MT-1620 al.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 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
MT-1620 al.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 中=c2 on one boundary φ=c1 on one boundary This has more than one boundary(multiply-connected do=0 on each boundary However, =C1 on one boundary and C2 on the other(they cannot be the same constants for a general solution [there's no reason they should bel) = Must somehow be able to relate c to c. Paul A Lagace @2001 Unit 12-2
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 φ = C2 on one boundary φ= C1 on one boundary This has more than one boundary (multiply-connected) • dφ = 0 on each boundary • However, φ = C1 on one boundary and C2 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 C1 to C2 Paul A. Lagace © 2001 Unit 12 - 2
MT-1620 al.2002 It can be shown that around any closed boundary τds=2AGk (12-1) Figure 12.2 Representation of general closed area 4A4E4 Where t= resultant shear stress at boundary A= Area inside boundary k= twist rate da dz Paul A Lagace @2001 Unit 12-3
MIT - 16.20 Fall, 2002 It can be shown that around any closed boundary: ∫ τds = 2AGk (12-1) Figure 12.2 Representation of general closed area τ where: τ = resultant shear stress at boundary A = Area inside boundary k = twist rate = dα dz Paul A. Lagace © 2001 Unit 12 - 3
MT-1620 al.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
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
MT-1620 al.2002 Figure 12.4 Exploded view of cut-out 3-D elements this face is stress free, thus normal =0 Look at a 2-D cross-section in the x-y plane Figure 12.5 Stress field at boundary of cross-section 4n=o,。 sInce o, X Paul A Lagace @2001 Unit 12-5
σ MIT - 16.20 Fall, 2002 Figure 12.4 Exploded view of cut-out 3-D elements this face is stress free, thus σnormal = 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 σtan = σres since σnormal = 0 Unit 12 - 5
MT-1620 al.2002 resul tant Ory COSY +Oxx siny geometrically.cosγ d s siny Thus: τds ds to ds dy ozx dx We know that W G kx dy W G-ky Paul A Lagace @2001 Unit 12-6
MIT - 16.20 Fall, 2002 τresultant = σzy cos γ + σzx sin γ geometrically: cos γ = dy ds dx sin γ = ds Thus: dx τ ds = ∫ dyds σzy ds + σzx ds ∫ ds = ∫ σzy dy + σzx dx We know that: ∂w σzy = G kx + ∂y σzx = G −ky + ∂w ∂x Paul A. Lagace © 2001 Unit 12 - 6
MT-1620 al.2002 W G k Gi-ky dx Gfow dx dx We further know that dw =w=0 around closed contour So we re left with τds=Gkd{xd Paul A Lagace @2001 Unit 12-7
MIT - 16.20 Fall, 2002 ⇒ = ∫ τ ds ∫ G kx + + ∫ dy ∂w G −ky + ∂wdx ∂y ∂x + + ∫ dx dy Gk ∫ ∂w ∂w = G {xdy − ydx} ∂x ∂y = dw We further know that: ∫ dw = w = 0 around closed contour So we’re left with: ∫ τds = Gk ∫ {xdy − ydx} Paul A. Lagace © 2001 Unit 12 - 7
MT-1620 al.2002 Use Stoke's Theorem for the right-hand side integral (Mdx+ Dyl In this case we have M N=X We thus get f0:1-y eddy Paul A Lagace @2001 Unit 12-8
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 ∫∫ 2dxdy Paul A. Lagace © 2001 Unit 12 - 8
MT-1620 al.2002 We furthermore know that the double integral of dxdy is the planar area ∫g女dy=Area=A Putting all this together brings us back to Equation (12-1) τds=2AGk Q.E.D Hence, in the general case we use equation(12-1)to relate C, and C2 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
MIT - 16.20 Fall, 2002 We furthermore know that the double integral of dxdy is the planar area: ∫∫ dxdy = Area = A d Putting all this together brings us back to Equation (12 - 1): ∫ τds = 2AGk Q.E.D. Hence, in the general case we use equation (12 - 1) to relate C1 and C2. 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
MT-1620 al.2002 Special Case of a Circular Tube Consider the case of a circular tube with inner diameter r and outer diameter r。 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 fres is directed along circles Paul A Lagace @2001 Unit 12-10
τ MIT - 16.20 Fall, 2002 Special Case of a Circular Tube Consider the case of a circular tube with inner diameter Ri and outer diameter Ro 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 τres is directed along circles Paul A. Lagace © 2001 Unit 12 - 10
