MIT - 16.20 Fall, 2002 Unit 10 St. Venant Torsion Theory Readings : Rivello 8.1, 8.2, 8.4 T & G 101, 104, 105, 106 Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 III. Torsion Paul A. Lagace ? 2001 Unit 10 - p . 2 MIT - 16.20 Fall, 2002 We have looked at basic in-plane loading. Let ’ s now consider a second “building block ” of types of loading: basic torsion. There are 3 basic types of behavior depending on the type of cross-section: 1. Solid cross-sections “classical ” solution technique via stress functions 2. Open, thin-walled sections Membrane Analogy Paul A. Lagace ? 2001 Unit 10 - p . 3 MIT - 16.20 Fall, 2002 3. Closed, thin-walled sections Bredt’s Formula In Unified you developed the basic equations based on some broad assumptions. Let ’ s … ? B e a bit more rigorous ? Explore the limitations for the various approaches ? Better understand how a structure “resists” torsion and the resulting deformation ? Learn how to model general structures by these three basic approaches Look first at Paul A. Lagace ? 2001 Unit 10 - p . 4 MIT - 16.20 Fall, 2002 Classical (St. Venant ’ s ) Torsion Theory Consider a long prismatic rod twisted by end torques: T [in - lbs] [m - n] Figure 10.1 Representation of general long prismatic rod Length ( l ) >> dimensions in x and y directions Do not consider how end torque is applied (St. Venant ’ s principle) Paul A. Lagace ? 2001 Unit 10 - p . 5 MIT - 16.20 Fall, 2002 Assume the following geometrical behavior: a) Each cross-section (@ each z) rotates as a rigid body (No “distortion ” of cross-section shape in x, y) b) Rate of twist, k = constant c ) Cross-sections are free to warp in the z-direction but the This is the “ St. Venant Hypothesis ” warping is the same for all cross-sections “warping” = extensional deformation in the direction of the axis about which the torque is applied Given these assumptions, we see if we can satisfy the equations of elasticity and B.C. ’s. ? SEMI-INVERSE METHOD Consider the deflections : Assumptions imply that at any cross-section location z: ? d α ? α = ? dz ? z = k z ( careful ! a constant Rivello uses φ !) rate of twist (define as 0 @ z = 0) Paul A. Lagace ? 2001 Unit 10 - p . 6 ααα ??? MIT - 16.20 Fall, 2002 Figure 10.2 Representation of deformation of cross-section due to torsion (for small α ) undeformed position This results in: consider direction of + u u (x, y, z) = r α (-sin β ) v (x, y, z) = r α ( cos β ) w (x, y, z) = w (x, y) ? independent of z! Paul A. Lagace ? 2001 Unit 10 - p . 7 MIT - 16.20 Fall, 2002 We can see that: r = y 2 x + 2 sin β = y r x cos β = r This gives: u (x, y, z) = -y k z (10 - 1) v (x, y, z) = x k z (10 - 2) w (x, y, z) = w (x, y) (10 - 3) Paul A. Lagace ? 2001 Unit 10 - p . 8 MIT - 16.20 Fall, 2002 Next look at the Strain-Displacement equations : ? u ε = = 0 xx ? x ? v ε = = 0 yy ? y ? w ε = = 0 zz ? z ? u ( consider : u exists, but ? x = 0 v exists, but ? v = 0 ) ? y ? No extensional strains in torsion if cross-sections are free to warp Paul A. Lagace ? 2001 Unit 10 - p . 9 MIT - 16.20 Fall, 2002 ? u ? v ε = + = ? zk + zk = 0 xy ? y ? x ? cross - section does not change shape (as assumed! ) ? v ? w ? w ε yz = ? z + ? y = kx + ? y (10 - 4) ? w ? u ? w ε zx = ? x + ? z = ? ky + ? x (10 - 5) Now the Stress-Strain equations : let ’ s first do isotropic 1 ε xx = E [ σ xx ? ν ( σ yy + σ zz ) ] = 0 ε yy = 1 [ σ yy ? ν ( σ xx + σ zz ) ] = 0 E 1 ε zz = E [ σ zz ? ν ( σ xx + σ yy ) ] = 0 ? σ xx , σ yy , σ zz = 0 Paul A. Lagace ? 2001 Unit 10 - p . 1 0 MIT - 16.20 Fall, 2002 ( ε = 21 + ν ) σ = 0 ? σ = 0 xy E xy xy ε = 21 + ν ) σ yz ( E yz (10 - 6) 1 / G ( ε xz = 21 + ν ) σ xz (10 - 7) E ? only σ xz and σ yz stresses exist Look at orthotropic case: 1 ε xx = E 11 [ σ xx ? ν 12 σ yy ? ν 13 σ zz ] = 0 1 ε yy = E 22 [ σ yy ? ν 21 σ xx ? ν 23 σ zz ] = 0 1 ε zz = E 33 [ σ zz ? ν 31 σ xx ? ν 32 σ yy ] = 0 ? σ xx , σ yy , σ zz = 0 still equal zero Paul A. Lagace ? 2001 Unit 10 - p . 1 1 MIT - 16.20 Fall, 2002 1 ε = σ yz G 23 yz 1 ε = σ xz xz G 13 Differences are in ε yz and ε xz here as there are two different shear moduli ( G 23 and G 13 ) which enter in here. for anisotropic material : coefficients of mutual influence and Chentsov coefficients foul everything up (no longer “simple ” torsion theory). [can ’ t separate torsion from extension] Back to general case … Look at the Equilibrium Equations : ? σ xz ? z = 0 ? σ xz = σ xz (, xy ) ? σ yz ? z = 0 ? σ yz = σ yz (, xy ) Paul A. Lagace ? 2001 Unit 10 - p . 1 2 MIT - 16.20 Fall, 2002 So, σ xz and σ yz are only functions of x and y ? σ xz + ? σ yz = 0 (10 - 8) ? x ? y We satisfy equation (10 - 8) by introducing a Torsion ( Prandtl ) Stress Function φ (x, y) where: ? φ ? y =? σ xz (10 - 9a) ? φ ? x = σ yz (10 - 9b) Using these in equation (10 - 8) gives: ? ? ? ? ? ? φ ? + ? ? ? φ ? ≡ 0 ? x ? y ? ? ? y ? ? x ? ? Automatically satisfies equilibrium (as a stress function is supposed to do) Paul A. Lagace ? 2001 Unit 10 - p . 1 3 MIT - 16.20 Fall, 2002 Now consider the Boundary Conditions : (a) Along the contour of the cross-section Figure 10.3 Representation of stress state along edge of solid cross- section under torsion outer contour is stress-free surface (away from load introduction) Figure 10.4 Close-up view of edge element from Figure 10.3 σσσσ xz (into page) σσσσ yz (out of page) x Paul A. Lagace ? 2001 Unit 10 - p . 1 4 MIT - 16.20 Fall, 2002 Using equilibrium: ∑ F z = 0 (out of page is positive) gives: ? σ xz dydz + σ yz dxdz = 0 Using equation (10 - 9) results in ? ?? ? φ dy ? ? ? + ? ? φ ? dx = 0 ? ? ? y ? ? x ? ? ? φ dy ? ? ? + ? ? φ dx ?? = d φ ? ? ? y ? ? x And this means: d φ = 0 ? φ = constant We take: φ = 0 along contour (10 - 10) Note : addition of an arbitrary constant does not affect the stresses, so choose a convenient one (0!) Paul A. Lagace ? 2001 Unit 10 - p . 1 5 MIT - 16.20 Fall, 2002 Boundary condition (b) on edge z = l Figure 10.5 Representation of stress state at top cross-section of rod under torsion Equilibrium tells us the force in each direction: F x = ∫∫ σ z x dxdy using equation (10 - 9): = y R ? φ dxdy ∫∫ y L ? y where y R and y L are the geometrical limits of the cross- section in the y direction Paul A. Lagace ? 2001 Unit 10 - p . 1 6 MIT - 16.20 Fall, 2002 =? ∫ φ y R dx [] y L and since φ = 0 on contour F x = 0 O.K. (since no force is applied in x-direction) Similarly : F y = ∫∫ σ zy dxdy = 0 O.K. Look at one more case via equilibrium: Torque = Τ = ∫∫ [ x σ zy ? y σ zx ] dxdy ? φ ? φ = ∫∫ x B x ? x dxdy + ∫∫ y y L R y ? y dydx x T where x T and x B are geometrical limits of the cross-section in the x-direction Integrate each term by parts: ∫ AdB = AB ? ∫ BdA Paul A. Lagace ? 2001 Unit 10 - p . 1 7 MIT - 16.20 Fall, 2002 Set: A = x ? d A = d x ? φ d B = d x ? B = φ ? x and similarly for y x Τ = ∫ [ ] x φ B ?∫ φ dx ] dy + ∫ [ ] y φ y R ?∫ φ dy ] dx x T y L = 0 = 0 since φ = 0 in contour since φ = 0 in contour ? Τ =? 2 ∫∫ φ dxdy (10 - 11) Up to this point, all the equations [with the slight difference in stress-strain of equations (10 - 6) and (10 - 7)] are also valid for orthotropic materials. Paul A. Lagace ? 2001 Unit 10 - p . 1 8 MIT - 16.20 Fall, 2002 Summarizing ? Long, prismatic bar under torsion ? Rate of twist, k = constant ? w ? ε yz = kx + ? y ? w ? ε xz = - k y + ? x ? φ ? φ ? y =? σ xz ? x = σ yz ? ? Boundary conditions φ = 0 on contour (free boundary) Τ =? 2 ∫∫ φ dxdy Paul A. Lagace ? 2001 Unit 10 - p . 1 9 MIT - 16.20 Fall, 2002 Solution of Equations (now let ’s go back to isotropic ) Place equations (10 - 4) and (10 - 5) into equations (10 - 6) and (10 - 7) to get: ? ? w ? σ yz = G ε yz = G ? ? k x + ? y ? ? (10 - 12) σ xz = G ε xz = G ?? ? k y + ? w ? (10 - 13) ? x ? We want to eliminate w. We do this via: ? ? x { Eq. (10 - 12) } ? ? ? y { Eq. (10 - 13) } to get: ? σ yz ? σ xz ? ? 2 w ? 2 w ? ? x ? ? y = Gk + ?? + k ? yx ? ? ? ? xy ?? Paul A. Lagace ? 2001 Unit 10 - p . 2 0 φφφ MIT - 16.20 Fall, 2002 and using the definition of the stress function of equation (10 - 9) we get: ? 2 φ ? 2 φ ? x 2 + ? y 2 = 2Gk (10 - 14) Poisson ’s Equation for φ (Nonhomogeneous Laplace Equation) Note for orthotropic material We do not have a common shear modulus, so we would get: ? ? + ? ? = ( ) + ( ) ? ?? 2 2 2 2 2 φ x k G w xy G G xz yz yz xz ? We cannot eliminate w unless G xz and G yz are virtually the same + ? φ y G Paul A. Lagace ? 2001 Unit 10 - p . 2 1 MIT - 16.20 Fall, 2002 Overall solution procedure : ? Solve Poisson equation (10 - 14) subject to the boundary condition of φ = 0 on the contour ? Get T - k relation from equation (10 - 11) ? Get stresses ( σ xz , σ yz ) from equation (10 - 9) ? Get w from equations (10 - 12) and (10 - 13) ? Get u, v from equations (10 - 1) and (10 - 2) ? Can also get ε xz , ε yz from equations (10 - 6) and (10 - 7) This is “St. Venant Theory of Torsion ” Application to a Circular Rod Figure 10.6 Representation of circular rod under torsion cross-section Paul A. Lagace ? 2001 Unit 10 - p . 2 2 MIT - 16.20 Fall, 2002 “Let”: φ = C 1 ( x 2 + y 2 ? R 2 ) This satisfies φ = 0 on contour since x 2 + y 2 = R 2 on contour This gives: ? 2 φ ? 2 φ ? x 2 = 2 C 1 ? y 2 = 2 C 1 Place these into equation (10-14): 2 C 1 + 2 C 1 = 2 G k Gk ? C 1 = 2 Note : (10-14) is satisfied exactly Paul A. Lagace ? 2001 Unit 10 - p . 2 3 MIT - 16.20 Fall, 2002 Thus: φ = Gk ( x 2 + y 2 ? R 2 ) 2 Satisfies boundary conditions and partial differential equation exactly Now place this into equation (10-11): Τ =? 2 ∫∫ φ dxdy Figure 10.7 Representation of integration strip for circular cross-section R + Τ = Gk ∫∫ - R - R - 2 2 y y 2 2 ( R 2 ? y 2 ? x 2 ) dxdy -R Paul A. Lagace ? 2001 Unit 10 - p . 2 4 1 ? ? MIT - 16.20 Fall, 2002 + Τ = Gk ∫ ? R R ?? ? ( R 2 ? y 2 ) x ? x 3 3 ?? ? y y 2 2 R R 2 2 ? ? dy ? 4 R / = Gk 3 ∫ ? R ( R 2 ? y 2 ) 32 dy + R / = Gk 4 1 ? y ( R 2 ? y 2 ) 32 + 3 R 2 y R 2 ? y 2 + 3 R 4 sin ? 1 y ? 3 4 ? 2 2 R ? ? R = 0 = 0 = 3 R 4 π 2 This finally results in π R 4 Τ = Gk 2 Paul A. Lagace ? 2001 Unit 10 - p . 2 5 MIT - 16.20 Fall, 2002 Since k is the rate of twist: k = d α , we can rewrite this as: dz d α Τ = dz GJ where: ? π R 4 ? J = torsion constant ? ? = 2 for a circle ? ? α = amount of twist and: GJ = torsional rigidity Note similarity to: dw dx M EI 2 2 = where: (I) J - geometric part (E) G - material part EI = bending rigidity Paul A. Lagace ? 2001 Unit 10 - p . 2 6 τττ MIT - 16.20 Fall, 2002 To get the stresses, use equation (10 - 9): ? φ Τ σ = = G kx = x yz ? x J Τ σ xz =? ? φ =? G ky = ? J y ? y Figure 10.8 Representation of resultant shear stress, τ res , as defined Define a resultant stress: τ = σ zx zy 2 σ + 2 Τ = J x 2 + y 2 = r Paul A. Lagace ? 2001 Unit 10 - p . 2 7 τττ MIT - 16.20 Fall, 2002 The final result is: τ = Τ r J for a circle Note : similarity to ?? σ x =? Mz ? Ι ? τ always acts along the contour (shape) resultant Figure 10.9 Representation of shear resultant stress for circular cross-section Cross-Section No shear stress on surface Paul A. Lagace ? 2001 Unit 10 - p . 2 8 τττ MIT - 16.20 Fall, 2002 Also note: 1. Contours of φ : close together near edge ? higher τ Figure 10.10 Representation of contours of torsional shear function 2. Stress pattern ( τ ) creates twisting Figure 10.11 Representation of shear stresses acting perpendicular to radial lines magnitudes of τ res Paul A. Lagace ? 2001 Unit 10 - p . 2 9 MIT - 16.20 Fall, 2002 To get the deflections, first find α : d α Τ = dz GJ (pure rotation of cross-section) integration yields: Τ z α = + C 1 GJ Let C 1 = 0 by saying α = 0 @ z = 0 Use equations (10 - 1) and (10 - 2) to get: Τ z u =? yzk =? y GJ Τ z v = xzk = x GJ Go to equations (10 - 12) and (10 - 13) to find w(x, y): Equation (10 - 12) gives: ? w σ yz = ? kx ? y G Paul A. Lagace ? 2001 Unit 10 - p . 3 0 MIT - 16.20 Fall, 2002 using the result for σ yz : ? w Gkx = ? kx = 0 ? y G integration of this says w(x, y) = g 1 (x) (not a function of y) In a similar manner … Equation (10 -13) gives: ? w = σ xz + ky ? x G Using σ xz = -Gky gives: ? w =? Gky + ky = 0 ? x G integration tells us that: w(x, y) = g 2 (y) (not a function of x) Using these two results we see that if w(x, y) is neither a function of x nor y, then it must be a constant . Might as well take this as zero Paul A. Lagace ? 2001 Unit 10 - p . 3 1 MIT - 16.20 Fall, 2002 (other constants just show a rigid displacement in z which is trivial) ? w(x, y) = 0 No warping for circular cross-sections (this is the only cross-section that has no warping) Other Cross-Sections In other cross-sections, warping is “the ability of the cross-section to resist torsion by differential bending ”. 2 parts for torsional rigidity ? Rotation ? Warping Ellipse Paul A. Lagace ? 2001 Unit 10 - p . 3 2 MIT - 16.20 Fall, 2002 ? x 2 y 2 ? φ = C 1 ? ? a 2 + b 2 ? 1 ? ? Equilateral Triangle φ = C 1 ? ? x ? 3y + 2 ? ? 3y ax + 1 a ? ax + ? 2 ? ? 3 ? ? 3 ? ? 3 ? Rectangle Paul A. Lagace ? 2001 Unit 10 - p . 3 3 MIT - 16.20 Fall, 2002 π π φ =∑ ? C n + D n cosh ny ? cos nx n odd ? b ? a Series: (the more terms you take, the better the solution) These all give solutions to ? 2 φ = 2GK subject to φ = 0 on the boundary. In general, there will be warping see Timoshenko for other relations (Ch. 11) Note : there are also solutions via “warping functions ”. This is a displacement formulation see Rivello 8.4 Next we ’ ll look at an analogy used to “solve” the general torsion problem Paul A. Lagace ? 2001 Unit 10 - p . 3 4