Course Stabilty of structures

Transcription

1 Curse Stabilt f structures Lecture ntes abut structural mechanics basic relatins 2D beams, 2:nd rder ther 2D beams, stabilit /Per J. G. 1

2 General bases in structural and slid mechanics analsis Differential equatin with displacement (u,v,w) as the unknwn. Bd lads and material prperties are knwn. Bundar cnditins Static (surface lads, Neumann ) Cnvective (springs, Rbin ) Kinematic (displacements, Dirichlet ), v z, w x, u The material is assumed t be cntinuus. (u,v,w) is the displacement f material pint (x,,z). (x,,z) is the name f a material pint, named after its lcatin at sme fixed reference time, i.e. Lagrange crdinates (cmpare Euler crdinates). The differential equatin relates displacement t lad. It is derived frm three relatins: 1) Kinematic: displacement strain (ε x =du/dx +, etc) 2) Material: strain stress (σ x =Eε x +, etc) 3) Equilibrium: stress lad (q x =dσ x /dx +, etc) Differential equatin + bundar cnditins give (u,v,w) fr all (x,,z) b slving the equatin in sme wa. Knwing (u,v,w), the stress can be calculated. 2

3 Analsis f a general structure/slid (All belw quantities are a functin f (x,,z) ) Gverning diff. equatin (r set f equatins) relating displacement t lad Г(u, v, w) = f(q x, q, q z ) Displacements u { v w Lads q x { q q z 1) Kinematics 3) Equilibrium Strains { ε x γ x ε { γ xz εz γ z 2) Material Stresses σ x σ { { σ z τ x τ xz τ z The gverning set f equatins is linear if : 1) Kinematics: Small defrmatins s that strain is prp. t displacem. 2) Material: Linear relatin between stress and strain. 3) Equilibrium: Is established fr the unladed (reference) gemetr. Secnd rder ther : Relatins 1) and 2) are the same in the linear First rder ther. Relatin 3): Equilibrium is established fr laded defrmed gemetr. 3 The displacements are assumed t be small ( f(x+δx)=f(x)+δx f (x) )

4 Beams t be discussed (in relatin t 1:st and/r 2:nd rder ther): 2D beam in bending- and bar actin. (Equatin, slutin and bundar) Trsin (St Venant and Vlasv trsin) (Equatin, slutin and bundar) 3D beam in bending, bar and trsin actin (Eq., slutin and bundar) 2D beam in bending and bar actin Basic assumptins: Linear elastic istrpic material. Crss sectin assumptins accrding t Bernulli/Euler beam ther. Equilibrium accrding 2:nd rder ther., v v tt (x)=v i (x)+v(x) q (x) An initiall straight r almst straight beam with varing crss sectin. q x (x) m(x) σ (x,) z L k x, k, k m x, u The quantities in the gverning set f equatins are a functin nl f x. The x-axis is lcated alng beam crss sectin (mechanical) center f gravit. We want: A gverning equatin fr bending actin with v(x) as the unknwn A gverning equatin fr bar actin with u(x) as the unknwn 4

5 2D beam in bending- and bar actin, Case A, v q q x z, w x, u Case A: N eigenstresses. Initiall straight beam withut imperfectins. N distributed mment lad. N elastic bed. Gverning f equatins, displacement vs lad Г v (v(x)) = q (x) Г u (u(x)) = q x (x) Displacements { u(x) v(x) Lads { q x(x) q (x) 1a) Beam mdelling assumptins 4) 3b) Equilibrium Displacement u(x, ) Crss sectin frces { N(x) M(x) 1b) Kinematics 3a) Static equivalence Strain ε x (x, ) 2) Material Stress σ x (x, ) 5

6 1a) Beam mdelling assumptins The Bernulli-Euler assumptins* ) : Plane crss-sectins remain plane and remain perpendicular t the centre axis-f the beam Small slpe f the beam axis s that sin(v )=v. u(x, ) u v (1a) 1b) Kinematics Small defrmatins are assumed ε(x, ) u (x,) (1b) 2) Material Linear elastic istrpic material with mdulus E(x) is assumed σ(x, ) Eε(x, ) (2) 3a) Static equivalence N A Equatins 1a) 3a) give σ(x)da, N EAu M EIv M A (x)σ(x)da (3a) (4) The tw relatins in Eq. (4) are the same fr 1:st and 2:nd rder ther. A and I are the cnventinal the crss sectin gemetr parameters. * ) a) The Bernulli-Euler assumptins crrespnd t prescribed zer shear strain. In turn this crrespnds t infinitel large shear stiffness, G, f the material. b) The Bernulli-Euler beam ther give the same result as full 2D plane stress analsis if nl bending and axial frce in the beam (n shear frce) and if the lad is applied in accrdance with the beam ther. 6

7 3b) Equilibrium fr a shrt part dx f the beam Equilibrium fr laded defrmed gemetr (2:nd rder ther) The displacements are assumed t be small ( f(x+δx)=f(x)+δx f (x) ) The lads mve with the particles and keep their glbal rientatin cnstant, v A shrt part, dx, f the beam in tw psitins Particle (x,) Particle (x+dx,) N M V q q x v (x) V+dV N+dN M+dM v(x) sin(v )=v cs(v )=1 u(x) x x+dx Laded psitin x, u dx Reference psitin Equatins f equilibrium: N(NdN) Vv (VdV)(v dv ) qxdx 0 (5a) V (V dv) Nv (N dn)(v dv ) qdx 0 (5b) dx dx M (M dm) V (V dv) (5c) N (Vv ) qx 0 V (Nv ) q 0 M V 0 (6a,b,c) 7

8 Eliminatin f V b means f (6c) gives N (Mv ) q M (Nv ) q x (7a,b) With N=EAu and M=EIv frm eq (4) we get the gverning equatins fr 2D Bernulli/Euler beams accrding t 2:nd rder ther: (EAu ) ((EIv ) v ) q (EIv ) (EAu v ) q x (8a,b) The equatins are cupled and nn-linear. In the analsis f bending it is nw assumed that the nrmal frce N(X)=EAu is knwn! In the analsis f axial frce it is mrever assumed that the 2:nd rder cntributin frm shear frce (EIv ) v = -Vv can be neglected! (This crrespnds t v =0, i.e. a straight but pssibl inclined beam.) (EAu ) q x (EIv ) (Nv ) q (9a,b) These are the tw uncupled and linear gverning differential equatins fr fairl well-knwn Case A. 8

9 Case B: The fllwing cnsideratins are added t thse in Case A: Initial stress (eg. due welding r temperature r misture gradients) Initial beam curvature (imperfectins) Distributed mment lad Initial stress ( fix-stress ) is taken int accunt in the material stress-strain relatin: The stress at zer strain is assumed t be σ σ(x, ) Eε(x, ) σ0(x,) (10) The stress σ is the stress at enfrced zer strain and is smetimes called the fix-stress. B eigenstress (r residual stress) is meant the stress in beam when n external lad is acting. Example: A unifrm temperature increase f a beam gives the fix stress σ =α ΔT E fr all pints in the beam. The eigenstress is an utcme f the analsis, taking int accunt beam gemetr and bundar cnditins. Bundar cnditins crrespnding t a staticall determinate beam wuld in this case give zer eigenstress. Fllwing the Case A derivatin f eq. (4), eq. (10) gives N EAu N M EIv M (11) where N M A A σ σ (x)da (x)da 9

10 Initial beam curvature and distributed mment lad are taken int accunt in the equilibrium equatins., v x A shrt beam part, dx, in three different psitins u i x+dx Equilibrium in the laded psitin (cmpare eq. (6)) gives N -(Vvtt ) qx 0 V (Nvtt ) q 0 M V m 0 v i v i V+dV N+dN M+dM v tt =v i +v Initial psitin: zer lad, zer strain, dx pssible nn-zer imperfectin (u i (x), v i (x)) and pssible nn-zer fix-stress σ (x,) Reference psitin: beam straight. (12a,b,c) Eliminatin f V b means f (12c) gives (12b) as M (Nv ) q (Nvi ) -m (13) with the assumptins N knwn and Vv tt =0, and with u tt =u+u i, v tt =v+v i, M=EIv and N=EAu, frm (12a) and (13b): N u M V q m v v i q x Laded psitin u tt =u i +u v tt =v i +v x, u (EAu') -q (EIv ) x N (Nv ) q m M (Nv) i (14a,b) These are the gverning equatins fr Case B. The right hand side f the bending equatin can be called a generalized transversal lad. 10

11 Case C: The fllwing cnsideratin is added t thse in Cases A and B: Elastic bed supprt fr u and v displacements and fr v rtatin., v k m z, w k x x, u It is assumed that the springs are linear elastic. It is assumed that the spring frces are zer in the initial psitin f the beam. The springs gives additinal distributed lad: Lad in x-directin: -k x u is added t q x Lad in -directin: -k v is added t q Mment arund z-axis: -k m v is added t m B eq. (14) is then fund k (EAu') -k (EIv ) x u -q ((N-k m x N )v ) k v q m M (Nv) i (15a,b) These are the gverning equatins fr the fairl general Case C. Nn-zer k x and k, but nt nn-zer k m, changes the character f the diff. equatins. 11

12 General slutin f gverning equatins fr bending actin Case B (including Case A as special case) (EIv ) (Nv ) q m M (Nv i ) (14b) Sme numerical slutin methd is in general needed in the case f varing cefficients (= varing EI and/r N). Fr cnstant cefficients, the equatin is EIv -Nv f where the generalize d tranversalladis f q m M (Nv) i (16) The slutin t (16) when f=0, i.e. the slutin t the hmgeneus equatin, is v v h h C 1 C cs( λx) C 1 2 csh( λx) C sin( λx) C Sme particular slutins: 2 3 x C sinh( λx) C 3 4 x C 4 frn 0 (cmpressi n), frn 0 (tensin), λ λ -N/EI N/EI.(17a,b) fr f(x) 0: fr f(x) cns t. a: fr f(x) sin(bx) : fr f(x) acs(bx): v v v v p p p p 0 2 ax /(2N) 4 2 asin(bx)/(bei b N) 4 2 acs(bx)/(bei b N). (18a,d) Several particular slutins can be added if the lad f(x) is divided int several lads. The ttal displacement is: v tt =v i +v h +v p (19) 12

13 Case C (including Case A and B as special cases) (EIv ) ((N-k )v ) m kv q m M (Nv i ) (15b) T find a slutin fr the general case sme numerical methd is needed. Fr cnstant cefficients, the equatin is EIv -(N-km)v kv f where the generalize d tranversalladis 13 f q m M (Nv) i Belw are slutins t (20) as interpreted frm analses given b (Heteni, 1946) and (Bazant, Cedlin, 1991) f the hmgeneus equatin (f=0). Verif befre use! First definitin f sme parameters: γ (Nk )/(2 k EI) α (k /(EI)) k α γ γ 1 m 1/4 2 1,2 r α ( 1 ) / 2 α (1 ) / 2 (21) Fr γ>1 (heav cmpressin) vh C1sin(k1 x) C2cs(k1 x) C3sin(k2 x) C4cs(k2 x) (22) Fr γ=1 (cmpressin) vh C1sin( x) C2xsin( x) C3cs( x) C4xcs( x) (23) Fr -1< γ<1 (small cmpressin r small tensin) vh C1sin(rx)sinh( ρx) C2cs(rx)sinh( ρx) C3sin(rx)csh( ρx) C4cs(rx)csh(ρx) Fr γ=-1 (tensin) vh C1sinh( x) C2xsinh( x) C3csh( x) C4xcsh( x) (25) Fr γ<-1 (heav tensin) vh C1sinh(k1 x) C2 csh(k1 x) C3sinh(k2 x) C4 csh(k2 x) (26) (20) (24) A particukar slutin fr f=cnst.=a: v p =a/k (27) Ttal slutin: v tt =v i +v h +v p (28)

14 Bundar cnditins 2D beam in bending- and bar actin The equatins fr the mst general case: (EAu') -k (EIv ) x u -q ((N-k m x N )v ) k v q m M (Nv) i (15a,b) 2+4=6 bundar cnditins are needed. Cmmnl, but nt necessaril, 3 are defined fr x=0 and 3 fr x=l., v z, w L x, u Kinematic cnditins ( Dirichlet ) u, v and/r v knwn Static cnditins ( Neumann ) N, V and/r M knwn Frm eq (11) and (12c), the static cnditins give fr the unknwn u and v: u (NN)/EA v (M-M)/EI (EIv) V M m (29a,b,c) These relatins are valid fr all x. NOTE: N, V and M are the beam crss sectin frces and mment in the actual laded psitin f the beam. 14

15 Cnvective cnditins (springs, Rbin ) n x =1 n x =-1, v z x, u k x k θ k At x=0 V M N N M V k θ k At x=l k x k x, k and/r k θ knwn Where fr fixed rientatin f the springs: k x =n x (N-v V)/u), k =.. and k θ =. This gives a nn-linear set f equatins fr slving the integratin cnstants. Hwever, recalling the assumptins In the analsis f bending it is nw assumed that the nrmal frce N(X)=EAu is knwn! In the analsis f axial frce it is mrever assumed that the 2:nd rder cntributin frm shear frce (EIv ) v = -Vv can be neglected! k x =n x N/u, k = n x V/v and k θ = n x M/v This crrespnds t assuming c-rtatin f the spring frces. Frm eq (11) and (12c), the cnvective cnditins give fr the unknwn u and v: EAu nxkxu nxn (EIv ) nxkvv M EIv -n k v M x θ m (30a,b,c) 15

16 Mixed (cupled) cnditins An example: inclined rller supprt at x=l M V, v z x, u α (α-v ) α usin( α) vcs(α) 0 Ncs(α) Vsin( α) 0 M 0 where N, V and M can be expressed in u and v b eq (11) and 12c) 16

17 Exampel, 2D beam bending 2:n rder ther, v q Q x, u L P, v q Q P x, u A shrt part dx 0 f beam at x=l N q M V Q P v (L) Differential equatin: EIv' ''' Nv'' q N(x) is assumed t be knwn. N(x) is cmmnl determined b 1:st rder ther, in this case giving N(x)=cnstant=-P. Since N<0: Gen. slutin: v C cs(cx) C sin(cx) C x C - qx /(2N) därc Bndar cnditins (4 needed) -N/(EI) Fr x=0 is v=0 (1) Fr x=0 is v =0 (2) Fr x=l is M=0, which since M=EIv gives v =0 (3) Fr x=0 is V=Q+qL, which since V=-EIv gives v =(Q+qL)/EI (4) Alternative t (4): Fr x=l is V=Q+v P, which since V=-EIv gives v -(P/EI)v =-Q/EI (alt. t 4) 17

18 The fur cnditins: With v, v, v and v frm the general slutin, the 4 cnditins are: It is in general cnvenient t intrduce the numerical values befre slving the equatin. The equatin gives C 1 -C 4 and thereb is the deflectin v(x) knwn. Als V and M are thereb knwn accrding t M EIv'' M and V -((EIv'')' M m) Knwing V, M ch N nrmal stress and shear stresses can be calculated accrding t cnvential equatins. 18

19 General abut elastic stabilt Stabilt analsis f 2D beam There are varius definitins f stabilit and methds f stabilit analsis in relatin t cnservative mechanical sstems. B such a sstem is here meant a structure made f a linear elastic material and laded b cnservative frces (frces fr which ptential energ can be defined). Cmmnl the cnditins fr neutral equilibrium are studied in elastic structural stabilit analsis. Neutral equilibrium ma r ma nt result in an unstable perfrmance and structural damage. (Fr beams/clumns the result is cmmnl structural cllapse. Fr plates there is ften a significant pstbuckling lad capacit.) A situatin f neutral equilibrium has varius names: critical lad buckling (knäckning) bifurcatin (förgrening) instabilit and varius methds f analsis are used: perturbatin analsis, Δπ=0 neutral equilibrium analsis, d 2 π/dδ 2 =0 bifurcatin analsis, tw r mre adjacent equilibrium slutins these methds can be shwn t be equivalent (in general) Ove Peterssn s ften used definitin (BYGG, 1971) f buckling lad (knäcklingslast) reflects bifurcatin analsis: 19

20 Equilibrium psitins f a cantilever clumn in P-δ space P The sstem is in equilibrium fr all pints at the curves. dπ/dδ=0 at all these pints. Stable equilibrum, d 2 π/dδ 2 >0 Neutral equilibrium, d 2 π/dδ 2 =0 Critical lad Buckling (Knäckning) Bifurcatin (Förgrening) Unstable P P Unstable equilibrium, d 2 π/dδ 2 <0 If a sstem is in stable equilibrium it can becme unstable nl if passing a pint f neutral equilibrium, i.e. if passing a bifurcatin pint. In the present analsis f 2D beams bifurcatin analsis shall be used t determine psitins f neutral equilibrium. 20 P δ P δ

21 Stabilit anals f 2D beam fr Case B (n elastic bed), v v tt (x)=v i (x)+v(x) q (x) An initiall straight r almst straight beam with varing crss sectin. q x (x) m(x) σ (x,) z L x, u The gverning equatins fr 2D beam 2:nd rder were (EAu') -qx N (EIv ) (Nv ) q m M (Nv) i (14a,b) Bending, eq 14b), is cnsidered. This equatin might have mre than ne slutin (bifurcatin) if the equatin is hmgeneus (=n transversal lad) the bundar cnditins are hmgenus (=n nnzer prescribed v, v, M r V) 21

22 Thus we have v ) (Nv) 0 (EI 14b) which fr cnstant EI and N has the slutin v C1cs( λx) C2sin( λx) C3x C4 frn 0 (cmpressi n), λ v C1csh( λx) C2sinh( λx) C3x C4 frn 0 (tensin), λ N/EI (17a,b) Fr N>0 nl ne slutin is pssible: C 1 =C 2 =C 3 =C 4 =0. Thus n bifurcatin fr tensile lad. Fr N<0 the utcme is different fr different bundar cnditins. The 4 bundar cnditins give 4 equatins with C= [C 1, C 2, C 3, C 4 ] T as the unknwn AC=0 (31) A is a 4x4 cefficient matrix with cefficients values as functin f λ, that is as functin f N. Thus A=A(N). One slutin t (31 ) is C=0. Mre slutins (=bifurcatin) are pssible if N is such that det(a(n))=0 (32) -N/EI The smallest cmpressin that satisf (32) is the buckling lad, N cr. Fr sme bundar cnditins the calculatin can be carried ut analticall, fr ther numerical calculatin is mre cnvenient. Matrix A(N cr ) has at least ne eigenvalue that is zer (since the right hand side f (31) is zer). The crrespnding eigenvectr has the same directin as C. Thus the shape, but nt the magnitude, f the deflectin v(x) is knwn frm (17a). Fr nn-cnstant EI and/r N sme apprximate slutin v must be used instead f 17a), e.g. v(x)=α 1 +α 2 x+α 3 x 2 +α 4 x 3 +, where α 1-4 is determined b sme residual methd (eg Pint cllcatin r Galerkin) t give a gd fit t 14b). 22

23 Example P L=1.0 m E=12000 MPa I=0.025* /12=2.6*10-10 m 4 k θ varing Buckling lad =??? L, EI k θ x Equatin: General slutin and it s derivatives: v C 1cs(λ x) C2sin(λ x) C3x C4 v C1λ sin( λ x) C2λ cs(λ x) C3 v 2 2 C1λ cs(λ x) C2λ sin( λ x) Bundar cnditins: frn 0, λ (1) v(0)=0 (kinematic cnditin) (2) EIv (0)-k θ v (0)=0 (mment spring, cnvective cnditin) (3) v(l)=0 (kinematic cnditin) (4) v (L)=0 (since M(L)=0, static cnditin) give 1 2 λ cs(λl) 2 λ cs(λl) λk λ /EI sin( λl) 2 0 θ sin( λl) k 0 θ L 0 /EI 1 C 0 C 1 C 0 C N/EI P/EI 23

24 1 2 λ cs(λl) 2 λ cs(λl) λk λ 0 /EI sin( λl) 2 θ sin( λl) k 0 θ L 0 /EI 1 C1 0 0 C2 0 1 C 3 0, AC=0 0 C 0 4 This equatin has ne slutin C=0. Additinal slutins exists if det( Α (λ)) 0 B caluculatin f det(a) (e.g. b means f Matlab) fr increasing λ (increasing P), the smallest λ (the smallest P) that give det(a)=0 is eas t find. Result fr sme different k θ /EI with P cr written as P cr =απ 2 EI/L 2 : k θ /EI [1/m] α 0 (1.00) *) (2.05) *) *) frm litterature P cr =απ 2 EI/L 2 24

25 Next Frida 25