Torsion of Circular Sections

Transcription

1 7 orsion of Circular Sections 7 1

2 Lecture 7: ORSION OF CIRCULR SECIONS BLE OF CONENS Page 7.1. Introduction Problem Description erminology, Notation, Coordinate Systems Deformation of orqued Shaft Stress and Strain Distributions he Stress Formula he wist Rate nd ngle Formulas orsion Failure Modes Principal Stresses and Maimum Shear Material Failure: Ductile Versus Brittle

3 7.2 PROBLEM DESCRIPION 7.1. Introduction he present lecture begins a set of three classes on torsion. his topic is presented now to provide timely theoretical support for the first eperimental Lab, which will be demo ed on Friday Sept 21. he first lecture covers circular cross sections. hese can be be treated by elementary (Mechanics of Materials) methods that nonetheless provide the eact elasticity solution. he net two lectures deal with hin Wall (W) cross sections of important in erospace. For these approimate methods are necessary. y (+) (a) (+) y (b) z z (+) (+) Figure 7.1. orsion of a circular shaft: (a) Single-headed arrow (a.k.a. harpoon ) representation and (b) double-headed arrow vector representation for torque. Both symbols are used in this course Problem Description his Section introduces the problem of a torqued member (shaft) of circular cross section (solid or annular), and presents the necessary terminology erminology, Notation, Coordinate Systems We consider a bar-like, straight, prismatic structural member of constant cross section subjected to applied torque about its long ais, as pictured in Figure 7.1(a). his kind of member, designed primarily to transmit torque, is called a shaft. is is directed along the longitudinal shaft dimension. he applied torque is shown as a moment about that ais. It is positive when it acts as indicated, complying with the right-hand screw rule along the direction. es y and z lie on a cross section conventionally taken as origin. Figure 7.1(b) shows an alternative picture of the torque as a double-headed arrow vector. We will use this representation whenever it leads to a more convenient visualization. n important use of shafts is to transmit power between parallel planes, as in automobile power trains, electrical machinery, aircraft engines or helicopter rotors. See Figure 7.2. In this lecture we restrict consideration to shafts of circular cross section that may be either solid or hollow (annular, tubular), as pictured in Figure 7.3(a,b). If the shaft is solid, its radius is R. If hollow, the eterior and interior radius are denoted by R e and R i, respectively. In addition to the (, y, z) Rectangular Cartesian Coordinate (RCC) system of Figure 7.1 we shall also use often the cylindrical coordinate system (,,θ) depicted in Figure 7.3(c) for a typical 7 3

4 Lecture 7: ORSION OF CIRCULR SECIONS Figure 7.2. ransfering power between planes via torqued shafts. (a) (b) y (c) P(,,θ) θ 2R 2R 2R z i e Cross sections considered in this lecture Point P referred to cylindrical coordinate system (looking at cross section from + down) Figure 7.3. Circular cross sections considered in this Lecture, and coordinate systems. (a) (b) Figure 7.4. Visualizing deformations produced by torque. 7 4

5 7.2 PROBLEM DESCRIPION cross section. he position coordinates of an arbitrary shaft point P are, y, z in RCC and,,θ in cylindrical coordinates. Here is the radial coordinate (not a happy notation choice because of potential confusion with density, but it is the symbol used in most Mechanics of Materials tetbooks) and θ the angular coordinate, taken positive CCW from y as shown. Note that y = cos θ, z = sin θ. (7.1) Deformation of orqued Shaft o visualize what happens when a circular shaft is torqued, consider first the undeformed state under zero torque. rectangular-like mesh is marked on the surface as shown in Figure 7.4(a). he rectangles have sides parallel to the longitudinal ais and the circumferential direction θ. pply torque to the end sections. Rectangles marked in Figure 7.4(b) shear into parallelograms, as illustrated in Figure 7.4(b). his diagram displays the main effect: shear stresses and associated shear strains develop over the cylinders = constant. Net subsection goes into more details Stress and Strain Distributions z y (a) Material element aligned with cylindrical coordinate aes θ radial direction (b) ll other stress components are zero τ θ longitudinal direction circumferential direction θ τ θ we will often call τ = τ = τ θ θ Figure 7.5. Etracting a (,,θ)-aligned material element. he only nonzero stresses are τ θ = τ θ. material element aligned with the cylindrical coordinate aes is cut out as shown in Figure 7.5(a). he element is magnified in Figure 7.5(b), which plainly shows that the only nonzero stress component on its faces is the shear stress τ θ = τ θ. It follows that the stress and strain matrices referred to the cylindrical coordinate system have the form [ τθ ] τ θ Hooke s law 7 5 [ γθ ] γ θ (7.2)

6 Lecture 7: ORSION OF CIRCULR SECIONS d = d dθ θ τ θ τ θ τ=τ θ θ 9 τ=τ θ d dθ θ Figure 7.6. Integration over the cross section of the elementary moment produced by the shear stress component τ = τ θ will balance the internal torque. he aes have been reordered as (,θ,)for convenience in identifying with plane stress and strain later. In the sequel we often call τ = τ θ = τ θ, and γ = γ θ = γ θ to reduce subscript clutter. Stresses and strains in (7.2) are connected at each material point by Hooke s shear law τ = G γ, γ = τ G, (7.3) in which G is the shear modulus. Because of circular symmetry τ and γ do not depend on θ. hey may depend on if either the cross section radius or the internal torque change along the shaft length, but that variation will not be generally made eplicit. We will assume (subject to a posteriori verification) that τ and γ depend linearly on, and we scale that dependence as follows: τ = R τ ma, γ = R γ ma (7.4) Here τ ma and γ ma are maimum values of the shear stress and strain, respectively, over the cross section. hese occur at = R, the radius of the shaft. (For an annular shaft, R = R e, the eterior or outer radius) 7.3. he Stress Formula Call the internal torque acting on a cross section = constant. he shear stress τ = τ θ acting on an elementary cross section area dproduces an elementary force df = τ dand an elementary moment dm = df = (τ d) about the ais. See Figure 7.6 for the geometric details. Integration of the elementary moment over the cross section must balance the internal torque: = (shear stress elemental area) lever-arm = (τ d) = R τ ma d= τ ma 2 d (7.5) R = τ ma 2 ( d dθ) = τ ma 3 d dθ = τ ma J R R R, 7 6

7 7.4 HE WIS RE ND NGLE FORMULS φ at γ =(γ ) =(γ ) ma θ ma θ ma L B B' φ B Figure 7.7. wist deformation of a torqued circular shaft. Deformations and rotations greatly eaggerated for visualization convenience. in which J = 2 d= 3 d dθ, (7.6) is the polar moment of inertia of the cross section about its center. Solving for τ ma gives the stress formula: τ ma = R J, τ = R τ ma = J (7.7) For a solid circular section of radius R, J = 1 2 π R4. For an annular cross section of eterior radius R e and interior radius R i, J = 1 2 π(r4 e R4 i ) he wist Rate nd ngle Formulas torqued shaft produces a relative rotation of one end respect to the other; see visualization in Figure 7.4(b). his relative rotation is called the twist angle φ. he twist angle per unit of -length is called the twist rate dφ/d. Often the design called for limitation on this rate or angle. It is thus of interest to have epressions for both of them in terms of the applied torque and the properties of the shaft. Geometric quantities used for working out those epressions are defined in Figure 7.7. Note that infinitesimal strains and rotations are assumed; deformations depicted in Figure 7.7 are greatly eaggerated for visibility convenience. For small angles BB R φ B L γ ma.tan arbitrary distance from the fied end, R φ γ ma whence φ = γ ma /R. Consequently the twist rate is But according to Hooke s law (7.3) dφ d = γ ma R = γ so γ = dφ d (7.8) γ = τ G = GJ = dφ d (7.9) 7 7

8 Lecture 7: ORSION OF CIRCULR SECIONS Fied end 6 Material: steel, G = 12 1 psi Solid circular section, R = 1 in L = 5 in B = 5 lbs-in Figure 7.8. Eample calculation of maimum shear stress and twist angle. which yields dφ d = GJ (7.1) his is the twist rate formula. o find the twist angle φ B, where subscripts identify the angle measurement endpoints, integrate along the length of the shaft: φ B = φ B φ = φ B = φ B = If, G and J are constant along the shaft: L dφ = L dφ L d d = GJ d. (7.11) φ B = GJ L d = L GJ (7.12) his is the twist angle formula for a prismatic shaft. Eample 7.1. Data for this eample is provided in Figure 7.8. Find: maimum shear stress τ ma, maimum shear strain γ ma and relative twist angle φ B. τ ma = R J = R 5 = π R = 3183 psi (7.13) γ ma = τ ma G 32 psi = = 265 µ (microradians) (7.14) psi φ B = L GJ = ( 1 2 π 14 ) =.133 rad =.76 (7.15) 7 8

9 7.5 ORSION FILURE MODES (b) (c) (a) Figure 7.9. hree failure modes of a torqued specimen: (a,b): material failure in solid shaft, (c): wall buckling in thin walled tubular shaft orsion Failure Modes Shafts should be designed against possible failure modes. hree practically important ones are sketched in Figure 7.9. Specimen (a) and (b) display two classical material failure modes in a solid shaft, whereas (c) illustrates a wall buckling failure that may occur for hollow shafts if the wall is too thin. Only the first two modes are discussed here, since the last one requires knowledge of shell buckling, which falls outside of the scope of this course Principal Stresses and Maimum Shear s discussed in 7.2.2, the stress matri at a point P(,,θ)referred to a cylindrical coordinate system with aes reordered as (,θ,)takes the simple form [ ] τ τ, τ = τ θ = τ θ = τ ma R, τ ma = R J. (7.16) We can observe that all points are in plate stress because all stresses with a subscript vanish, although the shaft is not a thin plate. It is also in plane strain because the transverse strains (that is, strain components with a subscript): ɛ, γ and γ θ, are zero. Since the shear stress is proportional to, we consider a point on the outer surface = R, where τ = τ ma. he Mohr s circle for stresses in the (,θ)plane passing through P(, R,θ)is shown in Figure 7.1(a). By inspection we see that (a) he principal stresses are σ 1 = τ ma (tension) and σ 2 = τ ma (compression). hese occur at ±45 from the longitudinal direction. (b) he maimum inplane shear is τ ma, which occurs on the shaft cross sections. Consideration of 3D effects does not change these findings because the zero principal stress is the intermediate one. hus the circle plotted in Figure 7.1(a) is the outer one, and the maimum overall shear stress τma overall is the same as the maimum inplane shear stress, that is, τma inplane. 7 9

10 Lecture 7: ORSION OF CIRCULR SECIONS (a) Mohr's circle for (,θ) plane at =R (outer shaft surface) radius of circle is τ ma τ τ ma (b) Material failure surfaces for two etreme cases: ductile vs. brittle 45 ο σ 2 = τ ma σ = τ 1 ma σ τ ma Ductile material failure surface (plane of cross section) Brittle material failure surface (=helicoid) Figure 7.1. By inpection of the Mohr circle, maimum normal and shear stresses at a point on the outer shaft surface determine the kind of material failure Material Failure: Ductile Versus Brittle We are now in a position to eplain the two material failure modes shown in Figure 7.9(a,b). If the material is ductile, it fails by maimum shear τ ma reaching yield, followed by post-yield plastic flow. he failure surface is the cross section, as sketched in Figure 7.9(a). he picture in 7.11(a) shows a mild steel specimen taken to failure by torque. If the material is brittle, it fails by maimum tensile stress, which is σ 1 = τ ma. Since this occurs at 45 from the longitudinal direction, the fracture is more complicated and will look (at least near the shaft outer surface) like a helicoid, as sketched in Figure 7.9(b). he photographs in Figure 7.11 show two fractured specimen fabricated of brittle materials: cast iron and chalk, respectively. (a) Ductile failure by shear (mild steel) (b) Brittle failure by tensile normal stress (cast iron) (c) Brittle failure by tensile normal stress (chalk) Figure ctual material failure observed in torqued specimen. If the material is not isotropic, the failure process can be more involved than the two etreme cases dicussed above. For eample, wood and laminate composites may fail by delamination across layers or fibers. 7 1