Lecture Pure Twist

Transcription

1 Lecture Pure Twist pure twist around center of rotation D => neither axial (σ) nor bending forces (Mx, My) act on section; as previously, D is fixed, but (for now) arbitrary point. as before: a) equilibrium of wall element: d q + d σ t = 0 b) compatibility (shear strain) d d u + v = γ = 0 small deflections ds dx c) tangential displacement (δv) in terms of η, ζ and φ (geometry) δv δη δζ = cos ( α ) ( ) + hp δφ + sin α δx δx δx δx N.B. h p =>h D from definition of problem further assumptions: 1) preservation of cross section shape => ζ = ζ(x); η = η(x) φ = φ(x) 2) shear though finite is small ~ 0 => d u = d v 3) Hooke's law holds => σ = E δu axial stress δx from equilibrium pure twist σ da = N x τ h p da = qh p ds = T p σ da = N x = 0 σ y da = -M.z ( ) da = q cos α ds = V y σ y da = M z = 0 τ cos α ( ) σ z da = M y ( ) da = q sin α ds = V z σ z da = M y = 0 τ sin α ( ) pure twist also => only φ is finite i.e. other displacements (and derivatives) ζ = η = 0 => δv δη δζ = cos ( α) + sin ( α ) + hp δφ becomes δx δv = hd δφ δx δx δx δx δx using negligible shear assumption d u = d v => d u = h D δφ and integration along s => ds δx 1 notes_13_pure_twist.mcd

2 u = δφ h D ds + u 0 ( x ) δx previously u = η' Y ζ' Z + u 0 ( x ) which showed u linear with y and z => plane sections plane. here - only if h is constant so it can come outside h D D 1 ds - is u (longitudinal displacement) linear. u is defined as warping displacement (function). stress analysis can be made analogous for torsion and bending IF the integrand h D ds thought to be a coordinate. calculation of stresses will involve statical moments, moments of inertia and products of inertia which will be designated "sectorial" new coordinate = Ω Ω wrt arbitrary origin and ω wrt normalized sectorial coordinate (as before like wrt center of area) dω = h D ds = dω the warping function then becomes: δφ u = Ω + u 0 ( x ) = φ' Ω + u 0 ( x) δx b) warping stresses as before: axial strain = du/dx => u' = φ'' Ω + u' 0 ( x) and σ = E u' = E φ'' Ω + E u' 0 ( x) σ da = 0 determines u' ( ) 0 x ( E φ'' Ω + Eu' 0 ( x )) da = 0 => Ω da Ω da and stress becomes: Eu' 0 ( x ) = E φ'' σ = E φ'' Ω + E φ'' A A = E φ'' ω Ω da Ω da that is: σ = E φ'' Ω = E φ'' ω where ω = Ω A A this defines the normalized coordinate in the same sense as y and Y etc.... as an aside: u' = φ'' ω => in this sense, ω is defined as the unit warping function displacement per unit change in rotation dependent only on s within a constant... u = φ' ω + constant dω = h D ds 2 notes_13_pure_twist.mcd

3 shear flow follows from integration of d q + d σ t = 0 along s as above and leads to : d ds q = d d x σ t => qs, x) = d ( σ t ds + q1 ( x) dx using the expression for axial stress σ = E u' = E φ'' ω s s ( ( ) d σ s t ds = q1 ( x ) E φ''' ω t ds = q 1 ( x ) + E φ''' qs, x) = q 1 x dx ω t ds where q 1 ( x) is f(z) and represents the shear flow at the start of the region. it is 0 at a stress free boundary which is convenient for an open section: q 1 ( x) = 0 as before if we designate the integrals which are the static moments of the cross section area: e.g. Qy and Qz: s Q ω = ω da = ω t ds 0 = "sectorial statical moment of the cut-off portion of the cross section" therefore: qs, ( x) = E φ''' Q ω designate torsional moment wrt D by Tω τ hd da = qh D ds = T ω now, since dω = h D ds = dω => qh D ds = q dω and using integration by parts parts: u = q v = ω u dv = (u v) ( b ) ( uv )( 0 ) v du du = dq dv = dω integration along s and as dq = δq/δs*ds q dω = q ω(s = b) q ω(s = 0) ω dq = q ω(s = b) q ω(s = 0) ω δq ds δs q ω(s = b) = 0 and q ω(s = 0) = 0 as q(s=b) and q(s=0) = 0 (stress free ends) now using equilibrium: d q + d σ t = 0 3 notes_13_pure_twist.mcd

4 q dω = 0 ω δq ds = d ω δs d x σ t ds substituting σ = E φ'' ω from above d σ = E φ''' ω => dx q dω = ω d σ t ds = E φ''' ωω t ds = E φ''' I ωω dx where I similar to Iz ωω = ωω da = ωω t ds I z = yyda = yy t ds N.B. sometimes this is represented by Iyy going back to the relationship for torsional moment, where we have derived relationships for qh D ds => T ω T ω = qh D ds = q dω = E φ''' I ωω therefore: φ''' = EIωω Tω if we think of a distributed torsional load (moment/unit length) m D ; md equilibrium over element dz => Tω + dtω/dx*dx T ω + m D dx + T ω + d T ω dx = 0 => T' ω = m D dx and just as M' y = V z the warping moment M' ω may be defined as M' ω = T ω M ω thus: φ'' = and the stresses are as follows: EI ωω M ω T ω σ = E φ'' ω = ω and... from qs, ( x) = E φ''' Q ω qs, ( x) = τ t = Q I ω ωω I ωω 4 notes_13_pure_twist.mcd

5 c) Center of twist calculation of the sectorial quantities assumes center of twist y D and z D are known. the second and third equilibrium conditions above require: 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 now for some geometry: determine distance from tangent to wall from h D in terms of the coordinates of the center, the angle (α) that the y axis would have to rotate to line up with the positive direction of the tangent and the perpendicular distance from the origin of the centroidal coordinates h C tangent hc y h D = h C a b a = y D cos ( ) b = z D sin ( ) ds z hd D yd zd a b α C parallel to tangent = α π cos ( ) = cos α π = sin ( α ) 2 2 sin ( ) = sin α π = cos ( α ) 2 h D = h C a b = h C y D cos ( ) zd sin ( ) = h C y D sin ( α ) + zd cos ( α ) multiply by ds and apply geometry: α y,η h D ds = h C ds y D sin ( α ) ds + zd cos ( α) ds ds cos ( ) = dz = ds sin ( α ) ds dz dy ds sin ( ) ( )) dy = ds cos α = dy = ds ( cos α ( ) z,ζ h D ds = h C ds y D sin ( α ) ds + zd cos ( α) ds = hc ds y D dz + z D dy sectorial coordinate ω = hds => h D ds = dω D = h C ds y D dz + z D dy = dω C y D dz + z D dy 5 notes_13_pure_twist.mcd

6 which is now integrated: ω D = ω C y D z + z D y and introduced into the equilibrium equations above where ω is ω D : ω y da = 0 and ω z da = 0 => ωd y da = 0 = (ω C y D z + z D y) y da and... ωd z da = 0 = (ω C y D z + z D y) z da now using second moment nomenclature (including treating ω as a coordinate) => ωc y da y D yzda + z D yyda = 0 becomes I zωc y D I yz z D I z = 0 recall that I yωc is referred to C for ω and... ωc z da y D zzda + z D yzda = 0 becomes I yωc y D I y + z D I yz = 0 which provides two equations in two unknowns y D and z D Given I yωc y D I y + z D I yz = 0 I zωc y D I yz + z D I z = 0 y D z D := Find ( y D, z D ) z D y D I yωc I z I yz I zωc I zωc I y + I yz I yωc 2 and... zd I y I z I yz 2 I y I z I yz and for principal axes I yz = 0 I yz := 0 y D := ( I yωc I z I yz I zωc ) z D := ( I zωc I y + I yz I yωc ) 2 I 2 I y I z I yz I y I z I yz I yωc y D I and... y z D I zωc I z 6 notes_13_pure_twist.mcd