MIT - 16.20 Fall, 2002 Unit 12 Torsion of (Thin) Closed Sections Readings : Megson 8.5 Rivello 8.7 (only single cell material), 8.8 (Review) T & G 115, 116 Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 Before we look specifically at thin-walled sections, let us consider the general case (i.e., thick-walled). Hollow, thick-walled sections: Figure 12.1 Representation of a general thick-walled cross-section φφφφ = C 2 on one boundary φφφφ = C 1 on one boundary This has more than one boundary (multiply-connected) ? d φ = 0 on each boundary ? However, φ = C 1 on one boundary and C 2 on the other (they cannot be the same constants for a general solution [there’ s no reason they should be]) => Must somehow be able to relate C 1 to C 2 Paul A. Lagace ? 2001 Unit 12 - 2 MIT - 16.20 Fall, 2002 It can be shown that around any closed boundary: ∫ = τ ds AGk 2 (12-1) Figure 12.2 Representation of general closed area ττττ where: τ = resultant shear stress at boundary A = Area inside boundary k = twist rate = d α dz Paul A. Lagace ? 2001 Unit 12 - 3 MIT - 16.20 Fall, 2002 Notes : 1. The resultant shear stresses at the boundary must be in the direction of the tangents to the boundary 2. The surface traction at the boundary is zero (stress free), but the resultant shear stress is not Figure 12.3 Representation of a 3-D element cut with one face at the surface of the body To prove Equation (12 - 1), begin by considering a small 3-D element from the previous figure Paul A. Lagace ? 2001 Unit 12 - 4 σσσ MIT - 16.20 Fall, 2002 Figure 12.4 Exploded view of cut-out 3-D elements this face is stress free, thus σ normal = 0 Look at a 2-D cross-section in the x-y plane: Figure 12.5 Stress field at boundary of cross-section Paul A. Lagace ? 2001 σσσσ tan = σσσσ res since σσσσ normal = 0 Unit 12 - 5 MIT - 16.20 Fall, 2002 τ resul tan t = σ zy cos γ + σ zx sin γ geometrically : cos γ = dy ds dx sin γ = ds Thus: dx ? τ ds = ∫ ? ? dyds zy σ ds + σ zx ds ? ∫ ? ds ? = ∫ σ zy dy + σ zx dx We know that: ? ? w ? σ zy = G ?? k x + ? y ? ? σ zx = G ? ? k y + ? w ? ? ? x ? Paul A. Lagace ? 2001 Unit 12 - 6 ? MIT - 16.20 Fall, 2002 ? = ∫ τ ds ∫ G ?? k x + ? + ∫ dy ? w ? G ? ? k y + ? w ? dx ? ? y ? ? ? x ? + ? ? + ∫ dx G dy k ∫ ? ? w ? w ? = G { xdy ? ydx } ? ? x ? y ? = dw We further know that: ∫ dw = w ? = 0 around closed contour ? So we ’ re left with: ∫ τ ds = Gk ∫ { xdy ? ydx } Paul A. Lagace ? 2001 Unit 12 - 7 MIT - 16.20 Fall, 2002 Use Stoke ’ s Theorem for the right-hand side integral: ∫ ? ? N ? M ? { Mdx + Ndy } = ∫∫ ? ? ? x ? ? y ? ? dxdy In this case we have ? M M = ? y ? = ? 1 ? y ? N N = x ? = 1 ? x We thus get: Gk ∫ { xdy ? ydx } = Gk ∫ ∫ [ 1 ?? 1 ) ] dxdy ( = Gk ∫∫ 2 dxdy Paul A. Lagace ? 2001 Unit 12 - 8 MIT - 16.20 Fall, 2002 We furthermore know that the double integral of dxdy is the planar area: ∫ ∫ d dxdy = Area = A Putting all this together brings us back to Equation (12 - 1): ∫ = τ ds AGk 2 Q.E.D. Hence, in the general case we use equation (12 - 1) to relate C 1 and C 2 . This is rather complicated and we will not do the general case here. For further information (See Timoshenko , Sec. 115) We can however consider and do the … Paul A. Lagace ? 2001 Unit 12 - 9 τττ MIT - 16.20 Fall, 2002 Special Case of a Circular Tube Consider the case of a circular tube with inner diameter R i and outer diameter R o Figure 12.6 Representation of cross-section of circular tube For a solid section, the stress distribution is thus: Figure 12.7 Representation of stress “flow ” in circular tube τ res is directed along circles Paul A. Lagace ? 2001 Unit 12 - 1 0 MIT - 16.20 Fall, 2002 The resultant shear stress, τ res , is always tangent to the boundaries of the cross-section So, we can “cut out” a circular piece (around same origin) without violating the boundary conditions (of τ res acting tangent to the boundaries) Using the solution for a solid section, we subtract the torsional stiffness of the “ removed piece ” (radius of R i ) from that for the solid section (radius of R o ) π R 4 π R 4 J = o ? i 2 2 Exact solution for thick-walled circular tube let us now consider: Paul A. Lagace ? 2001 Unit 12 - 1 1 MIT - 16.20 Fall, 2002 Thin-Walled Closed Sections Figure 12.7 Representation of cross-section of thin-walled closed section outer inner Here, the inner and outer boundaries are nearly parallel ? resultant shear stresses throughout wall are tangent to the median line. Basic assumption for thin, closed section: τ resultant is approximately constant through the thickness t. Paul A. Lagace ? 2001 Unit 12 - 1 2 MIT - 16.20 Fall, 2002 For such cases: A outer ≈ A inner ≈ A Hence: ∫ τ ds ≈ ∫ τ ds ≈ 2 GkA outer inner Note : basic difference from singly-connected boundaries (open sections). Figure 12.9 Representation of stress distribution through thickness in open cross-section under torsion very important difference ττττ res varies linearly through- the-thickness Now, we need to find the boundary conditions: Paul A. Lagace ? 2001 Unit 12 - 1 3 MIT - 16.20 Fall, 2002 Figure 12.10 Representation of forces on thin closed cross-section under torsion Force: dF = ττττ t ds contribution to torque: dT = h τ t d s (h = moment arm) Note : h, τ , t vary with s (around section) Total torque = ∫ dT = ∫ τ th d s Paul A. Lagace ? 2001 Unit 12 - 1 4 MIT - 16.20 Fall, 2002 But τ t is constant around the section. This can be seen by cutting out a piece of the wall AB. Figure 12.11 Representation of infinitesimal piece of wall of thin closed section under torsion z x-y plane Use ∑ F z = 0 to give: ? τ td z + τ B t B d z = 0 AA ? τ t AA = τ B t B in general: τ t = constant Paul A. Lagace ? 2001 Unit 12 - 1 5 MIT - 16.20 Fall, 2002 Define: “shear flow” = q = τ t = constant (we will use the concept of “shear flow ” Analogy: single 1-D pipe flow when we deal with shell beams) uh = constant velocity Returning to ∫ dT = ∫ τ th d s since τ t = constant gives: ∫ dT = τ t ∫ h d s But, hds = 2dA via geometric argument: Paul A. Lagace ? 2001 Unit 12 - 1 6 MIT - 16.20 Fall, 2002 hds dA = 2 ? height 2 x base ? ? = Area T of Triangle ? Finally: ? dT t ∫ = τ ∫ 2 d A ? T = 2 τ t A T ? τ resultant = 2 At Bredt ’ s (12 - 2) formula Paul A. Lagace ? 2001 Unit 12 - 1 7 MIT - 16.20 Fall, 2002 Now to find the angle of twist, place (12 - 2) into (12 - 1): T ds = G k 2 A 2 At ∫ T ds ? k = ∫ 2 4 AG t This can be rewritten in the standard form: d α T k = = dz GJ 4 A 2 ? J = t ds ∫ ( Note : use midline for calculation) valid for any shape ….. Paul A. Lagace ? 2001 Unit 12 - 1 8 MIT - 16.20 Fall, 2002 Figure 12.12 Representation of general thin closed cross-section How good is this approximation? It will depend on the ratio of the thickness to the overall dimensions of the cross-section (a radius to the center of torsion) Can explore this by considering the case of a circular case since we have an exact solution: π R 4 ? π R 4 J = o i 2 versus approximation: 4 A 2 J ≈ (will explore in home assignment) t ds ∫ Paul A. Lagace ? 2001 Unit 12 - 1 9 MIT - 16.20 Fall, 2002 Final note on St. Venant Torsion: When we look at the end constraint (e.g., rod attached at boundary): Figure 12.13 Overall view of rod under torsion Here, St. Venant theory is good in this local region, violation of assumption of St. Venant theory Built-in end At the base, w = 0. This is a violation of the “ free to warp ” assumption. Thus, σ zz will be present. ? resort to complex variables (See Timoshenko & Rivello ) Paul A. Lagace ? 2001 Unit 12 - 2 0