Torsion

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134 CHAPTER 5. TORSION were assumed to remain valid for a bar of arbitrary cross-section, the displacement field, eqs. (5.4) and (5.3), and the corresponding strain field, eqs. (5.5) to (5.7), would also describe the kinematics of bars with arbitrary sections. The only non-vanishing stress component would be the circumferential shearing stress given by eq. (5.12). Unfortunately, this assumption can lead to grossly erroneous results because it implies a solution that violates the equilibrium equations of the problem at the edge of the section. Consider, for instance, the torsion of a rectangular bar as depicted in fig. 5.10. The circumferential shearing stress, τα, given by eq. (5.12) is shown at an edge of the section, and it is resolved into its Cartesian components, τ12 and τ13. The shear stress component, τ13, normal to the edge of the section and its complementary component shown acting on the outer surface of the bar must clearly vanish since the outer surfaces of the bar are stress free. Hence, the only allowable shearing stress component along the edge is τ12, and the stress distribution in eq. (5.12) therefore violates equilibrium conditions along the edges of the section. i 3 i 1 i 2 Figure 5.10: Shearing stresses along the edge of a rectangular section. 5.3.1 Saint-Venant’s solution The solution to the problem of torsion of a bar with a cross-section of arbitrary shape was first given by Saint-Venant. Kinematic description Consider a solid bar with a cross-section of arbitrary shape denoted A. A closer look at the problem and experimental tests reveal that for a bar with an arbitrary section, each cross-section rotates like a rigid body, but the cross-section is allowed to warp out of its own plane. This type of deformation is described by the following assumed displacement field 13 r 12 13 u1(x1, x2, x3) = Ψ(x2, x3) κ1(x1); (5.50) u2(x1, x2, x3) = −x3φ1(x1); u3(x1, x2, x3) = x2φ1(x1). (5.51) The in-plane displacement field, eq. (5.51), describes a rigid body rotation of the cross-section, as was the case for the circular cylinder (see eq. (5.3)). However, the out-of-plane displacement field does not vanish. Instead, it is assumed to be proportional to the twist rate, κ, and has an arbitrary variation over the crosssection described by the unknown warping function Ψ(x2, x3). This warping function will be determined by
5.3. TORSION OF BARS WITH ARBITRARY CROSS-SECTIONS 135 enforcing equilibrium conditions for the resulting shearing stress field. It will be further assumed that the twist rate is constant along the axis of the bar, i.e. κ1(x1) = κ1. This restriction results in what is known as the uniform torsion problem. The strain field associated with the assumed displacement field can be calculated by applying the straindisplacement equations (eqs. (1.51) and (1.58)) to the assumed displacement field, eqs. (5.50) and (5.51) to yield ε1 = 0; (5.52) ∂Ψ γ12 = ∂x2 ε2 = 0; ε3 = 0; γ23 = 0; (5.53) − x3 ∂Ψ κ1; γ13 = + x2 κ1. (5.54) ∂x3 The vanishing of the axial strain, eq. (5.52), is a direct consequence of the uniform torsion assumption, whereas the vanishing of the in-plane strains, eq. (5.53) stems from the rigid body rotation assumption for the in-plane motion of the section. The only non-vanishing strain components, γ12 and γ13, depend on the partial derivatives of the unknown warping function. The stress field For bars made of a linear elastic, isotropic material, Hooke’s law, eq. (1.91), is assumed to apply. The stress field is then found from the strain field as σ1 = 0; (5.55) τ12 = Gκ1 σ2 = 0; σ3 = 0; τ23 = 0; (5.56) ∂Ψ ∂x2 − x3 ∂Ψ ; τ13 = Gκ1 + x2 . (5.57) ∂x3 This stress field must satisfy general equilibrium equations, eqs. (1.7), at all points of the section. Neglecting body forces, and in view of eq. (5.55), two of the three equilibrium equations are satisfied and the remaining one reduces to ∂τ12 + ∂x2 ∂τ13 = 0. (5.58) ∂x3 Hence, using eqs. (5.57) in (5.58), the warping function must satisfy the following partial differential equation ∂ 2 Ψ ∂x 2 2 + ∂2 Ψ ∂x 2 3 = 0. (5.59) at all points of the cross-section. The relevant boundary conditions can be developed by requiring that equilibrium conditions must also be satisfied along the outer edge of the section that defines the contour C. Fig. 5.11 shows a portion of the outer contour, C, and a curvilinear variable, s, that measures length along this contour. As illustrated in fig. 5.10, the normal component of shear stress must vanish at all points on C, i.e. τn = 0, (5.60) whereas the component of shearing stress, τs, tangent to the contour does not necessarily vanish. In terms of Cartesian components, the normal component of shearing stress, see fig. 5.11, is dx3 τn = τ12 sin β + τ13 cos β = τ12 + τ13 − ds dx2 = 0. (5.61) ds Introducing eq. (5.57) then yields the following boundary condition for the warping function ∂Ψ ∂x2 − x3 dx3 ds − ∂Ψ dx2 + x2 = 0. (5.62) ∂x3 ds
Page 1 and 2: Chapter 5 Torsion In the previous c
Page 3 and 4: 5.1. TORSION OF CIRCULAR CYLINDERS
Page 9 and 10: 5.2. STRENGTH UNDER COMBINING LOADI
Page 11: 5.3. TORSION OF BARS WITH ARBITRARY
Page 15 and 16: 5.3. TORSION OF BARS WITH ARBITRARY
Page 25 and 26: 5.4. TORSION OF A THIN RECTANGULAR
Page 27 and 28: 5.5. TORSION OF THIN-WALLED OPEN SE
Page 29 and 30: 5.5. TORSION OF THIN-WALLED OPEN SE
Page 31 and 32: 5.6. THE MEMBRANE ANALOGY 153 the s
Page 33 and 34: 5.6. THE MEMBRANE ANALOGY 155 b t i

Torsion

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