Lecture 5 - 2003 Twist closed sections
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Lecture 5 - 2003 Twist closed sections As this development would be almost identical to that of the open section, some of the development is simply repeated (copied) from the open section development. pure twist around center of rotation D => neither axial (σ) nor bending forces (Mx, My) act on section ----------------------------- from equilibrium -------------------------------pure twist ⌠ σ dA = Nx ⌡ ⌠ τ⋅h dA = p ⌡ ⌠ σ⋅y dA = -M.z ⌡ ⌠ σ⋅z dA = M y ⌡ ⌠ q⋅ h ds = T p p ⌡ ⌠ σ dA = 0 ⌡ ⌠ ⌠ τ⋅cos( α ) dA = q⋅cos( α ) ds = Vy ⌡ ⌡ ⌠ σ⋅y dA = 0 ⌡ ⌠ ⌠ τ⋅sin( α ) dA = q⋅sin( α ) ds = Vz ⌡ ⌡ ⌠ σ⋅z dA = 0 ⌡ a) equilibrium of wall element: pure twist => . ξ = η = 0 => δv δx = δψ δx ⋅cos( α ) + δη δx ⋅sin( α ) + hp⋅ δφ δx δv becomes δx = hD⋅ δφ δx b) compatibility (shear strain) d d u+ v=γ ds dx here is first change. we cannot set γ = 0 as we did in the open problem τ δφ d d d u = γ ⋅ − v => u = ⋅ −hD⋅ G ds δx dx ds and integration along s => => s ⌠ τ δφ ⌠ ds − ⋅ hD ds + u0( x) u= G δx ⌡ ⌡ 0 for open sections u=− δφ ⌠ ⋅ hD ds + u0( x) as γ is small => = 0 δx ⌡ other assumptions: section shape remains etc. same 1 notes_14_twist_closed.mcd s ⌠ τ ds G ⌡ 0 See: Torsion of Thin-Walled, Noncircular Closed Shafts; Shames Section 14.5 particularly; equations 14.17, 14.18 and 14.21 (Bredt's formula) also: Hughes 6.1.19, 6.1.21, 6.1.22 and section 6.1 M x = 2⋅ q⋅A q := Mx 2⋅ A and ...... M x := G⋅ J⋅ δφ => δx q := G⋅ J δφ ⋅ 2⋅ A δx A in these relationships is the "swept area" i.e. per Shames; "total plane area vector of the area enclosed by the midline s." near 14.21 δφ s G⋅ J⋅ s ⌠ τ ⌠ 1 q δx ds = ds = G G t 2⋅ A⋅G ⌡ ⌡ 0 0 s 2 s ⌠ 1 J ⌠ δφ 1 ⋅ ds = ⋅ ds⋅ t 2⋅ A t δx ⌡ ⌡ 0 4⋅ A J= ⌠ ⌡ 0 b 1 t integral 0 to b => circular (all way around) defining J from 14.21 (Bredt's formula) ds 0 s ⌠ τ ⌠ s δφ ds − hD ds ⋅ u= + uo( x) = G δx ⌡ ⌡ 0 0 s s ⌠ δφ J ⋅ 1 ds − ⌠ + uo( x) hD ds ⋅ t 2⋅ A δx ⌡ ⌡ 0 0 as with open sections define "sectorial" coordinate = Ω, by its derivative Ω wrt arbitrary origin and ω wrt normalized sectorial coordinate s ⌠ 1 ds t ⌡ definition: J 1 dΩ = hD − ⋅ ⋅ds = dω 2⋅ A t s s s ⌠ ⌠ J ⌠ 1 0 ⋅ ds = hD ds − 2⋅A⋅ Ω = hD ds − 2⋅ A t b ⌡ ⌡ ⌠ 1 ⌡ 0 0 0 ds t ⌡ 0 the warping function then becomes (as previously): u = − δφ δx ⋅Ω + u0( x) = −φ'⋅ Ω + u0( x) s ⌠ 1 ds t ⌡ s ⌠ the warping function Ω has a "correction" to the hD ds term of ⌡ 0 −2⋅A⋅ 0 ⌠ ⌡ b 1 t ds 0 otherwise everything is identical. hD and hc still have same meaning in Ω and ω 2 notes_14_twist_closed.mcd b) warping stresses as before: axial strain = du/dx => u' = −φ''⋅ Ω + u'0( x) and axial stress: σ = E⋅ u' = −E⋅ φ''⋅Ω + E⋅ u'0( x) ⌠ σ dA = 0 ⌡ E⋅ u'0( x) = E⋅φ'' determines u'0( x) ⌠ Ω dA ⌡ ⌠ ⌡ (−E⋅ φ''⋅Ω + E⋅ u'0(x) ') dA = 0 and stress becomes: σ = −E⋅ φ''⋅Ω + E⋅φ''⋅ A ⌠ Ω dA ⌡ that is: σ = −E⋅ φ''⋅ Ω − = −E⋅ φ''⋅ω A => ⌠ Ω dA ⌡ where A = −E⋅ φ''⋅ω ω=Ω− ⌠ Ω dA ⌡ A axial stress: σ = −E⋅ φ''⋅ ω shear stress shear flow follows from integration of d d q = − σ ds dx d d q + σ ⋅t = 0 along s and leads to : ds dx ⌠ d q( s, x) = − σ ds + q1( x) dx ⌡ => using the expression for axial stress σ = E⋅ u' = −E⋅ φ''⋅ω ⌠ q( s, x) = q1( x) − ⌡ s 0 s ⌠ s d σ ⋅t ds = q ( x) − ⌠ − φ'''⋅ω⋅t ds = q E⋅ ( x ) + E⋅φ''' ω⋅t ds 1 1 ⌡0 ⌡ dx 0 where q1( x) is f(x) unlike open section we cannot set it = 0 q1( x) ≠ 0 3 notes_14_twist_closed.mcd we can superpose an open and closed problem setting the "slip" i.e. γ at an arbitrary cut = 0 this is equivalent to collecting all the s variation into the open solution and the x variation into the constant −Tω ⌠ Qω = ω dA = ⌡ qopen(s , x) = τ open⋅t = ⋅Q Iωω ω s ⌠ ω⋅t ds ⌡ 0 the ω derived above is the value with the constant of integration set to zero, i.e starting from open end. Tω q( s, x) = q1( x) + qopen(s , x) = q1( x) − ⋅Q Iωω ω ⌠ ⌠ γ ds = 0 = ⌡ ⌡ no slip => ⌠ 0= ⌡ => => ⌠ q ds = t ⌡ q1( x) = Tω q( s, x) = − ⋅ Qω − Iωω ⌠ ds = G ⌡ τ Tω ⋅Q q1( x) − Iωω ω t q t⋅ G ⌠ ds = q1( x)⋅ ⌡ Tω ⌠ ⋅ Qω ds Iωω ⌡ ⌠ ⌡ 1 t N.B. these integrals are circular i.e. no slip results are for complete way around the closed section ds = 0 1 t ds − Tω ⌠ ⋅ Qω ds Iωω ⌡ so we can say: ds ⌠ 1 ds t ⌡ ⌠ Q ds ω ⌡ thus a "correction" ⌠ Q ds ω ⌡ ⌠ ⌡ 1 t is applied to Qω for ds the closed section. the ω is for the closed section (with it's correction applied) 4 notes_14_twist_closed.mcd c) Center of twist as for an open section, the second and third equilibrium condition above requires: ⌠ σ⋅y dA = 0 ⌡ ⌠ σ⋅z dA = 0 ⌡ for pure twist ⌠ ⌠ using σ = −E⋅ φ''⋅ω this requires ω⋅y dA = 0 and ω⋅z dA = 0 as E ≠ 0 and φ'' ≠ 0 ⌡ ⌡ as shown above this relationship is identical with the new "corrrected" ω so the shear center and center of twist can be calculated the same way. yD = (Iyωc⋅Iz − Iyz⋅Izωc) and ... I ⋅I − I 2 y z yz and for principal axes Iyz = 0 y D := yD → zD = I ⋅I − I 2 y z yz Iyz := 0 (Iyωc⋅Iz − Iyz⋅Izωc) zD := I ⋅I − I 2 y z yz Iyωc Iy (−Izωc⋅Iy + Iyz⋅Iyωc) (−Izωc⋅Iy + Iyz⋅Iyωc) zD → and ... 5 I ⋅II − I 2 y z yz −Izωc Iz notes_14_twist_closed.mcd
