 dS 11 16 x 3 11 11 0 a 0 To proceed further we need certain integration formulas We start with 0 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 dA 7X Y23 0 3 Jfa dA This requires 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 ffS12 dA 0 F2 U1 kl qt E 275 2 If is a harmonic function i e V2lp 0 Green s theorem requires n v dS I 0 c Now is a harmonic function For the formulation to be consistent 11 14 must satisfy …