Kul Finite element method I, Exercise 04/2016. Demo problems

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1 Kul-49 Finite element method I, Eercise 4/6 Demo problems Determine displcement u Z = nd rottion θ Y = t the mid-po of the Bernoulli bem of the figure Use two bem elements of equl length () nd () Po moment () with mgnitude M is cting on node Young s modulus of the mteril E nd second moment of re I re constnts Z, M X, Answer u Z = θ Y = 8 M he frme of the figure consists of rigid body () nd two Bernoulli bem elements () nd () Determine the non-ero displcements nd rottions s functions of force F he bems re identicl nd cn be ssumed rigid in the il directions so tht the il strins vnish Displcements re confined to the XZplne Problem prmeters E nd I re constnts Answer ux = ux = 4 F F 4 Z, X, =, 4 Determine the rottion θ Y = t node of the structure loded by po moment (mgnitude M) cting on node Use Bernoulli bem elements () nd () of equl length nd po moment element () Assume tht the bems re rigid in the il directions Young s modulus of mteril E nd the second moment of re I re constnts Answer θ Y = 8 M X, Z, M he demo problems re published in the course homepge on Fridys he problems re relted to the topic of the n weeks lecture (ue 5- hll K 5) Solutions to the problems re eplined in the weekly eercise sessions (hu 5- hll K 8) nd will lso be vilble in the home pge of the course Plese, notice tht the problems of the midterms nd the finl em re of this type

2 Determine displcement u Z = nd rottion θ Y = t the mid-po of the Bernoulli bem of the figure Use two bem elements of equl length () nd () Po moment () with mgnitude M is cting on node Young s modulus of the mteril E nd second moment of re I re constnts Z, M X, Solution In hnd clcultions, it is enough to complete the figure by the mteril coordinte systems nd epress the nodl displcements/rottions in terms of the -symbols nd/or vlues known priori he components in the mteril coordinte systems cn lso be deduced from the figure (in simple cses) Virtul work epression of the bem -plne bending element is given by h h u y yy θ y h y θ y u h h h h f ( h h δθ = + = ) δu 6h 6h u 6 6h h 6h 4h h δθ As il loding vnishes, br deformtion cn be omitted he br mode cn be tken o ccount, but clcultions give u X = Element contribution of po force/couple is Xi Xi δθ Xi Xi u F M = δuyi FYi + δθyi MYi δu F δθ M Zi Zi Zi Zi Nodl displcements/rottions of the structure re eros ecept those for node u = θ Y = Z Element contribution of bem : u = uz = θy = θy = = 6 6 = δ δ 6 4 δ Element contribution of bem : u = uz = θy = θy =

3 6 6 δ 6 δ δ = = Element contribution of po moment : u = θ = Z Y X X δθ X X u F M = δuy FY+ δθy MY= + δ M= δ M δu F δθ M δ Z Z Z Z = ( ) δ M Virtul work epression of the structure is the sum of element contributions e e = = = ( ) + δ M 4 = ( ) δ 8 M Principle of virtul work = δ nd the bsic lemm of vritionl clculus in the form δ R = δ R = imply 4 = 8 M M = 8

4 he frme of the figure consists of rigid body () nd two Bernoulli bem elements () nd () Determine the nonero displcements nd rottions s functions of force F he bems re identicl nd cn be ssumed rigid in the il directions so tht the il strins vnish Displcements re confined to the XZ-plne Problem prmeters E nd I re constnts F 4 Z, X, =, 4 Solution In the plnr problem, torsion nd out-plne bending deformtion modes cn be omitted Since bems re inensible in the il directions, il displcements t the nodes of the bems coincide Virtul work epressions of the bem -plne bending element nd po force/moment elements re given by (notice tht the force/moment element is written in the structurl coordinte system) h h u y yy θ y h y θ y u h h h h f ( h h δθ = + = ), δu 6h 6h u 6 6h h 6h 4h h δθ Xi Xi δθ Xi Xi u F M = = δuyi FYi + δθyi MYi δu F δθ M Zi Zi Zi Zi Nodl displcements/rottions of the structure re eros ecept those for node nd Rottions t nodes nd vnish s element is rigid body (element does not stretch nor bend nd the geometry of the structure implies pure trnsltion for element ) Verticl displcement of these nodes vnish s bems nd re inensible in the il directions Hence u = u X = X Element contribution of bem : u = ux = = δ δ 6 6 = Element contribution of bem : u = ux =

5 = δ δ 6 6 = Element contribution of bem : (notice tht rigidity in the sense is erpreted s θ = θ = ) y y = 6 6 = Element contribution of po force 4: u X = X X X X u F θ M F = δuy FY+ δθy MY= + = δf δu F δθ M Z Z Z Z Virtul work epression of the structure is the sum of element contributions e 4 W = W = W + W + W + W e δ δ δ δ δ δ W = + + F = (4 F) δ δ δ δ δ Principle of virtul work = δ nd the bsic lemm of vritionl clculus in the form δ R = δ R = imply 4 F = = 4 F

6 Determine the rottion θ Y = t node of the structure loded by po moment (mgnitude M) cting on node Use Bernoulli bem elements () nd () of equl length nd po moment element () Assume tht the bems re rigid in the il directions Young s modulus of mteril E nd the second moment of re I re constnts Solution X, Z, M In the plnr problem, torsion nd out-plne bending deformtion modes cn be omitted As bems re ssumed to be rigid in the il direction, il displcements t the nodes coincide nd the br mode virtul work epression vnishes Virtul work epressions of the bem -plne bending element nd po force/moment elements re given by h h u y yy θ y h y θ y u h h h h f ( h h δθ = + = ), δu 6h 6h u 6 6h h 6h 4h h δθ Xi Xi δθ Xi Xi u F M = = δuyi FYi + δθyi MYi δu F δθ M Zi Zi Zi Zi Nodl displcements/rottions of the structure re eros ecept those for node Displcement t node vnishes s both bems re inensible in the il directions herefore, the only non-ero displcement/rottion component of the structure is θ = Y Element contribution of bem : θy = θy = = = δ δ Element contribution of bem : θy = θy =

7 δ = = δ Element contribution of po force : θ Y = X X δθ X X u F M = δuy FY+ δθy MY= + δ M= δ M δu F δθ M Z Z Z Z Virtul work epression of the structure is the sum of element contributions e e = = + + W = M = (8 + M) δ δ δ δ δ Principle of virtul work = δ nd the bsic lemm of vritionl clculus in the form δ R = δ R = imply 8 + M = = M 8

8 Kul-49 Finite element method I; Formule collection GENERA i ix iy iz I I j jx jy j = Z J = i j k J k kx ky k Z K K Coordinte systems: { } X i = Y h Z Strin-stress: ε ν ν σ ε yy ν ν = σyy E ε ν ν σ γy σy γ y = σy G γ σ E G = ( + ν ) or σ ν ν ν ε ε E σ yy ν ν ν = εyy [ E] εyy ( ν)( ν) + σ ν ν ν ε ε σy γy σy = G γ y σ γ ν E [ E] σ = ν ν ( ν ) / ν ν E [ E] ε = ν ν ( + ν)( ν) ( ν ) / Strin-displcement: ε u, ε yy = uy, y ε u, γ y u, y + uy, γ y = uy, + u, y γ u + u,, EEMEN CONRIBUION (constnt lod) Br (il): F EA u f h = F h u EA = h R ii ii f h i, in which R ii ii i X i = Y h Z Br (torsion): M GI θ mh = rr M h θ Bem (): F 6h 6h u 6 M y yy 6h 4h 6h h y h θ f h = F h 6h 6h u 6 M y 6h h 6h 4h y h θ PRINCIPE OF VIRUA WORK

9 e = + = = δ = δwdω e E Ω Br: δw = δu EAu,, δw = δuf Br (torsion): δ w = δφ, GIrrφ, δ w = δφm Bem (): Bem (y): δw = δ w w δw = δwf, yy, δw = δ v, v, δw = δvf y Bem (Bernoulli): A S, Sy u, u δ w = δ v E S I I v δφ GI φ w, S w δ y Iy Iyy,, y,, rr, u f δφ Sy fy + S f δw = δv fy+ δw, Sy f w f v δ δ S f Plne-stress (y):, δu, u, δw = δv, y te [ ] σ v, y δu + δv u + v, y,, y, δ w u f = δ v f y Plne-strin (y): δu, u, δw = δv, y te [ ] ε v, y δu, y + δv, u, y + v, δ w u f = δ v f y Kirchhoff-plte (y): w, w, t δw = δw, yy [ E] σ w, yy δ w, y w, y Reissner-Mindlin plte (y): δw = δwf

10 δθ, θ, t w, y δφ w, y φ δ w = δφ, y [ E] σ φ, y tg δ w, + δθ w, + θ δφ, δθ, y φ, θ, y δw = δwf Body (y): ε σ γ y σ y yy yy y y δ w = δε σ δγ σ δε σ δγ σ u f δ y δw = v f δ w f or, u,, y + δ,, y +, u u v u v δw = δv, y [ E] v, y δv, + δw, y G v, + w, y δw w δw + δu w + u,,,,,, APPROXIMAIONS (some) u = N ξ = h Qudrtic (line): N ξ + ξ N = N = 4 ξ( ξ) N ξ(ξ ) u = u (br) u Cubic (line): N ( ξ) ( + ξ) N h( ξ) ξ N N = = ( ξξ ) N hξ ( ξ ) u u u θ y = ( = ) (bem bending) u u u θ y iner (tringle): N = y y y y VIRUA WORK EXPRESSIONS uxi FXi θ Xi MXi Rigid body (force): = δuyi FYi + δθyi MYi δu Zi F Zi δθ Zi M Zi Br (il): EA u u = δu h u u fh = δ u Br (torsion): θ GIrr θ δθ h θ = θ mh = ) δθ

11 Bem (): 6h 6h u y yy 6h 4h 6h h θ y h 6 6 y θ y u δθ = δu h h u 6h h 6h 4h δθ u 6 δθ y h f h = δu 6 δθ y h Bem (y): y 6h 6h uy 6h 4h 6h h θ y h 6h 6h uy 6h h 6h 4h θ u δθ = δ u δθ uy 6 δθ fyh h = ) δ uy 6 δθ h CONSRAINS Frictionless contct: n u A = Jo: ub = ua Rigid body (link): ub = ua + θa ρab θb = θa

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