----331 12 Engineering Theory of Prismatic Members 12-1. INTRODUCTION St. Venant's theory of flexure-torsion is restricted to the case where--1. There are no surface forces applied to the cylindrical surface. 2. The end cross sections can warp freely. The warping function ¢ consists of a term due to flexure (Os) and a term due to pure torsion (t). Since k is independent of x-, the linear expansion F1M12 1 t -ll------Y2 (12-1)A '2 13 is the exact solutiont for alt. The total shearing stress is given by 6 1S = C, + 1-f (12-2) where at is the pure-torsion distribution (due to 4,) and af represents the flexural distribution (due to Of). We generally determine caf by applying the engineering theory of shear stress distribution, which assumes that the cross section is rigid with respect to in-plane deformation. Using (12-1) leads to the following expression for the flexural shear flow (see (11-106)): Q3 Q2qB = qA -2 3 (12-3) The warping function will depend on xl if forces are applied to the cylindrical surface or the ends are restrained with respect to warping. A term due to variable warping must be added to the linear expansion for al. This leads to an additional term in the expression for the flexural shear flow. Since (12-1) t A linear variation of normal stress is exact for a homogeneous beam. Composite beams (e.g.. a sandwich beam) are treated by assuming a linear variation in extensional strain and obtaining the distributions of ao from the stress-strain relation. See Probs. I I -14 and 12-1. 330 SEC. 12-2. FORCE-EQUILIBRIUM EQUATIONS satisfies the definition equations for FI, M2, M3 identically, the normal stress correction is self-equilibrating; i.e., it is statically equivalent to zero. Also, the shear flow correction is statically equivalent to only a torsional moment since (12-3) satisfies the definition equations for F2, F3 identically. In the engineering theory of members, we neglect the effect of variable warping on the normal and shearing stress; i.e., we use the stress distribution predicted by the St. Venant theory, which is based on constant warping and no warping restraint at the ends. In what follows, we develop the governing equations for the engineering theory and illustrate the two general solution procedures. This formulation is restricted to the linear geometric case. In the next chapter, we present a more refined theory which accounts for warping restraint, and in-vestigate the error involved in the engineering theory. 12-2. FORCE-EQUILIBRIUM EQUATIONS In the engineering theory, we take the stress resultants and couples referred to the centroid as force quantities, and determine the stresses using (12-1), (12-3), and the pure-torsional distribution due to MT. To establish the force-equilibrium equations, we consider the differential element shown in Fig. 12--1. The statically equivalent external force and moment vectors per unit -dxI/2 clxl1/2 --F. tr.* . 2' l)'_ +d xt ) dil _ dx ---v_-( 2-) Fig. 12-1. Differential element for equilibrium analysis. length along X, are denoted by b, nii. Summing forces and moments about 0 leads to the following vector equilibrium equations (note that F_ = -F+, M_ = -M+): dF+ b dx1 (a)dM+ -dxT+ + (71 x F+) = dxl333 ,,,l...cl' THFORY OF PRISMATIC MEMBERS CHAP. 12 -.332 rlNUILL--i, We obtain the scalar equilibrium equations by introducing the component expansions and equating the coefficients of the unit vectors to zero. The re-sulting system uncouples into four sets of equations that are associated with 3 plane, and twist.stretching, flexure in the XI-X 2 plane, flexure in the X-X Stretching dF + b =0 x Y Flexure in X 1-X2 Plane dF2 + b2 = 0 dxl (12-4)dM 3 + m3 +f F2 00 dx Flexure in X1-X 3 Plane dF3 + b3 = 0 dx 3 dM 2 ± rn2 -F3 == 0 dx1 Twist Mt + ml 0 (dxt This uncoupling is characteristic only of prismatic members; the equilibrium we shallequations for an arbitrary curved member are generally coupled, as show in Chapter 15. The flexure equilibrium equations can be reduced by solving for the shear force in terms of the bending moment, and then substituting in the remaining equations. We list the results below for future reference. Flexure in X1-X2 Plane dM3 nx-113F2 d= dir 1 d2 M3 din M + b2 =0 dxl dxl (12-5) Flexure in X1-X3 Plane dMz F3-d + M2 d2 M 2 dn2 b3 = oJd2M2t + dx2 dx, SEC. 12-3. FORCE-DISPLACEMENT RELATIONS Note that the shearing force is known once the bending moment variation is determined. The statically equivalent external force and moment components acting on the end cross sections are called endforces. We generally use a bar superscript Also, we use A, B to denote the negativeto indicate an end action in this text. and positive end points (see Fig. 12-2) and take the positive sense of an end X2 1A 2 FA2 ~ .X1 ' B3 IX3 rU -i, -Fig. 12-2. Notation and positive direction for end forces. force to coincide with the corresponding coordinate axis. The end forces are related to the stress resultants and couples by FBj -[Fj]x =L MBj = [Mj].-,=L (j = 1,2,3) (12-6)FAj =-[rEjr, = 0MAJ ICj3r= MIAj = -[Mj],x=o Aminus sign is required at A, since it is a negative face. 12-3. FORCE-DISPLACEMENT RELATIONS; PRINCIPLE OF VIRTUAL FORCES We started by selecting the stress resultants and stress couples as force parameters. Applying the equilibrium conditions to a differential element re-Tosults in a set of six differential equations relating the six force parameters. complete the formulation, we must select a set of displacement parameters and relate the force and displacement parameters. These equations are generally called force-displacement relations. Since we have six equilibrium equations, we must introduce six displacement parameters in order for the formulation to be consistent. Now, the force parameters are actually the statically equivalent forces and moments acting at the centroid. This suggests that we take as displacement334 335 ENGINEERING THEORY OF PRISMATIC MEMBERS CHAP. 12 parameters the equivalent rigid body translations and rotations of the cross section at the centroid. We define Ctand -cas = u/lj = equivalent rigid body translation vector at the centroid (12-7 = C)ji = equivalent rigid body rotation vector By equivalent displacements, we mean fJ (force intensity) (displacement) dA = F · + M -c (12-8) A Note that (12-7) corresponds to a linear distribution of displacements over the cross section, whereas the actual distribution is nonlinear, owing to shear de-formation. In this approach, we are allowing